The Berkeley Review MCAT Physics Part 1

PhysicsParti

Sections I-V

Section ITranslational Motion

Section IIForces, Circular Motion,

and Gravitation

Section IIIWork and Energy

Section IVMomentum and Torque

Section VPeriodic Motion

and Waves

1RERKELEY•*Kr

D R.E«V»I«E«l/ir

Specializing in MCAT Preparation

The

ERKELEYR • E • V • I • E • W

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October 2010-

So you plan to become a doctor! It is quite an admirable quest and sacrifice.Before getting into the crux of this book, I'd like to thank you. First, I'd like tothank you for choosing such a career path, and a life dedicated to servinghumanity. Second, I'd like to thank you for choosing us to help you achieve yourgoal. My sincerest hope is that this book helps make your path into medicalschool just a bit easier and more successful.

What we've set out to build here is the best physics MCAT preparation book ever.And by best, that means a book that clearly reviews material, presents clever andmemorable mnemonics, provides unique test-taking strategies, offers plenty ofpractice questions, presents thorough answer explanations, and ultimately servesas a printed friend.

This book is the latest edition of a project that started in 1992. It has undergonefive previous revisions, although this sixth and latest one represents its biggestchange to date. This book is the collaborative work of seven different authorswho collectively have over 10,000hours teaching MCAT preparation to studentslike yourself. Our goal was to write the book like a teacher would explain thesubject matter one-on-one to a student. We sought a diverse group of authors forthe practice questions and passages, so that you get exposed to a plethora ofdifferent voices and styles. Your MCAT will be a compilation of passages bydifferent authors, so that is what we aimed to produce here.

As you prepare for the MCAT, you need to keep in mind that MCAT-levelphysics is different than college-level physics. Most notably, it is in a multiple-choice format and has a strict time limitation. To do well on MCAT physics, youneed to think quickly, have a good conceptual understanding of the material, andnot get hung up on the details. Pick a best answer quickly and then move on tothe next question. To help you to do this, we have included plenty of mnemonicsand visualization tricks. Even more importantly we present and emphasize whatwe will call headstarts for saving time on questions. Because you don't have toshow your work on the MCAT, you arebetter off rewriting your equations into aneasier to use state. We rewrite many equations so that anyone can easily do themath in their head. You'll notice that we do this in nearly every chapter.

To get the most out of these books, you should practice the methods, algorithms,and techniques we present in the text on the practice questions at the end of eachchapter. You should emphasize doing as much in your head as possible, withoutwriting on scratch paper or this book. After taking the chapter exams, you shouldspend a significant amount of time making sure you understand each and everypassage and question. Repeat any questions you miss until you get them right,and then read and review the answer explanations.

If you are feeling driven, rewrite some of our questions by changing the numbers,the terminology, or the conditions and see how you do on your own questions.Doing such a thing has an amazing impact on your ability to take a multiple-choice test. Even if you don't choose to do that, there are tons of questions to testyourself with.

Each chapter has an average of 18 multiple-choice questions intertwined in thetext followed by a 25-question review exam and a 52-question practice test. Takeas much time as you need on the text questions and the 25-question review exams,but do your best to finish the 52-question practice tests within 70 minutes each.Developing speed is a vital part of your preparation, so make sure you do this atleast one time per chapter.

We wish you the best of luck staying motivated throughout your preparation.Keep the following three things in mind as you go:

1) The MCAT is a thinking exam, so keep things simple.

2) Avoid careless mistakes; be fast but not rushed.

3) The answer is given; all you need to do is find it

Once again, thank you for sharing this stage of your pathway to medical schoolwith us.

Sincerely and with great respect,

The Berkeley Review

Table of Contents

1. Translational Motion

Units and Dimensions page5

Vectors page 10

Speed, Velocity, and Acceleration page 16

Free Falling Bodies page 20

Projectiles page 27

Translational Motion Review Questions page 35

Detailed Answer Explanations page 40

Translational Motion Practice Exam page 45


2. Forces, Circular Motion, and Gravitation

Newton's Laws of Motion page 67

Law of Gravitation page 75

Uniform Circular Motion page 79

Centripetal Force page 86

Kepler's Laws of Orbital Motion page89

Friction page 91

Inclined Planes and Pulley Systems page 94

Forces, Circular Motion, and Gravitation Review Questions page 99


Forces, Circular Motion, and Gravitation Practice Exam page 109


3. Work and EnergyWork page 131

Kinetic Energy page 134

Gravitational Potential Energy page 139

Conservation of Energy page 141

Power page 144

Work and Energy Review Questions page 145


Work and Energy Practice Exam page 155


4. Momentum and TorqueLinear Momentum page 177

Impulse page 181

Torque and Equilibrium page 186

Momentum and Torque Review Questions page 191


Momentum and Torque Practice Exam page 201


5. Periodic Motion and Waves

Harmonic Oscillation page 223

Springs and Pendulums page 226

Wave Properties page 232

Periodic Motion and Waves Review Questions page 241


Periodic Motion and Waves Practice Exam page 251


Physics Chapter 1

86 100

by

Translational MotionSelected equations, facts, concepts, and shortcuts from this section

O Important Equations

_ 1 ot2d = ^-ar + v0t + d0 (if you're lacking vO

d

vt = vo + aAt (if you're lacking d)

vt2 = v02 + 2a(dt - do) (ifyou're lacking t)do + V° t Vt t (if you're lacking a)

© ImportantConcept

X-motion is independent of Y-motion, and flight time only depends on the y-component

All four launches start at the same height

Horizontal launches in all cases execpt vox = 0

x-Displacement at collision

All four objects hit at the same time

Time depends only on h, and not on vx

© Quick Free-fall CalculationTrick

The distance traveled in the y-direction of freefall from rest is found using Ah = 4.9t2 = 5t2The speed at impact in the y-direction whenfalling from rest is found using vt = 9.8t» lOt

By using the results when t = 1, t = 2, t = 3, t = 4, etc..., you can quickly estimate a best answer

Ah tfall Vf

1.25 m 0.5 s 5m/s

5 m Is lOm/s

20 m 2s 20m/s

45 m 3s 30m/s

80 m 4s 40m/s

125 m 5s 50m/s

320 m 8s 80m/s

How far... t=1.7s?

O A. 5.8 m

O B. 14.2 m

O C. 25.0 m

O D. 36.9 m

Given just under 2s

.-. just under 20m

B is the best choice

How long... d = 114 m?

O A. 2.92 s

O B. 3.36 s

O C. 4.82 s

O D. 5.66 s

Given just under 125m

/. just under 5s

C is the best choice

How high... vinity = 42 m/s?

O A. 21.8 m

OB. 42.9 m

O C. 71.4 m

O D. 90.6 m

Given vimpact =40+m/s/. t = 4+s and /. d = 80+m

D is the best choice

Physics Translational Motion

Translational MotionThe three basic types of motion that we will be considering are translationalmotion, rotational motion, and vibrational motion. Translational motioninvolves the movement of an object from one place to another place. As a result,the center of mass changes its position when an object undergoes translationalmotion. When this occurs, the center of mass is said to be displaced.Translational motion looks at the change of position of an object as a function oftime. Rotational motion examines how an object rotates about an axis over time.Vibrational motion examines how an object oscillates back and forth about somecentral point or axis over time. In both vibrational and rotational motion, thecenter of mass does not change over time. It is possible for an object to undergomore than one type of motion at a time. In this section, we will primarily beexamining translational motion. However, there will be times when we willdiscuss the combination of translational motion with both rotational andvibrational motion. These studies fall within the realm of kinematics, which is thestudy of motion without considering the various forces that cause it.

Units and Dimensions

UnitsThe system of units used on the MCAT is predominantly the metric system. Theunits of length, mass, and time in the metric system are the meter (m), kilogram(kg), and second (s).These metric units are referred to as the Systeme Internationald'Unites or the SI units. The metric system is based on atomic standards. TheEnglishsystem of units uses the foot (ft), pound (lb), and seconds (s).

Table 1.1

Multiples of the Meter

gigameter 1 G = 109 m

megameter lM = 106m

kilometer lkm = 103m

meter lm

centimeter 1 cm = 10"2 m

millimeter 1 mm = 10"3 m

micron 1 /im = 10'6 m

nanometer 1 nm = 10~9 m

Angstrom 1 A = 10"10 m

picometer 1pm = 10"12 mfemptometer 1 fm = lO"15 m

Multiples of the Second

year lyr =3.16x!07sday 1 day = 8.64 x104 s

hour 1 hr = 3600 s

minute 1 min = 60 s

millisecond 1 ms = 10"3 s

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Length Conversion Factors

mile 1 mi = 1609.38 m

yard 1 yd = 0.9144 m

foot 1 ft = 0.3048 m

inch 1 in = 2.540 cm

Mass Conversion Factors

kilogram 1 kg = 2.204 lb

gram 1 g = 10"3 kgmilligram 1 mg = 10"3 g = 10~6 kg

Typical Masses

Sun 2.0 x1030 kg

Earth 6.0xl024kg

RedBlood Cell 9 x10"14 kg

Proton 1.66 x10"27 kg

Electron 9.11 x10'31 kg


Exclusive MCAT Preparation

Physics

Test TipMagnitude Estimation

Translational Motion Units and Dimensions

Note that the kilogram and the pound do not represent the same physicalquantity. The kilogramis a unit of mass, whereas the pound is a unit of force— adistinction that will be addressed in Section II.

There will be occasions when a question will require conversion betweendifferent sets of units. For example, you may need to convert a distance of 100miles into meters. Table 1.1 shows that 1 mile is equal to 1609.38 meters. This isthe same as saying the following:

1609.38 m

1 mile

In order to convert 100 miles to meters, the conversion factor (shown above) ismultiplied by 100 miles. The miles will cancel out, leaving 160,938 m. This valueis commonly expressed in scientific notation as 1.61x 105 m.

(1609.38 m](100 miles) =160/938 mV lmile I '

If you did not know the specific value for this conversion factor, do not worry.Conversion factors for most units will usually be given on the MCAT. Youshould, however, have at least an approximate idea of these factors (e.g., knowthat therearearoundonethousand, or maybe abit more, meters in onemile.)

Example 1.1aTable salt consists of sodium chloride. The approximate distance between thenuclei of a sodium atom and a chlorine atom is 2.8 A. How far apart are thesenuclei in miles? (1 A= lO'10 mand 1 mile = 1609.38 m)

A. 1.74 x 10"13 milesB. 4.51 x 10"7milesC. 1.74 x 10-3 milesD. 4.51 x 103 miles

Solution

This is a numeric answer, which traditionally involves multiplying and dividingvarious numbers. This arithmetic can, however, become very time-consuming.You should try to avoid doing full arithmetic on the MCAT whenever you can.

Magnitude Intuition: The numbers in questions with numeric answerssometimes vary byorders ofmagnitude. As a first approach, use the magnitudeintuition technique. It can help to eliminate incorrect answers without doingmath. Simply cross outanswer choices that seem too big or too small. Here, weare asked about the separation of two atoms in a molecule. Which of the choicesmight you eliminate without resorting to math? Probably choice D. If you alsothinkaboutchoice C,and happento knowthat thereare thousands of feet in onemile, you'll see that choice Cis also too big. Is there a way to differentiate fairlyeasily between choices Aand B? Yes, use the magnitude estimation technique.

Magnitude Estimation: Simply write out the numbers in scientific notation andestimatethe order ofmagnitude of the result. Here, the actualcalculation is

distance in miles =(2.8 A)|10~10ml/-1 1A Hi

1 mile

60938 x 103m 1.6x 10"13m

To estimate the order ofmagnitude, look only at the powers often. Doing sowillgive 10-13 as the resulting order of magnitude. If another choice is within oneorderofmagnitude (e.g., lO'12 or 10'14), you will needto check theother numbersin the calculation (i.e., 2.8/1.60938). This, too, is easy. Simply askyourself if these

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numbers will result in a number bigger than ten or smaller than one. If they givea number bigger than ten, then the actual orderofmagnitude is 10"12. If theygivea number smaller than one, then the actual magnitude is 10"14. Rarely will thesenumbers give an answer smaller than 0.01 or greater than 100. Here, the nextclosest answer choice is six orders of magnitude away, so there is no need to domore than multiplying powers of ten.

The best answer is choice A.

Example 1.1bHow fast is a car traveling in miles per hour if it has a speed of 0.02906 km/s? (1mile = 1609.38 m)

A. 0.018 miles/hrB. 1.1 miles/ hrC. 65 miles/hrD. 780 miles/hr

Solution

Using the Test Tips fromExample 1.1, you will find the orderofmagnitudeto be101 and the remaining numbers (i.e., 29.06 x 3.6/1.60938) to be equal to somenumber between one and ten. The order of magnitude of the actual answer mustbe 101.

The best answer is choice C.

Common DimensionsPhysical measurements consistof countings. You watch the crests of waves passa certain point and you count them. Physical measurements can also beassociated with marks. A classic example is the marks on a ruler. Digital displaysprovide electrical marks. When a physical measurement is made, it willconsist oftwoparts. First, therewillbe the numberof itemsthat are counted. Second, therewill be some statement about the type of units that those items are counted in.For example, "11.2 seconds" not onlyindicates that 11.2 items havebeencounted,but also that those items have been counted in units of seconds.

An enormous number of physical quantities can be expressed in terms of justthree fundamental units. Those units are length [L], time [T], and mass [MJ. Asymbol withsquare brackets around it is often used to indicate a given physicalmeasurement associated with a particular unit. We will see that other unitsinvolve combinations of these three fundamental units. We often refer to ourcommonunits (meters, kilograms, and seconds)as MKS units.

Example 1.2aWhat are the dimensions of area?

A. m

B. m2C. m3D. m2/s

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FhySlCS Translational Motion Units and Dimensions

SolutionSince an area is simply a length multiplied by a length, the dimensions would be[L] x [L] or:

area = [L2]=m2

The best answer is choice B.

Example 1.2bWhat are the dimensions of density?

A. kgB. kg-m3C. kg/m2D. kg/m3

Solution

Density is an example of a derived unit. It is simply a ratio of a unit of massdivided by a unit of volume. Since the dimensions of volume would be a lengthtimes a length times a length, or [L3], the dimensions of density would be [M]dividedby [L3] or:

density =-!^-=-[L3]

The best answer is choice D.

Velocity (v) describes an objects change in position over time. It takes intoaccount both speed and direction. Velocity and speed are often confused withoneanother. Velocity requires the identification of a direction; speed does not. Ifone were to say, "I was driving my car at 65 miles per hour," then that wouldrefer to a speed. However, if one were to say, "I was driving my car north at 65miles perhour," then that would refer to a velocity. Both speed and velocity areconcerned with covering a given distance. Therefore, the dimensions of bothvelocity and speed are the dimensions of length divided by the dimensions oftime, as shown in equation (1.1).

velocity = — (1.1)PI

The rate of change of velocity with respect to time is referred to as acceleration(a). As shown in equation (1.2), the physical dimensions of acceleration are theunits of velocity divided by the units of time.

acceleration = - ¥L (1.2)[T]

Substitution of equation (1.1) into equation (1.2) gives the units of acceleration.Theyare lengthdividedby time squared, as shownin equation(13).

acceleration = D_=_LJ_ (1.3)m nf

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PhysiCS Translational Motion

Dimensional consistency is associated with some rules. We need to be able todistinguish between a physical equation and a mathematical equation. Aphysical equation says something about the real world. Everything in a physicalequation is a physical quantity. The equation consistsof sums of terms. A singleterm will be a physical quantity, which will have some physical dimensions.Velocity and acceleration are two such examples. Two rules involvingdimensional consistency are (a) that every term must be of the same physicaldimension and (b) the argument of a mathematical function has no dimensions.For example:

0 incosO ort/xi/2 in e 1/2


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Physics Translational Motion Vectors

Vectors

Scalar and Vector QuantitiesVectors are used because of the importance of direction. For instance, a vectordescribes velocity while a scalar (just a number) describes speed. A distinctioncan also be made between distance and displacement. If we say that two citiesare 100 miles apart, then that is a distance. No direction has been indicated.However, if we say that LosAngelesis 100milesnorth of San Diego, then that isa displacement. Displacement is a distance with a well-defined direction.

Some physical quantities, such as time, mass, temperature, electric charge, andchemical concentrations, haveno direction at all in space.Theseare referred to asscalar quantities. We measure scalar quantities on some type of scale mat hasnothing to so with three-dimensional space or directions. Quantities that do havedirectionare displacement, velocity, acceleration, force, and torque, to name buta few. These are referred to as vector quantities. Some common scalar and vectorquantities in physics are shown in Table1.2.

Table 1.2

Scalars Vectors

Symbol

d

V

t

E

m

Name

distance

speedtime

energy

mass

Symbol

d

V

a

F

X

Name

displacementvelocityacceleration

force

torque

How can we specify directions in space? Let's first consider a two-dimensionalspace. The paper that you are reading is in two-dimensional space. We canrepresent two-dimensional space by drawing two perpendicular lines (Figure 1-la). We can label those lines withan x-axis and a y-axis as shown. A compass canalsobe used to specify a direction in space (Figure 1-lb). We can label those lineswith an north, east, south and west as shown. The Earth has imaginary linesaround it called longitude and latitude (Figure 1-lc). These lines are measuredfrom some standard starting point. Ifwe could open up the Earth and lay it outflat, we would find that the lines of longitude (running north and south) andlatitude (running eastand west) essentially form a grid (Figure 1-lb).

y-axis A

x-axis

(a)

North

AWest < 1 >

YSouth

(b)

Figure 1-1

Longitude

East

(c)

What exactly is a vector? Wecan represent a vectoras an arrow with a head anda tail (Figure 1-2). This vector will give us a direction, and its length will alsogiveus the sizeofsomething. We canimagine a scale on the vector in which eachunit represents some quantity. The scale is represented by the perpendicularlines in Figure 1-2.

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Physics

y-axis

Translational Motion

A typical vector(segments represent units)

Figure 1-2

A vector can be represented in a two-dimensional perspective (most common) ora three-dimensional perspective. We would need to add another axis to our x-and y-coordinate system. We could call this new axis the z-axis.

A physical vector represents a physical quantity. Physical vectors have threeproperties: a counting number, a unit, and a direction." For example, we mightsay "100 miles per hour in the northeast direction. The "100" is the countingnumber, the "miles per hour" are the units, and "Northeast" is the direction. Wecanrepresent this on the graph shown in Figure1-3. In thisgraph, each mark onthe vector represents 10 miles per hour (there are 10 marks). Since physicalvectors have units, we also see that they have dimensions. The dimension of adisplacement vector is a length (L), of a velocity vector is lengthdivided by time(L/T), andofacceleration is lengthdivided by time squared (L/T2).

x-axis

Figure 1-3

We are now in a position to add two vectors. Supposewe walk 3 mileseast andthen 4 milesnorth. Theseare displacement vectors. We alsocould have arrived atour destination had we walked in a straight line. Let's represent our vector of 3miles eastby A, of 4 miles north by B, and of the straight line between the twovectors as C. The straight line, C, can be representedas the sum of vectorsA andB(e.g., C=A + B). Since the triangles shown in Figure 1-4 arerighttriangles, thevector Cmust beequal to5miles. Remember, inmathematical terms, a2 +b2 = c2.

[The vectors A,B,and Cwould generallybe denotedwitha small arrowabove the letters.Another wayto indicate vectors is to type the vector letters inbold face. In thissection onvectors, we will bold face and italicize the letters representing vectors.]

C=A+B / ..

c/

A

Figure 1-4 (a)

B


B

*

*

/ C=B+A'C

Figure 1-4 (b)

11

Vectors


PhysiCS Translational Motion Vectors

In adding these vectors, it would not have made any difference if we first wenteast and then went north (Figure l-4a) or if we first went north and then wenteast (Figure l-4b). In other words, C = A + B is the same as C = B + A.

Suppose we do not choose a right angle and instead we add vector B at somerandom angle to vector A. We will still get the resultant vector C. In other words,C still equals A plus B. If we interchange the order of the vectors (as we did inFigure 1-4), then we will still get a situation where C=B + A (Figure 1-5).

C = A + B

Figure 1-5

What do we do, if we wish to add several vectors? Suppose we wanted to addvectors A, B, C, and D, as shown in Figure 1-6. We will follow the sameprocedure. We take vector B and move it so that its tail touches the head ofvector A. We then take vector C and move it so that its tail touches the head ofvector B, and so on. We will eventually obtain the resultant vector, E. It will notmatter what order we take the vectors in, as we will always get the sameresultant vector, E.

Figure 1-6

Of most applicability to what we do in physics is the splitting of a vector intocomponent vectors associated with parallel axes, such as the x-axis and y-axis.Let's consider the components of a vector. We will define our coordinatesaccording to the x-axis and y-axis. Supposewe have a vectorA (Figure 1-7). Wecan arrive at vectorA by means of a right triangle which is squared offwith ourcoordinates. This means that we go out in the x-direction by an amount Ax andup in the y-direction by an amount Ay. If these amounts were thought of asvectors, then we could write A=Ax +Ay. But that would notbe the way towritethis expression, because we must regard Axas the size of the vector (therewouldbe no arrow over the A) in that direction. Asimilar argument would exist for Ay.

So, how do we describe direction? We could say A=Ax (in the x-direction) +Ay(in the y-direction). Even though this is correct, it will take too long to write outevery time. We must shorten it by using some form of notation. That notation iscalledx (readas x-hat) and y (readas y-hat). Thesenotations refer to the x and ydirections, respectively. They are unit vectors. What we find is that we can writeA = Axx + Ay£. Recall that we mentioned that vectors contain a countingnumber, a unit, and a direction. In this case, the Ax has the counting number andthe unit in it but doesnot havethe direction. Thedirection is reserved forx or y.

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y-axis

A =Axx +Ayy

x-axis

Figure 1-7

In order to find the magnitude of the components of a vector, we need tounderstand the definitions of the sine, cosine, and tangent functions for a righttriangle (see Figure 1-8). A right triangle has a hypotenuse, an angle 6, a sideopposite to the angle 6, and a side adjacentto the angle8.

y-axis

adjacent x-axis

Figure 1-8

Sin 9 =

Cos 8 =

Tan 8 =

opposite

hypotenuse

adjacenthypotenuse

oppositeadjacent

Table1.3shows the sine, cosine, and tangent values for the commonangles thatare encountered on the MCAT. You should also know that cos20 + sin28 = 1

Table 1.3

Angle Cosine Sine Tangent

0° V4/2 =2/2 =1.oo o/2=o °/i=o30° V3/2 =1.73/2 =0.86 VT/2=i/2=050 0-50/086 =0.58

45° V2 /2=1.41 /2=o.71 V2/2 =1.41/2=0.71 a71/0.71 =1-<)060° VT/2 =i/2==0.50 V3/2 =1.73/2=0.86 a86/0^0fi-7390° o/2=o V4/2 =2/2 =1.oo 1/ o=undefined

Using trigonometry relationships, we can find the magnitude of the vectorcomponents shown in Figure 1-7. We see thatAx =Acos8 andAy =Asin0.

y-axis

Ax = Acos8

Figure 1-9

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A2= A2cos28 + A2sin29

A2 =A2(cos28 +sin2e)

1 = cos^ +sin^

13

Vectors

Many people recall

SOHCAHTOA


Physics

Test TipGraphical Estimation

Translational Motion Vectors

Example 13aAn elephant pulls a large tree trunk 4 km in a direction 50° east of North. Whatare the components of the elephant's displacement in the easterly and northerlydirections?

A. 3.06 km east and 2.57 km northB. 2.57 km east and 3.06 km northC. 3.94 km east and 0.69 km northD. 0.69 km east and 3.94 km north

Solution

Graphical Estimation: Most physics problems can be visualized, or at least canbeschematically drawn. The graphical estimation technique canallow youto solvequestions by considering a picture of the problem. Draw a reasonably carefulpicture of the problem, and use it to eliminate invalid choices or to identify thebest one. Here, we shouldestablish a coordinate system,so that we know whichway thevector should point. "50° eastofNorth" means that the50° angle shouldstart on the North axisand open up towards the Eastaxis,as below:

North

y-axis

x-axis

Note 1: Because 40° <45°, Rx >RyGiven that Rj^ >Ry, choices BandD are eliminated.

Note 2: If 8=45°, then ^ =RyBecause 8=50°, ^ =Ry, sochoiceA is a better answer than choice C.

Notice that in drawing the R vector, we put hash marks along its length. Eachmark represents 1kmand serves asa quick way tokeep track oflengths.

Now, we want to know the values of the East and North components of R. Toestimate them, we also draw hash marks on them. We do so for two reasons:First, it allows us tocompare the North andEast components. From ourpicture,the East component islonger. This rules outchoices Band D. Second, it can giveus a reasonable ideaabout the magnitude of these components. We see that theNorth component isaround 2.5 km, according to the hash mark positions. Thisfavors choice A over choice C.

Had we solved thisproblemthe traditional way, we would find:

Cos 45° = 0.71 and Cos 30° = 0.86, so Cos 40° = 0.75

Sin 45° = 0.71 and Sin 30° = 0.50, so Sin 40° = 0.65

East Rx = RCos8 = (4 km)(Cos40°) = 4 x 0.75 = 3.0 km

North Ry = RSin8 = (4 km)(Sin40°) =4 x 0.65 = 2.6 km

which is a fine method, assuming you know the sine and cosine values, and thatyou can multiply numbers quickly. The key to calculation questions on theMCAT is being able to do them quickly and get close enough to choose a bestanswer. "Ifs about speed more than precision!"


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Example 1.3bIn map making, cartographers sometimes rely on measurements taken by asurveyor. A personreports that she walked3200 m northand then 2400 m westbefore reaching the measurement site. What is the magnitude and direction ofthe shortest path between her starting and ending points?

A. 4000 m 37° west of NorthB. 4000 m 53° west of NorthC. 5600 m 37° west of NorthD. 5600 m 53" west of North

SolutionSince we are given the components of the vector displacement, all we need tofind is the magnitude and direction of the resultant vector. Using the methods ofthe previousexample, we have

Rx= 2400m

*-r

X

1 H\ Northy-axis

*Ae-7 = 3200 mV

West x-axis East

Notice that the length of the resulting vector cannot be 5600 m, because thatcould only result if the 2400 m and 3200 m component vectors were connectedhead-to-tail in a collinear fashion. This rules out choices C and D. Notice that theRx component is smaller than the Ry component. This means that 8 must besmaller than 45°. Recall that at 45°, the x and y components of a vector are equalin magnitude. This rules out choice B.


You can also use the Pythagorean theorem (c2 = b2 + a2) to find that themagnitude of the resultant vector is 4000m. You may have noticed that thetriangle isactually a 3-4-5 right triangle. This observation could have saved youfrom doing a lengthy calculation. To determine the direction inwhich the personmust travel, we use the tangent function, where tan8 = opposite/adjacent.Substituting the appropriate values we get tan8 = 2400/3200 = 0.75. If youremember that tan 45° = 1 and tan 30° = 1/V3, then you will find 8 < 45°, whichaccording to the answer choices means 8 = 37°.

We can add vectors using x-components and y-components. We know that theresultant vector will have components in the x-direction and y-direction. In otherwords, for a system where the vectors A and B are added to form the resultantC, Cx = Ax + Bx and Cy= Ay+ By.

It ispossible to add vectors, to subtract vectors, to multiply vectors by scalars, totake components of vectors, and to find the magnitude or length of a vector, ifwe are given the components. If given the components, we can also find theangle of a vector. Vectors and their manipulation are an integral part incalculating physical quantities.

Copyright©by The Berkeley Review 15

Vectors


1 iiySiCS Translational Motion Speed, Velocity, and Acceleration

Speed, Velocity, and Acceleration

SpeedThere are many examples ofmotion, such as a rock falling off the edge ofa cliff.When a rock falls off the edge ofa cliff, it will travel a given distance defined bysome starting point and some ending point. The difference between those twopointsis called a displacement. It is importantto realize that the displacement ofan object might notbe thesame as the actual distance that an object travels. Theactual distance an object travels between its initial and final positions is called themagnitude of the displacement. As we learned in the section on vectors,displacement isa vector quantity. Thus, a sign (positive ornegative) is necessaryto specify the direction ofa displacement.

Motion can be measured by considering either speed or velocity. The averagespeed (denoted either asv orv) ofthe rock (or anyobject) issimply the distance(d) traveled divided by the elapsed time (t). The dimensions ofspeed are [L]/[TJ(e.g., milesper hour, metersper second,and so on).

y-d -distance traveled . .t elapsed time

Example 1.4In "running a gel," a microbiologist applies a voltage across an agar gel to inducestrands of DNA to move across the gel. Instead of leaving the voltage oncontinuously, he applies it intermittently. During the firsthour, theDNA travels4 cm. During the second hour, the machine is off and the DNA does not move.During the third hour, the DNA travels another 6 cm. The average speed inthisthree hour period is:

A. 3.3 cm/hour.B. 4 cm/hour.C. 5 cm/hour.D. 6 cm/hour.

Solution

The average speed of the DNA is found using equation (1.4). We must not forgetthat the time we need to be concerned with is the total elapsed time of the "run"and not just the time that was actually spent with the voltage on. With that inmind,we find that theaverage speedfor the run is 3.3 cm/hour.

y-distance traveled „di+d2 +d3_4cm +0cm +6cm_10crn ^33*™elapsed time tx +12 +13 lhr + lhr + lhr 3 hr hr


In Example 1.4 welooked at theaverage speed of the DNA. Since the DNA wasnot continuously moving, the speed of the DNA at nearly every instantof timewas not3.3 cm/hour. The instantaneous speed will tell us the speed ofthe DNAat each moment of the run. However, because speed is a scalar quantity, it willnot tell us the direction of the DNA motion.

Copyright©by The Berkeley Review 16 The Berkeley Review

1 fiySlCS Translational Motion Speed, Velocity, and Acceleration

VelocityWe have mentioned that velocity is a vector quantity. It depends on an object'sspeed and its direction of motion. The averagevelocity of a rock falling off a cliffis just the displacement (d) traveled divided by the elapsed time (t). Thedimensions of velocity (like speed) are [L]/[Tj. Note the difference between"distance traveled" in equation (1.4)and "displacement" in equation (1.5).

—_ d _ displacementt elapsed time

(1.5)

Example 1.5Themicrobiologist realized,near the end of the run in Example 1.4, that his DNAwas about to run off the gel plates. If the hall between his office and the lab is 50mlong, and he runs at 4 m/s, how longdoes it takehimto reach the lab?

A. 0.08 sec

B. 4 sec

C. 12.5 sec

D. 200 sec

SolutionAllwe need to do is substitute the appropriate values into equation (1.5). It takesthe microbiologist 12.5 secondsto movebetweenthe twolocations.

v =d t =4=-^m-=12.5st v 4m/s

The magnitude estimation technique could have helped you here. If you've everhad to run about 50m, you know it probably won't take0.08 or 200 seconds.


In most cases, it helps to assign an object's displacement to a coordinate system.For example, if the displacement is along a straight line, we could assign thatstraight line tobeanalogous to thex-axis. We could even assign thedisplacementto be along the y-axis. The displacement of an object on the x-axis would simplybe the difference in positions along the x-axis, or Ax = X2 - xi (where xi is theinitial position and X2 is the final position). Ananalogous situation for the y-axisis Ay = y2 - yi-Theelapsed timebetween the initial and final positions is just At=tfc - ti. Therefore, we can formally write the average velocity as:

-=Ax=X21x1At t2-t1

As we will see throughout physics, the idea of an averagevelocity will not be asimportant as a quantity called the instantaneous velocity, v. The instantaneousvelocity of an object is that object's velocity at a particular instant in time. Theinstantaneous velocity is given by equation (1.7).

(1.7)


1 hySlCS Translational Motion Speed, Velocity, and Acceleration

AccelerationAn object is accelerating whenever the velocity of that object is changing. Forexample, as a rock falls off a cliff, its velocity will increase from zero to some newvalue. The average acceleration, a, of an object is defined as the change in itsvelocity divided by theelapsed time for that change. This canbe seenin equation(1.8). The dimensions of acceleration are [L]/[T]2 and the units are usuallyexpressed in m/s2 or even ft/s2. If the acceleration ofan object is in a directionopposite to the velocity, then that object willbe decelerating.

—_ V2 - vi _ change in velocityt2-ti elapsed time

Example 1.6Peering from behind a bush, a field biologist notices that as a ground squirrelrushes over to get an apricot, it slows from 1.25 m/s to 0.25 m/s over a timeinterval of2 seconds. What is the average acceleration ofthe ground squirrel?

A. 0.50 m/ sec2 in thedirection of thevelocityB. 0.75 m/ sec2 in thedirection ofthevelocityC. 0.50 m/ sec2 opposite thedirection ofthevelocityD. 0.75 m/sec2 opposite thedirection ofthevelocity

Solution

Note thatthe instantaneous velocity ofthe squirrel is decreasing. In this case, weneed to be aware of the directions of our vectors. This means that the accelerationvector must point in a direction opposite to that of the velocity. This rules outchoices A and B. We mustnow do a bit of math to select the correct choice, sincethe two remaining numbers are not all that different.

Thesquirrel'svelocity decreases from 1.25 m/s to 0.25 m/s over2 seconds, so weget:

j„ V2-vi-0.25 m/s-1.25 m/s _ Q.5m/C2t2-h 2s-0s

The negative sign means the squirrel isdecelerating. The bestanswer ischoice C.

The instantaneous acceleration, a, of an object is the change in velocity of thatobject divided bya very small increment oftime for thatchange. This isshown inequation (1.9).

(1.9)

Copyright ©byThe Berkeley Review 18 The Berkeley Review

Physics Translational Motion Speed, Velocity, and Acceleration

Uniformly Accelerated MotionNow that we have developed the relationship between the position, velocity, andacceleration of an object with respect to time, we can examine four equations ofuniformly accelerated motion. It will be important to keep in mind that thesefour equations are valid only when the acceleration of an object is constant.These constant acceleration equations are of such great importance that we willlist themtogether, as shown in equations (1.10) through(1.13). Notethat we haveintroduced a new notation. A letter subscripted with an "o" (e.g., Xq, v0, or to)refers to the original condition,while a letter standing alone(e.g., x, v, or t) refersto the final condition. The "x" subscript indicates that all motion, whetherposition, speedor acceleration, is alongthe x-axis.

no vt included Ax = voxAt + V£ax(At)2

no dx included vx = vox + axAt

no a included Ax = ^(vox + vtx)At

no t included vtx2 = v0x2 +2axAx

(1.10)

(1.11)

(1.12)

(1.13)

Example 1.7A motorcycle drives past a stop sign and constantly accelerates during thistrespass of the law. After driving another 60 meters, a highway patrolman"clocks" themotorcycle's speedat 20 m/s. If this "clocking" occurs 4seconds afterthe motorcycle passes the stop sign, the speed of the motorcycle upon passingthe sign must have been:

A. 1 m/sB. 5 m/sC. 10 m/sD. 20 m/s

SolutionDraw a picture as shown below. We want tofind v0. We know that v =20 m/s,x0 = 0 m, x = 60 m, and t = 4 s. Note that we do not have an acceleration. Wetherefore need an equation that does not contain acceleration. If we chooseequation (1.12) and make the appropriate substitutions, we will find that thevelocity ofthe motorcycle asit passed the stop sign is10 m/s.

= ?vf= 20m/s

qP5 60mXf =60m &.ZL =(vl +2Q)£L .-.30 =̂ +20 50^=10^.

-x-f^)"=(-^^(4s)=(vi +20)(2s)

Xj = 0m

The best answer is choice C. The constant value of the acceleration in thisproblem is2.5 m/sec2.

Copyright©by The Berkeley Review 19 Exclusive MCAT Preparation

Physics Translational Motion Freely Falling Bodies

Freely Falling BodiesAfter careful measurements, Galileo Galilei concluded that all objects willaccelerate towards the center of the Earth at the same rate (if air resistance isignored). The average acceleration of gravity near the surface of the Earth isabout 9.8 m/s2 or32.2 ft/s2. "Average" acceleration isused, because very precisemeasurements of theacceleration of gravity will tell us that it is not exactly thesame at all locations on the Earth. For example, variations in the acceleration ofgravity will occurbecause of changes in elevation or even in latitude.

g = 9.8 m/s2 = 32.2 ft/s2

Justas we used the x-coordinate to represent the horizontal position ofan object,we canuse they-coordinate to represent the vertical position ofan object. We cannow modify the equations for uniformly accelerated motion in the x-direction tofit uniformly acceleration motion in the y-direction. These equations are shownin (1.14) through (1.17). Remember, we are assuming that the gravitationalacceleration isconstant. It iscommon when doing problems with these equationsto replace "a" with "-g." The minus sign indicates that we have chosen "down" tobe thenegative y-direction. As withthe"x" subscripts usedforx-axis motion, the"y" subscripts indicate positions, speeds, and accelerations along the y-axis.

no Vty included Ay =voyAt +Vfc ay(At)2

no dy included vy =voy +ayAt

no a included Ay =Vi (voy +vty)At

no t included vty* = voyz + 2 ayAy

(1.14)

(1.15)

(1.16)

(1.17)

Note again that the acceleration due to gravity is a constant, independent of themass or shape of the objects being dropped. Then why do we observe thatdifferent falling objects have different accelerations? Suppose we drop a piece ofpaper and a piece ofmetal. Initially, the surrounding air is exactly like the air inthe Earth's atmosphere. The two objects do not fall with the same accelerationbecause of air resistance. The piece of metal would hit the ground before thepiece of paper. However, if they were to fall through a vacuum, we would findthat both objects fall together, with the same acceleration.

If the acceleration is constant, then the distance [y] an object travels from aninitial position to a final position is just one-half times the acceleration times thetime squared. This is shown inequation (1.18). We arrived at this expression byusing equation (1.15) and setting y0 = 0, v0 = 0, and t0 = 0.

y =Vi ayAt2 (1.18)

We can apply dimensional analysis to this equation to see ifwe are correct. Bothsides of the equation should be the same. If we let [y] = [L] and substitute theappropriate parameters into equation (1.18), we will get equation (1.19). Indeed,we find that [L] = [L].

[L] =[^ [t]2 =[L][t]2

(1.19)



We can demonstrate the use of equation (1.18) by timing a falling steel ball. InFigure 1-10 we have a measuring stick with a magnet at the top. The steel ball isattached to the magnet. At a distance of roughly 0.3 m below the magnet is alight beam; at a distance of roughly 1.2 m below the magnet is another lightbeam. Releasing the steel ball from the magnet starts the timer. The times atwhich the light from the first light beam and second light beam are interruptedby the falling steel ball are recorded. From this experiment, it is possible tocalculate the time it takes the steel ball to reach the first and second light beams.The answer that we should get is 0.25 seconds for the first beam and 0.50secondsfor the second beam. The solution to this demonstration is shown below:

Timer

Magnet

Upper light beam

Lower light beam

•Ball

0.3 m

I 1.2 m-

y--aA1? l=-(9.8)(t!-g2tj = 0.25 s

y^-iaAt2 4=i(9.8)(t2-t0)2t2 = 0.50 s

Figure 1-10

From this experiment, we discover a useful shortcut for future free-fall problems.When the drop time is doubled, the drop distance increases by a factor of 4, aspredicted by equation (1.18). Knowing an object will fall approximately 1.25 m in0.5 seconds (neglecting wind resistance), we can approximate that the objectwould fall 5 m in 1.0 s, 20 m in 2 s, and so on. Assuming g = 10 m/s2, we findthat in the absence of wind resistance, Ay = 5t2. We can generate the followingapproximate data for fall time and fall distance, as shown in Figure 1-11.

1.25 m

0.5 s 5.0 m

;;1.0 s

20.0 m

2.0 s

Time Approximate Ah

0.25 s 0.31 m

0.5 s 1.25 m

1.0 s 5.0 m

2.0 s 20.0 m

3.0 s 45.0 m

4.0 s 80.0 m

5.0 s 125 m

6.0 s 180 m

7.0 s 245 m

8.0 s 320 m

Figure 1-11

Copyright © by The Berkeley Review 21

Approximate vt

2.5 m/s

5 m/s

10 m/s

20 m/s

30 m/s

40 m/s

50 m/s

60 m/s

70 m/s

80 m/s

Freely Falling Bodies

Test TipNumeric Estimation



We can do a similar demonstration with an inclined plane (Figure 1-12). If a ballis rolled down the inclined plane, we find that the shallow inclination of theplane has the effect of reducing the impactof gravity and thereby increasing thetimeof the descent fortheball. As theballrolls the lengthof the track, it picks upspeed. A metronome (timer) is used in order to give equal increments of time.When the first click of the timer is heard, the ball is released from its startingposition. At each subsequent clickof the timer, the ball will have rolled 20 cm, 80cm, 180 cm, and 320 cm, respectively. In this case, the net acceleration due togravity (which is partially offsetby the normal force) is uniform.

datlstdick =20x(l)2 =20cm

dat 2*1 click =20 x(2)2 =80 cm

80 dat3rdclick =20x(3)2 =180cm180

dat4«h dick = 20 x (4r = 320cm

Figure 1-12

Example 1.8A clumsy fruit jugglerdrops an apricot from the top of a cliff. If the cliffis 310 min height and the fruit is dropped from rest, how long does it take to reach thebottom,and how fast is it moving upon impact?

A. 5.62 sec and 39.0 m/ s downwardB. 7.95 sec and 77.9 m/ s downwardC. 31.6 sec and 310 m/s downwardD. 63.3 sec and 620 m/s downward

Solution

Inthis problem, we start by defining ourcoordinates. Let's choose the positive y-direction as upwards and the negative y-direction as downwards. We first wantto inventory our known variables. We can define y0 = 0 m, because mis is thepointfrom which the apricot was dropped. Therefore, when the apricot hits thebottom, y = -310 m. The apricot starts from rest, so v0 = 0 m/s. The accelerationdue togravity is-9.8 m/s2, butthe answer choices are far enough apart from oneanother that we can use -10 m/s2. In our system, the sign for g is negative. Thevariables thatwedo notknow are the time, t, and thefinal velocity, Vf.

Which equations should we use? Equation (1.14) does not have the final velocityterm, which will allow us to calculate the time it takes the apricot to hit thebottom of the cliff. As we shall see, the best answer will be +7.95 seconds.

T" ©

310 m


Ay =voyAt+-ay(At)2

-310 = (0)At+-(-10)(At)2

-310 = -5(At)2/. At2 = 62

At * 7.9 s



Short cut solution: We could have also arrived at this answer by using our newshortcut method. Knowing that in 1.0 s, an object will fall approximately 5 mallowsus to build a chart of values. When the drop time is doubled, the distanceincreases by a factor of 4. In 2.0 s, an object will fall approximately 20 m. In 4.0 s,an objectwill fall approximately 80 m. In 8.0 s, an object will fall approximately320 m. Fromthis information, we can guess that it will take slightly less than 8.0s to fall 310 m, assuming wind resistance to be negligible. This method is veryfast and would allow us to solve this question in about ten to fifteen seconds.

Giventhat eachanswer has a time and a final velocity, we could have also solvedthis question by determining the final velocity (rather than determining the timeof flight). We could have found Vf in the following manner.

Vfy2 =voy2 +2ayAyVfy2 =2 (-10X-310) =6200 m/s

vfy = 80 m/s-a little bit

This also concludes that choice B is the best answer. (Note: vfinai = lOt)

Square Root Estimation: Do not ever calculate a square root on a multiple choiceexam. Use square root estimation. Here, we want the square root of a number thatisapproximately 62 sec2. Find a number whose square root you know and thatisclose to the number whose square root you want to find. Here, 62 is close to 64,whose square root is 8. Therefore, the time must be around 8 seconds;specifically, a bit less than 8 seconds, because62is less than 64.

This singles out choice B as the answer. Notice that we don't even need tocalculate the speed to find the right answer. As a rule, always do as little work aspossible in seeking an answer. Full-blown arithmetic is not your friend on theMCAT, as it will slow you down. Keep in mind: Speed beats Precision!

Since the acceleration of gravity is a fairly important topic, let's take a look at itfrom a different vantage point. Let's set up a coordinate system in which thepositive y-direction is upwards and the negative y-direction is downwards.Suppose we throw a ball straight up into the air with a voy = 40 m/sec. Whendoes this ball stop going up? Another way to ask this question is, how long doesthe ball take to get to the top of the flight? When the ball gets to the top, it has abrief instant where vy = 0, because it changes direction. This means that vy = 0 atthe top. What is the acceleration due to gravity? It is equal to 9.8 m/s2. Usingequation (1.18), we find that it takes the ball approximately 4 seconds to reach thetop. If we were to plot velocity as a function of time; then we would see that thevelocity will fall uniformly to about t = 4 seconds (Figure 1-13).

vy ~ voy + g*

wherevy= 0;voy = -gt

t ='oy 40 m/s

"6 9.8 m/s2 ~



Test TipSquare Root Estimation



When the ball stops, how high will it be above the ground? Since we weren'tgiven the time, wemustuse equation (1.17). Letthe height at the topbe ytop anda = g (which is about -10 m/s2). Making the appropriate substitutions, we findthat the height to the top is about 80 meters.Figure 1-14 shows the distanceas afunction of time. Initially, we start with a certain slope, v0. As the ball gets closerand closerto the top, the velocity decreases until it (and the slope)reachzero.

vry voy2 =2ayAh

Vfy = 0, voy =40, and ay = •10

-voy2 =2ayAh

Ah =voy

"2a7

1600

-20

-402 m2/s22x(-10) m/s2

i^=80mTime

Figure 1-14

When will the ball return to the ground? Before trying to solve the problem, it isbest to reason it out. If the ball takes about 4 seconds to reach the top, then itmust take another 4 seconds to reach the ground, because the vertical portion ofa projectile's flight (neglecting air resistance) is symmetric. The total time in theair is 8 seconds. We can arrive at this answer by using equation (1.14). Solvingthis expression gives the desired answer.

y=voyt +Vi ayt2

0=voyt-^gt20=t(vOy-^gt)

t=

0= VQy - Vl gt

2voy 2(40 m/s)8 " 9.8 m/s2

= 8s

The graphs in Figure 1-13 and Figure 1-14 consider only the upward portion of aprojectile's flight We can finish Figure 1-13 by drawing Figure 1-15. If we startwith some velocity, v0, then when the ball reaches the top at t = 4 seconds, thevelocity willbe zero.Theball will change direction and hit the ground 4 secondslater. As the ball comes down to the ground, the velocity increases. However,since the direction of the ball changed, we place a negative sign in front of thevelocity. Completion of Figure1-14gives the parabola shown in Figure1-16.

Time (s)

Figure 1-15

Copyright ©by The Berkeley Review 24 The Berkeley Review


Knowing that the upward half of a flight is simply the symmetric opposite of thedownward halfof a flight can prove to be useful when quickly solving multiple-choice questions. Consider the following turbo-solved questions.

Example 1.9 (Turbo-solved question)Approximatelyhow high above the ground is a spherical projectile nine secondsafter it has been launched straight up with an initial velocityof 60 m/s?

A. 11.1m

B. 44.4 m

C 138.2 m

D. 320.7 m

Turbo Solution

As we've seen in the previous examples, the time it takes for a projectile to reachits apex is equal to its initial vertical velocity divided by 9.8 (t = v0/a). Thismeans that it takes roughly 6 seconds to reach its highest point and roughly 12seconds for the total flight. The height it reaches at the 9 second mark will beequal to the height is reaches at the 3 second mark. From here we solve usingequation (1.14) Ah = v0t + Vi at2, where v0 =60 m/s anda =-10 m/s2.

Ah = v0t + % at2= 60(3) + % (-10)(3)2

Ah = 180 + Vi (-10)(9) = 180 + Vl (-90) = 180 - 45 = 135 (choice C)

Example 1.10 (Turbo-solved question)What is the approximate vertical launch speed for an object that is at the sameheight above the ground at 2.7 seconds as it is at 5.3 seconds?

A. 27.3 m/sB. 39.2 m/sC. 53.5 m/sD. 78.4 m/s

Turbo SolutionIf a projectile is at the same elevationat 2.7secondsand 5.3 seconds, then it mustbe traveling upwards at 2.7 s and downwards at 5.3 s. We can conclude that itreached its apex at 4 s. We know from other examples that the time to reach theapexis v0/a, so its initial launch speed must be t x a =4 x10= 40m/s, choice B.

Example 1.11 (Turbo-solved question)For an objectlaunched straight up with a vertical speed of 45 m/s, what is NOTtrue?

A. The projectile will be traveling upwards at the 3.0s mark.B. The projectile will be the same distance above the ground at the 2.0s mark as

me 7.2 s mark.C. The projectile will have the same speed at the 3.2s mark as the 6.0s markD. The object will reach its apex at the 6.6 s mark.

Turbo Solution

Choices B and C are complementary answers, because they average out to 4.6 s.It also fits with choice A where at times less than 4.6 s, the projectile will betraveling in the upwards direction. Because the apex must be reached at the 4.6smark, choice D must be a "NOT true" statement.



Test TipTurbo Solution


PhysiCS Translational Motion Perturbations

Perturbations

Wind ResistanceUp until now,all of our examples have disregarded the impact of air resistance.In Chapter 7 we shalladdress the conceptof drag forces associated with motionthrough a fluid medium in greater detail. For now, we shall simply say that adrag force such as wind resistance opposes the direction of an object's velocity.The magnitude of such a resistive force typically depends on the contact area ofthe object with the surrounding medium, the viscosity of the surroundingmedium, and the speed of the object It is unlikely that you will have do anycomputations to determinethe magnitude of the drag force, but you should havea good conceptual understanding of the impact of wind resistance on the flightofan object.

Consider dropping a solidball and a hollowed-out ball of the same volume andmaterial from a platform 20 m above the ground. We predict that in the absenceof wind resistance, an objectshould take about 2.02seconds to reach the groundbelow. However, what we observe is that the solid ball takes 2.07 s and thehollowed-out ball 3.11 s. Both balls took longer than the 2.02 s that we assumefrom calculation, but to a notably different degree. The solid ball was slowedslightly by a drag force while the hollowed-out ball was slowed much more. Bothballs have the same shape and volume, so they each must have the same contactarea. They are each falling through the same still air, so the viscosity of themedium cannot explain the difference in the drop times.

The explanation for the difference in fall times has to do with the impact of thedrag force on the acceleration of the balls. For a solid ball, the impact of windresistance over a short distance is trivial compared to its weight. However, thehollowed-out ball is much less massive, so the magnitude of the drag forcequickly becomes comparable to its low weight which means that while it mayfeel an acceleration of -9.8 m/s2 when it is first released, that value drops inmagnitude rapidly as it gains speed upon descent. Our empirical conclusion isthat objects with a large contactarea-to-mass ratio experiencea more significantimpact of wind resistance than their denser counterparts.

Example 1.12Whichof the following falling objects will exhibitthe greatest terminalvelocity?

A. A ping-pong ballB. A parachuterC. A rock

D. An envelope

Solution

Terminal velocity is the greatest speed an object can reach. This occurs when theacceleration acting on an object has a magnitude of 0. For objects falling throughair, this results from the drag force perfectly offsetting the gravitational force(equal in magnitude but opposite in direction). The greatest terminal velocitywill be associated with the densest object having minimal contact area, becausethe resistive force must overcome the greatest weight per surface area. A ping-pong ball is hollow, so its density is low, resulting in a small terminal velocity.Both a parachuter and an envelope have large surface areas, so their large contactareas generate a great drag force, resulting in small terminal velocities. A rock isquite dense, so it has a small surface area-to-mass ratio, resulting in a trivial dragforce until the speed becomes significant. A rock will reach a high speed beforethe drag force is able to offset the gravitational pull. Choice C is the best answer.



Projectiles

Lefs consider two-dimensional motion. What takes place in the x-direction andwhat takes place in the y-direction can be treated as two independent andseparate quantities. Let's apply this to an object that is subject to the accelerationof gravity. Gravity pointsdownward in the y-direction. There is no accelerationdue to gravity in the x-direction. This means that ax =0and ay =-g.

Because we can treathorizontal and vertical motion separately, wecan solve forthe motion of a projectile by using the equations for uniform horizontal motionin a straight line at constant velocity and the equations for vertical motion in aline with constant acceleration.

Motion in the x-direction has zero acceleration and so it is motion with constantvelocity, vx = vOX/ where vox is the initial velocity. The displacement ofx will beequal to voxt. Similarly, we find that vy =voy -gtand that y=voyt -Vi gt2.

Lefs first consider what happens when we throw anobject horizontally from anelevation. Whether we drop the object from rest or throw it horizontally, the y-direction aspect of each pathway is identical, because they start from rest in they-direction. At some point in your physics education, a physics instructor hasundoubtedly told you that if you fired a bulletfrom a gun horizontally and thesame time you dropped a bullet from the same height, the two bullets would hitthe ground at the same time. This is based on the idea that the y-direction isindependent of the x-direction.

Figure 1-17 shows four trajectories representing four different launches. Thearrow labeled Launch 1 represents the trajectory ofan object being droppedfromrest with an initial height of h. The arrow labeled Launch 2 represents thetrajectory of an object being thrown horizontally at an initial speed of 4units/time from aninitial height ofh.The arrow labeled Launch 3 represents thetrajectory of an object being thrown horizontally at an initial speed of 8units/time from an initial height of h. Finally, the arrow labeled Launch 4represents the trajectory ofan object being thrown horizontally at an initial speedof12units/ timefroman initial height of h.

All foilr launches start at the same hei

Horizontal launches in

"^^all cases execpt vox =0

Launch 1 Launch 2\ Launch 3\ Launch 4 \^vOx = 0 vox =4 \ vox =8 \ Vox =12 \vOy = 0 voy =0 ^ voy = 0 \ voy =0 \

1 '

Copyright©by The BerkeleyReview

0 x 2x 3x

x-Displacement at collision

Figure 1-17

All four objects strike theground at the same time. Launch 3 has twice the rangeofLaunch 2 and Launch4 has three times the range of Launch 2.

27

Projectiles


Physics Translational Motion Projectiles

Example 1.13What is the flight time for an object fired from a horizontal cannon at 57.5 m/sfrom a cliffthat is 143m above the ground below?

A. 2.4 secondsB. 3.9 secondsC. 5.4 secondsD. 8.1 seconds

SolutionThe initial velocity in the x-direction is extraneous information, because the timeof flight depends solely on the y-direction components of the trajectory. Theobject as no initial velocity in the y-direction, so it can be solved in the samefashion as if the objectwere dropped from rest starting at an elevation of 143 m;The turbo-method is to note that a dropped objectcovers 5 m in 1 second, 20 m in2 seconds, 80 m in 4 seconds, and 320 m in 8 seconds. 143 m falls between 80 mand 320 m, so the time of flight must lie between 4 seconds and 8 seconds. Onlychoice C fits in the range, so choice C is the best answer. We could have alsofound the answer using equation (1.14).

Ah = v0t+ Vl at2

-143 = 0 + Vi(-9.8)t2 = Vi(-lO)t2 = (-S)t2

143 / 5=t2, so t2 =28.6 and t=5.something (choice C)

The point of Example 1.13 is to show that even though the objectwas launchedhorizontally, we can treat the y-componentof its flight likean object in free fall.

Let's consider an exampleinvolving a projectile. The projectile is set into motionfrom the originwith a definite directionwith respect to the horizontal,defined as6o, and a definite starting speed, v0. The projectile's path will follow a parabola,or a piece of a parabola (Figure 1-18). Two properties of the parabola are theheight (h) and the range (r). These quantities can be related to the startingconditions (v0 and 60).

y-axis

x-axis

Figure 1-18

The initial velocity, Vq, can be divided into a horizontal component, as shown inequation (1.20), and a vertical component, as shown in equation (1.21). Motion inthe x-direction will be uniform motion and follow equation (1.22), while motionin the y-direction will follow the rules of an object that is thrown verticallyupward, as shown by equation (1.23).

Copyright ©by The BerkeleyReview 28 The Berkeley Review


v0x = voCos0o

v0y =v0Sin60

x = v0xt

y-Voyt-O-Jgr2

(1.20)

(1.21)

(1.22)

(1.23)

Note that the projectile's pathway is a parabola. The time toget to the top of theparabola is given by equation(1.24), and the time to return to the ground is equalto that time (ttop = tdown)- The range of the projectile can be given by equations(1.25) or (1.26), whilethe height is givenby equation(1.27).

ttop =̂g

2VoxV0vr =

g

r =vl Sin20

g

h=<2g

(1.24)

(1.25)

(1.26)

(1.27)

Equations (1.25) and (1.26) will work only if the initial heightand the final heightare the same. Equations (1.24) and (1.27), however, are general equations and donot depend on the initial and final heights being the same.

Let's consider multiple launches where the launch angle, Oq, changes while thelaunch speed, Vq, remains constant. We find a series of paths that the projectilecan take (Figure 1-19). If the projectile were launched at 90°, then it would gostraight up and fall back down again. If the projectile were launched at 45°, thenit would travel a great range in the x-direction, but not reach as high a maximumheight as it did with the 90° launch. If we launched the projectile at 0°, then itwould immediately fall below the x-axis. All of the paths of our projectile will fallwithin an envelope (represented by the dashed line in Figure 1-19). Notice thatwe have shown only launch angles of 45° and greater. Launch angles less than45° will fit in the envelope as well, but no part of their trajectory will touch theedge of the envelope. This is because a 45° launch angle corresponds to themaximum possible range, and no other angle can reach the maximum range.

y-axis

x-axis

Figure 1-19


Projectiles



The maximumheight that a projectile can possiblyobtain is h (possible only witha launch angle of 90°), while the maximum range is 2h (possible only with alaunch angle is 45°). As the launch angle is reduced, the maximum heightdecreases, due to a reduction in the vertical launch speed. If the projectile werelaunched at 45°, then its maximum height would be h/2. This concept can bedemonstrated using water droplets (projectiles) being shot from a hose. Themaximum range of the droplets will take place when the angle of the spout theycame from is at 45°. Consider the diagram in Figure 1-20, which considers threelaunch angles, 30°, 45°, and 60°. When the water leaves the hose at 45°, themaximum range is 40 units. The maximum height occurs at 20 units (in the x-direction) and corresponds to 10 units in the y-direction. Larger angles result in agreater maximum height. Launch anglesof30° and 60° result in the samerange.

x-axis

Figure 1-20

Knowing the relationship between maximum height and range for variouslaunch angles can help save time of multiple-choice questions. Table 1.4 showsthe data associated with three launches with identical initial launch speed, butdifferent launch angles.

Table 1.4

Angle 'initial 'initx initv Range

30° 20 m/s 17.2 m/s 10.0 m/s 34.4 m

45° 20 m/s 14.2 m/s 14.2 m/s 40.0 m

60° 20 m/s 10.0 m/s 17.2 m/s 34.4 m

With a launch angle of 30°, the object has a relatively short flight time and thusdoesn't reach a great maximum height. However, because of its relatively largex-direction velocity, it is able to cover a good amount of x-direction distanceduring its time in the air. With a launch angle of 60°, the object has a relativelylong flight time and thus reaches a fairly high maximum height. However,because of its relatively small x-direction velocity, it doesn't cover all that muchx-direction distance during its time in the air. It ends up that the reduction in itsx-direction speed is offsetby its increased flight time, because launches at anglesof 30° and 60° generate the exact same range. This is confirmed by equation(1.26), which shows that complementary angles (angles that add to 90°) result inthe same range. Consider the following turbo-solved question.

"max Time

2.04 s

R/h

5.1m 6.7

10.0 m 2.84 s 4.0

14.6 m 3.44 s 2.4



Example 1.14(Turbo-solved question)What is the range for a projectile launched at a 45° angle thathasa flight time of3 seconds?

A. 14.2 m

B. 44.1m

C. 72.2 m

D. 144.4 m

Turbo Solution

With a flight time of 3 seconds, it takes 1.5 s to reach its apex. A 1-second droptime covers 5 m and a 2-second drop time covers 20 m, so a 1.5-second dropcovers about 11 to 12 m, so the maximum heightis about 11-12 m. The range fora 45° launch is four times the maximum height, so R = 44-48 m, choiceB.

Example 1.15The fruit juggler, dissatisfied with simply dropping fruit, decides to throw anapricot off a cliff. The apricot leaves the top of the cliffat a speed of 35 m/s andat an angle of 50° above the horizontal. If the cliff is 310m tall, how far downrange is the fruit when it hits the ground? (Neglect air resistance, and note thatthe Cos 50° is 0.643 and the Sin 50° is 0.766.)

A. 180 m

B. 251m

C. 390 m

D. 1423 m

Solution

Lefs first solve this question using traditional physics methods. To do so, weneed to establish the velocity of the apricot in the x-direction and in the y-direction from trigonometric relationships. We will use the standard x-axis andy-axis, as shown below.

R = ??

v^ =(35 m/s)(cos 50"

v0 =(35m/s)(0.643)

vn « 22.5 m/s

vOy= (vo)(sin 6)v« =(35m/s)(sin50°)

v„ =(35m/s)(0.766)

v„« 26.8 m/s

Although these numbers are specific, you can still make use of the numericestimation. To continue, lefs take an inventory of the variables that we have andthose that we do not know. We separate them according to horizontal andvertical components.

Copyright ©by The Berkeley Review 31

Projectiles



Horizontal component Vertical component

vx= 22.5 m/s x = ??

t = ??

y = -310 m vy = ??Voy =26.8 m/s y=??ay =-9.8m/s2

If we choose equation (1.14), we will be able to calculate the flight time of theapricot from when it leaves the student's hand to when it hits the ground. Theonly variable in that equation that we do not know is the time, t. Doing theproblem this way requires using the dreaded Quadratic Equation! This solutionis reallya chance for you to see how to apply this equation. Let'ssolve for t:

y=v0yt +^ayt2

-310 m = (26.8 m/s)t + Vi (-9.8 m/s2)!2

(-4.9 m/s s2)!2 + (26.8 m/s)t + 310m = 0(in the form of a quadratic equation)

To precisely solve for the time, we must use the quadratic equation. Only thepositive time is valid,because that willbe a time after the apricot was thrown.

x=--b±Vb2-4ac

2aQuadratic Equation

t_-26.8 m/s ±̂ (26.8 m/g)2 -4(- 4.9 nys2)(31o"m72(-4.9n/s2)

_-26.8 m/s ±V(718.24 mVs2) +6076 "^/J-9-8"/s2

_-26.8 m/s ±̂ 6794.24 "^/^ _-26.8 ™/s ±82.43 m/s-9.8 % " -9.8 %

t = - 5.58 s or t = + 11.15 s

Since we know the velocity of the apricot in the x-direction, and we have justcalculated the time of flight, we can now solve for the distance that the apricotwas thrown using the x-direction speed and the time offlight. It isapproximately251 meters.

x = vxt = (22.5m/s)(11.15s)

22.5 x 11.15 = 225 + 22.5 + 2.25 + 1.125 = 250.88 m

It is approximately 251 meters, so the best answer is choice B. However, thereality is that we just spent way too much time solving a multiple-choicequestion to be useful on the MCAT. The purpose of showing the long-windedsolution is two-fold. First was to reacquaint you with what you learned inphysics before. Second was to make the turbo-solution method that much moreimpressive in terms of its utility on the MCAT.

The conclusion so far: NEVER use the quadratic equation to solve anything onthe MCAT!



Turbo Solution

This is a challenging question even for the turbo-method, because we'll needmultiple steps.

Start bybreaking the overall pathway into two component portions: the upwardssegment (from start until it reaches its apex) and thedownwards segment (fromits apex until it strikes the ground). For the upwards portion of the flight, weknow that v0 = 26.8 m/s, so the time it takes the apricot to reach its apex isvGy/g =26.8/9.8 s- 2.8 s.

Fromthat, we can approximate that the apricot climbs about 40m from launch toits apex(based on the approximation that a 1-sclimb leadsto a heightof 5 m, a 2-s climb leads to a height of 20m, and a 3-s climb leads to a heightof 45 m). Thismeans that at its highestpoint, the apricot is roughly 350 mabovethe ground.

Weknow from our trick in Figure 1-11 that it takes8 s to fall 320 m, so falling350m will take just over 8 s. To solve more precisely for the time to fall 350 m to theground below, we can use equation (1.14), where 350 =Vi ayt2. Plugging in -10for g leads to70 = t2, so t« 8.3 s orso(given that82 =64 and 92 = 81).

We now know that the total flight time is roughly 2.8 s + 8.3 s = 11.1 s. Todetermine the range requires multiplying the flight time by the x-directionvelocity, as it did in the long winded solution.

dx = vxt = (22.5 m/s)(ll.l s) = 23 x 11 = 253 m

This method, using approximations and information from previous exampleswe've consciously decided to retain for future reference, has allowed us to avoidthe quadratic equation and get a pretty good answer for a multiple-choicequestion in a reasonable amount of time. Becauseafter all, speed beats precision!

The turbo-method can only be developed through much practice, so as you moveon through the homework passages and questionsat the end of this chapter, tryto use it as oftenas possible. One of the primary goalsof this bookis to help youdevelop the speed needed to do well on the physical sciences section of theMCAT.

Example 1.16aWhat is the range for a projectile that is launched at an angle of 45° with respectto horizontal with an initial speed of 25 m/s?

A. 50.0 m

B. 33.3 m

C. 72.2 m

D. 62.5 m

Solution

The easiest way to solve this question is to plug into the range equation thatcontains launch velocity, acceleration, and launch angle, equation (1.27). If weplugin 25 m/s for the initial speed, 9.8 m/s2 for g, and 45° for 9, then we get thefollowing solution:

p_v02(sin2e) _252(sin90°) _625.625g 9.8 ~ 10

m

The range is approximately 62.5 m, so choice D is the best answer.


Projectiles


PhysiCS Translational Motion Projectiles

Example 1.16bWhat is the maximum height attained during flight for a spherical projectilelaunched from a 45° angle with respect to horizontal at an initial velocity of 30m/s?

A. 22.9 m

B. 40.0 m

C. 55.5 m

D. 91.8 m

SolutionThe quickest way to solve this question is to start by plugging into the rangeequation, equation (1.27). Once you have determined the range, then you mustdivide the range by 4 to get the maximum height for a projectile that travels aparabolic pathway from a 45° launch angle.

p_v02(sin2e) _302(sin90°) _900_90mg 9.8 ~ 10

The range is approximately 90 m, so the maximum height is approximately22.5m. The best answer is choice A.

Example 1.17Whatare the rangeand maximum height for a spherical projectile launchedfroma 30° anglewith respect to horizontal with an initialvelocity of20m/s?

A. R = 32.8 m; hmax = 8.2 mB. R = 45.8 m; hmax = 6.8 mC. R = 26.7 m;hmax = 13.4 mD. R = 34.4 m; hmax = 5.1 m

Solution

For a 30° launch, the range is a little less than 7 times the maximum height Thiseliminates choices A and C, which have range-to-maximum height ratios of 4and 2 respectively. To decide between choices B and D, we need to first considerthe range. If the launch angle were 45°, then the rangewould be 40 m (202/10).Because the range is less whenthe projectile is launched at a 30° angle than whenthe launch angle is 45°, the best answer cannot be choice B. Only choice Dremains, so it must be the best answer by processof elimination.

p_v02(sin2e) _202(sin 60°) ,400x0.86 ,340 =34mg 9.8 ~ 10 10

The range is approximately 34m,so the maximum height is approximately 5 m.The best answer is in fact choice D.


25 Translational Motion Review Questions

I. Cannon Comparison

II. Uniform Acceleration

III. Acceleration due to Gravity

Questions Not Based on a Descriptive Passage

(1-7)

(8 -14)

(15 - 21)

(22 - 25)

The main purpose of this 25-question set is to serve as a review of the materialpresented in the chapter. Do not worry about the timing for these questions.Focus on learning. Once you complete these questions, grade them using theanswer key. For any question you missed, repeat it and write down your thoughtprocess. Then grade the questions you repeated and thoroughly read the answerexplanation. Compare your thought process to the answer explanation and assesswhether you missed the question because of a careless error (such as misreadingthe question), because of an error in reasoning, or because you were lackinginformation. Your goal is to fill in any informational gaps and solidify yourreasoning before you begin your practice exam for this section. Preparing for theMCAT is best done in stages. This first stage is meant to help you evaluate howwell you know this subject matter.

Passage I (Questions 1 - 7)

A cannon is located on the edge of a cliff 100 m above aflat, sandy canyon floor. Three cannon balls are launchedfrom the cannon with the same initial speed of 50 m/s, but atdifferent angles. The cannon balls are identical except fortheir color. A red ball is launched horizontally, a white ball islaunched at a 30° angle above the horizontal, and a blue ballis launched at a 60° angle above the horizontal.

In all questions you may ignore air resistance and thelength of the cannon, unless otherwise stated.

1.

2.

For reference: Sin 30° = 0.50

Sin 60° = 0.86

g=10m/s2

Cos 30° = 0.86

Cos 60° = 0.50

About how long will it take the red cannon ball to strikethe canyon floor?

A. 3.5seconds

B. 4.5seconds

C. 5.5seconds

D. 6.0seconds

If it were possible to launch all three cannon ballssimultaneously, in which order would they hit theground?

A. White, blue, red

B. Blue, white, red

C. Red, white, blue

D. Red, blue, white

3. In comparing the maximum heights and ranges for thewhite and blue cannon balls, which of the followingstatements is TRUE?

A. The blue ball has a smaller maximum height but alonger range.

B. The blue ball has a larger maximum height and alonger range.

C. The white ball has a smaller maximumheight and ashorter range.

D. The whiteball has a smaller maximum heightbut alonger range.

Copyright ©by TheBerkeley Review®

4.

5.

6.

7.

36

Approximately how far above the canyon floor will thewhite ball be when it reaches its apex?

A. 30 m

B. 120 m

C. 131m

D. 225 m

When the white cannon ball reaches its apex, the:

A. acceleration is zero and its velocity is not zero.

B. acceleration is zero and its speed is zero.

C. acceleration is not zero and its velocity is zero.

D. acceleration is not zero and its speed is not zero.

What is the acceleration of the red cannon ball while it

is insidethe cannon barrel, if the barrelis 1 m long?

A. 25 m/s2

B. 125 m/s2

C. 250 m/s2

D. 1250 m/s2

Which of the three cannon balls hits the groundwith theGREATEST kinetic energy?

A. Red

B. White

C. Blue

D. They all hit the ground with the same kineticenergy.

GO ON TO THE NEXT PAGE

Passage II (Questions 8-14)

Students who wish to study uniformly acceleratedmotion make the following measurements on a moving car:A car of mass 1500 kg accelerates uniformly along a straighthorizontal road from rest to a maximum velocity of 15 m/s in10 seconds. After this, the car remains at a constant speed of15 m/s for the next 10 seconds. It then decelerates uniformlyto 5 m/s during the following 5 seconds, and remains at thisconstant speed for 5 seconds. The students use just one trialto observe and collect the information they desire, becausethe pathway and velocity of the car are determined prior tothe start of the experiment. Figure 1 shows four separatepossible graphs that are presented by the students as part oftheir analysis.

Time of car's motion

III

\0 Timeof car'smotion


IV


Figure 1.

8. Which graphs in Figure 1 best represents therelationship between the magnitude of velocity of thecar and the time of the car's motion?

A. I

B. II

C. Ill

D. IV

9. Which of the graphs in Figure 1 best represents therelationship between the displacement of the car and thetime of the car's motion?

A. I

B. II

C. Ill

D. IV

Copyright ©byThe Berkeley Review® 37

10. How far will the students say the car has traveled in thefirst 10 seconds?

A. 7.5 m

B. 15 m

C. 75 m

D. 225 m

11. What is the value of the second acceleration?

A. 1 m/s2 in thedirection opposite thevelocity.B. 1.5 m/s2 in thedirection opposite thevelocity.C. 2 m/s2 in thedirection opposite thevelocity.D. 1 m/s2 in thedirection of the velocity.

12. If the acceleration in the first time period were insteadtwice as large in magnitude, how would the speed at theend of the period compare to 15 m/s?

A. The speed would be half as much.

B. The speed would be the same.

C. The speed would be twice as much.

D. The speed would be four times as much.

13. In order to change the maximum velocity obtained bythe car, the students should:

A. allow the car to travel a greater distance during thefirst acceleration period.

B. use a car with a larger mass.

C. allow the car to coast before decelerating.

D. perform this experiment on a planet with a reducedforce of gravity.

14. The effect of friction between the car's tires and the

road:

A. means the students need to re-do this problem,taking friction into account.

B. cannot be determined, unless we know thecoefficient of friction.

C. is not present in this problem.D. has already been taken into account.


Passage III (Questions 15 - 21)

In an attempt to study kinematics and in order toaccurately determine the acceleration due to gravity, a groupof students conducts two experiments. Both experimentsinvolvedropping balls of varying mass out of windows. Airresistance is treated as negligible in the followingexperiments:

ExperimentI

The students wish to measure the value of g. Fourstudents participate in this experiment. Three studentseach drop a ball out of different windows. The windowsare all at the same height, 20 meters above the ground.One ball is made of lead, with a mass of 0.25 kg andstrikes the ground in 2 seconds. A second ball is alsomade of lead, with a mass of 0.5 kg. The third ball ismade of plastic and has a mass of 0.25 kg. The fourthstudent drops a lead ball, mass 0.25 kg, from a window40 meters above the ground.

Experiment 2

For this experiment, three students each drop a ball outof different windows at the same height above theground, 20 meters. All the balls have the same mass.Ball I is dropped straight down. Ball II is thrown straightdown with an initial velocity of 5 m/s. Ball III is thrownout horizontally with an initial velocity of 5 m/s.

Note: Unless otherwise stated, ignore the effects of airfriction.

15. For Experiment 1, in comparing the effect of mass andmaterial on the value of g, the students found that thevalue of g was affected by:

A. using balls with different mass, but made of thesame material.

B. using balls made of different materials, but havingthe same mass.

C. using a ball either with a different mass or beingmade of a different material.

D. Neither massnor material will affect the valueof g.

16. From the data in Experiment 1, what did the studentsobtain for the value of g?

A. 9.6 m/s2

B. 9.8 m/s2

C. 10.0 m/s2

D. 10.2 m/s2

Copyright ©by TheBerkeley Review® 38

17. In Experiment 2, which balls hit the ground at the sametime?

A. Ball I and Ball II only.

B. Ball I and Ball III only.

C. All of the balls strike at the same time.

D. None of them strikes at the same time.

18. In Experiment 2, how far away from the bottom of thebuilding will Ball III land?

A. 2.4 meters

B. 5.0 meters

C. 10.0 meters

D. 20.4 meters

19. In Experiment 1, what effect does doubling the heighthaveon the velocityof the ball when it hits the ground?

A. The velocity increase by a factor of 2.

B. The velocity increases bya factor of V2.C. The velocity increases by a factor of 4.D. The velocity is independentof height.

20. Howfast will Ball II hit the ground?

A. 20.4 m/s

B. 30.5 m/s

C. 40.0 m/s

D. 201.0 m/s

21. Compared with Experiment 1, Experiment 2 moredirectly demonstrates:

A. the independence of the horizontal and verticaldirections.

B. how horizontal motion increases the effects ofgravity on a moving object.

C. nothing new,comparedto Experiment 1.D. the importance of gravity in causing something to

fall.


Questions 22 through 25 are NOT based on a descriptivepassage.

22.

23.

24.

25.

When a projectile undergoing parabolic motion is at itsmaximum height, which of the following statements istrue?

A. The speed is zero.

B. The speed and the velocity are zero.

C. The speed equals the initial speed.

D. The speed is at a minimum.

All of the following can affect the initial x-directionvelocity of a projectile fired from an spring-loaded gunEXCEPT the:

A. angle at which the projectile is launched.B. distance the spring is compressed from

equilibrium position.C. radius of the 100-gram projectile.D. the thickness of the coils in the spring.

its

While an air cart is moving up and then down aninclined air track the magnitude of the acceleration:

A. the cart experiences on the way up is only half asmuch as the magnitude of the acceleration itexperiences on the way down.

B. on the way up is necessarily different than themagnitude of the acceleration on the way down.

C. has the numerical value of the magnitude of theacceleration due to gravity, g.

D. on the way up is the same as the magnitude of theacceleration on the way down.

For a disk that travels up an inclined surface for 11.25m and then changes directions and returns down thesame 11.25 m with a total transit time of 6 seconds,what are the average speed and average velocity for theentire motion?

A. speed = 1.88 m/s; velocity = 1.88 m/sB. speed = 1.88 m/s; velocity = 0C. speed = 3.75 m/s; velocity = 3.75 m/sD. speed = 3.75 m/s; velocity = 0


1. B 2. C 3. D 4. C 5. D

6. D 7. D 8 C 9. B 10. C

11. C 12. C 13. A 14. D 15. D

16. C 17. B 18. C 19. B 20. A

21. A 22. D 23. C 24. D 25. D

YOU ARE DONE.

Answers to 25-Question Translational Motion Review

Passage I (Questions 1-7) Cannon Comparison

l.

4.

Choice B is the best answer. To arrive at the answer, we should use a kinematics equation for the y-direction. Remember,when you are asked to calculate the time of flight, you usually need to use the y-direction information. The red ball waslaunched horizontally, so its initial velocity in the y-direction is zero. The kinematics equation we need is: Ay = v0 t + '/£at2.If we choosedownward to be the positive y-direction, then Ay= 100m, v0 = 0, and a = g = 10 OL, Therefore,

s2

Ay = !4at2, 100 = !£(10)t2, 10 = '/it2, 20 = t2 so t = V20 sec.

Solving for t requires us to estimate the square root of 20. This should lie between 4 and 5, the square roots of 16 and 25.Only choice B lies between 4 and 5. We could also have used our trick knowing that in 1 second it would fall 5 m, in 2seconds it would fall 20 m, and in 4 seconds it would fall 80 m. To fall 100 m, we needjust more than 4 seconds. The bestanswer is choice B.

Choice C is the best answer. The angle of the launch should tell us the order in which the cannon balls strike the ground.The ball with the greatest maximum height will be the ball that remains in the air the longest and thus is the last tostrike theground. The blue ball should have the longest time of flight, because it was launched at the steepest angle. This implies thatthewhite ball should finish in the middle, while the red ball should hit the canyon floor first. Because the white ball is in themiddle, it should land second, which rules out choices A and D.

A mathematical solution requires Ay = v0yt + '/$at2, with none of the terms equal to zero, which would require using thequadratic equation. Solving the quadratic equation for this problem would take quite a bit of time, so an intuitive andqualitative solution involvingvisualization is bestfor this question. The best answer is choice C.

Choice D is the best answer. The blue ball is launched at a steeper angle than the white ball and should therefore have thegreater maximum height than the white ball. This permits us to rule out choice A.

White

Blue

White

Now we must consider the range. Before we actually determine which ball has the longer range, let's look at choices BandC. Note that they are saying the same thing, which means neither of them can be the best answer. Choices B and C eliminateone another. Thus, choice D must be correct by process of elimination. To verify this, consider that the blue and white ballare launched atcomplementary angles, so if they were to land at the same elevation atwhich they were launched, they shouldexhibit the same range. However, because they land at a lower elevation, they remain in the air longer than the typicalscenario, so the white ball is able to travel farther than the blue ball, having a greater x-direction velocity than the blue ball.A picture is helpful to convince you that the blue ball does indeed have a shorter range. If you consider the picture here,you'll note that the two paths cross on thehorizontal line extended from launch point. The best answer is choice D.

Choice C isthe bestanswer. The ball is launched from a height of 100 m, so the maximum height must be greater than 100m, which eliminates choice A. The cannon is on a cliff, so for the sake of simplicity, we can let the initial position of thecannon height = 0. We can then calculate how far above the cliff the ball rises and add another 100 m to figure outhow farabove the canyon floor it rose. To calculate how far above the cliff the ball rose without knowing the time, we need to use-,% -„2Vfy = vgy + 2aAy, noting that at the apex, the y-velocity is zero.

I =v?_ x%a„ A„-V5y_(50sin30°)2Vfv = v6y + 2aAy Ay = -^- =2g 2(10)

Ay = 2500 _ 2508(10) 8

30 m

The ball rose about 30 m above the cliff, which means it rose about 100 m + 30 m= 130m above thecanyon floor.

Copyright©by The Berkeley Review® 40 MINI-TEST EXPLANATIONS

We could have also solved this using our shortcut method to determine the apex height above the cliff made by the projectile.The ball is launched at a 30° angle, so if it were to land at the same elevation as the point from which it was launched, itsmaximum height would be about 7 times its range. The range is found s follows:

R= (502 x sin 60°)/10 = 2500 x 0.86 * 10 = 250 x 0.86 = 215

hapex*R/7 =215/7 =210/7 =30m habove canyon floor = 100m + 30 m = 130m

This method still requires math, so it doesn't necessarily save us that much time. In fact, the numbers are almost the same.The only advantage may be that this method works well with your visualization. The best answer is choice C.

Choice D is the best answer. First of all, you should realize that the acceleration for a projectile is always g, assuming thatthere is no air resistance to consider. Based on that fact, we could throw away choices A and B. The only thing we mustdetermine now is whether the speed or velocity is zero. At the apex, the speed and velocity of a projectile are not zero. It istrue that the y-component of the velocity is zero at that point, but the projectile still has speed in the x-direction, so the x-component is not zero. The x-component of the velocity is actually a constant throughout the flight, so neither the speed norvelocity is zero at the apex. The best answer is choice D.

Choice D is the best answer. We are told that the barrel is 1 m in length, but we are not given any information about thetime it spends in the barrel. The cannon ball starts from rest and exits the cannon with a speed of 50 m/s. To calculate theacceleration, we need to use the kinematics equation that involves the initial and final speeds, the acceleration, and thedistance, but does not include time:

v2 =vg +2aAx a= -£-= 5Q2_=2500 =i250m/s22Ax 2(1) 2

If you can't think of the applicable kinematics equation, you might also consider the viability of the answer choices. Theobject exits with a speed of 50 m/s and is in the barrel a very short time, so the acceleration must be significant. Because theobject goes from 0 to 50 m/s, it had an average velocity of 25 m/s in the barrel. To cover 1 m at an average speed of 25 m/srequires0.04 s. The acceleration is Av/At= 50/0.04 = 5000/4. The number is bigger than 1000, so choices A, B, and C are alleliminated. The best answer is choice D.

Choice D is the best answer. The kinetic energy of the cannon ball at impact depends on the kinetic energy of the ball atlaunch and the change in potential energy experienced by each ball. Since all three cannon balls start at the same height, theyall have the same initial potential energy. They also have the same initial speed, which means that they also all have the sameinitial kinetic energy. All three cannon balls have the same final potential energy as well, because they all finish at the samefinal height (i.e., 0). Using conservation of energy: KEf + PEf = KE[ + PEj. Since all three balls have the same initial kinetic,the same initial potential energy, and the same final potential energy, they must also have the same final kineticenergy. Thebest answer is choice D.

Passage II (Questions 8-14) Uniform Acceleration

8. Choice C is the best answer. For uniformly accelerated motion, the velocity has a linear dependence on time. A graph ofvelocity as a function of timefor this kindof motion should always giveus straightlines. Recall thatfor thefirst 10seconds,the velocity is increasing, so we should get a straight line with a positive slope at the start of the graph. For the next 5seconds, the velocity is constant, meaning we should geta horizontal line.For the next5 seconds, the car is slowing down, soweshould get a straight line with a negative slope; andfor the final time increment, the car travels with a constant velocity,so we shouldget anotherstraight line. Graph III showsall of this. The best answer is choice C.

9. Choice B is the best answer. For uniformly accelerated motion, the car's position depends on the square of the timetraveled, x « t2, so any graph of position over time should look like a parabola. This eliminates choice C. For constantvelocity, a graphof position as a function of time should give a straight line. Recall that for the first 10 seconds, the car isaccelerating, so the graph of position as a function of time should give us a parabola that curves up at the start. Thiseliminates choices A and D. Then the car travels with constant velocity, so the graph of position as a function of time shouldlook like a straight line with a positive slope. Next, the car slows down, so the graph should look like another paraboliccurve, but one that curves to the x-axis while still showing an increasing y-position over time. Finally, the car again travelswith constantvelocity, so we should get another straightline with a positiveslope. Graph II best fits this description. Graphrv is the only other possible candidate, but its parabolas curve the wrong way. That graph indicates that the car is slowingdown at the beginning of the trip and then speeding up at the end of the trip. This is opposite of what we were told in thisproblem. The best answer is choice B.

Copyright ©byThe Berkeley Review® 41 MINI-TEST EXPLANATIONS

10. Choice C is the best answer. In order to solve for the distance the car has traveled in the first 10 seconds, we must firstsolve for the acceleration of the car during this time:

a=Av = 15m/s-0 = L5m/s2At 10 sec - 0

Now, we solve for the position using:

x=x0 +v0t +-1- at2 =0+(0)t +1 at22 2

x =I at2 =1 (1 .S-ro-XlO sec)2 =75 m2 2 s2

Doing a question in two steps is not optimal on the MCAT if it could be done in one step, because of the limited time aspect.It is not only important to have a way to solve these questions, but perhaps even more important to have a algorithm thatallows you to select the best way to solve a question. It starts by taking an inventory of the terms we are given. We are givena time of 10 seconds, the initial and final velocities, and asked for distance. We are not given the acceleration nor are wesolving for the acceleration. This means that the easiest route for us to take to the solution is to choose the kinematicsequation that does not include an a-term. This question could have been solved more quickly using the equation:

x=(vf +vo)/2x t=(15+°)/2x 10= 15/2x 10 =7.5 x 10 =75 m

When it comes to kinematics, there are often multiple routes to an answer, but only one route is faster than the other routes. Itis often best to choose the kinematics equation based on what the question does not give you out of velocity, acceleration,distance, and time. The best answer is choice C.

11. Choice C is the best answer. We can cross out choice D, because we know the acceleration must slow down the car. Thismeans that the acceleration must point in the direction opposite to the direction of the velocity. To be more quantitative, wecan solve for acceleration by using vf = v0 + aAt, which can be rewritten as Av = aAtand the manipulated to form:

a=Av=5m/s-15m/s =.2m/s2At 5 sec

The minus sign indicates that the acceleration is in theopposite direction of the velocity, since we defined the velocity valuesas positive in the calculation. As a side note, remember that signs, positive and negative, only indicate directions in physics(e.g., directions of motion, forces, or energy processes). The best answer is choice C.

12. Choice C is the best answer. The final speed depends upon the acceleration. Therefore, a change in the acceleration, whilekeeping the time period fixed, will change the final speed. This rules out choice B. A biggeracceleration should also mean abigger final speed, ruling outchoice A. Actually using the equation a = Av/At shows that doubling the acceleration and fixingthe time period will double the final speed. Why worry about the conceptual arguments, when we could have immediatelyplugged into the equation? It is always good to have a conceptual feel for what is happening. Not only can it successfully getyou through many problems on the MCAT, it can also help you to decide which equation is appropriate to use, should youneed one. The best answer is choice C.

13. Choice A is the best answer. Velocity can be related to acceleration by v2 =v2, +2aAx, so a longer distance to travel givesthe cara greater velocity. Choice B is wrong, because there is no dependence on mass for any kind of uniformly acceleratedmotion. Choice C iswrong for either of two reasons-in the absence offriction, "coasting" implies the car does not speed up.In the presence of friction, coasting means the car would slow down. Choice D is wrong for the same reason that choice B iswrong. The best answer is choice A.

14. Choice D is the best answer. When solving for the accelerations that the car experiences, we make no statements about theforces that areacting on the car. Theacceleration is the result of all of the forces acting on the car, so if friction is present, itscontribution is already accounted for. Choice A is wrong, for the reasons just mentioned. Choice C is wrong, because we donot know whether friction is present or not. Choice B is wrong, for the reasons mentioned. We do not need to know theparticulars of friction to be able to measure the acceleration of the car. The best answer is choice D.

Passage III (Questions 15 - 21) Acceleration due to Gravity

15. Choice D is the best answer. Acceleration due to gravity is taken to be constant when the projectile motion is near thesurfaceof the Earth and the force acting on the projectile is assumed to be mg,so ma = mg, meaning thata = g. If there is noair resistance, the accelerations and motions of the balls are independent of their masses or the kind of material of which eachthe ball is made. The best answer is choice D.

Copyright© by The Berkeley Review® 42 MINI-TEST EXPLANATIONS

16. Choice C is the best answer. We are given the change in heightand the time of flight, so can solve for the value of g using:y= y0 + v0t + Vigt2. Because the initial velocity is zero, this equation reduces to:

17.

18.

19.

20.

21.

Ay =igt


2->g=2^=2(20m) =10m/s2t2 (2sec)2

Choice B is the best answer. Motion in the horizontal and vertical directions are independent of one another. Both Ball I andBall III start at the same vertical height and with the same initial vertical velocity (0), so both balls should strike the groundat the same time. Ball II has a downward nonzero initial vertical velocity, so it will strike the ground before the other twoballs. The best answer is choice B.

Choice C is the best answer. The range we want is the horizontal displacement of the ball, which is the result of the verticalspeed and the time offlight. Here, it is: Ax =vxt = (5 m/s)(t). But what is the time offlight? It is equal to the time the ballspends in the air, which is just the amount of time it would take the ball to fall the 20 m in free-fall. To get the free-fall time,we use:

«-? -=V¥=V 2(20 m)

9.8 m/s22 sec

You may have already known that the time in air would be 2 seconds, because you have taken the time to know that it takesan object 1 s to fall 5 m, 2 s to fall 20 m, and 4 s to fall 80 m. Now we solve for the horizontal displacement, using:Ax =vxt = (5 m/s)(2 s) = 10 m. Notice that we wrote down the range equation first. After all, that is what we want~therange. In solving equations, always write down what you want first. If there are still missing variables, then you need to go tootherequationsto find them. But in many cases, you will not need to. The best answer is choice C.

Choice B is the best answer. We are asked to relate the height and velocity at impact for an object in free-fall, so we haveall of the kinematics variables except for time. Becausewe need an equation that does not contain t, the velocityand heightcan be related using the equation: v2 = v2, - 2gAy -» v= V2gAy. According to the relationship, when the height doubles,the velocity increases by V2. This same result could also have been found if you had applied conservation ofenergy, wheremghjnjt = Vimvf2, instead ofusing a kinematics equation. The best answer ischoice B.

Choice A is the best answer. We are not given the time it takes to reach the ground, so we need to use a kinematicsequationthat does not include t. We can relate velocity and height using:

v2 = v§-2gAy -* v= V(5 m/s)2 +2(9.8 ml s2)(20 m) =20.4 m/s

Do not expect that you are going to calculate exactly 20.4 using this math. Approximate gas 10 m/s2 and estimate the squareroot Never plod through excess math, unless you have to: v2 » 52 +2(10)(20) =25 +400 =425. We know that 202 is 400and212 is 202 + (20 + 21) = 441. The squarerootof 425 lies between 20 and 21, making choice A the bestanswer.

Another way to getto theanswer is to approximate how long it takes theball to fall to theground. If you know this, then youcan use vQ = at + v\. In this case, t turns out to be less than 2.0seconds (thetime it would take infree fall from a height of 20m with no y-velocity at the start), so v0 is less than (9.8)(2.0) + 5.0= 24.6. Of the choices, only 20.4 is less than 24.6. Thebest answer is choice A.

Choice A is the best answer. Whatdiffers between Experiments 1 and 2 is the addition of horizontal motion and the useofone mass in Experiment 2. If you recall, mass is not a factor in free fall, when air resistance is ignored. So how does thehorizontal motion fit in? It turns out to be independent of the vertical motion. Experiment 2 shows this. Experiment 1 doesnot The best answer is choice A.

Questions 22 - 25 Not Based on a Descriptive Passage

22. Choice D is the best answer. The speed at any time during the motion can be written as:

=VvJ + v<

When the projectile reaches its maximum height, itsvertical speed is zero. There is still horizontal velocity, however, so thetotal speed cannot bezero. This eliminates choices A and B. Ignoring any effects of wind resistance, the horizontal velocityisa constant throughout theentire flight. Theprojectile's speed at its maximum height is the horizontal speed only, so it is ata minimum. This eliminates choice C and makes choice D the best answer. The best answer is choice D.


23. Choice C is the best answer. The projectile's launch angle affects its initial x-direction velocity, because vox = v0 cos6. Thedistance the bungee is stretched affects the potential energy of the system, which in turn affects the amount of kinetic energytransferred to the ball. This affects the ball's initial velocity. The elasticity of the bungee cable also affects the potentialenergy. As the bungee gets more rigid, it requires more energy to stretch, so it has more energy to transfer to the ball. Allobjects of the same mass have the same initial velocity, regardless of radius, assuming resistance to be negligible. The bestanswer is choice C.

24. Choice D is the best answer. The acceleration that the air cart experiences is related to the acceleration due to gravity-it issome fraction of the acceleration due to gravity (which depends on the angle of incline). This assumes that the air track is africtionless surface. Since this acceleration is a constant, it is reasonable to assume that the acceleration the air cartexperiences is also a constant throughout the problem. However, this question can still be answered just by thinking aboutwhat is happening to the air cart, without knowing that the air cart's acceleration is related to gravity. If the air cart slowsdown on the way up the track, this means that the accelerationmust be pointing down the track. The air cart speeds up goingdown the track, consistent with an acceleration that points down the track. Further, the initial and final speeds are the same,implying a symmetry to the motion—the acceleration responsible for slowing the air cart down is the same accelerationresponsible for speeding the air cart up. Only a single acceleration is needed to affect the motion of the air cart. The bestanswer is choice D.

25. Choice D is the best answer. To calculate average speed:

Average speed -Total distance traveled- 22.5 m _ 375m/sTotal elapsed time 6 sec

Note that the total distance traveled by the disk is 2(11.25 m) = 22.5 m, since the disk has to go up the track and then backdown the track. To calculate the average velocity:

-_Ax_x2-xi _QAt t2 - ti

The beginning and end positions of the disk are thesame, so there is no net displacement The average velocity is zero. Thebest answer is choice D.

Copyright ©byTheBerkeley Review® 44 MINI-TEST EXPLANATIONS

52-Question Translational Motion Practice Exam

I. Projectile Range on Halfnia (1-7)

II. Sling Shot Experiment (8 -12)

III. Air Track Experiment (13-17)

Questions Not Based on a Descriptive Passage (18 - 21)

IV. Light Clock Experiment (22 - 27)

V. Kicking Field Goals (28 - 32)


VI. Golf Ball and Club Analysis (37 - 42)

VII. Two Falling Cars (43 - 48)


Translational Motion Exam Scoring Scale

Raw Score MCAT Score

42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3


Galileo Galilei, Renaissance scientist and interplanetaryexplorer, designs an experiment to study the effect ofthrowing a 1-kg stone horizontally from the top of his 20-mrocket. He observes its motion in four separate trials. Hetosses two stones A and B from the top of the rocket while onEarth, and he tosses two stones C and D from the top of therocket while on Halfnia, a planet with half the gravitationalforce of the Earth and the same atmospheric composition.The relevant data from the four separate trials aresummarized in Table 1.

Trial g [m/s2] v0 [m/s]

Stone A 9.8 10.0

Stone B 9.8 5.0

Stone C 4.9 10.0

Stone D 4.9 5.0

Table 1

Figure 1

The range is defined as the distance the object travels inthe x-direction, the direction perpendicular to theacceleration. It can be found using standard kinematicsequations. (Neglect air resistance.)

1.

2.

Of the two tosses done on Earth, A and B, which stonestays in the air longer?

A. Stone A remains in flight longer.

B. Stone B remains in flight longer.

C. Both stones remain in flight for the same amountof time.

D. The time varies with the radius of the stone.

If, in Trial A, Galileo threw a 2-kg stone instead of the1-kg stone from the original experiment, the newstone's range would be:

A. double that of the stone in the original experiment.B. half that of the stone in the original experiment.

C. increased bya factor of il from that of the stone inthe original experiment.

D. the same as that of the stone in the originalexperiment.

Copyright©by The Berkeley Review® 46

3.

4.

5.

6.

7.

When Galileo compares the flight times of Stone A andStone C, he notices that Stone A's flight time is:

A. il times that of Stone C.

B. ^Ifi times that ofStone C.C. the same as that of Stone C.

D. double that of Stone C.

In trial D, Galileo records the stone's acceleration justafter launch and just before impact. He notes that theacceleration just before impact is approximately:

A. zero.

B. the same as it was just after launch.

C. V20 of what it was justafter launch.D. double what it was just after launch.

If Stone C were thrown instead from a rocket that is

four times as tall as Galileo's, the stone would remaininflight:

A. twice as long.

B. V2 times as long.

C. just as long.

D. four times as long.

Next, Galileo compares the range (i.e., horizontaldistances traveled by the tossed stones) on each planet.He found that the stone in Trial D traveled:

A. 2 times as far as in Trial B.

B. V2 times as far as in Trial B.

C. Vy^ times as far as in Trial B.

D. the same distance as in Trial B.

How should the ranges of Stone A and Stone Dcompare?

A. Stone A's range is longer.

B. Stone D's range is longer.

C. Their ranges are equal.D. It would vary with the temperature of the

atmosphere.



A group of students study the optimum way to align aslingshot for launching 200-gram spherical glass balls. Aslingshot is built by attaching a flexible cord (in this casemade of bungee cable) to two fixed poles. The cable is pulledback on an axis perpendicular to the vertical poles. Theprojectile is placed in the cavity in the middle of the bungeecable (the point farthest displaced from rest). When thebungee cable is released, it will accelerate towards itsequilibrium position, and from that point undergo dampenedharmonic oscillation, until coming to rest. The potentialenergy of the bungee cord will convert to kinetic energy,launching the ball along the axis of displacement.

The experiment involves several trials in which theangle, the initial potential energy, and the mass of theprojectile are varied. The distance from the base of the cliffthat the projectile strikes the water is recorded throughobservation. Markers are placed in the water that showdistances from the base of the cliff. The trials are conducted

when the crosswind is negligible. The cliff is 100 metersabove the water at high tide, when the experiment isconducted.

9.

200-gram ball

Slingshot

Figure 1.

What interval should be used for the time

measurement?

A. From release of cable to sound of contact.

B. From release of cable to sight of contact.

C. From the projectile leaving cable to sound ofcontact.

D. From the projectile leaving cable to sight ofcontact.

The ball will travel farthest under what conditions?

A. Hot day with off-shore breeze.

B. Cold day with off-shore breeze.

C. Hot day with on-shore breeze.

D. Cold day with on-shore breeze.

Copyright ©by The Berkeley Review® 47

10.

11.

If the ball is placed too high in the cavity, it will leavethe slingshot with backspin. How does this backwardsrotation affect flight time and the maximum height?

A. Both flight time and the maximum height increase.

B. Flight time increases and maximum heightdecreases.

C. Flight time decreases and maximum heightincreases.

D. Both flight time and the maximum height decrease.

If a 200-gram ball is used in the slingshot with a launchangle of 45°, what will be observed when a heavier ballis used?

A. The lighter ball strikes the water sooner.

B. The heavier ball strikes the water sooner.

C. Both balls strike after the same amount of time,

because g is constant.

D. Both balls strike after the same amount of time, if

they are made of the same material.

12. Which of the following graphs BEST depicts therelationship between launch angle and range in theexperiment?

A. B.

C.

-

-

45—i

90

D.

—r—

45


90

Passage III (Questions 13-17)

In order to study the relationships between speed,velocity and acceleration, a group of students perform a setof experiments using air carts of various masses, a long air,track and photogates. Neglect friction in the followingexperiments:

Experiment I

The air track is tilted so that it makes a small angle withthe horizontal. An air cart of mass 0.05 kg is given aninitial velocity of 5 m/s as measured by the photogates.It travels 5 m in 2 seconds up the air track. As the cartmoves up the track, it is observed to slow down andmomentarily come to rest. It then begins to slide downthe track, going faster the farther down the air track itgoes. When it reaches its starting point, it is againmoving at 5 m/s.

Experiment 2

The air track is kept at the same angle as in Experiment1. An air cart of mass 0.05 kg is allowed to travel up anddown the air track a number of times. The students

record the following information for these runs:

Total Transit Time (s)(Time to travel up and

down the air track.')

4

5

6

Distance car moves

up the air track (m)

5.00

7.80

11.25

Experiment 3

The track is kept at the same angle as in Experiment 1.Air carts of different masses are given the same initialvelocity. Total transit times and distances traveled aremeasured.

For Experiment 1, whichof the following graphsBESTrepresents the velocity vs. time for the air cart?

A. l B.

13.

C.

\. Time


14. For Experiment 1, what is the magnitude of theacceleration that the students measure for the air cart?

A. 2.5m/s2

B. 1.25 m/s2

C. 9.0m/s2

D. 10m/s2

15. If the apparatus is changed to increase the frictionacting upon the carts, what might the students findregarding the carts' UPWARD motion in subsequentexperiments, as compared to previousexperiments?

A. The carts experience a bigger magnitude ofacceleration and move farther up the tracks.

B. The carts experience a bigger magnitude ofaccelerationand do not moveas far up the tracks.

C. The carts experience a smaller magnitude ofacceleration and move farther up the tracks.

D. The carts experience a smaller magnitude ofacceleration and do not move as far up the tracks.

16. For Experiment 2, what might accountfor the differenttimes and different distances measured?

A. The initial velocity in each case is different.

B. The acceleration in each case is different.

C. Both A and B.

D. Neither A nor B.

17. ForExperiment 3, when measuring theacceleration, thestudents found that:

A. the larger the mass of the air cart, the larger theacceleration.

B. the larger the mass of the air cart, the smaller theacceleration.

C. the larger the mass of the air cart, the longer thetravel time.

D. the mass of the air cart and the acceleration areunrelated.



18. Consider a ball that is thrown straight up with a speedv0. What are the magnitudes of its acceleration and

speed when it reaches its maximum altitude?

A. a = 0, v = v0

B. a = 0,v = 0

C. a = g,v = vQ

D. a=g,v = 0

19.

20.

Suppose the moon is in uniform circular motion aboutthe Earth. The moon would then be an object that:

A. doesn't accelerate, but travels with constant speed.

B. doesn't accelerate, but travels with constant

velocity.

C. accelerates and travels with constant velocity.

D. accelerates and travels with constant speed.

Which of the following graphs represents the maximumheight reached by a projectile versus its initial y-direction velocity?

A. B.

'oy

'oy

D.

00

'33JS

as

E

'oy

'oy


21. Consider the graph of distance as a function of time.This graph could represent a jogger that:

A. sped up, and then slowed down.

B. slowed down, and then sped up.

C. traveled with increasing velocity.D. traveled with increasing acceleration.


Passage IV (Questions 22 - 27)

A light is mounted behind a circular plate that has fourevenly spaced slits along its perimeter, as shown in Figure 1.

Rotation

Figure 1. Spinning plate with uniformly spaced slits

The wheel rotates at a constant rate so that light passesthrough consecutive slits at regular intervals. In front of thisapparatus is a strip of photographic paper to record the lightpulses. In an experiment, a spherical projectile is released toroll down a ramp and then free fall in front of the light sourceso that its fall may be recorded on the photographic paper.The ramp is elevated to an angle of 20° from abovehorizontal. Figure 2 shows the apparatus and the results ofthe experiment as collected on the photographic paper.

Release pointResults of

experiment

Flash 1

Flash 2

Film strip forrecording motion

of falling ball

ffiRR£ £Flash 3

22.

23.

Light

•OS-Spinninj

Plate Flash 4

Figure 2. Experimental apparatus

The falling ball shows the greatest change in heightbetween the:

A. time it leaves the rampand the first light pulse.B. first and second light pulses.

C. second and third light pulses.D. third and fourth light pulses.

The best description of the conversion of energy fromstart to finish for the ball in the experiment is from:

A. kineticenergy to kinetic energy.B. potential energy to kinetic energy.

C. kineticenergy to potential energy.D. potential energy to potential energy.


24. Why does the difference in height between consecutivemarks on the photographic strip vary over time?

A. The marks are unevenly spaced, because the lightpulses are not separated by uniform time intervals.

B. The marks are unevenly spaced, becausegravitational force increases the rate at which theball falls.

C. The marks are unevenly spaced, becausegravitational force increases the acceleration of theball as it falls.

D. The marks are unevenly spaced, because of windresistance experienced by the ball as it falls.

25. Not correcting for wind resistance when evaluating theresults of the experiment would lead to a value for thegravitational force constant that is too:

A. large. The standard deviation in the raw data wouldbe affected by the presence of wind resistance.

B. large. The standard deviation in the raw data wouldnot be affected by the presence of wind resistance.

C. small. The standard deviation in the raw data

would be affected by the presence of windresistance.

D. small. The standard deviation in the raw data

would not be affected by the presence of windresistance.

26. If the ball is 0.2 kg and it strikes the ground at 20meters/second, what is the kinetic energy of the ball asit strikes the ground? Assume that the collision iselastic.

A. 40 J

B. 80 J

C. 2 J

D. 4 J

27. Which of the following changes would result in morepictures of the ball on the photographic paper?

I. Fewer slits on the circular plate

II. Using a lower release point on the ramp

III. Adding an adhesive substance to the ramp.

A. I only

B. II only

C. I and II only

D. II and III only


Passage V (Questions 28 - 32)

A football team is holding tryouts for place-kickers. Aplace-kicker's job is to kick the football from the ground sothat it travels in a trajectory that allows it to clear a horizontalbar that is 3.03 meters above the ground. The bar has alength of roughly 7 meters, and it is bordered by vertical barsat each end of the horizontal bar. The object thus constructedis referred to as the uprights, and is shown in Figure 1 below.

END ZONE

Figure 1. Uprights for kicking field goals

A field goal is considered "good," if the ball verticallyclears the horizontal bar between the uprights, whether it is inan upward or downward trajectory. Good kickers can makethe kick from as far away as 50 meters. The great HansWiinderlegg can kick field goals from 100 meters away.

If we neglect the impact of air resistance during flightand any rotational motion associated with the flying football,the kicking of a field goal can be described by the lawsgoverning projectile motion. For a projectile that starts andends at the same vertical height, its range can be determinedusing Equation 1, where g is the gravitational force constantand v0 is the initial velocity of the ball leaving the foot.

p _ vj$sin26g

Equation 1

When the ending position has a different vertical heightthan the starting position, Equation 1 does not apply. Forfield goal attempts, the starting position (ground) and theending position(the horizontal bar) are at different heights.

28. During two separate field goal attempts, the football iskicked with the same speed but the angle at which thefootball leaves the ground is changed from 40° to 50°.Which of the following statements about the flight of akicked football is true from the time it leaves the

ground to the time it first strikes the ground?

A. Both the range and the height increase.B. The height increases but the range remains the

same.

C. The range increases but the height remains thesame."

D. The range decreases but the height increases.


29. Suppose that while a kicked football is in flight duenorth through the air, a crosswind is blowing from theeast to the west is blowing. This causes the x-directiondisplacement of the football to:

A. increase.

B. decrease.

C. remain the same.

D. We need to know the magnitude of the crosswindbefore we can answer this.

30. In the presence of air resistance, which of the followingstatements is true about the flight of the football?

A. The range remains the same; the time of flightdecreases.

B. The range remains the same; the time of flightincreases.

C. The range decreases; the time of flight remains thesame.

D. The range increases; the time of flight remains thesame.

31.

32.

All of the following statements about the kicking of thefootball and its flight are true EXCEPT:

A. it is best if the football has a vertical velocity in thedownward direction as it crosses the horizontal bar.

B. the mass of the football does not affect its range.

C. the football changes direction throughout its flight.D. the horizontal velocity of the ball is affected by air

resistance.

During two separate field goal attempts, the kickerkicks the football with the same horizontal speed, butwith a larger vertical speed on the second kick. Thekicker made this change most likely because on thesecond kick, the goal posts were:

A. farther away, but the kicker wanted the same flighttime.

B. the same distance away, and the kicker wanted theball to reach a lower maximum height

C. farther away, and the kicker wanted a longer flighttime.

D. closer, and the kicker wanted a smaller launchangle.



33. Which of the graphs below best represents distanceversus time for a car traveling with uniform, non-zerovelocity?

A. B.

34.

time time

D.

time time

Air resistance has what affect on the flight of a balllaunched horizontally from a height of 4.17 m?

A. The ball travels farther and remains in flight forless time than it would without air resistance.

B. The ball travels less distance and remains in flightfor less time than it would without air resistance.

C. The ball travels farther and remains in flight for alonger time than it would without air resistance.

D. The ball travels less distance and remains in flightfor a longer time than it would without airresistance.

35. If a penny is released at a point that is 2.5 meters abovethe ground,how long will the flight last?

VF second

B. —second2

C. —second4

D. 1 second

Copyright ©byTheBerkeley Review® 52

36. What is the average speed during the first 8 seconds offree fall for an object that starts at rest?

A. 10m/s

B. 20m/s

C. 40m/s

D. 80m/s


Passage VI (Questions 37 - 42)

A professional golfer is hired to compare theeffectiveness of three different golf clubs. Clubs I and II are112 cm in length while the Club III is 103 cm in length.Clubs II and III have the same head while Club I has a head

made of an alloy with a different coefficient of restitution.The coefficient of restitution describes how well the club

head transfers kinetic energy to the ball it strikes. Itrepresents the fraction of the kinetic energy that is transferredduring a collision. The golfer strikes fifty identical balls witheach club from a tee that mounts the ball exactly 0.85 cmabove the ground. The distance each ball travels beforecoming to rest is measured and recorded. Table 1 shows theresults.

ClubLongest

Distance (m)Shortest

Distance (m)Average

Distance (m)

I 289 267 279.6

II 319 300 311.4

III 277 256 268.3

Table 1

The golf balls used in the test have dimples distributedacross their surface in a distinctive aerodynamic pattern toincrease flight time. These dimples create additionalcollisions with air if the ball is spinning. The head of thegolf club is designed such that its face is at an angle fromvertical ranging from 28° to 66°, so the ball leaves the clubhead with backspin. Coupled with the dimpled surface of theball, this serves to increase the range by increasing the ball'sflight time. As a result of the increased flight time, theoptimal launchangle for maximizing range is less than 45° tothe ground.

37. Which of the following are likely differences betweenthe golf clubs found in a golfer's bag?

I. The length of the club's shaft and size of its head

II. The angle that the face of the club head makes withthe ground when the shaft is perpendicular to theground

III. The weight of the club head

A. I only

B. I and II only

C. I and III only

D. I, II, and III

38. What is the speed of a ball after it rebounds off a fixedwall with a coefficient of restitution of 0.80 if it first

struck the wall while traveling at a speed of 50 m/s?

A. 60 m/s

B. 50 m/s

C. 45 m/s

D. 40 m/s


39. The range of a launched ball is maximized when theangle at which it leaves the ground is:

A. slightly greater than 45° and the ball is notspinning.

B. equal to 45° and the ball is not spinning.C. slightly greater than45°and theballhasbackspin.D. slightly less than45° and the ball has topspin.

40. Dimples on the surface of a golf ball launched withbackspin will:

A. increase the air pressure above the spinning ball.B. increase the air pressure below the spinning ball.C. create drag above and below the ball.D. prevent the ball from spinning.

41. All of the following changes will increase the range of agolf ball EXCEPT:

A. launching the ball at a 50° angle rather than a 40°angle.

B. using a longer club while swinging it with the sameangular speed.

C. using a club head made of a material with a greatercoefficient of restitution.

D. using a dimpled golf ball rather than a smooth golfball.

42. Which of the following graphics accurately depicts theflights of a dimpled and undimpled golf ball?

•Dimpled Ball Undimpled Ball


Passage VII (Questions 43 - 48)

Two cars, a Chevrolet Chevette and a Dodge Aries, areparked on a steep hill. Their owners, in a moment ofdissatisfaction, decide that these cars would make good testsubjects for a study on projectile motion. They each parktheir cars pointing down the hill. They each place their car inneutral and the one at a time, release the parking break andwatch. Both cars travel down the slope, towards a cliff andthe bottom of the hill.

Upon reaching the base of the hill, the cars proceed offof the cliff in a perfectly horizontal fashion. The Chevetteleaves the cliff with a speed of 25 m/s. The Aries leaves thecliff with a speed of 31 m/s. They both leave the cliff at thesame instant and then proceed to fall into the valley below.

The car owners measure the point of impact in the valleybelow from the base of the cliff. They find that the Arieslanded approximately 30 m farther from the base of the cliffthan the Chevette landed.

43. If the Chevette falls for 5 seconds, before hitting thevalley, how high is the cliff?

A. 5 m

B. 25 m

C. 125 m

D. 250 m

44. On which slope would the cars slide down withdecreasing acceleration and increasing speed, if theywere in neutral?

45.

Hill I Hill II

A. Hill I only.

B. Hills I and III only.C. HillsII and HI only.D. Hill III only.

Hill III

The Chevette, after falling for 5 seconds, reaches avertical speedof 50m/s. After 10 secondsof falling, itsacceleration:

A. doubles.

B. quadruples.

C. stays the same.D. is halved.


46.

47.

48.

If instead of falling, the Aries were to leave the cliff atan angle 6 with the horizontal and land on another cliff,at the same altitude but a distance r away, then whichformula best represents the relationship between thecar's flight range and launch parameters?

* _ Vq2 sin28mg

sin26

g

_v^sin29

B. r -V3?C. r-

g

D.gsin26

Which of the following are TRUE regarding the cars'impact with the valley floor?

A. The Chevettehits first and with the highestspeed.B. The Aries hits first and with the highestspeed.C. They hit the ground at the same time, but the Aries

hits with the highest speed.

D. They hit the ground at the same time with the samespeed.

If the two cars were traveling at the velocities at whichthey left the cliff, the Aries would take a time t& todrive 12.5 m whereas the Chevette would take a time tcto drive 12.5 rh. Assuming their velocities are constant,what is the ratio of Ia to tc?

A. 31:25

B. 1:1

C. 25:31

D. 15.5:2


Questions 49 through52 are NOT based on a descriptivepassage.

49. Which of the following graphs BEST represents thespeed of a rolling mass as a function of the distance ittravels down the inclined plane?

A. B.

Distance (m) Distance (m)

Distance (m) Distance (m)

50. What is the velocity at impact with the ground for anobject that is dropped from a height of 3 meters?

A. 0.30 m/s

B. 3.00 m/s

C. 7.75 m/s

D. 9.80 m/s

51. How long will it take a projectile to strike the groundbelow after it has been launched horizontally at 23.2m/s from the top of a roof 80 m above the ground?

A. 0.81s

B. 1.54 s

C. 2.11s

D. 3.96 s


52. What are the range and maximum height during flightfor a spherical projectile shot from an air-cannon withan initial speed of 40 m/s and a launchangle of 45° withrespect to the horizontal?

A. R » 160 m; hmaximum « 40 mB. R« 160 m; hmaximum*80 m

C. R« H2m; hmaximum*^ m

D. R« H2m; hmaximum ~56 m

1. C 2. D 3. B 4. B 5. A 6. B

7. A 8. D 9. A 10. A 11. B 12. B

13. C 14. A 15. B 16. A 17. D 18. D

19. D 20. C 21. A 22. D 23. B 24. B

25. D 26. A 27. D 28. B 29. C 30. C

31. B 32. C 33. C 34. D 35. A 36. C

37. D 38. C 39. B 40. B 41. A 42. C

43. C 44. D 45. C 46. C 47. C 48. C

49. D 50. C 51. D 52. A

YOU ARE DONE.

Answers to 52-Question Translational Motion Practice Exam

Passage I (Questions 1-7) Projectile Range on Halfnia

1. Choice C is the best answer. The key concept in solving this question is the fact that motion in the x-direction can betreated independently from motion in the y-direction. To determine the time a stone remains in flight, we need to use one ofthe kinematics equations for free fall in the y-direction, because gravity acts in the y-direction. The relevant equation neededto solve this problem is:

y=Yo +voyt +^ayt2.

This equation allows us to solve for time basedon the initial height of the projectile. Because the stone is thrown horizontally(i.e., solely in the x-direction), we can say that: vov = 0 and ay = -g. Also, y0 = h. In this case, h = 20 m (the height of therocket). The ground is at 0 m, by this definition. To determine how long it takes for the stone to hit the ground, we set y = 0and then plug in our other definitions. This gives us:

y: h = —gt2, where t is the amount of time to fall from height h2

We need consider only the y-direction, because the stone continues to travel along the x-direction at a constant x-directionvelocity until it strikes the ground. Thus, it is the height that dictates how long a stone remains in flight. To solve, use theheight equation:

t= n/lf

Time of flight depends only upon h and g. These quantities are identical in Trials A and B (both carried out on Earth), sotheir flight times must be identical. The best answer is choice C.

Choice D is the best answer. The x-displacement (range) is found using distance = rate x time. Solving for flight time from

the kinematics equation h = —gt2, we find the equation for range to be: R = vox n/ •This equation does not contain a2 IV g )

variable for mass. Therefore, changing the mass of the stone has no effect on the range R. Gravity affects all stones,regardless of mass, in the same way. The best answer is choice D.

Choice B is the best answer. The time in flight depends only on the y-direction for the projectiles in question. To determinethe relative times of flight, we can use the following equation:

h=Igt2 .-. t=n/2lL2 V g

Trials A and C start from the same height h, but their values of g are different. Note that gearth = 2 ghalfnia- On Halfnia,

thalfnia =V—^—>where ghalfnia =4.9 *?-, while on Earth, tearth =V-^—, where gearth =9.8 *&-.ghalfnia gearth

Substituting, we get: tearth =,/ 2h ^ I 2h )=l/F ('halfnia)V2 ghaifnia V2 IghalW V2

The Earth-bound stone (Trial A) remains in the air for a factor of 1/ A- less than the Halfnia-bound stone (Trial C). Even if

you did not set the math up correctly, you should note that the stone falls faster on Earth, where gravity is greater than it is onHalfnia, meaning that the flight time on Earth must be less than the flight time on Halfnia. The best answer is choice B.

Choice B is the bestanswer. Near the surface of the Earth, g isessentially a constant (equal to 9.8 m/s2). In fact, itdoes notdrop to 9.7 m/s2 until you reach an altitude of about 33,000 meters (about 21 miles). An altitude of 20 meters makes nosignificant difference in the value of g. The best answer is choice B.

Choice A is the best answer. The relevant time-of-flight equation is derived from h = —gt2, which can be rewritten as: t =2

2b-. Ifwe substitute 4h for h in the equation, we get: t =a/ ——- = 2 \l 2b_ , which is twice as long. Interesting falloutg V g V g

from this fact is that when an object is dropped from any height, it falls three times farther during the second half of its flighttime than during the first half of its flight time. The same holds true for an object launched vertically upward. The bestanswer is choice A.

Copyright ©byThe Berkeley Review® 56 REVIEW EXAM EXPLANATIONS

6. Choice Bis the best answer. The fundamental equation we need to determine the x-displacement (range) is distance =rate xtime. We are given the x-direction velocity, but we must determine the time the stone remains in flight To determine thetime in flight, we need to use one of the kinematics equations for free fall in the y-direction. Tosolve this question, we needan equation for range: R= voxt. Keep in mind that time is found from h=i-gt2.

2

x: R = voxt, whereR is the range (i.e., horizontal distance) traveled by the stone,and

y: h=J-gt2, where t is the amount of time to fall from height hand travel arange R.

To solve, use the height equation:

1=1/211

To solve for range, you must first solve fortime and then plug into the range equation. This yields the following equation:

Applying this equation to the two trials, B and D, yields the following:

RB =Vox (V^H, where v0b =5^ and gB =9.8 -"1 and RD =vox fl/^E], where voD =5™- and gD =4.9 *$-.\ » 8B / s s2 \ Vgo / s s2

Note that vqb = v0D and gB = 2 go. so we can write:

Rd =v0B /-2h- =V2 v0B l/^k =^2 RB> which is choice B.

You could also have also solved this question intuitively by noting that both stones have the same initial speed in the x-direction, but Stone D stays in the air longer than Stone B (due to reduced gravity). We know that the stone in Trial D willtravel farther than thestone inTrial B, so choice D is eliminated. They startfrom thesame height, which relates toflight timeby a square root factor, so there must be a square root in the relationship. This eliminates choice A. To decide betweenchoices B and C, the math above was required. The best answer is choice B.

7. Choice Ais the best answer. Let's look at the formula for range: R=vox [<y 2lL ], as it applies to Trials Aand D. Ra =vox

(V^)» where VoA =57"and gA =9'8 —' Rd =v°x (V^)' Where VoD =5—and 8D =49- Note that voA =2VqD andgA =2gD, so we can write: Ra =2vox (* —\ =V2vox \\j^\ =V2Rd, which is greater than Rp. Conceptuallyspeaking, thisquestion requires you to compare the effects of bothx-direction speed and gravity on the range of a projectile.Stone A has twice the initial speed of Stone D, but it remains in the air for less time. The key point here is that twice the

gravitational pull results in aflight time that is reduced by i/ *-. The doubling ofastone's speed has agreater impact on itsrange than doubling the gravitational pull, so Stone A has a greater range than Stone D, making choice A the only possibleanswer. The best answer is choice A.

Passage II (Questions 8 - 12) Sling Shot Experiment

8. Choice D is the best answer. The basic problem when using sound to measure points in time is that sound takes longer totravel from the source to the receiver than light does. It is best to use sight for the measuring method if possible. The flightstarts when the projectile leaves the cable, and it ends when we see it strike the water. The best answer is choice D.

9. Choice A is the best answer. A projectile travels farthest when the air surrounding it offers the least resistance. Theprojectile is being fired from the shore out to sea, so it is traveling in the offshore direction. Its x-direction velocity isincreased by an offshore breeze and decreased by an onshore breeze. This eliminates choices C and D. Because the air is lessdense on a hot day, there is less air resistance on a hot day, so the projectile travels farther. The best answer is choice A.

10. Choice A is the best answer. Backspin, as you may have observed when playing tennis or golf, causes an object to rise,which is referred to as lift. Lift is the result of greaterair pressure below the ball than above the ball. With backspin, the airmolecules collide more violently with the bottomof the ball than they do with the top. This creates a greaterpressure on thelower portion of the ball, causing it to rise which in turn causes its maximum height and flight time both to increase. Thebest answer is choice A.

Copyright ©by TheBerkeley Review® 57 REVIEW EXAM EXPLANATIONS

11. Choice B is the best answer. It is easy to assume that mass makes no difference in projectile motion, because this concepthas been presented so frequently in physics. In this case, however, there is more than just simple projectile motion toconsider. There is also a transfer of momentum and kinetic energy. If the conditions are identical, then the elastic cable hasthe same potential energy at full displacement Each projectile leaves theslingshot with thesame kinetic energy. However, ifthe mass of the projectile is greater, then the velocity of the projectile must be less. This means that the heavier projectileleaveswith a lower initial velocity in the y-direction, so it remains in flight for a shorter time. The heavierobject strikes thewater sooner. This makes choice B the best answer. The best answer is choice B.

12. Choice B is the best answer. The trick to this experiment is that the starting point and ending points are at different heights.The projectile's path is not symmetric, so the graph is notsymmetric. Under conditions where the starting and ending pointsare at the same height,45° provides the greatest range,and the graph is symmetrical.This eliminates choiceA. The questioncomes down to how the extra flight time affects the travel of the ball in the x-direction when the angle is not 45°. The bestcompromise of maximizing the angle to maximize the flight time, while simultaneously minimizing the angle to maximizethex-direction velocity, is achieved at a value of slightly less than45°.

Start Height



Launched at an angle > 45"

Launched at an angle = 45°

Launched at an angle < 45°

Air Track Experiment

13. Choice C is the best answer. The air car starts off with an initial velocity, slows down to zero, and then speeds up in theopposite direction. Graph C is the only graph that indicates that the velocity becomes negative at some point in the problem-that is, that the air cart changes direction. The best answer is choice C.

14. Choice A is the bestanswer. To calculate the acceleration, we use: a =— = Q~5 m/s - .2.5 m/s2. The cart has an initialAt 2 sec - 0

velocity of 5 m/s. It slows to zero as it goes up the air track, and it takes 2 seconds for this to happen. Note the presence ofthe minus sign. This tells us that the acceleration is pointing in a direction opposite to the motion of the air cart. However, thequestion asks only for the magnitude of the acceleration, so we drop the minus sign. The best answer is choice A.

15. Choice B is the best answer. The choices contain two components: the magnitude of acceleration and the distance the cartsmoveup the track. When confronted with multiple concepts in a problem, think about the easier conceptfirst. For example,you might easily expect that a cart cannot go as high up the incline if there is increased friction. Since this turns out to betrue, cross out choices A and C. What about the acceleration? If the cart slows down faster, it must have a larger magnitudeof acceleration. If it does not make it up the inclineas far as it would without friction, it must slow down faster. Another wayto think about this is to think of the forces acting on the cart Gravity pulls the cart down the incline. This force therefore hasa component pointing down the incline. When the cart moves up the incline, friction tries to slow it down. This frictionalforce must therefore point down the incline, as well. This results in a larger total force down the incline than gravity alonewould provide. Thus, the resulting acceleration is greater in magnitude. The best answer is choice B.

16. Choice A is the best answer. By keeping the angle that the air track makeswith the horizontal a constant, we do not changethe acceleration-the acceleration is constant for all of Experiment 2. Any differences in time of travel can be attributed todifferent initial velocities—the faster the air cart goes initially, the farther and the longer it will be able to travel. The bestanswer is choice A.

17. Choice D is the best answer. The definitions of speed, velocity, and acceleration reveal that they are all unrelated to themass of an object. It's the old "a-bowling-ball-falls-off-of-a-cliff-at-the-same-speed-as-a-pebble" trick. The best answer ischoice D.

Copyright ©byTheBerkeley Review® 58 REVIEW EXAM EXPLANATIONS


18. Choice Disthe bestanswer. Using the multiple concepts approach, we must consider speed and acceleration at the top ofavertical path. The ball should be motionless at the instant reaches its apex, so v = 0; this rules out choices A and C. Sincethere is still (and only) gravity acting on the ball, there must be an acceleration. This rules out choice B. If there were noacceleration, the motionless ball would stay that way. A good thought to consider is what would happen to the ball if youheld it in that position motionless, and then let go. If it moves from rest, there is an acceleration acting on the ball. The bestanswer is choice D.

19. Choice D is the best answer. Using multiple concepts, we must consider the velocity/speed and the acceleration of themoon. Since the moon's direction of motion changes, it must have a changing velocity. Recall that velocity tells us bothhow fast and which way an object moves. This rules out choices B and C. If velocity changes, then the moon must beaccelerating; that is part of the definition of an acceleration. This rules out choice A. The best answer is choice D.

20. Choice C is the best answer. The projected maximum height is dependent on its initial velocity. We can eliminate choice D,because this curve showsthe maximum height reaching a limiting value, even with increasing initial velocity. The remainingthree graphs all show that the maximum height continues to increase with increasing initial velocity. To choose which ofthese three graphs is correct, we need to look at a kinematics equation that relates velocity to height. We are using heightandgravity, but do not know the time, so the best equation to use for this problem is:

vy = voy+ 2aAv

From this equation, we see that Ay oc vj .This is the equation ofa parabolic curve. The bestanswer ischoice C.

21. Choice A is the best answer. You can compare the velocities at different times by comparing the slopes of the curveon thegraph that plots distance as a function of time. A steeperslope means a greater velocity. Since the slope becomes less steep,then moresteep, and then less steep again, the jogger must be slowingdown, then speeding up, and then slowingdown again.The best answer is choice A.

Passage IV (Questions 22 - 27) Light Clock Experiment

22. Choice D is the best answer. The falling ball will show it greatest change in height per unit time when it is falling at itsgreatest speed. Because the ball in free fall is constantly being accelerated downward due to gravity, the velocity downward isconstantly increasing. Its greatest change in distance per unit time occurs in its last interval where it has the greatest velocity.Of the choices given, the last interval occurs between the third and fourth light pulses, making choice D the best answer. Theanswer can be verified by observing the increasing distance between consecutive marks in the results drawn in Figure 2 of thepassage. The best answer is choice D.

23. Choice B is the best answer. Before the ball is released, it has potential energy due to gravity (mgh). At the end of theexperiment, the ball is in motion, so it has kinetic energy (]/imv2). Therefore, the conversion of energy during the experimentis from potential energy to kinetic energy. The best answer is choice B.

24. Choice B is the best answer. The light pulses are evenly spaced in time, because the slits are uniformly distributed on thedisk that rotates at a constant rate. This eliminates choice A. Acceleration is constant in free fall, so choice C is eliminated.Wind resistance is negligible at the start of free fall compared to gravity, so there is a net acceleration downward. The speed atwhich the ball falls increases as it continues to fall, due to the acceleration from gravity. The best answer is choice B.

25. Choice D is the best answer. Wind resistance opposes the direction of the motion (it always opposes the velocity of amoving object). This means that there is an upward force on the ball as it undergoes free fall, resulting in a reduced overalldownward force. The time it would take for the ball to fall is increased because of wind resistance leading to lower calculatedvalue for the acceleration, so the value of g would be underestimated if you failed to correct for wind resistance. Thiseliminates choices A and B. The effect of wind resistance would be uniform across the data, because it increases as velocityincreases. This means that no variation in the standard deviation should be observed beyond that of human error, so rule outchoice C in favor of choice D. The best answer is choice D.

26. Choice A is the best answer. Kinetic energy is Vimv2, where m is mass and v is speed. Plugging the values from thequestion into the equation yields the following calculation of kinetic energy of the ball as it strikes the ground: !^(0.2)(20)2 =(0.1)(20)(20) = 2 x 20 = 40, choice A. The best answer is choice A.


27. Choice D is the best answer. Fewerslits on the circular plate would increase the time between light pulses, so more distancewould be covered in the interim period. This would reduce the number of pictures of the ball on the strip. Statement I isinvalid, because it would lead to fewer, notmore, marks on the photographic strip. Starting the ball rolling from a lower pointon the ramp would give the ball less speed in the y-direction (downward speed) as it leaves the ramp, so that the ball would befalling more slowly. This may result in more pictures of the ball on the photographic strip. Statement II is valid, because itleads to more marks on the photographic strip. Adding adhesive to the ramp would also give the ball less speed in the y-direction (downward speed) as it leaves the ramp, so the ball would be falling more slowly. This may result in more picturesof the ball on the photographic strip. Statement III is valid, because it leads to more marks on the photographic strip. OnlyStatements II and III could lead to more marks on the photographic strip.The best answer is choice D.

Passage V (Questions 28 - 32) Field Goal Kicking

28. Choice B is the best answer. The only difference between the two field goal attempts is the angle at which the football iskicked. As the angle at which the football leaves the ground increases, the maximum height the football reaches will alsoincrease. This can be seen from:

vv2 =vL - 2gAy

At the maximum height, vy is zero. Solving for Ay, the height, we get:

.2

2g 2g

This means that as the angle is increased up to 90°, the maximum height the football reaches is also increased. Thiseliminates choice C. You could have also reached this conclusion intuitively. The range of the football (when it starts andfinishes at the same height) can be written as:

R_v^sin28ot

This equation tells us that the range will increase as 0 goes from 0° to 45°, reaching a maximum at 45°, and then it willdecrease as 9 goes from 45° to 90°. Stated another way, two projectilesfired with the same v0 at complementary angles give

the same range. (Take note that sin 80° is equal to sin 100°). The range remains the same, so choice B is the best answer.This question may have been a bit tricky, because the discussion of field goal kicking talked about the ball clearing ahorizontal bar at a different height than the ground. This question did not deal with the parameters described in the passage,but instead presented the more simplistic case of the same height at the start and at the finish of the football's flight. The bestanswer is choice B.

29. Choice C is the best answer. A crosswind is a wind that blows in a direction that is perpendicular to the direction of motionof the football; i.e., if the football has velocity components in the x-direction and the y-directions, then the crosswind isblowing in the z-direction. Because the vectors describing force and velocity can be treated as independent of one another, acrosswind in the z-direction has no effect on the x-direction displacement, which is what we are measuring. Since the z-direction is independent of the x-direction, it does not make any difference what the speed of the crosswind is or if thecrosswind blows in the positive z-direction or the negative z-direction. If the speed of the crosswind is big enough, it maydeflect the football in the z-direction sufficiently enough that the football does not pass between the vertical posts, but it willhave no effect on the horizontal range. The total distance that the ball travels would increase, but the x-directiondisplacement is the same. The best answer is choice C.

30. Choice C is the best answer. In the presence of air resistance, the time of flight experiences no net change. On the way up,air resistance acts in the same direction as the acceleration of gravity. These two effects provide a larger acceleration thangravity alone, so the ball reaches its peak in a shorter amount of time. The peak it reaches is lower than the peak it wouldreach without wind resistance. However, on the way down, air resistance now acts in a direction opposite to that of theacceleration of gravity. This means that the ball takes longer to reach the ground than it would in the absence of windresistance. The time for the ball to come back down is longer than the time for it to reach its peak. The effects of the twodirections (up versus down) cancel one another for the overall flight, because they each act on the projectile for the sameamount of time. This eliminates choices A and B. Air resistance also acts in the horizontal direction. The horizontal

velocity no longer remains constant in the presence of air resistance. In fact, it decreases with time. The range of thefootball therefore decreases. (To see that increasing the time in the air does not necessarily increase the range, try throwing afeather! The feather remains in the air a long time, due to air resistance, but it does not travel very far horizontally. For bestresults, make sure the feather is not still attached to a bird.). The best answer is choice C.

Ay =voy_v2sin2(


31. Choice Bis the best answer. Choice B is correct. During the parabolic (or nearly parabolic) flight ofthe football, its heightwill be the same at two distinct points during its path: once on the way up (with a vertical velocity in the upward direction)and once on the way down (with a vertical velocity in the downward direction). When comparing these two points ofequalheight, the football is further displaced horizontally at the point where it has a downward vertical velocity than at the pointwhere it has an upward vertical velocity. The objective when kicking a field goal is to have the football clear a verticalheight as far from the initial point as possible, so it is best to have the football with a downward vertical velocity as it clearsthe bar rather than an upward vertical velocity. Choice A is valid. The football is kicked, so we assume that its kineticenergy is all gained from the transfer of kinetic energy from the foot of the kicker. We further assume that the kicker kickswith the same motion in each field goal attempt, so the football is assumed to have the same kinetic energy in each case. Asthefootball gets heavier and the kinetic energy is kept constant, the initial velocity of the football mustdecrease. This meansthat a heavier football has less initial velocity than a lighter football, so the heavier football will have less range. This makeschoice B invalid. If this doesn't seem clear, you can think of the difference between kicking a bowling ball (heavy) and atennis ball (lighter). You can kick the tennis ball farther. You can also break your foot kicking a bowling ball, so maybe thisbest left as a thought experiment rather than an empirical study. The initial velocity of the football is in a different directionthan the acceleration (due to gravity), so the football will constantly be accelerated in a different direction than its tangentialvelocity. The football changes direction during its entire flight, which is to say that it doesn't travel in a straight line, butrather, it travels in a parabolic path. Choice C is valid. All non-zero components of velocity are affected by air resistance.Because the ball is given positive x-direction velocity initially, there will be a negative x-direction force due to air resistance.Choice D is valid. The best answer is choice B.

32. Choice C is the best answer. If the football leaves the ground with a greater vertical velocity, then it will go higher beforechanging direction and returning to the ground. Thus, the football will havea greater maximum height and be in the air for alonger time. This eliminates choices A and B. If the football remains in the air longer while having the same x-directionvelocity, then it will travel farther in the horizontal direction, resulting in a greater range. It must be that the goal posts mustbe farther away on the second attempt! This makes choice C the best answer. Increasing the vertical component of theinitial velocity while keeping the horizontal component the same results in both a greater launch angle and a greater totalvelocity. This eliminates choice D. The best answer is choice C.


33. Choice C is the best answer. Having uniform velocity means that acceleration is zero. Thus, d = v0t. This is the equationof a linefor these axes, which eliminates choices B and D. ChoiceA, incidentally, says that the distance doesn't change withtime; this would be true only if the velocity were zero. The graph that shows a linear increase in position over time is choiceC. The best answer is choice C.

34. Choice D is the best answer. Air resistance provides a force in the opposite direction of the ball's velocity. It behaves in amanner similar to friction. Because the force is in the opposite direction of the motion, it reduces the speed at which the balltravels, and thus reduces the distance the ball travels. This eliminates choices A and C. Because there is a resistive forceupwards, the ball experiences a downwards acceleration that is less than g, so it does not reach as great of a velocity in the y-direction as a ball that experiences no air resistance. In the end, the ball does not land as soon, so it is in flight a longer time.The end result is a shorter distance and longer flight time. The best answer is choice D.

35.

36.

Choice A is the best answer. The flight time is dependent on the y-direction, not the x-direction. Because the ball isreleased when perpendicular to the ground, the initial y-direction velocity is 0. The time for the flight can be found using the

formula dy = !/iat2. By manipulating that equation to isolate t, we find that the time can be determined according to the

equation: t =2dv

a V 10your trick of knowing that it takes 1 s to fall 5m, so in 0.5 s the object would fall 1.25 m. A height of 2.5 m lies between 1.25m and 5 m, so the flight time must lie between 0.5 s and 1 s. Only choice A fits in that range. The best answer is choice A.

Choice C is the best answer. We are given the time of flight and know that a = g, because we are in free fall. We arc notgiven the distance the ball falls in that time, so we need to use an equation that does not include distance: Vt = v0 + at. Giventhat v0 = 0, we have vt = at, so the final velocity is vt = (10)(8) = 80 m/s. As tempting as it may be to pick choice D at thispoint, we need to make sure we read the question carefully. They are asking for average speed, not the final speed. The

average speed in the sum of the initial speed and the final speed divided by 2, vaveragC =(vi +vf)/2- P'u88mo mtne numberswe get: vaverage =(0 +80)/2 =8°/2 =40 m/s- The best answer is choice C.

. Plugging in 10 for a and 2.5 for dy yields: 2x2.5 5- = \l JL, You could have also used0 V 2


PassageVI (Questions37-42) Golf Ball and Club Analysis

37. Choice D is the best answer. Different golf clubs are designed to launch golf balls different distances. This means thateach club isdesigned with a specific optimal distance (range) in mind. Changing the length of the club's shaft and mass of itshead will change the speed and kinetic energy of the head of the club just at the point when it makes contact the stationarygolf ball. This results in a change in the amount ofkinetic energy transferred to the golf ball which affects the launch speedfor the ball, so Statement I is a valid statement. Changing the angle that the face of the club head makes with the groundwhen the shaft is perpendicular to the ground will change the angle at which it strikes the ball, which in turn will changeangles at which the ball is launched. Launch angle definitely affects the distance the golf ball travels, so Statement II is avalid statement. Changing the weight of the club head will change the speed and kinetic energy of the head of the club whenit strikes the ball. Just as was the case when we considered Statement I, this results in a change in the launch speed of thegolf ball, so Statement III is a valid statement. All three statements are valid statements for explaining why different clubswould have different dimensions, so the best answer is choice D.

38. Choice C is the best answer. According to the passage, a club head with a coefficient of restitution of 0.80 will transfer80%of its kinetic energy to the ball it strikes. So, in the case of a ball rebounding off of a stationary wall made of the samematerial as the golf club's head, the ball will lose twenty percent of its incident kinetic energy and rebound with eightypercent of its incident kinetic energy. This means that its speed will decrease, so choices A and B are eliminated. The ballrebounds with eighty percent of its initial kinetic energy, noteighty percent of its initial speed, so given that0.80 x 50 m/s =40 m/s, choice D cannot be the right answer. This leaves choiceC as the only remainingchoice. Few of us would feel securechoosing answer choice C based on that pathway, so we should do the math to be certain of our choice. The final kineticenergy of the ball is eighty percent of the initial kinetic energy O/imballVfinal2 =0-80 xV^mbalivinitial2), so:

'/imbalivfinal2 =0.80 x VimballVinitial2 •"• Vfjnal2 =0.80 xVjnitial2

vfinal2 = 0.80 x(50)2 = 0.80 x 2500 = 2000

vfina| = V2000 = V400 x 5 = 20H

40 = 20 x 2 = 20V4 < 20VJ, so the speed isgreater than 40 m/s.


39. Choice B is the best answer. This is one of those commonly known facts that in the context of the passage, we mightdisregard or ignore. The optimal launch angle for a non-spinning object that starts its flight and finishes it flight at the sameelevation is 45°. This makes choice B the best answer and eliminates choice A. For an object with a backspin, there is a liftforce which gives the projectile additional flight time. In order to take advantage of this additional time in flight, the objectshould be launched at an angle that is slightly lower than 45° to increase in its x-direction velocity. In other words, becausethe parameters give it a y-directional advantage (backspin), we can reduce the launch angle slightly to convert some of thaiadvantage into an x-direction advantage and thereby share the advantage equally between the x- and y- directions. Thiseliminates choice C. Choice D can be eliminated using the same logic, knowing that topspin reduces the flight time, so anincrease in the launch angle would be needed to regain some of the lost flight time. The best answer is choice B.

40. Choice B is the best answer. As stated in the passage, the dimples on a golf ball increase the interaction of the ball with theair through which it is moving. Backspin creates lift, which must be explained by either an increase in the upward force or adecrease in the downward force acting on the ball. Choice A is eliminated, because it would result in an increase in thedownward force, which would decrease the range. Choice B is viable, because it would result in an increase in the upwardforce, which would increase the range. Choice C is eliminated, because drag above and below the ball would not increasethe range. Choice D is eliminated, because the passage discusses the coupling of backspin and dimples on the surface of theball as increasing the flight time. Because the two effects are coupled, it's unlikely that dimpling would reduce spinning. Thebest answer is choice B.

41. Choice A is the best answer. The optimal angle for a dimpled golf ball launched with backspin is less than 45°. As a result,launching the ball at a 50° angle will result in a shorter range than launching the ball at a 40° angle. This makes choice A thebest answer. Using a longer club while swinging it with the same angular speed gives the ball greater kinetic energyfollowing collision. The result is that the ball is launched with more speed, so the range increases. This eliminates choice B.Using a club head made of a material with a greater coefficient of restitution transfers more kinetic energy to the ball duringcollision, so it is launched at a greater speed. Because the ball is launched with a greater speed, it has a greater range. Thiseliminates choice C. Using a dimpled golf ball rather than a smooth golf ball gives the ball greater lift as it travels throughthe air. This increases its flight time, resulting in a greater range. This eliminates choice D and leaves only choice A as aremaining answer. The best answer is choice A.

Copyright ©by The Berkeley Review® 62 REVIEW EXAM EXPLANATIONS

42. Choice C is the bestanswer. Thedimpled ball has a greater flight time than the undimpled ball, as stated in the passage, sothe dimpled ball should fly farther through the air. This means that the solid line should travel a greater length than thedashed line. Choices B and D are eliminated. The pathway should not be symmetric (perfectly parabolic), because the ballexperiences wind resistance as it flies through the air. We must consider wind resistance, because we areconsider the impactof lift caused by spinning in the air. The ball will travel farther in its first half of flight (during its ascent) than in its secondhalfof flight(during its descent). This is because it slows as it goes, so it has a greateraverage speedduring its first half offlight than during its second half of flight. The pathway will not be a symmetric parabola, so choice A is eliminated. Thebest answer is choice C.

Passage VII (Questions 43 - 48) Two Falling Cars

43. Choice Cisthe best answer. Since we want the vertical distance fallen, we use h= Vigt2. (Note that the horizontal velocitydoes not explicitly enter when calculating vertical motion.) A bit of math gives us h = V£(10)(25) = 5 x 25 = 125, choice C.If you don't know the equation, you could use a bit of physical intuition to rule out choice A. 5m is around 15 ft, which takesmuch less than5 sec to fall through. That wouldat least makeit better than a random guess. But, perhaps the easiestway isto know that an object falls about 5m in 1 s, about 20 m in 2 s, about 80 m in 4 s, and about 320 m in 8 s. A fall time of 5seconds will result in a distance slightly more than 100 m, which of the choices best fits 125 m. When the answer choicesare spaced as far apart as they are in this question,we have the freedom to use some of our techniques to save time.The bestanswer is choice C.

44.

45.

46.

47.

48.

Choice D is the best answer. Sliding down any of these slopes increases the speed. Thus, we must look at the changingacceleration to answer this problem. We know the magnitude of acceleration must be smaller at the bottom than at the top ofthe correct slope. Using limiting cases, we know there would be a lot of acceleration if the slope were steep and none if theslope were horizontal. We need a slope that becomes progressively less steep as the car descends. This occurs only for theslope of Hill II. You could solve this intuitively by considering yourself skiing, running, bicycling, or by some means goingdown a hill of each of the shapes.

Although this question is intuitive, we should also consider it from a test-logic perspective to hone that skill. We are lookingfor a change in the acceleration, so a linear graph is unlikely. This eliminates choices A and B. Graphs II and III areopposites of one another, so it is not likely that both can be correct. This eliminates choice C and leaves choice D as the lastone standing. The best answer is choice D.

Choice C is the best answer. This question is somewhat deceptive. They present information about a change in thevelocity over time, and then ask you about acceleration. The information they gave you about velocity is irrelevant and youmust know about acceleration from background information, albeit common background information. Since the accelerationis usually constant in free fall problems (unless air resistance or the like comes into play), it won't change during flight. Thevelocity will change, but not the acceleration. The best answer is choice C.

Choice C is the best answer. This problem takes too long to solve analytically, so answer it using units (Book I, pg. 8) andlimiting cases. We want dimensions of length. In choice A, m appears in the denominator but not the numerator, so thatequation leads to a value with kg in its units, which rules out choice A. In choice B, the square root sign leads to units ofsquare root of meters, which is not the correct unit for length. Choice B is thereby eliminated. The difference betweenchoices C and D is the position of the sin 28. In the limit that 0 goes to 90°, we would expect that the range, r, should go tozero (a vertical launch will result in no horizontal displacement). Plugging in 90° for 0 gives r - 0 for choice C but r asundefined in choice D, eliminating choice D. The best answer is choice C.

Choice C is the best answer. This is a multiple concepts question, that requires us to consider both falling time and impactspeed. Since vertical and horizontal motion are independent in standard free fall problems and since the cars fall through thesame height starting from the same vertical speed (zero), their falling times are identical. This rules out choices A and B,leaving us to consider whether the Aries hits with an equal or higher speed than the Chevette. Speed is the magnitude of thetotal velocity, which includes both horizontal and vertical motion. Since the Aries has a higher horizontal component ofvelocity and an identical vertical component, it has the higher total speed. The best answer is choice C.

Choice Cis the best answer. Since t=^/v, v=^*/t» so lA:tC (which is also written as lA/tp) equals vc:va- The Ax dropsout of the ratio, because it is the same for both objects. Plugging in the speeds of 25 m/s for the Chevette and 31 m/s for theAries yields choice C. This could also have been determined by deducing that faster motion equates to less time over thesame distance, so the time ratio be the inverse of the speed ratio. The best answer is choice C.



49. Choice D is the best answer. The speed increases as the ball travels a greater distance, but all of the graphs show this basictrend. The acceleration is constant, so the speed increases uniformly. This means that speed goes up linearly over time.However, because the average speed of the object is greater with each subsequent interval of time, it covers more distance ineach consecutive time segment. This means that in a set period of time, the speed goes up linearly, while the distance goes upexponentially. The graph should bend toward the distance axis (x-axis in this case), because it is the axis of greater change.We can also solve this using one of the kinematics equations. Using the equation vf2 = v02 +2aAx, we see that v a Vx. GraphD best represents this relationship. The best answer is choice D.

50. Choice C is the best answer. First, we start by taking an inventory of the variables that we are given. Did they give us all"DAT" on this question? We are given a vertical distance of 3 m, acceleration is g in free fall, but we are not given the time

nor are we asked for the time. This means that we need the kinematics equation that does not contain t: vf" = v02 +2aAy. Theinitial velocity is zero, so itdrops out. This leaves us with: Vf2 =2gAh =2(10)(3) =60. The speed at impact is the square rootof 60, which lies somewhere between 7 and 8 (the square roots of 49 and 64), albeit closer to 8. Only one of the answerchoices lies between 7 and 8, so the correct answer must be 7.75 m/s. The best answer is choice C.

51. Choice D is the best answer. This is a perfect question for our turbo-solution method. Starting with the fact that an objectwill fall approximately 5 m in 1 s, we know that it will fall 20 m in 2 s, 80 m in 4 s, and 320 m in 8 s (based on ourknowledge that when the time in the air is doubled, the distance it falls increases by a factor of 4. It ends up that weessentially know that it will take approximately 4 s to fall 80 m, so the best answer is choice D. But in case you didn't take tothe turbo-solution method, then let's solve this problem using traditional kinematics equations.

First, we start by taking an inventory of the variables that we are given. Did they give us all "DAT" on this question? We aregiven a vertical distance of 80 m, acceleration is g in free fall, and we are asked for the time. This means that we need the

kinematics equation that contains d,a, and t: yt = y0 + vovt + '/£ayt. We can use Ah for (yt - y0) and v0y = 0,sothe equationreduces to: Ah =!^ayt2 =Vigt2. This results in the following calculation:

Ah = i4gt2 SO 80 = i^(10)t2 = 5t2

16 = t2 t = 4

The time is approximately 4 s. The best answer is choice D.

52. Choice A is the best answer. For a launch angle of 45°, the range is 4 times the maximum height. This fact allows us toeliminate choices B and D. Now we must solve for the range of the projectile. Given the speed of the object at launch and thelaunch angle, the equation we need for range is:

^ _ v^sin20oo

With an angle of 45°, the sin28 term is sin 90°, which is equal to 1. This means that the range is simply the speed squared2

divided by g, R = v /0. From here, we get the following math:

R=V2/g =(40)2/i0 =(40 x40)/l0 =40 x4=160 mThe range is 160 m, so the maximum height must be 40 m. The best answer is choice A.


Forces, CircularMotion, and Gravitation

Physics Chapter 2

Fancy counterweight

by

the Berkeley Review

Forces, Circular Motion,and Gravitation

Selected equations, facts, concepts, and shortcuts from this section


F = ma Fkinetic friction = /<kN

A8 = V2at2 + co0t (if you're lacking cot) (ot = (o0 + at (if you're lacking 0)

^centripetal *~ m r ^gravitational"" *-*mim2

6t = e0 + !£(cd0 + a>t)t (if you're lacking a) (ot2 = co02 + 2aA0 (if you're lacking t)

Newton's First Law

Object will remain in motion untilstriking (and breaking) the computer.

© Important Concept

Newton's Laws

Free falling starIf caught, place into pocket until rainy day.

Newton's Second Law

weight = ma

tmspider8

Newton's Third Law

Mveight = *(Fi2 = F2i)

© Vector Estimation Trick

Whether a component of force depends on sine or cosine can be deduced byconsidering what happens at the extreme conditions (vertical and horizontal).

FNet = T - mg sin 8 - /^nig cos 0

= mg cos 0

When 0 is large: Sin 0 is large and Cos 0 is smallSteep hill /. resultant must be large

Resultant must be a sin function: mg sin 0Normal force would be small .*. a cos function

When 0 is small: Sin 0 is small and Cos 0 is largeSmall hill ,\ resultant must be small

Resultant must be a sin function: mg sin 0Normal force would be large .*. a cos function

Physics Forces, Circular Motion, and Gravitation Newton's Laws of Motion

Forces, Circular Motion, and GravitationIn Chapter 1, we studied ways of describing and analyzing motion withoutsaying what caused it. That is called kinematics. We now want to study the forcesthat cause and change that motion. This is called dynamics. Newton's three lawsform the cornerstone of these topics. In common English the words force, power,and energy are used fairly interchangeably. In physics these words are quitedifferent from one another, and we shall see how.

Newton's Laws of Motion

In 1687 Isaac Newton published the three laws of motion and the universal lawof gravitation in his book, Principia Mathematica. This treatise is considered to beone of the cornerstones of science as we know it today.

Newton's First LawIf we were to slide a penny on a horizontal table top, it would slow down andeventually come to a stop because of the frictional forces acting on it. Thisobservation can be summarized in Newton's First Law:

An object at rest will remain at rest, and an object in motion will continue to move withuniform velocity in a straight line, unless acted upon byan external force.

It turns out that Newton's First Law is not true for all reference frames. Forexample, if you are at an airport waving good-bye to someone leaving in a plane,you will be at rest in the frame of reference of the airport. However, your friendleaving in the plane will be in a different reference frame, as he will beaccelerating down the runway. Viewed from his position, you will appear toaccelerate towards the back of the plane. Therefore, in order for Newton's firstlaw to be a true statement, we need to define a reference frame that allows it tobe true. Thus, an inertial reference frame is a frame of reference in whichNewton's First Law is valid. In our example, the plane accelerating down therunway is not an inertial reference frame, while the person waving good-bye atthe airport is. All the reference frames that we will consider in our subsequentdiscussions will be inertial reference frames (unless you are told otherwise).

Newton's Second LawIf you were to push the penny on the horizontal table top with a certain force,then that penny would experience an acceleration in the same direction as theforce. This observation can be summarized in Newton's Second Law:

A force acting on an object will give that object an acceleration in the direction of theforce. Tlie acceleration of the object is directly proportional to the resultant force appliedto the object andinversely proportional to themass of the object.

We can express Newton's Second Law in a mathematical form as shown inequation (2.1). Since forces (F) and acceleration (a) have both magnitude anddirection, they are vector quantities. Mass, however, is not a vector. Rather, it is ascalar quantity.

a = -£- or F = mam (2.1)

When we use Newton's Second Law, we must use the same system of units forthe force, the mass, and the acceleration. As we have mentioned, SI units arewidely used in scientific works. The SI unit for mass is the kilogram (kg), while

Copyright © by The Berkeley Review 67 Exclusive MCAT Preparation

Physics

Test TipQualitative Change

Forces, Circular Motion, and Gravitation Newton's Laws of Motion

that of acceleration is the meter/second squared (m/s2). The SI unit for force iscalled a Newton (N). If we substitute the units for force, mass, and accelerationinto theequation F= ma,wewill find that1N = (1 kg)(l m/s2).

1N =-<ikg)p£)In the English system the unit of force is the pound (lb), acceleration is thefoot/second squared (ft/s2), and massis the slug.

Example 2.1aA single force acts on an orbiting asteroid. What happens to the magnitude ofthat asteroid's accelerationas the magnitude of the net force increases?

A. The acceleration increases.B. The acceleration decreases.

C. The acceleration remains the same.D. It depends upon the mass of the asteroid.

Solution

We want to know about acceleration. We are told that force increases. How doesforce relate to acceleration?

Fnet on the asteroid = masteroidaasteroid

Since m is most likely fixed, and in most problems of this nature mass is assumedto be fixed, the acceleration must increase if the force increases.

Qualitative Change: A question about qualitative change asks how changing onevariable will affect another. There is no arithmetic or algebra to haggle with,making it a fairly simple question type. To solve it, you usually need to identifyonemainconcept or equation that relates to the subject at hand.Thenidentify thetwo relevant variables in that concept or equation, and decide whether they aredirectly or inversely related, or related at all. Here, the two relevant variables areF,which you are told about and a,which you are asked about.

F is directly proportional to a, so the best answer is choice A.

Example 2.1bA donned helmet and less massive hockey puck collide. Each is subjected to aforce equal in magnitude, but opposite in direction to the other. Which objectexperiences a greater magnitude of acceleration?

A. The helmet experiences a greater magnitude of acceleration.B. The puck experiences a greater magnitude of acceleration.C. Bothexperience the same magnitude of acceleration.D. It depends upon the duration of contact.

Solution

We want to know about acceleration. During the collision, the magnitude of theforce on the helmet equals the magnitude of the force acting on the puck.However, because the two objects have different masses, they cannot experiencethe same acceleration.The lighter puck feels a greater acceleration, because of therelationship between mass, force, and acceleration (F = ma, where at as m I).

mhelmetahelmet = n*puckapuck /. if mhelmet > ^puck/ thenahelmet < apuck

This makes choice B the best answer.


Physics Forces, Circular Motion, and Gravitation

We have mentioned that the mass of an object is quite different from the weightof an object. What exactly is the mass of an object? The mass of an object isdefined as the amount of matter of which a given object is composed.The massof an object is exactly the same, whether it is here on the Earth or sitting on theMoon. We will see in a future section that mass can also be defined in terms an

object's inertia. The weight of an object is defined as its downward force due togravity in a particular environment. For example, the gravity on Earth is aboutsix times greater than the gravity on the Moon. Even though an object's mass isthe same on the Earth as it is on the Moon,its weight is about six times greater onthe Earth than it is on the Moon. We can relate the weight of an object to its massby equation (2.2). Note that both w and g are vectors.

W = mg (2.2)

Example 2.2aA 1-kg Moon rock is weighed on the surfaces of both the Earth and the Moon.The acceleration due to gravity is six times greater on the Earth than on theMoon. What is the ratio of the mass of the rock on the Earth to the mass of therock on the Moon?

A. 6:1

B. V6:lC. 1:1D. It cannot be determined from the information given.

SolutionThis is a typical ratio problem. The answer is of the form "mass of object onEarth" : "mass of object on Moon". Just replace the "to" with a ":". So, how dothe massescompare?They are the same! Remember, mass is an intrinsic propertyof matter. What does vary about the object? The weight. It depends upon thelocal acceleration due to gravity. Trickquestion, you say? No. Ifs a question thattests your presumptions. Maybe you said to yourself, "Object on the Moon? Ohyeah, ifs lighter. There's a six in the answer, so it must be choice A." There areno trick questions in physics, but there are careless assumptions. The questionasks for mass, not weight.


Example 2.2bA sample of DNA has a mass of 2 mg. A sample of RNA has a mass of 3 timesthat of the DNA. What is the ratio of the weight of the DNAto that of the RNAifboth were weighed on the Moon?

A. 1:29.4

B. 1:6

C. 1:4.9

D. 1:3

SolutionWhether they are on Earth or the Moon, both samplesare subjected to the samegravity. So, the ratio of theirmasses will be thesame as the ratio of theirweights.The question states that the mass of the RNA sample is three times the mass ofthe DNA sample, so the ratio of the weights is equal to 1:3.


Copyright©by The BerkeleyReview 69



1 liySiCS Forces, Circular Motion, and Gravitation Newton's Laws of Motion

Newton's Third LawIf you exert a force on a penny to move it across a horizontal table top, then thepenny will exert a force of equal magnitude, but in the opposite direction. Thisobservation can be summarized in Newton's Third Law:

Ifone object exerts aforce on a second object, the second object will exert a reaction forceon the first object which is equal in magnitude, butopposite in direction.

Applications of Newton's Laws of MotionShown in Figure 2-1 is a train of cars. These cars are being pulled by some forceF. We will assume that there is no friction in our system. The cars have massesmi, 1TI2, and 1113. There are tensions in the cords that connect the cars. The tensionbetween cars 1 and 2 is T12, while the tension between cars 2 and 3 is T23.

Car 3 Car 2 Car 1

m3 123 ITl2 TK n h Fi

u u TT" {J w "TJ

Figure 2-1

Let's now imagine all three cars as a single mass, m. In other words, m = mi + m2+ 013. We will look only at horizontal forces. Newton's Second Law of motionsays that the force on this whole system is just F (this is the only force which ishorizontal) and that it is equal to the total mass of the system (m) times theacceleration of the system. In other words, F = ma.

The cars are tied together, so they must all have the same acceleration. Let'sisolate the first car and examine it. There are two forces acting on that car. Thereis the force F to the right (which we will take to be in the positive direction) andthere is a tension pulling it to the left (T12). The net positive force is just thedifference between the two that will equal the mass of the first car times theacceleration (F - T12 = m\a). Let's consider the second car. There are also twoforces on this car. The cord, T12, pulls the car forward, while the other cord, T23,pulls it backwards. These forces must be equal to the mass of the second cartimes the acceleration (T12 - T23 = ir^a). What about the third car? There is onlyone force, the cord T23, pulling the car forward. This must be equal to the mass ofthe third car times the acceleration (T23 = m3fl).

F = ma

F - T12 = mia (2.3)

T12 - T23 = m2a (2.4)

T23 = m3a (2.5)

We have four equations. How many unknowns would we expect? It would benice if the number of equations and the number of unknowns were to come outequal, because then we could do something about it. Let's add the bottom threeequations together as shown below:

F - T12+ T12 - T23 + T23 = (mi + 1TI2 + m3)a = ma (2.6)

As you can see, we did not need F = ma as a separate equation. It is alreadycontained within equations (2.3), (2.4), and (2.5). How many unknowns arethere? That depends on what is given and what is solved for.



Let's assume that F, mi, rri2, and 1T13 are given. In this case, the unknowns are theacceleration and the two tensions (T12 and T23). To solve this problem, use F =ma to find the acceleration. Next, plug the acceleration into the equation (2.5) andsolve for T23. Once you have T23, you can use equation (2.4) and solve for T12.You can also find T12 from equation (2.3). You should note that T12 > T23.

Example 2.3aThe pulley system shown below, sometimes referred to as an Atwood machine, hastwo masses, mi and m2. Which of the following formulas represents theacceleration of this system?

A. a =

B. a =

C. a =

D. a =

im- -m2)g

(mj + m2)

(mi + m2)g

(mj - m2)

g

(nil + m2)

m2§

(mj + m2)

Solution

Consider the diagram shown in Figure 2-2a, in which we have a string hangingover a pulley with masses on either end. Assume that mass mi is greater thanmass m2. If we let these masses go, we see that mi goes down, and m2 goes up.

What forces are acting on this system? At every place in the string there is atension T. What we need to do is look at the masses separately.

Let's look at mi first, as shown in Figure 2-2b. There is a downward force ofgravity acting on mi. It is the weight, Wi. There is also a string pulling up on miwith a tension T. If we choose the downward direction to be positive, then thenet downward force can be given by equation (2.7).

mi > m2


T

nil

Wi

Wi -T = m^i

(b)

Figure 2-2

Wi -T = miai

W2 -T = ITt2a2

a2 = -ai

71

T

m

Wr

W2 - T = m2a2

(c)

(2.7)

(2.8)

(2.9)



Physics

Test TipChecking Units

Test TipLimiting Cases

Forces, Circular Motion, and Gravitation Newton's Laws of Motion

Let's look at m2 now. The second mass is being pulled down by gravity and upby the string. This will giveequation (2.8). The tensions appear in both equations(2.7) and (2.8). The accelerations are not related in a similar way. If mi goesdown one meter, then m2 goes up one meter. They must do it in the sameamount of time.At any moment in time the velocitiesare the same, except for thedirection. Acceleration is the rate of change of velocity, so the acceleration willalso have to be the same except that when one acceleratesupward the other willaccelerate downward. In other words, we get equation (2.9).

Substitute equation (2.9) into the equation (2.8). This will give the equation (2.10).Assume that mi, m^ and of course, gravity is given. Find the tension and theacceleration of either mass. If we subtract equation (2.7) from equation (2.8), wewill get equation (2.11). We can solve this equation for the acceleration of mi, asshown in equations (2.12) through (2.14).

W2-T = -m2(ai)

(Wi-T)-(W2-T)=(mi+m2)aiWi - W2 = (mi +1T12)ai

a Wlai= lW2_mig-m2g

mj + m2 mi + m2

(2.10)

(2.11)

(2.12)

(2.13)

ai=(r?^)s (2'14)vmi + m2 /

There are faster and easier ways tohandle these formula identification problems:

Checking Units: The units ofan equation mustmatch on bothsides of theequalsign. Lookingover the four possiblechoices, you see that choiceC has bad units(kgdon't cancel out). Cross outCasa possible choice. Unitchecking canbea niceway to eliminate some, but usually not all, of the choices. Units may be hard toevaluate, though, especially in complex equations with assorted constants. So, ifyou figure that it will take you more than about ten seconds to calculate units,don't waste your time.

Limiting Cases: The limiting case technique looks at extreme physical behaviorof the given apparatus or concept. There are typically two limiting cases, andthey are usually easy to identify conceptually. For this pulley system, the twocasesare no motion and maximal motion. For example, if m2 is 0 kg, how wouldyou expect the system tobehave? You would expect that the remaining mass miwould accelerate with an acceleration of g. To test the three remaining possibleanswers quickly, plug zero in for m2 and see which equations give g for theacceleration. Only choices A and Bpass the test. So, cross out choice D. Asyoustill have two choices, look at the other limiting case. Specifically, if mi = m^what would you expect the resulting motionto be?Thereshould ne no motion atall. The system would be balanced, so the acceleration would be zero. Testchoices A and Bforthis. Only choice A works, making choice A thebestanswer.Had you immediately done this test before considering units or the maximalmotion,you would haveeliminated allbut choice A.Thislimiting case, and mostlimiting cases, takes less than 30seconds to apply and usually leads you to thebest answer, which is a substantial savings in time and effort over the algebraand trigonometry used before.

As a final reminder, notice that we immediately cross out any choice we find anyreason to invalidate. Process of eliminationis a great way to answer a multiple-choice question, or at least betteryourchances ofcorrectly guessing.


Copyright © by The BerkeleyReview 72 The Berkeley Review


Example 2.3bWhich equation represents the acceleration of a package sliding down africtionless inclined plane that makesan angle 8 with the horizontal ground?

A. a = g sin9B. a = g cos8C. a = gD. a = mg

Solution

We need to find an acceleration, so we should immediately think of Newton'sSecond Law, F = ma. If we solve this equation for the acceleration, we get a =F/m. In the question, F and m are not given, so we need to find them. We aretold that the incline is frictionless, so the only two forces acting on the packageare Fnormal pointing perpendicular from the surface of the inclineand the weightW of the package pointing towards the center of the Earth.

Let's draw a picture as shown below and give it some coordinates and somecomponents. Recall that W = mg. At the surface of the Earth, g will (for ourpurposes) be a constant. We know g, but we are not given m, so we need eitherto determine m or to get rid of it. We can get rid of it by breaking the W vectorinto component vectors in the x- and y-directions as shown in the picture. Thatcomponent vectors turn out to be mgsinG down the incline and mgcosO opposingthe normal force.

The normal force is cancelled out by the cosine component of the weight, so weare left with a resultant force of mg sin8. Substituting mg sin8 for F in therelationship F = ma gives mg sin6 = ma. Cancelling out m on both sides of theequation and thereby solving for the acceleration gives a = g sinB. Note that aswe've set our x-axis and y-axis, the acceleration in the y-direction is zero.

N = mg cos8

F = maX

F_m

mg sin6

m

av =

av = = g sin0


Let's try it by checking units and looking at limiting cases. Testing the units of eachchoice invalidates choice D, because the right side of the equation has units offorce while the left side of the equation is acceleration. Choices A, B, and C allhave correct units, so we need to look at those choices in terms of limits. Whatlimits are good? How about "vertical slope" and "no slope"? If there is a verticalslope (i.e., 0 = 90")/ what should happen? Free fall. The acceleration should bemaximum, which would be g. Which choice gives that? Both choices A and Cresultin a = g when8 = 90°. If there is no slope (i.e., 8 = 0°), what shouldhappento the package? Nothing. The acceleration should be zero. Which answer choicegives a = 0 when 8 = 0°? Only choice A fits this restriction. If we had consideredthis limiting case immediately, the question would have taken less than 20seconds to solve.




Physics Forces, Circular Motion, and Gravitation Newton's Laws of Motion

Example 2.4aWhich graph BEST represents the acceleration versus incline angle of a carsliding down an icyhill?

Incline Angle (8) Incline Angle (8) Incline Angle (8) Incline Angle (8)

Solution

Basic Relationship: You may not remember the right equation, but you willoften have a good sense of the basic relationship between variables. For theinclined plane, what would you expect of the acceleration if the plane angleincreases? You would expect the acceleration to increase. Looking at the choices,A is wrong. It says that theacceleration is independentof the incline angle. Crossout choice A.

Limiting Cases for Graphs: As with formula identification problems, look forlimiting cases. Again, expect two limiting cases. The limiting cases are for whenthe planeis eithertotally horizontal or totally vertical. If the angle were0°, for thehorizontal limit, you would expect that the car would not accelerate. Theremaining choices, B, C and D, all agree with this. This limit wasn't much help.How about the otherlimit? If the plane is vertical, what do you expect? Thecarshould free fall with an acceleration of g. As for the graph, the curve shouldreach some limit, in this case g, as the angle increases. Choice C does this.Choices B and D do not. These choices have the acceleration increasingindefinitely as the angle increases. Not physically nice. Notice that limiting caseproblems don't require that you know any equations! All you need is theknowledge of how to read a graph and the knowledge that you've beenintuitively doing physicsfor quite a long time.


Example 2.4bWhich graph BEST represents the relationship between the magnitude ofupward acceleration ofan elevator and the tension in the cable used to supportand move the elevator?

D.

co

•a

2

Tension Tension Tension Tension

Solution

Even when the elevator is stationary, there must be some tension in the cable tosupport it in place. Thismeansthat tensionshould not reachzero on the graph.Choices B and D are eliminated. As the tension in the cable increases, the elevatorwill be pulled up, meaning that it will experience an upward acceleration. ChoiceC shows the wrong relationship,so it is eliminated. The best answer is choiceA.



Law of Gravitation

Newton's law of gravitation states that even extremely small pieces of matter inthe universe attract each other. Consider the two spheres shown in Figure 2-3.The force on sphere 1 due to sphere 2 is a force of attraction, and by Newton'sThird Law, the force on sphere 2 due to sphere 1 is an equal and oppositeattraction. The size of the force is proportional to the two masses, and it isinversely proportional to the square of the distance between them.

'12 '21Mo

Figure 2-3

This relationship, given by equation (2.15), is also referred to as the inversesquare law. In order to turn this proportionality into an equation, we mustinclude a constant of proportionality. The constant that we add is called G and itis equal to 6.67 x 10"11 N-m2/kg2. It turns out that this constant is associated withall the matter in the universe.

(2.15)

Newton's Law of Gravitation is almost exactly correct by modern standards.Deviations in the exactness of this law are so small that almost no one can evermeasure them. The only place where we would see a slight exception is wherethere would be an extremely large mass and a very small mass, and a very shortdistance r between them. Such an example involves our Sun and Mercury.

How does the law of gravitation work on Earth? The Earth is composed of tinypieces of matter. If we were interested in the force of gravity on a personstanding on Earth, we would find that the bits of matter contained in the Earthwould pull on thebits of matter that compose the person. We would need to addup all the vectors in these "different directions" in order to find out what theEarth does to the person. This is a problem left for integral calculus. Newton wasnot interested in what the Earth did to him, but rather what the Earth did to theMoon (which is also made up of a large number of tiny bits of matter). Each bitof matter in the Earth affects each bit of matter in the Moon such that you havemany vectors. Newton invented integral calculus to add up all of these vectors.

It took Newton a rather long time to deal with the problem concerning the Moon.What he found is that if you imagine that the Earth and Moon are sphericallysymmetric, then all of the adding of the tinybits ofmatter canbe eliminated. Youcan replace the Earth and Moon by a tiny massive point at their centers (seeFigure 2-4). In other words, spherical bodies act under gravity as though all oftheir respective masses were concentrated into one pointat the center.

^^I^^Earth Moon

M Moon

Figure 2-4


Law of Gravitation


Physics Forces, Circular Motion, and Gravitation Law of Gravitation

What does the Earth do to you? As shown in Figure2-4, we can replace the Earthwith a point mass (Me). The distance from you to this point mass is the radius ofthe Earth (rg). You also have your own mass (m).The law of gravity says that theforce exerted onyou bytheearth is: F = GMgrn/rE2. The force on you is equal toyour mass times little g (F = mg). These two equations must agree. This meansthatGMg/rE2 hasgottobelittle g.It turnsout that g = 9.8 m/s2, asshown below.

F =GMEm

rE

F = ma = mg

GME (6.67 x10"11 N-m2/kg2)(6.0 x1024 kg)g

rE

g =9.8 m/s2

Figure 2-5

(6.4xl06m)2

What would you weigh if you were inside a system that was accelerating? Let'sconsider an elevator. When an elevator starts to go up, you might have noticedthatyoufeel heavier for a moment. When theelevator starts toslow down asyoureach the top, you start to feel lighter for a moment. Those changes in yourperceived weight are the result of the acceleration of the elevator floor. In thediagram in Figure 2-6, we have chosen the positive direction to be up. In theelevator we find a person standing on a scale. If the elevator is accelerating up,then a is in the upward direction. Let's analyze the forces involved. The scale ispushing up on the person (lefs call this S), while gravity is pulling down on theperson (lefs call this W). The "up"force minus the "down"force is the net force inthe upward direction. This must be equal to your mass times your acceleration.In other words, we can say thatS - W= ma. Your apparent weight can be readfrom the scale. It is S. Your apparent weight isequal to your actual weight plusyour mass times the acceleration, or S = W + ma. Your mass is connected withyour ordinary weight, because W = mg or m = W/g. Knowing this, we cansubstitute for m inS= W+mato getS= W+ (W/g)a. Factoring out W, wegetS=W(l +a/g). Your apparent weight isrelated toyour ordinary weight when youare accelerating in an elevator by this correction factor, (1+ a/g).

S-W = ma .-. S = W + ma

W = mg so m = —g

S=W+(^)a =w(l+|.)

Figure 2-6

Suppose the cable in the elevator were cut and we had a free-falling elevator. Inthis case, our correction factor would become (1 - g/g), which equals (1 -1) orzero. You become weightless. In contrast, if the elevator is accelerating upwardwith the acceleration being equal to the acceleration ofgravity, then wehave(1 +g/g) or 2. Most elevators do notaccelerate that fast, because a 100-pound personwould press against the floor of the elevator as a 200-pound person. Rocketsleaving the Earth's atmosphere can reach 4 or 5 times the force of gravity (4 or 5g's). Thisis what is referred to as "anacceleration of (so many) g's."



Example 2.5aOn May 4, 1989 the space shuttle Atlantis lifted off from Cape Canaveral andcarried a satellite named Magellan into space. After being deployed from thecargo bay of the shuttle, Magellan began its 15-month journey to Venus, thesecond planet from our Sun. As shown in the diagram below, Magellan's orbitaround Venus is egg-shaped. The satellite's closest position to the planet isreferred to as its perigee, while its farthest position is referred to as its apogee.

Magellan • A

At which point(s) in the orbital path are the magnitudes of the accelerationandforce the smallest on the satellite?

A. The acceleration is smallest at A. The force is smallest at C.B. Both are smallest at A.C. Both are smallest at C.D. Both are constant at all points.

SolutionKnowing that Newton'sLawof Gravitation predicts that the force gets smallerasthe distance between two objects increases, the force should be smallest at pointA. As for acceleration, we know that acceleration and net force are directlyproportional to each other. Thus, theacceleration mustbe thesmallest at pointA,as well.

Multiple Concepts: When asked to considermorethan one concept in an answer,you can speed up the solution process by using the multiple concepts method.Simply consider one of the concepts at a time, picking the one that's easiest foryou. Then use your understanding of this one concept to eliminate invalidchoices. For our satellite, you might consider forces first. What do you knowabout the gravitational force acting on the satellite? Does it vary as the satelliteorbits? Yes. So, cross out choice D. Where is it biggest? When it is closest toVenus, point C, according to Newton. Therefore, it's probably thesmallest whenit isas far away from Venus as possible, at pointA. This leaves only choice B. Thereason we considered force (specifically, the maximal force) first is that mostpeople think about acceleration after considering forces on an object. Thinkingabout acceleration would have required us to think about forces, anyway.Considering forces first saved a bit of time and effort. The reason we consideredthe biggest force, instead of immediately considering the smallest force, is thatmost people immediately think big when considering the size of something. It'sunimportant whether big or small entered your mind first. Just remember whatthe question is asking for.



Law of Gravitation

Test TipMultiple Concepts


Physics

Test TipRatio Method

Forces, Circular Motion, and Gravitation Law of Gravitation

Example 2.5bAt which point in its orbit is Magellan moving the fastest?

A. A

B. B

C. C

D. Its speed is constant.

Solution

Magellan will increase its speed and acceleration as it approaches Venus. Itsmaximum speed occurs at the instant it is closest to Venus, because once pastthat point, the acceleration will oppose its velocity.The greatest speed will occurat Point C. The best answer is choice C.

Example 2.6An astronaut weighs 650 N on the surface of the Earth. If he were orbiting at adistance of 2 Earth radii from the center of the Earth, his weight would:

A. appear to be double.B. appear to remain the same.C. appear to be smaller by a factor of two.D. appear to be smaller by a factor of four.

Solution

First thought: "Would the force changeat all?" Yes. So, cross out choice B. Wouldthe force be bigger or smaller? It would probably be smaller.After all, if he werevery far from the Earth, there wouldn't be much of any Earth-bound force onhim. So, cross outchoice A. To choose between Cand D, try thefollowing:

RatioMethod: This problem isa quantitative change problem (e.g., "ifthisdoubles,then that quadruples").A fastway to handle these is to take ratios.

Decide on the BasicRelationship. Here, we are asked told about distance andasked about force. For gravity, force and distance are related by theequation:

P_GMEarthMastronautr2

where r is the distance between the centers of the Earth and the astronaut

Write out the equation twice (once for each situation) and divide them. Thetwo situations given are "on the surface of the Earth" and "orbiting." Ourratio looks like this:

QtfEarthrflastronaut/? /r>_£orbit 'Iprbit _ \&Earthf_ _]£ =l

22 4^surface ^Earthniastronaut/? (2REarth)2surface

Note that you need to write out only the variables you care to know (e.g., F)and the changing variable you are given (e.g., r). The other stuff almostalways drops out, so don't waste your time writing them down.

Solve for the variable in the situation you are asked about. We see that theforce while in orbit is one-fourth of that on the surface of the Earth. Noticethat writing the "new" situation on top of the "old" situation saved us about 5seconds of algebra over writing it the other way around.




Uniform Circular Motion

In order to discuss circular motion, we need to make a connection between arclength (S), radius (r), and angle (8) for motion in a circle. The distance we travelaround the curved path of the circleshown in Figure 2-7, an arbitrary distance inthis case, is S. Note that we have gone through a certain angle 0 from our startingpointas we traverse S. What we want is a relationship between 9,S,and r.

Start curve here

Figure 2-7

If we define the distance we traverse around a circle as S, as shown in Figure 2-7,then we see that S is proportional to the angle 6.The distancearound a completecircle, its circumference, is 2nr (Sfuii = 2nr). If we go around only half of the circle,thedistance is equal to rcr (Shaif = nr). Ifwe go around only a quarterof thecircle,

the distanceis equal to —r (SqUarter = —r).

We see that the arc length is the angle (as measured in radians) times the radius,as shownby equation (2.16). Equation (2.16) is valid, provided we use an anglethat is measured in radians. If we rearrange equation (2.16) to read 0 = S/r, asshown by equation (2.17), then we have a ratio of two lengths (S and r), and theangle 0 would be dimensionless. We must pay close attention to how we defineanangle ina problem (i.e., witha circle as opposed to with an inclined plane).

(2.16)

(2.17)

It is easy to convert radians into degrees or degrees into radians. The angle 0 inradians is equal to the angle 0 in degrees times n divided by 180 degrees, asshownby equation(2.18). Similarly, we could writeequation (2.19).

Oradians —©degrees I )\180°/

0degrees —oradians I

(2.18)

(2.19)

Since a circle contains 360°, we find that 2nrad = 360°. This means that 1 rad isequal to 57.3° (from360°/2n)

Let's assume that we are moving around a circle, and we want to know how fast0 is changing. Let's consider a small change in 0, A0, corresponding to a smallchange in distance around the circle, AS. It will takea small, finite amountof time(we are looking at At) to go this small distance AS. If we divide both sides ofequation (2.17) by At, we will get equation (2.20). Note that AS/At is just thevelocity, v.Equation (2.20) simplifies toequation (2.21). The quantity A0/At is therate of change in radians and is referred to as omega (co), which is the angularvelocity in radians per second.This is shown in equation(2.22).

£K(f)(i)=«





A©.:At

_y_

r

C0 =. V

r


(2.21)

(2.22)

Asyou can see, we have twoseparate relations of similar form. One is given byequation (2.20), and the other is given by equation (2.22). As we move around thecircle uniformly (atconstant linear speed), we havea constant angularspeed.

We can repeat the process we just went through and analyze the change in thevalue of co that comes about in some small time At. We find that Aco = Av/At.Dividingboth sides of the equationby At gives equation (2.23). Note that Aco/Atis just the angular acceleration, a, and is in radians per second squared. Thisgives equation (2.24).

fch=^)fe)=fc)k) (2.23)

_act =

(2.24)

The relations that we have just mentioned are closely correlated to the relationsfor motion in a straight line as was discussed in the last section. Recall that weconsidered a linear displacement, a velocity, and an acceleration. Here we havean angular displacement (0), and an angular velocity (co), and an angularacceleration (a). For every equation for linear motion with uniform accelerationfromthe last section, thereis a equivalent equationin circular motion.

Let's consider thedirection ofthese quantities. Let's first look at thedisplacementduring circular motion. Ifwe go part way around a circle as shownin Figure 2-8,the displacement is the vector shown. This vector is referred to as a chord of thecircle. If the displacement is quite small, then there is littledifference between thechord and thearc. Ifwemake thedisplacement small enough (take thelimit), thechord is tangent tothat point onthe circle. Inother words, for very small angulardisplacements, thedisplacement is tangent to the circle in which wearemoving.

Tangent

Start Here

Figure 2-8

When we are considering a velocity, we are considering a small displacementdivided by a small time. This means that if we are moving around in a circle,thenat anyparticular moment thevelocity is directed tangentto that circle. Whatabout the acceleration? Let's assume that we are moving uniformly around thecircle. This means that the angular acceleration is zero and that the velocityvectoralwayshas the same length. In other words, Iv I = v = constant. Only thevector's direction,which is rotating, is changing (as shown in Figure 2-9).



Figure 2-9

What we can do is compare velocity vectors at two closely spaced points on thecircle, as shown in Figure 2-10. To compare the velocity vectors shown in thecircle below more accurately, we shall bring their tails together. These velocityvectors have the same length. One vector is v-initial (t>i), and the other is v-final(vf). We find that Ai> = vf - v\. This Az; is just the tiny vector that closes the top ofv\ and v{. Even though the size is constant, the direction can change. Keep inmind that a change in the velocity vector is caused by an acceleration vector, soAt? tells us about the acceleration causing the circular motion. This means that ifyou are going around in a circle in a steady circular motion, the acceleration isperpendicular to the velocity. The accelerationis perpendicular to a tangent, so ithas to point towards the center. This is calledcentripetal acceleration.

Avcaused by a, so

Starting here

Blowup view of vxand Vf to show Av

Figure 2-10

We now want toshow that the magnitude of the acceleration isv2/r. Recall thatacceleration is Av/At. Ifyou lookat Figure2-10, you cansee that the accelerationpoints towards the center of the circle, because the acceleration is perpendicularto the velocity vector. Our picture is like the picture thatwe have used to defineanangle inradians. Ifthe angle in radians isvery small, then the arc length (let'suse Av) divided by the radius (let's use v) is the angle A0, equation (2.25). If wedivided bothsidesofequation(2.25) by At, we will get equation (2.26).

Av = vA0

Av _vA0 _ /AQA?" At \At •a

(2.25)

(2.26)

The expression A0/At isthe rate ofchange ofthe angle with respect totime ortheangular velocity, omega (co), asshown inequation (2.27). Recall that co =v/r (see(2.22)). If we substitute co = v/r into (2.27), we will get the equation (2.28). Whatwe haveshown is that the centripetalacceleration towards the centerhas the sizev2/r. Sometimes the centripetal acceleration willbe written as ac.

ac = vco (2.27)

ac"8-f (2.28)




Physics Forces, Circular Motion, and Gravitation Uniform Circular Motion

Note that if we substitute cor for v in equation (2.27), then we get equation (2.29).Thisis a differentway to write the relationship for the centripetal acceleration.

ac=(cor)(co) = co2r (2.29)

We know that co is equal to the change in angle divided by the change in time(i.e., co = A0/At). Suppose we are interested in uniform motion through onecomplete revolution. In this case, we find that A0 = 2rt rad, where At is the timefor one complete revolution. The time to complete a full revolution is the period(T). We can now treat omegaas shown in equation (2.30) to find that co = 2ti/T.

co = 2*L(2.30)

Linear Motion versus Rotational MotionAlthough rotational motion (a.k.a. uniform circular motion) often strikesstudents as being confusing at first, it is actually quite similar to linear motion.With linear motion, you concern yourself with the positions, velocities, andaccelerations of objects along one or more axes (i.e., along some line orcombination of lines). With rotational motion, you concern yourself with thesame ideas. However, the motion is now described in terms of rotations andangles, instead of lines. The equations describing the where, when, and why ofrotational motion are thus similar to the linear ones.

Forexample, the definition ofa linearvelocity is v = Ax/At and the definition ofangular velocity is co = A0/At. Notice the similarity in the two equations.

The linear equation translates as: "Linear velocity is equal to the linear distancean object moves in a time At, divided by At." The rotation equation translatessimilarly: "Angular velocity is equal to the angle through which an object rotatesin a time At, divided by At." To get from one equation to the other, substituteangular variables wherever you see linear ones. This is a quick way to rememberrotational equations and concepts. Table 2.1 lists the common variables andequations for linear motion (on the left side), and their angular correspondents(on the right side).

Linear Rotational

X 0

V CO

a a

m I (rotational inertia or moment of inertia)

F x (torque)

P L (angular momentum)

v = v0 + at co = co0+ at

Ax =v0t+ i^at2 A0 =coot+ l/2 at2v2= v2, + 2aAx co2 = cOo + 2aA02F = ma 2x = Ia

KEiincar = 1/2 IT1V2 KErotetional -= *'2 1*0

p = mv L=Ico

+/- signs for up/downor left/right directions

+/- signs for clockwise andcounterclockwise directions

Table 2.1



Wewon't discuss some of these equations until sections III, IV, and V.The pointof presenting this list now is not to give you six more equations to memorize, butrather to show you that you can figure out many rotational motion problemsusing what you already know about linear motion. Think about the angularvariables and their relationships with each other in the same way you thinkabout the linear ones.

Example 2.7aA well-greased laboratory centrifuge operates at 1500 rpm, and is later turnedoff. If it decelerates at 136 rev/min2,how longwill it taketostopspinning?

A. 15 minutes

B. 11 minutes

C. 9 minutes

D. 7 minutes

Solution

Weare given an angular acceleration a and a change in angular velocity co. Wewant a time t. How do we relate these? Using the angular versions of thekinematics equations,specifically, we find that:

co = coo + at

Now co = 0 when the centrifuge stops, easing the solutionfor t, which is

-con _-1500rev/rrun -1500.^-1360 +140^- n mint- woa -136 rev

min =

136 136mm

The negative value of the acceleration signifies that the acceleration is opposedtothevelocity. Doing the division yieldsapproximately 11 minutes, choice B.

Guessing Equations From Units: If you happen to forget a specific equation,when confronted with a numeric answer, try guessing the equation from units givenin thequestion. Simply combine the numbers given sothat the resulting unitsarethose of the choices. Here, you want time, which has units of minutes. You aregiven two numbers, with units of rpm (revolutions/minute) and rev/min2. To getunits ofmin, youcancombine themonly by dividing theangular velocity by theangular acceleration. Ifyou do this, you now getto have fun with long division.Or do you?

Numeric Estimation: Complex arithmetic is not your friend on any exam. Let'sestimate theanswer. Try tomake numbers look similar, inordertoease themath.Here, 1360 is not all that different from 1500. If we split 1500 into 1360and 140,the answer is 10plus 140/136 minutes. We chose 1360 because it looks like 136.Because our denominator number (i.e., 136) is a bit less than 140, our real answershould be a bit bigger than 11 minutes. Thiseliminates choices Cand D,pointingtowards choiceB.Choice A just looks too big.


Example 2.7bHowlongdoesit takefora 33-rpm record to rotate once, at constant speed?

A. 0.03 secondsB. 0.55 secondsC. 1.82 secondsD. 3.33 seconds



Test TipGuessing Equations



Physics

Test TipPhysical Intuition

Forces, Circular Motion, and Gravitation Uniform Circular Motion

Solution

We are given the angular frequency in revolutions per minute, and asked for theperiod, T, in seconds. First we need to convert the frequency into Hz and thentake the inverse of frequencyto get the period (T = 1//). The math is as follows:

aq rev / . „ 1 min /00 I min * / 1

T = l/. = 60s/a. =60/-.s/ =20/11s/ =i8S/1 If / 33 rev I 33 I rev / 11 / rev ~ l-° I rev

The number is great than 1and less than 2, so only choice C is possible.

Example 2.8aA child riding ona merry-go-round jumpsoffa woodenhorse and startswalkingtowards the center of the ride. As he moves, the acceleration he feels:

A. increases,because his tangential speed decreases.B. increases, because his radius from the center decreases.C. decreases, becausehis tangentialspeed increases.D. decreases, because his radius from the center decreases.

Solution

A formal approach is to consider therelevant equation. Thefirst thoughtis:

acentipetal=—-

As the child moves inward (toward the center), r decreases. It would seem thatthiswould make acincrease. The problem withthat thought is that thetangentialspeed of the child also decreases as the child moves in. Remember, when at asmaller radius from the center, the child doesn't have to move as far to completeone revolution as he would if here were near the rim of the ride. Thus v shoulddecrease. This would tend to make ac decrease. So, which is it? Increase ordecrease? Well, this is always a problem, when two variables change and wewant to know about a third variable. Because the velocity term is squared, it ismore significant that the radius in this question, so overall we expect thecentripetal acceleration to decrease. To verify this, lefs consider equation (2.29)rather than equation (2.28), sothat only one variable changes onthe right side ofthe equation.

2acentripetal = to r

Recall that co is constant anywhere on a spinning object. Thafs why we took theextra time to define angular variables. They are variables that are constant overthe entire surface ofa spinning object, butv, a,andr are not. Now, we only haveto worry about r. Thus, ac must decrease because r decreases.

Physical Intuition: The MCAT stresses physical intuition. Most of you have ahighly developed sense of physical intuition. Intuition is much faster than anyequation and much faster than math. Use intuition first, for any problem. Here,we are asked about acceleration. If you ever rode on a merry-go-round,especially oneof those thatwere completely powered by children pushing it incircles, you know that it's unpleasant being near the outer edge. Why? Becauseyou usually get thrown offthe ride. Near the center, now that's the sweetspot!No real worriesaboutacceleration or scraped knees. Near the center, you are at asmall radius and have a small acceleration.


60s=33/60reV/s


1 JiyS1CS Forces, Circular Motion, and Gravitation Uniform Circular Motion

Example 2.8bWhat happens to the centripetal acceleration of a space shuttle, if it goes into ahigher orbit but keeps its tangential speed fixed?

A. The acceleration increases, because the orbital radius increases.B. Theacceleration increases, because the angular velocity decreases.C. The acceleration decreases, because the orbital radius increases.D. Theacceleration decreases, because the angular velocityincreases.

Solution

In this question we are given information about the radius and the tangentialvelocity of an object in a circular orbit. We are asked about acceleration, so theequation we need is:

acentipetal=~

The tangentialvelocity is said to remain the same, so the questioncomesdown tothe inverse relationship between acentnpetal and r- If r increases, then accordingto the equation the centripetal acceleration should decrease. This eliminateschoices A and B. The angular velocity, co, actually decreases, because thetangential velocity is constant while the radius increases. This means the shuttlehas the same linear speed, but has to cover a greater distance to complete a fullrevolution. As a result, the shuttle does not orbit a full revolution as quickly asbefore. This can be verified by considering equation (2.22) which shows that asradius increases while v remains constant, the value of co decreases.

co= ^r

The reason why we did not choose to use acentnpetal = o*2! (equation (2.29)) isbecauseboth variables on the right side of the equation (co and r) change. Again,though, the variable that is squared (co in this case) has a greater impact on thecentripetal acceleration than the term that is not squared (r in this case).


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Physics Forces, Circular Motion, and Gravitation Centripetal Force

Centripetal Force

Along with centripetal acceleration we can have a centripetal force as shown byequation (2.31), where the "Fcentripetal" represents the centripetal force. Notethat to obtain this equation we have used Newton's Second Law, F = ma, andsubstituted equation (2.28) for the acceleration. Equation (2.31) applies to theinward force and the inward acceleration on an object that is moving. As thatobject moves, it is subject to Newton's Third Law.

^centripetal =mTr- (2.31)

Imagine a mass that is movingaround on a string as shown in Figure 2-11. Thereis a component of the force exerted by the string that is pulling the mass intoward the axis. ByNewton's Third Law, the mass is pulling outward against thestring. This outward force on the string is called the centrifugal force. Thecentrifugal force and the centripetal force are not exerted on the same object. Theobject which is moving in a circle has a centripetal force acting upon it, and byNewton's Third Law it is exertingan opposite force outward against whatever itis that is pushing it in.

Figure 2-11

If you are driving a car down a straight road, you see that there is no sidewaysforce acting on the car. As you come into a left-hand curve in the road, the carturns to the left. What makes the car do that? The road pushes your car towardthe center of the turn. Your carin turn triesto push the road out and away fromthe center. Butthe road does not move. Let's look at this in another way. If youare standing on a floor, then the floor is pushing up on you to support yourweight. You are alsopushing down on the floor, but the floordoes not move. Thefloor is receiving a force from you, just as the road is receiving an outward forcefrom the car and in turn is exerting an inward force on the car. This is whatmakesyou go around the corner. If the road were icy, then you wouldn't be ableto makethe turn,because the road couldn't 'push' you around the turn.

Let'sconsider Figure2-11 again. If we spin the ball around the axis in such a waythat we form a cone with the string, then we can analyze the motion in terms ofits forces. This is an example that you might want to pay close attention to. InFigure 2-12a we see the cone in a three-dimensional perspective. The mass isgoing around in a circle. In Figure 2-12b we see the same picture, but this timefrom a side-view.There will be some angle 0 as shown. The cord will pull on themass with a tension T. The force of gravity will pull down on the mass with aforce equal to W. The radius from the ball to the axis is large R. To study theforces that act on the mass, we need to draw the mass all by itself. Wecan dividethe tension T into a vertical component (TsinO) and a horizontal component(TcosO).



(a) (b)

Figure 2-12

If the mass stays in the same horizontal plane during its revolution, then it is notrising or falling. This means that there is no vertical acceleration, so the sum ofthe forces in the vertical direction must add to equal zero, as shown by equation(2.32). There is a single horizontal force; and since there is nothing to cancel itout, there mustbe an acceleration acting on theball. The horizontal force pointstoward the center of the rotation. This force is supplying our centripetalacceleration, and its size is given by equation (2.33). If we substitute equation(2.28) for the acceleration, we will get (2.34).

TsinO = W

TcosO = ma

Tcos0 = i-(f)

(2.32)

(2.33)

(2.34)

These tworelations tell us everything we need to know aboutthe net force actingon the ball and its acceleration relative to the other parameters. If we canmeasure 6 and the length of the cord, then we can find the radius R.

Example 2.9aWhen a planet orbits the Sun, a force acts upon it that depends upon its orbitalradius. How does the resulting tangential speed of the planet relate to the radius,assuming that the orbit is essentially circular?

A. Thespeed is inversely proportional to the square root of the orbital radius.B. Thespeed is directly proportional to the orbital radius.C. The speed is inversely proportional to the orbital radius.D. Thespeed is inversely proportional to the square of the orbital radius.

Solution

Lefs thinkabout forces in this scenario. There is only one real force acting on theplanet, that due to gravity.When you sum the real forces acting on an object, youget the net forceacting on that object. In mathematical terms:

X^on object= ^net onobject = mobject aobject

which for the planet becomes:

QrtplanetMsun _ mpianetV2

And thanks to the magic of algebra, we find

vaVT,


Centripetal Force


PhySiCS Forces, Circular Motion, and Gravitation Centripetal Force

which is choice A.Again, when askedabout relationshipsbetween two variables,as in quantitative change problems, avoid the wasted time of writing out anyvariable that will not change in the problem. Here, only v and r are relevant andare therefore the only variables you need write down.


Example 2.9bWhy does it become unsafe to takea turn in your car past a thresholdspeed?

A. Because the centripetalforce can exceed the force holding the car to the roadB. Because the mass of the car increases during turns and raises its static frictionC. Becausethe increased velocityreduces the centripetal forceD. Becausethe increased velocityreduces the tire's coefficientof friction

SolutionWhen a car is making a turn, the lateral frictional force of the road against yourcar tires causes your car to turn. There is a maximum static force that the roadcan exert against your tires before they break free and begin to slide sidewaysacross the surface of the road. The maximum centripetal force the car canwithstand before the tires break free is equal to the maximum static force on thetires. Up to the threshold point the followingrelationship holds true.

f'static max s ~~~—r

According to the equation, if r and m remain constant, then the threshold forcecan only be reachedby raising the tangential velocity of the car. Once you exceedthe maximum v in the above equation, then the car will start to slide and thefrictional force will be kinetic rather than static. Because kinetic friction is not asstrong as the maximum static friction, the car will continue to slide unless thetangential velocity decreases or the radius of the turn increases by enough toovercome the kinetic friction of the slide.

But the reality is that We didn't need to go this deep into the concept to answerthis question. The mass does not change as the car makes a turn, so choice Bshould not have been considered. According to the equation (2.28), an increase inthe velocity will increase, not decrease, the centripetal force. Choice C iseliminated. There is no centrifugalforce, so choiceD is eliminated. Only choiceAremains, so it must be the answer of choice.


2

Copyright© by The Berkeley Review 88 The Berkeley Review

Physics Forces, Circular Motion, and Gravitation Kepler's Laws of Orbital Motion

Kepler's Laws of Orbital MotionThe three fundamental lawsofplanetary motion formulated by Johannes Keplerwere discovered empirically in conjunction with Tycho Brahe. Kepler's ThirdLaw, sometimes called the "harmonic law," was saved from obscurity as a meremathematical oddity when Isaac Newton explained how it could be derivedfrom his physical laws of motion and gravitation. The harmonic law allowsastronomers to determine the relative distances of planets from the Sun bysimply observing their orbital periods, and it allows NASA technicians todetermine preciseorbits for the satellites they send into space.

Kepler's First LawAll orbital paths are elliptical. When a smaller object orbits a more massiveobject, it moves in an elliptical path, with the larger object at one focus of theellipse. All of the orbits in Figure 2-13 are elliptical, but they varyin eccentricity.

Earth satellite

Figure 2-13

Kepler's Second LawAn orbiting object moves faster and is subject to a greater force when it iscloser to the object about which it orbits. Technically, Kepler claimed that "Theradius vector of an orbiting planet sweeps out equal areas in equal times." Forpractical purposes, it means that an orbiting object experiences more force andmoves faster as its orbital radius decreases. Its speed (v), acceleration (a), force(F), momentum (p), and kinetic energy (KE) are maximal when the orbital radiusis minimal. The opposite is true when the orbital radius is maximal.

planet at tj: smallest r

largest V, a, F, p, KEof any point of the orbit

Figure 2-14

planet at t2: largest rsmallest v,a,F,p, KE

of any point of the orbit

Kepler's Third LawT squared is proportional to R cubed. When the orbital radius R of an orbitingobjectincreases, so does its orbital period T. Mercury has an orbital period of 88while the Earth's period is 365 days. Mercury and Earth both orbit the Sun, butEarth's mean distance from the Sun is 2.6 times greater than Mercury's.Expressed mathematically, Kepler's Third Law says:

T2 =/4tt2\d3GM

•W(2.35)

where M is the mass of the orbited object (e.g., the Sun). The most importantparts of this equation are the T2 and the R3. Incidentally, this equation comesfrom using the force relationship in Example 2.9aand replacingv with 2rcR/T.

Copyright ©by The Berkeley Review 89 Exclusive MCAT Preparation

PhysiCS Forces, Circular Motion, and Gravitation Kepler's Laws of Orbital Motion

Example 2.10When Johannes Kepler, 17th-century astronomer and smart guy, predicted arelationshipbetween the mean orbital radius and orbital period of a planet aboutthe Sun, he found that the period squared is:

A. proportional to thecube of the orbital radius.B. proportional to thesquareof the orbitalradius.C. proportional to the orbitalradius.D. proportional to the squareroot of the orbital radius.

SolutionAccording to equation(235), the period squared is proportional to the radius ofthe orbit cubed: T2 « R3. The best answer is choice A.

Example 2.11aWhich of the following changes to a planet or its orbit will change its orbitalperiod the LEAST?

A. Doubling the average orbital radius.B. Halving the average orbital radius.C. Doubling the mass of the planet.D. Doubling the mass of the star it orbits.

SolutionWe are asked to think about how changes in the orbital radius and mass of aplanet could affect its orbital period. From Kepler's Third Law, we know that ifthe orbital radius changes, then so does the period. Kepler's Third Law also saysthat the star's mass M determines the orbital period. However, it makes nomention of the planet's mass. In the derivation of Kepler'sThird Law, that factordrops out. This means that the planet's orbit is independent of its mass (i.e.,changes is the planet'smass willnot change the orbit in any way). Noticethat wedidn't worry about the specific way in which changes in planetary radius andsolar mass could affect the orbital period. Your first task is always to ask yourselfwhether a given change will change the variable you care about. This approachcan save time and effort.


Example 2.11bWhich of the following orbits results in the LARGEST variation in orbital speedfor a satellite over the course of one full cycleof motion?

A. A perfectly circular orbit.B. A slightly elliptical orbit.C. A highly elliptical orbit.D. All three orbital paths give a similar variation.

SolutionLefs start by considering that a perfectly circular orbit has uniform angularvelocity, so there is no variation in speed. This eliminates choice A and therebyalsoeliminates choice D.By applyingKepler'ssecondlaw, we know that the areaswept per unit time in an elliptic orbit is constant, so the more elliptic an orbit,the more its speed must vary to keep the area changing at a constant rate. Thegreatestvariation in orbital speed willbe seen with a highlyelliptical orbit.




Friction

As we begin to discuss friction we will first concern ourselves with the frictionbetween solids. Later, we will consider frictional forces in fluids. Those frictionalforces are called viscous forces.

The frictional force between one solid and another solid depends on thecomposition of the two materials. The friction of glass on rubber and the frictionof wood on concrete are quite different. Friction alsodepends on how tightly thesurfaces are being pressed together. Friction generally opposes the motion of anobject and is usually denoted by the lower case f. The force is equal to someconstant times the tightness between the objects. The "tightness" between objectsis called the normal force. "Normal" in this case refers to being perpendicular. Theconstant ofproportionality depends on what the twomaterials are and is usuallyreferred to as a coefficient offriction (u). There are two types of coefficients; staticfas), where there is no sliding and kinetic fak), where there is sliding. Thefrictional force for static friction is given by equation (2.36), while the frictionforce for kinetic friction is given by equation (2.37).

^staticmax - MsN

fkinetic = MkN

(2.36)

(2.37)

The kinetic frictional force turns out to be less than the static frictional force (i.e.,Mk < Ms)- What this tells us is that if an objectis at rest, then it is harder to start itmoving than it is to keep it going. In other words, fk < fs(max)- In Table 2.2 wesee some typical values for the coefficients of static and kinetic friction.

Type of Surface

Rubber on Concrete

Us Mk

0.7-1.0 0.8-1.1

Steel on Steel 0.8 0.6

Glass on Glass 0.9 0.4

Wood on Wood 0.25-0.50 —

Lubricated Metals 0.1 0.06

Synovial Joint Fluid ~ 0.003

Table 2.2

Example 2.12aAn experiment is conducted to test the smoothness of two different horizontalsurfaces, A and B. Various mass were slid on these surfaces and the resultingkinetic frictional forces measured. A graph of the resulting data is shown to theleft. Whichsurface had a larger coefficientof kinetic friction?

A.

B.

C.

D.

mass

Surface A.Surface B.

Both surfaces have the same coefficient.It depends upon the gravitational acceleration.


Friction


Physics

Test TipComparing Graphs

Forces, Circular Motion, and Gravitation Friction

SolutionThis type of problem requires a basic understanding of kinetic friction and theability to read a graph. You know that greater friction can arise from a largercoefficient of friction or a larger normal force on the rubbing object. So how dowe compare the two sets ofdata?

Comparing Graphs: One way to compare graphs is to draw a line perpendicularlyfrom one axis and comparethe data where it intersects the graphs. If you draw avertical line from the middle of the mass axis, it intersects both sets of data. Thoseintersectionpoints correspond to specific values of fk for each set of data. Now,compare those frictional values.At any given value of m, surface A gave rise to agreater frictional force. The same value of m implies the same normal force forboth surfaces. The surface A must have a higher coefficientof kinetic friction.


Example 2.12bIf the same experiment were conducted on the Moon, which exerts a muchsmaller gravitational force, the slopesof the data curves would:

A. both decrease.B. both remain the same.C. both increase.D. One would increase and one would decrease.

SolutionThe force due to friction, both kinetic and static, depends on the normal force.Given that two surfaces are horizontal, the normal force is equal in magnitude tothe weight mg, but in the opposite direction. Because the gravitational forceconstant on the Moon is less than it is on Earth, the frictional force will be less onthe Moon than the Earth.This will lower the slopes for both A and B.


Example 2.13aA mover pushes a 150-lb bed set across carpet, with a coefficient of static frictionof 0.8between the bed and the carpet. What minimum horizontal force must sheuse to get the bed moving?

A. Less than 120 lbs.B. At least 120 lbs.C. At least 150 lbs.

D. More than 150 lbs.

Solution

First, we must draw a diagram, as shown below. If she is to move the bed, thenthe force she uses must be greater than the largest possible value of the staticfriction.


Direction bed is

being pushed (P)

Fpush >Static = MSN

FPush >MSN =(0.8)(150 lbs) =120 lbs



Example 2.13bAback-packer grabs her75 N pack and drags it with a constant speed at an angleof 30° to the horizontal. The force she applies to the pack is 40 N. What is themagnitude ofthekinetic frictional force that is acting onthe pack?

A. 20newtons

B. 28 newtons

C. 35 newtons

D. 40 newtons

Solution

The picture for this situation is shown below. Note that the pack isbeing movedat constant speed. This tells us that the acceleration of the packis zero. In otherwords, Fx = max = 0 and Fy = may = 0. If we solve for the components in the x-direction, we will get (40 N)(Cos 30°) - fk = 0. Thus, the maximal value for thecoefficient friction, f^, is 35 N.

P = 40N 40Nsin30°

N >440N cos30°

IFx = max = 0

40N(cos6)-fk = 0

fk= 40N(cos30°)

fk= 40N(0.86)

32N<fk<40N

Note that in solving for the friction, we did not use the weight of the pack. Wesimply needed the free-body diagram. If you were you having trouble recallingthe cos 30° term, then you should know thatifs notequal toViandifs notequalto 1. So, the answers can't be A or D. We should remember that

Cos 30° =^3/2which isaround 0.86. We know that40 x 0.8 is 32, so40 x 0.86 isgreater than 32,so the answer can only be C.


Although it may sound strange and possibly incorrect when you first considerthe idea, it is possible for a static friction to accelerate an object from rest. Thereare two everyday examples that can help you to accept this concept: walking anddriving. When you start walking from rest, you are clearly going from v = 0 tohaving a velocity in the direction you are moving. This means that you haveaccelerated. This is achieved by pushing off against the ground in a lateraldirection. So, there must be a force accelerating you in a lateral direction.According to Newton's third law, there is an equal and opposite force as yourpush-off force. Thatopposing force is a static friction of the ground against yourfoot, as long as your foot does not slip. This explains why it is harder to startwalking on an icy floor than a carpeted floor, because us for ice is much lowerthan us for carpet.

Rolling car tires can accelerate a car from a full stop, so that is also a case of staticfriction being able toaccelerate an object. Likewise, when you brake, it isactuallythestatic friction of the tiresagainst the road that slows yourvehicle.


Friction


Physics Forces, Circular Motion, and Gravitation Inclined Planes and Pulleys

Inclined Planes and Pulleys

Simple Machines and Mechanical AdvantageHumans create machines to make life easier. Two simple machines thatoccasionally surface on the MCAT are the inclined plane and the pulley system.They make life easier, because they redirectand/or reduce the amount of forcewe need to apply in accomplishing some task. But as we shall see in Section III,they do not lessen the overall amount of work that must be done. It still requiresthe same amount of energy to carry out the task.

One way to characterize a machine is through its mechanical advantage. Simplydefined as:

.. . . . A, . Weight of the object to be supportedMechanical Advantage = - * —

Applied forceneeded to support the object

A machine with a bigger mechanical advantage will require less applied forcethan a machine with a smaller mechanical advantage when it is used to supportsome object. Othersimple machines include levers (as we will see in Section IV)and hydraulic lifts (which we willsee in Section VII.)

Example 2.14A spiral staircase provides what sort of mechanical advantage over a verticalhoist?

A. A mechanical advantage based on differing gear radiiB. A mechanical advantage based on a pulley systemC. A mechanical advantage based on a lever and fulcrum systemD. A mechanical advantage based on an inclined plane

SolutionTo lift the object using a vertical hoist would require a force strong enough toovercome the weight of the object, mg. If, however, we take the object up onestep at a time, although we ultimately move the object for a longerdistance, lessforce is necessary. The force varies with the average slope of the stairs, so thestairs are in essence a stepwiseversionof an inclined plane.This is why a ramp isoften used to help move cargo to and from transport vessels.




Inclined PlanesIn the last section, we considered friction in terms of a system which washorizontal. In a horizontal system, both velocity and acceleration in the y-direction are zero. However, when we consider inclined planes we must beaware of the vertical component(s) of our system. If we choose our coordinatesystem to be perpendicular (the y-axis) and parallel (the x-axis) to the plane ofthe incline, then we will find that any acceleration on an inclined plane will beparallel to the surface of that plane. There will be no acceleration in a directionperpendicular to the plane. Why? Because an object sliding on an inclined planewill remain in contactwith that plane as it moves.

Example 2.15How much tension musta rope have if it is used to support a precious paintingwith a mass of m on a smooth, frictionless surface, inclined at an angle 6 withrespect to the horizontal ground?

A. mgB. mgsinOC. mgcosOD. mgtanO

Solution

We start by drawing a free-body diagram showing all of the forces and then sumthe forcesacting on the mass.

IFx = max %tN

\T x^y^Smtx

Fx =T - mg sin0 = 0

T = mg sinO mg

Notice that T < W, meaning that the person or object using the rope to supportthe objectdoes not have to exert a force equal to the weight of the object. This iswhy a ramp is used to help load cargo into a truck.

Notice that the y-axis forces did not enter into the problem at all. Would theyever? Yes, if the question specifically asked you or if the problem involvedcalculating the magnitude of friction. As a rule, just write down the equation thathas what you want. For example, only the x-axis equation deals with the kind oftension being asked about here. If you still have other unknown variables aftertrying to solve that equation, then resort to other equations. Use the limiting casestechnique described in Example 2.3a.

The correct choice is B.

The normal forcehelps you hold the painting in place (it does the cosine 0 portionof the supporting), so you need to apply less force than if you were supportingthe painting alone. The mechanical advantage of the inclined plane in Example2.15 relates the force you must apply in the form of tension to hold the paintingin place to the actual weight of the painting. For an inclined plane, themechanical advantage is equal to 1/sin 6.


Inclined Planesand Pulleys


Physics Forces, Circular Motion, and Gravitation Inclined Planes and Pulleys

Pulley SystemsOne type of a simple machine is a pulley. Pulley systems are used to change thedirection of a given force. A system of multiple pulleys can be used to reduce aforce needed to lift a heavy object (by providing a mechanical advantage).

Consider the pulley system shown in Figure 2-15 in which pulley A and pulley Bare in contact with one another through a common cord. Let's assume that we arenot trying to accelerate the system. In this case, we will consider constant speedand can therefore look at the system in terms of an equilibrium. In order to seewhat is happening with the pulley, we need to examine specific parts of thesystem. The cord that winds through the pulley system is non-expanding,flexible, and it is ideal. In other words, the force (tension) along that cord is thesame at every point. Note that there is a bracket that holds pulley B to the ceiling.That bracket will have a different force associated with it, which we will call Tr.

Figure 2-15

In order to study the equilibrium of the system, we need to consider the twopulleys separately. Assume there is no motion in the x-direction. Let's look atpulley A, as shown Figure 2-16. There is a downward force on this pulley that isequal to the weight W. There are two cords pulling up on the pulley, each withforce Tsin0. Since there are two cords, the total upward force is 2Tsin9. Becausethe system is in equilibrium, we know that 2Tsin0 = W or that T = W/2sin0.

£Fy = may =0

Fy =2Tsin6-W =0

2Tsin0 = W

WT =

2 sine

Figure 2-16

Note that in most problems the angle 0 will be 90" (i.e., the cords will becompletely vertical). In this case the equation tells us thatT = W/2.



Let's look at pulley B, as shown in Figure 2-17. It has a bracket holding it to theceiling with some tension Tb- Note that this pulley has two cords-one comingfrom pulley A and the other leaving pulley B. Since thesystem is in equilibrium,we know that Tb = 2Tsin0.

£Fy = may =0

Fy =TB-2Tsin0 = O

TB = 2Tsin0

T =T

B

2sin0

Figure 2-17

Let's assume the angle 0 in Figures 2-15, 2-16, and 2-17 is equal to 90° (i.e., all thecords are completely vertical). Now let's consider the whole system once again.We know that the upward forces must be in equilibrium with the downwardforces. We do not have to consider any forces in the cord between the twopulleys (because that force is the same). The total upward force is just T + Tb,while the total downward force is just T + W.Since we are in equilibrium,T + Tg= T + W. For this particular pulley problem, we find that Tb = W.

Notice that the force T that we need to exert on the cord is just W/2. In otherwords, T = W/2 or 2 = W/T. The mechanical advantage is defined as the weightyou can lift divided by the force you have to exert (W/T). For the problem wehave just considered, the mechanical advantage happens to be equal to 2.

We mentioned that T = W/2 or W/T = 2. Note that in the complete systemshown in Figure 2-17 the number of cords that support pulley A (the movingpulley) is 2. What you will find in other pulley systems is that the mechanicaladvantage is equal to the number of cords supporting the moving pulley.

Example 2.16aA rope at a loading dock is wrapped around a lower pulley, as schematicallyshown. How much force (Fa) must be applied to just support the mass? (Assumethat the ropes and pulleys are massless and frictionless.)

A. 2mgB. mgC. mg/2D. mg/3


Inclined Planes and Pulleys


PhysiCS Forces, Circular Motion, and Gravitation Inclined Planes and Pulleys

SolutionTo relate the applied force, Fa/ to the weight, consider an object that interactswith both. Lefs look at Pulley B. It has two sections of rope pulling upwards onit and one mass m pulling downwards. If the force just supports the object, thenthe system is not accelerating. The applied force is essentially the tension in therope. For massless and frictionless ropes, the tension is taken to be the samethroughout the rope. Equating these tensionsto the weight gives:

2FA = mg or FA =m^2

A simple trick is to count the number of vertical ropes supporting the lowerpulley,and divide the weightofthe object by that number.


Example 2.16bWhat is the mechanical advantage of the pulley system in Example 2.16a?

A. 0.5

B. 1

C. 2

D. 3

Solution

The quickest route to the solution is to count the number of vertical ropes thatimpact the hanging weight In this example, there are two vertical ropes attachedto Pulley B,so the mechanical advantage of this pulley system is 2.

You could have also concluded this by comparing the applied force, Fappije£i, tothe weight of the hanging object. The applied force is essentially the tension inthe rope. For massless and frictionless ropes, the tension is taken to be the samethroughout the rope. The tension in the rope is found by dividing the weight,mg, by 2sin 0, where 8 = 90° when the ropes are vertical. Sine 90° = 1, so T = W/2.The ratio of the of the weight of the object to the force applied to support theobject is W : W/2, which reduces to 2 : 1, confirming that the mechanicaladvantage is 2.



25 Forces, Circular Motion, and Gravitation Review Questions

I. Gravitational Force

II. Circular Motion

III. The Thrills of Skiing


(1-8)

(9-15)

(16 - 22)

(23 - 25)



According to the universal law of gravitation, any twobodies with mass are attracted to each other with a force

given by Equation 1.

F = gMulr2

Equation 1

In Equation 1, G=6.67 x 10"11 Nm2/kg2, Mand marethe masses of the two bodies, and r is the distance betweenthe two bodies. Equation 2 gives the gravitational potentialenergy.

Tj = -GMmr

Equation 2

When one discusses the force that a planet exerts onperson it is customary to write the gravitational force as:

F=mg,

Equation 3

where m is the mass of the person and g is the accelerationdue to gravity. These three equations describe the interactionbetween two bodies that have mass, and they apply in thepresence or absence of other forces. Gravitational force canbe considered as acting independently of other forces.

1. If you could stand on the surface of a shrinking planet,what would happen to your weight, assuming that theplanet did not lose any mass?

A. Your weight would increase, because the planetbecomes denser.

B. Your weight would decrease, because the planetbecomes denser.

C. Your weight would increase, because the planetbecomes less dense.

D. Your weight would decrease, because the planetbecomes less dense.

2.

3.

If the distance between two masses is tripled, then themagnitude of the gravitational potential energy is:

A. increased by a factor of 3.

B. increased by a factor of 9.

C. decreased by a factor ofV3~.D. decreased by a factor of 3.

The acceleration due to gravity on the surface of theEarth is g. Planet X has a mass 10 times the mass of theEarth and a radius 2 times that of the Earth. What is the

acceleration due to gravity on the surface of Planet X?

58/25g

10g

2g

Copyright ©byThe Berkeley Review®

A.

B.

C.

D.

100

When a person stands on the surface of the Earth, wheredo they appear to weigh the most? (Assume the Earth isa perfect sphere, i.e., ignore those darn mountains.)

A. The Equator

B. The North Pole

C. Midway between the Equator and the North Pole

D. The person's weight will be the same at any pointon the earth's surface.

Which of the graphs below best represents gravitationalpotential energy, U, versus the radius of the planet, r?

A.

U

D.U

As two massive bodies move towards one another, whatis observed?

A. Their speeds increase uniformly.

B. Their speeds increase by a decreasing magnitude.C. Their speeds increase by an increasing magnitude.

D. Their speeds remain constant until they collide.

As Comet Shoemaker-Levy 9 moved towards Jupiter in1994, which of the following statements was trueregarding their interaction with each other?

A. The comet felt a smaller magnitude of gravitationalacceleration than Jupiter.

B. The comet felt the same magnitude of gravitationalacceleration as Jupiter.

C. The comet felt a greater magnitude of gravitationalforce than Jupiter.

D. The comet felt the same magnitude of gravitationalforce as Jupiter.

If a moon orbited a planet in uniform circular motion,the force of gravity from the planet acting on the moon:

A. would do work on the moon, otherwise it wouldcease to orbit the planet.

B. would do no work on the moon, otherwise themoon would change orbital speed.

C. would do work on the moon keeping speed constant.

D. would do no work on the Moon, which changes theorbital speed of the Moon.



A physics student performs two experiments to studycircular motion.

Experiment 1

The student attaches a ball of mass m to a light,inflexible string of length L. She then holds the end of thestring and swings the ball in a nearly horizontal circle withconstant speed v.

Experiment2

The student takes the ball and string from Experiment 1and swings the ball in a vertical circle such that the ballnearly hits the floor. The path is shown in Figure 1.

9.

10.

Ball swings inclockwise circle

Figure 1.

Three identical bugs (marked A, B, and C) are standingon a turntable as it begins to spin. As the spinningincreases, which of the bugs is most likely to slip first(with respect to the turntable surface)?

A. Bug AB. BugBC. BugCD. All three of the bugs are equally likely to slip.

Suppose the maximum tension in the string is 25 N. Ifthe mass of the ball is 0.5 kg and the length of the stringis 1.0 m, what is the maximum speed the ball may havein Experiment 1? (Assume that the ball swings in acompletely horizontal circle.)

A. 50 m/s

B. 25 m/s

C. 7.1 m/s

D. 5.0 m/s


11. How long does it take the ball to complete onerevolution in Experiment 1?

A. jiL2vB. 2jtLv

C. v/jtL2D. 2jt.L/v

12.

13.

14.

If, in Experiment 1, the speed of the ball were increasedwhile the length of the string remained constant, thenthe:

A. tension in the string must have increased.B. tension in the string must have decreased.C. centripetal force must have decreased while the

tension must have increased.

D. centripetal force must have increased while thetension must have decreased.

At point A in Experiment 2, the tension in the string ismeasured to be three times the weight of the ball. Whatis the speed of the ball?

A. 2fgLB. V3gLC. V2gLD. 2gL

How long will it take the ball in Experiment 2 to reachthe ground if it were released at point A with a speed ofV^L?

A.

B.

C.

D.

Y2gL

4L

2VF

15. If the ball in Experiment 2 is swung in uniform circularmotion, then how are the magnitudes of tensions atpoints A, B, C, and D in Figure 1 related?

A. TA = Tc,TB<TD

B. Ta = TB = TC = TD

C. Ta<Tb<TC<TdD. TA<Tc,TB = TD



Snow skiing is a popular winter sport. In downhillskiing, skiers are taken to the top of large mountains bymeans of a lift chair or to the top of small hills by means of atow rope. With a tow rope, the skier grabs on to the tow ropeand is pulled up the hill (Figure 1). Once at the top of the hill,the idea is for the skier to get back down, and enjoy the ridewhile doing it! Inexperienced skiers tend to ski straight downthe mountain; their motion is much like that of a blocksliding down an inclined plane (Figure 2). Experienced skiersknow how to control their rate of descent by making a seriesof S-shaped turns as they come down a snow-covered slope.

Figure 2.

Cross-country skiing is different from downhill skiing.Essentially, specially designed skis allow the skier to "walk"across the snow on a predominantly horizontal surface. Thebase of the ski is coated with a surface that glides easily inthe forward direction, but has great resistance in the reversedirection. This action is different from using snowshoes,however. Cross-country skis allow the skier to bend his legatthe ankle, so that the skier can push off against the snow oneleg at a time causing the skis to glide across the snow.

Friction plays a significant role in both types of skiing.In downhill skiing, we are focused on kinetic friction. Incross-country skiing, both static friction and kinetic frictionare responsiblefor a skier's ability to move along.

Friction is present in the following problems. Where anumerical answer is called for, assume the angle ofinclination of the hill is 20°, cos 20° is 0.94, sin 20° is 0.34,and that the coefficient of kinetic friction is 0.2. Take the

mass of the skier to be 60 kg. [Note: /<k = coefficient ofkinetic friction; /*s - coefficient of static friction; a =acceleration.]


16. Occasionally, a skier has to "side step" up a hill. Shedoes this by turning her skis perpendicular to the hill,and then she makes series of small sideways steps upthe hill. This enables her to climb the hill. The

maximum steepness of a hill that a skier can side step isdetermined by:

I. the weight of the skier.

II. the coefficient of static friction between the skis

and the snow.

III. the coefficient of kinetic friction between the skies

and the snow.

IV. the initial velocity of the skier.

A. I only

B. I and III only

C. II, III, and IV onlyD. II only

17. Suppose a skier skis down the following curved hill:

18.

As a skier skis down the hill, which of the followingstatements accurately describes acceleration and speed?

A. The speed decreases, but the acceleration increases.

B. The speed increases, but the acceleration decreases.C. Both the speed and the acceleration increase.D. Both the speed and the acceleration decrease.

Referring to Figure 1, which of the following equationsbest represents the minimum tension the tow rope mustprovide to keep the skier moving up the hill?

A. T = mg(sin0 - /*kcos8)

B. T = mg(sin0 + fi^cosQ)

C. T = mg(sin8 + ^scos0)

D. T = mg(sin0 - /<scos0 - a)

19. Suppose a skier falls and begins sliding down the hill.Her speed at the bottom of the hill depends on:

I. her weight

II. her speed at the time of the fall.

III. the length of the hill.

TV. the steepness of the hill.

A. I only

B. Ill only

C. I, II, and III only

D. II, III, and IV only

GO ON TO THENEXT PAGE

20. A skier is given a strong push so that he slides up thehill in Figure 2 for a certain distance (fi^ = 0.1). When

he gets to the highest point, he slides back down. Howdoes the acceleration of the skier on his ascent compareto the acceleration on his descent? (Do not consider theacceleration during the initial push.)

A. The acceleration on the descent is smaller in

magnitude than on the ascent.B. The acceleration on the ascent is smaller in

magnitude than on the descent.C. Both accelerations are the same.

D. The accelerations have the same magnitude, butdifferent directions.

21. After leaving the following quarter-circular slope atpoint P, what will the resulting trajectory of the skierbe? (Note: point P is at the bottom of the circular path.)

22. Two skiers find themselves on a hill separated by adistance d as shown:

Assume the skiers experience the same coefficient offriction. As the two skiers ski down the hill, thedistance d between them:

A. increases.

B. decreases.

C. remains the same.

D. cannot be determined without knowing their initialvelocities.



23. If a person starts at the rim of a spinning platform andis pushed radially toward the central axis by a movingexterior wall, then what happens to the normal force feltby that person due to the wall?

A. It remains a constant.

B. It decreases, since r decreases.C. It increases, since r decreases.D. It decreases, since angular speed decreases.

24.

25.

If the tension in a rope pulling a 150-kg block in astraight line is 750 N, what will be the velocity of theblock after it has been pulled 200 meters, if thehorizontal surface it traverses has a fi^= 0.4?

A. 0.2 m/s

B. 5 m/s

C. 21 m/s

D. 100 m/s

A screw is an example of what type of mechanicaladvantage system?

A. Hydraulic pressB. Inclined planeC. Lever and fulcrum

D. Pulley

1. A 2. D 3. A 4. B 5. D

6. C 7. D 8 B 9. C 10. C

11. D 12. A 13. A 14. D 15. D

16. D 17. C 18. B 19. D 20. A

21. A 22. D 23. B 24. C 25. B

YOU ARE DONE.

Answers to 25-Question Forces, Circular Motion, and Gravitation Review

Passage I (Questions 1-8) Gravitational Force

5.

Choice A is the best answer. As the planet shrinks without losing any mass, its density must increase. Based on that, wecan immediately rule out choices C and D. Your weight is equal to the force of gravity acting on your body mass, which is:

p-gMmr2

The two masses do not change, but the planet's radius gets smaller as it shrinks. As the radius gets smaller, the force ofgravity acting on your body mass increases, so your weight increases. The best answer is choice A.

Choice D is the best answer. The gravitational potential energy is given by Equation 2:

u =-GMm.r

From that relationship, we know that U« 1. So, if r is tripled, then the magnitude of Umust be decreased by a factor of 3.Choice D is therefore the best answer. Choices A and B definitely cannot be correct, because the potential energy decreasesas the distance increases. As a point of interest, the gravitational potential energy is a negative number! If the distancebetween the two masses is tripled, the gravitational potential energy becomes a less negative number. The best answer ischoice D.

Choice A is the best answer. The acceleration due to gravity on the surface of the Earth is g. Planet X has a mass 10 timesthe mass of the Earth and a radius 2 times that of the Earth. To determine the acceleration due to gravity on the surface ofPlanet X, we need to find out how the radiusand mass of a planet are related to the acceleration due to gravity.

GMm=mg g= GMr2

(Note: M = mass of planet, r = radius of planet) Now we plug in the values for Planet X.

,10Megx = G^

(2reyThe best answer is choice A.

=(lftjoSk,. gx =iggx =G^=(iOVjMl., gx sigt choice A.U/ T2 2 (2re)2 \4) r2 2

Choice B is the best answer. A person standing on the surface of the Earth weighs the most at the North Pole. When aperson is at the Equator, he or she is in uniform circular motion because the Earth rotates. This motion creates a force thatcancels out some of the impact of gravity. When a person is standing on the North Pole, they are not in uniform circularmotion, so the force they feel is purely gravitational. Thisproduces a difference in weight.

At the North Pole Newton's second law gives: N = mg. A scale reads normal force. At the Equator Newton's second lawgives:

Fc = mg - N = mP— .•. N = mg - m^—r r

You appear to weigh a little bit less at the Equator becauseof the rotation of the Earth! The best answer is choice B.

Choice D is the best answer. The graph that best represents Uversus r can beeasily found byconsidering sign convention.Since U is given by U=-GMm., u must be negative at all times. This eliminates choices A and B. As the value of rincreases, the magnitude of U decreases, so choice C is eliminated. The best answer is choice D.

Copyright ©by TheBerkeley Review® 104 MINI-TEST EXPLANATIONS

6. Choice C is the best answer. Choice D can be eliminated immediately, because the two masses will speed up due to theacceleration they experience from the gravitational pull. As two masses draw closer to one another, the gravitational forceincreases according to distanced squared. Because the force is increasing, the magnitude of acceleration that each mass feelsis also increasing. This means that the two masses will speed up by a greater amount as they get closer and closer to oneanother. Their speeds will increase by an increasing magnitude over time. The best answer is choice C.

7. Choice D is the best answer. By Newton's third law, the two bodies must experience equally strong forces, ruling outchoice C. Because the comet is less massive, in order to have the same force, it must have a larger acceleration. This rulesout choices A and B and leaves choice D as the lone remaining choice. The best answer is choice D.

8. Choice B is the best answer. As a moon orbits a planet in uniform circular motion, the force of gravity from the planet onthe moon does no work on the moon, otherwise it would change orbital speed. The force of gravity exerted on the moon by aplanet always points towards the planet. Assuming that the moon is in uniform circular motion the gravitational force isalways perpendicular to the moon's path. Whenever a force is perpendicular to a path it does no work (i.e. Fd = 0). Theonly way to change the speed of an object is to do work (work - KE theorem). The best answer is choice B.

Passage II (Questions 9 -15) Circular Motion

9. Choice C is the best answer. The three bugs will stay on their circular paths as long as the static friction between them andthe table is sufficient to supply the needed centripetal acceleration. The greater the necessary acceleration, the greater thenecessary friction. Typically, objects revolving further from a rotational axis move faster (have a greater tangential speed)and consequently have a larger centripetal acceleration than closer objects. Bug C therefore requires more friction than theothers and is likely the first to slip. The best answer is choice C.

10.

11.

Choice C is the best answer. In Experiment 1, the ball does not change elevations, so the only force we shall consider istension. The ball is in uniform circular motion and Newton's second law becomes:

2FC= mac = m— •'• T = m^—

. v2 = 50 nr ... v = 7.1 JH-, choice C.25 N = 0.5 kg1.0 m

Even if you had little clue what to do on this question, you hopefully noticed that choice C is the square root of choice A andchoice B is the square root of choice D. Because force relates to velocity by a square factor, choices C and D are betteroptions than choices A and B. Pick choice C, and feel a warmjubilation of correctivity. The best answer is choice C.

Choice D is the best answer. To determine how long it takes the ball to complete one revolution in Experiment 1 requiresdividing distance by rate. The ball travels a distance of 2jcL with speed v. The time it takes is the distance divided by thevelocity.

time = distance . 2gLvelocity v

Choice D is the best answer. The other three answer choices could have been eliminated on the basis of units. Answer

choiceA has units of m3/s. The units of answer choice B are m2/s. The units of answer choice C are 1/ms. Pick choice D,and you'll be tickled by the feather of correctness. The best answer is choice D.

12. Choice A is the best answer. If, in Experiment 1, the speed of the ball is increased while the length of the string remainedconstant the tension in the string must have increased. Newton's second law for the ball gives:

T=m*ir

Since m and r are constants, the thing that could have possibly changed is the tension. If v increases, so does T. In this case,the tension is the centripetal force. Keep in mind that "centripetal" only refers to a direction! Pick choice A for the sensationof correcthood The best answer is choice A.

13. Choice A is the best answer. At point A in Experiment 2 the tension in the string is measured to be three times the weight ofthe ball. To calculate the speed of the ball at this point, we must use Newton's second law. The radius being used is L.

2 ,22FC = Ta + mg = m^—, where T^ = 3mg .\ 4mg = m^—


v2 =4gL .-. v = V4gL = 2YrgL, choice A.

MINI-TEST EXPLANATIONS

14. Choice D is the best answer. How long will it take the ball in Experiment 2 to reach the ground if it is released at point Awith a speed of V2gL? When the ball is released at point A its velocity is completely horizontal. So in the y-direction theinitial velocity is zero. That means the ball essentially free falls in the y-direction. To find the time we need to use thekinematics equations.

Ay =lp^ =2L=* t2 =4L .-. t=2<\[L2 8 V g


15. Choice D is the best answer. To determine the relative magnitudes of tension, we shall start by determining the point atwhich the tension in the string is the greatest To do this, we must first look at the forces acting on the ball at the variouspoints in its motion. Following this, apply Newton's second law at each point. Newton's second law gives the relationships

2

shown in the picture. The tension at point C is the greatest because m— is the same at all four points. As such, Tc has the

greatest magnitude and Tb = Tr> This eliminates choices A and C. We know that Tc has the greatest magnitude, so Tc >Ta- This makes choice D the best answer, so picking it should be sheer D-light.

v2At Point A: 2FC =TA +mg =m-^r2

whichgivesTa = m-7- - mg

At Point D:TD =m-y


At Point B: TB = m-y-

-q At Point C:ZFc =Tc-mg =m-^2

8 which gives Tc =m-y- +mg

Passage III (Questions 16 - 22) The Thrills of Skiing1

16.

17.

ChoiceD is the best answer. Start by drawing a force diagram. Keep in mindthat the skier is not sliding, but is stepping, sothe frictional force of interest is due to static friction and not kinetic friction. When the skier is side-stepping up the hill, atany instant in time there are only two forces acting on the skier in the direction of motion~the static friction force and acomponent of theforcedue to gravity. The skierdoes notwishto slidedownthe mountain, so theseforces mustbe equal.

psN = mgsin6

/<smgcos0 = mgsin6

/<scos8 = sin6

sin6Vs =

cos6= tan8

The two forces affecting motion relate according to: mgsinB = //mgcos8. We can see that the angle of inclination andcoefficient of static friction are related, but that the mg term cancels out This means that the mass of the skier and thegravitational force constant do not affect the maximum steepness of the hill. For a given angle, whether or not the skier canstep up the mountain depends only on the coefficient of static friction. The best answer is choice D.

Choice C is the best answer. In the absence of friction, the acceleration of any object sliding down an inclined plane isgiven by: a = g sin 6. As the skier skis down the curved hill, the angle of inclination is increasing, so the acceleration of theskier is increasing. Because acceleration is the rate of change of velocity, and the acceleration is positive in the direction weare considering, the velocity is also increasing. This question could have also been solved by visualizing yourself sliding,skiing, running, or hopping down the hills in question. The best answer is choice C.


18. Choice B is the best answer. The skier is being towed up the hill, so they are sliding across the snow. This means that thefriction we are dealing with is kinetic in nature, not static. This eliminates choices C and D. From here we need to draw aforce diagram and write down the net force equation on the skier:

ma = T -mgsinG - f^

ma = T -mgsin8 - pi^N

because N = mgcos8

ma = T -mgsin0 - /^mgcosS

Minimum tension in the rope to keep the skier moving means that the skier is not accelerating, a = 0; the skier is towed upthe mountain with a constant velocity. Solving for T, we get: T = mg(sin8 + fi^cosQ). This makes choice B a better choicethan choice A. Note that choice C gives us the minimum tension required to get the skier moving from rest. The best answeris choice B.

19. Choice D is the best answer. To determine the speed at the bottom of the hill, we can use either a kinematics equation or theenergy relationship. If we use the relationship between kinetic energy and gravitational potential energy, then we know thatthe mass does not impact the speed at the base of the hill. This eliminates choices A and C. If we choose kinematics, we needan equation with speed, acceleration, and the distance the skier slides down the hill. The kinematics equation:

v2 = v02 +2aAx

relates the initial velocity, the distance traveled, and the acceleration to the final velocity. The acceleration depends on thesteepness on the hill—the angle of inclination. Therefore, initial velocity, how far she slides down the hill, and the steepnessof the hill are all needed to determine her final velocity. The best answer is choice D.

20. Choice A is the best answer. Friction and the gravitational force both act on the skier as he goes up the hill and as he slidesback down. However, as he moves up the hill, friction and gravity act in the same direction, so these two forces add together.As he slides down the hill, gravity pulls him downward, while friction pulls upward. Now these two forces subtract. The netforceactingon the skier as he moves up the hill is greater than the net force acting on the skier as he movesdown the hill, sothe accelerationon the ascent is greater than on the descent.The best answer is choice A.

21. Choice A is the best answer. Point P is at the bottom of this circular path, meaning that the velocity of the skier, uponreaching this point, is horizontal and to the left After the skier leaves the slope, gravity pulls their trajectory downwards.Only choice A shows a horizontal departure followed by a descent Choice C is wrong, because it says the skier somehowgoes up before coming back down. This cannot happen if the velocity, upon reaching point P, is horizontal and to the leftThe best answer is choice A.

22. Choice D is the best answer. To answer this question, you need to know the velocity and acceleration of both skiers at thetime they are separated by the distance, d. They both are on the same slope feeling the same coefficient of friction, so intheory they experience the same acceleration. But, we need to know if bothof the skiershad the samevelocity at the time ofobservation. Let's call the skiers Skier 1 and Skier 2. Because each skier experiences the same acceleration, since they are onthe same hill and have the same coefficient of friction, the positions of the skiers as a function of time can be written as:

Skier 1: xi =x0i +v0it +'/iat2 and Skier 2: X2 = x02 + v02t+ !4at2

The distance d between the skiers is the difference of these two positions:

d = Ax = X2 - xi = (x02 + vQ2t) - (x0i + v0it)

If the skiers had zero initial velocity or the same initial velocity, then distance d would remain the same. If the skiers haddifferent initial velocities, then distance d would change. For the latter case, whether d would increase or decrease dependson which skier had the greater initial velocity.

Intuitively, if the lower skier were initially stationary and the upper skier were initially moving downhill, then theirseparation distance would decreaseas time wenton. On the otherhand, if the lower skier were initially moving downhill andthe upper skierwere initially stationary, thentheirseparation distance would increase with time. Either way, wecannot makean assessment of their separation distance withoutfirst knowing their respective velocities, so choices A, B, and C must beeliminated, leaving only choice D as a possible answer choice. The best answer is choice D.



23.

24.

25.

Choice B is the best answer. Providing that the platform spins with a constant speed, the normal force the person would feelis proportional to the centripetal acceleration. This acceleration can be written as:

,2ac= y or ac = arr

where o is the angular speed, to is the same for all points on a rotating body (choice D is incorrect). Note that the accelerationdepends on r, so it cannot remain constant as the person moves toward the center (choice A is incorrect). Since co is the samefor all points, and r is decreasing as the person moves radially toward the center, acceleration, a, must decrease (thereforechoice C is incorrect). The best answer is choice B.

Choice C is the best answer. We must first solvefor the acceleration of the block, which we do by usinga force diagram todetermine the applicable forces:

ma = T-f

N

mg

ma = T-//N

_ T - i*mg _ 750 - (0.4)050 kg)(9.8 m/s2)150 kg

a =m

a_ 750-(0.4x150x10) _ 750-600 _ 150 _ {150 150 " 150 "

Since the block ismoving, we use the coefficient ofkinetic friction. We can relate velocity toacceleration and distance usingv2 =v02 +2aAx. Then we find that:

v2s2(lm/s2)(200m) =400m2/s2 .-. v » 20 m/s

Before working out the math, you should ask yourself which of the four choices this could be. If you didn't immediatelyguess C, then ask yourself what the square root ofsome number around 400 is. Choice Cworks best Don't get bogged downin math. Doas littlemath as possible to getyouto thebestanswer quickly. The best answer is choiceC.

Choice B is the best answer. There is no fluid in a screw, so the mechanical advantage of a screw is not similar to ahydraulic press. Choice A is therefore eliminated. There is no balancing of torque with a screw, so it is not like a lever andfulcrum. Choice Cis therefore eliminated. There are no cables ina screw, soit isnot a pulley system. Choice Disthereforeeliminated. The mechanical advantage ofa screw is similar to that of an incline plane. The screw turns many times, butmoves a very short distance. This issimilar tomoving an object by a significant distance up a ramp, but having its elevationchange by only by a small amount The best answer is choice B.


I. Terminal Velocity

II. Space Lab

III. Kepler's Laws


IV. The Effects of Friction

V. Gravitation and Planetary Orbits


VI. Threshold Angle Study

VII. Inclined Planed and Pulley


(1-6)

(7-11)

(12 -17)

(18 - 21)

(22 - 26)

(27 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)

Forces, Circular Motion, and Gravitation Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3


The concept of free-fall is governed by several factors.When an object is in free-fall within a vacuum, its velocity issimply a result of the initial velocity and the acceleration dueto gravity. However, it is rare that an object on Earth willtravel in a vacuum. Usually, it moves through the air. Thestudy of motion through the air is known as aerodynamics.The air offers resistance to the motion and acts in the

opposite direction to the motion. When the resistance is greatenough, it can generate a force equal in magnitude to thegravitational force. When this occurs, the object is said tohave reached terminal velocity. A simplified way to viewterminal velocity is:

vt •v2mg

CpA

Equation 1

The formula gives the terminal velocity, v*, of a falling

object, where m is the mass of the falling object, g is theacceleration due to gravity, C is dimensionless and dependson the objects shape, p is the density of the surrounding air,and A is the cross-sectional area of the object

The dynamics of free-fall for an object can change if theobject is able to deform during flight Liquid droplets willvary in shapeas their velocity and the surrounding air densitychanges. This is taken into account when designing objectsfor flight.

1.

2.

The terminal velocity, vt, of an object is defined as thevelocity of an object:

A. just after it hits the ground.B. just before it hits the ground.C. when theobject's acceleration goes to zero.D. when the object's acceleration is in a direction

opposite that of gravity.

Which of the following statements are valid about anobject falling through the air?

I. A cube is moreaerodynamic than a sphere.

II. As the initial velocity of an object is increased, itwill reach a greater terminal velocity.

III. An object will reach a smaller terminal velocitywhen falling through a more viscous medium.

A. I onlyB. Ill only

C. I and III onlyD. II and III only


3. Which has a greater terminal velocity when fallingthrough the air, a 1000-kg boulder (with a cross-sectional area of 5 m2) or a 1.0-gm pebble (with across-sectional areaof 5 cm2)?

A. The boulder.

B. The pebble.

C. Both the pebble and the boulder have the sameterminal velocity.

D. There is insufficient information given todetermine the relative terminal velocities of the

boulder and pebble.

4.

5.

6.

What are the dimensions of the drag coefficient, C?

A. It has no dimensions.

B. meters/second

C. kilogram meters2/second2D. meters^/second^

If the raindrop starts its fall from rest, which graphBEST represents the relation between the drop'sinstantaneous speed and time?

A. B.

vt vt

time (s)D.

time (s)

time (s) time (s)

If the raindrop freezes into a round hailstone, how doesits new terminal velocity compare with the terminalvelocity it would have had if it had remained as araindrop?

A. vt stays the same.

B. vt gets smaller.

C. vt gets bigger.

D. The relationdepends upon how m changes.



One of the more impressive long-range projects beingdeveloped by NASA is the building of a permanent spacestation that would orbit the Earth. This space station wouldserve as both home and laboratory to scientists and civilians.One problem with living in space is the lack of a stronggravitational force. Although people can adjust to living in aweightless condition, there are some biological consequencesto prolonged weightlessness, including calcium depletionfrom bone and loss of muscle mass. These conditions are

believed to be reversible if the exposure to weightlessness isfor a short time, but long exposures to weightlessness mayhave irreversible effects.

To avoid this possibly permanent and undesirable effect,one space station design would provide an artificial gravity.A typical design has the space station built like a wagonwheel. Residents would live and work along the inner rim ofthe station, as schematically depicted in Figure 1. Residentswouldalso have workspace closer to the center of the station,but with the same orientation as along the rim. The artificialgravity is provided by a normal force acting on the residentsand points towards the central axis of the station. The stationrotates at a constantangularvelocity about this centralaxis.

7.

8.

Figure 1.

Given that the radius of the space lab is 1000 meters,what should the linear speed be at its rim so thatsomeone at the rim feels the same acceleration assomeonestanding on the planet Earth?

A. 10 m/s

B. 100 m/s

C. 1000 m/s

D. The speed is irrelevant.

Thespace station would mimic the Earth's gravitationalpull:

A. by providing a massequivalent to thatof the Earth.B. by spinningas fast as the Earth.C. by providing a largecoefficientof friction.D. by providinga suitable normal force.

Copyright ©by The Berkeley Review®

9.

10.

11.

111

If the space station were brought back to Earth andallowed to rotate as in Figure 1 about an axis parallel tothe ground, and if the acceleration due to gravity is g atall points on the station, at which point would thenormal force on the residents in the rotating station beGREATEST?

A. At the point closest to the Earth.

B. At the point farthest from the Earth.

C. At the point of fastest rotation along the rim.

D. The normal force will be the same at all points.

Suppose two residents of a rotating space station playcatch with a ball. They are positioned as shown below:

direction of

rotation

Person 1 is located a distance ri away from the center

of the station and Person 2 is located a distance r2 away

from the center of the station. The space lab is spinningin the direction indicated. Person 1 throws a balldirectly at Person 2. As viewed by an observerstandingoutside the space station and at rest with respect to thecentral axis, the resultant linear velocity of the ball isBEST represented by which of thefollowing arrows?A. B.

C. D.

If the space station is located a distance of twice theradius of the Earth away from the surface of the Earth,the acceleration due to gravitational attraction from theEarth on the space station is:

A. one-ninth of that on Earth.

B. one-quarter of that on Earth.C. one-half of that on Earth.

D. equal to that on Earth.



In the early seventeenth century, Johannes Keplerpublished three empirical laws that describe orbital motion ofplanets in the solar system. Although his laws dealspecifically with planets orbiting the Sun, they are in generalvalid for the orbital motion of any object. Kepler's First Lawstates that the planets orbit the sun in an elliptical orbit withthe Sun at one focus. Kepler's Second Law states that animaginary line drawn from the Sun to an orbiting planetsweeps out equal areas in equal time. Kepler's Third Lawstates that the square of the orbital period of a planet isproportional to the cube of the average orbital radius.

About a hundred years later, Newton later showed thatKepler's laws could be derived from his universal law ofgravitation, which is:

p_ GMmr2

where F is the gravitational force, M is the mass of oneobject, m is the mass of the other object, r is the distancebetween the centers of mass of the objects, and G is theuniversal gravitation constant. Figure 1 shows a Planet Porbiting the Sun.

Planet P

Perihelion Aphelion

Figure 1

12. At which point in Planet P'sorbit is the magnitude of itsmomentum the GREATEST?

A. Point A

B. Point B

C. Point C

D. Point D

13. Suppose Planet P is actually the Earth. As the Earthorbits from point B to point D, the gravitational forcefrom the Sun does:

A. work on the Earth, thusspeeding it up.B. workon the Earth, thus slowing it down.C. nowork on theEarth, thusspeeding it up.D. no workon the Earth, thus slowingit down.

Copyright©by TheBerkeley Review® 112

14. If a new planet were discovered orbiting the Sun at amean distance of four times the mean Earth-Sun

distance, it would have a period of:

A. 2 Earth years.

B. 4 Earth years.

C. 8 Earth years.

D. 16 Earth years.

15.

16.

Assuming this solar system consists only of the Sun andthe planet, the net force on the planet points:

A. towards the Sun.

B. away from the Sun.

C. in the directionof the planet'svelocity.D. in the directionopposite the planet's velocity.

If the mean orbital radius of the Earth were doubled, thegravitational pull from the Sun would:

A. decrease by a factor of 4.B. decrease by a factor of 2.

C. decrease bya factor of V2.D. increase by a factor of 2.

17. The true orbits of the planets in our solar systemdeviate from perfect ellipses. What could account forthe deviation?

I. The gravitational forces between the planets

II. The Sun itself is not stationary

III. The planetschangingmassas theyorbit

A. I onlyB. II only

C. I and II onlyD. II and III only



18. The gravitational force between two objects:

A. occurs only when the objects have very differentmasses.

B. is greater on the more massive of the two objects.

C. is not an attractive force.

D. increases in magnitude as the two objects approacheach other.

19. Newton's Third Law states that "To every action, thereis an equal and opposite reaction." This means that arope will pull on a mass with the same magnitude offorce that the mass pulls back on the rope! Why is amass able to move from rest when a tension is appliedto an attached rope?

A. The force exerted on the block is actually largerthan the force on the rope.

B. It has nothing to do with the rope and everything todo with the friction force.

C. The person pulling exerts a larger force on therope.

D. The action-reaction forces act on different bodies.

20. Mythhas it that Galileo dropped two round stones fromthe top of the Leaning Tower of Pisa. Taking intoaccount air resistance, if he dropped the stones fromrest and from the same height, and if the stones wereequal in size but different in mass, which stone hits theground first?

A. The lighter stone hits the ground first.

B. The heavier stone hits the ground first

C. Both stones hit the ground at the same time.

D. The less dense stone hits first.


21. Which of the following changes would give theGREATEST force on an object next to a large body?

A. Doubling the mass of the large body.

B. Doubling the distance from the large body.

C. Halving the mass of the large body.

D. Halving the distance to the large body.



A person has to pull a large, 150-kg block over a roughsurface by means of a rope connected to the block. There arethree ways the rope may be connected to the block, as shownin Figure 1 below:

II8=30°

HI

0 =30°

Figure 1. Sliding options for block

Once the block reaches a speed of 40 m/s, the ropedetaches and the block continues to slide in same direction,but at a gradually decreasing speed. The process is repeatedseveral times to confirm the results. In one instance, the largeblock immediately strucka smaller blockof mass 0.10 kg.

For the following questions, take the coefficient ofkinetic friction, ^, between the surface and the block to be0.4 and the coefficient of static friction, pis, to be 0.6. Notealso that sine 30° = 0.5 and cosine 30° « 0.87.

22. For configuration I, in order to get the block to beginmoving, the tension in the rope must be at least:

A. 600 N

B. 750 N

C. 900 N

D. 1000 N

23. When the block reaches a speed of 40 m/s, the ropesuddenly breaks. If the block is allowed to slow to astop, how long will it travel before coming to rest?

A. 5 seconds

B. 10 seconds

C. 15 seconds

D. 20 seconds


24.

25.

26.

Suppose the collision between the large block and thesmall block described in the passage lasted only 0.01seconds. What average force did the large block exerton the small block? Assume the speed of the largeblock just before collision is 40 m/s and that the changein speed of the small block is 80 m/s.

A. 40,000 N

B. 800 N

C. 200 N

D. 12N

If the block is to be pulled with a constant velocity,which way should the person choose to connect therope to the block so that the horizontal component oftension in the rope is the LEAST?

A. I

B. II

C. Ill

D. II and III

Consider the case where instead of allowing the blockto slow to a stop, after the rope detaches the large blockcollides immediately and elastically with a smallerblock of mass 0.1 kg. Immediately following thecollision, the velocity of the large block is:

A. relatively unchanged.

B. half the magnitudeof the initial speed.C. one-quarter the magnitude of the initial speed.D. one-sixteenth the magnitude of the initial speed.



When Isaac Newton was developing the laws ofgravitation, he studied among other things, the interaction ofthe planets with the Sun, and also the interaction of the Moonwith the Earth. Newton realized that any satellite—the Moonaround the Earth or the Earth around the Sun-is really aprojectile. For example, the Moon continuously falls towardthe Earth, but because of the orbital speed of the Moon andthe curvature of the Earth, it never reaches the ground.

Although he never named the force explicitly, he knewthat something had to attract the Moon if it were to remain inorbit around the Earth, and something had to attract the Earthif it were to remain in orbit around the Sun. Newton coined

the phrase "centripetal force" for any inward-pointing force,directed toward the center of an object's motion. Gravity isthe centripetal force that keeps the Moon in its orbit about theEarth, and it is the centripetal force that keeps the Earth in itsorbit about the Sun.

For the following problems, Re = radius of the Earth,Mg = mass of the Earth, Res = orbital radius of Earth around

the Sun.

27.

28.

Suppose you are standing on the surface of the Earthwith a baseball of mass m. How fast would you have tothrow the baseball horizontally, so that it never landedon the Earth?

A. v

B. v

C. v

D. v

•v" VGME

/GMEni

•v

GME

Re

Re

ReGMgm

If the Moon were moved four times as far away fromthe Earth as its present orbit, what would happen to the

gravitational force exerted upon it by the Earth?

A. The force would increase by a factor of 4.

B. The force would increase by a factor of 2.

C. The force would decrease by a factor of 16.

D. The force would decrease by a factor of 4.


29.

30.

31.

32.

115

We wish to put a space station somewhere between theEarth and the Moon so that there is a minimal net

gravitational force on the space station. Where shouldwe place it? (Consider just the Earth and the Moon inthis question.)

A. Closer to the Moon than the Earth.

B. Closer to the Earth than the Moon.

C. Exactly half-way between the Earth and the Moon.

D. There is no such location.

Placing the Moon in a new orbit that is four times as farfrom the Earth as its present orbit would:

A. have no effect on its orbital velocity.

B. decrease the orbital velocity.

C. increase the orbital velocity.

D. More information is needed to determine the effect

on velocity.

A new planet is discovered, IOORes away from theEarth. With only this information, what else can wedetermine about this planet?

A. Both its mass and its orbital velocity

B. Mass only

C. Orbital velocity only

D. Nothing else can be determined

Given that the Earth is 80 times more massive than the

Moon and has a radius 4 times as big, how does theacceleration due to gravity on the Moon compare to theacceleration due to gravity on the Earth?

A. gM= 0.05gE

B. gM= 0.2gE

C. gM = 20gE

D. gM= 40gE


Questions33 through 36 are NOT based on a descriptivepassage.

33. How long will a person on a circular track take tocomplete one lap, if the radius of the track is 1000mand they have a tangential speed of 3 m/s?

A. (2ji x 1000)/(3)

B. (3)/(2ji x 1000)

C. (2jtx 3)(1000)

D. (2jc)/(3 x 1000)

34. A pulley system would be most likely employed to:

A. raise a small mass a short distance.

B. raise a large mass a short distance.C. raise a small mass a long distance.D. raise a large mass a long distance.

35. What is the net acceleration for the mass on the rightside of the Atwood machine shown below?

A. a-(ml-m2)gmj + ni2

B. a-(ml + m2)Smi - m2

C. a = i—ml + m2

D. a = JB2i_mi + m2


36. Which of the following statements is/are true?

I. Static friction can speed an object up from rest.

II. Kinetic friction can prevent an object from slippingon a surface.

III. Kinetic friction can slow an object down to rest.

A. I and III only

B. Ill only

C. II and III only

D. I, II, and III



A student conducted an experiment to determine therelative coefficients of static friction for a series of three

materials. To do so, she used three blocks of equal mass anddimensions, each made of a unique material. Each block wasplaced on a smooth surface that could be elevated in such away as to change its angle of inclination. The smooth surfacecould be made of any of the three possible materials.

The student placed the block on each surface as far fromthe hinged point as possible. She gradually increased theangle of inclination in increments of 0.2° until the blockbroke free and began to slide down the incline. Once theblock broke free, it slid down the surface with increasingspeed. The student carried out nine trials in total. Theexperimental apparatus is shown in Figure 1 below.

£>

Initial flat surface Elevated surface

Figure 1

The three materials used were polished polycarbonate,rubber, and redwood. Table 1 lists the threshold angles thestudent found for each of the block-surface combinations.

Block\SurfaCe Redwood Polycarbonate Rubber

Redwood 14.2 9.6 26.2°

Polycarbonate 9.6° 4.2° 17.0s

Rubber 26.0° 16.8° >30°

Table 1

The kinetic friction exerted on an object is found usingEquation 1. Equation 2 describes the static friction exerted onan object. It should be noted that the static friction increasesas an external force is applied until it reaches a thresholdvalue, at which time the friction is said to become kinetic.

37.

Fkinetic= /% N

Equation 1

Fstatic * A<s N

Equation 2

How can it be explained that when a rectangular blockof polished polycarbonate is placed on a smooth surfaceof redwood, the threshold angle is the same, no matterwhat face of the block is placed on the redwoodsurface?

A. Friction depends on the contact area of the objectto the surface.

B. Friction depends on the position of the center ofmass of the object.

C. Friction depends on the total surface area of theobject.

D. Friction depends on the mass of the object.

Copyright ©byThe Berkeley Review® 1 17

38. Which of the three materials studied has theGREATEST coefficient of static friction, //s?

A. Redwood

B. Polished polycarbonateC. Rubber

D. All three materials have equal /.ts values.

39. Which of the following graphs best represents the forcedue to static friction as incline angle increases for aredwood block atop a smooth rubber surface?

40.

41.

42.

1 "1

u«Incline Ancle'

'j-

Incline Angle

Incline Angle Incline Angle

Which of the following is NOT true in terms offriction?

A. A material's coefficient of static friction, /<s, is

greater than its coefficient of kinetic friction, /z^.

B. Friction occurs only between two flat surfaces.

C. Friction can accelerate on an object.

D. Kinetic friction takes energy from a moving object.

Which of the following changes to the system shown inFigure 1 would reduce the threshold angle needed toinitiate sliding?

I. Changing the direction in which the ramp ispointing by repositioning its base

II. Adding a lubricant to the top of the inclined surface

III. Adding a lubricant to the top side of the block

A. II only

B. I and II only

C. II and III only

D. I, II, and III

The frictional force acting on a block when it begins toslide down an inclined surface is best described by:

A. nk mg cosO.

B. /<k mg sinO.

C. /<s mg cosO.

D. ;<s mg sin9.



A mass M is held motionless on a smooth ramp, whenthe tensiondue to a hanging mass m and an applied force Faare in balance with the force of gravity, as shown in Figure 1.The applied force and tension necessary to balance gravityvary with the angle of the inclined plane.

Figure 1

In an experiment, a student varies the hanging mass andsets the system to three different angles. The student thenmeasures the applied force, at each angle, needed to establishequilibrium at each angle. A spring of known springconstant is used as the applied force. Table 1 lists the datacollected from seven separate trials. The student found thatas the hanging mass approached the mass of the block,equilibrium could not be established at smaller test angles.

m(kg) e ^applied

2.0 30° 4.9 N

2.0 45° 15.2 N

2.0 60° 22.1 N

3.0 30° No equilibrium

3.0 45° 5.4 N

3.0 60° 12.2 N

4.0 60° 2.4 N

Table 1

43. What would be observed for the system in Figure 1 if m= 2.5 kg and 0 = 45°, if the applied force were absent?

A. The box on the ramp would slide down the rampwith increasing speed.

B. The box on the ramp would slide down the rampwith constant speed.

C. The box on the ramp would not move.D. The box on the ramp would slide up the ramp with

increasing speed.

44. How does the magnitude of the tension in the ropecompare with the magnitude of the applied force?

A. Fa is greater than the tension.

B. Fa is less than the tension.

C. Fa is equal to the tension.

D. There is insufficient information to tell.


45. How would a static frictional force on the ramp and airresistance affect the applied force needed to keep themasses stationary?

A. Both forces lessen the magnitude of Fa needed.

B. Neither force lessens the magnitude of Fa needed.

C. Friction lessens the magnitude of Fa needed, but

air resistance does not.

D. Air resistance lessens the magnitude of Fa needed,

but friction does not

46.

47.

48.

Suppose Fa is accelerates the mass at an increasing rate

as it increases. Which of the following graphs wouldNOT represent Fa as a function of time?

A. B.

FA

In the absence of any friction, what would be thedownhill acceleration of mass M, if the rope were cut?

A. gsin8 +FA/M

B. gsin9-FA/M

C. gcosO - FA/M

D- (g-FA/M)/sine

Why are there no entries in Table 1 for a 4.0-kghanging mass with an incline angle of 30s?

A. The spring cannot be realigned to that angle.Equilibrium cannot be established because the boxwith mass M would accelerate up the ramp.

Equilibrium cannot be established because the boxwith mass M would accelerate down the ramp.The cable deforms with that much tension.

B.

C.

D.



49.

50.

51.

52.

Starting from rest, a disk rotates about its axis atconstant angular acceleration. After 6.0 seconds, it hasrotated through 72 rad. What is its angular acceleration?

A. 2rad/s2

B. 4rad/s2

C. 6rad/s2

D. 8rad/s2

On a rainy day, the coefficient of friction between thetires of a car and the road drops to roughly 70% of itsnormal value. The maximum safe speed for rounding acurve is:

A. unchanged from its normal value.

B. reduced to 84% of its normal value.

C. reduced to 70% of its normal value.

D. reduced to 50% of its normal value.

A 2 kg ball at the end of a lm string is spun in a verticalcircle. The tension in the string is 52 N when the ball isat the bottom of the circle. What is the ball's speed?

A.

B.

C.

D.

4m/s

5m/s

6m/s7m/5

If a wheel completes 80 rev in 10 s and finishes with anangular velocity of 12 rev/s, what was its initial angularvelocity, assuming it underwent uniform angularacceleration?

A. 2 rev/s

B. 4 rev/s

C. 8 rev/s

D. 12 rev/s


1. C 2. B 3. D 4. A 5. D 6. B

7. B 8. D 9. A 10. D 11. A 12. D

13. A 14. C 15. A 16. A 17. C 18. D

19. D 20. B 21. D 22. C 23. B 24. B

25. B 26. A 27. A 28. C 29. A 30. B

31. C 32. B 33. A 34. B 35. A 36. A

37. D 38. C 39. D 40. B 41. A 42. A

43. A 44. D 45. C 46. A 47. B 48. B

49. B 50. B 51. A 52. B

YOU ARE DONE.

Answers to 52-Question Forces, Circular Motion, and Gravitation Exam

Passage I (Questions 1-6) Terminal Velocity

1. Choice C is the best answer. Terminal velocity is, by definition, the speed a falling object attains when it is no longeraccelerating. An object accelerates for some time after release. At the end (or termination) of the acceleration time period, itsspeed no longer changes. Air resistance is the force causing this to happen, because the force upward due to air resistance isequal to the force downward due to gravity. Resistive force increases as the falling object's speed increases. At some point,resistive force balances the force due to gravity. When the forces cancel, the acceleration is zero. When acceleration is zero,speed does not change. This makes choice C the best answer. Choice B can be eliminated, because although the terminalvelocity may have been reached just before striking the ground, we do not know that it reached terminal velocity withoutknowing the distance it falls. If the object's acceleration were in the direction opposite to gravity, then the object would slow,which it does not, so choice D is eliminated. The best answer is choice C

2. Choice B is the best answer. According to the equation, less surface area for an object results in greater terminal velocity. Acube has more surface area, so it experiences a greater resistive force when moving through the air. This makes a cube lessaerodynamic, so statement I is invalid. According to the equation, the initial velocity of an object is never considered whendetermining its terminal velocity. A greater initial velocity causes the object to reach terminal velocity in a shorter time, but ithas no effect on the value of the terminal velocity. Statement II is invalid. According to the equation, there is no variable forviscosity in the terminal velocity, so this requires intuition. Viscosity is defined as resistance to flow, so in a more viscousmedium, there is more resistance, causing an object to move more slowly. It is true that an object reaches a smaller terminalvelocity in a more viscous medium, making Statement III valid, and making choice B the best answer. Viscosity and densitycan be interconverted, in theory. The best answer is choice B.

3. Choice D is the best answer. To answer this question, think about the variables that go into the terminal velocity equation:

vt=2mg

CpA

We know g, A, and m. p is the same for both the pebble and the boulder, because both fall through the same air. The dragcoefficient, C, depends upon the shape of the object. Boulders and pebbles can have different shapes. We are not told theirvalues for C, so we do not have enough information to complete the problem. The best answer is choice D.

Choice A is the best answer. In the passage, C is referred to as being dimensionless. This question is more a test of whetheryou pick up on seemingly unimportant details than it is a straight-forward physics problem. However, if you did not read thepassage, you could still figure out that C is dimensionless by manipulating the terminal velocity equation.

vt =2mg

CpATherefore, C =

|mass][i^gth/time2]

2mg

(vt)2pA

-, which is dimensionless.which has units of

( [lengfi%meI )^sWh3Hi-^gth2]The best answer is choice A.

Choice D is the best answer. Choice A is incorrect, because it suggests that the raindrop has some high speed when it is firstdropped and then slows down as its speed eventually goes to V[. This is not true. The drop starts from rest and speeds up, aswe would expect. Choice B is incorrect, because although it says that the drop starts from rest and eventually gets to Vt, at v =

Vt, the speed abruptly stops changing. This does not happen. Most things in nature happen smoothly. To stop abruptly, thedrop would have to hit something, like the ground. Just hitting air molecules would not do this. Choice C is incorrect,because it implies that after some time, the speed exceeds vt. By the definition of terminal velocity, this cannot happen under

these conditions. Choice D is correct, because it starts from rest and smoothly approaches vt. The best answer is choice D.

Choice B is the best answer. If the terminal velocity changes as a raindrop freezes into a hailstone, then one or more of thevariables in the formula must change. The formula has variables m, g, and p that do not change, because it is the sameamount of object, the same amount of gravity, and the same amount of air. Also, C does not change, because C depends uponthe shape, and the droplet and the stone have approximately the same shape, with the droplet having perhaps less resistance.The A term is all that remains. As water freezes into ice, it expands. Thus, the hailstone has a bigger A than the raindrop. If Aincreases, then vt goes down. The best answer is choice B.

.©Copyright © by The Berkeley Review' 120 REVIEW EXAM EXPLANATIONS

Passage II (Questions 7-11) Space Lab

10.

11.

Choice Bis the best answer. The net acceleration on the person is centripetal. Any time an object moves in acircular path,the net acceleration of that object is centripetal. The centripetal acceleration of the Space Lab will simulate the Earth'sgravitational acceleration. Centripetal acceleration isgiven by the equation:

,2ac=vir

We want ac to be equal to g, the acceleration due to gravity a person feels standing on the surface of the Earth. Solving theabove equation for v,we get v=VaJ= Vp7 =-/(9.8 2L)(i000m) = 98.99 HI - 100 ™-. The best answer ischoice B.

V s s s

Choice Dis the best answer. The space station creates an "artificial gravity" by providing a centripetal acceleration whosevalue is very nearly g. This centripetal acceleration comes from a netcentripetal force, which in turn is the vector sum of allof the forces that point in or out along the radius of the space station. The normal force of the walls of the space stationprovides this inward pointing-force. This principle is the same as theone that governs the operation of certain carnival rides(e.g., rotating cylinders where the floor is removed). The best answer is choice D.

Choice A is the best answer. To see how the normal force relates to the Earth's gravity and to see whether it varies in thespace station, let's draw forces for the points farthest from and closest to the Earth's surface.

Farthest fromthe Earth

Schematic side-viewof the space station

Closestto the Earth

The top point (farthest from the Earth) has a normal force of N+ mg = mv2/R or N= mv2/R - mg. Repeating for the bottompoint (closest to the Earth) gives a normal force of N= mv2/R + mg. The normal force is strongest at the bottom point,comparing these two points. This rules out choices B and D. Choice C is also wrong, because the rotation rate is the same atany point on the circular rim. The best answer is choice A.

Choice D is the best answer. To an observer outside the space station, the ball has two components to its linear velocity.One component, v1? is tangent to the circle of radius r^ the ball is on the space station and is therefore moving in a circularpathalong with the people. The second component, v2, is along the downward direction that the ball was thrown, from thefirst person to the second person. An observer outside the space station would see the vector sum of these two components:


Choice A is the best answer. To solve for the acceleration (call it g') exerted on the space station from the Earth, we use:

„. _ GME

where r is the center-to-center distance between the Earth and the space station. This is a spin-off of the gravitational forceequation, r for this problem is 3RE, where RE is the radius of the Earth. (Note that they stated the space station was 2REaway from the surface of the Earth. Remember to add in the radius of the Earth to get the center-to-center distance.) We donot know the mass of the Earth, nor do we know the radius of the Earth, so the best way to solve this problem is to recognizethat g, the acceleration due to gravity at the Earth's surface, can be written as:

GME£ = -/-- 1S

g =GME

R2EWe then take the ratio of g' to g: (3RE)2/GME=9

(ReVThe best answer is choice A.


PassageIII (Questions 12 -17) Kepler's Laws

12. Choice D is the best answer. At which point in Planet P's orbit is the magnitude of its momentum the greatest? Themomentum is the greatest when the speed of the planet is the greatest. The speed is greatest at point D. This is aconsequenceof Kepler's second law. Apicture might help us to see that point Dis the best answer. In order to sweep out equal areas Ajand A2 in the same amount of time, the planet must move faster at point D. In fact, the closer a planet gets to the Sun duringits orbit, the faster it travels.

A

Planet P


13. Choice A is the best answer. Suppose Planet P is actually the Earth. As the Earth orbits from point B to point D, thegravitational force from the Sun does what to the Earth? In order to satisfy Kepler's second law, Planet P must speed upas ittravels from point B to point D, ruling out choices B and D. Does the Sun do any work on the planet? The answer is yes,based on the work-kinetic energy theorem. The work-kinetic energy theorem states: Wxotal = AKE. Since the planet speedsup, it changes its kinetic energy. Since it changes its kinetic energy, there must have been work done on it. The only forcethat could have done this work is the force of gravity from the Sun. The best answer is choice A.

14. Choice C is the best answer. If a new planet were discovered orbiting the Sun at a mean distance of four times the mean

Earth-Sun distance, what would its period be? Kepler's third law states T2 a R3. We can set up a ratio using Kepler's thirdlaw between the new mystery planet and Earth.

12 /d \3 / ir, \3_ Kpsl (j ft-./rp^HjJBsf , —-Ues' IRes / Ip«V(1 year)2(4)3 =(1 year}/64 =8Years

The orbital period of our new mystery planet is eight Earth years. The best answer is choice C.

15. Choice A is the best answer. There is one force acting on this planet, that of gravity pulling it towards the Sun. The netforce must point towards the Sun. The best answer is choice A.

16. Choice A is the best answer. What would happen to the gravitational pull from the Sun, if the mean orbital radius of the

Earth were doubled? Newton's universal law ofgravitation is: F = GMm

The gravitational force is proportional to the inverse of the square of the distance. If we double the distance, the force must

decrease by a factor of4: Fnew =GMm =1/GMm j_JJpy jne jjest answer js choice A.(2r)2 4l r2 / 4

17. Choice C is the best answer. The true orbits of the planets deviate from perfect ellipses. What could account for thedeviation? Maybe perfectly elliptical orbits would be possible under highly idealized conditions. One of those conditionscould be an assumption that the Sun is the only force pulling on the planets. The gravitational attraction between the planetsthemselves could disturb the idealized elliptical orbits. This makes Statement I a valid statement. Having the Sun stationaryat one focus of the ellipse also seems like an idealized condition. In reality, the Sun and planets orbit the center of mass of thesolar system. The motion of the Sun itself as it moves through the Milky Way galaxy would seem to be able to cause adeviation from perfectly elliptical orbits. This makes Statement II a valid statement. However, the planets do not shed or gainmass as they orbit, the Sun, so although a changing mass could explain the deviation from elliptical orbit, Statement III is nota true statement. The best answer is choice C.


18. Choice D is the best answer. The gravitational force works between every pair of masses in the Universe, ruling out choiceA. When two objects interact via force, each object feels the same magnitude of force but in opposite directions. This isNewton's third law and the reason that choice B is wrong. Gravity is always an attractive force. Choice C is wrong. You aremore attracted to the close Earth than you are to the distant Jupiter. Gravity is stronger when you are closer to the object. Thebest answer is choice D.


19. Choice D is the best answer. Although the rope and the mass exert equal but opposite forces on each other, whenconsidering the motion of the mass, we lookat only those forces acting on the mass. We do not lookat those forces exertedby the mass. Choice A is wrong; it contradicts Newton's third law. Choice B is wrong; the mass moves from rest, because thetension in the rope isgreater than the friction force. Choice C is wrong; by Newton's third law, any tension in the rope mustcome from the person pulling on the rope, so we might expect those forces between the person and the rope also to be equalbut opposite. The mass accelerates because it experiences a net force. The best answer is choice D.

20. Choice B is the best answer. As the tale goes, the two stones hit the ground at the same time. However, this is notnecessarily true. The reason is rooted in the concept of air resistance. In the case of extreme air resistance, where a fallingobject experiences a drag force equal in magnitude to the gravitation force, the object reaches a terminal velocity. Terminalvelocity is the result of air resistance slowing the stone's acceleration to zero.

Even if the objects do not reach their terminal velocities before they hit the ground, the denser stone still should hit first,because air resistance is unable to hold it back as much as air resistance can hold back the less dense stone. Air resistancedepends on the contact area of the object with the medium. When the object has a high mass and low contact area, then theweight is more significant than air resistance. Conversely, when the object has a low mass and high contact area, then theweight is comparable to air resistance. This means that the stone that hits the ground first will be theone that has the highestmass-to-contact area ratio. This supports choice B.

So why do physicists insist that they hit at the same time? It is likely that two nearby stones have close to the same densities,so they should experience similar impact from air resistance. The two stones should hit the ground at times that are separatedby a fraction of a second, making the difference in time too small to detect, thus perpetuating the Galileo legend. The bestanswer is choice B.

21. Choice D is the best answer. The gravitational force between two objects depends upon each of their masses and thedistance from one center of mass to the other center of mass, according to the following equation:

F = GMmr

From this equation, wesee that F« Mand thatF <* Vr2- The force increases if the mass of the large object increases or if thedistance between the two objects decreases. This rules out choices B and C. Of the two remaining choices, choice A leads toa doubling of F, and choice D leads to a quadrupling of F. This means that choice D results in a greater increase in themagnitude of the force. The best answer is choice D.

Passage IV (Questions 22 - 26) The Effects of Friction

22.

23.

24.

Choice C is the best answer. In order to get the block moving in the first place, the applied force must exceed the staticfriction force. Therefore, the tension in the rope must at least equal this static friction force: T = u,mg = (0.6)(150 kg)(10

m/s2) = 900 newtons. The best answer is choiceC.

Choice B is the best answer. When the block is free of the rope, the net force acting on it is the kinetic friction force. Wecare only about the forces in the x-direction, because the forces in the y-direction, gravitational and normal, cancel oneanother. We solve for acceleration using: ma = -/<mg.

a = - u.g =- (0.4)(10 m/s2) = - 4m/s2

To solve for the time, we use: v = v0 + at.


t= y_1Vo.= o^40mZs_=1Oseca -4 m/s2

Choice B is the best answer. The average force exerted by each object on the other object in a collision is best found byconsidering the impulse of the collision. The relationship is: Faverage * At = mAv, where Faverage is the average force duringthe collision, At is the time of collision, m is the mass of the object of interest, and Av is the object's change in velocity.Manipulating the equation leads to Faverage = mAv/At. The math can besolved as follows:

F =


mAv

At

(0.1 kg) (0-80 m/s)

0.01 seconds= 800 newtons


25.

26.

Choice B is the best answer. This question can be answered by summing up the forces for each of the three systems, andthen comparing the resulting tensions. This would take you at least five minutes to do so, assuming you are good at this kindof thing. Just don't do it. Answer this one using physical intuition. In fact, approach every problem intuitively, and do moreonly if intuition alone can't isolate the best choice. Here, we want to know which configuration leads to the minimumhorizontal component oftension. It is probably not setup I, because that's the middle situation. To decide between II and III,we need to think a little more. Let's draw the forces acting on the block in the two cases:

IiN

yT

[If /<* = 30

mg

The tension competes with friction horizontally. How do you think the frictions compare in setups II and III? It depends onthe normal force. How do the normal forces in these two setups compare? The normal force is less in setup II, because therope is tending to lift the blockoff the ground, therefore decreasing thecontact pressure between the block and ground. Sincesetup II has a smaller normal force, and because thefriction is proportional to the normal force, setup II hasa smaller frictionforce. Since the tension counteracts the friction horizontally, the horizontal component of tension must be smaller for setupII. You could have guessed this answer, if you've ever moved a refrigerator or other large object. To move the object mosteasily,you lift a bit and pull. In other words,your arm pulls in a fashion similar to setup II. The best answer is choice B.

Choice A is the best answer. When two masses collide, they exert equal but opposite forces on each other, according toNewton's third law. The larger mass therefore experiences the smaller acceleration, resulting in a smaller change in velocity.If the two masses are vastly different, as they are in the passage, the speed of the larger mass is relatively unaffected by thecollision. You could also answer this question using the following equation:

'mi - ITI21v'l =

\mi + IT12/

If mi »iri2, then m\ + iri2 is roughly equal to mi, and m\ - m2 is also roughly equal to mi. So this equation reduces to:

mlv'i a —vi, which implies that v'i = V]

mi

A simpler way to think about it, if you understand the relative mass difference here, is to picture a coasting luxury car hittinga carton of milk. Is the car slowed much, if at all? The best answer is choice A.

Passage V (Questions 27 - 32) Gravitation and Planetary Orbits

27. Choice A is the best answer. In order for something to orbit the Earth and never touch it, the gravitational force exerted bythe Earth on the object must just equal the centripetal force. If the object is thrown very near the surface of the Earth, theradius of its orbit is the radius of the Earth. Setting these two equations for force equal to each other, we get:

mv2 1_ GMEm

re> (RE)2and

gme

Re

This problem can be solved more immediately by using the limiting cases test tip. First, if the mass of the Earth were to getvery small, what would you expect of this escape velocity? It should probably get very small, since there would not be muchof an Earth to get away from. This rules out choices B and D. Now, the difference between the remaining choices is thebaseball mass. Should it matter? No. The mass of the forced object usually drops out of the resulting equations when gravityis involved. This rules out choice C. The best answer is choice A.

28. Choice C is the best answer. The gravitational force has a 1/R2 dependence on distance:

FccXR2

If the distance increases by a factor of4, this introduces 42 in the denominator, so the force is reduced by a factor of 16. Thebest answer is choice C.


29.

30.

31.

Choice A is the best answer. If the space station is somewhere between the Earth and the Moon, it experiences agravitational attraction from both the Earth and the Moon. Recall that gravitational force iswritten as:

p- GMm

" R2This equation says that the gravitational force is proportional to mass, and it is inversely proportional to the square of thedistance between the two masses. Since the Moon is less massive than the Earth, the space station must be closer to theMoon so thatthenet gravitational force felt bythespace station is zero. The best answer is choiceA.

Choice Bis the best answer. For an orbiting body, the gravitational and centripetal forces must be equal. Ifthe gravitationalforce is reduced, then so is thecentripetal force. The centripetal force ingeneral is written as:

U/where m is the mass of the object undergoing the centripetal motion. It may not be clear from this expression what willhappen to the velocity when the radius of the orbit gets larger, sowe setthe gravitational and centripetal forces equal to eachother and do a little manipulating:

GMm

R:= m which leads to GM = v2R

where Mis the mass of the body exerting the gravitational force on m. For this problem, Mis Me, the Earth's mass. Noticethat the mass of the orbiting object cancels out Although we have moved the Moon farther away from the Earth, it is stillorbiting the Earth. The productGM is a constant, so that as R gets larger,v has to get smaller. The best answer is choice B.

Choice C isthe bestanswer. From the relationship between velocity and orbital radius (v2R = constant), we see that we candetermine the velocity of the planet. However, the mass of the planet(the orbiting body) cancels out of the equation, so wedo not have any way to determine that. The best answer is choice C.

32. Choice B is the best answer. To determine the acceleration due to gravity, we use:

8e =GME

and gM =GMM

RM

which come from setting:

mg =fimM. at the planet's surface.R'

Setting up a ratio between the acceleration due to gravity acting on the Moon and the acceleration due to gravity acting onthe Earth, we get:

Mm Mm

Questions 33 - 36

8m

Se

Rm

Me

Rl

8m Rm

gE 80 Mm

(4RMfIf you happened to remember that you weigh about 1/6 of your Earthbound weight when you are on the Moon, you wouldget to the answer by converting this fraction into a decimal value. The best answer is choice B.

*i = !^= 02Se 80

Not Based on a Descriptive Passage

33. Choice A is the best answer. This question is essentially asking you to identify the correct formula. One quick way toanswer this is to consider units. The 2n term has no units. Use the units given in the problem for the speed and radius.Checking the units for choice A gives the correct unit of time. Since the other choices are algebraically different, theyprobably have (and do have) different units. Any time you see problems of this format, use the same tips you use forformula-identification type problems to solve these. The basic equation for time is distance/rate. The distance traveled aroundthe circle is 2jtt and the rate is given as 3 m/s. Plugging the values from the question into the equation yields (2jt x 1000) 13,choice A. The best answer is choice A.


34. Choice Bis the best answer. In apulley system, the ropes can shorten only asmall amount before the wheels collide, so thesystem is used to move an object only a short distance. As implied by using a machine that generates a mechanicaladvantage, the mass moved by die pulley system must be great, or the pulley system would not be necessary. This meansthat the best answer, the answer that the author ofthe passage wants you to pick, is choice B. The best answer is choice B.

35. Choice A is the best answer. This is a repeat ofExample 2.3a. Hopefully you recall from that example that attempting tosolve the question by traditional means (mathematics and the equating of forces) is not the most time-efficient method nornecessary on a multiple-choice exam. Plugging in extreme values and seeing which equation matches the predicted results isthe ideal route to finding the best answer. Choice Cis the first to be eliminated, because the kg unit doesn't cancel out. Ifweconsider the case where mi is greater than m2, then we predict by observation that the system will accelerate downward onthe left side. The magnitude of a should increase as mj increases and m2 decreases. Choice D definitely doesn't fit thisobservation, so it is eliminated. If m2 is greater than mi, then we would expect the mass on the left side to accelerateupwards, so the sign of a has to be different when m2 >mi than it was when m\ > m2. Only choice Afits this restriction. Thebest answer is choice A.

36. Choice A is the best answer. It may seem counterintuitive, butstatic friction can push a stationary object to start moving.Walking is a perfect example where your foot pushes laterally against the ground and the static friction force pushes back.The pushing back force is what gets you start walking. That pushing back force is static friction. Statement I is a validstatement, eliminating choices B and C and telling us that Statement III must be true. Slipping implies that the object isstationary and then starts to slide. This is a matter of semantics, because the object starts at rest, so the friction we care aboutis static friction, not kinetic friction. It is static friction, not kinetic friction, that can prevent an object from slipping on asurface. Statement II is invalid, so choice D is eliminated. The best answer is choice A.

Passage VI (Questions 37 - 42) Threshold Angle Study

37. Choice D is the best answer. When a rectangular block made of any material is placed on a smooth surface made of anymaterial, the threshold angle is the same, no matter which face of the block is placed on the surface. This is because theweight, normal force, and friction force do notdepend on the total areaof directcontact between the two surfaces. As such,friction does not depend on the contact area between the object and the surface, nor does it depend on the position of theobject's center of mass. Neither of these two factors would affect the force, so choices A and B are eliminated. You couldhave also considered the fact that the faces of a rectangular block have different surface areas, so the contact area must notimpact the friction force. The block has the same total surface area, no matter how it is aligned, so choice C can also beeliminated. The force due to friction depends on the coefficients of friction of the materials and the normal force, which inturn depends on the angle and the weight. Because the mass of theobject affects its weight, the best answer is choice D. Thebest answer is choice D.

38. Choice C is the best answer. The material with the greatest coefficient of static friction, fis, will require the greatest

threshold angle to be put into motion. This is because as the angle increases, the normal force decreases, so the net forceacting on the object is increasing until it finally breaks free. According to the data in Table 1, the greatest threshold angle isfound with rubber against rubber. The best answer is choice C.

39. Choice D is the best answer. When the surface is flat, the angle is defined as 0°. The only two forces acting on the objectare gravity (downward) and the normal force (upward). There is no other force to consider, so the force due to static frictionis zero at that point. This means that the graph should start at the origin, which eliminates choices A and B. The staticfriction increases with the angle of inclination until the object breaks free and begins to slide down the surface. This can beseen in the graph in choice D. If choice C were true, then there would be no threshold value, because the maximum staticfriction force does not occur right as the object breaks free. This eliminates choice C. The best answer is choice D.

40. Choice B is the best answer. The coefficient of kinetic friction, fifc, for an object on a surface must be less than the

coefficient of static friction, pis, for an object on a surface, because objects remain in motion once the threshold force isapplied. If kinetic friction were greater than static friction, then the object would experience an increase in resistance tomotion once it started to move, which would in turn stop the motion. Because the block stays in motion, the kinetic frictionis less than the static friction it overcame to start moving. Choice A is a true statement, and thus is eliminated. Friction canoccur between any two surfaces, not just between two flat surfaces. This means that choice B is not true, making it the bestanswer to this question. Friction is a force, so it can and does accelerate an object. Choice C is a true statement, and thus iseliminated. Kinetic friction does work on an object as it moves, so it drains energy from an object in motion. Choice D is atrue statement, and thus is eliminated. The best answer is choice B.


41.

42.

Choice Ais the best answer. Anything that can decrease the maximum static friction force will decrease the threshold angleneeded to initiate sliding. Changing the direction in which the ramp is pointing by repositioning its base has no effect on themagnitude ofthe forces acting on the block. This means that the threshold angle would remain the same, so Statement I isinvalid, which eliminates choices Band D. Adding a lubricant to the top of the inclined surface, the surface across which theblock slides, would decrease the maximum static friction force. This would result in a reduced threshold angle, which makesStatement II a valid statement This does nothing to help with the elimination ofanswer choices, given that Statement II canbe found in each of the answer choices. Adding a lubricant to the top side of the block does not affect the surface of theblock that makes contact with the surface on which it is sliding. This means that the threshold angle would remain the same.This makes Statement III invalid, which eliminates choice C. The best answer isIIonly. Thebest answer ischoice A.

Choice A is the best answer. The force offriction acting on a moving object is defined as FkjneUc friction =/<k-N. Becausethe object is in motion, we need //fc, and not jis. This eliminates choices Cand D. The question is reduced to determiningwhether the normal force is mg cos 9or mg sin 0. When the surface is perfectly flat, the normal force is equal to mg. Theangle isdefined as0°, soNmust equal mg cos 0for the normal force toequal the weight, mg. The bestanswer ischoice A.

Passage VII (Questions 43 - 48) Inclined Plane and Pullev

43. Choice A is thebest answer. According to Table 1, when the hanging block has a mass of3.0 kg and the ramp isset atanangle of45°, an applied forced of5.4Nisnecessary to prevent the block from sliding down the ramp. This means that if thehanging mass were reduced to 2.5 kg, thus reducing the tension pulling the block up the ramp, the block would require aneven larger applied force to hold it in place. If theapplied force were removed, then the system would not be in translationalequilibrium,and a net force would exist. A net force would cause the block to accelerate, so choices B and C are eliminated.Because the applied force pushes the block up the ramp, we can assume that the block would accelerate down the ramp uponthe removal of the applied force. The best answer is choice A.

44. Choice D is the bestanswer. The tension in the rope and Fa add up to balance the component of gravity down the slope.We know only that their sum balances a piece of gravity. Since this tells us nothing about how the tension and Fa compareto each other, we have insufficient information to know their relationship. We could have also considered the data in thetable where sometimes mg (equal to the tension) is greater than the applied force, Fa, andsometimes mg is less than Fa- Noconclusioncan be reached from the data, confirming that choice D is the best answer. The best answer is choice D.

45. Choice C is the best answer. Using multiple concepts, we must consider the effects of friction from the slope and resistancefrom the air. Physical intuition tells us that a rough surface helps to keep the mass from sliding down the slope; this rulesout choices B and D. Since air resistance arises when an object moves through the air, a stationary mass will not besubjected to air resistance. This rules out choice A. The best answer is choice C.

46. Choice A is the best answer. If the acceleration is increasing, it does so because Fa is increasing. In all of the choicesexceptchoice A, Fa is increasing over time. Becauseof the word "NOT" in the question, we want the graph where Fa is notincreasing with time. This is an odd question that may seem hard to interpret, even with multiple readings. The point of thisquestion is to emphasize that you are looking for a "best" answer, not necessarily a correct one or one that makes perfectsense. The best answer is choice A.

47. Choice B is the best answer. Using limiting cases, we expect acceleration to come only from Fa, if the slope is horizontal(i.e., 0 = 0°). Since cos 0° = 1 and sin 0° = 0, this rules out choice C. Choice D drops out, because the sin© in thedenominator says the acceleration increases indefinitely as the slope becomes horizontal. Choice A drops out because thepositive sign of both terms says that the two forces (gravity component and Fa) point in the same direction. Since theyoppose one another, one must be positive and the other negative. The best answer is choice B.

48. Choice B is the best answer. Table 1 shows that when the angle is 30° and the hanging mass is 3.0 kg, no equilibrium can beestablished. Because there is an entry for 30°, the angle must be attainable, so choice A is eliminated. Other table entriesexhibit greater tension in the cable than 30°, so choice D is eliminated. Based on the 3.0 kg mass at an angle of 30° notreaching equilibrium, it is safe to assume that a 4.0-kg hanging mass could not reach equilibrium. The problem is that thegravitational force of the hanging mass exceeds the y-direction magnitude of the force on the block on the ramp. As a resultthe hanging mass would fall, causing the block on the ramp to accelerate up the ramp. The best answer is choice B.



49.

50.

51.

52.

Choice B is the best answer. This is a case where we need to apply one of the equations of circular motion. We start bytaking an inventory of the variables that we are given. Did they give us all "DAT" on this question? We are given totaldistance (72 rad), asked for acceleration (a), and are given time (6 s). The equation we need is A9 =!£at2 +cu0t, where co0= 0, so the equation reduces to:

A0 = to2

72 rad = ^a(6)2 = ^a(36) = 18a

a = 72/18rad/s2= 4rad/s2


Choice B is the best answer. First, we can eliminate choice A immediately, because a drop in the coefficientof friction willdefinitely impact the maximum safe speed for the turn. To solve this question we need to consider the two forces at work.The car is able to turn because static friction pushes the tires of the car sideways. This means that the maximum speed isassociated with the maximum static friction of the surface of the road against the tires. The speed is found by setting themaximum static friction, /*smg> equal to the centripetal force, mv2/r.

2/<smg = m /r

v2//<sg- /r

We can disregard g because it's a constant and r because the radius of the turn does not vary. This leaves us with thefollowing relationship:

HS oc v2

/'s wet =0-7 Us dry •'• vwet2 =0.7 Vdry2The square root of a number less than 1 is larger than the number itself, but less than 1.The best answer must be bigger than0.7, but less than 1. Only choice B fits within that range. The best answer is choice B.

Choice A is the best answer. We must consider the forces acting on the ball at the bottom of its circular path. There istension pointing to the center of the circle (in this case upwards), the weight downwards, and centripetal force downwards.

T - mg = ma,.

T - mg =mv

52 - 20 =.2

mv

32 =mv 2V

Fnet= 32N mg = 2 x 9.8 = 20 N16 = v2 .-. v = 4

Dividing thecentripetal force, 32 N,by the 2 kg/m, results in v2 = 16 m2/s2, so v = 4 m/s. The best answer is choice A.

Choice B is the best answer. We need to determine the appropriate equation of circular motion. We are given the totalangular distance (80 rev), the final angular velocity (12 rev/s), the time (10 s), and asked for the initial angular velocity. Weare not given the angular acceleration, a, so we need an equation that lacks a: The equation we need is A0 = (co0 + co0t,

where co0 = 0, so the equation reduces to:

A8 =(winit +wfinal)/2 xt

80x2/10rev/s =coinit+12 .

80 rev =(winit + 12)/2 x 10

160/l0 rev/s = 16 rev/s =cojnjt + 12

Winit12 16- 12 =4 rev/s

We could also have determined that the average angular velocity is 8 rev/s. Given that the final angular velocity is 12 rev/s,the initial angular velocity had to be 4 rev/s. The best answer is choice B.


the

Work and

Energy

"climb

JULBvw

m acceleration zone

"drop

Physics Chapter 3

Roller Coaster Systema

by

= Work

id— Position

Berkeley Review

Work and EnergySelected equations, facts, concepts, and shortcuts from this section

KE = &mv2

W = Fii x d


^gravitational = mgh

AKE = KEfinal - KEjnitial = Wtotai

^spring —/2kx

W = Pxt

© Important Concepts

Conservation of Mechanical Energy (PEjnjtjai + KEjnjtiai = PEfinal + KEfjnai)

Starts

motionless

Pulley wheel

Lift plate

Initial

PE = mgyj

KE = 0

• Finishes with

(& "" *<^ speed

Work-Energy Theorem (0 = W + AKE + APE)

Final

PE = 0

KE =1/2mvf2

Point a Point b Point c

PE = mgha PE = 0 PE = 0

KE = 0 KE=1/2mvb2 KE = 0

W = 0 W = 0 W = ^smgdb-

b Rollers cOOOOOOOOOOOOOOOOOO

© Strategy for Solving Energy Questions

Start by considering what you are given and how it relates to what they are asking for

Qu. How much work is required to stop a 10-kg box moving at 10 m/s across a horizontal plank?

Sol. Height doesn't change, so APE is out. They ask for work but don't give distance or force. This must besolved using kinetic energy, because they gave you mass and speed. W = KEfinai - KEjnitial

Qu. A 2-kg cart starts on a 2-m high table before it slides down a 30°-ramp with \i\ - 0.20 and then comesto rest 1.31 m from the ramp on a horizontal surface. How much work was done to stop the cart?

Sol. Speed starts and ends at 0, so AKE is out. They ask for work but don't give the frictional force. Thismust be solved using potential energy, because they gave you mass and height. W = PEfinai - PEinitial

Physics Work and Energy

Work and EnergyIn Chapter 2, we studied forces and how they impacted motion. In this chapter,we shall focus on the energy required to move against these forces and howenergy can manifest itself within physical systems.

Work

In physics, work is a form of energy. It is concerned with the force exerted on anobject as that object undergoes a displacement. For example, suppose we pull abox across the floor, as shown in Figure 3-1. As the box is being pulled, a force isbeing applied to it to move it a certain distance. Work is being done, becauseenergy is being transferred from the cause of the force to the box being moved.

\

/ F||L* Ax •_^i •

' Ti :

Figure 3-1

The specificquantity of work done on the pushed object can be calculated, if weknow the applied force and the resulting displacement. In Figure 3-1, thedisplacement is simply Ax to the right. As for the applied force, not all of theforce contributes to displacing the object; the component parallel to Ax alonemakes the object move. The force component that is perpendicular to thedisplacement may aid in lifting the object, but it will not contribute to the workdone in displacing it to the right. The contributing component F|| can also bewritten as Fcos 8, allowing the work to be written in one of two mathematicallyequivalent ways:

W = FiiAx (3.1)

W=(Fcos OXAx) (3.2)

The M.K.S. unit of work is the Newton-meter, but it is typically referred to as aJoule. You may occasionally see energy expressed in ergs, too. Don't worry, this isjust the Joule equivalent in the cm-gm-sec unit system.

To see the usefulness of equations (3.1) and (3.2), keep the following in mind: Ifthe applied force is parallel to the displacement, the force does maximal work onthe object. If the applied force is perpendicular to the displacement, the forcedoes no work on the object. If there is no displacement, there is no work doneon the object, regardless of the applied force. For an example, what work does akinetic frictional force do on a sliding object? In this case, the applied force isantiparallel to the displacement, giving 6 = 180 (that is, cos 180° = -1). The workdone by friction on the object is:

»» on object by friction —"tkincticAX (3.2a)


Work


Physics


Work and Energy Work

What does the negative sign imply? When negative work is done on an object,that object loses energy to its surroundings. Here, friction takes energy awayfrom the sliding object. As an aside, pay attention to whether the work is doneon or by the object or system of interest. Generally, if positive work is done onthe object, the object gains energy. If positive work is done by the object, theobject loses energy. If the work isnegative, thentheopposite would be true.

Example 3.1aA child grabs thestrap on hispack and pulls it for 1.5 m across theground. Thetension in the strap is 25 N and is directed at an angle of 30° above the ground.What work does the child do in pulling his pack?

A. 12.5 JoulesB. 18.8 JoulesC. 32.5 JoulesD. 37.5 Joules

SolutionThis example tests the definition of work given by equation (3.2). All we need todo is enter the appropriate values and obtain the answer. We find that the workdone by the child in pulling his pack is 32.5 J. Remember, we must report thework in Joules.

W = (25 N)(cos 30°)(1.5 m) = 32.5 J

Numeric Estimation: If you don't immediately remember the cosineof 30°, youshould at least know that it is less than 1. This rules out choice D. Also, it'sprobably at least 0.5. This rules out choice A.Tochoose between choice B, whichcorresponds to cos30°, and choice C, you need to know its approximate value.Cos30° = V3/2 =0.866, which isclose to0.8 or 0.9. Using either ofthese simplernumberswillpoint towardsa numbera bit bigger than 30.


Example 3.1bWhat force is needed to do 100Joulesof work on a box, while pushing it uphill atan angle of 60° with respect to the horizontal ground?

A. 100N

B. 87 N

C. 50 N

D. There is insufficient information to determine the work.

Solution

Work is defined as being equal to the amount of force applied parallel todirection of motion times the distance the object is displaced. In this case we aregiven the work done, the angle at which the force is applied, and then askedabout the applied force. However, we are told nothing about the distance that theobject is displaced.

W= F|| xdisplaced= FappUed * cosO x displaced

^applied = / cos 0xdisplaced

Because we don't know the distance the object is displaced, we cannot solve forthe force that was applied. This makeschoiceD the best answer.



Example 3.2aThe workdoneon a blockby friction as it slideson a roughsurface is:

A. negative and directly proportional to u^.B. negative and inversely proportional to u^.C. positive and directly proportional to u^.D. positive and inversely proportional to u^.

Solution

Multiple Concept: Let's consider the sign of the work first. We want to knowwhat kind of work is done on the block.A positive amount of work done on theblockmeans that the block is gaining energy. A negativeamount means that theblock losesenergy, as it rubs against and heats the rough surface.Thus, the workmust be negative, and the best answer must be either choiceA or B.

Now, let's consider the friction. If there is no friction, then there won't be anywork done on the block by friction.Thus, if u^ decreases, we expect that the workdecreases.This relationship is a direct proportionality (i.e., W « jjk.)


Example 3.2bThe work done on a road surface, as a car skids across it, is:

A. negativeand directly proportional to the skidding distance.B. negative and inversely proportional to the skidding distance.C. positiveand directly proportional to the skidding distance.D. positive and inversely proportional to the skidding distance.

Solution

Multiple Concept Again, we should look at the sign of the work first. We wantto know what kind of work is done on the road surface. A positive amount ofwork done on the road surface means that the car is losing energy to the road. Anegativeamount means that the car gains energy from the road, as it skids acrossand heats the surface.Thus, the work must be positive,and the best answer mustbe either choice C or D.

Now, lefs consider the distance over which the work is being done. If there is nomovement, then there won't be any work done on the road by friction. Thus, ifthe skidding distance increases, we expect that the work increases. Thisrelationship isa direct proportionality (i.e., W« displaced-)



Work

Test TipMultiple Concept

Test TipMultiple Concept


Physics Work and Energy Kinetic Energy

Kinetic EnergyWhen work is done on an object, there is a transfer of energy to that object.Energy comes in many forms. Forexample, the energy of motion is referred to askinetic energy. The energy of an object due to its position in a gravitational fieldis referred to as gravitational potential energy. The energy stored in a spring iscalled elastic energy, while the energy stored in fuel (e.g., gasoline or glucose) iscalled chemical energy. There arealsomagnetic, electrical, thermal, and nuclearforms of energy, too.

The two forms of energy that we will consider in this section are kinetic energyand gravitational potential energy. We will encounter the other forms of energyin future sections. First, let's consider kinetic energy.

Kinetic energy is the energy an object has because it is in motion. The faster anobjectmoves, the more kineticenergy it has. Suppose that some mass is acted onby a constant force and that a displacement results in a direction parallel to theforce. This is how we defined work in equation (3.1). The acceleration of thisobject can be determined from Newton's Second Law of motion, F = ma. If wesubstitute this expression into equation (3.1), we will get equation (33):

W = (ma)(Ax) (3.3)

Recall that the displacementof the objectcan be given by equation (1.15).

vtx2 = Vox2 + 2 axAx (1.15)

Equation (1.15) can be manipulated to isolate Ax.

vtx2-v0x2 = 2axAx

Ax =V**2-v°*22a

Substitution into equation (33) gives equation (3.4).

W=(ma)(Vf2~Vi2) (3.4)2a

Initially, the object had a velocity of vj. As the force displaces the mass thevelocity will increase to some final value vf.

When we multiply the two terms on the right side of equation (3.4), theacceleration term drops out. Weare left with W = V^mfrf2 - vj2). If we define theterm %mv2 tobe the kinetic energy of the object, thenwe've justshown that thework done by the force on the object is equal to the change in its kinetic energy(AKE), as shown by equation (3.5).

AKE = V^mvf2 - MmVj2 (3.5)

What are the units of kinetic energy? The units for kinetic energy must be thesame as those for work, which are Newton-meter or Joules. The dimensions ofkinetic energy are [M] x [LT1]2, which equals (ML2T"2].

The conclusion from this calculation is that the total work done on a system willequal its change in kineticenergy, as shown in equation (3.6).

AKE = KEfinai - KEinitial = Wtotal (3.6)



Let's explore this further. Suppose we have a box of sugary, preservative-ladenkids snacks that rests on a smooth, flat table, as shown in Figure 3-2. A force isexerted on thebox until it moves to some new position a distance Ax away.

Ax

Figure 3-2

According to equation (3.6), we can say that the final kinetic energy (Vimvf2-) isequal to the initial kinetic energy (Vimv,2) that we had when we started, plus thework that was done on the object (FAxcosG). This is shown in equation (3.7).Note that the cosine of 9 is 1, because the angle between the force (F) and thedisplacement of the box (Ax) is zero. Equation (3.7) might look familiar. If youmultiply this equation by 2 and divide by m, then it should remind you ofequation (1.15). Note that amount of work you need to do to slide the box varieswith the kinetic friction, because it will require a greater force to move the boxacrossa surface that exerts greater friction on the sliding box.

Vimvf2 = V&mvi2 + FAx (3.7)

Consider a different situation. Suppose we have the same smooth table as shownin Figure 3-3. In this case, we still move the box some distance Ax. However, theforce will be directed at some angle 0, rather than the horizontal direction. Thisforce can be resolved into its horizontal component (FcosG) and its verticalcomponent (Fsin0).

Ax

Figure 3-3

Assuming we pull the box with the same magnitude of force as before, thekinetic energy change is smaller than when the force is directed entirely alongthe displacement. This is because the acceleration on the box is reduced, so itdoesn't reach as high of a speed as before. A reduced speed equates to lesskinetic energy. We must now be a bit more general about how we write equation(3.7). The more general work and kinetic energy relationship is equation (3.8):

Vfcmvf2 = Vfcmvj2 + (Fcos6)(Ax) (3.8)

The cos 9 allows us to account for the orientation of the applied force F. To seethe implications of equation (3.8), consider a case where the applied force isdirected perpendicular to the displacement (6= 90 ). In this case, cos 90° = 0, andthere is no change in the kinetic energy. This is reasonable, since there is no workdone on the object when the applied force is perpendicular to the displacement.In other words, if you push down on a box, you don't move it and thus it gainsno speed. Realize that it is only the component of an applied force that lies in thedirection of the displacement that imparts a change in an object's kinetic energy.


Kinetic Energy



As another example, considerwhen there's an applied force but no displacement(i.e., Ax = 0).This occurs when one holds a barbell steadily overhead, as in Figure3-4. "Holding steadily" implies that there is no change in the kinetic energy,which equation (3.8) verifies when we set Ax = 0. This also means that aweightiifter does no work when holding a weightsteadily. This might seem a bitstrange, as weightlifters and do tend to get tired. Both expendenergy (typically,chemicalenergy),but neither do "work" as it is defined in equation (3.1).

MCATStrong

Figure 3-4

Suppose we have a blocksliding down an inclined plane, as shown in Figure 3-5.The block initially starts at rest (vi = 0). When it gets to the bottom of the inclinedplane, it will have picked up some speed (vf). Let's assume that there is nofriction as the block slides down the plane a distance Ax.

Figure 3-5

/Final

What is the work done in this situation? There are two forces that act on theblock. The force of gravity (weight = mg) acts downward and the normal force(N) acts perpendicular to the plane-block interface. We can resolve the force ofgravity into two components. One component acts downward along the axis ofthe normal vector (mgcos0), and the other componentacts perpendicular to thenormalaxisalongthe plane (mgsinO). The force perpendicularto the plane doesno work, because it is perpendicular to the displacement, Ax. The normal forcedoesnot do anyworkeither, because it is alsoperpendicular to the displacement,Ax. Theonlyforce thatdoes anyworkis the component ofgravity that is parallelto the planeoftheincline. In thiscase, the work canbe given by equation (3.9).

W = (mgsin0)(Ax) (3.9)

We can get the same answer by another method. For example, we can say thatthe work is the total force ofgravity (mg) times the distance (Ax), times the cosineof the angle between them. That angle would be (j) as shown by equation (3.10).The angles ty and 9 are complements (see Figure 3-6), meaning that their sum is90°. The cosine of <J) is the same as the sine of 0. This means that the work is thesame as mgsin0(Ax).

W = (mgcos<|>)(Ax) (3.10)



Figure 3-6

For the triangle shown in Figure 3-6, the height (h) is opposite the angle 0Trigonometry tells us that sin0 = h/Ax, which can be rearranged to demonstratethat h = sin0(Ax ). Substitution into equation (3.9) gives equation (3.11), whichsays that the work done on a body is equal to mgh.

W = mgh (3.11)

If the work done on a body is mgh, then that is the amount by which its kineticenergy will increase.What this is saying is that the final kineticenergy is equal tothe initial kineticenergy, plus the work done (or mgh).Thisexpands on equation(3.6), and can be written as KEf= KE, +mgh. Upon substitution, we get equation(3.12).

V^mvf2 = Vfcmvj2 + mgh (3.12)

Note that if we start the block from rest at the top of the incline in Figure 3-5,then the viterm inequation (3.12) is 0. This gives a relationship of^mvf2 = mgh.Multiplying by 2 and dividing by m gives equation (3.13).

vf2 =2gh (3.13)

Equation (3.13) says that the speed gained as the block slides down the inclinedepends only on how far down the incline (drop in the y-direction) the blockhasgone, and not on what angle defined the plane of the incline itself or the actualdistance the block travels. Remember, in order to get this result, we assumed thatthere was no friction. If there is friction, then the speed is less because some ofthepotential energy is converted into heatby friction, leaving lesskinetic energy.


Kinetic Energy



Example 33aA student applies force to a stalled car over a distance Ax to increase its kineticenergy. Which graph BEST represents the relationship between the kineticenergy and the pushing distance?

A.

KE

Ax

Solution

First, will the kinetic energy change at all, if the student pushes over a longerdistance? Yes, so cross out choice A. The remaining choices all say that thekinetic energy increases as the pushing distance increases. But how should itincrease? We need to relate energy to distance. Work does this for us:

W = FAx

Work is energy and, here,it is being converted into kineticenergy.Thus,

KE = W = FAx

The relationship between KE and Ax is a linear one, so the graph should be thatof an ascending line and not a curve.


Example 33bReferring to the car in Example 3.3a, which graph BEST represents therelationship between thecar's speedand the pushing distance?

A.

Ax

Solution

Asthecaris pushed farther and farther, it builds up kinetic energy, as wesawinthe previous example. As the kinetic energy increases, its v2 increases. Therelationship is v2 a Ax, which is not linear, so choices A and Bare eliminated.

This is where we need to be careful, because mistakes are often made. It is best toplug in numbers and see how the two values relate to one another. If v isdoubled, thenAx mustincrease by a factor of4.Thismeansthat Ax changes morethan v changes, so the graph should bend towards the Ax-axis. Remember,"graphs curve towards the axis of greatest change." Although this point isstated numerous times in the general chemistry book, it is worth repeatinghere.




Gravitational Potential EnergyAs the blockslidesdown the inclined plane, it is gainingkinetic energy.Where isthat energy coming from? The kind of energy that is lost to kinetic energy iscalled the potential energy against gravity, or gravitational potential energy.

Some textbooks use the symbol U to represent the gravitational potential energy.In these notes, we use the abbreviation PE. As the block travels down the incline,some of its gravitational potential energy is lost. Alternatively, this means that asthe blockis moved up the plane, it will gain gravitationalpotential energy.

If we were to lift an objectwith a constant force some distance above the ground,then we could say that the force applied to that objectwould be F = mg. As theobject is moved, it will undergo a displacement from yi to yf. We know that thework done by a constant force on an object is the force times the component ofdisplacement parallel to the direction of the applied force. Making theappropriate substitutions into equation (3.2)gives equation (3.14).

W = FCosOAy = F(yf- yj)CosO° = mg(yf- yj) (3.14)

All the work that we have done to raise the object has been used to increase thatobject's gravitationalpotential energy, PE. This allows us to write equation (3.15).

APE= mg(yf-yj) (3.15)

where yf- yj is equal to Ah, the change in the vertical height.

Equation (3.15) is important, and we must understand how to apply it. Thegravitational potential energy of an object depends only on the verticalseparation of that object from the ground. It does not depend on any horizontalchange in position.

If an object were to have a change in horizontal direction as it was movingupward, then the change in gravitational potential energy would still be equal tomgAh. For example, if we were to pull an object up an inclined plane, then wewould need to consider the angle of the plane and the distance the object wasmoved to determine the height change. The work necessary to pull the objectupsuchan inclined plane canbe found using equation (3.9), W = (mgsin 0)(Ax). Thechange in the object's gravitational potential energy in that situation would begiven by equation (3.16), a spin-off of equation (3.9).

APE = mg(Axsin0) (3.16)

The Axsin0 term comes from the change in the height as the object moves up theinclined plane (see Figure 3-7).

Final

Figure 3-7


Gravitational Potential Energy


PhysiCS Work and Energy Gravitational Potential Energy

Beaware that the gravitational potential energy does not depend on the path anobject takes as it moves upward. It only depends on the difference betweeninitial and final heights of the object.

Example 3.4aA longshoreman unloads two identicalboxes, sliding one down a rough inclinedramp and dropping the other one. Both end up 2 meters below their staringpoint How does the change in gravitational potential energy compare for thetwo boxes?

A. The dropped box has a bigger change in gravitational potential energy.B. The slid box has a biggerchangein gravitational potential energy.C. Bothboxeshave the samechangein gravitationalpotentialenergy.D. It depends upon the slope of the inclined ramp.

Solution

The change in gravitational potential energy depends only upon the change inheight, and not the distance the object travels or the angle of inclination. Bothboxes start at the same initial heightand finish at the same final height, so theyeachchange by the same height, 2 meters. Because they each undergo the samechange in height and have the same mass, they experience the same change ingravitational potential energy.


Example 3.4bTwo balls are lifted. If one ball experiences a greater change in potential energy,thenwhich ofthefollowing statements could possibly explain thedifference?

I. The balls have different masses.II. Theballsare liftedto different heights.

III. The balls reach their final points by different pathways.

A. I onlyB. II onlyC. I and II onlyD. I, H, and in

Solution

The change ingravitational potential energy for each ball can be calculated usingequation (3.15), APE = mgh. If the two balls experience a different change ingravitational potential energy, it must be associated with either a difference intheir masses or a difference in their change in heights. The change ingravitational potential energy is independent of the pathway between the initialand final heights. Statements I and II are valid, but Statement HIis invalid.




Conservation of EnergyThe energy in the universe is a conserved quantity. It is believed that the amountof energypresent in the universe today is the same as it was billionsof years ago.We can restate this as follows:

Energy cannot be created nor can it be destroyed, only changed intodifferent forms.

Hie total energy in theuniverse is constant.

So how does energy conservation apply to an object or system? Mathematically,it applies as:

^initial ofsystem + ** doneon system —̂ final ofsystem (3.17)

This says that the energy of a system is constant, unless some work is done onthe system. In that case, the final energy of the system will be greater if the workis positive and less if the work is negative (as might occur with friction). Atypical form of the energy E in such problems is:

E = KE + PE (3.18)

This energy is the sum of kinetic and potential energies and is often called themechanical energy of the system, as it usually appears in mechanics problems.Note that a system or object possessing mechanical energy can do work on othersystems or objects.

Energy conservation is one of the most useful ideas in the physical sciences. Withit, you can solve many problems that would be difficult if approached usingforces or kinematics. Remember, forces and kinematics involve vectors. Energyis a scalar quantity, as are most equations relating to it. Scalarsare your friends.

Energy Conservation: Outlined below is a suggested general approach to solvingproblems using energy conservation. It should always be considered beforeforces or kinematics when solving a quantitative problem.

1. Identify the problem as a candidate for energy conservation when you are toldor asked about some initial/final situation for the system in question.

2. Draw out, on scratch paper or in your head, the initial and final situations forthe system.

3. Find the initial energy Ein;tiai of system ofthe system by asking the following:

* Do things move? If so, include kinetic energy.

* Do things change configuration or position? If so, include potentialenergy.

* Are there forces external to the system and acting on it? If so, includework. For example, friction is an external force.

* Is there heat? If so, include work or Q = mcAT.

* Are electric fields or voltages involved? If so, include work.

* Are chemical reactions involved? If so, include heat or electromagneticenergy.

* Basically, consider everything that could influence the energy.

4. Repeat step 3 in finding the final energy Efjnai 0f system-


Conservation of Energy

Test TipConservation Questions


Physics

Test TipPairing the List

Work and Energy Conservation of Energy

5. Repeatstep 3 in finding the work done on the system.

6. Plug everything into equation (3.17) and solve for your unknown.

This approach can alsobe used to see qualitative relationships between energiesor other physical quantities in a problem.

Example 3.5aA spring is used to launch a toy car on a smooth horizontal surface. Which of thefollowing quantities must be known in order to calculate the car's final speed?

I. The spring constantII. The amount of spring compression

III. The altitude of the surface

A. I and II onlyB. II and III onlyC. I and III onlyD. I, H, and III

Solution

Paring the List: Questions regarding lists can occasionally undermine yourconfidence. One way to tackle them is to pare the list one item at a time. Thisapproachis thesame as thatused for the multiple concept type problems. Pick theitem on the list that is easiest for you to deal with. For example, we know thatonly changes in altitudewillcontribute to changing the speed of the car.Since thecar moves on a horizontal surface, Statement III must not matter. Crossing outanswerchoices that include Statement IIIleaves us with onlychoice A as thebestanswer. We are done with not too much hassle. We didn't even need to thinkaboutStatements I and II. Just to complete the thought, the final speed of thecardepends on its kinetic energy, which in turn depends on the initial potentialenergy of the spring system. The potential energy of aspring is: Ugpnng =V^kAx2.


Example 3.5bThe toy car, after attaining a maximum speed, skids to a stopona rough sectionof the surface. Which of the following quantities must be measured in order tocalculate the length of this skid?

I. The mass of the carII. The coefficient of kinetic friction between the road and the wheels

III. The initial speed of the car

A. I and II onlyB. I and III onlyC. Hand III onlyD. I, II, and III

Solution

The workdone to stop thecarmust equal the loss in kinetic energy. Because thecar finishes at rest, it loses all ofits kinetic energyby the end of theskid.

^rnvinitiai2 = j*kmg x Ad

Mass cancels out, so Statement I is unnecessary. Only the initial velocity and u^are needed to determine the length of the skid. The best answer is choice C.



Example 3.6aA spring of stiffness k is compressed a distance x, and a toy car is placed againstit If the car has a mass m, which equation represents the final speed of the carafter the spring is released and accelerates the car?

A. v •iB. v =

kxim

m

kx2

•V

mx"

D. vmx"

SolutionThis is aformula identification question, and it can be solved formally using energyconservation, or quickly using limiting cases. The initial energy is the potentialenergy in the spring, the final energy is kinetic for the car, and no other work isdone on the car/spring system. Thus, the energy conservation equation becomes:

i^kAx2 + 0 = iAmvfinal2

Solving algebraically forv gives choice A.

If you want to consider limiting cases and avoid all of the above algebra, askyourself how each relevant variable will affect the car's final speed. If the springwere initially compressed more, the final speed would increase. Thus, x shouldbe in the numerator, and choices B and D should be crossed out. Between A andC, the positions ofk and m are switched. Consider the one of them that is easierfor you to understand. For instance, if you know that a less massive car wouldget more speed from a released spring, then you know that the m should go inthe denominator (i.e., there must be an inverse relationship between v and m).This is choice A. Alternatively, if you know the equation for a compressedspring's potential energy, youknow that thek andxgotogether. This also pointsto choice A. The best answer is choice A.

Example 3.6bA match head is rubbed a distance d over a rough surface in order to increase itstemperature to its flash point. If the match head material has a mass m and aspecific heat c, which equation represents the average force needed to raise thematch head's temperature by AT?

A.

B.

C.

D.

p _ mcATd

F = mcATd

F=-d_mcAT

F-mATdc

SolutionTo solve this question, you must equate the workdone on the match to the heatgenerated. Thework done is the product of the force and distance traveled.

F x Ad = mcAT _mcAT /AdOnly choice A gives units of newtons. Thebestanswer is choice A.

Copyright ©by The BerkeleyReview 143

Conservation of Energy


Physics Work and Energy Power

Power

Power is the rate of transferring energy. It is also the rate of doing work. Ratesinvolve time. Therefore, power is simply how much work can be done per unittime. This is shownin equation(3.19). Power is symbolizedwith the letter P andhas the units of Watts (W) where 1 Watt is equal to 1 Joule/second.

(3.19)

In a few countries, power is alsoexpressed in terms of horsepower. Wecan easilyconvert to Watts, if we know that 1 horsepower (hp) equals 746watts.

Example 3.7aIf an electrical generator plant increases its daily amount of output energy by100%, the plant'saverage outputpowerincreases by:

A. 25%B. 50%C 100%D. 200%

Solution

How does energy relate to power? They are positively and linearly related. A100% increase in the energy output means the energy output doubles. Thus, ifenergyoutput in oneday doubles, so does the power output.

The correct choice is C.

Example 3.7bIfa motor that generates 25 kj in 4 minutes and 10 seconds increases its poweroutputby50%, thenwhatis thenewpoweroutputof themotor?

A. 25 W

B. 50 W

C. 100W

D. 150W

Solution

The motor initially generates 25 kjin250 seconds, which equates to 25,000 J/250seconds. The math works out cleanly to yield 100 W. However, the question asksfor the power output after a 50% increase. If the motor initially produces 100 W,it will increase to 150W after a 50% increase (150% x 100W = 150W).


Copyright©by TheBerkeley Review 144 The Berkeley Review

25 Work and Energy Review Questions

I. Television on a Ramp

II. Roller Coaster Friction

III. Motion of Cars


(1-7)

(8 -14)

(15 - 21)

(22 - 25)


Passage 1 (Questions 1 - 7)

Movers from a moving company must move a heavy bigscreen television (m = 100 kg) from the ground to a platformthat is 1.0 m above the ground. They decide to use a plank2.0 m long as a ramp to assist in the move. The schematic inFigure 1 shows the ground, platform, and ramp and the finalposition of the big screen television. The goal of the moversis to move the television from point D to point A.

Figure 1

The television is pushed up the frictionless ramp at aconstant speed. The movers provide a force that is alignedwith the center of mass, so that the television does not rotate

at any point during its path. All work energy is provided bythe movers, during all points along the move from the groundto the platform. There is friction from point D to point C.

1. How much work is done by the movers on thetelevision as it slides up the ramp, if the total force theyexert is 500 N?

A. 250 J

B. 500 J

C. 1000 J

D. 2000 J

2. By how much would the potential energy of thetelevision change if a 3 m ramp were used to get it up tothe platform instead of a 2 m ramp?

A. 500 J

B. 1000 J

C. 2000 J

D. 3000 J

3. When the television reaches the platform, it keepssliding until coming to a second ramp. As the carelessmovers relax and drink a soda, the television slidesdown the ramp. If the ramp is frictionless with heightof h, and the length of the ramp is L, what is the speedof the television at the base of the ramp?

A. h2/2gB. Y^L

C. V2g¥D. V2g(L-h)

Copyright© by The Berkeley Review®

4.

5.

6.

7.

146

Suppose the television goes down the ramp in Figure 1,passing point C with a speed of V20 m/s and coming torest at point D. If the piano has a mass of 100 kg, howmuch work did friction do as the television slid from C

to D in Figure 1?

A. -225 J

B. -500 J

C. -1000 J

D. 1000 J

As the television is pushed up the ramp from C to B inFigure 1, the normal force is:

A. less than its weight.

B. more than its weight.

C. equal to its weight.

D. doing work on it.

Which of the following graphs BEST represents thespeed versus time as the television slides from B to C toD in Figure 1, and then comes to rest?

A. B.

—

Time

C.

—

Time

ft

-r.

Time

Time

What power must be generated by the movers as thetelevision slides up the ramp with an average speed of2.0 m/s? Assume that the movers provide a total forceof 500N.

A. 500W

B. 1000W

C. 2000 W

D. 4000 W



An overpaid engineer designs a new ride for the localcarnival. In the pre-launch position a cart of mass M is placedon the tracks at height a above ground, where it compressesa spring of force constant k by a distance X. After the cart islaunched, it completes a loop of radius R, goes down the hillfrom Q to Y, and is brought to rest from Y to Z, as in Figure1. The tracks are frictionless, except for the stretch from Y toZ. Point Z is defined as the zero height reference.

i ^ acceleration zone

Figure 1.

For all questions, ignore air resistance, and take theacceleration due togravity tobe g = 10m/s2.

8.

9.

10.

What is the total energy of the system (cart and spring)in the pre-launch position?

A. Mga

B. i/£kX2

C. Mga + VikX2D. Mg(a + 6)

If the cart is barely able to complete the loop, what is itsspeed at point P, the top of the loop?

VRgV2pgV2(a + P)gV2(J

A.

B.

C.

D. a)g

The cart is brought to rest because of the frictional forceacting from Y to Z. The work done by friction to stopthe cart under normal conditions is W. Under rainyconditions, how much work is done by friction to bringthe cart to rest, assuming the cart reaches point Y withits normal speed and the frictional force is reduced byhalf? (Assume the tracks are long enough to bring thecart to a halt.)

A. W/2B. W

C. V2WD. 2W

Copyright ©byTheBerkeley Review®

11.

12.

13.

14.

147

Due to an error, the cart reaches point Q with such alarge speed that it leaves the tracks horizontally andlands at point Z. How long was the cart airborne?

i. YkX'M

2 (a + P)gB.

C2(6-a)

g2a

;-Vd. i/za

If the mass of the cart is 100kg and it reaches point Ywith a speed of lOm/s, how much work must friction doto stop the cart in a distance of 20 m?

A. -200 J

B. -500 J

C. -2000 J

D. -5000 J

Which of the following pictures best represents thedirection of the net force F acting on the cart when it ishalf way up the loop?

A. _ B.

Suppose the spring has a force constant k = 500 N/m,and it is compressed 2.5 m in the pre-launch position.When the cart is launched, about how much work doesthe spring do on the cart?

A. 200 J

B. 500 J

C. 750 J

D. 1600J



When analyzing the motion of an object, it is possible todo so using kinematics equations or energy equations. Bothtypesof equationsgive informationabout positionand speed.Kinematics equations give information about the motion ofan object from moment to moment and account for direction,while energy equations typically detail the initial and finalpoints of the motion and tell you about speed of an object butnot the direction.

Consider the following three scenarios:

I. A car travels at a constant speed on 10 m/s for 50 mand then the driver applies the brakes, stopping thecar completely after 10 m of braking distance.

II. A car travels at a constant speed on 15 m/s for 50 mand then the driver applies the brakes, stopping thecar completely after 22.5 m of braking distance.

15.

16.

17.

III. A car travels at a constant speed on 20 m/s for 50 mand then the driver applies the brakes, stopping thecar completely after 20 m of braking distance.

If the initial velocity is 10 m/s, the work required tostop this car in only 5 meters would be:

A. greater than the work in scenario 1, because thefriction force would have to increase.

B. less than the work in scenario 1, because thefriction force would have to increase.

C. equal to the work in scenario 1, because the car islosing the same amount of kinetic energy.

D. equal to the work in scenario 1, because the frictionforce would necessarily remain the same.

If the initial speed of a car doubles while the brakingremains constant, the distance required to stop this car:

A. remains the same.

B. increases bya factor of V2.C. increases by a factor of 2.D. increases by a factor of 4.

Chemical potential energy stored in gasoline isconverted into kinetic energy as the car increases itsspeed from 0 m/s to 10 m/s. Then car accelerates from10 m/s to 20 m/s. The energy required to go from 10m/s to 20 m/s, compared to the energy required to gofrom 0 m/s to 10 m/s, is:

A. half as much.

B. the same.

C. two times as much.

D. three times as much.

Copyright©by TheBerkeley Review® 148

18. For the car described in the first scenario of the

passage, what is its acceleration as it comes to a stop?

A. 1 m/s2 in the direction of travel.

B. 5 m/s2 in the direction opposite the direction oftravel.

C. 10 m/s2 in the direction of travel.

D. 20 m/s2 in the direction opposite the direction oftravel.

19. What work is being done on a car moving with aconstant speed along a straight, level road?

I. No work of any kind is being done.

II. Work is being done on the car.

III. Work is being done by the car.

A. I only

B. II only

C. Ill only

D. II and HI only

20. If the mass of a car doubles, but its speed remains thesame, the distance required to stop the car:

A. remains the same.

B. increases by a factor ofV2.C. increases by a factor of 2.D. increases by a factor of 4.

21. Two cars traveling on a level road have the samekinetic energy. Car 1 has twice the mass of Car 2. Howdo the speeds of the two cars, vi and V2, and the workrequired to stop each car, Wi and W2compare?

A. v1 is twice V2; Wi is twice W2.

B. vi is half V2; Wj is twice W2.

C. v1is^2_; Wi is thesame asW2.V2

D. v1is^-; Wj is thesame asW2.4


Questions 22 through 25areNOT based on a descriptivepassage.

22.

23.

An Olympic springboard diver first jumps onto theedge of the spring-loaded diving board before jumpingfrom the board into the air. If the diver jumps up fromtheboard as it ascends, then she can attain roughly:

A. four times the increase in potential energy than shewould attain jumpingoff a non-deforming board.

B. twice the increase in potential energy than shewould attainjumping off a non-deforming board.

C. half the increase in potential energy than she wouldattain jumping off a non-deforming board.

D. the same change in potential energy than she wouldattain jumping off a non-deforming board.

The efficiency of an engine's conversion of fuel intouseful work would be maximized by all of thefollowing changes EXCEPT:

A. lowering the exhaust temperature.

B. minimizing the number of heat transfer points.

C. reducing the number of moving parts.

D. burning the fuel faster.

24. Why is the following lift fitted with a counterweight?

Support beam

Lift plate

A. To reduce the work needed to return the lift plate tothe base after the cargo has been removed

B. To reduce the work needed to lift the cargo

C. To lower the system's center of mass

D. To establish a mechanical advantage during boththe ascent and descent of the lift plate


25. A car is traveling around a curved portion of a flathighway at a constant speed v. The curve has a radiusR. What is the minimum coefficient of static frictionbetween the tires and the road necessary for the car tomakethe curve without skidding?

A. \i=fRg

B.

C.

D.

mg

H=v'Rg

R2gp.=

1. C 2. B 3. C 4. C 5. A

6. A 7. B 8 C 9. A 10. B

11. D 12. D 13. D 14. D 15. C

16. D 17. D 18. B 19. D 20. A

21. C 22. B 23. D 24. B 25. C

YOU ARE DONE.

Answers to 25-Question Work and Energy Review

Passage I (Questions 1 - 7) Television on a Ramp

1. Choice C is the bestanswer. The work done on the television as it slides up the ramp takes into account the total force of500 N and distance that it slides up the ramp. Work is defined as W= F||d. Plugging the numbers into the work formula

6.

gives:

W = (500N)x(2.0m)= 1000 J

You could have also solved this question by calculating the change in the gravitation potential energy from where thetelevision starts and where it ends. The television starts and finishes at rest, so there is no change in kinetic energy toconsider. Any work acting on the system is equal in magnitude to the change in potential energy. The change in potentialenergy is equal to mgAh, which in this case is (100 kg)(10 m/s2)(l m) = 1000 J. As expected, this matches the result whenUsing the work equation. There areoften multiple ways to solve energy calculation questions. The bestanswer is choice C.

Choice B is the best answer. The change in potential energy is independent of the pathway, so whether it's a 2 m ramp or a3 m ramp, the change in height for the object is still 1 m. This means that the change in potential energy for the system isfound by APE = mgAh.

APE = mgAh =(100 kg)( 10.0 m/s2)( 1.0 m) = 1000 JOnly a change in mass or a change in elevation can change the potential energy of an object on the surface of the Earth. Thebest answer is choice B.

Choice C is the best answer. The fastest way to determine the speed of the television at the base of the ramp is to use theconservation of energy equation.

Etop of ramp =Ebottom of ramp •'• mght0p of ramp =-mv2 => v= V^gh, which is the same result as free fall!

It is important that you use the change in height and not the length of the ramp. The best answer is choice C.

Choice C is the best answer. As the television slides from point C to point D, the only force that does work on it is friction.Since we do not know the coefficient of kinetic friction, we do not know the magnitude of frictional force, so we cannot usethe relationship: W = F||d. Instead, we can use the work kinetic energy theorem, where the change in gravitational potentialenergy is zero, because the height does not change. This means that work done by friction is equal to the change in kineticenergy for the system.

W=AKE =KEf -KE[ =- mv? -1 mv? =-1 mv? m-1(100 kg)(V20 m/s)2 =-1000 J2 r 2 ' 2 ' 2

Kinetic friction always does negative work on a sliding object. The best answer is choice C.

Choice A is the best answer. As the television is pushed up the ramp from point C to point B, the normal force is less thanits weight. To see this, a free body diagram of the piano would be helpful.

appliedforce

If we now apply Newton's second law perpendicular to the plane, we get: IF = N - mgcosO = 0, which implies N = mgcosO.This tells us right away that the normal force is less than the weight of the television, because cos6 is less than 1. We werenot given the angle 9, but it is not necessary to solve this problem. The best answer is choice A.

Choice A is the best answer. The television will speed up as it goes down the ramp (from point B to point C), and then slowdown once as it skids across the rough surface (from point C to point D). This eliminates choices C and D, and leaves onlychoices A and B. As the television slides down the plane, it has a constant acceleration, which means that the speed-versus-time graph must be a straight line during this part of its motion. This eliminates choice B. The best answer is choice A.


7. Choice Bis the best answer. The power that must be generated by the Acme people as the piano slides up the ramp with aspeed of2.0 m/s, assuming they provide a total force of500 N, isgiven by:


p=w =F||V _(500N) x (2 0^ =1000 wt

Passage II (Questions 8 - 14) Roller Coaster Friction

8. Choice C is the best answer. When the cart is in the pre-launch position, its total energy is the sum of its gravitationalpotential energy and the spring potential energy. In this case, the cart starts ata height a above the ground (where itfinishes)and therefore has an initial gravitational potential energy = Mgct. The cart also compresses the spring by a given distance Xand therefore has a spring potential energy = ftkX2. The total potential energy of the cart system at the start is thereforeequal to:

Mga+ V$kX2


Choice A is the best answer. To calculate the speed, we need to consider thesum of theforces acting on the cart when it isat the top of the loop. The forces in the x-direction do not affect whether or not the cart will remainon the tracks,so we shallconsider only the forces acting in the y-direction. The sum of the forces in they-direction when the cart is at the top of theloop is:

22Fy = N + mg= mX—,

R

where at the top of the loop both the normal force and the force of gravity point down and the centripetal force must offsetthe net downward force in order to keep the cart on the tracks. If the cart barely makes it around the loop, then the normalforce is zero. The equation for sum of the forces reduces to:

2Fy = mg= m¥—, where solving for v yields: v = VRgR

The most difficult part of this question is to realize that the normal force is zero, if the cart barely makes it around the loop.The best answer is choice A.

10. Choice B is the best answer. We should use the work-kinetic energy theorem, which says Wj = AKE. Since the cart reachespoint Y with its normal speed, it has its normal kinetic energy. Since it arrives with its normal kinetic energy, it has the sameAKE. If it has the same AKE as a cart in normal conditions, the work required to stop the cart is the same. It is spread outover a greater distance. The best answer is choice B.

11. Choice D is the best answer. The cart is airborne from point Q to point Z, because it has such a large speed that it leaves thetracks horizontally and lands at point Z. If the cart is launched horizontally, its initial velocity in the y-direction is zero. Tofind the time of flight, we need to use one of the kinematics equations:

Ay =v0yt+^gt2,where the initial velocity in the y-direction iszero, so the equation reduces to: Ay = + !/£gt2.

Because Ay = a, and we can set the sign convention of positive as down, we solve for t as t = /%/ — ,V g


which is choice D.

12. Choice D is the best answer. The work-kinetic energy theorem is best employed to solve this question. The elevation doesnot change once the cart passes point Y, so the work necessary to stop the cart must equal the kinetic energy of the cart atpoint Y. The only force doing work to stop the cart in this region is the force of friction. The work-kinetic energy theoremgives:

Wstop =AKE /. Wstop =i(100)(0)2 -i(100)(10)2 =-5000 J2 2

Distance is not needed to calculate the work. Had they given us the force due to friction (or the components necessary tocalculate it, then we could have calculated the work using Wst0p = Fxd. Thatinformation was notgiven, so the only methodpossible to determine the work done was to look at the kinetic energy that was lost. The best answer is choice D.


13. Choice D is the best answer. When the cart is on the loop, the only forces acting on it are the force of gravity and thenormal force. The normal force points to the left, and the force ofgravity points straight down, as in the following picture:

N-*

resultant

As the force diagram shows, the resultant vector points down and to the left, and the picture that is closest to this is choice D.Pick choice D for the sensation of correcthood. The best answer is choice D.

14. Choice D is the best answer. The work done by the spring when it launches the cart is equal to the change in its potentialenergy when it goes from compressed to equilibrium. This relationship can be expressed as:

Ws =-APES =-(^k(Xf)2 - i/ik(Xi)2)

Ws =-(-L(500)(0)2 - i(500)(2.5)2) =(250)(2.5)2 =(250)(6.25), avalue greater than 750, eliminating choices A, B, and C.2 2

Choice D must be correct. The actual value is found to be 1562.5 J, closest to 1600, making choice D the best answer. Youdo not have to solve all math questions exactly. It is often better to solve for a window in which only one of the answerchoices fits. The best answer is choice D.

Passage HI (Questions 15-21) Motion of Cars

15. Choice C is the best answer. Work is defined as the force F required to move an object a distance d:

W = F||d,

where F|| is the component of F along the direction of motion. However, work can also be written as the change in kineticenergy:

W=AKE =KEf - KEj, where: KE = ^mv2

Since the initial and final velocities are the same, whether the car stops in 10 meters or in 5 meters, the work required to stopthe car must be the same in both cases. Incidentally, the force required to stop the car in only 5 meters must be greater thanthe force required to stop the car in 10 meters. The force that causes the car to stop is the force of friction, and this force isrelatively constant while the car is slowing down. The best answer is choice C.

16. Choice D is the best answer. The force required to stop the car is constant, and this force is the force of friction. By the lawof the conservation of energy, we can write:

W= AKE .-. -fs = KEf-KEi .*. - p.mgs = - i^mvj2

where f is the friction force, s is the distance traveled by the car, m is the mass of the car, and p. is the coefficient of friction.From this equation, we get:

v2 =2 p.gs

That is, the distance traveled is proportional to the square of the velocity. Thus, if the initial speed doubles, then the distancetraveled by the car before it stops increases by a factor of 4. The best answer is choice D.

17. Choice D is the best answer. Going from 0 m/s to 10 m/s requires an amount of work equal to:

W= AKE=^mvf2- i/imvj2

.-. W0-»io = ^m(10)2 - I4m(0)2 = ^m(lO)2 = '/4m(100)

In going from 10 m/s to 20 m/s, we can write the work required as:

Wio-*20 = !4m(20)2 - !4m(10)2 = '̂ m(400) - J4m(100) = i^m^OO)

Going from 10 m/s to 20 m/s requires three times as much work (and thus three times as much energy) as going from 0 m/sto 10 m/s. If you noticed that the speed doubled, you may have simplified the problem down to determining the ratio ofI(2v)2 - v2]: v2, which becomes (4v2 - v2): v2 = 3v2 : v2 = 3 : 1. The best answer is choice D.

Copyright © by The Berkeley Review' 152 MINI-TEST EXPLANATIONS

18.

19.

20.

Choice B is the best answer. We want to find a, and we are given the stopping distance, Ax, and the initial and finalvelocities, vj= 10m/sandvf= 0. These relate through:

vf2 = vj2 + 2aAx

which simplifies to:

-100 _ c

(2)(10)"The negative sign implies that the acceleration is in the direction opposite to that of the initial velocity (i.e., it is slowingdown). If you were drawing a blank as to which equation to use, you can at least narrow your choices to Band D, if yourealize that the acceleration and velocity point in opposite directions when an object slows down. Notice that there are nounits in this calculation. You will not be graded for including units. Don't waste time with units if you know the equation,andif youknow the numbers are all in thesameunitsystem (e.g., M.K.S.). The best answer is choiceB.

Choice D is the best answer. Since the car is traveling with a constant speed, no net work is being done on the car.However, some kind of work is being done by the car-this is why it moves~and some kind of work is being done on the carby friction. These two kinds of work must have the same magnitude, but a different sign. If one work were larger than theother, then the kineticenergyof the car would be changing. The best answer is choice D.

Choice A is the bestanswer. From the previous equation (v2 = 2 p.gs), we see that the mass of the car cancels out of theequation relating theworkrequired to stop thecar to thechange in kinetic energy of thecar.Thus, if the initial speed remainsthe same, then the distance required to stop the car also remains the same. The best answer is choice A.

a =

2Ax

21. Choice C is the best answer. Sincethe twocarshave thesame initial kinetic energy, the work required to stopboth cars is:

W = AKE = KEf-KEi = -KEi

To compare the speeds of the two cars, note that mi = 2m2- Make a ratio of the two kinetic energies:

KEj 12mv12L = 2

KE2


Questions 22 - 25

:. v2 =2v2. Solving for vi: vi =^-

Not Based on a Descriptive Passage

22. Choice B is the best answer Jumping off a flat, non-deforming platform would result in the diver achieving some arbitraryheight above the board based strictly on their jumping ability. The spring board allows the diver to initially jump to thatheight, travel back down to land on the edge of the board, which will proceed to compress some distance down (like aspring). The board will then accelerate the diver upwards so that when she returns to the point where the board is flat again(at its equilibrium position), she has an upward velocity. If she jumps at that precise moment, then she will have both thekinetic energy associated with the conversion of her spring potential energy plus the energy associated with her second jump.In an ideal situation, she can perfectly add the potential energy gain of her first jump to the potential energy gain of hersecond jump. She ideally should be able to gain twice the potential energy change she would attain if she were to jump off anon-deforming board. This question may seem odd in terms of what they are asking, but hopefully you can eliminate choicesC and D, as it would make no sense to do something that would reduce the height of the diver above the board. There's noreason why the energy would be four times as great, so choice A is not feasible. The best-case scenario would be for thedouble jump to result in double the gain in potential energy of a single jump.

If visualizing numbers helps, then let's say her first jump gives her 100 J of gravitational potential energy at her highestpoint. She then drops back onto the board converting her potential energy into kinetic energy and then compressing the boardwith 100 J of spring potential energy. The board then accelerates her back up transferring the spring potential energy backinto 100 J of kinetic energy for the diver. If she jumps a second time at just the right moment (when the board is flat and hasno potential energy), then she can add another 100 J of kinetic energy to the 100 J she has from her first jump, resulting in200 J of total energy. At her highest point following her second jump, she ideally has 200 J of potential energy. Given that asingle jump would result in 100 J of potential energy, the perfect situation would give her twice as much potential energy.Because there is loss of energy at different transfer points, it won't work out to be exactly double, but roughly double is afeasible goal. Given that her mass is constant and gravity is constant, her change in potential energy must be the result of herreaching a greater height This is why springboard divers can do more tricks than platform divers, because they descend froma great height. The best answer is choice B.


23 Choice Dis the best answer. To maximize the efficiency of conversion from raw fuel into useful work, the loss of energyin transfer steps should be minimized. Lowering the exhaust temperature results in less heat being released (lost withoutbeing converted to work). If less heat is lost (wasted without being converted into work), then it serves to reason that moreheat was converted to work, so choice Ais eliminated. Minimizing the number ofpoints atwhich heat is transferred helps toincrease the efficiency, because less heat is wasted in the heating of additional engine parts. Choice B is eliminated.Reducing the number of moving parts reduces friction within the engine (responsible for converting mechanical work intoheat), so less of the energy is lost as heat This increases the efficiency, so choice Cis eliminated. When the fuel is burnedfaster, there is a surge of heat, so the environment heats up to a greater extent by the rapid release of heat This results inenergy being wasted asescaping heat. Thebest answer ischoice D.

24. Choice B is the best answer. As cargo is lifted, the counterweight lowers, which helps to reduce the work needed to lift theload. However, once the cargo is pushed onto the platform, the counterweight must be lifted to get the lift plate to go backdown, so work is required. The counterweight reduces the amount ofwork necessary to lift cargo, but that reduction in workduring the cargo's ascent must be applied later. Ultimately, the same amount of work is done over acomplete cycle, but itisspread out over the lift and drop of the plate, not applied just during the lift Choice A is the exact opposite of what isobserved with the system, so it isout Lowering the system's center ofmass would offer no advantage, not to mention that ahanging counterweight raises the center of mass. Choice C is eliminated. There is no mechanical advantage for the totalprocess, because the reduction in force during the lift is canceled out by the increase in force during the descent ofthe liftplate. This eliminates choice D. The reason for the counterweight is to make it easier to lift cargo. The best answer ischoice B.

25. ChoiceC is the best answer. Whenthe car is traveling on thecurve, it is in uniform circularmotion with the force of staticfriction pointing towards thecenterof thecircle. Applying Newton's second law gives:

2Fc =jiN =ni^-, where p.mg =n^-, and p. =^?-R R Rg

The coefficient of friction requireddepends on the speedof the car, the radiusof the curve, and g. Noticethat the massof thecar cancelsout of the problem. The best answer is choice C.


52-Question Work and Energy Practice Exam

I. Waterfall Generator

II. Gravitational Potential Energy

III. Thrill Ride Energetics


IV. Energy Conversion

V. Rainfall Generator


VI. Box Relocation Program

VII. Rollercoaster Energetics


Work and Energy Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-5)

(6 -10)

(11 -16)

(17 - 20)

(21 - 26)

(27 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)


Hydroelectricity is the production of electricity byhydroelectric generators, machines that convert mechanicalpower from a hydraulic turbine or water wheel into electricpower. One way to accomplish this is by placing a hydraulicgenerator at the bottom of a dam. Water flows from thereservoir and falls over the dam. At the bottom of the dam isa water wheel. When the falling water strikes the wheel, thewheel will rotate, causing a conducting coil in the generatorto rotate through a magnetic field. The rotation of theconducting coil through the magnetic field produces aninduced voltage given by:

e = e0sincot,

Equation 1

where eQ = BAw, B = magnetic field, A = area of coil, co =2ji/, / = frequency of rotation. This voltage provides theenergy that allows a current to flow. B, A, and / determinehow much current is produced.

Let the heightof the dam be 150 meters and the widthofthe dam be 500 meters. The water is moving at a speed of 5m/s and has a depth of 2 meters as it goes over the dam. Takethe density ofwater to be 1000 kg/m3.

1.

2.

If the water strikes the base of the dam with a

horizontal speed of only 47.2 m/s, then what percent ofthe potential energy at the top of the damn is convertedinto kinetic energy at the bottom of the dam?

A. 11.1%

B. 47.2%

C. 75.8%

D. 96.4%

If the generator has a 20% efficiency in converting thewater's gravitational potential energy into usefulelectrical energy, then how much energy can 5 kg ofwater deliver?

A. 0.2x5 x 9.8 x 150joules

B. 5 x 9.8 x 150/0.2 joules

C. 0.2x5x9.8/150joules

D. 0.2x150/(5x9.8) joules


3.

4.

5.

The useful energy the generator outputs is less than theinitial energy of the water, because:

A. the potential energy of the water decreases as itfalls.

B. of frictional losses in the water and generator.

C. energy conservation.

D. the water gains kinetic energy as it falls.

For the entire process described in the passage, theBEST description of energy conversion is:

A. kinetic energy to potential energy, electricalenergy, and heat.

B. potential energy to electrical energy and kineticenergy.

C. kinetic and potential energy to electrical energy,kinetic energy, and heat

D. electrical energy to kinetic energy.

Which of the following factors would improve currentproduction?

A. Release the water from a point somewhere belowthe top of the dam.

B. Decrease the magnetic field through which the coilrotates.

C. Build a wider dam.

D. Use a coil with a smaller area.



When studying the motion of a spacecraft orbiting theEarth, we may treat it like any satellite orbiting a body.Newton's work showed that any satellite--the Moon orbitingthe Earth or a spacecraft orbiting the Earth-may be treatedlike a projectile. For example, a spacecraft in orbit about theEarth continuously falls toward the Earth, but because of theorbital speedof the spacecraftand the curvature of the Earth,it never reaches the ground.

One way to analyze the motion of a spacecraft is toconsider the forces acting on it. In order for a spacecraft toorbit the Earth, something must attract the spacecraft to thecenter of the Earth. There must be some inward-pointingforce, a centripetal force due to gravity, that keeps thespacecraft in orbit about the Earth.

A second way to analyze the motion of the spacecraft isto look at conservation of energy. As the spacecraft movesaway from the surface of the Earth, kinetic energy isconverted to gravitational potential energy. We may useconservation of energy to determine what is called "escapevelocity." This is the minimal velocity the spacecraft musthave to escape the Earth's gravitational pull. We may useconservation of energy to determine a new orbital distance, ifwe know a new velocity. However, because the spacecraftgets reasonably far away from the Earth, we must write thegravitational potential energy as Equation 1:

u= GMgnr

Equation 1

Here, m is the mass of the spacecraft, and r is thedistance away from the center of the Earth. The reason forwriting the gravitational potential energy like this is becausethe force due to gravity acting on the spaceship is notconstant the farther the spacecraft gets from the Earth. Sincethe force is not constant, the work done by this force on thespacecraft (and hence the potential energy) will not beconstant. Note the presence of the negative sign forgravitational potential energy.

Because of this negative potential energy, the Earth issometimessaid to be at the bottom of a "gravity well."

For the following problems, Re = radius of the Earth,Me = mass of the Earth.

6. If this astronaut could get infinitely far away from theEarth, his potential energy would be:

A. zero, because it is a maximum.

B. zero, because it is a minimum.

C. infinitely large, because it is a maximum.

D. infinitely large, because it is a minimum.


7.

8.

An astronaut of mass m is in a spaceship that takes offstraight up from the surface of the Earth. The spaceshiphas an acceleration keptequal to9.8 m/s2. Theeffectiveweight of the astronaut just after take-off is:

A. 0.

B. mg.

C. V2mg.D. 2mg.

When the spacecraft is launched from the surface of theEarth, it would be better to launch it:

A. slightly towards the East.

B. slightly towards the West.

C. directiy upward.

D. it makes no difference in which direction it islaunched.

9. The magnitude of the escape velocity of a rocket isnecessarily larger if:

A. the rocket has a larger gravitational potentialenergy upon liftoff.

B. the rocket has a smaller gravitational potentialenergy upon liftoff.

C. the planet has a larger mass.D. the planet has a smaller radius.

10. Which graph BEST represents the relationship betweenthe magnitude of gravitational potential energy as afunction of orbital radius?

A. B.

U



A common thrill ride at many amusement parkscombines the thrill of free fall with the periodic motion of apendulum. The rider is fit into a harness, which is connectedby cable to the apex of an arch. From the low point of thependulum pathway, the rider is hoisted vertically anddisplaced laterally to a starting point. Upon initiation, therider undergoes free fall until the cable connecting theharness to the support brace is pulled taught At that time,the rider takes on a pendulum likemotion andoscillates untilthe ride operator terminates the ride. Figure 1 shows aschematic representation of the ride.

Support Brace

Figure 1. Thrill ride apparatus

The harness and rider are hoisted by a pulley system tothe starting platform. At this point, Equation 1 represents thepotential energy of the rider and harness, where Ah is thedifference in height between the ground and the platform.

U = mgAh

Equation 1

The cable is made of a light metal that exhibits minimaldeformation. The height at which the harness passes abovethe ground at its nadir depends on the mass of the rider.

11. As the rider moves from the free fall starting pointthrough one-half a cycle to a point of no motion on theright, how does the energy of the system change?

A. Chemical energy is converted into kinetic energyand then into gravitational potential energy.

B. Kinetic energy is converted into gravitationalpotential energy and then back into kinetic energy.

C. Elastic potential energy is converted into kineticenergy and then into gravitational potential energy.

D. Gravitational potential energy is converted tokinetic energy and then into gravitational potentialenergy.


12. Which change would increase the velocity of theharness and rider at the system's lowest point?

A. Using a lighter harness.

B. Using a heavier harness.

C. Startingfrom a higher platform.

D. Using a shorter cable.

13. Assuming that friction is negligible, how does theheight at the first point following the initial drop atwhich the rider is motionless, compare to the height ofthe free fall starting point?

A. The heights are approximately equal.B. The initial height is greater than the first still point,

because there is slack in the cable initially.

C. The initial height is less than the first still point,because there is slack in the cable initially.

D. The initial height is always less than any still point

14. Why does it require a continually increasing amount ofenergy to lift the rider, as the rider is hoisted from restto the free fall starting point?

A. The gravitational force constant decreases as therider rises to a greater height

B. The gravitational force constant increases as therider rises to a greater height

C. The mass being lifted decreases as the rider rises toa greater height, because less cable must be lifted.

D. The mass being lifted increases as the rider rises toa greater height, because more cable must be lifted.

15. Which of the following serve to slow the rider downduring the course of the thrill ride?

I. Gravitational pull on the harness

II. Friction as the cable swings

III. Air resistance as the rider travels through the air

IV. Elasticity in the cable

A. I and II only

B. II and III only

C. I, II, and IV only

D. I, II, III, and IV

16. If the change in height for the ride is doubled, then thevelocity at the bottom of the path would:

A. remain the same.

B. increase by a factor ofV2.C. double.

D. quadruple.



17. As a mining cart travels down the rails of a mine, whatis true of its relative energies?

A. PEmaximum = KEmaximum

B. PEjnitiai = PEt + KEt

C. KEmaximum > PEfinal + KEfmai

D. TEmaximum > KEmaximum

18. The amount of work required to bring a car traveling atspeed V to a complete stop on a dry road is Wp. The

amount of work required to bring the car traveling atspeed V to a complete stop on an icy road is Wi.

Assuming the car experiences less friction on the icyroad, which of the following statements is NOT true?

I. Wi>WD

II. Wi = WD

III. w^Wd

A. I and III only

B. II and III only

C. I and II only

D. II only

19. A child must traverse a certain distance over the roughsurface at the base of a downward sloping slide beforecoming to a complete stop. If the slide designers wishedto stop the child in half that distance, the work requiredby friction would necessarily:

A. double.

B. quadruple.

C. remain the same.

D. halve.


20. Negative work is a concept that impliesthat:

A. kinetic energy is increased.

B. theforce causing the motion is decreasing.C. the force is perpendicular to the motion.D. the applied force is opposite to the direction of

motion.



Humans have utilized different natural phenomena toextract useful energy for several millennia. During themiddle ages, humans used the gravitational energyof fallingwater to power water wheels. Water falls from a givenheight ontoa wheel with paddles, causing the wheel to turn.This rotational energy, while originally used to power grainmills,is now usedprimarily to powerelectrical generators.

Another fuel source for generating electricity is theenergy released when burning fossil fuels. Although theygive off toxic byproducts, fossil fuelsdo supply substantiallymore energy per unit massthan the water. In recent decades,a new sourceof energy was discovered within the nucleus ofatoms. Certain radioactive atoms decay into smaller nuclei,releasing energy in the form of the kinetic energy of thesedaughter nuclei. When these nuclei interact with a waterbath, they boil it The resulting steam is used to powerelectric generators.

These fuel sources are not perfectly efficient atconverting their total energy into useful work, where

Efficieny(%) = rj = WorkQutx 100Energy In

Table 1 lists common energy sources and their relevant data.

SourceTotal energy/kgof source (J/kg)

Efficiency (%)

Falling Water 98 10

Fossil fuel 1.4 xlO7 20

Nuclear fission 7.6 x 1013 25

Table 1

21. From how high does the water fall, in supplying energythe water wheel?

A. lm

B. 10 m

C. 100 m

D. 1000 m

22. If 2 kg of fossil fuel is used to power a 100 W lightbulb, how long will the bulb stay lit?

A. 5.6xl04s

B. 1.4 x105 s

C. 2.8xl05s

D. 2.8xl07s


23. Assuming negligibleair drag, if 10 kg of fossil fuel canspeeda car from rest up to a speedv, whatwould be thefinal speed of the car if the fuel efficiency weredoubled?

A. v

B. V2v

C. 2v

D. 4v

24. How many Joules of energy are not converted intouseful energy when burning 1 kg of a fossil fuel?

A. 1.75 xlO7 J

B. 1.12 xlO7 J

C. 2.8xl06J

D. 1.4xl06J

25. Which graph best represents the relationship betweenthe mass of fossil fuel used and the operating time, ifthis fuel is used to power a 1200 W hair dryer?

A. B.

26.

m

Given that the specific heat of water is 42000 J/kg-K,how many kg of water can 1 kg of nuclear fuel heatfrom 0°C to 100°C? (Assume perfect efficiency)

A. 4.5xl06kgB. 9.0xl06kgC. 1.44xl07kgD. 1.8xl07kg



For years, researchers have searched for better methodsto generate power to supply the needs of society. Oneproposal from BennettTechnologies, a company responsiblefor such great products as the glass baseball bat and springloaded toilet seat, involves taking advantage of the potentialenergy of water vapor in clouds. Falling rain is collected andfunneled into a stream that turns a turbine as it falls. Theidea involves using the kinetic energy of the falling rain torotate turbines, which in turn generate electrical flow. Thebasicschemeof the Rainfall Generator is shown in Figure 1.

Catch

Funnel Shoot Tube

Turbine I

Turbine II

Figure 1. Rainfall Generator

The catch collects falling rain and the funnel channelsthe water into a stream that passes through the shoot tube.The water falls straight down to the blades of the awaitingturbines. Only two turbines are shown in Figure 1. Thedesign can include more.

The energy extracted by the Rainfall Generator isassumed to be the difference between the potential energy ofthe water and the frictional work lost to friction of the turbine

wheels. The ideal design would be lightweight and haveminimal friction. A standard turbine works by turning anattached bar magnet within a coiled wire, thereby inducing acurrent in the coiled wire. Because of friction, turbines arenot one hundred percent efficient at converting mechanicalenergy into electrical energy.

27. Which changes would result in faster rotation of theturbine cranks?

A. Placing the turbine cranks closer togethervertically.

B. Placing the turbine cranks further apart vertically.

C. Increasing the mass of each crank.

D. Increasing the number of paddles on the wheelfrom eight to ten.


28. The conversion of energy in the Rainfall Generator canbe described as:

A.

B.

C.

D.

potential energy to kinetic energy to mechanicalenergy to electrical potentialenergy.potential energy to mechanical energy to kineticenergy to electrical potential energy.potential energy to chemical energy to mechanicalenergy to electrical potential energy.

potential energy to chemical energy to kineticenergy to electrical potential energy.

29. How much potential energy does a 1000 L cloud withdensityof 2.20 g/L have 500 metersabove the ground?

A. 1000 x 2.20x9.8x500 J

B. 10S°-x 9.8 x500 J2.20

C. 1000 x 0.0022 x 9.8 x 500 J

D. 100Q x9.8x500 J0.0022

30. Why is the Rainfall Generator impractical?

A. Too much surface area is required to catch enoughwater to make it practical.

B. The turbines are likely to oxidize in the rain.

C. The rainfall must be falling straight down.

D. The raindrops must be large.

31. Which of the following statements accurately reflectsdesign features of the Rainfall Generator?

I. The paddles are placed on the rim of the wheel tomaximize the torque applied to the crank.

II. The design takes advantage of the height of clouds.

III. Water's polarity keeps the magnetic field uniform.

A. I only

B. II only

C. I and II only

D. II and III only

32. Assuming that air resistance is negligible, what is thevelocity of a raindrop when it strikes the catch of theRainfall Generator, given that the cloud is L metersabove it?

A.

B.

C.

D.

=4gL2= -/2gT


Questions 33 through 36 are NOT basedon a descriptivepassage.

33. Which of the following factors will NOT impact thekineticenergy of an object sliding down a slope?

A. The height of the slope

B. The length of the slope

C. The coefficient of kinetic friction (ji^)

D. The coefficient of static friction (^s)

34. At the lowest point of a pendulum's swing, which ofthe following is at a minimum?

A. Potential energy

B. Kinetic energy

C. Speed

D. Tension in the cable

35. A ball is dropped by a person from the top of abuilding, while another person at the bottom observesits motion. Both people:

A. agree on the change in potential energy as well asthe kinetic energy of the ball.

B. agree on the change in potential energy, butdisagree on the kinetic energy of the ball.

C. agreeon the kineticenergyof the ball, but disagreeon the change in potential energy.

D. disagree on the kinetic energy of the ball and thechange in potential energy.


36. How does the work necessary to increase the speed of avehicle from 0 m/s to 10 m/s compare to the worknecessary to increase the speed of a vehicle from 10 m/sto 20 m/s?

A. It takes the same amount of work to carry out eachchange in speed.

B. It takes twice as much work to carry out the 10-20m/s speed increase as the 0-10 m/s speed increase.

C. It takes three times as much work to carry out the10-20 m/s speed increase as the 0-10 m/s speedincrease.

D. It takes four times as much work to carry out the10-20 m/s speed increase as the 0-10 m/s speedincrease.



In order to move boxes with a mass between 80 and 100

kg, a storage facility set up a lift, frictionless ramp, and rollerspanning from the delivery dock to the rear of the facility.The lift contained a flat metal lift plate onto which a boxcould be placed. It is connected to a counterweight byvertical cables around a pulley wheel hanging from theceiling. A mass of 50 kg was chosen for the counterweight.Figure 1 shows the apparatus.

Pulley wheel

Lift plate

b Rollers cooooooooooo

Figure 1. The amazing box-moving apparatus

Once the box reaches the top of the ramp at point a, it ispushed onto the frictionless ramp. From there, it acceleratesdown the ramp until reaching point b, at which time it levelsout and rolls across horizontal rollers. The rollers exist from

point b to point c. The rollers serve to slow the box downuntil it comes to rest At point a, the potential energy of thebox is represented by Equation 1, where Ah is the differencein height between the ground and the top of the ramp.

PE = mgAh

Equation 1

The length of the roller region and the friction of therollers are adjusted based on how far the box needs to travel.

37. In order to increase the distance the box travels across

the rollers, what adjustments should be made?

A. A wax can be applied to the rollers.

B. An adhesive can be applied to the rollers.

C. The wheels can be tightened so that they don't rollas freely.

D. The wheels can be loosened so that they roll morefreely.

38. At what point in the system is the temperature thegreatest?

A. Point a

B. Point b

C. Point c

D. At the base of the lift


39. When the box has a mass of 90 kg, what is true of thetotal work associated with raising the box compared tothe work needed to raise the counterweight and returnthe lift plate to the base?

A. The work needed to raise the box exceeds the workneeded to lift the counterweight.

B. The work needed to raise the box is less than the

work needed to lift the counterweight

C. The work needed to raise the box is equal to thework needed to lift the counterweight

D. No work is required to raise the box.

40.

41.

42.

Increasing 0 while keeping point a at the same spotwould have what effect on apparatus?

I. The speed of the box at point b would increase.

II. The work done by friction in the region from pointb to point c would increase.

III. The number of rollers needed between point b andpoint c would remain the same.

A. Ill only

B. I and III only

C. II and III only

D. I, II, and III

Which graph accurately relates the height at point a tothe speed of the box at point b?

A. CCJ

C

8,c

ICO

/cS /

J3BO

'53X

•i—i

J300

'5X/

Speed at point b Speed at point b

C3 D-« /c

'5s

'5a. /

CCJ 3 /Xi00

'5X

j=op

"53X J

Speed at point b Speed at point b

C.

How much work is done by friction in the region frompoint b to point c?

A« mboxgAn

B. mboxgAdc_b

C mboxgcosO Adc_b

D- ^mboxviift2



Engineers invest countless hours when designingvarying speed thrill rides for theme parks. A focus of theirdesign is the energetics of the system at given points alongthe path. A functional rollercoastermust have kinetic energyat all points during the ride. Figure 1 shows the energyassociated with an empty passenger cart as it traverses therails of Rollercoaster A. Figure 2 shows the energeticsassociated with the same empty passenger cart as it traversesthe rails of a different rollercoaster, Rollercoaster B.

= PE

Distance traveled (m)

Figure 1. Energetics of Rollercoaster A

= KE = Work

gm

%1

1 f1 /1 /1 /ifIf

II

1 1

/ \

- =PE =KE =Work

\ .-"•+ ^-v /V v V* * /

\ I A Wc •• \J jr \ / \\

Distance traveled (m)

Figure 2. Energetics of Rollercoaster B

All of the work shown in the energy diagrams isattributed to work done by friction. No additional energy isadded to the system once the cart is in motion.

43. Each rollercoaster path has a loop where the cartreaches its slowest nonzero speed at the top. Whichrollercoaster has the loop with the bigger radius?

A. Rollercoaster A, because the intermediate kineticenergy is higher in Figure 1 than it is in Figure 2

B. RollercoasterA, because the intermediate potentialenergy is higher in Figure 1 than it is in Figure 2

C. Rollercoaster B, because the intermediate kineticenergy is higher in Figure 1 than it is in Figure 2

D. Rollercoaster B, because the intermediate potentialenergy is higher in Figure 1 than it is in Figure 2


44. The energy of the empty passenger cart in Figure 1 atthe halfway point of its path is best described as:

A. E = mgh

B. E=&mv2

C. E= mgh + Vimv2

D. E= Vfcmgh + Vfcmv2

45. How would the addition of passengers to the cart affectthe energy diagram in Figure 1?

A. The graph would have the same shape, but higheramplitude at each point

B. The graph would have the same shape, but smalleramplitude at each point

C. The graph would have the same amplitude at highpoints, but a different shape.

D. The graph would be identical to Figure 1.

46. Which of the following statements correctly describesthe differences between Rollercoasters A and B?

I. Rollercoaster A experiences the greater initial drop.

II. Rollercoaster B has less hills than Rollercoaster A.

III. Rollercoaster B reaches a greater maximum speed.

A. I only

B. HI only

C. I and II only

D. II and III only

47. Which of the following design features is undesirablefor a rollercoaster?

A. The top of the loop higher than the startingpointB. Minimal friction on the track prior to the stopping

region

C. An aerodynamicpassengercartD. Incorporation of spring potential energy at the start

of the ride

48. Which rollercoaster experiences braking beforeentering the loop?

A. Rollercoaster A, because it shows an increase inpotential energy in the middle of Figure 1.

B. Rollercoaster A, because it shows an increase inwork in the middle of Figure 1.

C. Rollercoaster B, because it shows an increase inpotential energy in the middle of Figure 2.

D. Rollercoaster B, because it shows an increase inwork in the middle of Figure 2.



49.

50.

Two identical spacecraft are orbiting a planetSpacecraft A is at an orbital radius of R, whileSpacecraft B is at 3R. What is the ratio of thegravitational potential energy of Spacecraft A to that ofSpacecraft B?

A. 9tol

B. 3tol

C. lto3

D. lto9

By how many meters can a 10.0-kg mass be lifted usingan engine with an efficiency of 0.85 that burns 1.0grams of a hydrocarbon fuel with a heat of combustionofl5.2kJ/g?

A. 0.13 m

B. 12.9 m

C. 71.4 m

D. 132 m

51. How much work is need to lift a 100 kg mass by 4 m, ifthe lift plate has a mass of 10 kg and the counterweighthas a mass of 60 kg?

Support beam

A. 1960 J

B. 5880 J

C. 7840 J

D. 23,520 J

Lift plate


52. How much work is done by gravity on a 50-kg box thatis pushed up a 5 m ramp to a new elevation 2 m higherthan its initial height?

A. 2500 J

B. 1000 J

C. -1000 J

D. -2500 J

1. C 2. A 3. B 4. C 5. C 6. A

7. D 8. A 9. B 10. A 11. D 12. C

13. A 14. D 15. B 16. B 17. D 18. A

19. C 20. D 21. B 22. A 23. B 24. B

25. B 26. D 27. B 28. A 29. C 30. A

31. C 32. D 33. D 34. A 35. A 36. C

37. D 38. C 39. B 40. A 41. D 42. A

43. B 44. C 45. A 46. B 47. A 48. D

49. B 50. D 51. A 52. C

YOU ARE DONE.

Answers to 52-Question Work and Energy Practice Exam

Passage I (Questions 1 - s] Waterfall Generator

1. Choice C is the best answer. The water falls from a height of 150 m. If all of the water's initial potential energy wereconverted into kinetic energy, then we could say that mghjn}tjai = V£mvbase2. The vertical speed starts at 0, so we know thatVmax base2 = 2gh. Plugging in we get:

Vmax base2 =2(10)(150) =2 x 1500 =3000

Rather than solve for v, we should leave the value in terms of v2, because KE « v2. We now need to solve for the actual v2and then determine the ratio of the actual v2 to the maximum v2.

vactual2 =47.22 =502 =2500

^^mvj=Vac,ua,2/vmax2 - 2500/3000 - 5/6 =0-833 =83.3%We know that the 2500 value we used was a bit too high, so the actual percentage should be a little less than 83.3%. Theclosest answer choice is choice C, which makes it the best answer. But for some of us, 83.3% is too far away from 75.8% tofeel comfortable. If that's the case and you can do math quickly, there is a nice trick for solving squares. Using binomialexpansion, we know that (x + y)(x - y)= x2 - y2, so97 x 3 = 502 - 472 .-. 472 = 502 - (97 x 3)= 2500 - 291 = 2209. Theratio of 2250: 3000 is % = 75%. The best answer is choice C.

2. Choice A is the best answer. This is essentially a formula-identification problem. Let's solve it using the limiting cases testtip. If the generator is more efficient, should more or less energy be produced? More. Cross out choice B. If the water fallsfrom a higher starting point, should more or less energy be produced? More. Cross out choice C. Finally, if there is morewater being used by the generator, should more or less energy be produced? More. Cross out choice D. If you wanted aformal method instead, then know that:

(useful work) = (efficiency)(input energy)

W = (efficiency )mgh

W = 0.2x5x9.8x 150


3. Choice B is the best answer. While choices A, C, and D are true, they do not explain why the generator is not 100%efficient Any machine that takes in one form of energy and converts it to another can never convert all of the input energyinto useful work. A major reason for this is friction. Friction is typically the death, or at least mortal wound, of potentiallyuseful energy. In the generator, friction probably originates in its moving parts. In the water, friction is related to the water'sviscosity. The best answer is choice B.

4. Choice C is the best answer. As the water leaves the top of the dam, it is moving, so it has initial kineticenergy. It is 150meters above the hydraulic generator, so it also has initial potential energy. The final outcome is the conversion of themechanical energy to electrical energy. The best answer is choice C.

5. Choice C is the best answer. From the passage, we see that the voltage is proportional to the magnetic field and to the areaof the coil. If either of these values is decreased, then the induced voltage is decreased. This, in turn, would decrease theamountof current produced. Releasing the water from a point below the top of the dam would decrease the initial potentialenergy of the water, so it is not moving as fast when it strikes the water wheel. If a wider dam were built, maintaining thesame heightand waterspeed, it would increase the mass of the wateras it leaves the dam. This increases the potential energyof the water, which would give the water greater final kinetic energy. Provided we could harness this extra energy, the poweroutput would be increased. The best answer is choice C.

Passage II (Questions 6 - 10) Gravitational Potential Energy

6. Choice A is the best answer. As the spacecraft gets farther away from the Earth, r gets larger and the numerical value for Ugets smaller. However, U is negative, so as r gets larger, U gets less negative—itapproaches zero, becoming more positive. Infact, zero is the largest value U can ever be, so the gravitational potential energy is a maximum when it is zero. The bestanswer is choice A.


7. Choice D is the best answer. Because the floor of the rocket is accelerating up and pushing on the astronaut's feet, thenormal force is greater in magnitude than it would be if the rocket were stationary. Given that the normal force is theapparent weight, we can conclude that the astronaut apparently weighs more than normal, so the normal force must begreater than mg. This eliminates choiceA and B. Suppose theastronaut just happens to be standing on a scale. Draw a free-body diagram of the astronaut Thereare only twoforces acting on theastronaut: theforcedue to gravity, mg,and thenormalforce, N. Recall that the normal force is the force exerted on the astronaut by the scale. By Newton's third law, this is theforce that the astronaut exerts on the scale. By solving for the normal force, we determine the force that the astronaut exertson the scale: hence, we can determine the astronaut's effective weight.

N

mg

scale


ma = N - mg

N = ma + mg

m.a = 9.8 7S2 = g

N = m(a + g) = m(g + g) = 2 mg

8. Choice A is the best answer. The Earth spins from West to East. Launching the spacecraft to the East means we can usesome of the Earth's spin to contribute to the escape velocity. The best answer is choice A.

9. Choice B is the best answer. Once the rocket has "escaped" the effects of the planet's gravity, its gravitational potentialenergy will be zero. Mathematically, this is when the rocket is infinitely far away from the planet When the rocket hasgreater gravitational potential energy, it is higher in elevation (farther from Earth). Under such circumstances, not as muchvelocity is necessary to escape Earth's gravitational pull as when the rocket is closer to Earth. This means that the greatestescape velocity (and thus greatest kinetic energy) is needed when the rocket has small gravitational energy (i.e., is close toEarth). If you guessed choice C, note that it says nothing about the planet's radius, which is important at liftoff. Choice Dsays nothing about the planet's mass. Both must be known to tell whether the needed escape velocity is greater. The bestanswer is choice B.

10. Choice A is the best answer. This is a typical plot: the inverse proportionality. That is, U is proportional to 1/r, as the givenformula states. It's not choice B, as many people guess. That would be U is proportional to -r. You should know this type ofplot. The best answer is choice A.

Passage III (Questions 11 - 16) Thrill Ride Energetics

11. Choice D is the best answer. The rider starts motionless at a high elevation (on the left as shown in Figure 1), therefore thesystem initially has gravitational potential energy and no kinetic energy. This makes choice D the best answer. Through thecyclic motion of any pendulum that starts at an apex, the system converts from gravitational potential energy at the start intokinetic energy at its lowest point, and back into gravitational potential energy at its next still point (point of no velocity). Thebest answer is choice D.

12. Choice C is the best answer. The velocity of the harness and rider at the lowest point of the swing can be found by settingthe maximum kinetic energy equal to the change in potential energy.

V£mv2 = mgAh .*. }4v2 = gAh

v2 = 2gAhv = V2gAh

Upon solving this, we find that the velocity at the lowest point depends on the gravitational force constant and the change inheight only, but not the mass. Because mass does not affect the results, choices A and B are eliminated. Starting from ahigher point increases Ah, which in turn increases the speed of the rider and harness at the lowest point so choice C is thebest answer. Using a shorter cable reduces Ah, which in turn decreases the speed of the rider and harness at the lowest point,so choice D is eliminated. The best answer is choice C.


13. Choice A iscorrect. The question compares the initial height ofthe rider to the height after the rider has completed the firsthalf of the first cycle. The rider cannot achieve a state of higher potential energy, because no energy has been added to thesystem during the initial free fall orduring the swing back to an apex, so choices Cand Dcan both be eliminated. The initialslack in the cable has no effect on the height of the starting platform or the height that the rider reaches, so choice B iseliminated. Because thereis minimal friction, theheights should beessentially equal. The best answer is choiceA.

14. Choice D is the best answer.The gravitational force constant does notchange overa few meters in height so choices A andB are eliminated. As the rider and harness are raised, it becomes more difficult to lift the system, because the total massbeing lifted is increasing as the rider and harness are elevated. This isbecause the cable starts outdangling, but it rises as therider and harness are lifted, so the mass of cable being lifted is increasing as the rider is elevated. Choice D is the bestanswer. Choice C could also have been eliminated, becausethe question states that a continually greater amount of energy isrequired to raise the rider and harness as their elevation increases, which could not beexplained by a decrease in mass. Thebest answer is choice D.

15. Choice B is the best answer. Gravitational pull on the rider provides the potential energy and accelerates the system duringdescentand decelerates the systemduring ascent It has no neteffecton the riderduring the courseof the ride, so it does notslow the rider down when considering the entire ride. But this question is ambiguous, so we cannot eliminate Statement Icompletely yet It gets a "?" for the time being. Both friction and air resistance oppose the motion, so thus they both slowthe riderdown, making Statements II and III valid. This eliminates choicesA and C. Elasticity in the cable will cause it tostretch, most notably at the rider's lowest point resulting in a greater Ahduring the course of the ride. This will changetheabsolute value of the energy of the system, but it does not drain the system of energy. Elasticity in the cable serves only toconvert gravitational potential energy into springpotential energy. This makesStatement IV invalid. This eliminates choiceD. The best answer is choice B.

16. Choice B is the best answer. The overall change in height that occurs during the ride (from the initial platform to the bottomof the swing) can be doubled by either raising the starting heightor by lowering the nadir (lowest point). When the change inheight, Ah, is doubled, the initial potential energy of the system is doubled. Because the energy is proportional to thevelocity squared (PEjnjtjal = mgAh and KEmax = ^mvmax2 •'• Ah « vmax2), when the value of Ah doubles, the velocity willonly increase by afactor ofV2. Thebestanswer ischoice B.


17. Choice D is the best answer. The mining cart is going down hill, which implies that it starts with maximum potentialenergy, PEmax> when the cart is at its highest point Once the cart begins to move, its potential energy is converted intokinetic energy. As the cart travels the track, the work done by friction increases. This means that not all of the potentialenergy is converted into kinetic energy, as some of it is wasted as heat due to friction. Because not all of the potential energyis converted into kinetic energy, the maximum potentialenergy, PEmax> is greater than the maximum kinetic energy, KEmax-This eliminates choice A. As a general rule, the total energy of a system is constant in the absence of an energy drain such asfriction or heat loss. This would mean that the initial potential energy of an object at rest is equal to the total energy of theobject at any given time. However, because of friction, the kinetic energy at a given time is less than the initial potentialenergy (PEjnjtia] > PEt + KEt). This eliminates choice B. At the bottom of the ramp, the cart has come to a point of minimalpotential energy, and the cart has reached its maximum speed.This means that the final speed is the cart's maximumspeed,which means that its final kinetic energy is the cart's maximumkinetic energy. The maximum kinetic energy (which occurswhen the cart reaches its highest speed) is less than the sum of the potential energy and kinetic energy of the cart at thebottomof the hill (KEmax < PEfinal + KEfinal)- This eliminateschoice C. This means that the maximum total energy (at thestart when the cart is at its highest point) is greater than the maximum kinetic energy, because not all of the potential energyis converted into kinetic energy. This is because some energy is lost to friction over the course of the ride and there is alwayssome potential energy. This makes choice D a true statement which makes it the best answer. The best answer is choice D.

18. Choice A is the best answer. According to the work-kinetic energy theorem, the work required to stop the car is equal to thechange in kinetic energy experienced by the car. In equation form, this is:

Wrotal =AKE = Vimvf2 - Vimvj2

Since the car has the same final and initial speeds on the dry and icy roads, it follows that the work required to bring them toa stop is also the same under both conditions. That means both Statements I and III are false, while Statement II is true. Becareful on this question, because it is asking for what is not true. The best answer is choice A.


19. Choice C is the best answer. The rough surface dissipates the kinetic energy the child has upon reaching the base of theslide. Regardless of the stopping distance, the rough surface must still dissipate all of the child's kinetic energy, if it is tobring the child to a stop. Therefore, the work required by friction is the same for any length of rough surface. What needs tochange is the roughness of the surface. How should it change? If the child is to stop in a shorter distance, the surface must berougher. For this problem, the coefficient of friction is what has to double. However, the work is unchanged. The bestanswer is choice C.

20. Choice D is the best answer. Negative work removes energy from the system, so the kinetic energy should be decreasing,not increasing. This eliminates choice A. Whether or not the force is changing in magnitude does not define the sign of thework, so choice B is irrelevant, and thus eliminated. When a force is perpendicular to the direction of motion, work is zero,so choice C is eliminated. Negative work implies that energy is being removed from the system, so it must be associated witha force that is slowing the object down. This means that the applied force is opposite to the direction of motion, makingchoice D a true statement The best answer is choice D.

Passage IV (Questions 21 - 26) Energy Conversion

21.

22.

23.

24.

Choice B is the best answer. From Table 1, we see that 1 kg of fallen water supplies 98 Jvof energy. The supplied energycomes from the water's loss of gravitational potential energy. Using the equation PE = mgAh, along with PE = 98 J for 1 kgof water, we find the following math:

PE = mgAh .-. Ah == PE/r

The water fell 10 m total. The best answer is choice B.

'mg =98/l-9.8 =9%.8 =10

Choice A is the best answer. Use magnitude estimation to solve this, because the answer choices vary in their power of 10.Time relates to the energy through W = Pt, so t = W/P, where W is the useful work the supplied energy can do and P is the

power needed. Here, W= (2 kg)(1.4 x 107 J/kg)(2 x 10'1) and P= 100 watts, so t = 2 x 1.4 x 107 x 2 x lO'VlO2. The 2 x10"1 accounts for the imperfect efficiency of the fossil fuel source. Multiplying outonly the powers of 10 gives and order ofmagnitude of 104. The remaining numbers, when multiplied should yield a number between 1and 10, meaning that the finalanswer must have an order of magnitude of 104. Avoiding detailed multiplication of numbers, and instead estimating frompowers,will save you much grief. The best answer is choice A.

Choice B is the best answer. Twice the efficiency means twice the useful energy from the fuel. Doubling the energy allowsthe car to reach a higher final speed, ruling out choice A. Since this energy becomes kinetic, and because kinetic energy isproportional to the square of the speed, the increased efficiency should change the final speed by a factor of V2, not by afactor of 2. The best answer is choice B.

Choice B is the best answer. We need to determine the wasted energy, which for a fossil fuel should be 80% of the supplied

fuel (according toTable 1,20% of the fuel is useful). Since 1kg offuel supplies 1.4 x 107 J of energy, the wasted energy issimply 80% ofthat value (.8 x 1.4 x 107 J). Now, (.8)(1.4) is greater than 1and less than 1.4. This means the answer's orderofmagnitude is 107, which eliminates choices Cand D. 1.75 is too large, so choice A is eliminated, leaving only choice B.The best answer is choice B.

25. Choice B is the best answer. Twice as much fuel should last twice as long, since the hair dryer uses energy at a constant rate(i.e., the hair consumes constant power). The relationship between the amount (i.e., mass) of fuel and the operating timeshould therefore be linear and increasing. Choices C and D are eliminated, because they are not linear graphs. Choice A iseliminated, because it is not increasing. Only in choice B is there a linear increase. The best answer is choice B.

26. Choice D is the best answer. Starting with the equation Q = mCAT, we can solve and for m which gives us the relationship

m= Q/cAT' where Q is the supplied energy, C is the specific heat of water, and AT is the temperature increase of thesurrounding water. Note that "perfect efficiency" means we do not need to prorate the supplied energy. Using magnitude

estimation gives a magnitude of 107. The remaining numbers should be 7'̂ /4,2» which is between 1and 10. This rules outchoices Aand B. Since the ratio looks similar to 8/4, the final answer should have a number in it close to 2. The bestanswer is choice D.


Passage V (Questions 27 - 32) Rainfall Generator

27. Choice B is the best answer. The turbine cranks turn faster when a greater torque is applied. If we placed the turbine crankscloser together, then the water falling from one crank to the next would not have as much distance (and thus less time) togain momentum (and kinetic energy) by the time it were collected in the second crank. This reduces the force of the waterstriking the second turbine, and thus reduces the torque created by the second turbine, sochoice Aiseliminated. If we placedthe turbine cranks further apart then the water falling from one crank to the next has a greater distance (and thus greatertime) to gain momentum (and kinetic energy) by the time it is collected in the second crank. This would increase the forcewith which water strikes the paddles of the second turbine, and thus increases the torque on the turbine wheel, which makeschoice B thebestanswer. Increasing themass of thecrank increases the resistive force(creates a greater inertia for thewheelto overcome when it tries to turn), which in turn would reduce the angular momentum of the wheel. This eliminates choiceC. Increasing the number of paddle wheels reduces the amount of water that each paddle collects, butit increases the pointsof applied force. The two effects would likely negate one another, because the amount of collected water would beessentially the same, although the wheel would be heavier which would reduce its angular velocity. This eliminates choiceD. The best answer is choice B.

28. Choice A is the best answer. The concept of the Rainfall Generator is based on the harnessingof the gravitational potentialenergy of water vapor in the sky. This means that the conversion of energy begins with gravitational potential energy. Asthe raindrops begin to fall, gravitational potential energy is converted into kinetic energy. This makes choice A the bestanswer. The raindrops fall into the catch and then pass through a chute tube before they strikes the bladesof a turbine, thusturning the turbine and generating mechanical energy. The spinning turbine converts mechanical energy into electricalcurrent into, which is stored as electrical potential energy. The best answer is choice A.

29. Choice C is the best answer. The equation for gravitational potential energy is PE = mgh. To determine the gravitationalpotential energy for the system, the mass and the heightof the cloud must both be known. The height is 500 m and g is 9.8

m/s2, soonly mass must bedetermined. The mass can befound by multiplying the volume of the cloud by its density. Thevolume is 1000 L and the density is 2.20 g/L. They give us units that are not in the standard mks system, so you must convertyour units from 2.20 g/L to 0.0022 kg/L. Remember to convert units to the mks system (meters, kilograms, and seconds) orfeel the pain of lost points. The best answer is choice C.

30. Choice A is the best answer. Since the catch is designed to collect water droplets and create water flow to turn the turbines,the size of the droplets and the angle at which they strike the catch should not affect the quantity of water, so choices C andD can be eliminated. The turbines can be made of a material that does not readily oxidize or isolated from the rain, so whileoxidation of metal is a problem in a moisture-rich environment design tricks can be applied to avoid any significant issueswith oxidation. The major problem with the design in the great volume of water needed to make it practical. The systemwould require a catch that is far too large to be stable in wind and to be supported against gravity. It would be extremely top-heavy. The design is absurd and impractical because of the great size that is needed for the catch. The best answer is choiceA.

31. Choice C is the best answer. Statement I is valid, because having the paddles at the edge of the wheel maximize themoment arm, which in turn maximizes the torque (x = F-r-sin 0) of the system. Statement II is valid, because the initialenergy of the system is the gravitational potential energy of the clouds, which results from the great height of the clouds.Statement III is invalid, because the polarity of water does not affect the uniformity of an external magnetic field. The fieldmay influence the manner in which the water droplets fall, but this would not change the uniformity. The best answer ischoice C.

32. Choice D is the best answer. The velocity of the falling raindrop as it strikes the catch can be determined either through thekinematics equations or by equating gravitational potential energy with kinetic energy. The answer choices do not look likemanipulated forms of the kinetics equations, so we should approach this question by setting the change in the potentialenergy equal to the kinetic energy: mgAh = Vimv2. This relationship reduces to:

gL=^v2

Multiplying both sides of the equation leads to:

v2 =2gL .-. v= V2gTThis formula should look familiar from your previous physics experiences. You could have eliminated choices A and C withapplying math derivations, because they don't have the correct units of m/s. The best answer is choice D.



33. Choice D is the best answer. The kinetic energy of an object sliding down a slope will depend on the initial gravitationalpotential energy of the object and the work done by friction as it slides down the surface. The gravitational potential energyis found using the equation, U = mgh. This means that the heightof the slope will ultimately impact the kinetic energyof theobject as it slides down the slope, which eliminates choice A. The work done by friction depends on the frictional force andthe distance over which it slides, which eliminateschoice B. Because the frictional force is p^N, the amount of work done byfriction depends on p^, which eliminates choice C. The kinetic energy for a sliding object will not depend on the staticfriction, which is associated with a stationary object. The best answer is choice D.

34. Choice A is the best answer. The system initially has all potential energy and no kinetic energy at the start, so the systemstarts at maximum potential energy and minimum speed and therefore kinetic energy. The system reaches its greatest speedand thus maximum kinetic energy and has its minimum potential energy at the lowest point of the swing during the course ofthe ride. This makes choice A valid and eliminates choices B and C. At the lowest point, the mass is hanging straight down,so the tension in the cable needed to offset the gravitational pull is the greatest. The tension in the cable is at a maximum andnot minimum at the lowest point, so choice D is eliminated. The best answer is choice A.

35. Choice A is the best answer. The potential energy of the system is gravitational in nature, so the change in potential energyis found using AUgravitational = mgAh. No matter where the observers stand and what they define as the zero point reference,they will measure the same Ah. This means that both observers agree on the change in potential energy, eliminating choicesC and D. The kinetic energy of the ball results from the conversion of gravitational potential energy, so if both observersagree on the change in potential energy, then they will agree on the object's kinetic energy. The best answer is choice A.

36. Choice C is the best answer. Going from 0 m/s to 10 m/s requires an amount of work equal to:

W= AKE='/imvf2- i^rnvj2

.-. W0->io= ^mClO)2- i/im(0)2 = i^m(10)2 = i/2m(100)

In going from 10 m/s to 20 m/s, we can write the work required as:

Wio^20 = '/2m(20)2 - i/2m(10)2 = ^m(400) - i/2m(100) = !^m(300)

Going from 10 m/s to 20 m/s requires three times as much work (and thus three times as much energy) as going from 0 m/sto 10 m/s. If you noticed that the speed doubled, you may have simplified the problem down to determining the ratio of|(2v)2 - v2]: v2, which becomes (4v2 - v2): v2 = 3v2: v2= 3 : 1. The best answer is choice C.

Passage VI (Questions 37 - 42) Box Relocation Program

37. Choice D is the best answer. To increase the distance the box travels on the rollers, there needs to be less resistance as itrolls. Adding an adhesive increases the kinetic friction, so choice B is eliminated. Tightening the wheels increases theresistance, so choice C is eliminated. As is often the case, we can get it down to two choices with relative ease, but decidingbetween the remaining two choices often proves to be challenging. Applying a wax to the wheels decreases the kineticfriction, but the box is not sliding, it's rolling. Addition of wax is irrelevant for this system. If the box were meant to slide,they wouldn't use rollers. The only way that the resistance will be reduced is to allow the rollers to rotate more freely. Thebest answer is choice D.

38. Choice C is the best answer. The highest temperature is found at the point where the greatest amount of heat is generated byfriction. At point a, the box has been lifted, so no friction has occurred yet. Choice A is eliminated. The ramp isfrictionless, so no heat is generated as the box slides down the ramp. As a result, it has not heated up as it passes point b, sochoice B is eliminated. At point c, it has been slowed to a rest, so it has experienced some friction. Point c is the warmestspot so far. At the base of the lift, nothing has happened yet, so the temperature is still ambient. Choice D is eliminated,leaving choice C as the best answer. The best answer is choice C.

39. Choice B is the best answer. The box has a mass of 90 kg and the counterweight has a mass of 50 kg. This means that thelowering of the 50 kg counterweight offsets lifting the 90 kg box. It is in essence like lifting a 40 kg box. Once the box isoff loaded, then the counterweight must be raised. There is nothing to offset this, so 50 kg must be lifted. The same height istraveled up as down, so the only difference in the work comes from the force applied. More force is applied to lift thecounterweight than to raise the box, so choice B is the best answer. Choice D is a throwaway answer. That could only betrue if the counterweight and box had the same mass. The best answer is choice B.


40. Choice A is the best answer. If point a remains at the same spot, then the height at the top of the ramp remains the same.This means the box starts its descent with the same initial potential energy. As a result, it will have the same kinetic energyat the bottom of the ramp. The mass of the box remains the same, so the maximum speed of the box must also remain thesame. By increasing the angle, the slopeof the ramp is steeper, which means that the acceleration is greater (a = g sin 0).But, it travels a shorter distance, so it is accelerated for a shorter amount of time, so it reaches the same speed at the bottomof the ramp (point b). Statement I is invalid. This eliminates choices B and D. Statement III must be true based on theremaining choices (beingthat it is presentin bothchoices A and Q, so we needonly considerStatement II. The workdoneby friction is equal to the change in kinetic energy between points b and c, which happens to equal the change in potentialenergy from point a to point b. This means that Statement II is invalid, so choice A is the best answer. The best answer ischoice A.

41. Choice D is the best answer. As the height increases, the box starts from rest with greater potential energy, so it reaches agreater speed at the base, point b. This eliminates choice A. The relationship between speed at the base and initial heightcomes from the conservation of energy theorem. Equating mgAh to Vfcmvb2 yields Vb = V2gh. This means that theirrelationship is not linear, so choice B is eliminated. To decide between the remaining choices, it is easiest to plug innumbers. If h were doubled, then the velocity at point b would increase by a factor of square root of 2, about 1.4. Thismeans that the height experiences the greater change, and graphs that are not linear or asymptotic bend toward the axis ofgreater change. The best answer is choice D.

42. Choice A is the best answer. The box starts at rest and finishes at rest, so there is no kinetic energy to consider. This meansthat the work done by friction is equal to the work done lifting the box. The work needed to lift the box is equal to its changein potential energy. This makes choice A the best answer. The work due to friction could also be found by evaluating thechange in kinetic energy from point b to point c. But, we do not know the speed of the box at point b, so that method is notpossible. It is also not an answer choice, although choice D shows some random variables in the kinetic energy equation.Choice D is eliminated, because it is sheer lunacy, and you should never make lunacy your choice. Choice B is eliminated,because the equation has the wrong distance in it Gravitational potential energy depends on the height, which the distancefrom point b to point c does not equal. Choice C would actually be correct if it were sine instead of cosine in the equationand the distance was from point a to point b. It isn't sine nor is it the correct distance, so that thought is useless for thisquestion, but is good food for thought The best answer is choice A.

Passage VII (Questions 43 - 48) Rollercoaster Energetics

43. Choice B is the best answer. The question gives you a hint in that it says the top of the loop occurs when the cart has itsslowestspeed (not counting zero speed at the end of the ride). The point at which the cart has its slowest speed is also thepointat which the cart has its lowest kinetic energy, so we need to look at the potential energy when the kineticenergy is atits lowest intermediate point. This eliminates choices A and C. The answer choices give us a hint to consider theintermediate potential and kinetic energies, confirming that we are on the right track (no pun intended). At the top of theloop, the system has a local maximumfor potential energy, so the greatest intermediate potential energy is associated withthe rollercoaster traversing the biggest loop. The intermediate potential energy is higher in Figure 1 than Figure 2, whichmeans that the cart on Rollercoaster A has traveled around a bigger loop than the cart in Rollercoaster B. The best answer ischoice B.

44. Choice C is the best answer. While the halfway pointon the path is not labeled, we can safely assume that it is after thestarting point and prior to the ending point In the middle of the graph in Figure 1, the cart has both kinetic energy andgravitational energy. The energy of the cart is a sum of both energies, which eliminates choices A and B. Gravitationalpotential energy is equal to mgh, so choice C is the bestanswer. If you are thinking "there'sgot to be more to this questionthan just knowing the equations," then realize that in order to generate a bell curve, there must be some straightforwardquestions as well as challenging questions. Get the answer quickly, don't try to find something tricky that isn't there, andmove on. The best answer is choice C.

45. Choice A is the best answer. Adding passengers to the cart wouldincrease the mass of the system. This would mean that itwould start with a greater initial potential energy (PEiniuai = mghjnjtiai). The cart system would also have more kineticenergy at different points on the track, despite having the same speed. This means that all of the points on the graph wouldbe higher than they are in Figure 1. The energy lost to friction depends on the normal force, which in turn depends on themass of the cart system. Every energy term increases as mass increases, so the graph should be the same shape with greateramplitude at each point. This eliminates choices B, C, and D and leaves choice A the best answer. The best answer ischoice A.


46. Choice B is the best answer. In Figure 2, the potential energy at the first nadir (first low point, reading from left to right) islower in value than the first nadir in Figure 1. Correspondingly, the kinetic energy at the first apex (first high point, readingfrom left to right) in Figure 2 is greater than the first apex in Figure 1. This means that the cart has dropped by a greaterdistance and picked up more speed on Rollercoaster B than Rollercoaster A, which makes Statement I invalid. Thiseliminates choices A and C. By default, Statement III must be valid, because it is present in both of the remaining answerchoices. It is true because the maximum kinetic energy is greater in Figure 2 than it is in Figure 1. Considering the cart isthe same in each trial, it has the same mass. The only way for the cart to have a greater kinetic energy is to have a greaterspeed. This confirms that Statement III is valid. Both Figure 1 and Figure 2 show two apexes for potential energy after thestart of the ride. One of the apexes is associated with a loop. The other apex could be another loop or a hill. As such, thereis no way to determine the validity of Statement II. If it isn't definitely valid, we can't pick an answer choice containing it.The best answer is choice B. Eliminating Statement II is likely what the test writer wants us to do, because the graphs lookvery similar and therefore probably have the same elements. Sometimes you just have to pick what you think the test writerwants you to pick. The best answer is choice B.

47. Choice A is the best answer. If the top of the loop in a rollercoaster is higher than the start of the ride, then it is at a pointofhigher potential energy than the initial potential energy. This is not possible unless energy is added to the system. Thismeans that choice A is an undesirable feature. Many rollercoasters have this feature, but they need to tow the cart to a higherelevation prior to the loop. Having minimal friction on the track prior to the stopping region is a good idea, because thesystem loses minimal energy that way. If all of the potential energy can be converted into kinetic energy, then the ride canachieve greater speeds. This makes choice B a desirable design feature, which eliminates it. Having an aerodynamicpassenger cart is a good idea, because the system loses minimal energy to wind resistance over the course of the ride. Again,if all of the potential energy can be converted into kinetic energy, then the ride can achieve greater speeds. This makeschoice C a desirable design feature, which eliminates it. Incorporation of spring potential energy at the start of the rideallows the ride to be built at a lower starting elevation (because the cart requires less gravitational potential energy). Whilethis may not necessarily seem like a desirable feature, it is not an undesirable feature either. This eliminates choice D andreaffirms choice A as the best answer. The best answer is choice A.

48. Choice D is the best answer. Figure 1 shows that the work done by friction is minimal until the very end of the ride, afterthe loop has been completed. This means that a brake is not applied until after the loop, so choices A and B are eliminated.Figure 2 shows that the work done by friction has a jump in the middle of the ride. This means that a brake is being appliedto slow the cart at some point on Rollercoaster B. The application of the brake occurs prior to the cart reaching its lowestkinetic energy, which is associated with the top of the loop. The best answer is choice D.


49. Choice B is the best answer. Whenever you compare a physical variable in two different situations, you can use the ratiotechnique to solve the problem. Here, we want to know about U, when we are told about r. How do they relate?

50.

51.

®Copyright © by The Berkeley Review1

IJ«I,. ^cci**

173

r Ub 1/rB ta 1

This ratio is the mathematical way of comparing UA to UB, the ratioof the two potential energies. Don't worryabout writingout the other variables or the minus sign, because they drop out. The answer is 3 to 1. The best answer is choice B.

Choice D is the best answer. The first thing we need to do is determine the amount of useful mechanical energy that can beobtained from 1 g of the hydrocarbon fuel. This is simply the product of the heat per gram and the conversion efficiency.

Useful energy = 1 g fuel x 15.2 kJ/g x 0.85 = 15.2 x (1 - 0.85) = 15.2 - 2.3 = 12.9 kJ

If all 12.9 kJ are applied to lifting a 10-kg object, then we can say that the change in potential energy is found as follows:

W = APE .-. 12,900 J = mgAh = 10x9.8 x Ah «98Ah

Ah=12,900/98=12,900/l00=129m

The actual value is slightly greater tban 129 m. The closest answer to 129 m is 132 m. The best answer is choice D.

Choice A is the best answer. 110 kg is being lifted by 4 m while 60 kg is dropping by 4 m. The net result is that 50 kg isbeing lifted by 4 m. Using the equation:

^gravitational = ™°^ =50(9.8)(4) - 50 x 10 X4 = 2000

The real value is slightly less than 2000, making 1960 J the best option of the answer choices. The best answer is choice A.

REVIEW EXAM EXPLANATIONS

52. Choice C is the best answer. How much work is done by gravity as a box is pushed up a 5 m ramp to a new elevation 2 mhigher than theinitial height? The work done bygravity is independent of theforce provided by the people or the pathway.The work done by gravity is positive when an object moves downhill and is negative whenan object moves uphill. In thisproblem, thebox is moved uphill, so thework turns outto benegative. Because the term "against" is used, the value shouldbepositive. The amountof workdone by gravity is given by:

Wg =-APE =-mgh =-(50 kg)(10.0 m/s2)(2.0 m) =- 1000 JChanging the sign topositive, tocorrespond with "against gravity," makes the best answer choice C.Thebestanswer is C.


the

Momentum

and TorquePhysics Chapter 4

t=o t=i t = 2

(ItMomentum and Energy are conserved if the balls are not deformed and do not stick.

by

Berkeley Review

Momentum and TorqueSelected equations, facts, concepts, and shortcuts from this section


(nw)jnitial =(mv)fmal p=mv J=Amv=Faverage * t^iiisionx = F||xr x = rFsin6

© ImportantConcept

Collisions between Balls of Roughly Equal Mass (miviinitiai + m2V2initial = Hiivifmai + m2V2fmal)

vinit _ /-i^ t-.; i ci.i.. /-*Sk/ikVexit.Initial State:©-^-^Q) Final State: Qfc-SV mlVliniUal =(mi +m2)vfinalstationary stick and move together

V* •♦ Vif VofInitial State:@^2!U-Q Final Stole: Q_*^_*. mivlinitial =mivlflnal +m2v2flnal

stationary don't stick and move separately

Initial State:Q—i2ii-^Q Final State: © Q v exit

mlvlinitial = m2v2finalstationary impact ball stops after collision

Initial State:Q » && Final State: X Final State: e<$> X uN% *&/' inelastic I x-direction <&j/ N^,

>^w>^ collision wcancels out S elastic >*X Q© © colIision ©

mlvl initial + m2v2initial = miVlfmal + m2V2yf,nai

Initial State:0^^x-^^Q Final State: +0&-Q X Q-^HU.mlvl initial + m2v2 initial = mlvlfinal + m2v2yfinal

© Strategy for Solving Rotational Equilibrium Questions

(Tleft side =bright side)A large mass ata short distance offsets a small mass at a long distance and theoffsetbar plays just asmall role

xnet =0 implies rotational equilibrium

xleft - Tright

"^weight on left = ^bar heavier on right "*" "^weight on rightM1gr1=MBargrBar + M2gr2

Physics Momentum and Torque

Momentum and TorqueIn Chapter 3, we studied kinetic energy. In this chapter, we shall focus onmomentum and the transfer of momentum and kinetic energy through collisions.In Chapter 2, we studied circular motion and angular velocity and acceleration.In this chapter, we shall address torque, which causes angular acceleration.

Linear Momentum

Momentum is defined as the product of mass and velocity; p = mv. It is a vectorquantity. Any object with a velocity must have momentum (as well as kineticenergy). You may not find momentum itself to be interesting. However, thetransfer of momentum is quite interesting and is often the subject of questions.

Conservation of Linear MomentumThe conservation of linear momentum states that the momentum of a system isconserved. We can define a system to be any part of the universe that we areinterested in considering. For example, let's consider the collision between twomarbles, as shown in Figure 4-1. Each marble has a mass, and they each havesome initial velocity as they approach one another. When the two marbles hiteach other there will be a force from Marble 2 on Marble 1 (F2 on l) and, similarly,there will be a force from Marble 1 on Marble 2 (F\ on2).

Before marbles

collide

During thecollision

After marbles

collide

vl initial

r2onl

Figure 4-1

v2 initial

f m" jv2 final

Consider Marble 1, whose change in momentum is mvignai - mvlinjtia|. Thechange in momentum is due to the interaction of Marble 2 with Marble 1. It is F2on1 exerted over the duration of the collision that leads to the marble's change invelocity, which manifests itself also as a change in momentum. Because theforces exerted and felt by the two marbles are identical, according to Newton'sthird law, and the time of contact is the same for both marbles, they eachexperience the same change in momentum. We can write expressions for theinteractions between Marble 1 and Marble 2 as shown in equation (4.1).

F2 on lAt = miv1finai - mivlinitiai (4.1a)

Fl on2At = m2V2final " m2V2initial (4.1b)

F2onlAt = - F2onlAt, so we can equate the two as shown in equation (4.2).

n*lvlfinal ~ mlvlinitial = - (m2v2final - m2v2initial) (4-2>


Linear Momentum


Physics Momentum and Torque Linear Momentum

The directions of the two forces are opposite of one another, which accounts forthe negative sign. Equation (4.2) can be rewritten into a more useful form, asshown in equation (43).

mivlinitial + m2v2initial = miVifinai + m2V2Bnal (4.3)

Equation (43) describes any generic system involving the collision between twoobjects. It appliesas longas there is no net externalforce actingon the system.

Example 4.1aA sharpshooter fires a gun at a wooden block of mass M. If the bullet, of mass m,becomeslodged inside the block, then which of the following is NOT true?

A. If the bullet is much lighter than the block, the block will not move much.B. If the bullet is much heavier than the block, the block will move at

approximately the impact speed of the bulletC. If the bullet has the same mass as the block, the block will not move after the

collision.

D. The block can never mover faster than the bullefs impact speed, after thecollision.

Solution

This problem isa negation question (i.e., which choice is physically wrong?). Thechoices talk about the speeds of the block and masses of the bullet. Let'sapproach it using physical intuition. If the bullet is negligibly small, it probablywon't affect the block. True. This rules out choice A.

If the bullet is extremely massive comparedto the block, the blockwon't slowthebullet much, and the block will move at nearly the impact speed of the bullet.True. Cross out choice B.

If thebullet andblock areequal in mass, should theblock move at all? Probably.What happens to a person when you collide with him? You both move. Thus,choice C is not physically valid and is the answer. Ifyou convince yourself thatachoice is theright one, donotwaste time by looking overtheremaining choices.


Example 4.1bIf the bullet and the block in Example 4.1a, have the same mass, the final speedof the block is:

A. zero.

B. halfof the impactspeed.C. equal to the impactspeed.D. twice the impactspeed.

Solution

In this scenario, the massof the total system doubles when the bullet embeds inthe block, so we know that the impact speed should decrease. This eliminateschoices C and D. The speed should not drop to zero, so choice A is alsoeliminated. With the mass doubling, it is balanced out by the speed cutting tohalf of its original value. This makes choice B the best answer.

mbulletvbullet initial =(^bullet +Mblock)vSystem final = 2mbulletvSystem final

Vsystem final =(mbullet/2mbullet)Vbullet initial =^vbullet initial



Example 4.2aA northbound cart is moving at 5 m/s when it collides with a southbound cart,moving at 1 m/s. If the northbound cart is twice as heavy as the southboundcart, what is their final velocity after they collide and become stuck together?

A. 2 m/s north

B. 3 m/s north

C. 2 m/s southD. 3 m/s south

Solution

Here is a general approach to momentum conservationproblems:

1. Identify the problem as a candidate for momentum conservation wheneverthe problem involves a collision or an explosion.

2. Draw out, on scratch paper or in your head, the initial and final situationfor the system. Make sure that all of the interacting objects are included,and that there are no external forces acting along the direction of interest.Draw coordinates to indicate +/- directions for the velocities.

3. Find and sum the initial momentum of the system by considering whetheran objectis moving initially. If so, it has momentum.

Pinitial of object = mobjectvinitial of object

4. Repeat Step 3in finding the final momentum ofthe system, pfinal ofsystem-5. Plug everything into equation (4.3) and solve for your unknown.

Both colliding carts are moving initially, giving an initial momentum of:

Pinitial of system = (rnNB-cartVinitial of NB-cart) +(mSB-cartVmitial of SB-cart)

Pinitial = (mNB-cart)(+5 m/s) + (mSB-cart)(-l m/s) = (2m)(+5 m/s) + (m)(-l m/s)

where 2m represents the northbound cart's mass, since it is twice as heavy as thelighter cart, and the "-" sign represents southbound motion. There are no forces,other than their interaction, so we can apply conservation of momentum todetermine the momentum of either cart. Solving for the final momentum gives:

Pfinal of system =(^NB-cart + mSB-cart)vfinal of both carts = (3m)vfinai ofboth cartsBecause momentum is conserved in the absence of an external force, we can setthe initial momentum to the final momentum (like equation (4.3)).

(2m)(+5 m/s) + (m)(-l m/s) = (3m)vfinai of both carts

(10mkg-m/s) - (lm kg-m/s) = 9m kg-m/s = (3m)vfinai 0fbothcarts

9m kg-m/s/ 3m =Vfinal of both ^^ .-. Vfinai 0f both carts =3m/s

The "+" sign means that the carts are northbound after the collision. This isintuitive because the larger cart was initially moving faster and moving north.

With momentum problems, the important predictors of the final velocites arethe objects' relative masses and relative velocities. Here, we did not know thespecific masses of the carts, only their relative masses. It did not matter whetherthey were massive or light—only that one was twice as heavy as the other.Likewise, their relative velocity before collision is all that matters in determiningtheir final velocity. Had the carts been moving at 14 m/s north and 8 m/s northoriginally, they would still have had a relative speed of 6 m/s and the resultingvelocity changes would have been the same. The best answer is choice B.


Linear Momentum

Test TipMomentum Conversion


Physics Momentum and Torque Linear Momentum

Example 4.2bWhen a gun is fired, it recoils a bit as the bullet leaves with a speed of 300 m/s.If a gunman fires a bullet from the gun, which is 100 times as massive as thebullet what is the recoil speed of the gun?

A. 0.3 m/sB. 3 m/sC. 30 m/sD. 300 m/s

Solution

In this scenario, there is initially no velocity for either the bullet or the gun, so theinitial momentum of the system is zero. This is often the case with recoilproblems. Intuitively, we know that the bullet and the gun must share the samemagnitude of momentum, but in the opposite direction. Given that the gun is 100times as massive as the bullet, it would follow that the speed is the bullet is 100times that of the recoiling gun. Intuition tells us that the final speed of the gunshould be 3 m/s, making choiceB the best answer. The math is as follows:

0= mbulletvbullet final+ mgunvgun final •*• nibulletvbullet final ="mgunvgun final

1(^^/mgunHullet final =VlOOVgun final =11100(300 m/s)=3m/s

Vgun final =(mbullet / mgun)vbullet final =11 lOOVbullet final



Impulse

Impulse and Momentum TransferImpulse (J) results from an impact force acting on an object during a collisionthat changes the object7s momentum. During the collision, the object experiencesan impact force (which acts like a normal force) from the other objectwith whichit collides. This impact force accelerates the colliding object and in doing sochanges its velocity, and thereby changes its momentum as well. Suppose weapply an average impact force to a mass for a given amount of time. Newton'ssecond law (F = ma) tells us that this average force will cause that mass toexperience an average acceleration. This means that impulse can be quantifiedusing force, mass, and acceleration.

Recall that in chapter 1 we have defined the average acceleration as shown inequation (1.10):

—_ V2 - vi _ change in velocitya

t2 - ti elapsed time(1.10)

Substitution of the average acceleration a into F = ma gives equation (4.4), whichcan be rearranged to give equation (4.5). Equation (4.5) isolates momentum onone side of the equation, showing that impulse relates to momentum change.Equation (4.5) can be expressed in the more familiar form as shown in equation(4.6):

F=mp^lV t2 - ti

F(t2 - ti) = m(v2 - vi)

FAt = mvfinal - mvinitial

(4.4)

(4.5)

(4.6)

The left-hand side of equation (4.6) is called the impulse (of the force), and is theproduct of the average force and the impact time. The right-hand side ofequation (4.6) is the change in the object's momentum. Because the impulseequals the change in momentum, the dimensions of impulse ([MLT~2] x [T] =[MLT"1]) are the same as those of momentum ([M] x [LT"1] = [MLT"1]).

In collisions involving two objects where momentum is conserved, both objectsexperience the exact same magnitude of impulse. As a result, lighter objectsgenerally experience a greater change in velocity than heavier objects. Thisrelationship can prove useful when determining the change in momentum forone object if you know the change in momentum for the other object involved inthe collision.

Example 4.3aA driver and a passenger of equal massare in a carwhen it crashes into a garagedoor. The driver has an airbag and a seat belt on, while the passenger hasneither. What could account for the passenger's more severe injuries?

A. The passenger has a smaller impact time than the driver.B. The passenger has a bigger impact time than the driver.C. The passenger feels a bigger impulse than the driver.D. The passenger feels a smaller impulse than the driver.


Impulse


FhysicS Momentum and Torque Impulse

Solution

Injuries occur when a person experiencesa large, blunt force. We are looking todetermine why the passenger experiences a greater force during the collision.The choicesdiscuss time and impulse, so this is an impulse problem.

Fon objection object= TOobjectAVobject

Both people have the same mass and the same change in speed, since they are inthe same crashing car. Thus, they experience the same magnitude of impulse.This rules out choicesC and D. However, the one who takes longer to slow downwill experience a smaller average force during the crash. This is the driver, withhis cushioning airbag and seat belt. The passenger has a shorter impact time anda consequently larger impact force.


Example 4.3bWhy do racecar drivers seem to walk away unharmed from horrendous highspeed crashes when their car tumbles repeatedly?

A. The car experiences a smallerchange in momentum when it collides multipletimes before coming to rest rather than when it abruptly comes to rest

B. The car experiences a smaller impulse when it collidesmultiple timesbeforecoming to rest rather than when it abruptly comes to rest.

C. The car experiences a greater average force when it collidesmultiple timesbefore coming to rest rather than when it abruptly comes to rest.

D. The car experiences a longer total collisiontime when it collidesmultipletimes beforecomingto rest rather than when it abruptly comes to rest.

Solution

If you've ever witnessed(mostlikely via television) a hair-raisingcollision wherethe racecar flips and tumbles several timesbeforeskidding to rest withouthittinga wall, then you no doubt seem amazed that more often than not the driverwalks away seemingly unharmed.Theanswer can be found using equation (4.3).Injuries are the result of the force experienced by the driver. In a crash thatinvolvesrepeated tumbling, the momentum is lost in small piecesrather than allat once. This converts the total impulse into smaller fragments of impulse,reducing the force acting on the driver during each individual collision duringthe tumbling to rest process. The collision is spread over more time, so choice Dexplains this best.

Even if you didn't visualize this scenario, you could have eliminated choices Aand B, because impulse and momentum change are equal to one another. If oneis smaller, then the other one would also have to be smaller. Whether the carcomes to rest abruptly or extended over time, it experiences the same change inmomentum and thus the same total impulse. Choice C is invalid and choice D isvalid, because multiple collisions increase the total collision time, which therebyreduces the average force.




Types of CollisionsBecause so much of the study of momentum involves the transfer of momentumvia collision, we shall consider common types of two-object collisions. Figure 4-2shows the first type ofcollision we shall consider, the collision ofa small movingobject with a large stationary barrier. We generally don't consider the impact onthe massive barrier and focus juston thesmaller object's change in momentum.

Case 1:

(perpendicular to barrier)Case 2:

(angled to barrier)

Omit o

ovrebound ***"''<U«

.

Figure 4-2

In Case 1, the momentum change is greater than it is in Case 2, because thevelocity is complete in the direction perpendicular to the barrier, so nocomponent of the velocity is retained. In Case 2, because of the angle at whichthe ball strikes the barrier, it retains the downward component ofitsvelocity, soitschange in momentum is just in its perpendicular component.

The second type of collision we shall consider involves a moving object with astationary object of comparable mass. The objects after collision could sticktogether (which occurs with an inelastic collision), could completely transfermomentum resulting in the first object coming to rest, or they could collide insuch a way that the incoming object recoils after collision. Figure 4-3 shows allthree cases for this type of collision.

Initial Stat' • (f)-

Outcomes:

Case 3:

Case 4:

vinitial

initiallystationary

•*•©

'exit

(stick and move together after collision)

stationary

CD'exit

(impact ball stops after collision)

Case 5:^rebound ^n

®'exit

(impact ball rebounds after collision)

Figure 4-4


Impulse


Physics Momentum and Torque Impulse

Using conservation ofmomentum, wecansolve for the final velocities for bothofthe balls in the collisions in Cases 3,4, and 5.

Case 3 (inelastic collision):

mrvi initial = (mi + mil)vfinal

vn is small, impulse is small, and Ball I retains some of its initial momentum

Case 4 (elastic collision if mj = mrj):

mm initial = mnvn final

vn is medium, impulse is medium, and Ball I loses all of its initial momentum

Case 5 (only observed when mi« mn):

nuyi initial = nuyI final + mrjvil final

\q is negative, vrj is positive,BallI transfers all of its initialmomentum and gains some rebounding momentum

The third type of collision we shall consider involves a moving object withanother moving object of comparable mass.The objects after collision could sticktogether (which occurs with an inelastic collision) or they could collide in such away that the two objects recoil after collision (which occurs with an elasticcollision). Figure 4-5 shows the two extreme cases for this type of collision,perfectly inelastic and perfectly elastic.

Case 6:Resultantvector(s) depends on the collision (elasticor inelastic).

-©

inelastic collision XsP elastic collision(kinetic energy not conserved) (kineticenergy conserved)

Case 7: Whether they interchange velocities (elastic) orcome to rest (inelastic)depends on the collision.

/•5V VI initial ^ ^ ^VII initial /j> s?\ VI initial ^ ^ ^VH initial

v, final =0©@vnanal =0 ^ . pfinal x vnflnai f

inelastic collision elastic collision

(kineticenergy not conserved) (kinetic energy conserved)

Figure 4-5

Elastic collisions by definition have conservation of energy. As such, the objectsdonot deform as a result of the collision. The objectsbounce hard off of one another.

Inelastic collisions by definition do not exhibit conservationof energy.As such, theobjects undergo deformation as a result of the collision. The objects often sticktogether after the collision. At the very least, they slow down after collision.


I hyS1CS Momentum and Torque

Example 4.4aA 0.5-kg ball traveling at 10 m/s collides with a stationary 2.0-kg ball andrebounds in the opposite direction at 6 m/s. What is true ofthe speed of the 2.0-kg ball after collision?

A. The 2.0-kgball moves at 10 m/s.B. The 2.0-kg ball moves at 6 m/s.C. The 2.0-kg ball moves at 4 m/s.D. The 2.0-kg ball moves at 2.5 m/s.

Solution

This isa collision like the one seen in Case 5,where the impact ball rebounds offof an initially stationary ball. Intuitively we know that the 2.0-kg ball will bemoving faster than 2.5 m/s, thespeed it would have attained had the impact ballcome to rest. Because the impact ball rebounds, the 2.0-kg ball gets all of thetransferred momentum plus some recoil momentum. The recoil momentum issmall, its exit speed should be justa little greater than 2.5 m/s, making choice Cthe most probable answer. To solve precisely, we can apply the following math:

miight baliviight ball initial = flight balFUght ball final + mheavy balivheavy ball final

(0.5 kg)(10 m/s) =(0.5 kg)(-6 m/s) +(2.0 kg)vheavy ball final

(5 kg-m/ s) =(-3 kg-m/ s) +(2.0 kg)vheavy ball final

8kg-m/s = (2.0 kg)vheavy ball final

vheavy ball final =8k8*m/s/2kg =4m/sA good compromise between intuition and math is to note that the ratio of theobject's masses must equal the ratio of their changes in velocity. The incomingballhasa massof 0.5 kg and its changein velocity is 16m/s. This meansthat the2.0-kg ball must experience a change in velocity of 4 m/s. Staring at rest andundergoing a 4 m/s change in velocity resultsin a final velocity of4 m/s.


Example 4.4bIna head-on collision between two identical balls each traveling at a speedof20m/s in opposite directions results in both rebounding with a speed of 10 m/s.What is true of this collision?

A. Momentum was not conserved during the collision.B. The balls experience different magnitudes ofimpulse during thecollision.C. The collision is elastic.

D. The collision is inelastic.

Solution

The initial and final momenta for the system are both zero (the two balls haveequal momenta in opposite directions before and after collision. Choice A iseliminated. Both balls have identical magnitudes of momentumbefore and afterthe collision, so the impulse must be equal for both objects. Choice B iseliminated. The balls lose speed, so they lostkinetic energy during the collision.This makes the collision inelastic, rather than elastic.


Impulse


Physics Momentum and Torque Torque and Equilibrium

Torque and Equilibrium

Torque and Lever Arms (Moment Arms)There is one important quantity that we must consider when looking at a rigidbody, and that is torque (x). A torque is simply a measurement of the ability ofaforce to cause rotation about a given pivot point. Torque is usually measured inunits of Newton-meter. The point about which an object rotates is referred to asthe fulcrum. Consider Figure 4-6, in which a wrench is loosening a bolt. Notethat as the hand pushes down on the wrench, a force is being applied that isperpendicular to the wrench.

Figure 4-6

In this particular case, the torque (x) is proportional to the magnitude of theapplied force (F) and the distance between the center of the bolt and the pointatwhich the force acts (r). This length is called the lever arm or moment arm (ofthe force), and the center of the bolt is called the pivot point. It is the point aboutwhich the bolt will turn. So how do F, r, and x relate? From experience, weknow that pushing harder loosens the bolt readily, so x « F±. We also know thatthe farther down the handle we push, it is easier to loosen the bolt, so x « r.Mathematically, this looks like:

x = rF± (4.7)

Now, what if the force is not perpendicular to r? For this particular topic, you arebest served by considering your real life experiences to understand the concepts.Consider what happens when you push on the outer edge of a swinging door(Figure 4-7) with a force F. The door swings counter-clockwise, seen from thisoverhead vantage point. How big is the resulting torque? Is it just rF, or is itsmaller or larger than rF? Think about the components of this force and how youinstinctively push on a swinging door. Which component, F|( or F^, will cause therotation? It is caused by the component perpendicular to r (perpendicular to thesurface of the door). The parallel component pushes the door in towards thehinge, which is waste force in terms of turning the door. If you don't push in aperpendicular fashion, then the torque will be smaller than rF.

path of door's edgeN. i

Fapplied^hinge (pivot point) swinging door

(top view)

Figure 4-7



If the applied force is parallel to r, then there is no torque about the hinge. If theapplied force is perpendicular to r, then the torque will be maximal. From yourlife's experiences, you have probably learned that if you ever have open a dooror make anything turn, you instinctively do two things to make your task easier:Push as far away from the pivot point as you can, and push as close toperpendicular to r as you can. Because the only component of force thatgenerates a torque is perpendicular, the force component of interest is FsinG. Thisleads to equation (4.8).

x = rFsinG (4.8)

Example 4.5aAbrother and sister fight over whether a door should be opened or stay closed.They both push with the same magnitude of force, perpendicular to the door, asshown. Which sibling exerts more torque on the door, with respect to the hinge?

Top view

wall hinge

Fsister handle

brother

A. The sister.B. The brother.

C. They both exert the same torque.D. It depends upon the length of the door.

Solution

Torque has to do with both force and distance (i.e., distance from the pivotpoint). Here, the forces are the same in magnitude, and both are directedperpendicular to the door. Thus, the sibling pushing at a bigger distance from thehinge (our pivot point) is the sibling exerting more torque on the door. This is thebrother. Note that in comparison problems, fixed values, such as the door'slength, never enter into consideration and should therefore be invalidated aspossible choices.


Example 4.5bWhy is the top of a bolt wider than the shaft of the bolt?

A. To ensure that the force is applied parallel to the top of thebolt.B. To minimize the torque on the bolt.C. To minimize the moment arm on the bolt.D. To maximize the moment arm on the bolt.

Solution

A bolt gets threaded in and out of a nut, so the goal of the wider head is tomaximize the torque acting on the bolt when threading it into a nut. Themaximum torque occurs when the momentum arm is maximized, not minimizedan the force if perpendicular to the moment arm. Choices A, B, and C are allinvalid.






Rotational EquilibriumIn order for a rigid body to be at rest, both its translational acceleration androtational acceleration must,be equal to zero. Recall that a basic definition fortranslational motion is the movement of an object from one position to anotherposition, while the basic definition for rotational motion is the rotation of anobject about an axis.

Rotational equilibrium is defined as a system experiencing no rotationalacceleration, which implies that x = 0. Just as a rigid body moving in a straightline at constant speed hasa linear acceleration ofzero, a spinning rigid bodythatrotates at constantangular velocity is said to be in rotational equilibrium.

There are two important conditions that must apply for a rigid body to be inequilibrium. The first condition is that the netforce ona body must bezero. Thiscanbe expressed as shown in equation (4.9). The second condition is that thenettorqueon a bodymustalso be zero. Thiscanbe expressed in equation (4.10).

£F = Fi + F2 + F3 + F4 + F5 + ... = 0

£x = xi +T2 + T3+X4 +X5 + ... = 0

(4.9)

(4.10)

If we use the equilibrium conditions shown in equations (4.9) and (4.10), thenchoosing an axis for calculating a torque is entirely at our discretion. We canillustrate this with an example.

Example 4.6aA beampositioned on a fulcrum is6 m long. It has a weight of300 N attached toone end and a weight of250 N attached to the other end?

h*- •Xm- -H6.0m

1 1

A250 N A 300 N

Fulcrum)

To establish mechanical equilibrium, the fulcrum must be placed:

A. 2.87m from the 250-N weight.B. 3.00m from the 250-Nweight.C. 3.27m from the 250-Nweight.D. 5.67m from the 250-Nweight.

SolutionLefs start by applying our two conditions of equilibrium: £F = 0 and £x = 0.

Fnet = 0, so we will ignore forces and only consider the torques acting on the bar.As a point of reference, the fulcrum must exert a normal force of 550N upwards,to cancelout the two weightswhich collective have a total weight of550N.

Method #1Let us calculate the torques through the point where the fulcrum contacts theboard. We have two torques to consider, one for the 250 N weight and one forthe 300 N weight. Thetorque about the 250 N weight is: X250N = (W)r = (250 N)X.

The torqueaboutthe300 N weightis: x30on = (W)r = (300 N)(6- X).



Applying our conditions for rotational equilibriumwe find that:

Et=T250N+'«:300N= 0

It = (250 N)X - (300N)( 6 - X) = 0

250XN -1800 N-m + 300X N = 0 .-. 550XN = 1800 N-m

x=1800N.m/550N =180/55m =36/lim=33/nm+3/lim!S33inThe correct choice is C.

Method #2

Instead of choosing our axis through the fulcrum to calculate the torques, let'schoose it through the smaller weight. Once again, the net torque must be zero,although the torques now include the fulcrum and the 300-N weight. We get

LT = Tfulcrum + T300N = 0

I> = FnormalX - (300 N)( 6 - X) = 0

Recall that we calculated the normal force as 550 N. Making the appropriatesubstitution to solve for the distance, we find:

(550N)X-(300N)(6-X) = 0

550X N -1800 N-m = 0 .-. 550X N = 1800 N-m

x=1800Nm/550N =180/55m =36/lim=33/lim+3/lim=3.3m

This leads to the same solution of about 3.3 m. However, both methods involvemath and are thus time-consuming. Whenever possible, in the interest of savingtime, we should try toanswer thequestion conceptually.

Limiting Cases: Often, intuitively thinking about limiting cases can lead you toananswer without the need for much math. Initially, the distance X might beunknown, but what might it be if the two masses were equal in weight? We'dguess that the fulcrum wouldbe directly between the twomasses, because of itssymmetry, and thatXwould equal 3.00 m. Ifone mass isbigger than theother,we would expect the fulcrum to be closer to the bigger mass. This rules outchoices Aand B. Since 300 Nisnot much bigger than 250 N, we'd further expectthat the fulcrum should not be much closer to the 300N mass than it is to the 250N mass. This leads to choice C over choice D.

Example 4.6bFor the apparatus in Example 4.6a, what should be done to maintain balance asthe 250 N weight is moved towards the fulcrum?

A. Mass should beadded to the 300-N weight.B. Mass should be removed from the250-N weight.C. The300-N weight should be moved towards the fulcrum.D. The 300-N weight shouldbe moved away from thefulcrum.

Solution

Moving the 250-N weight towards thefulcrum reduces the torque onthe left sideof the fulcrum. To maintain balance (rotational equilibrium), the torque on theright side must be reduced equally. This is accomplished be either removingmass from the right side or moving the300-N weight closer to the fulcrum.


Copyright©by The Berkeley Review 189


Test TipLimiting Cases



Like the inclined planes and pulley systems we saw in Section II, the lever isanother example of a mechanically advantageous machine. Levers give us a powerto apply forces much greater than we normally would beable to.

Consider the rigid rod resting on the fulcrum in Figure 4-8. At one end of thisrod we have placed a weight. We apply a downward force at the other end of therod. The force that is exerted is very much less than the weight that is beingsupported, if the distance from the applied force to the fulcrum is much greaterthan the distance from the fulcrum to the weight.

'fulcrum •*+*• 'weight

Applied force

¥W

Fulcrum

Figure 4-8

The weight that is lifted, divided by the force being supplied, is equal to themechanical advantage of the system. This is shown in equation (4.11), below:

Mechanical Advantage =weight of object _dfuk

applied force needed to supportobject dweighi(4.11)

Example 4.7Many of the trails that you find at Yosemite National Park were constructed bythe Civil Conservation Corps, an organization established under PresidentRoosevelt's administration. One of the ways that they were able to move chunksof large granite was to make use of a fulcrum. Suppose a 2.5-m steel bar isplaced undera rock weighing 3000 N. A fulcrum is set 0.5 m away from the rockto be moved. What force must be exerted on the other end of the bar to move therock? (Neglect the mass of the bar.)

\+ 2.5 m

5 Applied force

A. 600 N

B. 750 N

C. 3000 N

D. 12000 N

SolutionWe can determine the necessary force by setting the two torques equal to oneanother: Fappiied(2 m) = Fweight(0.5 m).

F_(W)(dweight) _ (3000 N) (0.5 m) _ 75QNdfulcrum 2.0 m

The applied force is equal to 750 N, one-fourth of the weight of the rock. Thismeans that the mechanical advantage is 4, which happens to be rappiied : rrock-



25 Momentum and Torque Review Questions

I. Conservation of Linear Momentum

II. Masses and Torques

III. Clay Ball Collisions


(1-7)

(8 -14)

(15 - 21)

(22 - 25)

The main purpose of this 25-question set is to serve as a review of the materialpresented in the chapter. Do not worry about the timing for these questions.Focus on learning. Once you complete these questions, grade them using theanswer key. Forany question you missed, repeat it and write down your thoughtprocess. Then grade the questions you repeated and thoroughly read the answerexplanation. Compare your thought process to the answer explanation and assesswhether you missed the question because of a careless error (such as misreadingthe question), because of an error in reasoning, or because you were lackinginformation. Your goal is to fill in any informational gaps and solidify yourreasoning before you begin your practice exam for this section. Preparing for theMCAT is best done in stages. This first stage is meant to help you evaluate howwell you know this subject matter.


The conservation of linear momentum is one of the most

important principles of modern physics. The law ofconservation of momentum states that if no net external force

acts on a system, then the linear momentum of that system isconserved.

In an experiment involving conservation of momentum,a 10-g lead bullet with an initial speed of V0 = 200 m/s isfired into a 1-kg block of wood at rest on a horizontalfrictionless table of height D above the base of the table. Thebullet embeds itself in the block and together the bullet/blockslide off the table and land a distance R away from the baseof the table. Figure 1 shows a schematic of the set up.

1.

2.

3.

BulletBlock

D

•R-

Figure 1

How long is the bullet/block combination in the air afterthey leave the table?

A. J 2DV e

B. A / O

What is the speed of the bullet/block immediately afterthe bullet embeds itself in the block?

A. 20 m/s

B. 2 m/s

C. 0.2 m/s

D. 0.02 m/s

All EXCEPT which of the following will lead to asmaller final block speed after the collision occurs?

A. Giving the bullet a smaller initial speed.

B. Using an heavier block.

C. Using a lighter bullet.

D. Having an elastic, as opposed to an inelastic,collision.


Which of the following graphs BEST represents therelationship between the bullet mass and the final speedof the block, after the collision? (Note: 0 is the origin.)

A. B.

C.

oucoo

a

E

Bullet mass

Bullet mass

Bullet mass

Bullet mass

While the bullet/block is airborne, the block's kinetic

energy:

A. and momentum increase.

B. increases, but its momentum stays constant.

C. stays constant, but its momentum increases.

D. and momentum remain constant.

The energy transfer that occurs as the bullet/block slideacross the tabletop, if it were no longer frictionless,could best be described as from kinetic energy:

A. to heat.

B. to heat and gravitational potential energy.C. and heat to gravitational potential energy.D. and gravitational potential energy to heat and

kinetic energy.

Which of the following will increase the range R of theblock?

A. Increasing the initial bullet speed and decreasingthe table height.

B. Increasing either the initial bullet speed or the tableheight.

C. Decreasing either the initial bullet speed or thetable height.

D. Decreasing the bullet mass and increasing theblock mass.



A laboratory pan balance is constructed as shown inFigure 1.

40 cm-

•10cm »|« Xcm-»-|

& counterweight rod

pan

Figure 1. Laboratory balance

The rod has a mass of 0.5 kg. The counterweight has amass of 0.01 kg. With nothing on the pan, the rod is balancedat x = 1 cm. As low-massitems such as chemical samplesareadded to the pan, the counterweight can be moved to balanceout the torques and in doing so determine the weight of theitem added to the pan.

8.

9.

10.

What is the mass of the pan?

A. 0.1kgB. 0.5 kg

C. 1.0 kg

D. 5.0 kg

In general, when calculating the mass of the pan,neglecting the mass of the rod:

A. is possible, if the fulcrum is placed at the center ofmass of the whole balance.

B. is possible, if the center of mass of the rod is thepivot point

C. is not possible under any circumstances.D. is possible, if the chemical mass is larger than the

mass of the rod.

Based on the picture in Figure 1, when calculating themass of the pan, the fulcrum exerts:

A. no force on the rod, so the fulcrum can be ignored.B. a force on the rod, but no torque.

C. a torque on the rod, but no force.D. a force and a torque on the rod.


11. If an amount of chemical is now added to the pan, inorder to balance the rod, what happens to x?

A. It increases.

B. It decreases.

C. It remains the same.

D. Whether x increases or decreases depends on theamountof the chemical added to the pan.

12.

13.

14.

If there is no chemical in the pan, and thecounterweightis now moved from 1 cm to 5 cm, then the rod would:

A. no longer be in translational or rotationalequilibrium.

B. be in translational equilibrium, but not in rotationalequilibrium.

C. be in rotational equilibrium, but not in translationalequilibrium.

D. still be in both translational and rotationalequilibrium.

If the net force and net torque on an object are bothzero, what can you conclude?

A. The object will necessarily be at rest.B. The object may be in motion.

C. The object is necessarily in circular motion.D. Net force and net torque cannot simultaneously be

zero.

If the net force on an object is zero, what can youconclude?

A. The object is in translational equilibrium only.B. The object is in rotational equilibrium only.C. The object is in both translational and rotational

equilibrium.

D. The object is stationary.



Some Berkeley Review students conducted anexperiment where an array of clay balls were fired at eachother from many different locations at various speeds. Theclay balls had different masses as given below.

Clay Ball Mass (kg) Clay Ball Mass (kg)

I 0.100 III 0.200

II 0.150 IV 0.100

Several trials were performed where the balls in thisexperiment were fired from different positions using springlaunchers. Figure1 is a schematic of the differentpointsfromwhich the balls were fired (A, B, C, D, E, and F), and thecentral point at which each ball was aimed (X).

B0 = 30°

0 = 30° >X<

D

6 = 30°

"e"=3o"°

Figure 1.

All of the firing points are exactly 1.0 meters from pointX, and points A, X, and F are collinear and points C, X, andD are collinear. Additionally, the angles made by connectingthe points AXB and BXC are each 30 degrees.

15.

16.

In one trial Ball I was fired from point C, while Ball IIIwas fired from point E with the same initial kineticenergy. What did the path of the resultant look like?

TResultant Resultant /X-*—III ^F-X-^-m

II

C.

\ResultantX-*—III

D.

*^ Ill

No Collision

In a straight-line double collision first between amoving Ball I with a stationary Ball III, where Ball Icomes to rest and then between a moving Ball HI with astationary Ball IV, where Ball III comes to rest, what isobserved? (Assume no energy loss to friction.)

A.

B.

C.

D.

The final speed is greater than the initial speed.The final velocity is the same as the initial velocity.The final velocity is less than the initial velocity.The final momentum is less than the initial

momentum.


17.

18.

19.

20.

21.

In another trial, Ball II was fired from point A, whileBall IV was fired from point D. If both Ball II and BallIV had the same initial speed, what would the path ofthe resultant look like?

A. n. .iv B.II

C. II

IV

Resultant

X

Resultant\

IV

II

D.II <

X

Resultant/•

IV

IV

No Collision

What conditions are necessary to guarantee that if twoseparate balls are released from any two of the sixrelease points that they will collide at X?

A. The balls must have the same mass.

B. The balls must have the same initial speed.

C. The balls must have the same initial momentum.

D. The balls must have the same initial kinetic energy.

How would the results change if the objects were madeof compact material that did not deform, when equalmass balls are fired from positions A and F?

A. The balls would come to rest upon collision.

B. The balls would rebound directly backward withthe same speed.

C. The balls would rebound at 60° angles with thesame speed.

D. The balls would rebound directly backward withhalf of their original speed.

If the experiment were done using super rubber ballsinstead of clay balls, what quantities would beconserved in collisions?

A. Kinetic energy only

B. Momentum only

C. Kinetic energy and momentum

D. Kinetic energy, momentum, and torque

If Balls I and II were fired from points B and E,respectively, with the same momentum, they would:

A. collide at point X.B. collide at a point to the left of point X.

C. collide at a point to the right of point X.

D. not collide with one another.


Questions 22through 25 are NOT based onadescriptivepassage.

22. A ladder leaning against a wall makes an angle of 60°with thehorizontal, as shown below. A person needs toclimb this ladder.

23.

///////////

The ladder is morelikely to slip when the person:A. just steps on the very bottom end of the ladder.

B. stands on the ladder near the bottom end.

C. stands on the ladder near the middle.

D. standson the ladder near the top end.

When someone who is standing still on a skateboardstarts to walk forward, what happens to the skateboard?

A. It will remain stationary.

B. It will moveforward with the person.C. It will move in the direction opposite to person's

motion.

D. It will first move forward and then move

backwards.

24. A large truck, travelingdue south with speed v, collideswith a small car traveling due north with the samespeed v. After the collision, the car is moving due east,with the same speed it had prior to the collision. Whatcan you conclude about the final velocity of the truck?

A. It points southwest

B. It points southeast.

C. It points south.

D. It points west.


25. Consider a uniform circular platform mounted on afulcrum at its center of mass. Under which conditionwill angular momentum conservation of a platformNOT apply?

A. When theplatform has zero angular velocityB. When the platform has a constant, nonzero angular

velocity

C. When there areexternal torques on theplatformD. When there areinternal torques ontheplatform

1. A 2. B 3. D 4. A 5. A

6. A 7. B 8 B 9. B 10. B

11. A 12. A 13. B 14. A 15. D

16. B 17. C 18. B 19. B 20. C

21. C 22. D 23. C 24. A 25. C

YOU ARE DONE.

Answers to 25-Question Momentum and Torque Review

Passage I (Questions 1 - 7) Conservation of Linear Momentum

1.

5.

Choice A is the best answer. How long is the bullet/block combination in the air after it leaves the table? After thebullet/block leaves the table, it goes intoprojectile motion. Thiscalls for thekinematics equations. To find the time of flight,use:

Ay =voyt +iat22

Thebullet/block leaves thetable horizontally, which means voy = 0. This leaves us with:

Ay =Viat2 c> D=̂ gt2 .-. t=̂This question can also be answered by considering units. Choices B, C, and D do not have units of time. Another way tonarrow the choices is to consider limiting cases. If, for example, we increase the height D of the table, what should happen tothe time? It should increase, ruling out choices B and D. Units and limiting cases are typically faster than using algebra. Usewhatever is fastest, but still effective, for you. You'll have roughly 1.3 minutes per question on the physical sciences sectionof MCAT. The best answer is choice A.

2D

Choice B is the best answer. What is the speed of the bullet/block immediately after the bullet embeds itself in the block?This has conservation of momentum written all over it, because it is a perfectly inelastic collision. Conservation ofmomentum gives:

mV0 = (m + M)V

where m = mass of the bullet, M = mass of the block, VQ = initial speed of the bullet, V = speed of bullet/block after

collision.

y- mVp _(0.01)(200)„2-2m/cm + M 0.01 + 1.0 1

Any time there is a collision, your first thought should be conservation of momentum. Everything else is secondary! Thebest answer is choice B.

Choice D is the best answer. Even if you don't know the difference between an elastic and an inelastic collision, you cananswer this question using intuition and elimination: The block gets its speed from colliding with the bullet. A slower bulletwould not hit the block as hard, therefore giving the block a slower speed. This rules out choice A, since we are looking forthe choice that will lead to a constant or larger value for the block's final speed. Choice B is wrong, because a heavier blockwould not be as affected by the bullet as would a lighter block. Choice C is wrong, because a lighter bullet would not hit theblock as hard as a heavier bullet. This leaves choice D. An elastic collision does mean that no energy is lost to heat ordeformation during the collision, but it means even more if you consider this: Momentum is conserved in this collision, sothe bullet and block experience equal and opposite impulses (i.e., changes in momentum). If the bullet bounces, it has abigger change in its momentum than if it sticks in the block. Since the block has an equally big change in its momentum, theblock gets more momentum, and thus speed, when the bullet bounces. Thus, elastic collisions generally impart a biggerchange in momentum to the colliding objects than inelastic collisions do. The best answer is choice D.

Choice A is the best answer. You can answer this, if you know the formula that describes the collision and know how toplot it. However, graph-identification problems are easier, if you intuitively consider limiting cases. As a first limit, whatshould the final speed of the blocks be, if the bullet had no mass? Zero. This rules out choices B and C. Next, if we usedprogressively more massive bullets, what should happen to the final speed? Eventually, it should be very close to the initialspeed of the bullet That is, the final speed of the bullet would reach a maximum limit, as the mass of the bullet increased.The best answer is choice A.

Choice A is the best answer. Kinetic energy and momentum are defined as:

KE = '/imv2 and p= mv

The block speeds up as gravity pulls it downward. The mass remains constant, since it is the same block during its airborneflight. Thus, both the kinetic energy and momentum increase. If you had considered the energy and/or momentum to beconserved, you might have chosen B, C, or D. Remember, energy conservation takes all forms of energy into account. Theblock also has gravitational potential energy. KE + PE as a whole for the block may be effectively constant, but that doesn'tmean the KE alone has to be constant As for momentum, there must be no net external forces on the block, if its momentumis to be conserved. Here, gravity is the net external force, precluding momentum conservation. The best answer is choice A.


7.

Choice A is the best answer. The energy transfer that occurs as the bullet/block slides across the tabletop, if it were nolonger frictionless, would involve conversion of kinetic energy into another form of energy. As long as friction is involved,heat will always be an end product. Based on that we can eliminate choices B and C. Since the tabletop is flat, thegravitational potential energy does not change, thus eliminating choice D. You could have eliminated choices B, C, and D,because the gravitational potential energy is constant for this problem. The energy transfer that occurs is from kinetic energy(of the bullet/block) to heat The best answer is choice A.

Choice B is the best answer. Here is the algebraic way to approach the question. According to kinematics, the range Risgiven by:

R = Vxt

The speed in the x-direction is the speed of the bullet/block after collision, which is given by:

v _mVo_

The range R is related to the speed of the bullet through:m + M

p_ mV0tm + M

This equation can be used to rule out all choices except choice C. This problem could be answered more easily byconsidering limiting cases, which is a good way to approach any problem of this nature. First, if a faster bullet were used, theblock would fly farther off the table. This makes choices A and B likely candidates. Now, choice B must be the better ofthese two, because it has an "or" in it, meaning that only one part of that sentence need be true for the sentence to be correct.You aredone; don'twaste time thinking abouttheotherchoices. However, if you have time to waste and if it makes youfeelmore certain you havemade the rightchoice, then consider the other choices: Choice A is wrong, because it effectively saysthat a lower table gives rise to a greater range. Choice C is wrong for the above-mentioned reasons. Choice D is wrong,because a smaller bullet or heavier block would lead to a smaller velocity after the bullet and block collide. This wouldreduce the horizontal range from the table. Choice B calls the table height issue correctly. The best answer is choice B.

Passage II (Questions cS - 14) Masses and Torques

8. Choice B is the best answer. With the rod balanced, the net torque on the rod is zero, so the torque on the left side is equalto the torque on the right side. The picture below shows the balanced system with an empty pan. You may note that thecounter-weight is trivial in terms of mass and moment arm relative to the pan and rod, so the empty pan is essentiallybalanced by the asymmetric positioning of the rod on the fulcrum. Both the empty pan and rod's center of mass are 10 cmfrom the fulcrum, so the system is essentially balanced if the pan has the same mass as the rod. This means the best answer isthe one that is closest to the rod's mass, 0.50 kg.

We can verify the answer we chose using visualization by applying torque math. We simply need to set all of the magnitudesof torque on the right side equal to the magnitude of torque on tiie left side. Choosing the fulcrum as the pivot point, wecalculate the net torque about this point. Note that the weight of the rod exerts a torque on the rod. The rod's center of massis 10 cm away from the fulcrum. Let the mass of the pan be m. Then:

(10 cm)mg = (1 cm)(0.01 kg)(g) + (10 cm)(0.5 kg)(g)

Solving for m we get:

m_(10 cm)(0.5 kg) +(1 cmXO.Ol kg) _^ m=05kg10 cm

This confirms that the mass of the pan must be essentially equal to the mass of the rod. You may note that this particularsystem can only measure small masses, because the counterweight is so light The best answer is choice B.


9. Choice B is the best answer. The weight of the rod exerts a force on the rod. Theonly way this force would not cause therod to rotate is if the centerof mass were chosen as thepivot point In thatcase, therewould be no leverarm,so there wouldbe no torque. Choice A is incorrect: Moving the fulcrum to the center of mass of the balance does not mean we can ignorethe weight of the rod when calculating torque, unless we know that this pivot point is the at rod's center of mass. Choice Cisincorrect: We have mentioned the only circumstances in which we can ignore the mass of the rod. Choice D is incorrect:Neglecting the mass of the rod is not dependent on the mass of the chemical. Whether we can neglect the weight of the roddepends on where we select the pivot point. The pivot point would need to be at the center of mass of the rod. The bestanswer is choice B.

10. Choice B is the best answer. The fulcrum exerts a normal force on the rod. However, based on Figure 1, the fulcrum is alsothe pivot point. Therefore, the force the fulcrum exerts on the rod would not produce a torque (since there is no lever arm).The best answer is choice B.

11. Choice A is the best answer. Adding an amountof chemical to the pan wouldcause the pan to tip in the direction in whichthechemical is poured. That is, the rod would rotate counterclockwise upon the addition of the chemical to the left side(intothe panon the left side). To balance the rod, we must increase the clockwise torque. To do this, we can increase either theforce acting on the rightside or increase the length of the leverarm. Wecannotchange the mass of the bar or the massof thecounterweight, so we can only balance the torques by moving thecounterweight. We can movethe counterweight away fromthe fulcrum, so x will increase. The best answer is choice A.

12. Choice A is the best answer. We are told that the rod is balanced when the counterweight is at x = 1 cm. This means there isno net force or torque on the rod at that point Moving the counterweight to x = 5 cm would unbalance the system byincreasing the torque on the right side of the bar. Because the rod would start to rotate about its fulcrum in a clockwisedirection, there must be a net angular acceleration and, therefore, a net torque on the rod. This mean the rod is not inrotational equilibrium. This tells us that choices C and D are incorrectand should thereby be eliminated. As for translationalequilibrium, the rod experiences forces exerted by the pan, counterweight gravity, and the fulcrum. Because the hangingmasses now accelerate, they cannot pull on the rod with the same force as when x = 1 cm. There can now be a net force onthe rod's center of mass, removing translational equilibrium. In other words, because the center of mass for the entire systemis no longer on the fulcrum, there is a net force acting on the rod in the downward direction. The best answer is choice A.

13. Choice B is the best answer. If the net torque and net force on an object are zero, then the most we can conclude is that theobject is not accelerating. This means that the object can be either at rest or moving with a constant velocity, whicheliminates choice A and makes choice B the best answer. When an object is moving in a circle, it has a net centripetal forceacting on it. Because we are told that the net force and net torque acting on the object are both zero, choice C is eliminated.Since the net force and net torque can be zero simultaneously-this is a requirement for equilibrium-choice D is eliminated.The best answer is choice B.

14. Choice A is the best answer. The only thing we are able to conclude is that the object is in translational equilibrium. ChoiceB is incorrect: A net torque of zero is required for the system to be in rotational equilibrium. Choice C is incorrect: A netforce of zero does not imply a net torque of zero, because the component forces may be acting in an asymmetric fashionabout the system's center of mass, resulting in a net torque on the system. Choice D is incorrect: Translational equilibriummeans no linear acceleration, but the object could still be moving with a constant linear velocity (keeping in mind Newton'sfirst law that an object at rest will remain at rest and an object in motion will remain in motion unless there is a net forceacting on it). The best answer is choice A.

Passage III (Questions 15 - 21) Clav Ball Collisions

15. Choice D is the best answer. If the balls had the same initial velocity, then they would collide at point X. But rather thanhaving the same initial velocity, they have the same initial kinetic energy. Since Ball III is more massive than Ball I, Ball Imust have a greater initial speed than Ball HI. Because Ball I has the greater speed, it reaches point X before Ball HI reachespoint X. As a result of Ball I reaching point X first, it passes through point X and continues on its straight-line path beforeBall HI reaches point X. The result is that there is no collision between the two balls, so there is no resultant vector for thisscenario.

No Collision

3^ III



16. Choice B is the best answer. Momentum is conserved in both of the collisions, so the initial momentum before anycollisions equals the final momentum after the last collision. This means that Ball I enters with the same momentum thatBall IV exits. Given that Ball I and Ball IV have the same mass and they have the same momentum, they must also have thesame velocity. This means that the initial velocity of Ball I is equal to the final velocity of Ball IV, making choice B the bestanswer. The key feature in the velocity being transferred completely to Ball IV is the fact that Ball I and Ball HI are bothstationary after the collisions, meaning that all of the momentum is completely transferred during each individual collision.Had either ball been in motion after collision, then calculations would need to be done to determine the final velocity of ballIV. The best answer is choice B.

17. Choice C is the best answer. If the two balls are fired with the same initial velocity, then they will collide at point X. SinceBall II is heavier than Ball IV, Ball II has more momentum than Ball IV in both the x- and y-directions. Since both balls arefired from positions above the horizontal midline, they each have y-direction momentum in the downward direction. Thismeans that their y-direction momentum vectors simply add. The resultant vector must have momentum in the downward y-direction. Additionally, Ball II has a greater magnitude of momentum in the x-direction than Ball IV has, although Ball IVhas its x-direction momentum in the negative x-direction. After the collision, there is an x-direction component in thepositivex-direction, so the resultant vector must go off a little in the positive x-direction.

11^ ^rv

X

PII>PIV\ ResultantThe best answer is choice C.

18. Choice B is the best answer. If two balls are released from two of the six unique release points that are not collinear, thenthey will collide at point X only if they have the same initial speed. This happens, because all of the release points areequidistant from pointX, 1.0meteraway. Onlywith thesame initial speed can theyarriveat pointX at thesame time, whichis necessary for tiie balls to collide at point X. If the balls have the same mass and velocity, then they also have the samemomentum and kinetic energy. But this is not a necessary condition for a collision; it is a special case of the necessarycondition. The best answer is choice B.

19. Choice B is the best answer. If the balls are made of a material that cannot deform, then the collision will be elastic. If theballs could deform, then they would stick together and not rebound with any speed. Because they do not deform, energymust be conserved duringthe collision, so the ballsbounce back withtheiroriginal speed. This eliminates choiceA. If equalmass balls are fired from collinear points (which describes A, X, and F), then they will collide head on, and bounce directlybackwards. There is no angle associated with their rebounding, so choice C is eliminated. If they bounce off of one anotherand rebound back with equal magnitudes of velocity, then the initial momentum must equal the final momentum. This isconservationof momentum, which is not violated in any of the remainingchoices. There is no reason that the velocity wouldbe reduced to half of its original value, so choice D is eliminated. Because both speeds are equal, the choice B is valid. Thebest answer is choice B.

20. Choice C is the best answer. If the experiment were done using super balls instead of clay balls, our collisions wouldbecome elastic as opposed to being inelastic, as they were with the clay balls. In an elastic collision in the absence of a netexternal force, both kinetic energy and momentum are conserved, so the best answer is choice C. If the balls were spinningwhen they collided, then both rotational kinetic energy and angular momentum would be conserved as well, but not torque.That is why choice D is an incorrect answer. The best answer is choice C.

21. Choice C is the best answer. The two starting points are collinear with point X (the point at which they aimed), so there willbe a collision no matter what the initial speeds for each ball are. This eliminates choice D. If Ball I has the same initialmomentumas Ball II, then Ball I must be moving at a higher initial speed than Ball II, because Ball II is heavier than Ball I.Since the points B, X, and E all lie on a straight line, a collision will occur somewhere on this line, even though the initialvelocitiesof the balls differ. Because Ball I has the higher initial speed, it will pass point X before Ball II gets to point X, sothe collision will occur to the right of the point X. As one can now see, this corresponds to answer choice C. The bestanswer is choice C.



22. Choice D is the best answer. Let's choose thepoint where the ladder touches the ground to be the pivotpoint. As the personclimbs the ladder, he exerts a force—his weight-on the ladder. This force points vertically downward, and the angle betweenthis force and the ladder is 30° (i.e., 90° - 60°). The force and the ladder's angle do not change as the person climbs theladder. However, the distance between the pivot point (base of the ladder) and the person increases, as he climbs the ladder.Thisdistance is directly proportional to the length of the lever arm. For a given force and angle, a longer lever arm means agreater torque. When the person stands near the top of the ladder, the leverarm is about as long as it can be, so the torquefrom the person's weight on the ladder is as bigas it can be. This is when the ladder is most likely to slip. The best answer ischoice D.

23. Choice C is the best answer. Let the person and their skateboard be our system. Initially, the system is at rest—themomentum of the system is zero. When person walks forward, the skateboard must move backwards, in a direction oppositeto person's motion. These two motions give us a net final momentum of zero, which we must have by conservation ofmomentum. This is an example of recoil, where the two objects start at rest and the move apart with equal magnitudes ofmomentum. The person on the skateboard is likely much heavier than the skateboard, so the skateboard will acceleraterapidly to a high velocity, which explains why when you stepoff a skateboard it feels like you're going to fall off. The bestanswer is choice C.

24. Choice A is correct. Initially, the momenta of the truck and car are towards the South and the North, respectively. After thecollision, the car is moving with an eastbound momentum. By momentum conservation, the truck must acquire somewestbound momentum, ruling out choices B and C. However, also by momentum conservation, the total momentum alongthe North-South axis cannot change. Since the post-collision car no longer has momentum along this axis, the truck musthave it all. The truck has to be moving along the North-South axis and in the westbound direction. The best answer ischoice A.

25. Choice C is the best answer. When the platform spins with a constant angular velocity or does not spin at all, its angularmomentum is not changing. Therefore, choices A and B are invalid. From the definition of angular momentum conservationgiven in the passage, the momentum is conserved only in the absence of a net external torque. This is choice C. Notice thatthis constraint is similar to that for linear momentum. For linear momentum to be conserved, there must be no net externalforce acting on the system. The best answer is choice C.


52-Question Momentum and Torque Practice Exam

I. Ice Skating Collisions

II. Crane Operation

III. Bullet Momentum


IV. Torque and Force

V. Ball Collisions


VI. Automobile Safety

VII. Defining Torque


Momentum and Torque Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-6)

(7-11)

(12 -17)

(18 - 21)

(22 - 26)

(27 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)


Khoi and Jen are ice-skating enthusiasts. In preparationfor a major doubles figure skating momentum transfercompetition (complete with judges who accept bribes), theypractice three different momentum transfer maneuvers. Allthree momentum transfer compulsories take place on the ice,which is assumed to be frictionless. Note that Khoi is twice

as massive as Jen.

Compulsory I

Khoi skates towards a stationary Jen at a constantspeed of 2 m/s. They collide and move togetherfollowing collision.

Compulsory II

Jen skates towards a stationary Khoi at a constantspeed of 2 m/s. They collide and move togetherfollowing collision.

Compulsory III

Similar to Compulsory II, Jen skates towards astationary Khoi at a constant speed of 2 m/s exceptthey bounce off each other perfectly.

1.

2.

In Compulsory I, Jen feels:

A. a larger magnitude of acceleration and a largerimpulse than Khoi.

B. a larger magnitude of acceleration than Khoi, butthey both feel the same magnitude of impulse.

C. a larger magnitude of impulse than Khoi, but theyboth feel the same magnitude of acceleration.

D. the same magnitude of impulse and acceleration asKhoi.

If Khoi and Jen collide at the same speed on paths thatare separated by 120°, then which of the followingshows their resultantvector if they stick together?

A. a B.

fResultant AResultant

Resultant

D.No Resultant


3.

4.

5.

6.

For Compulsory III, if Khoi's post-collision velocity is4/3 m/s to the right, then what is Jen's final velocity?

A. 2/j m/sto the left

B. 2/3 m/s to the right

C. 4/3m/sto the left

D. 4/3 m/s to the right

In Compulsory I, Khoi and Jen have a final speed v.Khoi and Jen now double their masses and repeatExperiment I. What this new final speed in terms of v?

A. 50%ofv

B. 100% of v

C. 141% of v

D. 200% of v

In which compulsory, Compulsory I or Compulsory II,does Jen feel the largest magnitude of impulse?

A. Compulsory I

B. Compulsory II

C. Jen feels the same impulse in both Compulsories Iand II.

D. There is insufficient information to tell.

Which of the following is an assumption in CompulsoryI that makes the initial momentum equal to the finalmomentum?

A. The ice exerts minimal friction on the skaters.

B. The ice exerts no friction on the skaters.

C. The two skaters deform slightly after colliding.D. The two skaters do not deform after colliding.



Large cranes are commonly employed in theconstruction of tall buildings. A model of such a crane has apivot that supports a large beam, from which the buildingmaterial mi and a moveable counterweight m2 hang. A

tower supports the pivot point so that the system is stable.The base on which the pivot rests is wide, so that there issome margin for error when moving the counterweight.

The counterweight is typically more massive than thebuilding material and located closer to the pivot point. Mostcounterweights are made up of several smaller masses, eachof which can move independently of one another. Figure 1shows a highly simplified crane system. (Unless statedotherwise, assume the pivot, cable, and large beam aremassless and that the crane is balanced).

7.

8.

ni2

Counterweight / \PiVPivot point

mi

Load

Tower

Figure 1. Simplified crane

When lowering the building material at a constantvertical velocity:

A. the counterweight must be moved slowly towardsthe pivot point to maintain the crane's balance.

B. the counterweight must be moved slowly awayfrom the pivot point to maintain the crane'sbalance.

C. the counterweight need not be moved to maintainthe crane's balance.

D. the motion of the counterweight is unrelated tomaintaining the crane's balance.

If the mass of building material is increased by 50%,and balance of the crane is to be maintained, thedistance between the counterweight and the pivot pointmust:

A. decrease by 50%.

B. increase by 50%.

C. increase by 66%.

D. increase by 100%.


9.

10.

Where is the center of gravity of the crane in Figure 1?

A. Between the pivot point and the building materialB. Between the pivot point and the counterweightC. At the pivot point

D. Directly below the pivot point

How much force does the beam exert on the pivot?

A. 0

B. (mi + iri2)g downwards

C. (mi + m2)g upwards

D. (m2 - mi)g upwards

11. If the support cable for the building material were tobreak, what could be done to prevent the crane fromtoppling over?

I. Release the counterweights

II. Shift the counterweights towards the fulcrum of thecrossbar

III. Shift the counterweights away from the fulcrum ofthe crossbar

A. I and II only

B. I and HI only

C. II and III only

D. I, II, and III



In order to examine the relationships betweenmomentum, energy, and mass, a group of students conducttwo separate experiments.

Experiment 1

A block of wood is placed on a rough surface. A rubberbullet is fired at the center of the block of wood.

Following this, an aluminum bullet is fired at the center ofthe block of wood. The two bullets have the same size,

speed, and mass.

Experiment 2

The block of wood is now attached to a long, inelasticstring attached to a ceiling, as shown below in Figure 1.A rubber bullet is fired at the center of the block of wood.

Following this, an aluminum bullet is fired at the center ofthe block of wood. In both cases, the block goes intoharmonic oscillation after the collision. The two bullets

have the same mass, size, and speed.

Bullet= ~ .=*>

Cool lines behind the build COmake ii appear to be in motion Block of Wood

Figure 1. Bullet speeding towards dangling wood block(with cool lines behind the bullet to show its motion)

Aluminum is highly malleable and deforms upon impact,while the rubber bullet undergoes no permanent deformation.

12. For Experiment 1, which bullet is more likely to knockthe block over?

A. The rubber bullet hitting near the top of the block.B. The aluminum bullet hitting near the top of the

block.

C. The rubber bullet hitting near the base of the block.

D. The aluminum bullet hitting near the base of theblock.

13. For Experiment 2, assuming that both bullets strike theblock at the same point, which bullet will cause theblock to swing to the greatest height?

A. The rubber bullet

B. The aluminum bullet

C. Both bullets will cause the block to go the sameheight.

D. Neither bullet will cause the block to go up.


14. For Experiment 2, the work done on the block by thetension in the string is:

A. 0.

B. TL.

C. -TL.

D. mgL.

15. For Experiment 1, which bullet will most likely deformthe block?

A. The rubber bullet

B. The aluminum bullet

C. Both bullets have an equal chance of deforming theblock.

D. Neither bullet could deform the block.

16. For Experiment 2, after a bullet becomes embedded inthe block, it swings back and forth like a pendulum.During this swinging process:

A. momentum only is conserved.

B. energy only is conserved.

C. both momentum and energy are conserved.D. neithermomentum nor energy is conserved.

17. For Experiment 2, how much the height of the blockchanges depends on:

I. the initial velocity of the bullet.

II. the mass of the block.

III. the length of the string.

A. I only

B. I and II only

C. II and III only

D. I, II, and III


Questions 18 through 21 are NOT based ona descriptivepassage.

18. The total kinetic energy of colliding objects is given byKEtotal = KEobject! + KEobject2» where KE = &mv2.How does Ktotal before an inelastic collision comparewith KEtotal after the collision?

A. KEtotal before = KEtotal after.

B. KEtotal before > KEtotal after.

C. KEtotal before < KEtotal after.

D. It depends upon the initial speed of the ball.

19.

20.

Which change to a balanced system will NOT increasethe torque associated with a lever system on the sidewith the load?

A. The addition of weight to the load.

B. Moving the load farther from the fulcrum.

C. Extending the length of the bar on the side of theload.

D. Shortening the length of the bar on the side of theload.

An object of mass m is attached to a rod of negligiblemass and length L. The rod is attached to the ground atpoint O and lifted in such a way that the rod was nearlyvertical. The rod was then dropped to the ground, takinga semi-circular path back to the ground. If the rodmakes an angle 0 with the horizontal as it falls, what isthe torque about point O?

A. Lmg sinO

B. Lmg cos©

C. mg sinO

D. mg cosO


21. Why is it a good design that makes the counterweightmore massive than the maximum load on a panbalance?

A. So that adjustments to the counterweight's positionto compensate for variations in the mass of the loadare minimal.

B. So that adjustments to the counterweight's positionto compensate for variations in the mass of the loadare substantial.

C. To maximize the total mass of the systemD. To raise the center of mass of the system



Torque is a concept that can be observed in everydaylife. Tasks such as turning on the hot water tap and makingthe brakes stop your car require the application of a torque.Torque can be thought of as a twist or turn that causes achange in rotational motion. Torque is the product of a forceand the distance (from some rotational axis) over which theforce acts, as shown in Equation 1.

x = rFi = r±F

Equation 1

where Fi is the component of F perpendicular to r, andr± is similarly defined and often referred to as the lever arm.Because of this interplay between force and displacement,one can produce a fairly substantial torque even when theapplied force is small.

22.

23.

A worker finds it difficult to twist a stubborn bolt with

a wrench. He attaches a rope to the wrench as shownbelow, and then exerts the same force on the rope as hedid on the wrench. The rope is attached at the samepoint at which his hand originally pulled. How does thetorque with the rope compare to the torque without therope?

rope

wrench

<iQ>bolt

A. The torque has increased.

B. The torque has decreased.

C. The torque remains the same.

D. The torque depends on the mass of the rope.

r\

Applying a force parallel to a lever arm:

A. will have no effect.

B. may produce a rotational motional motion, but willnot produce a translational motion.

C. may produce a translational motional motion, butwill not produce a rotational motion.

D. may produce both a translational and rotationalmotion.


24. The four forces acting on the wheel below all have thesame magnitude. Which one produces the greatesttorque?

jl

25.

26.

A. I

B. II

C. Ill

D. IV

X3-

A beam has a mass that is uniformly distributed over itslength. The beam is placed on supports in one of twoways as shown below:

A" A

/T7K H

What can you conclude about Beam I and Beam II?

A. Both I and II are in translational and rotational

equilibrium.

B. I and II are in translational equilibrium; I is inrotational equilibrium.

C. I and II are rotational equilibrium; I is intranslational equilibrium.

D. I is in translational and rotational equilibrium; II isnot in any state of equilibrium.

A long plank of mass m and length L lies on theground. You wish to pick it up at one end, leaving theother end on the ground. What minimum force mustyou apply?

A. mg/2

B. mg

C. 2mg

D. 4mg



One day, an astronaut is spacewalking, and he getsbored. He decides to throw clay and rubber balls at a rigidbox of mass M. In one case, he throws a rubber ball of mass

m, with a speed V. It bounces off of the box with a recoil

speed It,, causing the box to start moving as well. This is

shown in Case I. Later, he throws a clay ball of mass m, witha speed V. It hits a similar box and sticks to it. The ball andbox now move together. This is shown in Case II.

Case I: Rubber ball

Before

Vii-V v2i = 0

After

vlf=T v'2f=?

Case II: Clay ball

Before

<>-•vn = V v2i = 0

After

a-vf =?

Figure 1. Collision experiment

The box has a mass of M, that is twice the mass of either

ball (M = 2m).

27. What is the final speed of the box in Case I?

A. V3

B. Y.2

C. 2V3

D. V

28. In Case II, what change would double the box's finalmomentum?

A. Doubling the ball's initial speed, V.

B. Doubling the box's mass, M.

C. Doubling the ball's mass, m.

D. Halving the box's mass, M.

Copyright ©by The Berkeley Review®

29.

30.

31.

32.

207

In Case II, the clay ball and box move off together witha speed vf;naj. Which graph BEST represents the

relation between vfjnai and V, the ball's initial speed?

A. B.

vf

V

Which ball makes the box move more?

A. They both move the box equally.

B. The clay ball moves the box more.

C. The rubber ball moves the box more.

D. Neither ball causes the box to move.

In collisions, one often talks about impulse:

I = Ap = FAt

This equations says, when an object is acted upon by animpulse, it has a change of momentum, Ap. Also, theobject is acted upon by an average force, F, for a time,At. With all of this in mind, which ball experiences alarger average force?

A. The clay ball.

B. The rubber ball.

C. They experience the same average force.

D. There is insufficient information to find the

answer.

The instant after the clay ball collides with the box, thebox has a speed vfjnai and is distance L away from theastronaut. After T seconds, how far is the box from the

astronaut?

A. VfinaiT

B. between VfmaiT and L + vfjna]T

C L + vfinaiT

D. more than L + vfjnaiT



33. What is the mechanical advantage associated with alever system made with a light-weight 2.0-m meterstick with the fulcrum at the 0.5 m mark?

A. MA = 4

B. MA = 3

C. MA = 2.5

D. MA = 1

34. When a projectile penetrates into a barrier, its:

A. momentum is converted into kinetic energy andheat.

B. potential energy is converted into heat and kineticenergy.

C. momentum is converted into potential energy andheat.

D. kinetic energy is converted into heat

35.

36.

What is the final velocity of a 0.2-kg ball traveling at 12m/s after if strikes a stationary 0.1-kg ball if they sticktogether after collision?

A. 3 m/s

B. 4 m/s

C. 6 m/s

D. 8 m/s

An object that has momentum must also have:

A. acceleration.

B. impulse.

C. kinetic energy.

D. potential energy.

Copyright©by TheBerkeley Review® 208 GO ON TO THE NEXT PAGE


The study of automobile safety is a very important field.Researchers study the effects of collisions on "crashdummies" positioned in vehicles. Different makes and sizesof cars moving at various speeds are collided with differentbarriers (moveable and immovable). As a direct result ofsuch studies, several important safety features have beenadded to cars in recent years. For example, shoulderharnesses have replaced lap safety belts as a means ofrestraining automobile occupants during collisions.

The basic goal of automobile safety is to reduce theimpact force an occupant of a vehicle experiences during acollision. Car manufacturers install air bags to protect boththe driver and the passenger in the front seat. Collapsiblebumpers help reduce the impact force of collision.

37. A safety device that is showing up in many new carmodels is a collapsible steering column. When thedriver collides with such a column:

A. the force on the driver is lessened because of the

decrease in time of the collision.

B. the force on the driver is lessened because of the

increase in time of the collision.

C. the force on the driver is lessened because the

momentum change of the driver is reduced.

D. the force on the driver is lessened because the

momentum change of the driver is increased.

38. For a given collision speed, heavier cars are safer incollisions than lighter cars. This statement is:

A. true, because acceleration is smaller.

B. false, because acceleration is smaller.

C. true, because impulse is larger.

D. false, because impulse is larger.

39. In a head-on collision between two identical cars,

which of the following is true when comparing elasticand inelastic collisions? (Assume that the initialmomenta and collision times are the same, regardless ofthe type of collision.)

A. An elastic collision is less damaging to the carsthan an inelastic collision.

B. An inelastic collision is less damaging to the carsthan an elastic collision.

C. Either type of collision is equally damaging.

D. The more damaging collision involves smallerimpulses.


40.

41.

42.

209

Two cars collide inelastically at a North-South East-West intersection. The skid marks point roughlynorthwest. Concerning the initial directions of the twocars, this tells you that one driver was traveling:

A. west; the other driver was traveling south.

B. west; the other driver was traveling north.

C. east; the other driver was traveling south.

D. east; the other driver was traveling north.

Consider a one-dimensional collision between a movingcar (Car I) and an initially stationary car (Car II). Whatstatement can you make about the mass of Car IIrelative to the mass of Car I, if Car II is to recoil withthe greatest speed?

A. The two masses should be equal.

The mass of Car II should be much greater than themass of Car I.

The mass of Car II should be much less than the

mass of Carl.

The relative masses of the cars are irrelevant.

B.

C.

D.

In any collision, the objects eventually slow down andstop. What does this mean in terms of momentumconservation?

A. Momentum is conserved only in the absence offriction.

B. Momentum is conserved in the presence of friction,if we extend what we mean by internal forces.

C. We have to consider whether friction will cause a

collision to be elastic or inelastic.

D. Momentum conservation works only for certaincollisions.



One way to analyze torque is to talk about force andlever arms, where x = rFi.

Torque is also defined is in terms of angular momentum,L. Angular momentum obeys L = Ico.where I is the momentof inertia, measured in kg-m2, and to is the angular velocity,measured in radians/sec. I is a measure of an object'sresistance to being torqued, just as mass is the resistance anobject puts up to being forced. I depends upon how anobject's mass is distributed with respect to a rotational axis:

I « mr2

where r is the distance from the axis of rotation out to the

mass. In terms of L:

x = ALAt

i.e., the torque is the change in angular momentum withrespect to time. If there is no net external torque on a system,then angular momentum is conserved CLmitial = Lfinal)-

This is similar to the conservation of linear momentum,when no net external force exists. A rotating object also hasrotational kinetic energy, which can be written as:

KE=i/£Io)2

A group of students perform two experiments on therelationship between torque and angular momentum.

Experiment IOne student sits on a stool that rotates freely. He holds a 5-kgmass in each hand. Initially, the student has an angularvelocity of 5 radians/sec with his arms in his lap.

Experiment2A series of experiments is done with a rotating platform.

(i) A student stands at the edge of a rotating platform ofradius 2 meters. She and the platform are initially atrest, and then she begins walking around the edge ofthe platform.

(ii) Two students of similar mass are now standing on theedge of the platform. Both begin walking around theedge of the platform with the same speed, but inopposite directions.

(iii) A single student now stands 1 meter away from thecenter of the platform. She begins walking with thesame angular speed she had in Experiment 2 (i).

43. In Experiment 1, when the student stretches out hisarms, he spins:

A. faster, because his mass decreases.

B. faster, because his moment of inertia decreases.

C. slower, because his mass increases.

D. slower, because his moment of inertia increases.


44. In Experiment 1, the student extends his arms. The newI, with his arms out, is 3 times I with his arms in his lap.His angular velocity, co,:

A. increases by a factor of 3 and his kinetic energy,KE, increases by a factor of 3.

B. decreases by a factor of 3; and his kinetic energy,KE, decreases by a factor of 3.

C. increases by a factor of VT; and his kinetic energy,KE, decreases by a factor of 9.

D. decreases by a factor of 3 and KE does not change.

45. In Experiment 1, with his arms outstretched, the studentdrops the weights. How can it be explained that thestudent's angular velocity does not change?

A. The center of mass does not change for the student,so angular momentum is constant.

B. The position of the weights does not impactangular momentum.

C. The falling weights continue to move in theircircular pathway as they fall, so they retain theirsame angular momentum.

D. The weights are rotating in opposite directions afterthey are dropped.

46. In Experiment 2 (i), when the student starts walking,what effect does this have on the rotating platform?

A. No effect

B. The rotating platform moves in the same directionthat she walks.

C. The rotating platform moves in the directionopposite to the one in which she walks.

D. The rotating platform will move first in the samedirection that the student walks; but as the studentspeeds up, its rotation rate will decrease.

47. In Experiment 2 (ii), with both students walking on theplatform in opposite directions, the platform:

A. does not move.

B. now rotates with twice the speed it had inExperiment 2 (i).

C. now rotates with half the speed it had inExperiment 2 (i).

D. changes direction of rotation, but keeps the samespeed it had in Experiment 2 (i).

48. Comparing Experiments 2(i) and 2(iii), the platform'sangular momentum in Experiment 2(iii) is less than thatin Experiment 2(i) by a factor of:

A. 4

B. 2

C. VT

D. The angular momentums are the same in bothexperiments.



49. Two equal mass objects travel in opposite directions atthe same speed prior to collision. After impact, if thecollision is:

A. perfectly inelastic, then their velocities areinterchanged.

B. elastic, then their velocities are interchanged.C. elastic, then their velocities are both zero.

D. perfectly inelastic, then they experience no changein their velocities.

50.

51.

52.

What is the final velocity of a 0.1-kg ball traveling at12 m/s after if strikes a stationary 0.3-kg ball thattravels at 6 m/s after the collision?

A. -6 m/s

B. -3 m/s

C. +2 m/s

D. +6 m/s

A 1 kg meter stick is attached to the ceiling by a stringat the 40 cm mark. If a 3 kg mass is attached to themeter stick at the end labeled 0 cm, what mass placed atthe 70 cm mark would balance the meter stick?

A. 1.50 kg

B. 3.67 kg

C. 4.00 kg

D. 6.00 kg

Two equal mass balls, one traveling west at 10 m/s andthe other traveling east at 20 m/s collide head on. Thewestbound ball recoils back at 6 m/s. What is true of

this collision?

A. The collision is elastic and the second ball

continues to travel east at 4 m/s.

B. The collision is elastic and the second ball

rebounds and travels west at 4 m/s.

C. The collision is inelastic and the second ball

continues to travel east at 4 m/s.

D. The collision is inelastic and the second ball

rebounds and travels west at 4 m/s.


1. B 2. B 3. A 4. B 5. C 6. B

7. C 8. B 9. D 10. B 11. A 12. A

13. A 14. A 15. B 16. B 17. B 18. B

19. D 20. B 21. A 22. C 23. C 24. B

25. D 26. A 27. C 28. A 29. D 30. C

31. B 32. D 33. B 34. D 35. D 36. C

37. B 38. A 39. A 40. B 41. C 42. B

43. D 44. B 45. C 46. C 47. A 48. A

49. B 50. A 51. B 52. C

YOU ARE DONE.

Answers to 52-Question Momentum and Torque Practice Exam

Passage I (Questions 1-6) Ice Skating Collisions

Choice B is the best answer. Use multiple concepts to answer this, considering the relative impulses and relativeaccelerations separately. When objects collide, they feel equal and opposite impulses (this is one way of stating Newton's3rd law). This is true because they exert forces of equal magnitudeon one another and the time of collision is the same forboth objects. You may also consider that when momentum is conserved, the momentum lost by one is equal to themomentum gained by the other. They each experience the same magnitude of change in momentum, so they each feel thesame impulse. Thus, choices A and C are incorrect. Since both people feel the same size of force, the lighter person willaccelerate more (F = ma). Choice B is the best answer. You could also solve this by considering their before and afterspeeds. Khoi goes from 2 m/s to 4/3 m/s in the same period of time that Jen goes from 0 m/s to 4/3 m/s. Over the sameduration of time, Jen experiences a greater change in velocity, so she has the greater acceleration. The best answer is choiceB.

Choice B is the best answer. If the skaters collide on non-parallel paths, then the resultant direction must be different thaneither of the initial directions. This fact doesn't help with this question, because they all show that effect. In this specificcase, the two skaters have the same speed, but not the same mass. This means than one skater has a greater magnitude ofmomentum than the other. The collision will not result in a symmetric resultant, so choices A and C are eliminated. Thesystem has momentum in y-direction that is not canceled, so choice D cannot be correct. The only answer that shows acollision resulting from two paths of unequal initial momentum is choice B. The best answer is choice B.

Choice A is the best answer. Using physical intuition, the lighter Jen, skating towards the stationary Khoi, should bouncebackwards following their collision. Since Khoi moves to the right, after the collision, Jen should move to the left. Thisallows us to cross out choices B and D. Conserving momentum to get the numbers,

Pinitial, Jen = Pfinal, Jen + Pfinal, Khoi

m(+2) =mfvfmai, jen) +2m(+4/3)

=> 2m =8/3m +m(vfinal) jen) => m(vfinali jen) =-2/3m .-. Vfiaai, jen =-2/3 m/s

Solving for Jen's final speed yields -2/3 m/s, the minus indicating that her direction is opposite that of the +4/3 m/s for Khoi.Notice that the specific mass is irrelevant; only the relative mass matters. The best answer is choice A.

Choice B is the best answer. Because only relative mass matters (and the ratio of their masses has not changed), the finalspeeds should remain the same. This makes choice B the best answer. Using conservation of momentum, we get thefollowing numbers:

Pinitial, Jen = Pfinal,Jen + Pfinal, Khoi •'• (mjen)vinitial,Jen = (mJen + mKhoi)vfinal

m(vinitial) =(m +2m)vfinai ,\ (V3)(vinitial) =Vfinal 0R 2m(vinitial) =(2m +4m)vfinai .'. (I/3)(vinitial) =v'finalThe best answer is choice B.

Choice C is the best answer. Impulse depends upon the relative collision speeds. For example, if you bump into a person

on the sidewalk or inside a 60 mi/hr ^us- y°ur impulses in these two collisions are notall thatdifferent. Now, Experiments Iand II are actually the same experiment, as far as relative collision speeds are concerned. One happens to be viewed in Jen'sreference frame and the other in Khoi's; the relative speeds are the same though. Thus, Jen should feel the same magnitude

of impulse in both experiments. In Experiment I, she goes from rest to a speed of 4/3 m/s. In Experiment II, she goes aspeed of 2 m/s to a speed of2/3 m/s. In both experiments, her speed changed by 4/3 m/s and she had the same mass, so herchange in momentum (mAv) was of equal magnitude. The best answer is choice C.

Choice B is the best answer. For the initial momentum to equal the final momentum, momentum must be conserved.Momentum is conserved in the absence of an external force. For that to be true, there can be no friction or wind resistance.This eliminates choice A and makes choice B the best answer. Choices C and D address deformation, which dictateswhether a collision is elastic or inelastic, but does not imply anything about conservation of momentum. Choices C and Dare not the best answer. The best answer is choice B.


Passage II (Questions 7-11) Crane Operation

7.

8.

9.

10.

11.

Choice C is the best answer. For balance to be maintained, rjmig must equal r2iri2g. Lowering the building material at aconstant velocity does not affect any variable in the equilibrium equation. Therefore, the counterweight does not need to bemoved(i.e., r2 must be left unchanged) to maintain the balanceof the system. The best answer is choice C.

Choice B is the bestanswer. For balance to be maintained, the net torque must be zero. That is, rjmjg = r2iri2g, where r\ isthe distance between the pivot and the building material and r2 is the distance between the pivot and the counterweight.Increasing the material mass by 50% means that the material becomes 1.5 times as massive. To maintain balance, thecounterweight must be moved to a distance 1.5 times that of itsoriginal distance from the pivot. It is now 50% farther fromthe pivot. As an aside, if the answers were phrased as a percentage of the initial value, then the best answer would have been"150% of r2". Be aware that these two different ways of phrasing a percentage answer lead to very different numbers, butare based on the same physics. The best answer is choice B.

Choice D is the best answer. If the center of mass is to either side of the pivot point, then the crane will be out of balanceand at risk for tumbling over. This rules out choices A and B. If you look at any stationary hanging object, you will noticethat its center of mass is always below the pivot point. Look at a coat hanger or a hanging houseplant for instances. Thus,the center of gravity for this crane should likewise lie below the pivot point. The best answer is choice D.

Choice B is the best answer. Because the beam is connected to the two masses, it pushes down on the pivot. This meansthat there is a downward force. Only choice B shows a downward force. The best answer is choice B.

Choice A is the best answer. If the cable holding the load were to break, then there would be a sudden loss of mass on thatside of the fulcrum, resulting in a lower torque on that side of the fulcrum. To compensate for that and to prevent the cranefrom tumbling over, the torque on the side without the load must be reduced. This can be accomplished by either reducingthe mass on the other side or by shortening the distance between the fulcrum and the center of mass of the counterweight.Dropping the counterweight can accomplish this, so Statement I is valid. This eliminates choice C. Moving thecounterweight towards the fulcrum also accomplishes this, so Statement II is valid. This eliminates choice B. Moving thecounterweight away from the fulcrum increases the torque, which is the opposite of what needs to be done, so Statement IIIis invalid. This eliminates choice D. The best answer is choice A.

Passage III (Questions 12 -17) Bullet Momentum

12. Choice A is correct. Both the rubber bullet and the aluminum bullet have the same speed and mass, so they strike the blockwith the same incident momentum. However, the rubber bullet will bounce off the block, while the aluminum bullet willpenetrate the block. All of the momentum of the aluminum bullet is transferred to the block; the block in turn supplies thenecessary impulse to stop it. The block also supplies the impulse necessary to stop the rubber bullet, but the block must alsosupply additional impulse to the rubber bullet to turn it around and send it back the way it came. If the collision between theblock and the rubber bullet is completely elastic, the impulse supplied by the block to the rubber bullet could be twice theimpulse supplied by the block to the aluminum bullet. A collision where the objects rebound generates a greater impulse(change in momentum) than when the two objects stick to one another. By conservation of momentum, this means that therubber bullet supplies twice the impulse to the block as compared to the aluminum bullet, so the rubber bullet is more likelyto knock the block over. This eliminates choices B and D. The next factor to consider is the torque exerted on the block bythe collision. It is the torque that is causing the block to pivot about its base and tumble over. The higher the bullet strikesthe block, the greater the moment arm (distance from the fulcrum to the applied force). This means that a bullet that strikeshigheron the block exerts more torque, and has a greater chance of knocking the block over. The best answer is choice A.

13. Choice A is correct. Because the rubber bullet rebounds off of the block, it gives the block the most momentum andtherefore the greatest velocity. The aluminum bullet gives the block the least momentum and therefore the least velocity.Because the block gains more velocity, and therefore speed, from the rubber bullet, it swings to a higher point beforemomentarily coming to rest and changing directions. Had you chose to look at kinetic energy, you might have run intotrouble. In the case of the rubber bullet, it rebounds at roughly the same speed, so it transfers limited kinetic energy. Thealuminum bullet loses nearly all of its kinetic energy when embedding in the block. At first glance, because the aluminumbullet loses more kinetic energy, we may be tempted to think that it transferred the most kinetic energy. However, the kineticenergy of the aluminum bullet gets converted into heat and goes into deforming the block, so there is not much left to betransferred and get converted into potential energy. So, although the rubber bullet loses a smaller amount of kinetic energythan the aluminum bullet, all of this kinetic energy can be transferred and get converted to potential energy. The best answeris choice A.


14. Choice A is the best answer.At all points of themotion, the tension is perpendicular to the direction of travel of thebullet +block system. A force can only do work on an object if some component of that force is parallel to the displacementTherefore, the tension can do no work on the block. The best answer is choice A.

15. Choice B is the best answer. Because of the elastic nature of rubber, the rubber bullet bounces off the block, carrying awaymost of its original speed, and therefore carrying away most of its kinetic energy. By contrast, the aluminum bulletpenetrates the block, surrendering all of its kinetic energy to the block. This kinetic energy becomes heatenergy and theenergy causing deformation of the bullet and the block (work is necessary to change the shape of both objects). The bestanswer is choice B.

16. Choice B is the best answer If we take as our system the bullet plus block, then momentum is not conserved because thereis an external force acting on this system, the force due to gravity. As a result, the velocity changes throughout itsoscillation, so its momentum is continually changing, which implies it is not conserved. The total energy of the system inharmonic oscillation is conserved. Gravitational potential energy is converted to kinetic energy and back again as the blockand bullet swing back and forth. The best answer is choice B.

17. Choice B is the best answer. The swinging of the block occurs after the collision, so we need to consider the block plusbullet after the collision. The block-bullet system has a maximum velocity right after the collision. If we apply conservationof energy to the swinging process, then we can relate the kinetic energy of the block-bullet system to gravitational potentialenergy using:

V£mv2 =mgAh .*. !£v2 =gAh

The mass cancels out of the equation, so the speed acquired as a result of the collision dictates the maximum height of theblock-bullet system. This means that the speed of the bullet at impact affects the height to which the swings, so Statement Iis valid. This eliminates choice C. However, the velocity after the collision depends on the mass of the block. Byconservation of momentum, the initial momentum of the bullet is converted to the final momentum of the block and bullet:

mbullet'Vbullet= (mblock + mbullet^Vbullet-block system

As the mass of the block increases, the velocity after the collision decreases. Hence, the mass of the block will affect themaximum height, so Statement II is valid. This eliminates choice A. The length of the string will impact the exact height ofthe block, but it does not impact the change in height, as that depends on energy, not position. Statement III is invalid, whicheliminates choice D.

Using physical intuition to confirm our answer, we know that as the bullet goes faster, the block will inherit more speed andthus swing higher. We also know that it is easier to move lighter objects, so a lighter block can swing higher than a heavierblock when pushed by the same force. The best answer is choice B.

Questions 18-21 Not Based on a Descriptive Passage

18. Choice B is the best answer. This question tests your knowledge of what an inelastic collision is. While total energy isalways conserved, not all of the energy has to be kinetic energy. When a ball collides with a surface and undergoes aninelastic collision, it deforms a little bit, which takes energy in the form of work. Also, when there is an inelastic collision,there is some heat released. The energy that leads to deformation and the energy that is released as heat come from thekinetic energy of the ball before the collision. Thus, the total kinetic energy after the collision will be less than the kineticenergy before the collision. This makes choice B the best answer. When collisions are elastic, all of the mechanical energy(i.e., kinetic + potential) before the collision goes into the mechanical energy after the collision. Use energy conservationwhen solving elastic collision problems. When collisions are inelastic, some of the initial mechanical energy goes into heatand deformation. Since it is practically impossible to calculate the exact amount of heat and deformation, you cannot useenergy conservation in an inelastic collision problem. The best answer is choice B.

19. Choice D is the best answer. Torque is calculated using: t = r x F. Increasing either the moment arm (r) or the load force(F) will increase the torque. The angle between the moment arm and the applied force also affects the magnitude of thetorque, but we ignore it in this question, because none of the answer choices address the angle. Additional weight increasesF, which in turn increases the torque, so choice A is eliminated. Moving the load further from the fulcrum increases r, whichin turn increases the torque, so choice B is eliminated. When the bar is extended on the side of the load, the bar becomesheavier andthecenter of mass moves further from the fulcrum, thus both Fwejght and r are increased. Thiseliminates choiceC. When the bar is shorted on the opposite side of the load, the bar becomes lighter and the center of mass moves closer tothe fulcrum, thus both Fwejgnt and r are decreased. This results in a reduction in the torque. This makes choice D the bestanswer. As a point of interest, extending the bar on the side of the fulcrum opposite of a load has the same effect as adding acounterweight. The best answer is choice D.


20. Choice B is the best answer. This problem can be answered formally, if you remember a trig identity relating sines andcosines and you evaluate the forces acting on the bar at different positions. But, if trigonometry doesn't come easily to you,then try considering units and limiting cases in solving this one. What should the units be? Torque is a length times a force,meaning that choices A and B have the correct units, whereas choices C and D do not. Next, is it sin6 or cosG? If the rodwere vertical, what torque would you expect because of the mass? None. So, as 0 goes to 90°, the trigonometric functionmust go to zero. This is what cosine 8 does, which is something the MCAT-designers expect you to know and which is whatmakes choice B the best answer. The best answer is choice B.

21. Choice A is the best answer. Torque is calculated using: x = rF±, so the magnitudeof the torque is affected by mass and thelength of the moment arm. As the mass of a counterweight increases, the length of the moment arm, r, necessary to achieve acertain torque is reduced. As such, heavier masses require a shorter moment arm. This means that a small adjustment in thepositioning of a large mass will alter the torque more than a small adjustment in the positioning of a small mass. In order tocompensate for load on a pan balance, the counterweight must be able to move, so by using a massive counterweight, itneeds only to be moved a short distance to compensate for variations in the load. This makes choice A a better answer thanchoice B. You don't want to maximize the mass of the system, nor raise its center of mass, so choices C and D areeliminated. The best answer is choice A.

Passage IV (Questions 22 - 26) Torque and Force

22. Choice C is the best answer. The lever arm, r, is the distance from the pivot point (the stubborn bolt) out to where the forceis applied at the end of the wrench. Attaching the rope to the wrench does not change r. The worker exerts the same force inboth cases, so F does not change. The rope has no impact on the torque, so choice D is eliminated. The angle between r and Falso does not change. Nothing that impacts torque changes, so the torque does not change. The best answer is choice C.

23. Choice C is the best answer. The definition of torque in equation form is:

x = rF±

where r is the length of the lever arm. If the force is exerted parallel to the lever arm, then there is no component of forceacting on the lever arm, and therefore no torque. However, there is a force acting on the object. Thus, translational motion isstill possible. The best answer is choice C.

24. Choice B is the best answer. All four forces have the same magnitude, so the difference in the torques must come from thedifferences in lever arms, or angles, or both. The wheel pivots about its center, which is its center of mass. The radius of thewheel is therefore the lever arm. Forces I and III are parallel to the radius (which is also the lever arm). When the force andthe lever arm are parallel, there is no torque, which eliminates choices A and C. Force IV makes an acute angle with theradius, so Force IV can produce a torque on the wheel. However, Force II makes a 90° angle with the radius. Sin 90° is 1, soa 90° angle will yield the largest torque for a given force and lever arm. Force II therefore produces the greatest torque. Thebest answer is choice B.

25. Choice D is the best answer. If you have ever tried to balance blocks or anything else, Beam I should appear to you to be inboth translational and rotational equilibrium. Beam II, however, looks as if it's going to flop over clockwise. It must not be inrotational equilibrium, since that would imply a net torque of zero and no possible change in its rotation rate. This rules outchoices A and C. More precisely, the center of gravity of Beam II is outside of its two support pivots, which gives rise to thenet gravitationally-induced torque. Now, when Beam II turns, it will turn about the right pivot, so its center of gravity willaccelerate, meaning that the net force on Beam II is no longer zero. It is, therefore, out of translational equilibrium. The bestanswer is D.

26. Choice A is the best answer. When you pick up one end of the plank, the other end will remain stationary. This end servesas the pivot point. In order to pick up the plank, the torques about this end must be equal. There are two torques acting on theplank: one from the weight of the plank, and one from your picking up the plank. The gravitational force on the plank iseffectively half the plank length (L) away from the stationary end, and the force (F) you apply is L away from the stationaryend. Setting these two torques equal:

Wig = LF2

F = B!£

An easier way to look at this question is to say that flat-out lifting something requires F = mg. However, when you lift justone side of the plank while leaving the other side on the ground, you are only partially lifting something (leaving some stilltouching the ground). Partially lifting the plank requires F < mg. Only choice A fits. The best answer is choice A.


Passage V (Questions 27 - 32) Ball Collisions

27. Choice C is the best answer. The rubber ball has viinitial =Vto the right, and vifinal = V/3 t0 the left- Tne box nas v2initial= 0. The box has twice the mass of the ball. If we call m the ball's mass, and M the box's mass, then M = 2m. In a collision,momentum is conserved as long as the momentum of all of the colliding objects is taken into account. Using momentumconservation to find V2final yields:

Pinitial total = Pfinal total, where Ptotal = Pobjectl + Pobject2

That is, the sum of the momenta of all the colliding objects. Given that p = mv, our momentum conservation equationbecomes:

mviinitial +MV2initial =mv Ifinal +Mv2finali which yields mV +0=-m (v/3) +2mv2fjnal

The minus sign in front ofv/3 indicates that the ball has changed directions and is recoiling at this point. Here, velocities thatpoint towards the right have a + in front of them, and velocities that point towards the left have a - in front of them. Solvingfor v2final gives us:

mV +m(V/3) =2mv2final ^> 4mV/3 =2mv2final •'. V2final =4mV/(3 x2m) =4mV/6m

v2final = +(2V/3)> where the + indicates that the box moves towards the right, as we would expect itto.


28.

29.

30.

Choice A is the best answer. At first glance, we might just use p = Mvf. If that were the case, then we could double themomentum p either by doubling M or doublingVf. However, this is not correct. Here's why:

mvu + Mv2j = mvif + Mv2f .'. mV + 0 = (m + M)vf .-. vf =

The valueof Vf actuallydepends upon M and m. Substitution into p = Mvfyields:

p= MmV

(m + M)

(m + M)

The results are that doubling V will double p, but doubling m will not double p (because we have m in both the numeratorand denominator). Choice C can be eliminated. In doubling M, we encounter the same problem we did when we doubled m.So doubling M will not double p. This eliminates choice B and consequently, choice D (because choice D includes choiceB). The best answer is choice A.

Choice D is the best answer. Choice A is incorrect, because it tells us that regardless of the initial speed V, the final speedof the box will remain the same. That's like saying, "No matter how fast a car crashes into you, you will have the same finalspeed." Choice B is incorrect, because it tells us that the higher the value of the initial speed V, the slower the final speed ofthe box will be. That's like saying, "The faster a car is going when it hits you, the slower you will move after the collision."As for choices C and D, they both have vf increasing as V increases.The question is whether the relationship is linear. Usingmomentum conservation:

Pinitial total = Pfinal total •"• m\\\ + Mvfl = mv1f + Mv2f .'. mV + 0 = mvf + Mvf = (m + M)vf

The clay ball and the box have the same final speed, because if they stick together after the collision, then they must be

moving together (i.e., moving at the same speed.) So, solving for Vf, we get: Vf =v/3- This is similar to the equation ofa line:y = mx + b. Here, b = 0 (the y-intercept), and m = V3 (the slopeof the line). Choice D is a linear graph, so choice D is the

best answer. The best answer is choice D.

Choice C is the best answer. Asking which ball makes the box move more is equivalent to asking which ball leaves the box

in the reverse direction with a greater final speed. The final speeds are vfmai = 2V/3 in Case I and vfina| = v/3 in Case II, sothe rubber ball gives the box a bigger push. The fundamental reason why is that the ball with a bigger momentum change(bigger impulse) is going to be the ball that moves the box more (impart the greater impulse onto the box). Which ball has abigger momentum change? If two balls have the same mass m and the same speed V, the one that bounces back always hasthe bigger momentum change. It takes a greater impulse to send something back than to stop it. The rule to remember is: Fortwo objects (one elastic, one inelastic) with equal initial speed and mass, the elastic one will exhibit the greater change inmomentum. The best answer is choice C.


31. Choice B is the bestanswer. To answer this question, we need toknow something about the change in momentum of eachball and the time of contact between each ball and the block. Although we do have enough information in the passage tocalculate the momentum change for each ball, we were not given enough information that would allow us to solve for thetime of contact. However, if you know some general rules about the differences between elastic and inelastic collisions, thenyou can answer this question. The change in momentum for an elastic collision is generally larger than the change inmomentum for an inelastic collision. This is because when an object collides and rebounds elastically, there is a directionchange in the object's velocity. The incoming and outgoing velocities may have the same magnitude, but the two velocitiesare pointing in different directions. This means that the object may experience a velocity change that is twice its initialvelocity: Av = v - (-v) = 2v. A large Av implies a large force: Ap= mAv = FAt.

Whatdo we knowabout the timeof contact? The timeof contactfor an elasticcollision is generallysmallerthat that for aninelastic collision. A small At implies a big force. Thus, the rubber ball experiences a larger average force during thecollision. The best answer is choice B.

32. Choice D is the best answer. Choices A and B are incorrect: When the collision occurs, theastronaut is a distance L awayfrom the box.Thus, our answer must include L. We know that if the box has a final speed Vf, it must also move a distancevfT in T seconds. The distance between the astronaut and the box at time T = 0 seconds after the collision of the ball with

the box is L + v/T. But the distance between the astronaut and the box afterT > 0 seconds is even greater than this. Why?When the astronaut throws the ball, he recoils a little due to momentum conservation between himself and the ball. Thus,aftera timeT, theastronauthas movedawayfrom his starting position and even fartherawayfrom the box.The best answeris choice D.


33. Choice B is the best answer.The mechanical advantage of a lever system is rooted in the asymmetric distribution of the2.0-m bar about the fulcrum. When the lever system works, it functions by balancing the torques on each side of the fulcrum.Torque is found using the relationship: x = rFj.. If the fulcrum is at the 0.5 m mark, then the two sides are 0.5 m and 1.5 m.The ratio of the moment arms is 1.5 : 0.5 = 3. This means that the ratio of the forces is 3 : 1 when the torques are equal.Mechanical advantage the ratioof the weight lifted to theapplied force. The mechanical advantage is 3. The best answer ischoice B.

34. Choice D is the best answer. Because the projectile gets imbedded in the barrier, the collision is inelastic. This means thatkinetic energy is not conserved, so the kinetic energy of the incident projectile is converted to heat. Choices A and C arethus eliminated. Momentum and kinetic energy are related concepts but they are notequivalent, so wecannot speak of beingable toconvert momentum into anyform of energy. Choice B is incorrect. Theprojectile is initially moving, sotheenergy itpossesses is kinetic energy, not potential energy. The best answer is choice D.

35. Choice D is the best answer.This collision is likeCase 3. Because the twoobjects sticktogether after thecollision, the totalmass of the system increases. If the masses were equal, then we'd expect that the final velocity would be 6 m/s. However,because the incidentball is heavier than the initially stationary ball, the final velocity must be greater than 6 m/s to offset thefact that the total mass increases to be less than double. The best answer mustbe choice D. The math for this problem is asfollows:

miviinitial + mnvninitial = (mi + mil)vfinal

(0.2 kg)(12 m/s) + 0 = (0.2 kg + 0.1 kg)vfmai = (0.3 kg)vfinai

2.4 kg-m/s =(0.3 kg)vfinai => 2'4 ^^/oj kg =vfinal .'. Vfin^S^The best answer is choice D.

36. Choice C is the best answer. An object with momentum has both mass and velocity, because p = mv. Having a velocitydoes not necessarily mean an object has an acceleration, because an object moving at constant velocity can have amomentum, but not have an acceleration. Choice A is eliminated. Impulse is the change in momentum (transferred duringacollision). An object in motion does not necessarily collide with anything else, so there is not alwaysan impulse. Choice Bis eliminated. Momentum is defined as the product of velocity and mass, so an object with momentum has a mass and avelocity. Kinetic energy has both mass and velocity, so choice C is the best answer. An object with kinetic energy (andmomentum) does not necessarily have any potential energy, so choice D is eliminated. The best answer is choice C.


Passage VI (Questions 37 - 42) Automobile Safety

37. Choice B is the best answer. Consider the concept of impulse to determine the average force of impact:

P= mAvAt

We see that the impact force is inversely proportional to the time of collision. As the time of collision increases, the averageforce willdecrease. Choice A is incorrect for the reason just mentioned. Choices C and D are also incorrect; thechange in themomentum of the driver is just (mvfma] - mvjniUal). This change is the same, regardless of how the driver stops. The bestanswer is choice B.

38. Choice A is the best answer. "Safer in collisions" refers to a reduced risk of injury and damage to the occupants in a vehicleduring a collision. Damage to the occupants occurs as a result of impact, so the goal of safety engineers is to reduce theaverage force of the impact that occupants experience when vehicles collide. If you recall that impulse creates a change inmomentum, then we can apply Amv = Faverage * impact-

Assuming all other factors to be equal except mass, for a given collision speed, a larger car mass means a smalleracceleration (longer braking time), and this will be less jarring to the occupants of the car, because the applied force ofimpact is spread out over a longer period of time. While a heavier car has a greater impulse, it is its lower magnitude ofacceleration during collision that increases occupant safety. Occupant safety isenhanced by minimizing the impact force thatpassengers experience (which is achieved by building in crumple zones around the passenger compartment) and bymaximizing the time over which the collision occurs (which isachieved by using airbags and collapsible surfaces). Safety ina collision actually has very little todo with car mass, relative to the other factors such as time and vehicle deformation. Thebest answer is choice A.

39. Choice A is the best answer. An elastic collision imparts a larger average force to the colliding objects than an inelasticcollision does, because there is recoil, which causes the impulse to be greater than if there were no recoil.This may seem toimply that more damage is done during an elastic collision. But by definition, an elastic collision is one in which both themomentum and the kinetic energy of a system are conserved. In order for the kinetic energy of a system to be conserved,there canbe no lossof kinetic energy to heat, sound, or deformation; no kinetic energy is lost to the changing of shapes of thecolliding objects. Think about throwing a rubber ball at a wall and then throwing a ball made of clay at the same wall. Therubber ball rebounds elastically with no change of shape. The clay ball inelastically collides with andsticks to the wall, so itsshape is altered. Although elastic collisions exert larger forces on the colliding objects, there is no damage done to thecolliding objects, because there is no loss of kinetic energy. Note that the question asked about damage to the vehicle. Inautomobile design, the designers worry more about damage to theoccupants than the vehicle, so theydesign the collisions tobe inelastic, so that the passengers in the car experience a small impulse, and ultimately a small collision force. The bestanswer is choice A.

40. Choice B is the best answer. Momentum is a vectorquantity. Conservation of momentum must conserve magnitude as wellas direction. The final direction must therefore reflect the initial direction. Since the skid marks of the wreak point northwest,one car had to be driving north initially and the other car had to be driving west initially prior to collision. The best answeris choice B.

41. Choice C is the best answer. Let's usephysical intuition in considering limiting cases, instead of resorting to algebra. If CarII were much more massive than Car I, would Car II recoil at all? No. It would just plow through Car I, in the same way youplow through when colliding with a gnat. This rules outchoice B, and it alsomeans that the relative mass does matter, rulingout choice D. Finally, as the question is asking about an extreme example, the relative mass difference should be extreme.This points to choice Cover choice A as the better answer. If the masses were roughly equal, the collision would not result inrecoiling. The best answer is choice C.

42. Choice B is the best answer. Momentum is conserved only when there is no net external force. When two objects collide,we consider the forces that the objects exert on each other. These are internal forces. Since by Newton's third law theseforces are equal in magnitude but opposite in direction, they sum to zero, and there is no net force acting on the system.Forces likefriction and gravity are usually considered to be external forces. In solving problems, we are often told to neglectfriction. However, by extending our system to include not only the objects thatare actually colliding butalso the surroundingair and ground, then we can conserve momentum in the presence of friction. That is, the two colliding objects exchangemomentum with each other, and they alsoexchange momentum with the air and the ground. Thus, if our system includes theairand the ground, then friction isan internal force and momentum is conserved in the presence of friction. The best answeris choice B.


Passage VII (Questions 43 - 48) Defining Torque43. Choice Dis the best answer. The student is not ridding himself ofany mass during this process. Therefore, choices Aand C

are incorrect. However, he is redistributing mass. The weights are moving farther from the rotational axis, which isapproximately along his spine. According to the passage, this should increase his moment ofinertia, thus slowing him down.There isa useful example to remember any time you must consider a situation in which an object rotates or can redistributeits mass: Think about an ice skater doing a pirouette (i.e., the fancy spin where she pulls her arms in and spins much faster).Thus, ifmore mass is close to the rotational axis, that object spins more easily. If not, itspins less easily. The best answer ischoice D.

44. Choice B is the bestanswer. From conservation ofangular momentum:

Linitial = Lfinal •"• Iinitial CO initial = Ifinal CO final

where:

Ifinal is 3Ijnitia! .". Iinitial CO initial = (3Iinitial) COfinalso:

cofinal = îilii^l-3

Making a ratio of final kinetic energy to initial kinetic energy, we get:

KFf. , Îfinaicoflnal ^3Iinitial)(^^)2ifinal _ 2 _2 \ 3 / _ 1

KEinitiai i,initialco?nilial înitialcomitial 3The best answer is choice B.

45. Choice C is the best answer. Just at the instant when the weights are dropped, the skater is spinning, so the weights are incircular motion about the pivot point until released. They fall once they are dropped, but not straight downward. They fly offfrom that circle in a tangential fashion, in opposite directions. Together, they exhibit no net momentum, so the skater cannotexperience a change in angular momentum. Dropping the weights does not change the moment of inertia for the system(although its total momentum is split between two systems now). Since the moment of inertia does not change when thestudent drops the weights, then by conservation of momentum, the angular velocity of the student does not change. The bestanswer is choice C.

46. Choice C is the best answer. The student and the platform constitute a system. Before the student begins walking, theangular momentum of the system is zero. When she starts walking, the angular momentum of the system must still be zero,by conservation of momentum. Thus, the platform will rotate in a direction opposite to the direction in which the studentwalks. The best answer is choice C.

47. Choice A is the best answer. The two students and the platform are the system. Since the angular momentum of the systemis zero before the students start walking, it must remain zero after they start walking. The students have similar masses andwalk with the same speed, but in opposite directions. This gives a net momentum of zero, so the platform does not move.The best answer is choice A.

48. Choice A is the best answer. When the student moves in closer to the center of the platform, her moment of inertiadecreases. Recall that I is proportional to r2. Since her distance from the center decreased by a factor of 2, her moment ofinertia decreased by a factor of 4. Her angular speed remains the same, as stated in the passage, so her overall angularmomentum is decreased by a factor of 4. If the student's angular momentum has decreased by a factor of 4, then the angularmomentum of the platform must also decrease by a factor of 4, to conserve angular momentum. Since we cannot change themoment of inertia of the platform, the angular speed of the platform must be reduced by a factor of 4. The best answer ischoice A.


49. Choice B is the best answer. In a perfectly inelastic collision, all kinetic energy is lost, so the balls would have no speedafter collision. Choices A and D are eliminated. In an elastic collision, kinetic energy is conserved, so each ball exits thecollision with the same kinetic energy it entered the collision with. This means that the balls exit with opposite velocities ofthat with which they entered. Hence, they interchange their velocities. The best answer is choice B.

Copyright© by The Berkeley Review® 219 REVIEW EXAM EXPLANATIONS

50. Choice A is the best answer. The question does not state whether the balls move together or whether the incident ballrecoils. From the answer choices, we know that the incident ball doesn't come to restafter the collision. To determine thevelocity after collision, we need to apply conservation of momentum:

miviinitial + mnvninitial = mryifmal + mrrvnfinal

(0.1 kg)(12 m/s) + 0 = (0.1 kg)vifinai + (0.3 kg)(6 m/s)

1.2kg-m/s = (0.1 kg)vifmal + 1.8 kg-m/s

-0.6 kg-m/s = (0.1 kg)vifinal

-0.6 kg-m/s =(0.1 kg)vifinai =» "°-6 k8'm/s/o.i kg =vfmal -'• Vfinal =-6m/sThe - 6 implies that the incident object changed directions (rebounded) after collision. This seems reasonable because the0.3-kg ball has a Av of6m/s which balances out the Av of-18 m/s for the 0.1-kg ball. The best answer ischoice A.

51. Choice B is the best answer. The3-kg mass on thelefthas a moment arm of40 cmthatwould normally beoffset bya 4-kgmass at 30 cm to the right side. However, because the bar is offset in such a way that it has a torque on the right side, themass doesnot needto be a largeas 4.00 kg. Thiseliminates choices C and D. Choice B is a better option than choice A, butwithout doing the math it is strictly an intuitive answer. We can solve for the unknown mass by setting the torques on theright equal to the torque on the left

M3kggf3kg = Mbargrbar + Munknowngrunknown

(3.0 kg)g(40 cm) = (1.0 kg)g(10cm) + MunknOwng(30 cm) .-. (3.0 kg)(40cm) = (1.0 kg)(10cm) + Munknown(30 cm)

120 kg-cm= 10 kg-cm + 30M kg-cm

110 kg-cm = 30M kg-cm

110kg-cm/30cm =M ., M= H/3 kg =3.67 kg


52. Choice C is the best answer. In an elastic collision, kinetic energy is conserved. Because the speed of one of the balls dropsform 10 m/s to 6 m/s and the speed of the second balldrops from 20 m/s to some unknown, but smaller value, kinetic energyis less after the collision than before the collision. This means that the collision is not elastic, so it must be inelastic. ChoicesA and B are eliminated. To determine the final velocities of each ball, we need to apply conservation of momentum.Intuitively, we know that the balls are of equal mass, so they should each experience the same magnitude of change invelocity. One ball changes by -16 m/s, so the other ball must experience the same Av. Starting with a velocity of 20 m/s andundergoing a Av of 16/s results in a velocity of 4 m/s in the same direction. The ball is moving eastbound before thecollision, so it will have a velocity of 4 m/s in the east direction after collision.

mrviiniual + mnvninitial = mjyifinal + mnynfmal

m(-10 m/s) + m(+20 m/s) = m(+6 m/s) + mvnf,nai

10m kg-m/s= 6m kg-m/s + mvnfmai .*. 10 m/s = 6 m/s + vnf,nai

vilfinal = + 4m/s

Because we started by defining eastbound as "+", a + 4 m/s refers to a velocity of 4 m/s east. The best answer is choice C.

Copyright ©by TheBerkeley Review® 220 REVIEWEXAMEXPLANATIONS

Periodic Motion

and Waves

the

Physics Chapter 5

• initial state t - 0 initii^l state

KE=0 IfflYF=mgsin9 a=amax F=-kAx | UU\

v =0 |es=

U=mgL(l-cos0) t=| U=ikAx2

ke ^ Hy_ i /T v=v f—L/T •"=

t = Jt

E = ]

8 KE = 0a = -j

v = 0

9V T=2jt./i PE =PEraax T=23t /m"

a = ~amax

by

Berkeley Review

Periodic Motion »nd WavesSelected equations, facts, concepts, and shortcuts from this section


^spring —~KX

USpring = '/2kx2

/=1/T

Fpenduium = - mg sin6

Upendulum = -mgL(l-COS0)

V= Xf /beat = l/2-/ll

T = 2rc1 spring ^Jt J l 1pendulum —^ \i o

1 n*-/_ 1 / opendulum —Ojr V L

** = 2L/n /. = nv/2L

/spring " 2jt Vm

© Important Concept

Harmonic Oscillators (such as pendulums and springs) follow predictableperiodic behavior, so their variables obey either sine curves orcosine curves.

Pendulum /

\arc

tc, 3ji \

Equilibriumtc 3tc 5jt 7tc2'2 '2 ' 2

= ^pendulum or xspring

0, 2rc, 4tc

sin9

- ^pendulum or vspring

tc, 3tc 0,2tc

Equilibriumtc 3tc 5tc 7tc2' 2 ' 2 ' 2

F = -kx

- "pendulum or aspring

time

© Mnemonics for Selected Common Equations

Pendulums "gLide", so you need g and L in the equations of pendulumsTo compress springs you need to "kick 'em", so you need k and m in the equations ofsprings

Often, when I see someone's guitar, I covet their musical tool. "I frequently envy their 2L." (/n =nv/2L)

Physics Periodic Motion and Waves

Periodic Motion and WavesIn Chapter 3, we studied the conversion between and potential energy andkinetic energy. One ofthe examples involved the oscillation ofa bob ona string.In this chapter, we shall look at this system in more depth. We shall introducethe concept of wavesand their properties.

Simple Harmonic OscillationHarmonic oscillation is associated with cyclic processes where an objectundergoesa repeated pathway at uniform intervals. Masses on springs and bobson pendulums are typical examples of harmonic oscillation. Figure 1 shows theequilibrium and starting positions of a mass on a spring system.

(a)

Springat equilibrium

(b)

|A/WV^ iSpring

displacement

applied

Figure 5-1

One cycle of motion starts at any given point and proceeds through its pathway,moving in one direction until coming to rest, then changing direction andmoving in the opposite direction until coming to rest again, and then moving inits original direction again until returning to its reference position. From thatpoint, it repeats the pathway again. As the object moves back and forth, it is saidto be vibrating. Vibratory motion is referred to as simple harmonic motion.Because the pathway is repeated many times over, it can be described by threeimportant characteristics:

Period

This is the time required to complete one full cycle of motion. For example, if werelease the object shown in Figure 5-lb, then one vibration will be the timerequired to return to that position. The period is given the symbol T and hasunits of time per vibration.

FrequencyThis is simply the number of vibrations that occur during a specified amount oftime. The frequency is usually given by the Greek symbol nu (v). It can also havethe symbol / or f. We will use f and / in this section instead of V, because Vlookstoo much like the symbol for velocity v. The unit of frequency is the Hertz (Hz),the number of vibrations per second or the number of cycles per second.

AmplitudeThis is the magnitude of the displacement of the object from equilibrium. Themaximum amplitude is the absolute magnitude of the greatest displacement ofthe object from equilibrium. The amplitude is given by the symbol A.

The frequency and the period for a cyclic process are inverses of one another andare related by equation (5.1). It is an important equation to remember:

/-V T (5.1)


Harmonic Oscillation


Physics Periodic Motion and Waves Harmonic Oscillation

When weletgoofthe object in Figure 5-lb, therestoring force ofthespring pullsthe object to the left and back towards the wall. The object accelerates due to therestoring force, building upmomentum asit goes. It approaches the wall with anincreasing momentum and easily moves past its original equilibrium position. Itslows down once it passes through its equilibrium position. Deceleration of theobject once it has passed its original equilibrium position, isdue tothe restoringforce now pushing on the object in the opposite direction. Once the object stopsfor a brief instant, it begins to accelerate back to the right, passes its originalequilibrium position, begins deceleration, stops, and repeats the whole processoveragain, untilit eventually runs out ofenergy and comes to a complete rest.

If the system loses energy as it goes, we refer to this as dampened harmonicoscillation. The dampening is caused by friction as the mass slides across thesurface. If we observe the energy, position and velocity of the system over time,simple harmonic oscillation follows the graphs shown in Figure 5-2 anddampenedharmonic oscillation follows the graphs shownin Figure 5-3.

s

— = xor8^,♦ f

t /t I

1 I§ 1

l 1

%

\\1

vorp^. ,^—

/ / time'.

S< 2\

*

*

rf /an

\ -l / 2\ * /

V2ti%

%*

*

5n\2 \

m+^

3rt /7tc/ 2

* y

= PE

In

Figure 5-2

A diminishes but / and T remain constant

= KE

time

= xor6

diminishes over time but / and T remain constant

-- = KE

Figure 5-3



Example 5.1aAn object swings back and forth while attached to rigid cable attached to theceiling. It takes the object 2.5 seconds to swing from one side to the other side.What is thefrequency of thisoscillatory system?

A. 5Hz

B. 2.5 Hz

C. 0.4 Hz

D. 0.2 Hz

Solution

We are told that it takes 2.5 seconds for the bob to swing from one side to theother side. This means that the bob should take 5.0 seconds (double the time) toswing back and forth, thereby completing one full cycle of motion. This meansthat the period of the harmonic oscillator is 5.0 seconds. However, we are askedfor frequency, no the period. Theperiodand frequency relate as follows:

/=VT- /5s = 0-2HzThe best answer is choice D.

Example 5.1bFor a system that undergoes dampened harmonic oscillation, what observation istrue?

A. The kinetic energy at consecutive maxima decreases while the potentialenergy at consecutive maxima remains constant.

B. The potential energy at consecutive maxima decreases while the kineticenergy at consecutive maxima remains constant.

C. The frequency of the system decreases as it loses energy, while the periodremains constant.

D. The speed at consecutive maxima decreases while the period at consecutivemaxima remains constant.

Solution

Dampened harmonic oscillation refers to a cyclic process that loses energy as itmoves. The loss in energy is generally attributed to friction (work energy lost asit moves against a resistive force). This drains the total energy of the system,causing both the kinetic energy and the potential energy at their respectivemaxima to be lower as time progresses. This eliminates choicesA and B.

Choice C cannot be true, because the frequency and period are inverses of oneanother. If one decreases, then the other one would have to increase. A decreasein frequency could not happen with a constant period.

Speed at consecutive maxima does in fact decrease, because the system loseskineticenergy as it undergoes its cyclicmotion. A loss in kineticenergy is in facta loss in speed. What seems odd at first about choice D is that the period isremaining constant. Dampening reduces the amplitude (displacement, which isassociated with potential energy), but it does not reduce or increase the timenecessary to complete a cycle. Over time, the object will cover less distance in agiven cycle, but because it's doing it at a reduced average speed, the time tocomplete a cycle remains constant.



Harmonic Oscillation


Physics Periodic Motion and Waves Springs and Pendulums

Springs and Pendulums

Spring SystemsIfan object experiences a deformation while being compressed or stretched, thenthat object is said to be elastic. A classic example of an object that is elastic is aspring. Materials that are elastic include rubber bands, wood, and even metal.Eachof these will deform, if a great enough force is applied to them.

Let'sconsider the deformation of a spring. If no force is acting on the spring, thenit is in equilibrium. If we push on the spring with a force F, then the spring willcompress farther and farther away from its original resting position. The forcewe are using to push the spring away from its original resting position isproportional to the distance (x) the spring moves. Figure 5-4 shows the positionof a mass on a spring over time, where the time is measured on a radian scale(which compares the cycle of the mass on the spring to the path around a unitcircle with which it shares the same period.)

initial state

t = 0

t= "

t = TT

t= 2n2

t = 2TT

'- M

PE = PErKE = 0

a = amaxV = 0

equilibrium line

-x

- M

Figure 5-4

PE = PE^KE = KEra = 0

v = V™™

PE = PErKE = 0

a = "ama>

V = 0

PE = PEaKE = KEra = 0

v = -V™™

PE = PErKE = 0

a = amaxV = 0



With aspring, you must push with agreater force as you continue to compress it,implying that the force with which you pushis proportional to the distance thatit compresses from equilibrium (f« x). Inorder to convert this proportionality toan equation, we need to add a proportionality constant k. This proportionalityconstant is called the force constant, and it depends on the rigidity anddimensions of the spring. This constant has units of Newton per meter (N/m)and is always a positive number. If k is small, then the spring can be movedeasily with a small force. However, if k is large, then it will require much moreforce to move the spring. The relationship between the force needed to displacethe spring and the force constant is shown in equation (5.2). This is calledHooke's Law (first proposed by Robert Hooke in 1676):

F = -kx (5.2)

The negative sign indicates that it is a restoring force. If we were to remove theforce on the spring after we have displaced it a certain distance, then therestoring force would return thespring to its original resting position (see Figure5-4). In other words, the force of the spring acts in the opposite direction of thespring's displacement.

Because the restoring force of the spring system increases as it gets furthercompressed, the work required to compress a spring increases the more thespring gets compressed. The work needed to compress a spring is found usingthe standard W= FaVerage x Ax, where Faverage is -kx/2 and the Ad is-x. Theaverage force is -kx/2, because the force goes from 0 to -kx as ifs deformed, sothe average force during the deformation is -kx/2. The work done on a springbecomes potential energy, which means that the potential energy of a springsystem cab be found using equation (5.3).

U spnng = i/*kx2 (5.3)

Example 5.2aIf a compressed spring stores 1700 ergs of potential energy and has a springconstantof200 dyne/cm, by how much is it compressed?

A. 2.9 cm

B. 4.1cm

C. 8.5 cm

D. 17 cm

SolutionSquare Root Estimation: Numeric problems with square roots are best solved byusing square root estimation. We want a distance, and we are givenan energy anda springconstant. Thepotential energystored in a compressed spring is:

PE = —ky21 Espnng 2

This gives:

x =2 (1700 ergs)

200 dynes/cm

V340Q ergs-cm _^yj200 dynes

cm" 4 cm

since 17 is close to 16. When confronted with a square root, try to make thenumberunder the square root symbola number whosesquare root you know.




Test TipSquare Root Estimation



Example 5.2bWhen a vertically oriented spring scale supports a 180-N block, the springstretches 0.3 m from rest. Neglecting any other masses associated with the scale,what is the value of the spring constant?

A. 30 N/mB. 294N/mC. 600 N/mD. 5880 N/m

Solution

To determine the spring constant, we need to consider the vertical forces actingon the system. Once the spring is fully stretched and no longer moving, then thenet force up equals the net force down. The force up is the restoring force of thespring whilethe force down in gravitational in nature. Fromthis we get:

mg =-kAx .-. k=mg/Ax =180N/o3m =1800N/3m =6ooN/m

The negative sign is ignored because the answer choices don't contain it. The bestanswer is choice C.

The spring constant plays a significant role in the potential energy and therestoringforce of the springsystem. It also impacts the frequency and the periodofa spring system. Equation (5.4) describes the period ofthe pendulum system,T,and equation (53) describes the frequency of tne pendulum system.

(5.4)

(5.5)

Example 5.3The spring scale and block from Example 5.2b are now set into oscillatorymotion, with a fixed frequency. If a second block is added to the scale, thefrequency of oscillation will:

A. increase.

B. remain unchanged.C. decrease.

D. depend upon the value of the spring constant.

Solution

The oscillation frequency of a mass/spring system is inversely proportional tothe square root of the mass. Thus, if the mass increases the frequency mustdecrease. Notice that we do not worry about spring constants or stray factors of2tc. These things are constant. Constants will never enter into a qualitative changeproblem such as this.


When it comes to spring systems, you should note that the equations contain kand m. To help recall the equations for springs (equations (5.4) and (5.5)), youshould note that to get a spring going, you need to kick 'em! There is a k and anm in "kick 'em' and spring equations.



Pendulum SystemsWe have seen how a spring can display simple harmonic motion by moving backand forth on a horizontal surface. An object like a pendulum swinging back andforth can also undergo simple harmonic motion. A pendulum is simply a masshanging on the end of a cord. The cord itself has negligible weight. Figure 5-5shows the details of a simple pendulum.

% arc

mg cos8

Figure 5-5

As shown in Figure 5-5, there are two forces that act on a simple pendulum, thetension (T) in the cord and the weight (mg) of the object at the end of the cord.Note that the tension in the cord is perpendicular to the arc described by themotion of the pendulum. The magnitude of the tension varies with the anglebetween the cable and the equilibrium line. Tension is greatest when the massivebob points straight down. The weight of the object at the end of the cord can bebroken down into two components. The component that is perpendicular to thearc has a magnitude of mgcosO while the component that is tangent to the archas a magnitude of mgsinO.

If we were to pull the pendulum to the right as shown in Figure 5-5and let it go,then it would swing back to the left. The component that acts to restore thependulum to its original position is the component that is tangent to the arcdescribed by the pendulum, mg sinB. The sign of the restoring force is negative,because that force is opposite to the direction in which the pendulum is beingdisplaced. In other words, the restoring force is given by equation (5.6):

Frestoring =- mg sinO (5.6)

As was the case with the spring system, a pendulum also has potential energy, aperiod, and a frequency. The respective equations for the potential energy,period, and frequency of a pendulum are given as equations (5.7), (5.8), and (5.9).

Upendulum =- mgL(l - COS0)

T=2ti -^V

3 2tt\| L

9 = 9max cos2n/t


(5.7)

(5.8)

(5.9)

(5.10)




When it comes to pendulum systems, you should note that the equations containg and L. To help recall the equations for pendulum (equations (5.8) and (5.9)),One of our physics teachers, likes to say that "when you release a pendulum, itglides."There is a g and an Lin "gLide' and pendulum equations.

t = 0

.-3

t = Tt

t= 3e2

t = 2n

initial statePE = PErKE = 0

a = amaxV = 0

PE = PErKE = KE

a = 0

v = v.

max

PE = PErKE = 0

a = "ama>V = 0

PE = PE

KE = KEra = 0

iron

PE = PE

KE = 0

a = amav = 0

max

Figure 5-6

When we let go of thebob at t = 0 in Figure 5-6, the gravitational restoring forcepulls the bob down and to the right towards the equilibrium point. The bobaccelerates counterclockwise around the arc, building up momentum as itdescends. It approaches the nadir (low point of the path) with an increasingmomentum and easily moves past its original equilibrium position at t = n/2.The bob decelerates and slows down once it has passed its equilibriumpositionand starts to climb again, moving against the gravitational restoring force nowpulling down on the bob. Once the object stops for a brief instant at t = n, itbegins to accelerate back down and to the left, passing through its equilibriumposition(at t = 3rc/2), whereit onceagain slow down and climb again. It stops att = 2rc, and then repeats the whole process over again. If it undergoes dampenedharmonic oscillation, it would eventually run out of energy and come to rest.



Example 5.4aA string and mass pendulum swings with a given period. If the string lengthwere doubled, the period of oscillation would be:

A.

B.

C.

V2 times the original period.2 times the original period.

-pr times the original period.72

D. i-times the original period.

Solution

We know that longer pendula have a greater distance to traverse per cycle thanshorter-stringed pendula, so they take longer to complete a cycle (i.e., have alarger period).Therefore, we cross out choices C and D.To differentiate betweenchoicesA and B, we remember that a square root is involved in the equation, sothere must be a square root in the answer. Choice A is a more logical selectionthan choice B. Try to use physical intuition to narrow your choices, first. It oftentakes less than five seconds to do this. If we had solved this formally, we wouldget choice A, in accordancewith equation (5.8).

Square Roots in Equations: If you encounterquantitative change problems, and therelevant equation involves a square root, then thestandard loreis that the answerwith the square root is correct. This is not always true, but it can work.


Example 5.4bThe acceleration due to gravity affects the frequency at which a pendulum willoscillate. If a pendulum were swung on the surface of the Sun, where theacceleration due to gravity is 28 times that on the Earth, the frequency ofoscillation would be:

A. 0.19 times that of the Earth.B. 4.80 times that of the Earth.C. 5.29 times that of the Earth.D. 28.0 times that of the Earth.

SolutionThis requires that you recall the equation for the frequency of a pendulum. Forboth a pendulum and a spring, the equation for the frequency involves a l/2rcterm and a square root of a restoring term over a dimension of the system. Forpendulums, our mnemonic is that "pendulums gLide", reminding us that thetwo terms that matter are g and L. This helps us to recall that the equation forthe frequency of a pendulum is:

/.ill

A larger value for g would imply a higher frequency, which makes sense giventhatit will be pulled back downwith greater force thanbefore and thus take lesstime to descend to equilibrium. This eliminates choice A. Because the equationcontains a square root, we know that thebestanswer is going tobe slightly morethan 5, the square root of 28. The best answer is choice C.



Test TipSquare Root Questions


Physics Periodic Motion and Waves Wave Properties

Wave Properties

Transverse and Longitudinal Wave MotionWaves can transfer energy,but they cannot transfer matter from one location toanother location. Be aware that the transfer of matter and the transfer ofinformation from one placeto another are very different properties.

The two types of waves that we will study are transverse waves and longitudinalwaves. Transverse waves cause vibrations perpendicular to the direction ofwave propagation. A good example of a transverse wave occurs at sportingevents when fans do "the wave." As one of our teachers says, "fans do thetrans... verse wave that is.". Fans move up and down, but the wave propagateslaterally. Radio and light waves are also transverse waves. Longitudinal wavescause vibrations parallel to the direction of wave propagation. You can say toyourself, "the particles move along... itudinal with the wave.". Sound waves areexamples of longitudinal waves.

Amplitude, Wavelength, Frequency, and VelocityBefore we begin to discuss specific characteristics of these types of waves, weneed to define a few important quantities.

AmplitudeThe amplitude of a wave is the magnitude of the maximum displacement of awave from its point ofequilibrium. This can be seen in Figure 5-7:

propagation of wave

Figure 5-7

WavelengthThe wavelength (represented by the Greek symbol lambda, X) is that distancebetween twoidentical points ofdisturbance in a wave. For example, in Figure 5-7 the wavelength is indicated as beingbetween two consecutive crests. It couldaseasily have been represented asbeing between two consecutive troughs.

FrequencyThe frequency of a wave is the number ofvibrations ofthe wave per unit time.Recall thatthe frequency ofavibration can begiven byequation (5.1).

VelocityThe speed ofa wave ishow far thewave travels per unit time. When thedistancea wave travelsis givenby Xand the time is given by the period T, the speed v ofa wave canbe expressed as shownin equation (5.11). Substituting equation (5.1)into equation (5.11) gives (5.12), which relates the wavelength, frequency, andspeed of a wave.Equation (5.12) canbe rearranged to giveX= v/f or / = v/X.

v =V' (5.11)

v = X/ (5.12)



Equation (5.12) is an essential equation in physics. Ifwehave a shortwavelength,then the frequency of that wave must be large. Conversely, if the wavelength islong, then the frequency of that wave must be short. This is an equation wellworth remembering, as you will use it on mostwave questions!

Since there are different typesofwaves, thereare different equations that onecanuseto calculate the speedof a wave. Forexample, wecancalculate the speed ofatransverse wave propagating along a string using equation (5.13).

v =(5.13)

In this equation, T is the tension in the string and p. is the mass per unit length,and v is speed, as it is in the more universally applicableequation (5.12).

Example 5.5aWhich graph BEST represents the relationship between the frequency andwavelength of waves emitted from a stereo speaker, if the wave speed is fixed?

A.

e0)

cr

wavelength(X) wavelength (X) wavelength (X) wavelength (X)

Solution

This is a graph identification problem. First, we need to determine if frequencyincreases or decreases with increasing wavelength. Based on v = /X, / and Xareinversely proportional to one another, so choice A is eliminated. The threeremaining choices say that frequency decreases as thewavelength increases.. Therelationship is not linear, so choice B is eliminated. Neither / nor Xshould reach0,so choice D is eliminated. ChoiceC represents an inverse relationship.

The correct choice is C.

Example 5.5bWhich graph BEST represents the relationship between the wavelength andspeed of wavesemitted from a tuning fork, if the wave frequency is fixed?

A. B. c. D.

^? V ^ y ^ ^"* ^<w^ ^r **-*'

•5bO / %/ fC X C X c / c0) X v y « / *0> X a* y *> / ^> X > / > / >(0 X « y « 1 <B$ 1 ^- £ Y. * E £

speed (v) speed (v) speed (v) speed (v)

SolutionFirst, we need to determine if wavelength increases or decreaseswith increasingspeed. Wavelength is directly proportional to speed, so choices A and D areeliminated. The relationship is linear, so choice C is eliminated.



Wave Properties



Superposition of WavesSo far we have been considering waves as if there were just one wave travelingthrough a medium at any given moment. Obviously, this is notalways the case.We hear many different sounds, and we see many different colors. Each soundthat we hear and each color that we see is associated with a particular wave, andthat wave has a defined frequency and a defined wavelength. Later, we will seethat the energy of those waves is also defined.

Most of the waves that we encounter in nature follow the principle ofsuperposition. The principle of superposition says that if two or more wavespassing through the same medium meet at a particular point in that medium,then the net displacement by those waves is the addition of the individualdisplacementsby the wavesas if they were traveling through the medium alone.

Even though the principle of superposition works rather well under mostconditions, it can fail if the wave amplitude is extremely large. Why? When awave's amplitude is too large, it can change the nature of the medium ittraverses, which, in turn, can change the nature of the wave propagation.Imagine, for example, sending sound waves through some solid material. Soundis a vibrational wave, which can cause the solid to heat up. If two waves ofsufficient amplitude constructively superposed in some region within the solid,the solid might heat up enough to change conhgurationally (e.g., melt orreorder). This would change the elasticity of that region and, therefore, distortthe waves that subsequently leavethat region.

We canexamine the principle ofsuperposition by considering two types ofwave transmission. We shall first consider the interference of waves that sharethe same frequency and wavelength, after which we will consider theinterference of waves with different frequencies and wavelengths. Figure 5-8shows the interference of two waves propagating along the same path, but inopposite directions.

(a) Constructive Interference (b) Destructive Interference(In-Phase Addition) (Out-of-Phase Addition)

f5\ fa ,.„ /3\_

/SWS\_ , &* rDoubledamplitude Cancelled amplitude

t = 2

/S^_f3u_ ,., fa\s/

fa R\ ... , , R\VST

Figure 5-8

Copyright ©by TheBerkeley Review 234 The Berkeley Review


Constructive Wave Interference

Suppose we send two wave pulses toward one another along a long cord, asshown in Figure 5-8a. Both of these waves are said to be in phase with oneanother. Note that that for simplicity we chose waves that have the sameamplitude. When these waves meet, they interfere with one another in aconstructive manner. Their amplitudes combine and, for a brief instant, theresulting wave combination shows a larger amplitude than either of theindividual waves. Once the individual waves pass through one another, theyretain theiroriginal amplitudesand continue theirpropagation.

Figure 5-9 shows constructive interference between two in-phase wavespropagating together alongthe sameaxis. A good example of the phenomenon iswhen you listen to two speakers playing the same music.

Wave 1

Wave 2

Doubled amplitude

Figure 5-9

Destructive Wave Interference

If we send two wave impulses toward one another along a long cord, as shownin Figure 5-8b, then when they meet they will interfere with one another in adestructive manner. Since the amplitudes of the individual waves are 180° out ofphase with one another, the resulting wave combinationwill show no amplitudeat all. This is because the displacement of the individual waves cancels out.However, once the waves pass through one another, they will propagate alongthe cord with their original amplitudes.

Figure 5-10 shows destructive interference between two in-phase wavespropagating together along the sameaxis. A good example of the phenomenon isanti-sound, where a device sends out a wave designed to cancel out an existingsound wave.

Wave 1

Cancelled amplitude

Wave 2

Figure 5-10

If two waves have different amplitudes and they approach one another, in a waysimilar to that in Figure 5-8b, then they will not completely cancel each other outwhen they destructively interfere with one another.


Wave Properties


Physics

loud

Periodic Motion and Waves Wave Properties

BeatsInterference patterns need not appear completely constructive or destructive.Those extremes occur only when the wavelengths of the interfering waves areidentical. If the wavelengthsof the waves are different, even by a small amount,the interference pattern can change drastically. A phenomenon known as a beatoccurs when two waves of differingwavelengths interfere with each other. Beatsare most noticeable when the two wavelengths are close to the same. How canwe graph a beat? Imagine two sound waves of 6 Hz and 4 Hz, as schematicallydrawn in Figure 5-11. When these two waves interfere with each other, theyproduce the third wave in the figure.

This wave is noticeably more complex than either of its original wavecomponents. It has a modulating amplitude that is sinusoidal, with a frequencyof6Hz-4Hz = 2 Hz. This is the beat frequency. Generally, if waves offrequency f\ and fi interfere, theywillgive rise to a beat frequency of:

/beat= 1/2-/1 (5.14)

Are beats useful? Yes. For instance, if you've ever played a stringed instrumentand tuned it using a tuning fork, you've relied on beats. In tuning a guitar, thefork (of known frequency) and the string (of unknown frequency) are strucksimultaneously. If a beat is heard, then the fork and string are at differentfrequencies. If no beat is heard, then the string is properly tuned to the frequencyof the fork. As the two component frequencies get closer to one another the beatfrequency gets smaller and more noticeable.



Standing WavesWe have seen that the constructive interference of two waves with similaramplitudes can momentarily produce a wave of a larger amplitude. Similarly,thedestructive interference between twowaves canmomentarily lead to a regionof no amplitude. In both of these cases, the waves continue on their way afterthey have interacted with each other. However, there are waves that can interferewith one another to produce a disturbance called a standing wave. These typesof waves do not appear to travel along their line of propagation.

What exactly is a standing wave? It is a wave that appears to have nopropagation velocity. It is a wave that results from the interference of two wavestraveling in opposite directions. Consider a rope of length L tied to a tree at oneend. The other end is held in your hand. If you begin to shake the rope up anddown, you eventually produce a continuous wave pattern as shown in Figure 5-12. This type of vibration is called the fundamental vibration or the firstharmonic of the standing wave. "Standing" refers to the wave's apparent lack ofpropagation velocity to the left or right. Actually, waves are still traveling to theright, as your hand shakes, and to the left, as those waves reflect off the tree.However, the way in which the left and right traveling waves interfere with eachother gives rise to this apparently stationary wave.

m=2L/i

N = nodeA = antinode

A. /1stHarmonic (L = / 2 •*• X=2L)

Figure 5-12

The specific standing wave in Figure 5-12 is called the firstharmonic, because itis the longest possible wavelength that can be generated that meets therestriction of a no movement at each end of the rope.

Harmonic frequencies are whole-number multiples of a fundamental frequency.Note that there is no vibration of the rope at either end. Areas of no vibration arereferred to as nodes. Nodes are often a restriction of our system and are found atboth ends of fixed-end strings. The area of the vibration where the amplitude isthe largest is referred to as an antinode. Antinodes are found symmetricallydisplaced between adjacent nodes.

If you shake your hand up and then down, then the period required for the wavepulse to travel down the rope and then back to you hand is given by equation(5.15). Recall that the period and the frequency are related to one another byequation (5.1). Substitution of equation (5.15) into equation (5.1) gives equation(5.16). This frequency defines the fundamental frequency. The fundamentalfrequency is usually given the symbol f\ to indicate that it is the fundamentalfrequency and not some other frequency. The wavelength of the first harmonic,as shown in Figure 5-12,is equal to 2L.

Ti =^A (5.15)

/l = v/ 2L (5.16)


Wave Properties



For waves on strings that are fixed at both ends, like our rope tied to a treeexample, there are other frequencies that will produce nodes at each end. Theseother frequencies are multiples of the fundamental frequency. In the secondharmonic, there are 3 nodes and 2 antinodes. This type of vibration is alsoreferred to as the first overtone.Figure5-13 shows the second harmonic.

N X2 =2L/2

2nd Harmonic (X= L)

Figure 5-13

Figure 5-14 shows the third harmonic. You may note the pattern between thewavelength for the particular harmonic and the length of the string. This patternleads to the equations we apply to fixed-end strings.

?N ^3 =2L/3

3rd Harmonic (A. = / 3L)

Figure 5-14

The trend we see in the wavelength in Figures 5-12, 5-13, and 5-14 allows us towrite a general expression for the wavelength of fixed-end strings, shown asequation (5.17). We can calculate the general harmonic frequency by substitutingequation (5.17) into equation (5.18), where n = 1, 2, 3, 4, 5,.... In each case, n is apositive integer.

X -2L/An- In

/n = nv/2L

(5.17)

(5.18)

Because harmonic frequencies are multiples of the fundamental frequency, f\,we can rewrite equation (5.18) in such a way that the harmonic frequency relatesto the fundamental frequency, as shown in equation (5.19).

/n = n/i (5.19)



Example 5.6aWhena string vibrates in its second harmonic mode, at what point on the stringis the string motion maximal?

A. One-quarter of the length away from an end.B. In the middle, between the two ends.C. One-third of the length away from an end.D. The amplitude is the same at any point on the string.

SolutionThis question is most easily answered if you have a picture of second harmonicmotion in your head. Most of these standing wave problems, here and withsound waves (SectionVI),are best solved by knowing what they look like.

2nd Harmonic (X= L)

The string vibrates the most at a distance of L/4 from either end of the string.The middle of the string for the second harmonic is a node, so choice B iseliminated.The string at the one-third position is displaced,but not in a maximalfashion, so choice C is eliminated. The amplitude is not the same at all positionsifyou have bothnodes and antinodes in thewave. The bestanswer is choice A.

Example 5.6bFor a fixed-end vibrating string in its second harmonic, at what point on thestring is the wavespeed the fastest?

A. At an antinode.B. At a node.C. In the middle, between the two ends.D. The speed is the same at any point on the string.

SolutionThis question is one where it is easy to make a careless mistake based onvisualization. It is not asking for the point at which the string itself is movingfastest (which would be an antinode). It is asking about the wave speed on thestring. The speed of thewave is constant in a uniform medium, meaning that thewave travels back and forth across the string at constantspeed, meaning that thespeed of the wave is the same at all points.



Wave Properties


PhysiCS Periodic Motion and Waves Wave Properties

ResonanceIf the amplitude of a vibrating system begins to decrease, then energy is beinglost. The system is said to be damped and the damping is generally due to sometype of friction. However, the amplitude, and hence the energy of the system,may increase if we supply a vibrating force that has a frequency which is thesame as the natural frequency of the system. This phenomenon is calledresonance.

What does it mean to resonate at a "natural frequency"? Most objects can vibratein some way. A cube of Jello, a guitar string, pendulum, and a covalently bondedmolecule all vibrate under different conditions and with different typicalfrequencies. These typical frequencies are simply their natural frequencies (i.e.>the frequency at which they wouldvibrate if you perturbed thembriefly and leftthem to shake on their own.) If you try to shake an object with some givenfrequency, it is likely to shakeonlywhen your shaking frequency is at or near theobject's natural frequency. Shaking it at some vastly different frequency willtypically not affect the object. But shaking at or near that resonant frequencydoes more than cause the object to vibrate. Your shaking gives energy to theobject, thereby increasing its vibrational amplitude (in some cases indefinitelyand disastrously).

Resonance phenomena show up in many different facets of physical science.Here are some examples:

When a child ona swing ispushed at the natural frequency of the swing,the childgoeshigherand higherwith eachpush.

When a singer hits and holds the proper note, she can shatter a wineglass,if the glasshas a matching resonant frequency.

After the Tacoma NarrowsBridge was completed in Washington 1940, asmall gale forced the bridge to start oscillating at one of its naturalfrequencies. As the gale continued, the oscillating amplitude grew untilthe bridge eventuallycollapsed.

When applying an oscillating magnetic field at the natural spinfrequency of a proton, the proton spin can resonate. This is called *HNMR

A carbonylbond in an organicmolecule will resonate and heat up whensubjected to incidentinfraredelectromagnetic radiation at 1700 cm'1.

When a surgeon wants to destroy gall stones or kidney stones in apatient, she can direct an ultrasonic wave at the stones, at their naturalresonance frequency. The vibration amplitude of the stones increasesuntil they shatter, making them easier to remove.

Thekey to identifying resonance phenomena is to recognize that a givenpassageinvolves some object that can vibrate, an applied vibrational force, and a matchbetween the natural and forcedvibration frequencies.


I. Pendulum Acceleration by Gravity

II. Strings and Things

IE. Feel the Vibes


(1-7)

(8-14)

(15-21)

(22-25)

The main purpose of this 25-question set is to serve as a review of the materialpresented in the chapter. Do not worry about the timing for these questions.Focus on learning. Once you complete these questions, grade them using theanswerkey. For any question you missed, repeat it and write down your thoughtprocess. Then grade the questions you repeated and thoroughly read the answerexplanation. Compare your thought process to the answer explanation and assesswhether you missed the question because of a careless error (such as misreadingthe question), because of an error in reasoning, or because you were lackinginformation. Your goal is to fill in any informational gaps and solidify yourreasoning before you begin your practice exam for this section. Preparing for theMCAT is best done in stages. This first stage is meant to help you evaluate howwell you know this subject matter.


Two physics students perform an experiment todetermine the value of g, the acceleration due to gravity. Fivesmall balls of measured mass 100 g each were attached tofive different light, inflexible strings, each forming apendulum of varying length. Each pendulum was hung fromthe ceiling and set into small amplitude oscillations. Thestudents measured the time t to complete 10 cycles for eachpendulum. Their data are summarized in Table 1.

Length(meters)

Time(seconds)

Trial 1 1 20.1

Trial 2 2 28.4

Trial 3 3 34.8

Trial 4 4 40.4

Trial 5 5 44.9

Table 1. Experimental results

Note: / = frequency, T = period, L = length of pendulum.Assume air resistance is negligible.

1. Which of the following graphs BEST represents therelationship between period T and frequency / for asimple pendulum?

A. B.

2. Which of the following statements is true for the massof an oscillating pendulum?

A. The mass' angular momentum and kinetic energyare conserved.

B. The mass* momentum and angular momentum areconserved.

C. The mass' momentum and kinetic energy areconserved.

D. The mass' momentum and angular momentum arenot conserved.


3. What is the approximate frequency of vibration forTrial 4?

A. 0.025 Hz

B. 0.25 Hz

C. 4.0 Hz

D. 40.0 Hz

4. The frequency of oscillation for a pendulum is givenby:

5.

6.

7.

2jc V LWhat formula could the students have used to calculate

the acceleration due to gravity?

A. -^~

B.

C.

4n'L

20nL2t2

20jtL

t

D 400ji2Lt2

If an additional trial were conducted with L = 9 m, whatwould be the expected period, T?

A. 3 seconds

B. 6 seconds

C. 30 seconds

D. 60 seconds

What is the tension in the string in Trial 4 if thependulum bob swings through equilibrium with a speedof4 m/s? (Assume g= 10m/s2 for this question.)A. 0.4 N

B. 0.6 N

C. 1.0 N

D. 1.4 N

If the pendulum bob in Trial 4 were initially raised 0.5m above the ground, as shown in the diagram below,what would be the approximate speed of the bob as itpassed through equilibrium?

A.

B.

C.

D.

V(2)(9.8)/(0.5) m/s

V(2)(9.8)(0.5) m/s

V(2)(100)(0.5)/(9.8)m/s

V(2)(9.8)/(0.5)(100) m/s



Consider two strings of the same material with the sameradius, but havingdifferent length. Each string is fixed to animmovable object at one end and the end of an oscillator atthe other end. Standing waves are established on both of thestrings using the oscillator set a frequency of 60 cycle/sec.The strings oscillate as shown in Figure 1.

String I

6n

8.

9.

3H

y

-3-

-6-

3 cm

Figure 1. Two oscillating strings

Which wave has the larger amplitude?

A. The wave on String I

B. The wave on String II

C. Both waves have the same amplitude.

D. Neither wave has any amplitude.

How does the speed of the wave on String I compare tothe speed of the wave on String II?

A. Wave I is twice as fast as Wave II.

B. Wave II is twice as fast as Wave I.

C. The wave speeds of the two waves are equal.D. There is no way to determine the relative wave

speeds.

10. For String II, the string moves the most at the:

A. 1 and 5 cm positions.

B. 2 and 4 cm positions.

C. 1,3, and 5 cm positions.

D. 0,2,4, and 6 cm positions.


11. Changing the velocity of the waves on the strings canbe accomplished by:

A. increasing the string tensiononly.B. increasing the stringdensity only.C. increasing the string tension or density.D. It is not possible to change the velocity of a wave

on a given string.

12.

13.

14.

The waves pictured in Figure 1 are best described as:

A. transverse traveling waves.B. longitudinal traveling waves.

C. transverse standing waves.D. longitudinal standing waves.

Consider a third string whose length is the same as thelength of String I and whose ends are also fixed. Thisthird string is now vibrated by some mechanism and itoscillates as shown below.

3n

y

-3Jcm

When comparing the vibration of this third string toString I, which of the following may be concluded?

A. The frequency is 3 times that of String I.

B. Thefrequency is V3 times thatof String I.C. The wavelength is 2 times that of String I.

D. The wavelength is Vi times that of String I.

At a particular time, the amplitude of String I is zero.What can be said about the energy of the standing waveat this point?

A. The energy of the wave is zero.B. The wave has only kinetic energy.

C. The wave has only potential energy.D. The wave has both potential energy and kinetic

energy.



Simple harmonic motion is different from other kinds ofmotion in that the acceleration is not constant It depends onthe position of the vibrating object in the way:a a -x.

Specifically, for a mass on a spring,we have:

a = --£-xm

In order to study simple harmonic motion, a group ofstudents are given the following set-up: an air cart of knownmass is attached to a spring. The spring is attached to an airtrack so that the cart and spring are horizontal. With the airtrack functioning, friction between the air cart and air trackcan be neglected. (Neglect air resistance as well.) When thecart is given a small initial amplitude, the cart will vibratewith a characteristic period given by:

t=*<VTThe potential energy of the spring can be written as:

U =±kx22

where x is the amount that the spring is stretched orcompressed away from equilibrium.

In Experiment 1, the students use a spring of unknownspring constant. The cart is displaced an unknown amountfrom equilibrium and then released. The students measure thetime it takes for the cart to move through 10 cycles.

In other experiments, a spring of known spring constant isused.

15. What does the data in Experiment 1 allow the studentsto measure or calculate?

A. Period, Frequency, Spring constant,andAmplitudeB. Period, Frequency, and Spring constantonlyC. Spring constantand Amplitude onlyD. Period, Frequency, andAmplitude only

16. Changing the initial displacement will affect which ofthe following?

I. Period

II. Spring constant

III. Maximum acceleration

IV. Maximum speed

A. I, II, and III onlyB. I, HI, and IV onlyC. Ill and IV onlyD. I, II, III, and IV


17.

18.

19.

Suppose the mass is pulled a smalldistance to the rightof the equilibrium position and then released when t =0. Which of the following BEST represents a graph ofthe velocity vs. time for the vibrating mass?

A.

Which of the following statements is true about thesimple harmonic motion of a mass on a spring?Consider magnitudes only.

A. As the mass moves from its initial amplitude backto equilibrium, the speed of the mass increases,while the acceleration of the mass decreases.

B. The mass reaches its maximum speed at the sametime it has its maximum acceleration.

C. The mass reaches its maximum speed when it is atits maximum displacement from equilibrium.

D. The equilibrium position is so called because atthat point, the mass has no acceleration andtherefore is at rest.

Which of the following changes could double thefrequency of oscillation?

A. Doubling the initial amplitude.

B. Quadrupling the mass.

C. Doubling the spring constant

D. Quartering the mass.


20. At whatposition x will the speedof the mass be half itsmaximum speed, vmax?

A. x = vmax<\ 4k.

B. x=./5m^V 4k

C« x = —-vmax2k

D- x = vmax

max

V 4k

21. Doubling which of the following would give theGREATEST change in the elastic energy stored in thespring?

A. The spring constant

B. The amplitude of motion.C. The frequency of motion.

D. The cart mass.


Questions22 through 25 are NOT basedon a descriptivepassage.

22. What is the ratio of the frequency of a pendulum on theEarth to that of an identical pendulum on the Moon?(The force of gravity is six times weaker on the Moonthan on the Earth.)

A. 36:1

B. 6:1

C. V6:l

D. 1:1

23. What is true of the velocity of a pendulum bob at itslowest point?

A. It increases as the mass of the bob increases.

B. It decreases as initial displacement angle increases.

C. It increases as the cord length increases, given thesame initial displacement angle of the pendulum.

D. It decreases as the cord length increases, given thesame initial displacement angle of the pendulum.

24. Increasing the wave amplitude of a standing wave will:

A. increase the number of nodes.

B. increase the number of antinodes.

C. increase the displacement of the cord at a node.

D. increase the displacement of the cord at anantinode.

25. All of the following affect the spring constantEXCEPT:

A. the material composing the spring.

B. the number of coils in the spring.

C. the direction of the coiling of the coils in thespring.

D. the diameter of each coil in the spring.

1. C 2. D 3. B 4. D 5. B

6. D 7. B 8 B 9. B 10. C

11. C 12. C 13. B 14. B 15. B

16. C 17. B 18. A 19. D 20. D

21. B 22. C 23. C 24. D 25. C

YOU ARE DONE.

Answers to 25-Question Periodic Motion and Waves Review

Passage I (Questions 1-7) Pendulum Acceleration by Gravity

1. Choice C is the best answer. Period is the inverse of frequency, T = Iff. The graph that best represents the relationshipbetween period T and frequency / for a simple pendulum is one of inverse proportionality. As frequency increases, theperiod decreases, which eliminates choices B and D. The relationship is not linear, sochoice A iseliminated. The graph thatshowsan inverserelationship is choice C. The best answer is choiceC.

2. Choice D is the best answer. Which statement is true for an oscillating pendulum? We have to consider whether or notkinetic energy, momentum, and angular momentum are conserved. As a pendulum swings back and forth, its speed andtherefore KE are constantly changing and not conserved. Therefore, we can rule out answer choices A and C. Is momentumconserved? In order to conserve momentum, there must be no net external force acting on the pendulum. We have theexternal forces of gravity and tension acting on the pendulum, so momentum is not conserved. Additionally, the momentumof the mass must change, because its velocity changes.This rulesout answer choice B. The best answer is choice D.

3. Choice B is the best answer. What is the approximate frequency of vibration for Trial 4? The time for 10 oscillations wasmeasured to be 40.4 seconds. The period is the time for 1 cycle, which in this case is about 4 seconds. Period is the inverse of

frequency, so: / is / =^ = i-= 0.25 Hz. The best answer ischoice B.T 4

4. Choice D is the best answer. The frequency of oscillation for a pendulum is given by f = —1/ —. However, we are given2ji V L

the time it takes to complete ten cycles, so we need the equation for period, not frequency. What formula could the studentsuse to calculate the acceleration due to gravity? The formula for period is the inverse of the formula for the frequency, so wesimply need to invert the equation they gave us. The period is equal to the time needed to complete 10 cycles divided by 10.

400jt2Lami i = -=- /. -=- = tJi'u —. aquanng doui siues yieius: —— = ^w—, vvnicn leaas to: g =T = 2jf\/ ^ and T = -*- /. -£-= 2jr/\/ ^-.Squaring both sides yields: -*— = 4jrris which leads to: g=

V g 10 10 V g ioo g


Choice B is the best answer. To determineperiodT for an additional trial with L = 9 m, it may seem like you need to knowa formula for period, but you don't. You can figure out which answer choice is best by looking at the table of experimentalresults. The period for Trial 1 is about 2 seconds. (Remember, the table gives the time for 10 oscillations, not oneoscillation!) The period for Trial 5 is about 4.5 seconds. When L = 9 m, the period should definitely be greater than 3seconds. A period of 30 seconds seems far too bigfor a length of 9 m, since the period for L = 5 m is only 4.5 seconds. Theonly reasonable answer is choice B, 6 seconds. The best answer is choice B.

Choice D is the best answer. What is the tension is the string inTrial 4, if the pendulum bob swings through equilibriumwith a speed of 4 m/s? The common mistake on this problem is to say that the tension is equal to the weight, which wouldgive the answer T = 1.0 N. When the pendulum bob swings through equilibrium, it is actually traversing on the arc of acircle, as in the figure.

rnet - m—

Applying Newton's second law in the y-direction gives:


ZFy =T-mg =mvl which leads to: T=mg +myl

So, T=m(g +£-) =(0.1)(10 +£-) =(0.1)(14) =1.4Nr 4


7. Choice Bis the best answer. This is aformula-identification type of problem. Try using limiting cases to answer this one.The answer choices contain a 9.8 (presumably g), a 0.5 (presumably the initial height), a 100 (presumably the mass of thependulum) and a2(who knows where that comes from?). If the gravitational pull were stronger, would the pendulum movefaster or more slowly through the equilibrium point, as compared to ag-value of 9.8 m/s2? Faster. This requires that the 9.8be in the numerator and means that choice Cbe wrong. How about the initial height? The higher it starts, the farther it falls,and the faster it moves through equilibrium. This means that the 0.5 must also be in the numerator, eliminating all answerchoices but choice B.

If you are wondering about the mass or the mysterious 2 in the equation, think about this problem in terms of energyconservation. The initial energy is all gravitational potential. As the bob passes through equilibrium, the energy is all kinetic.

Ej = Ef, which leads to: mgh = &mv2, which leads to: v= V2gh = V(2)(9.8)(0.5)The best answer is choice B.

Passage II (Questions 8 -14) Strings and Things

8. Choice B is the best answer. This is a definition problem. Theamplitude is the distance a wave disturbs the medium from itsequilibrium "position." We often consider maximum amplitude when viewing waves such as the ones in Figures 1and 2.Thequotation marks mean that not all waves have amplitudes in physical space (as water and string waves do.) Electromagneticwaves have an amplitude that represents the energy the wave carries and not the size of space the wave occupies. Here, thestring in Figure II has an amplitude of 6 mm, whereas the wave in Figure I has an amplitude of 3 mm. Choice B has a biggeramplitude. The best answer is choice B.

9. Choice B is the best answer. Both waves are generated using an external oscillator set at 60 Hz, so they each have thesamefrequency, 60 Hz. Both waves must obey v = Xf. However, from the picture, we see that Wave I has only half thewavelength of Wave II. Hence, Wave I can have only half the speed of Wave II. The best answer is choice B.

10. Choice C is the best answer. This question requires that you be able to read a graph and understand how the motion willcontinue as time goes on. The string is farthest from equilibrium (existing as an antinode) at the 1-cm, 3-cm, and 5-cmpositions. As time goes on, the standing wave will move through this point of equilibrium, looking flat at that instant, andthen move to look like the mirror image of Figure II. The string at 0 and 6 cm will not move, as those points are fixed. Thestring will not move at the 2-cm and 4-cm positions, as these are nodes of the standing wave. The best answer is choice C.

11. Choice C is the best answer. We know that speed of a wave in a string is determined by physical properties of the string-the tension and the density. Changing either one of these two factors will allow us to change the speed of a wave on thatstring. The best answer is choice C.

12. Choice C is the best answer. This is a definition problem. These strings are fixed at their ends; yet, they are vibrating acrosstheir lengths. This is how a standing wave appears, ruling out choices A and B. Furthermore, the strings vibrate in such amanner that they move away from equilibrium by moving off the x-axis. These are transverse waves. The best answer ischoice C.

13.

14.

Choice B is the best answer. We are asked about wavelength and frequency. First, the wavelength looks bigger. This rulesout choice D. Second, does the larger wavelength look as if it's two times bigger than that in Figure I? Figure I has threelobes: this picture has but one. The wavelength is now 3 times that in Figure I, and choice C is incorrect. Next, is thefrequency larger or smaller now? Smaller: v = fk , v is the same for the same string, and X. is now larger. This is choice B.

You could also have answered this by the ratio method. The relevant equation for a standing wave on a string (or anystanding wave that is "fixed" on its ends) is / = nv/2L, where L is the string length and n is the number of antinodes (i.e.,lobes). The ratio of /now : /Figure I is

/lv

now _

/Figure I 3v2L

2L

= 13

leading us immediately to select choice B. The best answer is choice B.

Choice B is the best answer. The standing wave exhibits the same behavior as a simple harmonic oscillator (SHO). Whenthe amplitude of a SHO is zero, the potential energy of the SHO is zero. However, it is at this time that the kinetic energy ofthe SHO is at a maximum. The string is passing through equilibrium, just as a mass on a spring moves with its maximumkinetic energy when passing through equilibrium. The best answer is choice B.


Passage III (Questions 15 - 21) Feel the Vibes

15.

16.

Choice B is the best answer. The students know the time it takes for the mass to complete 10 cycles of motion. Dividingthis time by 10 cycles gives the period, the time to complete one cycle. Frequency is the inverse of the period, the number ofcycles completed each second. Statements I and II must be in the final answer, ruling out choice C. The equation for periodwas given in the passage as:

T=2ju/^~V k

so once the period is known, then the spring constant can be determined. (Remember that the mass of the air cart wasknown.) This rules outchoice D. However, the period and the frequency of a simple harmonic oscillator are independent ofamplitude, so this cannot bedetermined from the students' data. The bestanswer is choice B.

Choice C is the best answer. For all simple harmonic motion, the period and amplitude are independent of eachother. Thisrules out Statement I, which rules out choices A, B, and D, leaving only choice C.

To be complete, we shall address the other choices. The spring constant is a physical property of the spring, and cannot bechanged just by changing the amplitude. In the passage, we are given the following relationship between acceleration andposition:

a = - -^-xm

so changing the initial displacement would affect the maximum acceleration. Changing the initial displacement would alsoaffect the total energy of the system, because it would change the maximum potential energy of the system. Recall:

U = ikx22

If the maximum potential energy is changed, then so is the maximum kinetic energy, which means that the maximum speedwould be changed. The best answer is choice C.

17. Choice B is the best answer. At t = 0, the mass is at its maximum displacement and is not moving. Because the mass is notmoving at first, the graphs must start at zero, which rules out choices C and D. As the mass speeds up, it moves to the left(remember it is initially pulled to the right). The velocity must start at zero and then start pointing towards the left. This ischoice B. As a side note, the graph of mass velocity versus time is sinusoidal, just as the graph of position versus time is. So,what would the graph of mass acceleration versus time look like? It would also be sinusoidal, but unlike velocity will notstart at zero. The force of the spring is a restoring force, so the graph for acceleration must be perfectly out of phase with theposition graph. Choice C would acceleration over time if it were labeled as such. But this question asked for velocity overtime. The best answer is choice B.

18. Choice A is the best answer. When the mass is initially released from equilibrium, it has an initial speed of zero. As itmoves back toward equilibrium, its speed increases. In terms of energy, we can say that the system loses potential energy andgains kinetic energy. As for acceleration, it is related to the distance away from equilibrium according to:

a =-Jixm

Thus, as x decreases, the magnitude of the acceleration decreases. How can the speed increase if acceleration decreases?Remember that as long as there is an acceleration in the same direction as the velocity, the speed will increase. It does notmatter how small the magnitude of the acceleration is, as long as it exists! The best answer is choice A.

19. Choice D is the best answer. The period of the spring system is given in the passage as:

V kFrequency is the inverse of the period, so we can take the inverse of the period relationship to determine the relationship forthe frequency:

f=-WX2jt V m

Doubling the frequency means we need an equation that looks like:

Choice D is the only answer given that will yield this result. The best answer is choice D.


20. Choice Dis the best answer. Afast way to solve formula-identification problems is to look atunits, k has units ofN/m orkg/sec2, from remembering F=kx. Looking at the choices, three have an m/k in them. The units of m/k are sec2. Since weneed units ofmeters (the units ofx), this m/k must combine with the vmax so as to cancel out the sec2. This suggests that them/k should be under a square-root sign. Testing the two resulting possibilities shows that only choice Dhas the correct units.Another (more formal) way tosolve for x istouse conservation ofenergy:

Etotal = KE + U

When the kinetic energy isa maximum, the total energy equals the kinetic energy, so we can say:1 w.2-7mvmax2 =lmvt2+ -kx^

2 2 2

Substituting max forvt, wecansolve for x:


•^Vma^lmtm^)2-*. ikx22 2 v 2 / 2

f +kx2 =Irmvmax -<- "/ 4mvmax2 = ^

2

mvmax

vl _ 3mv* ——vmax

4k

= J3mV 4k

'max

kx2

21. Choice B is the best answer. The passagegives a formula for the energystored in a spring:

U= ^kx2

This meansthat only choices A or B are possiblecandidates. It does not matter what mass is attached to the spring when it isstretched or compressed. It also does not matter how frequently it oscillates. Frequency does depend upon the spring constant(an importantfactor in U), but just changing the frequency does not guarantee that the potential energy will increase, becausethe frequency also depends upon the mass. As for k and x, doubling k will double U, but doubling x will quadruple U. Thebest answer is choice B.


22. Choice C is the best answer. The frequency of a pendulum's swing is proportional to the square root of the gravitationalacceleration (i.e., f « Vg). We want fEarth : fMoon- Since the Moon has weaker gravity, the ratio should exceed 1:1. Thisrules out choice D. The final answer involves a square root and a 6, making choice C the only possible choice. The bestanswer is choice C.

23. Choice C is the best answer. The mass of a pendulum bob exhibits no effect on the period of the pendulum. Velocity of thebob depends on time (period), so velocity is not affected by mass. This can also be concluded from the relationship of initialpotential energy (mgh) and maximum kinetic energy (V£mv2), where mass cancels out. Choice A is eliminated. As thedisplacement angle increases, the system has more initial potential energy (is at a greater Ah), so it achieves a greater kineticenergy at its lowest point, resulting in a greater maximum velocity. Choice B is eliminated. As cord length increases (with aconstant initial angle of displacement), the change in height of the pendulum over the range of motion increases. This isfallout from trigonometry. Given the same angles in a triangle, as the length of one side of the triangle increases, the othertwo sides must increase proportionally. The result is that as the cord gets longer, the Ah increases, so the change in potentialenergy of the system is greater. At the lowest point of the trajectory, potential energy is completely converted into kineticenergy, so the following relationship (based on conservation of energy) holds true:

mgAh = ^mv2 .*. v2 =2gAh .\ v= V2gAh

Considering that Ah isproportional toL,the velocity, v, isproportional toVE. The bestanswer ischoice C.

24. Choice D is the best answer. Increasing the wave amplitude affects its intensity, but not its wavelength or frequency.Because the frequency and wavelength do not change, the number of nodes and antinodes do not change. This eliminateschoices A and B. The cord is not displaced at a node, so choice C is eliminated. What occurs with increased amplitude is agreater displacement of the cord at all points where it is displaced, especially antinodes, where the greatest displacement isobserved. This confirms that choice D is your fine and caring choice of correctness. The best answer is choice D.

Copyright ©by The Berkeley Review® 249 MINI-TEST EXPLANATIONS

25 Choice Cis correct This question is one that depends on your physical intuition. You can solve it by recalling the spring-based systems you've dealt with over your lifetime. Some materials are easier to deform than other materials, so the materialthat makes up the spring will definitely impact the spring constant Choice Ais eliminated. The number of coils can impactthe spring constant, because agreater number of coils allows the spring to deform by agreater distance when the same forceis applied Choice Bis eliminated. The diameter of the coils impacts how easily the coils can be deformed, which impacts thespring constant Choice Dis also eliminated. The only choice that remains is choice C. Whether the spring has aclockwiseor counterclockwise coiling direction, it behaves the same when compressed or stretched by alinear force. The best answeris choice C.

Copyright ©by The Berkeley Review® 250 MINI-TEST EXPLANATIONS

52-Question Periodic Motion and Waves Practice Exam

I. Determining g

II. Pendulum Experiment

III. Standing Wave


IV. Dual Springs

V. Simple Pendulum Study


VI. Deep Ocean Waves

VII. Springs and Frequency


Periodic Motion and Waves Exam Scoring Scale


42-52 13-15

34-41 10-12

24-33 7-9

17-23 4-6

1-16 1-3

(1-6)

(7 -12)

(13 -17)

(18 - 21)

(22 - 27)

(28 - 32)

(33 - 36)

(37 - 42)

(43 - 48)

(49 - 52)


The value of acceleration due to gravity, g, varies withaltitude and location on the Earth. By measuring the period,T, of a swinging pendulum, one can calculate g. The periodof a pendulum relates to g accordingto Equation 1.

=2*VYgEquation 1

To reduce the effects of air resistance, a pendulum usedto calculate g is typically kept in an evacuated jar.Additionally, the amplitude of the pendulum is kept small tominimize nonlinear mechanical effects of its motion. Some

measured values of g are listed in Table 1:

Location g[m/S2]Equator, Sea Level 9.780

San Francisco 9.800

North Pole, Sea Level 9.832

Table 1. Measured values of g

Satellites can also be used to measure g, because theirorbital speeds vary subtly depending upon the strength of thelocal gravitational field of the Earth below. Although it isnot quite as precise as using a pendulum, it is popularbecause it allows the measurement of g at normallyinaccessible locations around the globe.

1.

2.

If a scientist wants to verify his results using a newtimer, but it has one-decimal place less of precision,what could he do to get the same precision for themeasured period of a pendulum that his first timer wascapable of delivering?

A. Measure for ten times as long.

B. Measure for one-tenth as long.

C. Use a pendulum that is ten times as long.

D. Use a pendulum that is one tenth as long.

If the pendulum system used to calculate the local valueof g were not placed in an evacuated chamber, thenhow would the results vary from the one that was?

A. The measured period would be longer; thecalculated value for g would be too low.

B. The measured period would be shorter; thecalculated value for g would be too low.

C. The measured period would be longer; thecalculated value for g would be too high.

D. The measured period would be shorter; thecalculated value for g would be too high.


3. A scientist operating a pendulum in an elevator seesthat the pendulum's period is shorter than can beaccounted for by gravity alone if she takesmeasurements when the pendulum apparatus is movingwith a constant:

A. upward velocity.

B. downward velocity.

C. downward acceleration.

D. upward acceleration.

The variation of g in Table 1 is most likely due to:

A. the differing altitudes of the locations.

B. the differing air densities at the locations.

C. the differing centrifugal forces at the locations.

D. random experimental error.

If the bob of a 1-meter long pendulum has a maximumspeed of 2 m/s, then what is the approximate maximumvertical displacement of the bob?

A. 0.1m

B. 0.2 m

C. 0.5 m

D. Insufficient information to tell

What is an example of a nonlinear mechanical effect?

A. Air resistance caused by localized eddies in theevacuated jar

B. Thermal expansion of the fulcrum due to friction

C. Induced electrical current in the moving bobD. Deformation of the cable caused by centripetal

acceleration



A student set out to study the effects of mass, cordlength, and medium on a pendulum. In the study, sheattached a 1.00 kg bob to a cord that was 70 cm long, andattached the top of the cord to a fixed point on the ceiling ofan enclosed box. Using a strobe light of adjustablefrequency, she recorded the period of the pendulum systeminitially displaced to the left wall, and then released. Figure1 shows the arrangement.

At Release At Low Point

Figure 1

Three separate experiments were conducted, whereeither the mass of the bob was varied, the cord length wasvaried, or the medium in the enclosed box was varied. Theresults are listed in the tables below.

Experiment I: Cord is 0.70 m; the medium is air @ 25°C

Trial Bob Mass Period (s)

1 1.00 kg 1.68

2 1.50 kg 1.68

3 2.00 kg 1.67

4 2.50 kg 1.68

Experiment II: Mass is 2.00 kg; the medium is air @ 25°C

Trial Cord Length Period (s)

1 60 cm 1.55

2 70 cm 1.68

3 80 cm 1.79

4 90 cm 1.90

5 100 cm 2.01

Experiment HI: Cord is 0.70 m; mass is 2.00 kg; T is 25°C

Trial Medium Period (s)

1 Vacuum 1.62

2 Helium 1.66

3 Air 1.68

4 Carbon Dioxide 1.79

5 Propane 1.89

7. What medium offers the greatest resistance?

A. Air

B. Carbon Dioxide

C. Helium

D. Propane


8. The cord in the experiment should be:

A. elastic, to allow the tension to establish a gradient.B. rigid, to prevent tension in the cord.

C. elastic, to prevent conversion of kinetic energy ofthe bob into potential energy by elongating thecord.

D. rigid, to preventconversion of kinetic energyof thebob into potential energy by elongating the cord.

9. How does the frequency of the pendulum relate to thecord length of the pendulum?

A. They are directly related.

B. They are inversely related.

C. They vary directly by a square root factor.D. They vary inversely by a square root factor.

10. How would the period of a pendulum change if themass of the bob were doubled, cord length weredoubled, and the medium was changed from air tocarbon dioxide?

A. The period more than doubles.

B. The period increases by a factor of roughly 1.5.

C. The period decreases by a factor of roughly 1.5.

D. The period remains the same.

11. Which of the following graphs represents velocityversus time for the pendulum system, starting with anangle of displacement of 45°?

12.

C;UUUL dUlj^

All of the following statements are true with respect tothe total energy of the system EXCEPT that:

A. the potential energy of the bob at its apex equalsthe kinetic energy when the bob at its nadir.

B. the kinetic energy of the bob increases uniformlyas the bob descends.

C. the potential energy and kinetic energy are equal toone another at a time that is one-eighth of theperiod after it had all gravitational potential energy,increasing the initial displacement angle willincrease the maximum kinetic energy of thesystem.


D.


A student compares two systems designed to generate astanding wave. A standing wave results from theinterference of two waves moving in opposite directionsalong a string, such that the resultant wave appears to havenovelocity. A standing wave can be generated byfixing oneend of a cord and connecting the other to a source ofvibration. If done correctly, the nodes and antinodes(apexes) remain atfixed points on the string. In the student'sexperiments, the cord is made of a flexible polymer. Amechanical oscillator capable of vibrating at a set frequencyis used. The standing wave is generated from the incidentwave and its reflected waves.

System 1

The cord is attached to the oscillator at the left

end and to a fixed point on a wall at the right end. Theoscillator frequency is adjusted until a standing wave isgenerated.

Oscillator Cord of length L Fixed

point

System2

The cord is attached to the oscillator at the left

end and to a ring on a pole at the right end. The oscillatorfrequency is adjusted until a standing wave is generated.

Oscillator Cord of length L MovableA

Ipoint

In System I, the wavelength is found by dividing twotimes the length of the cord (L) by an integer (n) thatdescribes the number of antinodess there are in a vibratingcord. The lowest value of n is n = 1, which corresponds tothe lowest possible frequency for the wave.

13. Which graph best depicts the periodic displacement ofthe cord at a node over time?

C.

c

«003

WWtime

D.

«0-


time

14.

15.

16.

17.

254

In System I, what occurs when the frequency isdoubled?

A. The number of nodes increases.

B. The number of antinodes decreases.

C. The wavelength increases.

D. The distance between nodes increases.

In System II, if the frequency of the oscillator is not aresonant frequency, thenwhathappens to theoscillator?

A. The reflected wave causes the oscillator to move.

B. The energy is fully absorbed by the ring encirclingthe rod.

C. The reflected wave is reflected by the oscillatoruntil it reaches a resonant frequency.

D. The oscillator deforms to assume the shape of thestanding wave.

Given that v = f-X where v is the wave speed, / is thewave frequency, X is the wavelength, and L is thelength of the cord, which equation describes thefrequencies that result in a standing wave in System II?

A f - unA. /„ - —

B. /„ =- "U

4Ln

4L-u

D. /„ = 1

4Lvn

If 98 Hz, 196 Hz, and 245 Hz are resonant frequencies,then which of the following statements must be true?

I. The lowest resonant frequency is 98 Hz.

II. 196 Hz and 245 Hz are consecutive resonant

frequencies.

III. There is a resonant frequency at 431 Hz.

A. II only

B. Ill only

C. I and II only

D. II and III only


Questions 18 through 21 are NOTbased ona descriptivepassage.

18. A pendulum is allowed to swing through one completecycle back to its maximum height, at which time thecable holding the bob is cut. What will be thesubsequent velocity of the pendulum bob?A. ^ B.

C. D.

19. At what point in the cycle of a pendulum does thekinetic energy equal the potential energy?

A. t = 0

B. to"/2

C. t=3jt/4

D. t=3*/2

20. Two tuning forks, one with a frequency of 400Hz andone with a frequency of 500Hz, are simultaneouslystruck. Aside from these two frequencies, what otherone will be heard?

A. 900 Hz.

B. 100 Hz.

C. 1.25 Hz.

D. 0.8 Hz.


21. Fora pendulum where a massive bob is displaced to anangle 9 and then released, what is the speed of the bobas it passes through its lowest point?

A.

B.

C.

D.

v = lT

V= L6g/

v =V2gL(rv=V2mgL

cos8)



An object of known mass is attached at each end to adifferent spring as shown in Figure 1. Both springs have thesame number of coils, the same material thickness, and thesame coil diameter. The spring on the left is made of steel.The material of the spring on the right is varied during fourtrials. When a steel spring is used on the right side, theobject oscillates symmetrically. When the right spring ismade of a material other than steel, the object also oscillatessymmetrically, but with a different period. Figure 1 showsthree intermediate positions during a half-cycle for the springsystem.

am

am

QCCQAA

t = 0

_ jtt =

t = Jt

Figure 1. Three positions of mass of springs

Observers record the positions of extreme displacementon each side for the four combinations. From the distances,

they calculate relative spring constants.

22. If the experiment were repeated with a heavier massand the same initial displacement, then how would theresults vary?

A. The frequency of the spring system would increase.

B. The springs would compress a greater distancefrom equilibrium at points of maximum potentialenergy.

C. The mass would have greater speed at equilibrium.

D. The mass would have greater momentum atequilibrium.

23. When two springs of different materials are used in theexperiment, what should be observed?

A. The mass oscillates in a symmetric fashion.

B. A greater displacement is observed on the side withthe spring having the smaller spring constant.

C. The potential energy is greater when the object is atrest on the side with the greater k than when theobject is at rest on the side with the smaller k.

D. The object reaches a greater maximum speed inone direction than the other.


24. When both springs are made of steel with a springconstant of k, the potential energy at any point equals:

i/^kx2.

kx2.

C. 2kx2.

A.

B.

D. i4k2x4.

25. How can the period of the spring system be found,where k represents the overall spring constant?

A. T = -L2:i

B. T = -L271

C. T = 2jt.

D. T = 2jt

26. Which of the following pictorial representationsaccurately depicts the spring system one-and-a-halfcycles after t = 0 in Figure 1?

27.

A.

B.

QQGQ

C.

qqqq

D.

KQQQQ

QQCQAA

A

QQGQ

M

QQQQAWhich observation is consistent with the experiment?

A. The mass oscillates at a higher frequency than itwould if it were attached to only one spring,instead of two.

B. The harmonic oscillation is more apt to be dampedwhen the springs are made of different materials.

C. The effective spring constant is equal to an averageof the two spring constants.

D. The springs not perfectly out of phase.



A group of students wishes to study simple pendulummotion. A simple pendulum consists of a string of length Land a ball of mass m attached to the bottom of the string.Thefrequency of motion of a simple pendulumis given by:

f = -L2n

and the positionor amplitude can be described by:

0 = 0ocoso) t or 0 = 60sino> t

Using a simple pendulum as described above, thestudents perform a series of experiments.

ExperimentIA pendulum of length L is attached to a ceiling. At a distance(3/4)L vertically below the point of attachment is a nail. Thependulum is pulled a small angle 0 to the right of the verticaland allowed to fall.

Experiment IIA small hole is drilled in the ceiling and the string of thependulum is passed through this hole. The pendulum ispulled aside a small angle 0 to the right of vertical andallowed to fall. As the pendulum oscillates, the string ispulled up through the hole.

t-QNote: Ignore air resistance.

28. For Experiment I, the period of motion is:

A. three times longer than when no nail is used.

B. more than three times longer than when no nail isused.

C. equal to the period when no nail is used.

D. less than the period when no nail is used.


29. For Experiment I, if the initial angle is 0, what can beconcluded about the angular displacement, 0', after halfa cycle has elapsed ?

A. 0' = 0

B. 0'>0

C. 0'<0

D. Not enough information is provided to answer thequestion.

30. For Experiment I, which of the following will have thehighest frequency of oscillation?

A. Using the apparatus as is.

B. Using the apparatus, but having a string twice aslong as in the original set-up.

C. Using the apparatus on the Moon.

D. Using the apparatus on the Moon and with a stringtwice as long as in the original set-up.

31. For Experiment II, what happens to the period and theamplitude of the motion as the string is pulled up?

A. The period increases; the amplitude remains thesame.

B. The period decreases; the amplitude increases.

C. The period remains the same; the amplitudeincreases.

D. The period increases; the amplitude decreases.

32. While Experiment II is being conducted, the ball at theend of the string suddenly breaks and half its mass fallsaway. What effect will this have on the period ofmotion?

A. The period doubles.

B. The period is unaffected.

C. The period increases by afactor ofV2.D. The period decreases byafactor ofV2.


Questions 33 through 36 are NOT basedon a descriptivepassage.

33.

34.

35.

36.

What change would increase the frequency of the firstharmonic generated by a cord fixed at both ends?

A. Moving the fixed points away from one another toa greaterdistance of separation.

B. Using a cord that is more dense.

C. Increasing the tension in the cord.D. Decreasingthe tension in the cord.

If the length of a pendulum's string is increased by300%, then by how much does its period change?

A. It would decrease by 100%.B. It would increase by 100%.C. It would increase by 200%.D. It would increase by 300%.

A standing wave on a string vibrates in its thirdharmonic. What is the ratio of the wavelength of thewave to the string length?

A. 1:3

B. 2:3

C. 3:4

D. 3:2

If the mass of the spring is no longer negligible, whathappens to the frequency?

A. The frequency increases.

B. The frequency decreases.

C. The frequency remains unchanged.

D. It depends on whether the mass is uniformlydistributed along the length of the spring.

Copyright ©byTheBerkeley Review® 258 GO ON TO THE NEXT PAGE


Ocean waves in deep water were studied extensively inthe late 1800s. Data were collected and analyzed foramplitude, frequency, and wavelength. One result derivedfor ocean wavesfar from shore was that the speed of a wavewith wavelength X is given by Equation 1.

Equation 1

Equation 1 applies when the wavelengths and amplitudesof waves are much smaller than the water depth butreasonably long (e.g., greater than 1 m.) Such waves aretypically generated by wind in the open sea. Studies showthat two waves of differing wavelengths could pass througheach other without any distortion in their wavelength oramplitude.

37. What is the ratio of wave speeds of two small-amplitude waves, if one wave has twice the wavelengthof the other?

A. 2:1

B. V2:l

C. 1:1

D. It depends upon their relative amplitude.

38. Quantities needed to calculate the speed of a deep-waterocean wave are the:

A. wave's period of motion and its frequency.

B. gravitational field strength and the wave'samplitude.

C. wave's wavelength and its period of motion.

D. wave's wave amplitude and its wavelength.

39. Which graph best represents the relationship betweenocean wave speed squared and wavelength?

A. B.

D.


40.

41.

42.

259

How can it be explained that when two deep-waterocean waves pass through the same pointsimultaneously, the waterlevel neither rises nordrops?

A. The wavelengths do not change when two wavespass through one another.

B. The wavelengths do change when two waves passthrough one another.

C. The two passing waves must have undergoneperfectlyconstructive interferenceat that point.

D. The two passing waves must have undergoneperfectlydestructive interferenceat that point.

Waves traveling across a large, deep lake are faster thanwaves traveling across the deep parts of the ocean.How can this be explained?

A. Lake water is less dense than ocean water; thewavelength is greater in a lake than in the ocean.

B. Ocean water is less dense than lake water; thewavelength is greater in a lake than in the ocean.

C. Lake water is less dense than ocean water; thewavelength is less in a lake than in the ocean.

D. Ocean water is less dense than lake water; the

wavelength is less in a lake than in the ocean.

If deep-water ocean waves elevate a stationary boatevery 4 seconds and the distance between every thirdcrest is 150 m, what is the average speed of the waves?

A. 0.08 m/s

B. 12.5 m/s

C. 37.5 m/s

D. 113 m/s



A block of mass mi is moving with a speed vi toward astationary second block of mass m2- The second block isattached to a spring with spring constant k as shown inFigure 1. The incident box collides with and sticks to thesecond box, and the two boxes proceed to move together andcompress the spring. The spring and mass system thenundergo simple harmonic motion, because the energy of theincident box is transferred to the new system.

Figure 1. Experimental apparatus

Note that the period of oscillation relates to the mass andspring constant in the following manner:

Neglect friction in the accompanying problems.

43. By how much will the spring be compressed?

A. x^J^-vjmi + ni2

kB. x =mi + iri2

C. x = -'^LJ!Lna-Vs?

vi

rnivi

vi

D. x =

Vk(mi +m2)

44. If friction were present between the masses and thesurface over which the masses slide, then it would:

I. decrease the amount the spring is compressed.

II. not affect the kinetic energy of the massesimmediately after collision.

III. decrease the subsequent speed of oscillation.

A. I only

B. I and III only

C. I, II, and III

D. None of them would be affected.

45. Considering the momentum and total energy of thetwo masses and spring, after the oscillations begin:

A. momentum and total energy are both conserved.

B. momentum is not conserved, but total energy isconserved.

C. momentum is conserved, but total energy is notconserved.

D. neither momentum nor total energy is conserved.


46. Which of the following graphs BEST illustrates thepotential energyversus time for the system in Figure 1,where t = 0 is defined as the time at which the incident

box first contacts the box on the spring?

A.

Time

Time

Time

47. If the initial velocity of mi increases by a factor of 2,

how will the frequency of vibration change?

A. It will increase by a factor of 2.

B. Itwill increase byafactor ofV2.C. It will decrease by a factor of 2.

D. The frequency is unaffected.

48. The set-up, discussed in the passage, is used in twodifferent experiments, and the post-collision massdisplacement is plotted as a function of time. If all thephysical parameters, except one, are the same in thetwo experiments, we can conclude that Experiment I:

x

A. used a more massive m2 than Experiment II.

B. used a less massive m2 than Experiment II.

C. used a weaker spring than Experiment II.D. used a spring with more coils than Experiment II.



49.

50.

51.

How can you explain that a bridge used by a marchingband will deteriorate faster than one used by foottraffic?

A. The bridge can resonate from foot traffic.

B. The bridge can resonate when the band marches.

C. The bridge has many oscillation frequencies.

D. The bridge cannot withstand random oscillation.

Suppose a string is 20 meters long. Which of thefollowing is a possible wavelength for standing waveson this string?

A. 3.5 m

B. 13.3 m

C. 30 m

D. 80 m

Addition of two waves with slightly differentfrequencies results in the production of beats or a beatfrequency. The result can be monitored using anoscilloscope. Consider the two figures shown below:

Keeping the lower frequency wave constant, weincrease the frequency of the other wave. You wouldexpect the display on the oscilloscope to go from:

A. Figure I to Figure II.

B. Figure II to Figure I.

C. There will not be a shift between Figure I andFigure II, but the amplitude will increase.

There will not be a shift between Figure I andFigure II, but the display will become brighter.

D.


52. Which of following graphs accurately relatesperiod, T, to the length of the pendulum, L?

A. B.

Length (m) Length (m)

Length (m) Length (m)

the

1. A 2. A 3. D 4. C 5. B 6. D

7. D 8. D 9. D 10. B 11. A 12. B

13. D 14. A 15. A 16. A 17. A 18. A

19. C 20. B 21. C 22. D 23. A 24. B

25. D 26. C 27. A 28. D 29. B 30. A

31. B 32. B 33. C 34. B 35. B 36. B

37. B 38. C 39. C 40. D 41. A 42. B

43. D 44. B 45. B 46. A 47. D 48. B

49. B 50. B 51. B 52. D

YOU ARE DONE.

Answers to 52-Question Periodic Motion and Waves Exam

Passage I (Questions 1-6) Determining g

1. Choice A is the best answer. In order to compensate for the loss in precision associated with using a timer with one lessdecimal place, the times being measured should be ten times greater. Measuring for a shorter time would not help here, sochoice B is incorrect. A shorter cable results in a shorter period, which won't help to increase precision, so choice D iseliminated. Since the period T is proportional to the square root of the length, using a pendulum that is ten times as longwould only lead to a period that is VTO times as long. This eliminates choice Cand leaves only choice A. To get the extraprecision, the scientist would measure for 10 periods and divide that time by 10 to determine the period. For instance, if theperiod is originally measure to be 2.71 seconds, then the researcher might find that ten full cycles takes 27.2 seconds(according to the new timing device which is only precise to the tenth place). The period in the second measurement is one-tenth of 27.2 seconds, resulting in a period of 2.72 seconds, a measurement as precise as was obtained with the originalwatch that measured to the hundredths place. The best answer is choice A.

Choice A is the best answer. In any environment where the pendulum swings in a medium, there exists drag. Thependulum bob would be slowed by the medium as it traveled its pathway, increasing the time required to complete a cycle.This means that the period would be larger, eliminating choices B and D. Because the period, T, is inversely proportional tothe square root of g, a larger T in the equation would result in a smaller calculated value for g. The best answer is choice A.

Choice D is the best answer. A shorter period corresponds to either a shorter pendulum or a larger effective gravitationalacceleration. Since the pendulum length is unchanged, the motion of the elevator must make the effect of gravity on thependulum seem stronger. This cannot happen when the elevator is moving at a constant velocity (i.e., zero net force), whichrules out choices A and B. When an elevator accelerates upward, you feel as if gravity is pulling you down harder. Thesame effect will make the pendulum "feel" as if gravity is pulling on it harder. This will decrease the period. Choice C iseliminated. The best answer is choice D.

Choice C is the best answer. All of the measurements are taken at around sea level, invalidating choice A. The pendulumresides in an evacuated jar, making local air density irrelevant and choice B wrong. Random experimental error (choice D) isincorrect, because the data, such as 9.780 and 9.832, differ more significantly than solely in the last decimal place. Thisleaves choice C as the best answer by elimination. The "centrifugal" force is an effective force one feels when moving alonga curved path. It's essentially the more familiar centripetal force, but in the reference frame of the forced object instead of thereference frame of the rotational axis. Regardless, a bigger centripetal force inwards means a bigger centrifugal forceoutwards. This lessens the normal force with the Earth and makes you feel lighter. The best answer is choice C.

Choice B is the best answer. The law of energyconservation tells us that the maximal kinetic energy the pendulum bob hasat the bottom of its path mustequal the maximal potential energyit has upon reaching its maximum height. Equating the twoenergies yields:


!^mvmax2_= mghc"ma\

llmnv —

2_ 'max

2g

'/Wmax2 = ghmax

2x2 ? a ^= ^-= 0.2 m2x 10 10

Choice D is the best answer. The passage states that, "the amplitude of the pendulum is kept small to minimize nonlinearmechanical effects of its motion." We are looking for a deviation that increases as the displacement angle increases. Thesystem is kept in an evacuatedjar, so air resistance is not the nonlinear mechanical effect that minimizing amplitude reduces.Choice A is eliminated. Thermal expansion of the fulcrum due to friction could potentially take place, but it's a constanteffect once a set temperature is reached. Choice B is an unlikely candidate for the best answer. There is no induced currentin this system, so choice C is eliminated. To induce current, the pendulum bob would need to be part of a conducting loopthat was swinging into and/or out of a perpendicular magnetic field. Deformation of the cable due to centripetal forces takesplace with a pendulum. At the lowest point of the bob's trajectory, it is moving fastest, causing the great extra pull on thecable. If the cable were to deform, then it would elongate and thereby change the period of the pendulum. This is anonlinear effect, because the force varies with velocity squared. By keeping the amplitude small, the initial potential energyis reduced, so the maximum speed of the pendulum's bob is kept small. This keeps the magnitude of the centripetal force toa minimum. The best answer is choice D.


Passage II (Questions 7 -12) Pendulum Experiment

Choice Disthebestanswer. From the data collected in Experiment III, it can be seen that the gas in the chamber affects theperiod of the pendulum. Although this does not appear in the equation for the period, it can be concluded that resistancefrom the medium slows the pendulum down as it cycles through its pathway. The medium dampens the harmonicoscillation. From the data, the longest period isobserved with propane, therefore the greatest resistance isbeing experiencedin the propane medium. This makes choice D the bestanswer. The data can also lead to the conclusion that the most viscousmedium is propane, because it offers the greatest resistance to flow (motion). The best answer is choice D.

Choice D isthe bestanswer. As the pendulum swings through its range of motion, it generates a radial force that is greatestwhen it has its greatest velocity (observed at its nadir-low point). This force can stretch (deform) the cord, soa rigid cableshould be used as the cord. If the cord is flexible, it will elongate at the lowest point, and relax at the highest point (wherevelocity is zero). This stretching and relaxing involves the interconversion of potential and kinetic energy, therefore the bestanswer is choice D. The tension will not exist in a gradient manner, so choice A should be eliminated. Tension in the cordcannot be eliminated without removing the bob. This eliminates choice B. The hope is that the changes in tension arepredictable. The best answer is choice D.

Choice D is the best answer. This question involves analyzing the data from Experiment II, and knowing the relationship ofperiod and frequency. The period is the reciprocal of the frequency, so theyare inversely related. From Experiment II, it canbe seen that the period increases with the cord length, but not in a one-to-one linear fashion. Based on the answer choices,the period is proportional to the square rootof the cord length. The question asks for the relationship between frequency andcord length, so the frequency (reciprocal of the period) is inversely related to the square root of the cord length. This makeschoice D correct. Background knowledge may be employed to arrive at an answer more quickly than data analysis. Therelationship is as follows:


T=2jn/L and T=IV g / 2jt V L

so /

10. Choice B is the best answer. From Experiment I, we see that the mass has no effect on the velocity, so doubling the mass ofthe bob will leave the velocity unchanged. From Experiment II, we see that increasing the length from 60 cm to 100 cm (lessthan doubling) increased the period from 1.55 to 2.01, a factor of roughly 1.3. It is safe to guess that a doubling of cordlength will increase the period by a factor of roughly 1.4 or 1.5. If you have background information, then you know from theequation that the factor isactually V2. From Experiment III, we see that changing the medium from air to carbon dioxide (amore viscous medium), increases the period from 1.68 to 1.79, an increase by a factor of roughly 1.06 or so. The summationof all of the affects leads an increase in the period by roughly a factor of roughly 1.5. The best answer is choice B.

11. Choice A is the best answer. The pendulum starts from rest, with all potential energy. Choices B and D can be eliminated,because they describe a system that starts with some initial velocity. Periodic motion follows a sine curve, where thevelocity is zero at the start (highest point on the left), maximum and moving left at its lowest point (Jt/2), zero once again atits highest point to the right (it), maximum and moving right upon its return to the lowest point (3jt/2), and finally zero at itshighest point to the left (2k). The velocity has direction associated with it. The bob changes direction, so it must have a timewhere the velocity is positive and a time where it's negative, shown only in choice A. The best answer is choice A.

12. Choice B is correct. At the lowest point, initial potential energy is completely converted into kinetic energy. Potentialenergy is maximized at its apex (highest point) and kinetic energy is maximized at its lowest point. Considering all of thepotential energy is converted into kinetic energy, choice A is a valid statement. The bob accelerates as it falls, but not withconstant acceleration, so its velocity does not increase uniformly during descent. As such, kinetic energy does not increaseuniformly, but instead follows a sine curve. Choice B is an invalid statement, which makes it the best answer. At maximumdisplacement, the energy is completely gravitational potential (because the bob is motionless). At zero displacement, theenergy is completely kinetic (because the bob is at its lowest point). This means that at a point in time that is half waybetween these two states, the kinetic energy must equal potential energy. Choice C is a valid statement. Increasing thedisplacement angle increases the initial potential energy, so the maximum kinetic energy will also increase. Choice D is avalid statement. The best answer is choice B.

Passage HI (Questions 13 -17) Standing Wave

13. Choice D is correct. At a node, the cord will not be displaced, which can be concluded from on the definition of a node.This means that at all times, the displacement of the cord is equal to zero. Tricky, tricky huh? The best answer is choice D.

Copyright ©by The Berkeley Review5 263 REVIEW EXAM EXPLANATIONS

14. Choice A is the best answer. The relationship of frequency to wavelength, v = Xf, is such that if the frequency doubles,then the wavelength must becut in half. The distance between nodes is half of a wavelength, so any change that decreasesthe wavelength will bring the nodes closer together, and thus increase the number of nodes in a fixed distance. This makeschoice A the best answer. The numberof antinodes should increase as well,so choice B is eliminated. Choices C and D arecomplementary answers, thus they should both be eliminated (they must both be wrong, because there is only one bestanswer per question). The choice for a happier tomorrow is choice A. Thebestanswer ischoice A.

15. Choice A is the best answer. If the frequency is not a resonant frequency, then when the reflected wave returns to theoscillator, it will have a non-zero amplitude associated with it at thatpoint This will move the oscillator to some extent, sochoice A is the best answer. The energy is not absorbed by the rod (otherwise, a reflected wave would not be produced,meaning that a standing wave could never exist). Because a standing wave is observed in System II, choice B should beeliminated. Because the oscillator is not stationary (likethe wall is in System I) and is able to move in any direction (unlikethe loop, which may only rise orfall), it isunlikely that the oscillator would always reflect the wave. Even if itdid, itwouldnot shift the frequency upon reflection. Choice C should be eliminated. The oscillator cannot change its shape, becausethere is notenough energy to move themolecules. If there were enough energy to deform theoscillator, then theoscillatorwould fall apartovertime. Choice D should be eliminated. The best answer is choiceA.

16. Choice A is the best answer The passage states that the wavelength of a standing wave on a string fixed at both ends isequal to twice the length of the cord, divided by the number of nodes (Xn =2L). Astring that is fixed at both ends showssimilarbehavior as an open pipe. However, a string system like System II behaves like a closed pipe, and not like an openpipe. Such a standing wave has a node at one end and an antinode at the other end, so Xn =4L. We know that v =f-X.

Manipulation gives / =**- =tr-1% Substituting the value for A, (i.e., 4L) into this rearrangement gives /n =-u-E- =—, choiceA. A. 4L> 4J_>

A. The best answer is choice A.

17. Choice A is the best answer The two closest resonant frequencies given are 245 Hz and 196 Hz, which differ by 49 Hz. 49Hz can be divided evenly into all of the values listed, so the first harmonic probably occurs at 49 Hz. Perhaps the actualanswer is 24.5 Hz, or some value that divides evenly into 49 Hz, but based on the data presented, it appears that 49 Hz is themost likely value for the first harmonic. This eliminates Statement I, because the lowest resonant frequency cannot be 98Hz. Statement III is also eliminated, because 431 Hz is NOT divisible by 49. If n = 10, the resonant frequency would be 490Hz. If n = 9, the resonant frequency would be 441 Hz. If n = 8, the resonant frequency would be 392 Hz. Only Statement IIcan possibly be true. This makes choice A the best answer. If the first harmonic is 49 Hz, then 196 Hz and 245 Hz areconsecutive resonant frequencies (/n = n/i). The best answer is choice A.

Questions 18-21 Not Based on a Descriptive Passage

18. Choice A is the best answer. When the pendulum is at its maximum height at the end of one cycle, it is stationary in roughlythe same position as which it started (depending on whether or not the first swing was dampened or not). Either way, todetermine the bob's velocity after the cable breaks, ask yourself two things: What is its velocity when it is at its maximumheight? What is the net force/acceleration acting on it? When at its maximum height, the bob is not moving (i.e., v = 0). If thecable breaks, the only force will be gravity pulling it down (being that breaking the cable removes the tension). Thus, it willacquire a velocity straight downward. The best answer is choice A.

19. Choice C is the best answer. At t = 0, the system is motionless, so it has KE = 0 and PE at a maximum. This eliminateschoice A. At n/2 the bob is at its lowest point, moving left. At 3ji/2 the bob is at its lowest point, moving right In bothcases, KE is maximum and PE is minimum. Because the pendulum has the same energetics at Jt/2 as it has at 3ji/2, choicesB and D are eliminated. Choice C is the best answer, because it is the only remaining answer. At 3je/4 the bob is half waybetween its lowest point and highest point, moving left This falls half way between points where KE is maximum and PE ismaximum. The best answer is choice C.


i*- Time

= PE

= KE

REVIEW EXAM EXPLANATIONS

20. Choice B is the best answer. If waves of two frequencies are emitted simultaneously and you are asked about otherfrequencies that might result, then you are probably dealing with a beat problem. This question requires that you know thebeatfrequency in terms of the two main frequencies. That relationship is:

/beat = '/I -H

where theabsolute valuesign means that we care only about thedifference between the two frequencies, and notaboutwhichone is larger. Applying this equation to our 400 Hz and 500 Hz tones yields a new tone of 100 Hz. Notice that the otherchoices are different ways that 400 and 500 can be simply combined. Essentially, you must know the correct formula. Butyou still could eliminate choices C and D, since those numbers would have no units, assuming that they arise from dividingone frequency by the other. The best answer is choice B.

21. Choice C is the best answer.This question can besolved using physical intuition. As the pendulum is displaced by a greaterangle, it has more potential energy, so it will have greater kinetic energy when the potential energy is converted to kineticenergy at its lowest point. The mass is the same, so this means that the velocity at its lowest point is greater. This eliminateschoice D. As the frequency increases, it cycles through its period faster, so it must move faster. This means that / cannot bein the denominator and choice B is eliminated. Choice A is eliminated, because it yields incorrect units. Only choice Cremains.

Should you want to solve this problem exactly, use conservation of energy. When the pendulum is displaced by an initialangle 9 from the vertical, it also gains potential energy that we can write as mgh. Just before the pendulum strikes the nail, itis at its lowest vertical position; the potential energy has been converted to kinetic energy:

mgh = '/£mv% which leads to: v = V2gh

However, we do not know what h is. In order to solve for h, we need to do a little trigonometry. From the following picture,we can solve for h:

cos 9=i^-p- |nj> h=L(1-cos 0)

• OThe best answer is choice C.

Passage IV (Questions 22 - 27) Dual Springs

22. Choice D is the best answer. If a heavier mass were used with the original displacement, then the system would still start

with the same initial potential energy (PEsprjng = !^kx2). This results in the same kinetic energy as the object passes throughits equilibrium position. Because kinetic energy is unchanged despite the greater mass, it must be moving more slowly witha heavier mass (to have the same value for Vimv2). Because the system is heavier, it will travel more slowly, and thereforewill take longer to complete a cycle. The frequency is inversely proportional to the period, so a greater period corresponds toa lower frequency. Choice A is eliminated. Given that the object starts at the same displacement, it must have the samepotential energy at all points in space, including maximum displacement, where it is at its maximum potential energy. Thismeans that the displacement distance at maximum potential energy is the same for the lighter and heavier masses, so choiceB is eliminated. Because the acceleration is reduced, the object will pass through its equilibrium position with a slowerspeed. This eliminates choice C. Momentum depends on both mass and velocity, so we must give a little more thought tochoice D than the other choices. The object is more massive, but it passes through equilibrium with less speed. The twofactors have an opposite effect on the momentum at equilibrium. Because the kinetic energy is the same for both the lighterand heavier masses, we can say they each have the same value for mv2. So, the mass is increased by a greater factor than thevelocity is decreased, because an increase in mass is exactly balanced by a decrease in v2. Because mass is more importantthan velocity, the heavier object has a greater momentum as it passes through its equilibrium point than the lighter object.The best answer is D.


23. Choice A is the best answer. When two springs ofdifferent materials are used in the experiment, we can treat the impact ofeach as independent of one another. As shown in Figure 1, when one is compressed, the other is elongated. As such, theyboth exert a restoring force on the mass in the same direction. Whether the springs are of equal spring constant or differentspring constants, the object still reacts to a uniform force, so it will oscillate in a symmetric fashion. Choice A is the bestanswer. The displacement from equilibrium is equal on both sides, because the restoring force is identical whether the springhas been elongated or compressed. If the object is displaced from equilibrium by one centimeter in either direction, themagnitude of the force exerted by each spring, independent of the other, is the same. This eliminates choice B. Because ofthe symmetry, the potential and kinetic energies are the same on either side of equilibrium. This eliminates choices C and D.Even if you were uncertain of the impact of the second spring, the choices distinguish themselves from one another in thatchoice A focuses on a symmetric system, while choices B, C,and Dfocus on asymmetry. The best answer is choice A.

24. Choice B is the best answer. The two springs are attached to the mass independent of one another, so they can be treatedindependently. Because they are attached in line, they each experience the same Ax when the mass is displaced. The totalpotential energy of the system is equal to the sum of the potential energies for the two springs. The potential energyassociated with each individual spring is !^kx2, soPEtotal = '̂ kx2 + ^kx2= kx2. The bestanswer is choice B.

25. Choice D is the best answer. The period of a harmonic oscillator is found by multiplying 2k (the distance around a unitcircle, which is being used here as a timing reference for the pendulum), by the reciprocal of the speed. This means that theequation should include 2k, not l/2jt, which eliminates choices A and B. The time to complete a cycle increases with aheavier mass, because heavier masses oscillate more slowly. This means that mass should be in the numerator. Thiseliminates choice C and makes choice D the best answer. You could also have considered the spring constant. A stifferspring, with a greater k, would pull the mass back to equilibrium faster, so the time to complete a cycle is reduced by anincreased k. This means that k should be in the denominator, confirming choice D's correctivity. The best answer is D.

26. Choice C is the best answer. The picture shown at t = k in Figure 1 represents the position of the system after a half-cycle.The object is fully displaced to the right and motionless, having started fully displaced to the left and motionless. Afteranother complete cycle, it will be in the same position, so the picture at t = k represents the system at times of jt, 3jt, 5ji, andso on, where 2k corresponds to a full cycle. Choice C is the best answer. Choice A represents times of 0, 2jt, 4jt, 6k, and so

on and choice Brepresents equilibrium, which occurs at times of^^ 12^ ^2' ^2- etc- Choice D has nothing todo withthe system in Figure 1, because both springs are elongated and the object is at equilibrium. The best answer is choice C.

27. Choice A is the best answer. With two springs, the effective spring constant is double what it would be with a single spring.Because the frequency increases with k, the mass will oscillate at a higher frequency with two springs than if there were onlyone. Choice A is the best answer so far. No conclusion can be drawn about the potential for dampening with differentsprings, because dampening was not studied in this experiment. Choice B is eliminated. The effective spring constant isequal to the sum of the individual spring constants, not an average of the two spring constants. Choice C is eliminated. Onespring is fully compressed while the other is fully elongated. As one spring is compressing, the other is elongating. Thismeans that the springs are perfectly out of phase with one another, so choice D is eliminated. The best answer is choice A.

Passage V (Questions 28 - 32) Simple Pendulum Study

28. Choice D is the best answer. Let's compare the periods with and without the nail, by tracking the motion. The motion of thependulum on the right of the nail is the same with or without a nail present. After all, the pendulum is not touching the nail atthese times. The half-periods on the right are the same. However, when the pendulum is to the left of the nail, the nail catchesthe pendulum string and shortens the length of pendulum that is actually moving. A shorter pendulum has a shorter period.Because the pendulum moves with a shorter period when it is to the left of the nail, the left half-period is shorter with the nailpresent than when no nail is present. The overall period will be shorter with the nail present than if no nail were present.

faster left

side swing

with nail

regular rightside swing

Don't think that you must immediately grope for an equation when you see a number in the list of possible answers. Use yourphysical intuition about how an object moves, first. The best answer is choice D.


29. Choice B is the best answer. Byconservation of energy, the initial height of the pendulum must equal its height a half-cyclelater, because at both of those points, the system has all potential energy and no kinetic energy. We know that h can bewritten in terms of L and 6:

h = L(l -cosO)

After completing a half-cycle, the length of the pendulum is only L/4, because the nail has restricted the motion of the fullstring. Because the length of the string has decreased, the angle (9) must increase, so that h remains the same when thesystem is at its highest point. In addition to the equation showing us this relationship, it can also be viewed in the followingpicture comparing the ball at the two positions. Note that the ball is at the same height on the right side and left side of thependulum.

Height at half-cycle, where the total s~\—_6energy of the system isall potential. ^**f 6


Initial height, where the total energyof the system is all potential.

30. Choice A is the best answer. This question tests your knowledge of which physical variables affect pendulum motion andhow they do it. If you increase the length of a pendulum, it will take a longer period of time to complete a full cycle, becausethe ball must traverse a greater distance per full cycle. The equation given in the passage demonstrates this, as does anyrelatively long swinging object. This rules out choice B. The equation also says that a smaller acceleration due to gravity (asyou might Find on the Moon) results in a smaller frequency, because the bob is not moving as fast through the full cycle.Conceptually, if there were no gravity, there would be no frequency, because the bob would remain stationaty. Thus choicesC and D are incorrect. The best answer is choice A.

31. Choice B is the best answer. You should recall that pendulums "glide." This means that the period of the pendulum dependson only g and L. Shortening the length of the string will impact the period. The period of the motion depends on thelength ofthe pendulum in this way:

T= 23f\/L

As the pendulum string is pulled up through the hole, the cable length of the pendulum gets smaller, so the period mustalsodecrease. This leads immediately to choice B. Incidentally, if you wonder why the amplitude, 0max increases, then recall thatthe length of the string and the angle (amplitude) are related by:

h = L(l -cosO)

Energy is conserved; the maximum height the pendulum reaches must be a constant. As the length of the string decreases, 8must increase. The best answer is choice B.

32. Choice B is the best answer. The frequency of a simple pendulum is given in the passage as:

tmUfl2k V L

l .and we know that the period T is given by -*-:

T = 2jt-\/^

As shown in the equation, the period depends on the length of the string and the acceleration of gravity, but it does notdepend on the mass of the bob suspended from the string. Although the ball has a different mass, it has no effect on theperiod. You have perhaps noticed this first hand if you ever jumped offa moving swing. The swing will continue to swingback and forth just as it was doing when your were seated in the swing. It is often helpful to consider analogous real lifesituations to verify what you expect from a pendulum system. The best answer is choice B.



33. Choice C is the best answer. The frequency of the standing wave (first harmonic in this case) depends on the wavelengthand wave speed. Either increasing the wave speed or decreasing the wavelength will increase the frequency of the firstharmonic. Moving the fixed points farther apart increases the length of the string (L), which in turn increases thewavelength, so frequency would decrease. This eliminates choice A. Using a denser string decreases the wave speed, so thefrequency would be reduced. This eliminates choice B. Increasing the tension in the string increases the wave speed in thatthere is less slack to oscillate as the wave passes. The increase in wave speed would result in an increase in frequency.Choice C is the best answer. Because increasing the tension increases the wave speed and frequency, decreasing the tensionhas the opposite effect, eliminating choice D. As a point of interest, the wave speed varies directly with the square root ofthe tension. The best answer is choice C.

34. Choice B is the best answer. According to Equation 1, the period of a pendulum's swing is proportional to the square rootof its length (i.e., T cc VE). Increasing the length by 300% (100% + 300%) means that the new length is 400% of the oldlength (Lnew = 4 x L0id). Because the new period is directly proportional to the square root of the new length (i.e., Tnew cc

V^oid), the new period must be V4 or twice the old period (200% xT0id)- The new period is 200% ofthe old period, whichmeans that the period would increase by 100% (100% + 100% = 200%). This question is a great example of how afundamental concept you may completely understand can be lost in cumbersome analysis. This was more of a question ofunderstanding math and percentages than it was a question on the understanding of the period of a pendulum. The bestanswer is choice B.

35. Choice Bis the best answer. We want the ratio of Xto L(also known as tyjj. We know that for standing waves on strings,Xn =2/nL. This means that for the third harmonic, X=2/3L, so Vl =2/3- Tnis means that the ratio Xto Lis 2:3, makingchoice B the best answer. The best answer is choice B.

36. VI. ifChoice B is the best answer. Frequency is related to the mass of the system by: / cc 7 */m. If the mass of the systemincreases, then the frequency will decrease. (Note: Getting the explicit dependence of the frequency of the system on themass of the spring is notan easy thing to do, but all we need to know here is how the frequency in general will be affected.)The best answer is choice B.

Passage VI (Questions 37 - 42) Deep Ocean Waves

37. Choice B is the best answer. Wave amplitude does not matter, because it does not show up in Equation 1, so choice D isincorrect. Since a square root is involved in the equation, it must be involved in the answer. Only choice B has a squareroot. The best answer is choice B.

38. Choice C is the bestanswer. Choice A is incorrect because the two quantities, period and frequency, are simply inverses ofeach other. You would need only one of the two for a calculation, if you needed either one of them. Since wave amplitudedoes not matter, choices Band D arc also incorrect. This leaves choice C as the only viable answer. Although wavelengthand period are not both in Equation 1, period is the inverse of frequency and the wavelength and frequency do relate to wavespeed in the basic equation v = fX. You could have also solved this by considering that the units of speed are m/s, so youneed a length measurement and a time measurement. The best answer is choice C.

39. Choice C is the bestanswer. The axes are a bit atypical here (v2 vs. X). To make the given equation look similar to theseaxes, we must square Equation 1and find v2 =&/gK. This means that v2 a X, so there is a linear relationship between "v2"and X. The best answer is choice C. Choice B represents the relationship between v and X. The best answer is choice C.

40. Choice D is the best answer. No matter what type of wave is involved, if two waves of the same type pass through oneanother, there is some interference. In thisquestion, the waves are deep-water ocean waves and the interference results in nodisplacement of the medium (a node). It is true that when two waves pass through oneanother, their wavelengths do notchange. However, this does not explain the observation of a node, soalthough it isa true statement, it is irrelevant (a dangverbal trick in the midstof our beloved physics... curse those test writers.) Choice A is eliminated. Choice B is eliminated,because the wavelengths do not change when they pass through one another. A node occurs when the two wavescancel oneanother, which is described as destructive interference, not constructive interference. Choice C is eliminated. The bestanswer is choice D.

Copyright © by The Berkeley Review' 268 REVIEW EXAM EXPLANATIONS

41. Choice A is the best answer. Wave speed depends on the medium, so a difference between ocean water and lake water mustbe responsible for the difference in wave speeds in each medium. Ocean water contains salt, so it is denser than lake water.This eliminates choices B and D. Basedon the relationship between the speed, frequency,and wavelength of a wave, v = fX,if the speed is greater, then either the frequency or wavelength mustbe greatertoo. The frequency of a wave is independentof the medium through which travels, so it is the wavelength that increases with greater speed. The best answer is choice A.

42. Choice B is the best answer. This question gives the periodand wavelength of a wave and asks you to determine the wavespeed. You can arrive at the answer either by using the equation v = fX or by evaluating units. The units are m/s, so thewavelength (m) must be divided by the period (s). The distance betweenevery third crest is 150 m, so the wavelength is 50m. The period is 4 seconds, so the speed is 50/4 m/s, which is 12.5 m/s. The best answer is choice B.

Passage VII (Questions 43 - 48) Strings and Frequency

43. Choice D is the best answer. First, use conservation of momentum to solve for the speed of mi + m2 after the collision:

miv^mi+m^M -* vfinal =-inm^(mi + m2)

Now, use conservation of energy to solve for how much the spring is compressed. All of the kinetic energy of the two massesis converted to the elastic potential energy of the spring:

I{mi+m2)v2 =ikx2 -* lfmi+m2)f miVl f =^kx22 2 2 Vmi + m2/ 2

44.

45.

46.

47.

-> x =miyi

Yk(mi + m2)The best answer is choice D.

Choice B is the best answer. The friction between two surfaces, when one surface slides over the other, is kinetic. Kineticfriction will eventually bring the two surfaces to rest with respect to each other. This slowing results in lower speeds andkinetic energies before, during, and after the collision (when compared with the frictionless surface). This rules out StatementII and rules in Statement III. Only choice B conforms to these conditions. Incidentally, this leads to a lesser compression ofthe spring, since the objects would be moving more slowly upon colliding. You did not need to consider Statement I toanswer this question. In fact, using a bit of logic and elimination can occasionally answer these questions without needing toconsider all of the options. Use elimination and save time. The best answer is choice B.

Choice B is the best answer. Momentum is conserved, if the total momentum does not change as time goes on. Here, we areasked about the momentum of the two masses after the oscillations begin (i.e., after the masses collide, stick together, andstart moving back and forth on the spring.) Since the velocities of these two masses change over time, their momenta mustchange as well. Momentum is not conserved, rulingout choices A and C. As for total energy, what type of energy is involvedwith the two masses? There is potential energy that is stored in the spring and kinetic energy of the masses as they move.Their elevation does not change with time, so their gravitational potential energy does not change. Over time, assumingfriction does not do work on the masses, the spring potential energy is converted into kinetic energy of the blocks, so noenergy has been released. This means that the total energy of the system is conserved, eliminating choice D. The bestanswer is choice B.

Choice A is the best answer. The potential energy for a spring is given by:

U = V£kx2 (which in this example starts at 0 andcannever be negative)

and x is a cosine or a sine function (remember what is meant by simple harmonic motion). When the masses first collide witheach other, the spring is initially relaxed, so it has no potential energy. As the spring becomes more compressed, it gainspotential energy. Because the potential energy depends on the square of the distance, and because the distance is a sine orcosine function, the graph of the potential energy is a sine- or cosine-squared curve. (Exactly which it is-cosine or sinefunction-depends on the initial conditions, and we were told to taket = 0 as the pointjust when the masses collided, beforethe spring was compressed.) The best answer is choice A.

Choice D is the best answer. The frequency of vibration depends on the physicalpropertiesof the system—the massesof thetwo objects and the spring constant of the spring:

=J_JX2ji Vm

Changing the initial speed of mi affects the amount that the spring is compressed, but it will not affect the frequency. Thebest answer is choice D.


48. Choice B is the best answer. The two experiments yield different plots either because of a difference in m2 or the springconstant. What is different about the two plots? Experiment I yields a shorter period and a largeramplitude than ExperimentII. Since the passage gives us a formula for the period, let's look to it for our answer. It says that a smaller mass or stifferspring would result in a shorter period (i.e., higher frequency) for the oscillating mass. Experiment I must fulfill either ofthese two criteria to produce its shorter period, ascompared to Experiment II. Choices Cand D target the spring constant, butthey both say the same thing. These two choices are therefore invalid. Choices A and B target the oscillating mass, andchoice B gives the correct qualitative relationship. Regarding the smaller amplitude in Experiment II, the larger m2 meansthat the speed of the two masses, immediately after collision, will be smaller than in Experiment I. The masses in ExperimentII will therefore not compress the spring as much, giving a smaller oscillation amplitude. When encountering any plotcomparison problem, immediately look for the outstanding difference(s) between the plots-the answer usually lies there.The best answer is choice B.


49. Choice B is the best answer. What is a chief difference between the vibrations created by a marching band and by the foottraffic of various pedestrians? Bands, unless they are really bad, march in unison at some frequency. Since we know that thebridge deteriorates faster from the band than from ordinary foot traffic, this must rule out choice D. Choice C may be true,but it is not the chief problem here. The band must be marching at a frequency that matches a natural frequency of thebridge-that is, the band must cause the bridge to resonate. Incidentally, resonance is a problem for any structure, and it mustbe avoided by damping such vibrations of the structure. This is one reason why your car has shock absorbers, and whymarching soldiers are sometimes ordered to break marching order when crossing old bridges. The best answer is choice B.

50. Choice B is the best answer. This string is essentially tied down at both ends. The formula for wavelengths that can fit on astring tied down at both ends is given by: X= 2L/n, where L is the length of the string and n is a positive integer: n = 1, 2, 3,etc... We are told that this string is 20 meters long, so possible wavelengths include:

X = 40 m, 20 m, 13.33 m, 10 m, 8 m, 6.67 m,...

Comparing these possible wavelengths with the ones listed, we see that 13.33 m is the only correct choice. The best answeris choice B.

51. Choice B is the best answer. Both pictures show the superposition of two waves of unequal frequencies. Adding two suchwaves produces beats, which is the difference in frequency between the two waves:

/beat=l/l-/2l

If the two wave frequencies are close together, there will be fewer beats. If the wave frequencies are far apart, there will bemore beats. From the picture, complex wave I shows more beats, so the two waves that add to yield complex wave I arefarther apart in frequency than the waves that make up complex wave II. As we increase the frequency of one of the waves,we should see the oscilloscope display go from complex wave II to complex wave I. Choice A is incorrect, for the reasonsjust mentioned. Choice C is incorrect. Although there will be an amplitude change, the pattern of beats will definitelychange. Choice D is incorrect. Whatever happens with the intensity of the pattern on the screen, there is still a change in thepattern of the beats. The best answer is choice B.

52. Choice D is the best answer. On graph questions, the first thing to consider is whether the terms increase together or if oneincreases as the other decreases. In this case, with a longer cable, it will take the pendulum bob longer to complete a fullcycle of motion. This means that if the length increases, the period increases, which eliminates choice B, The equation thatrelates the period of a pendulum to the length of the pendulum is:

T=2k^

The relationship involves a square root, so the graph cannot be linear. Choice A is eliminated. To decide between theremaining two graphs it is a good idea to plug numbers in. If L increases by a factor of 4, then T increases by a factor of 2.This means that the graph should "curve more" in the direction of L=axis, which is the x-axis in this case. Choice D curves inthe correct direction, with the graph bending towards the x-axis. The best answer is choice D.


Physics Book 1 Index

Topic

Acceleration 18-19.48Amplitude 223

Angular Acceleration 80

Angular Velocity 79

Antinode 237, 247Artificial Gravity 121

Atwood Machine 71

Beats (and Beat Frequency) 236

Centripetal Acceleration 81

Centripetal Force 86

Coefficient of Friction 91

Collisions (Momentum Transfer) 183

Conservation of Energy 141

Conservation of Momentum 177-179,192Constructive Interference 234-235

Dampened Harmonic Oscillation 224,240Destructive Interference 234-235

Dimensional Analysis 6

Free Fall 20

Frequency 223,232Frictional Force 91,114Fulcrum 186

Gravitational Force 75

Hooke's Law 227

Hydraulic Lift 94

Hydroelectricity 158

Impulse 181, 202Inclined Plane 95,118Keppler's Laws 89,112, 122Kinetic Energy 134

Lever Arm 186

Longitudinal Wave 232

Mechanical Advantage 94, 97, 190Mechanical Energy 141

Topic

Newton's First Law

Newton's Second Law

Newton's Third Law

Node

Normal Force

Pendulum

Period

Periodic Motion

Potential EnergyPower

Projectile RangePulley SystemsRadians

Resonance

Rollercoaster

Rotational EquilibriumSimple Harmonic OscillationSohCahToa

SpringStanding WaveSuperpositionTension

Terminal VelocityTorqueTranslational Motion

Transverse Wave

Vector Addition

VelocityVibrational Motion

Viscous Forces

WavelengthWind Resistance

Work

Work-Energy Theorem

Page

67

67

70

237, 24773,95

229-230,253223

223-224

139

144

28, 34, 5196

79

240

147, 164188,193223,244

13

226-227

237, 243, 254234

70,87

26,110186, 206

5

232

11-12

17

223

91

232

26

131

141,151

Physics Book 1 Test-taking SkillsTopic

Basic RelationshipChecking UnitsComparing GraphsFormula Identification

Graphical EstimationGraphical IdentificationMagnitude EstimationMagnitude Intuition

Page

74

72, 73, 19692

74, 143, 247, 24914

196, 2336, 17, 169

6

Limiting Cases

Topic Page

Multiple Concepts 59, 63,77, 127, 133, 142, 212Numeric Estimation 21, 132Pairing the List 142Physical Intuition 84, 212, 218, 231, 250, 265Qualitative Change 68,78,231Range Shortcut 30Square Root Estimation 23, 231Turbo Solutions 25, 31, 33, 51

63, 72-74, 127, 166, 189, 196, 197, 215, 218, 247

Tracking your ProgressIt is importantto be organized in your review and to recognize where you are strong and whereyou need to spend more time. On these pages, you should keep a scoring profile of yourperformance and a list of items you wish to return to at a future date and review again. Makinga checklist and keeping a log can prove very helpful in staying on task during your reviewperiod. It will also prove useful to write down any key terms and equations from each chapter,so you have a quick source to reference during your final days of reviewing. Keeping all of thisin your review books will allow you to have one source for all of your review.

Translational MotionDate Read: Date Reviewed: Text Questions to Repeat:

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Forces, Circular Motion, and GravitationDate Read: Date Reviewed:

Score on Review Questions: /25Review Questions to Repeat:

Key Equations for Forces, Circular Motion, andGravitation:

Text Questions to Repeat:

Score on Practice Test: /52Practice Test Questions to Repeat:

Key Terms for Forces, Circular Motion, andGravitation:

Work and EnergyDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: /- - Score on Practice Test: / CO


Key Equations for Work and Energy: Key Terms for Work and Energy:

Momentum and TorqueDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: * ^ *""-bB« kiReview Questions to Repeat: Practice Test Questions to Repeat:

Key Equations for Momentum and Torque: Key Terms for Momentum and Torque:

Periodic Motion and WavesDate Read: Date Reviewed: Text Questions to Repeat:

Score on Review Questions: j ~- Score on Practice Test: / co


Key Equations for Periodic Motion and Waves: Key Terms for Periodic Motion and Waves:

ERKELEYThe

B R'E-V'I-E'W®

PERIODIC TABLE OF THE ELEMENTS

1

H

2

He

1.0 4.0

3 4 5 6 7 8 9 10

Li Be B C N O F Ne

6.9 9.0 10.8 12.0 14.0 16.0 19.0 20.2

11 12 13 14 15 16 17 18

Na Mg Al Si P S CI Ar

23.0 24.3 27.0 28.1 31.0 32.1 35.5 39.9

19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36

K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

39.1 40.1 45.0 47.9 50.9 52.0 54.9 55.8 58.9 58.7 63.5 65.4 69.7 72.6 74.9 79.0 79.9 83.8

37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54

Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te 1 Xe

85.5 87.6 88.9 91.2 92.9 95.9 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3

55 56 57

LaT72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Cs Ba Hf Ta W Re Os Ir Pt Au Hg Ti Pb Bi Po At Rn

132.9 137.3 138.9 178.5 180.9 183.9 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)

87 88 89 RAc§

104 105 106 107 108 109 110 111 112 113 114 115 116 117 118

Fr Ra Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Uuq Uup Uuh Uus Uno

(223) (226) (227) (261) (262) (266) (264) (277) (268) (271) (272) (277) (287) (289) (291) (292) (292) (293)

58 59 60 61 62 63 64 65 66 67 68 69 70 71

t Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

140.1 140.9 144.2 (145) 150.4 152.0 157.3 158.9 162.5 164.9 167.3 168.9 173.0 175.0

90 91 92 93 94 95 96 97 98 99 100 101 102 103

§ Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr

232.0 (231) 238.0 (237) (244) (243) (247) (247) (251) (252) (257) (258) (259) (260)

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Transcript of The Berkeley Review MCAT Physics Part 1