How to Use this Book Kaplan MCAT Physics, along with the ...

How to Use this BookKaplan MCAT Physics, along with the other four books in our MCAT Subjectseries, brings the Kaplan classroom experience to you—right in your home, at yourconvenience. This book offers the same Kaplan content review, strategies, andpractice that make Kaplan the #1 choice for MCAT prep. All that’s missing is theteacher.

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To guide you through this complex content, we’ve consulted our best MCATinstructors to call out Key Concept, to offer Bridge to better understanding of thematerial, and Mnemonic devices to assist in learning retention. When you see thesesidebars, you will know you’re getting the same insight and knowledge thatclassroom students receive in person. Look for these as well as references to theReal World and MCAT expertise callouts throughout the book.



Introduction to the MCATThe Medical College Admission Test, MCAT, is different from any other test you’ve encountered in your academic career. It’s not like the knowledge-based examsfrom high school and college, where emphasis was on memorizing and regurgitatinginformation. Medical schools can assess your academic prowess by looking at yourtranscript. The MCAT isn’t even like other standardized tests you may have taken,where the focus was on proving your general skills.



Medical schools use MCAT scores to assess whether you possess the foundationupon which to build a successful medical career. Though you certainly need to knowthe content to do well, the stress is on thought process, because the MCAT is aboveall else a critical thinking test. That’s why it emphasizes reasoning, analytical thinking,reading comprehension, data analysis, writing, and problem-solving skills.



Though the MCAT places more weight on your thought process, you must have astrong grasp of the required core knowledge. The MCAT may not be a perfectgauge of your abilities, but it is a relatively objective way to compare you withstudents from different backgrounds and undergraduate institutions.



The MCAT’s power comes from its use as an indicator of your abilities. Goodscores can open doors. Your power comes from preparation and mindset becausethe key to MCAT success is knowing what you’re up against. That’s where thissection of this book comes in. We’ll explain the philosophy behind the test, reviewthe sections one by one, show you sample questions, share some of Kaplan’sproven methods, and clue you in to what the test makers are really after. You’ll get ahandle on the process, find a confident new perspective, and achieve your highestpossible scores.



Part IReview



1Units and Kinematics

A professor once said, “Biology is chemistry. Chemistry is physics. Physics is life.”Not surprisingly, this was the claim of a physics professor. At the time that I heardit, I considered it dubious at best and possibly even treasonous. I was, after all,premed and a biology major—like many of you—and I was fairly certain that lifewas biological and that physics, if it served any good at all, was just an excuse forold men to play with Slinky toys, magnets, and homemade batteries.



Then I went to medical school, and I grew up—and I came to understand what,exactly, my wise professor had meant. Indeed, I learned that when we treat patientsat the rehab hospital, we often talk about motion, forces, and bone strength. Anophthalmologist had me draw ray diagrams to help me better understand myopia andpresbyopia. When we talk about mitochondria functioning as the “batteries” of thecell, we mean that literally: A mitochondrion is very similar to a concentration cell, akind of battery that operates on the basis of an electrochemical gradient. Themitochondria convert electrical potential energy of NADH and FADH2 into thepotential energy of the high-energy phosphate bond in ATP.



Groan and roll your eyes all you want, but the irrefutable fact of life is this: Physicsis all around us. No, let me amend that: Stop your groaning and open wide your eyesto the understanding that physics affords us of life’s complexities. Open your eyesand be amazed—be delighted. Those old physics professors who pause midlectureto tinker around with Slinkies, magnets, and batteries are on to something. Theyknow that the physical world is at play and that it invites you to join in the game.They’re just trying to teach you the rules.



“Physics is life.” Fine, you say, but I’m not here just to learn about all the coolways I can understand life through physics. I’m here to prepare for the MCAT.You’re right. That is what you’re here for, and that’s what we’re going to help youdo. Nevertheless, the first order of business is to make sure that you begin yourpreparation for the MCAT—especially, perhaps, for the physics of the MCAT—with the right attitude. Far too many students struggle to make progress in their TestDay preparation because of a negative attitude. Nowhere do we see this more oftenthan with physics (and verbal reasoning). Many students believe that they areincapable of improving their understanding and are doomed to weak performance,but we know that this is simply not true. Kaplan’s integrated focus on sciencecontent, critical-thinking skills, and Test Day strategies is the key to buildingconfidence. That confidence leads to Test Day success.



All right, so let’s start your preparation for MCAT Physics with the basics. This firstchapter reviews the three systems of units encountered on the MCAT; namely, MKS(meter-kilogram-second), CGS (centimeter-gram-second), and SI (InternationalSystem of Units). Then we’ll review the fundamental mathematics necessary for thestudy of MCAT Physics, such as scientific notation, trigonometric functions, andvectors. Finally, we consider an important branch of Newtonian mechanics (Chapter2) that deals with the motion of objects: kinematics. Here, we’ll have some funconsidering how it’s possible to drive a car and travel many kilometers, yet have anaverage velocity of 0 meters/second. As a demonstration of motion in onedimension, we’ll drop a few (hypothetical) cats to see if it really is true that theyalways land on their feet, and speaking of cats—and motion in two dimensions—we’ll ask the intriguing (and not at all hypothetical) question, “How did that hairball endup in my shoe?”



ConclusionAll right! One chapter down! There’s a lot more to come, but this is a good start.We have reviewed the most important principles of mathematics relevant to theMCAT, including units, scientific notation, trigonometry, logarithms, and vectors.With this basic review, we’ve equipped you with the math, the language of physics,necessary to understand our first important topic for the MCAT Physical Sciencesection, kinematics. This study of objects in motion allows us to describe an object’s velocity, speed, acceleration, and position with respect to time. We nowunderstand how to use the five kinematics equations when objects experienceconstant acceleration, a relatively simple scenario presented on Test Day. Our goal in this chapter and throughout all of our review of the basic sciences is todemystify the concepts that have far too often been viewed by students asincomprehensible or irrelevant. Quite to the contrary, we believe that every studentwho commits herself or himself to consistent and thoughtful preparation canunderstand the mathematical and scientific concepts presented here and elsewhere inthe Kaplan MCAT program. Furthermore, we hope that you will come to anappreciation of the relevance that these concepts and principles have for yourperformance not only on the MCAT but also in medical school, residency training,and your career as a physician. While you may rarely need to make logarithmiccalculations as part of your daily medical practice, and while it is unlikely that yourpatient care will ever involve the kinematics equations, these basic concepts ofmathematics and of the physical and biological sciences form the foundation uponwhich the art of medicine rests.

CONCEPTS TO REMEMBERYou will be tested on only the metric units, especially those that constitute the

International System of Units (SI) for length (m), mass (kg), volume (L), temperature(K), force (N), time (s), velocity (m/s), acceleration (m/s2), energy (J), current (A),electric potential (V), and resistance (ohm).

Scientific notation allows you to express very large or very small numbers in aneasy-to-read and easy-to-manipulate format. Learn the rules for adding, subtracting,multiplying, and dividing numbers in scientific notation.

Knowing the trigonometric functions of sine and cosine for the classic angles (0, 30,45, 60, 90, 180, and 270) for right triangles will be essential for solving problemssuch as force vector addition or subtraction, inclined planes, and projectile motion.

Logarithms, like scientific notation, allow you to express a very large range of valuesin an easy-to-read and easy-to-understand format. The two logarithmic bases thatyou need to know for the MCAT are base-10 and base-e. pH and pOH, decibel, andRichter scales are base-10 (common log), while nuclear decay is base-e (natural log).Use the log rules to approximate logarithmic calculations on Test Day.

There are only two kinds of numbers: scalars and vectors. Scalars have magnitudeonly and include distance, speed, mass, pressure, energy, and work. Vectors havemagnitude and direction and include displacement, velocity, acceleration, force, and



electric and magnetic fields. Vector addition or subtraction is usually bestaccomplished by resolving the vectors into their X and Y components first.

Average velocity is calculated as the total displacement divided by total time. Averagespeed is calculated as the total distance divided by total time. The magnitude of theaverage velocity may be equal to or less than average speed, but it can never begreater. Instantaneous speed is always equal to the magnitude of the instantaneousvelocity.

Deceleration is nothing other than acceleration in the direction opposite of that of theinitial velocity. Such “negative” accelerations will cause objects to slow down andpossibly even stop and reverse direction.

On the MCAT, objects typically experience motion with constant acceleration, whichimplies a constant force. You must memorize and understand the kinematicsequations that relate velocity, acceleration, displacement, and time.

Objects in free fall experience acceleration equal to that of gravity (9.8 m/s2),discounting air resistance. Objects that achieve terminal velocity experience a netforce equal to 0 (and have acceleration equal to 0 m/s2) because the upward airresistance force exactly balances the downward force of gravity.

Projectile motion is motion in two directions. To solve projectile problems, you mustresolve the initial velocity into its X and Y components. For projectile motioninvolving gravity, only the Y component (vertical) velocity vector will change at therate of g; the X component (horizontal) velocity vector is constant, discounting theforce of air resistance.

EQUATIONS TO REMEMBER

Notes: 1. v0 and x0 are v and x at t = 0.2. When the motion is vertical, we use y instead of x.3. As illustrated below, in using these equations, we must remember that velocity and

acceleration are vector quantities.Practice Questions

1. A man walks three blocks (30 m) east, and then another five blocks (50 m) north tothe drugstore (point B). What is the magnitude of his final displacement from hisoriginal location (point A)?

A. 20 mB. 40 mC. 60 mD. 3,400 m



2. A submarine sends out a sonar signal (sound wave) in a direction directly downward.It takes 2.3 s for the sound wave to travel from the submarine to the ocean bottomand back to the submarine. How high up from the ocean floor is the submarine?(The speed of sound in water is 1,489 m/s.)

A. 1,700 mB. 3,000 mC. 5,000 mD. Cannot be determined.

3. A car is traveling at 40 km/hr and the driver puts on the brakes, bringing the car to restin a time of 6 seconds. What’s the magnitude of the average acceleration of the carin units of km/hr2?

A. 240 km/hr2

B. 12,000 km/hr2

C. 24,000 km/hr2

D. 30,000 km/hr2

4. If an object is released 19.6 m above the ground, how long does it take the object toreach the ground?

A. 1 sB. 2 sC. 4sD. 10s

5. At a place where g is 9.8 m/s2, an object is thrown vertically downward with a speedof 10 m/s while a different object is thrown vertically upward with a speed of 20 m/s.Which object undergoes a greater change in speed in a time of 2 seconds?

A. The first object, because the speed vector points in the same direction as theacceleration due to gravity

B. The second object, because it has a higher velocityC. Both objects undergo the same change in speed.D. Cannot be determined from the information given.

6. A firefighter jumps horizontally from a burning building with an initial speed of 1.5m/s. At what time is the angle between his velocity and acceleration the greatest?

A. The instant he jumpsB. When he reaches terminal velocityC. Halfway through his fallD. Right before he hits the ground

7. A circus clown is catapulted straight up in the air (the catapult is 1.5 m off the groundlevel). If her velocity as she leaves the catapult is 4 m/s, how high does she get, asmeasured from the ground?

A. 0.4 mB. 0.8 mC. 1.9 mD. 2.3 m

8. Which of the following expressions correctly illustrates the SI units of each one of thevariables in the formula?

mΔv = FΔt



A. 1b × mph = ft × 1bxsB. 1b × km = N × sC. kg × m/s = N × sD. g × m/s = N × s

9. If a toy magnet is able to generate one centitesla, how many teslas would 108 magnetsbe able to generate?

A. 1 kTB. 1 MTC. 1 GTD. 1 TT

10. Which of the following quantities is NOT a vector?A. VelocityB. ForceC. DisplacementD. Distance

11. A rifle is held by a man whose arms are 1.5 m above the ground. He fires a bullet at100 m/s at an angle of 30° with the horizontal. After 2 seconds, how far has theobject traveled in the horizontal direction?

A. 87 mB. 140 mC. 174 mD. 175.5 m

12. A BASE jumper runs off a cliff with a speed of 3 m/s. What is his overall velocityafter 0.5 seconds?

A. 3 m/sB. -5 m/sC. 5 m/sD. 10 m/s

13. A rock (m = 2 kg) is shot up vertically at the same time that a ball (m = 0.5 kg) isprojected horizontally. If both start from the same height,

A. the rock and ball will reach the ground at the same time.B. the rock will reach the ground first.C. the ball will reach the ground first.D. More information is needed to answer this question.

Small Group Questions1 . A baseball is thrown directly upward and hits the ground several hundred feet away.

Are there any points along its path at which the velocity and acceleration vectors areperpendicular? Parallel?

2 . A string with 10 beads attached to it at equal intervals is dropped vertically from aheight. The sound of each bead is heard as it hits the ground. The sounds will notoccur at equal time intervals. Why? Will the time between sounds increase ordecrease near the end of the fall? How could the beads be timed so the soundsoccur at equal intervals?



2Newtonian Mechanics

The dozen eggs you dropped yesterday morning on the kitchen floor that took anhour to clean up, the mailbox mangled by bat-wielding neighborhood ruffians, thesqueaking sound (and pinching pain) as you slid bare-legged down the metal slide: allNewtonian mechanics acted out in real life. The ways in which objects move fromone place to another are described by Newton’s laws of motion. While the results ofthese laws will not always be so tragic (after all, we know there’s no point in cryingover spilt milk), it is impossible to take even one step in the real world without beingsubjected to them. The related concepts of force, mass, and acceleration arefundamental not only to an understanding of mechanics but also to your success onthe Physical Sciences section of the MCAT. In fact, these concepts will account forthe largest aggregate of physics-related points on Test Day! Learn them well nowand then apply them over and over again in your online practice for MCATproficiency.



This chapter looks at each of Newton’s three laws and marvels that a world ascomplex as ours can really be quite well understood by three—not 3,000, not 300,not even 30—laws. Far more interesting than the laws themselves will be the use ofthe laws to make sense of all kinds of motion that we observe every day: falling,sliding, and rotating.



Because we’ve already learned how to divide forces into their geometriccomponents, also known as resolving forces (Chapter 1), we can understand how tohang pictures so that they are level, make predictions about who will win thetug-of-war contest, and even calculate the speed at which your friend (soon-to-beknown as “the crushed pelvis” in Room 238) will smash into the snow bank at thebottom of the double black diamond ski slope. Considering the application of forcesto objects that are free to rotate around a fixed point will tell us why wrenches andbolts are better than screwdrivers and screws and why sometimes chubby andskinny kids have difficulty finding balance on a seesaw. Lastly, we will think aboutcentripetal motion, which will help explain why race cars tend to crash on the curvesand why your puppy on her leash keeps running around you in a perfect circle.



ConclusionIn this chapter, we learned that different kinds of forces act on objects to cause themto move in certain ways. Applications of forces may cause objects to accelerate ordecelerate, according to Newton’s second law. If the vector sum of all the forcesacting on an object is equal to zero, the forces cancel out, and the object experiencesno acceleration, a condition known as translational equilibrium. This is expressed inNewton’s first law. Even when objects aren’t touching, they can still exert forcesbetween them, as described by Newton’s third law. Forces can cause objects totranslate (move with linear motion), and centripetal forces cause objects to move in acircle. Applying forces to objects that have pivot points can generate torques, whichcause the objects to rotate. Sometimes those torques cancel out to producerotational equilibrium. Newtonian mechanics is an MCAT favorite. Your carefulconsideration of the discussion topics in this chapter and your practice with thekinds of problems demonstrated here will pay off in many points on Test Day. Eggscrashing, milk spilling, signs hanging, cars speeding, clowns flying, children playing:life in motion—physics in life.

CONCEPTS TO REMEMBERNewtonian mechanics is all about the relationships among objects, forces,

accelerations, and energies. Newton’s three laws express the fundamentalrelationships among force, mass, and acceleration. The first law is actually a specialcase of the second law in which the object experiences a net force equal to 0 and,therefore, does not accelerate (that is to say, does not experience a change in eitherthe direction or magnitude of its velocity).

Newton’s third law does not require objects to be in direct physical contact. Manyequal and opposite forces can be exerted between objects that are separated by evengreat distances. Gravitational and electrostatic forces are two types that can existover large distances of separation. Normal force is a special kind of reactive forcebetween objects that are in direct physical contact.

Always draw a free-body diagram when working through problems of Newtonianmechanics. Be especially mindful to practice drawing free-body diagrams of objectson inclined planes. To resolve the forces into their X and Y components, you mustknow the trigonometric functions of sine and cosine for the classic right triangleangles.

The magnitude of gravitational force between two objects is inversely proportional tothe square of the distance separating their respective centers of gravity. Thisequation is in the same form as the electrostatic force equation, but the force due tomass is much smaller than the force due to charge.

“Little g” is an approximation of the acceleration due to gravity on an object close tothe surface of the earth. Although g is given as a constant, g = 9.8 m/s2, in reality, thetrue acceleration due to gravity depends on r and can be determined with thegravitational force equation.

Three types of motion are tested on the MCAT: translational, rotational, and circular.Translational and circular motion are caused by forces that cause objects to move



without rotating; rotational motion is caused by forces applied perpendicularly to afulcrum, generating torque and causing rotation around the fulcrum.

In uniform circular motion, the centripetal force vector and the centripetal accelerationvector point directly to the center of the circular pathway. The instantaneous velocityvector is tangential to the circular pathway at any given point.

All friction forces act in the direction opposite or opposing intended or actual motion.On the MCAT, you will be told when you can ignore friction and when you mustaccount for it. The forces of air resistance, static friction, and kinetic friction arecommonly encountered on the test. For an object in contact with a surface, themaximum value of static friction will always be greater than the (constant) value ofkinetic friction.

Translational equilibrium occurs when the vector sum of forces acting on an object isequal to 0. There is no net acceleration, and the object continues in a constantmotional behavior. Translational equilibrium does not require that an object bestationary, only that its motional behavior (stationary or moving) be constant.

Rotational equilibrium occurs when the vector sum of torques acting on an object isequal to 0. There is no net rotational acceleration, and the object continues in aconstant motional behavior. As with translational equilibrium, there is no requirementthat the object be stationary, only that its motional behavior (stationary or rotating)be constant.


Practice Questions1. A 1,000 kg rocket ship, traveling at 100 m/s, is acted on by an average force of 20 kN

applied in the direction of its motion for 8 s. What is the change in velocity of therocket?

A. 100 m/sB. 160 m/sC. 1,600 m/sD. 2,000 m/s

2. An elevator is designed to carry a maximum of weight of 9,800 N (including its ownweight) and to move upward at a speed of 5 m/s after an initial period ofacceleration. What is the relationship between the maximum tension in the elevator



cable and the maximum weight of 9,800 N when the elevator is accelerating upward?A. The tension is greater than 9,800 N.B. The tension is less than 9,800 N.C. The tension equals 9,800 N.D. It cannot be determined from the information given.

3. A 10 kg wagon rests on a frictionless inclined plane. The plane makes an angle of 30°with the horizontal. What is the force, F, required to keep the wagon from slidingdown the plane?

A. 10 NB. 30 NC. 49 ND. 98 N

4. A 20 kg wagon is released from rest from the top of a 15 m long plane, which isangled at 30° with the horizontal. Assuming there is friction between the ramp andthe wagon, how is this frictional force affected if the angle of the incline is increased?

A. The frictional force increases.B. The frictional force decreases.C. The frictional force remains the same.D. It cannot be determined from the information given.

5. An astronaut weighs 700 N on Earth. What is the best approximation of her newweight on a planet with a radius that is two times that of Earth, and a mass threetimes that of Earth?

A. 200 NB. 500 NC. 700 ND. 900 N

6. A 30 kg girl sits on a seesaw at a distance of 2 m from the fulcrum. Where must herfather sit to balance the seesaw if he has a mass of 90 kg?

A. 67 cm from the girlB. 67 cm from the fulcrumC. 133 cm from the girlD. 267 cm from the fulcrum

7. An object is moving uniformly in a circle whose diameter is 200 m. If the centripetalacceleration of the object is 4 m/s2, then what is the time for one revolution of thecircle?

A. 10π sB. 20π sC. 100 sD. 200 s

8. A distant solar system is made up one small planet of mass 1.48 × 1024 kg revolving ina circular orbit about a large, stationary star of mass 7.3 × 1030 kg. The distancebetween their centers is 5 × 1011 m. What happens to the speed of the planet if thedistance between the star and the planet increases (assume the new orbit is alsocircular)?

A. The speed of the planet increases.



B. The speed of the planet decreases.C. The speed of the planet remains the same.D. It cannot be determined from the information given.

9. An elevator is accelerating down at 4 m/s2. How much does a 10 kg fish weigh ifmeasured inside the elevator?

A. 140 NB. 100 NC. 60 ND. 50 N

10. A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed atthe bottom is 8 m/s. How many revolutions per second is the ball making when atthe bottom of the plane?

A. 6 revolutions/secondB. 12 revolutions/secondC. 20 revolutions/secondD. 23 revolutions/second

11. A 1,000 kg satellite traveling at speed v maintains an orbit of radius, R, around theearth. What should be its speed if it is to develop a new orbit of radius 4R ?

A. ¼ vB. ½ vC. 2 vD. 4 v

12. A 100 kg elevator initially at rest accelerates up at rate . What is the average forceacting on the elevator if it covers a distance, x, over a period of 10 s?

A. 2x + 1,000B. 100( + 1)C. 2 + 1,000D. 2 x

Small Group Questions1. Is the acceleration of a freely falling object affected by changes in ambient pressure?2. If the acceleration of an object is zero, are no forces acting on it?3. Only one force acts on an object. Can it have zero acceleration? Can it have zero

velocity?Explanations to Practice Questions

1. BThe average force on the rocket equals its mass times the average acceleration; theaverage acceleration equals the change in velocity divided by the time over which thechange occurs. So the change in velocity equals the average force times the timedivided by the mass:F = ma, so a = F/ma = Δv/ΔtΔv = aΔt = FΔt/mΔv = (20,000 N × 8 s)/1,000 kg



Δv = 160 m/s(B) is therefore the correct answer. 2. AThe forces on the elevator are the tension upward and the weight downward, so thenet force on the elevator is the difference between the two. When the elevator isaccelerating upward, the tension in the cable must be greater than the maximumweight so that the difference produces a nonzero acceleration. (A) is therefore thecorrect answer. For the sake of completion, when the elevator is moving at aconstant speed, its acceleration is zero, so the net force on the elevator is also zero,which means that the tension and the weight are equal. Similarly, when the elevator isaccelerating downward, the tension is less than 9, 800 N so that there is a nonzeroacceleration. 3. CThe weight of the wagon (W= mg) acts in a direction straight down. This force canbe separated into X and Y components, with the x-axis parallel to and the y-axisperpendicular to the plane of the incline.Wx = W sin 30°Wy = W cos 30°To keep the wagon from moving (i.e., to keep it in equilibrium), the sum of theforces must equal zero. In terms of the components,∑Fx =0 and ∑Fy= 0From a free-body diagram, it can be seen that the Y component of the weight iscounteracted by an equal and opposite force from the surface of the plane (thenormal force, N). Similarly, the X component of the weight, which tends to cause thewagon to roll down the plane, must be counteracted by an equal and opposite forceparallel to the plane. This is the unknown force, F. (Remember, the plane isfrictionless.) The magnitude of the required force can be determined by recognizingthat the sum of the forces in the x-direction must equal zero:

(C) best matches our result and is thus the correct answer. Note: We let “up the plane” be the positive x-direction. We could have also taken“down the plane” to be the positive x-direction, as long as we use the correct signswhen we calculate the components. The choice of positive direction is arbitrary. 4. BThe force of friction on an object sliding down an incline equals the coefficient offriction times the normal force. The normal force, which is given by mg cos θdecreases as the angle of the incline, θ increases. Therefore, the frictional force



decreases as the angle of the incline increases. (B) best reflects this and is thus thecorrect answer. 5. BThe weight of an object on a planet can be found by using Newton’s law ofgravitation with one minor simplification. Because the planet is much larger than theperson, r, which is the distance between the centers of the two objects (the planetand person), can be taken simply as the radius of the planet; the additional fewmeters to the center of the astronaut can be ignored. The astronaut’s weight on earthequalsF= GmME/rE

2 = 700 NOn a planet that is three times as massive as earth (Mp = 3ME), and has twice theradius (rP = 2rE), the force would be equal toF = GmMP/r2

p

F = Gm (3ME)/(2rE)2

F = 3GmME/(4rE2)

F = ¾ (GmME/rE2)

F = ¾ (the astronaut’s weight on earth)F = ¾ × 700NF 500 N(B) is therefore the correct answer. 6. BFor the seesaw to be balanced, the torque due to the girl must be exactlycounteracted by the torque due to her father. More generally stated, the sum of thetorques about any point must be equal to zero. Taking the torques about the fulcrumfor the girl and the father, obtain the following (Note, the f subscript represents thefather while the g subscript represents the girl):

(B) is therefore the correct answer. You can also solve this problem by finding thetorques about any point, not just the fulcrum. Doing so, however, is usually morecomplicated because the force of the fulcrum on the seesaw first has to bedetermined. Because both the weight of the girl and the weight of the father act in adownward direction, the fulcrum must exert a force of (30 × g) + (90 × g) = 120 g Nin an upward direction on the seesaw. Calculating the torques about the girl’slocation, we obtain the following:



This equals 2.67 m - 2 m = 0.67 m from the fulcrum. Once again, (B) matches yourresult. 7. AThis question is testing your knowledge of circular motion. To calculate the time forone revolution of the circle, start with the centripetal acceleration: a = v2/r. Given r =100 m and a = 4 m/s2, we have v = 20 m/s. Speed, v, and time of revolution (period), T, are related by v = 2πr/T (i.e., in a time equal to one revolution, the distancecovered is the circumference). Therefore,

T= 2π (100 m)/(20 m/s) = 10πs(A) matches your result and is therefore the correct answer. 8. BThe force on the planet due to the star is equal to

F = Gm starmplanet/r2

Because this force provides the centripetal force, it’s also true that F is equal toF= mplanetv2/r

Equating the two expressions for F gives the following:

Increasing the distance between the planet and the star, r, means that the speed, v,decreases. This means that (B) is the correct answer. 9. CYou may imagine that the fish is attached to a string and the tension in the string isgiven by ma = mg - T (because a and g point in the same direction). Then T = m ×(g - a) =10 × (10 - 4) = 60 N. A quicker way to solve this problem is to notice thatwhen the elevator accelerates down with magnitude g, the fish is in free fall andweighs 0 N. Alternatively, when the elevator goes up, one usually feels heavier. (C) istherefore the correct answer. 10. AIn 1 second, the ball passes 8 m. One revolution is 2πr = 1.3 m, so 8/1.3 is 6revolutions/s. Therefore, (A) is the correct answer. 11. BBecause of the opposing forces acting on the satellite, use this equation to solve theproblem:

Beyond this, the question is rather straightforward: GMe/R = v2 implies that thevelocity is inversely proportional to the square of the radius. (B) is therefore thecorrect answer.



12. AThis question mixes symbols with numbers. The best approach is to solve it andthen look at the answers:

Now you can find thatF = 100 × (10 + (2x/t2))= 1000 + 100(2x/t2)= 1000 + 100(2x/102)= 1000 + 200x/100)= 1000 + 2x.This matches with (A), the correct answer.



3Work Energy, and Momentum

The Greek myth of Sisyphus is a tale of unending, pointless work. In life, Sisyphuswas the son of a king and the founder and first king of Corinth. He was a powerfulman who seduced his niece, captured his brother’s throne, and even dared toexpose Zeus’s affair with the river god’s daughter. He was also an exceedinglytricky man who even managed, in death, to escape punishment once or twice.Having had enough of his trickster ways, Zeus finally punished Sisyphus with anunenviable task: For all of eternity, Sisyphus would have to roll a large, heavy rockup a steep hill. Just as Sisyphus would nearly reach the top of the hill, the rockwould roll back again to the bottom. Some say that the rock always managed to rollover his feet on its way down. The lessons here are many: Hubris will be punished,Zeus will not be mocked, and always wear steel-toed boots when moving heavyobjects.



The myth of Sisyphus is a tale of unending, pointless work—and quite literally it isthe story of work and energy. Pushing that boulder up the hill, Sisyphus exertedforces that worked on the rock, resulting in an increase in the rock’s energy. Whenthe rock escaped from his grasp and began to roll backwards, it experienced atransformation of energy from potential (gravitational) energy into kinetic energy andother forms of energy such as thermal and sound. Sisyphus’s punishment was notfound in the mere fact that energy can be transformed, although Zeus was evidentlysufficiently aware of mechanical physics to take advantage of this point. No, it wasthe obvious pointlessness and futility of the eternal task that was his punishment. Wehope to convince you throughout the Kaplan MCAT program that your preparationfor Test Day is in no way a Sisyphean task.



This chapter reviews the fundamental concepts of energy and work. A closeexamination of the different forms of energy will lead us to the understanding thatwork itself is just one of the two ways or processes in which energy can betransferred from one system to another. The work-energy theorem is a powerfulexpression of the relationship between energy and work. This fundamentalunderstanding will assist you in your approach to many problems on the MCAT,leading you to the correct answers. Furthermore, this chapter also covers the relatedtopics of impulse and momentum. Objects with mass are said to have momentum;momentum is a function of both the object’s mass and its velocity. As we’ve alreadystated in Chapter 2, objects tend to resist motion changes; that’s called inertia.Momentum and inertia are closely related, but they are not the same thing. Impulse,which derives from application of force to an object, causes that object’s motion tochange, resulting in a change in momentum. On the MCAT, momentum and impulseare most commonly tested through the presentation of collision incidents. We willdiscuss the different types of collisions that may be presented on Test Day, andthrough some examples, we’ll show how best to approach them. Finally we’lldiscuss the topic of mechanical advantage, and because we’re already talking aboutone king, we’ll talk about another: We’ll examine how a pulley or ramp might havebeen helpful in getting fat King Henry VIII up onto his horse.



Conclusion

Sisyphus and Henry VIII, two kings whose life stories have achieved mythologicalstatus, have served us well as vivid illustrations of the principles of energy and work.The conceptualization of energy as the capacity to do something or make somethinghappen is broad enough to allow us to understand everything from pushing a rockup a hill to melting an ice cube from stopping a car at an intersection to raising anoverweight king onto his horse. Inextricably tied to this notion of energy is theunderstanding that none of these events or phenomena could ever happen withoutthe transference of energy, either through work or heat. The work-energy theorem isa powerful expression that will guide our approach to many problems in the PhysicalScience section. A topic of Newtonian mechanics that is tested often on the MCATis the application of principles of energy and work to simple machines, such aslevers, inclined planes, and pulleys. These devices assist us in accomplishing workby reducing the forces necessary for displacing objects. Another topic that you canexpect to see on Test Day is momentum. Completely elastic, inelastic, andcompletely inelastic collisions are different instances in which momentum isconserved as objects collide and interact. Finally, we discussed the method fordetermining centers of mass and gravity, a calculation which will be helpful to youfor certain problems involving gravitational potential energy. Preparing for the MCAT is hard work. No one is denying that, and anyone who tellsyou otherwise is misguided at best. Nevertheless, hard work has its rewards, themyth of Sisyphus notwithstanding. You are well on your way to achieving successon Test Day if you continue to pay attention to the lessons that are provided in yourKaplan MCAT materials. This MCAT Physics Review book—and all the otherKaplan materials provided in your Kaplan program—is part of a set of tools—yoursimple machines, if you will—that will provide you with the mechanical advantage ofeasing your efforts toward a higher score. Analogous to our discussion of energy,work, and momentum, there’s much potential for accomplishing your goals ofearning a higher MCAT score, gaining entrance into the medical school that’s bestfor you, and becoming a great doctor. You just need to keep moving; the forces ofyour commitment and tenacity are more than sufficient to bring you to the place ofpersonal and professional accomplishment.

CONCEPTS TO REMEMBEREnergy is a property or characteristic of a system to do something or make something

happen, including the capacity to do work.Kinetic energy is energy of motion. Potential energy is energy stored within a system

and exists in forms such as gravitational, electric, and mechanical.In the absence of friction forces, total mechanical energy of a system will be

conserved. That is to say, the sum of a system’s potential energy and kinetic energywill be constant.

Work is a process by which energy is transferred from one system to another. Itinvolves the application of force through a displacement (or distance). The rate at



which work is done, or energy is transferred, is the power loss (or gain) of thesystem.

The work-energy theorem states that when net work is done on or by a system, thesystem’s kinetic energy will change by the same amount. When net work is done ona system, the system’s kinetic energy will increase; when net work is done by asystem, the system’s kinetic energy will decrease.

Momentum is a quality of objects in motion and is the product of mass and velocity.Inertia is the tendency of an object to resist changes in its motion. Impulse is achange in momentum achieved by the application of force through some period oftime.

There are three types of collision for which the system’s total momentum isconserved: completely elastic, inelastic, and completely inelastic collisions. Only in acompletely elastic collision is the system’s total kinetic energy also conserved. In theinelastic collisions, kinetic energy is transformed into another form, such as heat,light or sound.

Simple machines, such as inclined planes, levers, and pulleys, provide the benefit ofmechanical advantage, which is the factor by which the machine multiplies the inputforce or torque. Mechanical advantage makes it easier to accomplish a given amountof work since the input force necessary to accomplish the work is reduced. Thedistance through which the reduced input force must be applied, however, isincreased by the same factor.

For all machines that provide mechanical advantage, input work = output work,discounting energy “lost” due to friction. If the work of friction is accounted for,then the ratio of output work to input work is a measure of the machine’s efficiency.

The center of mass is the point within a two- or three-dimensional object at which allthe object’s mass could be represented as a single particle. The center of mass andthe center of gravity are at the same point within an object.


U = mgh (gravitational potential energy)E = U + K E = U + K = Constant (for conservative systems)W = Fd cos θ

Wnet= ΔK = Kf - Ki

p = mv J = Ft = Δp = mvf- mvi

mavai + mbvbi = mavaf + mbvbf (for completely elastic and inelastic collisions)

mavai + mbvbi = (ma+ mb)vf (for completely inelastic collisions only) force out



Practice Questions1. A weight lifter lifts a 275 kg barbell from the ground to a height of 2.4 m. How much

work has he done, and how much work is required to hold the weight at that height?A. 3,234 J and 0 J, respectivelyB. 3,234 J and 3,234 J, respectivelyC. 6,468 J and 0 J, respectivelyD. 6,468 J and 6,468 J, respectively

2. A tractor pulls a log that has a mass of 500 kg along the ground for 100 m. The rope(between the tractor and the log) makes an angle of 30° with the ground and is actedon by a tensile force of 5,000 N. How much work does the tractor do? (sin 30° =0.5, cos 30° = 0.866, tan 30° = 0.57)

A. 250 kJB. 290 kJC. 433 kJD. 500 kJ

3. A 2,000 kg experimental car can accelerate from 0 to 30 m/s in 6.3 s. What is theaverage power of the engine needed to achieve this acceleration?

A. 143 WB. 143 kWC. 900 JD. 900 kJ

4. A 40 kg block is resting at a height of 5 m off the ground. If the block is released andfalls to the ground, what is its total energy at a height of 2 m?

A. 0JB. 400 JC. 2 kJD. It cannot be determined from the information given.

5. If a human is lying on a flat surface and the weights of the head, thorax, and hips andlegs are determined to be 8 lbs, 20 lbs, and 20 lbs, respectively, what is the center ofmass of this person? (Assume that the head, thorax, and hips and legs are 10 inches,25 inches, and 35 inches long.)

A . 15 inches away from the top of the headB . 25 inches away from the top of the headC . 30 inches away from the top of the headD . 35 inches away from the top of the head

6.



In the pulley system shown, if the mass of the object is 10 kg and the object isaccelerating upwards at 2 m/s2, what is the tension force in each rope?

A. 50 NB. 60 NC. 100 ND. 120 N

7. Two billiard balls, one moving at 0.5 m/s, the other at rest, undergo a perfectly elasticcollision. If the masses of the billiard balls are equal and the speed of the stationaryone after the collision is 0.5 m/s, then what is the speed of the other ball after thecollision?

A . 0 m/sB . 0.5 m/sC . 1 m/sD . 1.5 m/s

8. A 2,500 kg car traveling at 20 m/s crashes into a 6,000 kg truck that is originally atrest. What is the speed of the truck after the collision, if the car comes to rest at thepoint of impact? (Neglect the effects of friction.)

A . 0 m/sB . 8.33 m/sC . 16.7 m/sD . 83 m/s

9. Which of the following are elastic collisions?I. A planet breaks into several fragments.II. Two balls collide and then move away from each other with the same speed but

reverse directionality.III. A volleyball hits the net, slows down, and then jumps back with one-half its initial

speed.A. I onlyB. II only



C. II and III onlyD. I, II, and III

10. A 1 kg cart travels down an inclined plane at 5 m/s and strikes two billiard balls,which start moving in opposite directions perpendicular to the initial direction of thecart. Ball A has a mass of 2 kg and moves away at 2 m/s, while ball B has a mass of1 kg and moves away at 4 m/s. Which of the following statements is true?

A. The cart will come to a halt in order to conserve momentum.B. The cart will slow down.C. The cart will continue moving as before, while balls A and B will convert the

gravitational potential energy of the cart into their own kinetic energies.D. These conditions are impossible because they violate either the law of conservation

of momentum or conservation of energy. 11. Tom, who has a mass of 80 kg, and Mary, who has a mass of 50 kg, jump off a 20

m tall building and land on a fire net. The net compresses, and they bounce back upat the same time. Which of the following statements is NOT true?

A. Mary will bounce higher than Tom.B. The magnitude of the change in momentum for Tom is 3,200 kg m/s.C. Tom will experience a greater force upon impact than Mary.D. The energy in this event is converted from potential to kinetic to elastic to kinetic.

12. What is the acceleration of a 20 kg box if a force of 100 N is applied, as depicted inthe ramp and pulley system shown?

A. 3 m/s2 down the rampB. 5 m/s2 down the rampC. 3 m/s2 up the rampD. 5 m/s2 up the ramp

13. Two identical molecules, A and B, move with the same horizontal velocities butopposite vertical velocities. Which of the following statements may be true after thetwo molecules collide?

A. The sum of the kinetic energies of the molecules after the collision is less than thesum of the kinetic energies of the molecules before the collision.

B. Molecule A will have a greater momentum after the collision than molecule B will.C. The sum of the kinetic energies of the molecules after the collision is greater than

the sum of the kinetic energies of the molecules before the collision.D. Molecule A will have a greater vertical velocity than molecule B will.

Small Group Questions1. The moon revolves around the earth in a circular orbit due to the gravitational force

exerted by the earth. Does gravity do positive work, negative work, or no work onthe moon?

2. A net force causes a particle’s speed to double and then double again. Does the net



force do more work during the first or second doubling?Explanations to Practice Questions

1. CBecause the weight of the barbell (force acting downward) is mg = 275 kg × 10 m/s2,or about 2,750 N, it follows that the weightlifter must exert an equal and oppositeforce of 2,750 N on the barbell. The work done in lifting the barbell is defined asfollows:W = FdW = 2,750 N × 2.4 mW ≈ 2,800 N × 5/2 mW ≈ 7,000 J (one J = one N × m)Using the same equation, it follows that the work done to hold the barbell in place isW = FdW= 2,750 N × 0 JW = 0 JBecause the barbell is held in place and there is no displacement, the work done iszero. (C) best matches our response and is therefore the correct answer. 2. CThe total force on the rope is 5,000 N. However, the force in the direction of motion(the horizontal direction) is only 5,000 N × cos 30° = 5,000 N × 0.866 ≈ 4,500 N.Therefore, the work done by the tractor isW = Fd cos θW = 5,000 × 100 × 0.866W = 4,500 × 100W = 4.5 × 105

This is the same as saying that the work done is 450 kJ, which best matches (C), thecorrect answer. 3. BThe work done by the engine is equal to the change in kinetic energy of the car:W = ΔKE = ½ mvf

2- ½ mvi2

vf = 30 m/svi = 0 m/sW = ΔKE = ½ mvf

2

W = ½ (2,000 kg)(30 m/s)2

W = 900,000 JThe average power supplied is therefore equal toP = W/tP = 900,000 J/6.3 sP ≈ 150,000 WP ≈ 150 kW(B) matches with our result and is thus the correct answer.



4. CConservation of energy states that the total mechanical energy of the block isconstant as it falls (neglecting air resistance). To calculate the total energy at anyheight, it is sufficient to know the total energy at a different height, say 5 m. At themaximum height of 5 m, the block only has potential energy equal to U = mgh = 40kg × 10 m/s2 × 5 m = 2,000 J. This means that if the total initial energy is 2,000 J, thetotal energy at any other time during the object’s fall is also 2,000 J, or 2 kJ. (C) istherefore the correct answer. 5. CTo calculate the center of mass of the person, we have to use the center of massformula:

where m is mass and x is the points along the x-axis (in our case, the length of eachbody part). Plugging in the numbers, we obtain the following:X = [(8 lbs × 10 in) + (20 lbs × 25 in) +(20 lbs × 35 in)]/(8 lbs + 20 lbs + 20 lbs)X = [(80 + 500 + 700) (lbs × in)] /48 lbsX ≈ 1,250 lbs × in/50 lbsX ≈ 25 inches away from the top of the headThis means that the center of mass of this person is around the belly button, rightwhen the thorax ends and the pelvis begins. (B) best matches our result and is thusthe correct answer. 6. BTo calculate the tension force in each rope, first draw a force diagram:

From the force diagram, notice that there are two tension forces pulling the mass up.The net force for this system is equal to Fnet= 2T - mg = ma. From this equation,solve for T:



2T = m (a + g)2T = 10 kg (2 m/s2 + 10 m/s2)2T= 10 kg × 12 m/s2

2T = 120 NT = 60 N(B) best matches the result and is therefore the correct answer. 7. ABecause we are working with an elastic collision, either conservation of momentumor conservation of kinetic energy can be applied here. The momentum and kineticenergy of the stationary ball after the collision equal, respectively, the momentum andkinetic energy of the moving ball before the collision. Because the two balls have thesame mass and the initial and final speeds are the same, 0.5 m/s, the momentum isconserved. Therefore, one of the balls must be stationary after the collision. Itsspeed is thus 0 m/s, making (A) the correct answer. A mathematical analysis revealsthe same result:mvli + mv2i = mvif + mv2f

vli + v2i = vlf + v2f

0.5 m/s + 0 m/s = vlf + 0.5 m/svlf = 0 m/sSo, v1 must be equal to zero, which confirms (A) as the correct answer. 8. BIn all collision problems, it is important to remember that the vector sum of themomentum before the collision is equal to the vector sum of the momentum after thecollision; also, if the collision is elastic, the kinetic energy is conserved. Our problemdeals with an inelastic collision, so it is sufficient for us to look at conservation ofmomentum in order to determine the speed of the truck after the impact:mv(car)i + mv(truck)i = mv(car)f + mv(truck)f

2,500 × 20 + 0 = 0 + 6,000 v(truck)f

v(truck)f = 50,000/6,000v(truck)f ≈ 8 m/s (truck)f(B) best matches our result and is thus the correct answer. 9. BOnly item II is correct. Elastic collisions occur when both momentum and kineticenergy are conserved. When a planet breaks into several pieces, each of them gainskinetic energy; the sum of the kinetic energies, of course, is greater than the kineticenergy of the initial object. When two balls collide and move away from each otherat the same speed, kinetic energy is conserved. All collisions must conservemomentum. The volleyball lost kinetic energy to the collision, so it is partly inelastic. 10. CThe law of conservation of momentum states that both the vertical and horizontalcomponents of momentum for a system must stay constant. If you take the initialmovement of the cart as horizontal and the two balls move in perpendicular



directions to the horizontal, it means that the cart must maintain its horizontalcomponent of velocity. Therefore, (A) and (B) are wrong. If the billiard balls moveas described, then kinetic energy is not conserved; the system gains energy in thisinelastic collision. (C) correctly describes how this scenario is possible. 11. AMary will not bounce higher than Tom. Because Tom and Mary land on the net atthe same time, the net does the same amount of work on both of them. The workdone by the net is equal to ½kx2. For either person, this is converted into kineticenergy (½m1v2 and ½m2v2), which is in turn converted into potential energy as m1ghand m2gh, respectively. The end result is that weight does not affect the height of therebound. If the spring is perfectly efficient at converting its energy into kineticenergy, then they would both rebound back to the point from which they jumped.The change in momentum for Tom is 80 kg × [vi -(-vi))]. , the square of his speedat impact, is 2 × g × h = 2 × 10 × 20 = 400; therefore, his momentum is 80 kg × 2 ×20 = 3,200 kg × m/s. The force upon impact is equal to the change in momentumdivided by the time of contact (which is the same for both individuals). Tom weighsmore, which explains his greater momentum. 12. DTo displace the box a distance x, a length of rope equal to 2x must be pulled throughthe pulley. If a force of 100 N is applied on the rope, the pulley system will magnifythe force on the box by a factor of 2. Therefore, a force of 200 N is exerted on thebox. Now you can set up the equations of motion. The forces acting on the box are200 N up the ramp and mg sin 30° down the ramp. So ma = 200 N - mg sin 30°implies that a = 200 N/20 kg - g sin 30° = 5 m/s2 upward. 13. AThe question does not specify any specific velocities, so assign any number that youlike. If both molecules initially have horizontal components of +5 m/s and verticalcomponents of +3 m/s and -3 m/s, then possible values after collision could behorizontal components of +8 m/s and +2 m/s and vertical components of +10 m/sand—10 m/s. This would conserve the momentum but not the kinetic energies of themolecules. Kinetic energy does not have to be conserved because you are not toldthat this is an elastic collision. While the magnitude of the vertical components mustremain the same for it to add up to 0, the magnitude of the horizontal componentscan vary as long as their sum adds up to +10. This implies that either molecule A orB may have a greater horizontal component and, therefore, greater overall speed andmomentum, so (B) is incorrect. (D) is wrong, because it would violate the law ofconservation of momentum.



4Thermodynamics

In the attempt to help students better understand the laws of thermodynamics, somehave resorted to the following three maxims:

1. The first law of thermodynamics says that you can’t win the game; the best you canhope for is a tie (conservation of energy).

2. The second law of thermodynamics says that a tie is only possible at absolute zero(the entropy of the universe is always increasing except at absolute zero).

3. The third law of thermodynamics says that you’ll never achieve absolute zero (asystem will asymptotically approach an entropy minimum as it asymptoticallyapproaches absolute zero).Now, we know that this is devastating news to some of you, especially to those ofyou who ascribe to a philosophy of optimism. We hate to burst your bubble—although we will, and then blame the “universe” for making us do it—but the fact ofthe matter is that the universe has stacked the cards against us. It’s our fault, really,because we keep insisting on doing things and building things that go against thenatural order of things. Rancid meat in the refrigerator because someone (we won’tname who) didn’t shut the door properly? You’ve only got yourself to blame. Ruston your car’s undercarriage because you thought the dealer’s special protectionpackage was an expensive rip-off? Don’t blame the universe. Find yourselfinconveniently dead? Who’s to blame? You are.



Yes, we are in an epic struggle to overcome the natural tendencies of the universe,and frankly, for as futile as our efforts will ultimately prove to be, we might as wellbe fighting this war from our loungers. What is this fight that we are so valiantly(some would say foolishly) fighting? You mean you don’t know? Look all around,and you’ll see the weapons that we brandish: batteries, refrigerators, heaters, lamps,ice cubes, double-pane windows, clothing, blankets, engines, and stoves, to namebut a few. You look confused. Don’t you see what all these things have in common?In one way or another, everything in this list is designed to help us control energyand exert advantage over it. We create batteries with electrochemical concentrationsthat we use to create current to run our electrical appliances. We design refrigeratorsto move energy from “cold” to “hot.” We make ice cubes to absorb the thermalenergy of our drinks to make them briskly refreshing. We weave clothing andblankets to keep us warm in the winter and cool in the summer. Nevertheless, it’s alosing battle: Batteries run “dead,” refrigerators break down, ice cubes melt, clothingand blankets eventually decay and disintegrate.



Thermodynamics is the study of the flow of energy in the universe, as that flowrelates to work, heat, the different forms of energy, and entropy. Classicalthermodynamics concerns itself only with observations that can be made at themacroscopic level, such as measurements of temperature,pressure , volume, andwork. Statistical mechanics, introduced in the 1870s through the work of Austrianphysicist Ludwig Boltzmann, provides a molecular understanding of and predictiveframework for the macroscopic observations made by classical thermodynamics.Although the MCAT will test the topic from a thermodynamic rather than statisticalunderstanding, we will briefly discuss the statistical understanding of statisticalunderstanding of entropy because it clears up so much of the confusion that hasarisen from the previously common (but now largely abandoned) characterization ofentropy as “disorder.”



This chapter reviews the laws of thermodynamics with a specific focus on thezeroth, first, and second laws. We will examine how the zeroth law leads to theformulation of temperature scales. Thermal expansion will be discussed as anexample of the relationship between thermal energy and physical properties likelength, volume, and conductivity. In the context of the first law, the conservation ofenergy, we will discuss the relationship between internal, energy, heat and work andcharacterize specific heat and of transformation. We will also review the variousprocesses by which a system goes from one equilibrium state to another, such asisobaric, isothermal, and adiabatic processes. Finally, we will discuss the second lawof thermodynamics, the concept of entropy and its measurement. The third law ofthermodynamics is not directly tested on the MCAT, so we will not formally reviewit.



ConclusionThis chapter reviewed the zeroth law of thermodynamics, which reflects theobservation that objects at the same temperature are in thermal equilibrium and thenet heat exchanged between them is zero. We may consider the zeroth law to be expost facto because it provides the thermodynamic explanation for the function ofthermometers and temperature scales, which had been developed many years priorto the law’s formulation. Examination of the first law of thermodynamics revealedthat the energy of a closed system (up to and including the universe) is constant,such that the total internal energy of a system, the sum of all its potential andmotional energies, equals the heat energy gained by the system minus the workenergy done by the system. Finally, we carefully investigated the second law ofthermodynamics and the concept of entropy. We found a better way to understandentropy as a measure not of “disorder” but of the degree to which energy is spreadout through a system, up to and including the universe. We now understand that theconstant energy of the universe is progressively and irreversibly spreading out andwill continue to spread out until there is an even distribution of energy throughout theuniverse. That wasn’t so bad now, was it? We know that few topics contribute more to thecollege physics student’s anxiety levels than thermodynamics. This is trulyunfortunate because like so many other physical principles, the laws ofthermodynamics are based on observations of common phenomena and so aresimply reflective of our everyday life experiences. With the exception of the third lawof thermodynamics, which states that absolute zero can never be actually reached,the laws of thermodynamics are recognizable from the mundane and extraordinaryoccurrences of our lives. There’s nothing to be anxious about; there’s nothing tofear. In fact, the next time you’re nearly hit by a falling turkey-sandwich-eating cat(Chapter 1), you won’t have to be scared at all; you’ll understand that the reasonwhy the cat is falling toward you is that the universe is causing energy to bedisbursed, and you’ll chuckle to yourself and think, “Silly cat!”

CONCEPTS TO REMEMBERThere are four numbered laws of thermodynamics: zeroth, first, second, and third.

These laws are principles that describe the nature and behavior of energy and itsrelationship to heat, work and temperature. Only the first three (zeroth throughsecond) laws are tested on the MCAT.

The zeroth law of thermodynamics states that objects are in thermal equilibrium whenthey are at the same temperature. Objects in thermal equilibrium experience no netexchange of heat energy . Temperature and heat are not the same thing. Temperatureis a qualitative measure of how hot or cold an object is; quantitatively, it is related tothe average kinetic (motional) energy of the particles that make up a substance.Heatis a quantity of energy that is transferred from a hot object to a cold one.

All materials have certain physical properties that change as a result of a change intemperature. Some of these temperature-dependent properties include length,



volume, and conductivity.Thermal expansion is the change in length or volume as afunction of the change in temperature.

The first law of thermodynamics is a statement of energy conservation: The totalenergy in the universe can never decrease or increase. For a closed system, the totalinternal energy is equal to the heat into or out of the system minus the work done byor on the system.

Specific heat is the amount of energy necessary to raise one kilogram of a substanceby one degree Celsius or one unit Kelvin. The specific heat of water is 1,000 cal/kg •°C or, 1 cal/g • °C.

When heat energy is transferred into or out of a system, the system’s temperature willchange in proportion to that amount of energy, as long as no phase change isoccurring. During a phase change, the heat energy is associated with changes in theparticles’ potential energy , not kinetic energy, so there is no change in temperature.

There are three special cases of the first law in which one of the conditions of the gassystem is held constant. For isovolumetric processes, the volume is constant, andno work is done. For adiabatic processes, no heat is exchanged. For closed cycleprocesses, the internal energy (and hence, temperature) is constant.

The second law of thermodynamics states that in a closed system (up to and includingthe entire universe),energy will spontaneously and irreversibly go from beinglocalized to being spread out.

Every real process is ultimately irreversible; under highly controlled conditions, certainequilibrium processes, such as phase changes, can be treated as reversible.

Entropy is a measure of how much energy has spread out or how spread out energyhas become.

EQUATIONS TO REMEMBERTC = TK-273ΔL = αLΔT ΔV= βVΔT ΔU = Q-W Q = mcΔT Q = mL W = PΔV

Practice Questions1. If an object with an initial temperature of 300 K increases its temperatures by 1 K

every minute, by how many degrees Celsius will its temperature have increased in 10minutes?

A. 10°CB. 27°CC. 100°CD. 270°C

2. Which of the following choices correctly identifies the following three heat transferprocesses?



1.Heat transferred from the sun to the earth2. A metal spoon heating up when placed in a pot of hot soup3. A rising plume of smoke from a fireA. 1.Radiation ; 2. Conduction; 3.B. 1. Conduction; 2.Radiation ; 3.C. 1. ; 2. ; 3. ConductionD. 1. ; 2. Conduction; 3.

3.Radiation A 20 m steel rod at 10°C is dangling from the edge of a building and is 2.5cm from the ground. If the rod is heated to 90°C, will it touch the ground? (α = 1.1× 10-5 K-1)

A. Yes, because it expands by 3.2 cm.B. Yes, because it expands by 1.6 cm.C. No, because it only expands by 1.6 cm.D. No, because it only expands by 3.2 cm.

4. What is the final temperature of a 5 kg silver pendant that is left in front of an electricheater that absorbs heat energy at a rate of 100 W for 10 minutes? Assume thependant is initially at 20°C and that the specific heat of silver is c = 236 J/(kg × °C)= 0.0564 cal/(g x°C).

A. 29°CB. 59°CC. 71°CD. 100°C

5. How much heat is required to melt completely 500 g of gold earrings, given that theirinitial temperature is 25°C? The melting point of gold is 1064.18°C, the heat offusion is 6.45 × 104 J/kg, and the specific heat is 129 J/(kg × °C).

A. 15 kJB. 32 kJC. 65 kJD. 99.27 kJ

6. Given the cycle shown, what is the total done by the gas in the cycle?

A. -10 JB. 0 JC. 7.5 JD. 17.5 J

7. In an adiabatic compression process, the internal energy of the gasA. increases, because the work done on the gas is negative.



B. increases, because the work done on the gas is positive.C. decreases, because the work done on the gas is negative.D. decreases, because the work done on the gas is positive.

8. The entropy of a system canA. never decrease.B. decrease when the entropy of the surroundings increases by at least as much.C. decrease when the system is isolated and the process is irreversible.D. decrease during an adiabatic reversible process.

9. The internal energy of an object increases adiabatic in an process. Which of thefollowing must be true regarding this process?

A. The kinetic energy of the system is changing.B. The potential energy of the system is changing.C. is Work done on the system.D.Heat flows into the system.

10. A certain substance has a specific heat of 1 J/(mol•K) and a melting point of 350 K.If one mole of the substance is currently at a temperature of 349 K, how muchenergy must be added in order to melt it?

A. More than 1 JB. Exactly 1 JC. Less than 1 J but more than 0 JD. Less than 0 J

11. The following graphs depict the change in pressure and volume of a gas. Whichgraph most likely represents a process in which work is done by the gas as theprocess moves from point A to point B?



12. The figure shown depicts a thick metal container with two compartments.Compartment A is full of a hot gas, while compartment B is full of a cold gas. Whatis the primary mode of heat transfer in this system?



A. RadiationB. ConvectionC. ConductionD. Enthalpy

13. Substances A and B have the same freezing and boiling points. If solid samples ofboth substances are heated in the exact same way, substance A boils beforesubstance B. Which of the following would NOT explain this phenomenon?

A. Substance B has a higher specific heat .B. Substance B has a higher of vaporization.C. Substance B has a higher heat of fusion.D. Substance B has a higher internal energy .

14. In experiment A, a student mixes ink with water and notices that the two liquids mixevenly. In experiment B, the student mixes oil with water; in this case, the liquidsseparate into two different layers. The entropy change is

A. positive in experiment A and negative in experiment B.B. positive in experiment A and zero in experiment B.C. negative in experiment A and positive in experiment B.D. zero in experiment A and negative in experiment B.

15. Which of the following processes is LEAST likely to be accompanied by a change intemperature?

A. The kinetic energy of a gas is increased through a chemical reaction.B. Energy is transferred to a solid via electromagnetic waves.C. A boiling liquid is heated on a hot plate.D. A warm gas is mixed with a cold gas.

Small Group Questions1. Using the conservation of energy, explain why the temperature of a gas increases when

it is quickly compressed, whereas the temperature decreases when the gas expands.2. Suppose the latent heat of vaporization for water was one-half its actual value. All

other factors being equal, would water in a kettle take the same time, a shorter time,or longer time to boil? How would the evaporative mechanism of the human body beaffected?

3. As living organisms grow, they convert relatively simple molecules into more complexones. Is this a violation of the second law of thermodynamics?

Explanations to Practice Questions 1. AThe Kelvin degree and Celsius degree are the same size; that is, a change of 1 K isequal to a change of 1°C. In other words, if the object increases in temperature by 1K every minute, it means that in 10 minutes, its temperature will change by 10 K,which is the same as 10°C. Notice that the actual temperature of the object in



degrees Kelvin and Celsius is different, but the change in the temperature is the same.(A) is therefore the correct answer. 2. ABecause there is essentially only empty space between the sun and the earth, the onlymeans of heat transfer is by radiation (i.e., electromagnetic waves that propagatefrom the sun to the earth). Therefore, the first process is radiation , allowing us tocross out (B) and (D). When a metal spoon is placed in a pot of hot soup, themolecules in the soup collide with those on the surface of the spoon, therebytransferring heat by conduction. (A) must be the correct answer, but let’s check thatthe last process corresponds indeed to convection. Fire warms the air above it, andthe warmed air is less dense than the surrounding air, so it rises. A rising column ofwarm air means that heat is being transported in the air mass, which is simply theprocess of convection. The smoke particles ride along with the upward moving airmass and create a plume of smoke. (A) is indeed the correct answer. 3. CFirst, find the change in length due to thermal expansion :ΔL = αLΔTΔL = 1.1 × 10-5 K-1 × 20 m × 80 KΔL ≈ 10-5 K-1 × 20 m × 80 KΔL ≈ 0.016 mΔL ≈ 1.6 cmBecause the rod is originally 2.5 cm above the ground and it increases by 1.6 cm,you can conclude that it will not touch the ground after the thermal expansionprocess is completed. (C) correctly reflects this answer. 4. CTo answer this question, first remember that watts are equal to joules per second; inother words,power is energy over time. In 10 minutes, the pendant absorbs thefollowing amount of energy:energy = power := x timeE = 100 W × 10 min × 60 sec/minE = 6 × 104 JNow that you know the heat added, Q = 6 × 104 J, you can find the final temperaturefrom this equation:Q = mcΔT= mc(Tf-Ti)6 × 104 = 5 × 236 (Tf-20)Tf≈ 70 °C(C) matches your result and is thus the correct answer. 5. DTo begin with, first determine how much heat is required to raise the temperature ofthe gold earrings all the way to the melting point of gold. Then, calculate the heatrequired to actually melt the earrings (the latent heat ). The total heat required to meltthe earrings completely will be the sum of the two heats. The heat required to raise



the temperature of the earrings from 25°C to 1,064.18°C isQ = mcΔTQ = mc(Tf-Ti)Q ≈ 0.5 kg × 130 J/(kg × °C) × (1,000°C-25°C)Q ≈ 65 J/°C × 1,000°CQ = 6.5 × 104 JThe result tells you to add about 65 kJ worth of heat to the earrings just to bringthem to their melting point. However, at this point, the earrings are still solid. Thenext step is thus to calculate how much heat is needed to melt the earrings. For this,use the heat of fusion (the latent heat) of gold:Q = mLQ = 0.5 kg • 6.45 × 104 J/kgQ ≈ 3.2 × 104 JQ ≈ 32 kJSo overall, it requires 65 kJ + 32 kJ = 97 kJ of heat to melt the gold earrings. (D) istherefore the correct answer. Notice how heavily approximated the numbers arebecause the answer choices are so spread out. 6. CThe total work done by the cycle is the sum of the work of paths A, B, and C.Knowing that the work done is equal to the pressure times the change in volume, orthe area under the curve, you can calculate WA, WB, and Wc and then add themtogether. WA is the area under the curve, which can be broken down into a triangle ontop of a rectangle.WA = (½)(5) × 3 + 2 × 5WA = 17.5 JWB can be calculated in the same way:WB= PΔVWB = 2 × (-5)WB =-10 JNext, we know that Wc is equal to zero because the volume is constant. The totalwork done therefore is the sum of WA + WB + Wc = 17.5 J + (-10 J) + 0 = +7.5 J,which matches with choice (C), the correct answer. 7. ATo answer the question, make sure you understand all the terms. An adiabaticprocess means that there is no exchange of heat; in other words, Q = 0. When a gasis compressed, positive work is being done on the gas, so the value for work doneby the gas will be negative, W < 0. Based on this, we can determine how the internalenergy of the gas changes by using the first law of thermodynamics:ΔU = Q-WΔU =-WBecause W is negative, ΔU will end up being a positive value. Since ΔU = Uf-Ui andΔU > 0, we can imply that the internal energy of the gas will increase, which matcheswith correct answer (A).



8. BThe entropy of a system (not isolated) can decrease as long as the entropy of itssurroundings increases by at least as much. On the other hand, the entropy of anisolated system increases for all real (irreversible) processes. Moreover, the entropyof a reversible isothermal process can be calculated using the expression ΔS = Q/T,so in an adiabatic process, the entropy remains constant. Based on this, (B) is thecorrect answer. 9. CIn an adiabatic process, no heat enters or leaves the system, so (D) is out right away.The internal energy will increase if either kinetic or potential energy is increased, so(A) and (B) are incorrect. The first law of thermodynamics tells us that ΔU = Q-W(where ΔU is change in internal energy, Q is heat transferred into the system, and Wis work done by the system); because the process is adiabatic, simplify this into ΔU=-W. Because ΔU is positive, we know that W must be negative, which means thatwork is done on the system. This question can also be answered with simple logic.Because the total energy of a system must always be constant, the only way toincrease the energy of an adiabatic system is to make the system do a negativeamount of work; in other words, work must be done on the system. 10. ATo find the amount of heat needed to bring the substance to its melting point, youcan use the specific heat: the product of the specific heat, the amount of thesubstance, and the change in temperature. This means that the heat is equal to (1J/[mol•K])(1 mol)(1 K) = 1 J. After the substance reaches its melting point,additional heat (which can be determined by the formula Q = mΔHf) is needed toactually induce the phase change. Therefore, the total amount of heat required isgreater than 1 J. 11. AWhen a gas is expanding or contracting at constant, pressure you can determine thework done in the process by finding the area under the pressure-volume curve byusing the formula W = PΔV (where P is pressure and ΔV is change in volume). In(A), ΔV is positive, meaning that the amount of work must also be positive and thatwork was done by the system. In (D), the amount of work is negative, so work wasdone on the system. In (B) and (C), ΔV is equal to zero. You can also answer thisquestion with simple logic; if the volume of a gas is decreasing while the pressurestays constant, something must be doing work on the gas in order to overcome thepressure , which will push in the opposite direction to prevent the volume fromchanging. If the volume is increasing while the pressure stays constant, the gas mustbe doing work on the environment for the same reason. 12. CIn this situation, heat will transfer from the warm gas to the metal and then to thecold gas. (B), convection , is the transfer of heat when two substances are mixedtogether; this cannot happen here because the gas will not naturally mix with the



metal. (A),heat transfer through radiation, is also implausible, not only because gasesare unlikely to emit heat in the form of waves but also because the radiation would beunlikely to penetrate the thick metal container. Enthalpy, (D), is not a form of heattransfer. Conduction, (C), is the most likely option; it happens when two substancesmake direct contact with one another. Here, gas A makes contact with the leadcontainer, which makes contact with gas B. 13. DSaying that substance B has a higher internal energy, (D), cannot explain thephenomenon because the internal energy is irrelevant; the heat involved in theprocess is related only to the specific heat, the heat of fusion , and the heat ofvaporization. All of the other choices could explain the phenomenon. The heatrequired to melt the solid is determined by the heat of fusion, (C). The heat requiredto bring the liquid to its boiling point is determined by the specific heat, (A). Theheat required to boil the liquid is determined by the heat of vaporization, (B). 14. BWhen the ink randomly intersperses throughout the water, the final state is moredisordered than the initial state, so the entropy change of the system is positive.When the oil separates from the water, the final state is just as ordered as the initialstate (because the oil and the water are still completely separate), so theentropychange is zero. You can also answer this question by noticing thereversibility of the two experiments. Experiment A has a positive entropy changebecause it is irreversible, while experiment B has no entropy change because thereaction is reversible. According to the second law of thermodynamics, the entropychange can never be negative in a thermodynamic process that moves from oneequilibrium state to another. 15. CIf a substance is undergoing a phase change, any added heat will be used towardovercoming the heat of transformation of the phase change. During the phasechange, the temperature will remain constant. Temperature is a measure of the kineticenergy of the molecules in a sample, so a change in kinetic energy , (A), is essentiallythe same thing as a change in temperature. The heat transfer by radiation described in(B) will definitely change the temperature of the solid as long as it is not in theprocess of melting. (D) describes heat transfer by convection, in which the warm gaswill transfer heat to the cold gas until they both reach a moderate temperature.



5Fluides and Solids

Hidden beneath the waves of the Mediterranean Sea at depths of more than 4,000meters lie three lakes. These lakes are not enclosed within the spaces of ancient rockformations. They are not hidden beneath the crust of the ocean floor. These lakeshave shore lines, “swash zones,” and beach ridges, and when deep sea exploratoryvessels sit on their surface, they float and cause pressure waves to emanate outward.How is it possible that these lakes of water sit on the ocean floor, creating a surrealbeach scene that would seem so alien and yet so familiar to anyone who has everwalked along the sandy shores of a lake or ocean?



Such lakes are unlike anything that we encounter in “our world” of dry land. Theyare without oxygen, and the surface pressure is 400 times that at sea level. Thesesuboceanic aquatic pools are actually brine lakes or brine pools. The water that fillsthese brine lakes is five to ten times saltier than the seawater that sits above it. In fact,the brine is nearly at saturation and is so dense that it does not mix with the seawaterthat (we can now say) floats above it. Brine pools, which are found also in the Gulfof Mexico and elsewhere along the deep floors of the world’s oceans, developedthousands of years ago when vast salt deposits from ancient oceans were exposedto seawater from rising oceans. The salt leached from the ocean floor and formedsupersaline layers of water that became the brine pools. Oceanographers and marinebiologists are just beginning the exciting work of investigating these alienenvironments to learn how life forms have adapted and thrived in one of the mostextraordinary microcosms ever discovered on earth.



Suboceanic lakes and rivers that ripple and flow as if disturbed only by a gentlebreeze present a particularly fascinating opportunity to observe the physics of fluidsand solids. This chapter covers the important concepts and principles of fluidmechanics and solids as they are tested on the MCAT. We will begin with a reviewof some important terms and measurements, including density and pressure. Ournext topic will be hydrostatics, that branch of fluid mechanics that characterizes thebehavior of fluids at rest in equilibrium. Dipping into hydrostatics, we’ll take a dip,so to speak, in the Dead Sea to figure out why inflatable arm “swimmies” areunnecessary for staying afloat. We’ll turn our attention, then, to flowing fluids toreview the concepts and calculations of hydrodynamics. We will demystify Bernoulli’s equation, figure out the whoopee cushion, and prepare ourselves to address whythe plastic sheeting over the space where the rear passenger-side window should bein your 1986 wood-paneled station wagon bubbles out when you drive and why themoldy cheap plastic curtain keeps getting sucked up against your legs when you’rein the shower. Finally, the chapter ends with a brief discussion of solids, examiningthe ways in which we can describe and quantify a solid object’s response toapplications of pressure.



ConclusionThe behavior of fluids and solids impacts every moment of our lives. Even if we arenowhere near an ocean or a lake, we are quite literally submerged in a vast expanseof fluid, a mix of gases known as the atmosphere, which exerts forces all over thesurfaces of our bodies. Whether we are taking a bath or dunking golden crowns intotubs of water, we experience the effect of buoyant forces exerted by the displacedfluid. Whether we are watering our gardens, taking a shower, or riding in ahovercraft, we experience the velocity, forces, and pressures of fluid on the move.In the world of medicine, one must consider fluids, flowing and at rest, whenevaluating the function of the respiratory and circulatory systems: Conditions asvaried as asthma and heart murmurs are related to the way in which the body causesfluids to flow. The balance of hydrostatic and oncotic pressures is important formaintaining the proper balance of fluid in the peripheral tissues of the body. Now that you have the basic concepts of hydrostatics and hydrodynamics, learn toapply them to MCAT passages and questions through your Kaplan practicematerials. Don’t be intimidated by the seeming complexity of buoyant forceproblems and applications of Bernoulli’s equation. Remember that all fluids, whetherliquid or gas, exert buoyant forces against objects that are placed in them as afunction of the mass of the fluid displaced. Remember that incompressible fluidsdemonstrate an inverse relationship between their dynamic pressure (as a function ofvelocity) and their static pressure. Congratulations on completing another chapter!Be sure to review the concepts contained here in this chapter before moving on tothe next.

CONCEPTS TO REMEMBERFluids are substances that have the ability to flow and conform to the shape of their

container. They can exert perpendicular forces, but they cannot withstand shearforces. The category of fluids includes both liquids and gases. On the MCAT,liquids are assumed to be incompressible and ideal conservative systems.

Solids are substances that do not flow and are sufficiently rigid to maintain their shapeindependent of any container. They can exert perpendicular forces and canwithstand shear forces.

The density of any fluid or solid is defined as the mass per unit volume. For aconstant mass, there is an inverse relationship between volume and density. Thus,the density of an object that experiences thermal expansion decreases as its volumeincreases.

Pressure is defined as a measure of the force per unit area. Pressure is a quantity; ithas no direction. The pressure exerted by a gas against the walls of its container willalways be perpendicular (normal) to the container walls.

Gauge pressure is defined as the pressure above atmospheric pressure due to theweigh of the fluid sitting above the point of measurement. Pressure can be measuredeither as an absolute pressure, which is the sum of the pressure at the surface plusthe gauge pressure, or simply as the gauge pressure.



Pascal’s principle states that an applied pressure to an incompressible fluid will bedistributed undiminished throughout the entire volume of the fluid system. Thisprinciple is the basis of hydraulic machines.

Archimedes’ principle states that the volume of fluid displaced by an object placed init will generate a buoyant force against the object that is equal to the weight of thefluid displaced.

On the MCAT, it is assumed that incompressible fluids will flow with conservation ofenergy as a result of the fluid’s very low viscosity and laminar flow. The continuityexpression is a statement of conservation of mass.

Bernoulli’s equation is an expression of energy conservation of for a flowing fluid:The sum of the static pressure and the dynamic pressure will be constant betweenany two points in a closed system. For horizontal flow, there is an inverserelationship between and pressure and velocity: As pressure decreases, velocity willincrease.

The elasticity of solid materials can be measured by any one of three moduli: Young’s, shear, and bulk. All three measure stress in the same way, as pressure (F/A). Thethree moduli differ in the way they measure strain.


P = Po +

Pg= P - Patm = (Po + ) - Patm

v1A1 = v2A2 = a constant

Practice Questions1. Objects A and B are submerged at a depth of 1 m in a liquid with a specific gravity of

0.879. Given that the density of object B is one-third that of object A and that thegauge pressure of object A is 3 atm, what is the gauge pressure of object B? (Usethe atmospheric pressure as 1 atm and g = 9.8 m/s2.)



A. 1 atmB. 2 atmC. 3 atmD. 9 atm

2. An anchor made of iron weighs 833 N on the deck of a ship. If the anchor is nowsuspended in seawater by a massless chain, what is the tension in the chain? (Thedensity of iron is 7,800 kg/m3, and the density of seawater is 1,024 kg/m3.)

A. 100 NB. 724 NC. 833 ND. 957 N

Two wooden balls of equal volume but differentdensity are held beneath the surface of a container of water. Ball A has a density of0.5 g/cm3, and ball B has a density of 0.7 g/cm3. When the balls are released, theywill accelerate upward to the surface. What is the relationship between theacceleration of ball A and that of ball B? (ρwater = 1 g/cm3.)

A. Ball A has the greater acceleration.B. Ball B has the greater acceleration.C. Balls A and B have the same acceleration.D. Cannot be determined from information given.4. Water flows from a pipe of diameter 0.15 m into one of diameter 0.2 m. If the speed

in the 0.15 m pipe is 4.88 m/s, what is the speed in the 0.2 m pipe?A. 1.3 m/sB. 2.7 m/sC. 3.66 m/sD. 6.5 m/s

5. A hydraulic lever is used to lift a heavy hospital bed, requiring an amount of work Wo.When the same bed with a patient is lifted, the work required is double. How can thecross-sectional area of the platform on which the bed is lifted be changed so that thepressure on the hydraulic lever remains constant?

A. The cross-sectional area must be doubled.B. The cross-sectional area must be halved.C. The cross-sectional area does not have to be changed.D. The cross-sectional area must be decreased.

6. The figure shown represents a section through a horizontal pipe of varying diametersinto which open four vertical pipes. If water is to be allowed to flow through the pipein the direction indicated, in which of the vertical pipes will the water level be lowest?

A. Pipe 1B. Pipe 2C. Pipe 3



D. Pipe 47. The speed of blood in the aorta is much higher than the speed of blood through the

capillary bed. How can this fact be explained using the continuity equation?A. The aorta is located higher than the capillary bed.B. The pressure in the aorta is the same as the pressure in the capillary bed.C. The cross-sectional area of all the capillaries added together is much greater than

the cross-sectional area of the aorta.D. The cross-sectional area of all the capillaries added together is much smaller than

the cross-sectional area of the aorta.8. A student wants to measure the shear modulus of Jell-O. When she applies a certain

amount of stress to a piece of rubber, she measures a strain of 3 (Young’s modulusfor rubber = 0.1 Pa). When she applies the same stress to a piece of Jell-O, shemeasures a strain of 6. What is the shear modulus of Jell-O?

A. 0.05 PaB. 0.1 PaC. 0.3 PaD. 0.5 Pa

9. A large cylinder is filled with an equal volume of two immiscible fluids. A balloon issubmerged in the first fluid; the gauge pressure in the balloon at the deepest point inthe first fluid is found to be 3 atm. Next, the balloon is lowered all the way to thebottom of the cylinder, and as it is submerged in the second fluid, the hydrostaticpressure in the balloon reads 8 atm. What is the ratio of the gauge pressure at thebottom of the first fluid to the gauge pressure at the bottom of the second fluid?(Atmospheric pressure is Patm = 1 atm.)

A. 1:3B. 3:4C. 3:5D. 3:8

10. An oddly shaped water-filled sculpture is designed to allow water levels to changedepending on a force applied at the top of the tank as shown. If a force, F1, of 4 N isapplied to a square, flexible cover where A1 = 16 and the area A2 = 64, what forcemust be applied to A2 to keep the water levels from changing?

A. 4 NB. 16 NC. 32 ND. No force needs to be applied.

11. Balls A and B of equal mass are floating in a swimming pool, as in the figure shown.Which will produce a greater buoyant force?



A. Ball AB. Ball BC. The forces will be equal.D. It is impossible to know without knowing the volume of each ball.

12. Bernoulli’s principle is the reason for the upward force that permits a lift force tocause air flight. What statement best summarizes the principle’s relationship to flight?

A. The speed of airflow is equal on the top and bottom of a wing, resulting innonturbulent flight.

B. The speed of airflow is greater over the curved top of the wing, resulting in lesspressure on the top of the wing and the production of a net upward force on thewing, in turn resulting in flight.

C. The speed of airflow on the flat bottom of the wing is greater than over the curvedtop of the wing, resulting in less pressure below the wing and the production of a netupward force on the wing, in turn resulting in flight.

D. The weight of the wing is directly proportional to the weight of the air it displaces.13. A low-pressure weather system can decrease the atmospheric pressure from 1 atm to

0.99 atm. By what percent will this decrease the force bearing on a rectangularwindow, 6 m by 3 m? The glass is 3 cm thick.

A. 1%B. 10%C. 1/3%D. 3%

14. Two fluids, A and B, have densities of x and 2x, respectively. They are testedindependently to assess absolute pressure at varying depths (see table). At whatdepths will the pressure below the surface of these two fluids be equal?

A. Whenever the depth of fluid A is 4 times that of fluid BB. Whenever the depth of fluid A equals that of fluid BC. Whenever the depth of fluid A is 2 times that of fluid BD. They will never be equal.

15. A water tower operator is interested in increasing the pressure of a column of waterthat is applied to a piston. She hopes that increasing the pressure will increase the



force being applied to the piston. The only way to increase the pressure is to alterthe speed of the water as it flows through the pipe previous to the piston. Howshould the velocity of the water change to increase the pressure and force?

A. Increase the speed.B. Decrease the speed.C. Release water intermittently against the pipe.D. The speed of water will not change pressure at the piston.

Small Group Questions1. Suppose you insert a straw into a tall glass of milk. You place your finger over the top

of the straw, capturing a small amount of air above the milk but preventing anyadditional air from entering or leaving, and then you remove the straw from the milk.The straw retains most of the milk. Does the air in the space between your finger andthe top of the milk have a pressure P equal to, less than, or more than theatmospheric pressure outside the straw?

2. Roofs of houses are sometimes blown off during a tornado or hurricane. UsingBernoulli’s principle, explain how.

Explanations to Practice Questions 1. CThe absolute and gauge pressures depend only on the density of the fluid, not that ofthe object.Pgauge = ρ liq ghPtotal = Patm + Pgauge

Because the two objects are under the same atmospheric pressure, at the same depthand in the same fluid , the pressure on object A will be the same as the pressure onobject B, making (C) the correct answer. 2. BThe tension in the chain is the difference between the anchor’s weight and thebuoyant force:T = W-FB

where T is the tension, W is the anchor’s weight, and FB is the buoyant force on theanchor. The object’s weight is 833 N, and the buoyant force may now be foundusing Archimedes’ principle. The buoyant force is equal to the weight of theseawater that the anchor displaces:FB = ρwVwgwhere pw is the density of water, Vw is the volume of water displaced, and g is theacceleration due to gravity. Because the anchor is submerged entirely, the volume ofthe water displaced is equal to the volume of the anchor, Vw = VA. To find thevolume of the anchor, use the following equation:VA= mA/ρa

Since WA = mg,VA= WA/(gρA)VA = 833 N/(10 m/s2 × 7,800 kg/m3)VA = 1.0 × 10-2 m3



This value can be then used to find the buoyant force:FB= FB ≈ 1,000 kg/m3 × 1.0 × 10-2 m3 × 10 m/s2

FB ≈ 100 NLastly, we can obtain the tension from the initial equation T = W-FB:T = 833 N-100 NT = 733 N(B) best matches our result and is thus the correct answer. 3. AUsing Newton’s second law, Fnet = ma, we obtain the following equation:Fbuoyant-mg = maThus,a = (Fbuoyant-mg)/ma = Fbuoyant/m-gBoth balls experience the same buoyant force because they both have the samevolume (Fbuoyant = ρVg). Thus, the ball with the smaller mass experiences the greateracceleration. In other words, since both balls have the same volume, the ball with the

smaller density has the smaller mass which is ball A. (A) is therefore thecorrect answer. 4. BIt is known that water flows faster through a narrower pipe. The speed is inverselyproportional to the cross-sectional area of the pipe, because the same volume ofwater must pass by each point at each time interval. Let A be the 0.15 m pipe and Bthe 0.20 m pipe, and write the following expression:

vAAA = vBAB

where v is the speed and A is the cross-sectional area of the pipe. Because v isinversely proportional to the cross-sectional area, and the area is proportional to thesquare of the diameter A = πr2 = πd2/4, we obtain the following:VB = VA(πdA

2/4)/(πdB2/4)

VB = VA × dA2/dB

2

VB = 4.88 × 0.152/0.202

VB ≈ 5 × 0.0225/0.04VB ≈ 5 × 2.25 × 10-2/(4 × 10-2)VB ≈ 2.5 m/s(B) most closely matches our result and is thus the correct answer. 5. AThis question tests our understanding of Pascal’s principle, which states that achange in pressure applied to an enclosed fluid is transmitted undiminished to everyportion of the fluid and to the walls of the containing vessel. Given the hydrauliclever, several equations can be written:ΔP = F1/A1 = F2/A2

V = A1d1 = A2d2



W = F1d1 = F2d2

We are told that the work required to lift the bed with the patient is double the workneeded to lift just the bed. In other words, the force required doubles when both thebed and the patient have to be lifted. To maintain the same pressure, double thecross-sectional area of the platform of the hydraulic lever on which the patient andthe bed are lifted. (A) matches our explanation and is thus the correct answer. Noticethat because (B) and (D) state almost the same thing; thus, neither could have beenthe correct answer. 6. BIt is not necessary to do any calculations to answer this question. The open verticalpipes are exposed to the same pressure (atmospheric) therefore, differences in theheights of the columns of water in the vertical pipes is dependent only on thedifferences in hydrostatic pressures in the horizontal pipe. Since the horizontal pipehas variable cross-sectional area, where the horizontal pipe is narrowest, water willflow the fastest and the hydrostatic pressure will have its lowest value (as seen inBernoulli’s Equation). As a result, pipe #2 will have the lowest water level, making(B) the correct answer. 7. CThe continuity equation states that the mass flow rate of a fluid must remain constantfrom one cross section to another. In other words, when an ideal fluid flows from apipe with a large cross-sectional area to one that is narrower, its speed decreases.This can be illustrated through the equation A1v1 = A2v2. In other words, the larger thecross-sectional area, the slower the fluid flows. If blood flows much more slowlythrough the capillaries, we can infer that the cross-sectional area is larger. This mightseem surprising at a fi rst glance, but given that each blood vessel divides intothousands of little capillaries, it is not hard to imagine that adding the cross-sectionalareas of each capillary from an entire capillary bed results in an area that is largerthan the cross-sectional area of the aorta. (C) matches this explanation and is thusthe correct answer. 8. ATo answer this question, first define Young’s and the shear moduli:

where F is the force applied, A is the cross-sectional area, L is the length, x is thehorizontal change in length, and h is the height. What these two equations have incommon is the idea that both Young’s and the shear modulus are a measure ofstress divided by strain. When the student measures a strain of 3 on the piece ofrubber with a Young’s modulus of 0.1 Pa, she applies a stress ofstress = Y × strainstress = 0.1 × 3stress = 0.3 N/m2



Because the student applies the same amount of stress to the Jell-O and measuresthe strain, determine the shear modulus through the following equation:S = stress/strainS = 0.3/6S = 0.1/2S = 0.05 Pa(A) matches our result and is thus the correct answer. 9. BThe first step in answering this question is defining the different types of pressures.Atmospheric pressure is the pressure at the top of the very first fluid , the pressuregiven by air (at sea level, it is equal to 1 atm). Gauge pressure is the pressure insidethe balloon, which excludes the atmospheric pressure; gauge pressure is the total(absolute or hydrostatic) pressure inside the balloon minus the atmospheric pressure.Gauge pressure depends on the depth at which the object is submerged in the fluid ,the constant of gravity, and the density of the fluid. Hydrostatic or absolute pressureis the total pressure in the balloon (i.e., the gauge pressure and the atmosphericpressure together). Because we are given the gauge pressure at the bottom of thefirst fluid as 3 atm, our task now is to calculate the gauge pressure at the bottom ofthe second fluid. The hydrostatic pressure at the bottom of the cylinder is 8 atm.One of these atmospheres is atmospheric pressure pushing on the fluids. The other 3atmospheres are accounted for by the first fluid that is pushing on the second fluid.Thus, the gauge pressure at the bottom of the second fluid is 8 - 1 - 3 = 4 atm. The ratio of the gauge pressures is therefore 3:4, making (B) the correct answer. 10. BThis is a basic restatement of Pascal’s principle that a force applied to an area willbe transmitted through a fluid. This will result in changing fluid levels through thesystem. The relationship is stated as F2 = F1A2/A1. Plugging in the numbers clearlyshows the answer is 16 N, (B). 11. DThe mass of these two objects is the same, but their volumes are different. As such,their densities will be different; ball B is much denser than ball A. The force ofbuoyancy will be equal to the weight of the water displaced by the object. Theweight of the displaced water is equal to the density of the ball multiplied by thevolume of the ball and the acceleration due to gravity. You know that B is moredense, but you do not know how much greater the volume of A is than B. It ispossible these two terms could cancel each other out. As such, (D) is the correctanswer. 12. BBernoulli’s principle states that airflow over a curved surface will be faster thanairflow over a flat surface. This is why wings have curved upper surfaces and flatlower surfaces. The fluid , in this case air, will move more quickly over the curvedsurface because it has further to travel to the end of the wing than the air at the flat



bottom surface. Increased air speed will mean lower pressure within the fluid. Thiswill result in higher pressure below the wing and an upward force. (D) is an incorrectrestatement of the principle of buoyancy that applies to fluids. 13. AThis question is a simple application of the following formula: pressure = force/area.The area is 18 square meters per window, and there are two windows. The thicknessof the glass has no bearing on the pressure. If pressure decreases 1 percent and areadoes not change, and pressure is directly proportional to force, the force will bedecreased 1 percent if the atmospheric pressure decreases 1 percent.(B) is included to throw you off on incorrect decimal placement. (C) and (D) areincluded to tempt you into considering the thickness of the glass, but it will notimpact the pressure calculation. 14. CThe chart is distracting. If you know the equation for absolute pressure, this will befairly easy to figure out. P = Po + where = density. Po, pressure at the surface,is always assumed to be atmospheric pressure, or approximately 105 Pa. Theacceleration due to gravity, g, will be approximately 10 m/s2, and p will be x or 2x.As such, the only variable to consider is h. The pressures below the surface will beequal where the depth of fluid A is two times that of fluid B. In this case, the 2 timesgreater depth will cancel out the 2 contributed from the 2x of greater density. Thiscan be illustrated algebraically:

ρAhA = ρBhB

xhA = 2xhB

hA = 2hB

15. BThis is a basic interpretation of Bernoulli’s equation that states, at equal heights,velocity and pressure of a fluid are inversely related. Decreasing the speed of thewater, (B), will increase its pressure. An increase in pressure over a given area willresult in increased force being transmitted to the piston. Releasing waterintermittently against the pipe, (D), may produce greater force at that instant, but thisis unsustainable, and you have no information in the answer choice as to the speedof the water to be released.



6Electrostatics

It’s early on a cold winter morning. You’ve already exceeded the legal limit forhitting the snooze button. Your eyes flicker open just long enough to catch the graystreaming in through your window. The air is cold and dry; it seems as if all themoisture in the room has crystallized into the frost that covers your window likemoss. Knowing that the inevitable can no longer be delayed, you ooze out of bed—your bones and muscles still sleeping. The nerve connections between your legs andyour brain seem to have been snapped as if in an ice storm. The only indication youhave that they are actually working is your perception that your bed is receding fromyour view faster than your hope to be able to return to it. Shuffling, shuffling towardyour door, you wonder if someone has replaced the carpet with flypaper. Finally,after completing what seemed to be a journey of religious obligation, you reach thedoor. Reluctantly yet instinctually, your hand levitates and floats toward the handle,your fingers curling in resigned anticipation of their forced labor. Closer, closer tothe golden orb they move. Suddenly, a miniature flash of light moves in the spacethat separates your fingers from the doorknob. And then, a pain—not searing like aburn, not throbbing like a cut, but undeniably, unmistakably the intenselyuncomfortable tingling of an electrical shock. It’s going to be a great day! youthink, as you turn around and march—careful to pick up your feet this time—backinto bed.



Electrostatics is the study of stationary charges and the forces that are created byand act upon these charges. Few things bring as much pleasure—and as much pain—to human existence as electrical charge. Without it, we would not be able to domany of the activities that we enjoy or even now consider essential to basic living.But living with electrical charge can also be dangerous and even deadly: Magnify thesmall shock you just suffered between the doorknob and your hand, and you have alightning bolt strong enough to kill you. In this chapter, we will review the basicconcepts essential to understanding charges and electrostatic forces. We will reviewCoulomb’s law to discover that this expression of force is remarkably similar to theequation for gravitational forces. Next, we will describe the electric field that allcharges set up around them, which allows them to exert forces on other charges inmuch the same way that fans can be set either to blow air away from them or suckair toward them. After we’ve discussed how charges set up these fields of “potentialforces,” we’ll observe the behavior of charges that are placed into these fields. Inparticular, we will note the motional behavior of these “test” charges inside a field inrelation to the electric potential difference, or voltage, between two points in space.And in relation to the movement of a test charge, we will observe the change inelectrical potential energy as the charge moves from a position of some electricpotential to another. Lastly, we will describe the electric dipole and solve a probleminvolving one of the most important molecular dipoles known to us: the H2Omolecule.



ConclusionFrom electrical shocks to lightning bolts, from batteries to DVD players, our worldis filled with charge. We may experience charge as pleasant or deadly, but we mustalways be mindful that it is yet another mode through which forces can be generatedand energy be delivered. In this chapter, we reviewed the very notion of charge,reminding ourselves that charge comes in two varieties, positive and negative. Welearned that charges establish electric fields through which charges can exert forceson other charges. We relied on similarities between electrical and gravitationalsystems to understand better Coulomb’s law and the nature of the forces that existbetween charged particles. Don’t forget that electrical forces can be repulsive as wellas attractive, which is one of the differences between electrical and gravitationalsystems. Charges contain electric potential energy, which we defined as their energyof position with respect to other charges. Charges move within an electric field fromone position of electric potential to another; they will move spontaneously throughan electric potential difference, or voltage, in whichever direction results in adecrease in the charges’ electric potential energy. Finally, we considered thegeometry of the electric dipole and derived the equation for calculating the electricpotential at any point in space around the dipole. We want to congratulate you for completing this chapter. We know from experiencethat premed students—actually, just about all students—find the topic ofelectrostatics to be particularly difficult. Review these basic concepts discussed hereuntil you are comfortable with them. But remember, there’s a lot yet to review, sodon’t get stuck on this one topic. The next two chapters will help you build greaterunderstanding of the behavior of charges. Furthermore, you will have manyadditional opportunities to consider the application of these basic concepts ofcharge in topics related to chemistry (electrochemistry), organic chemistry(electrophiles and nucleophiles), and even biology (membrane potentials and nerveaction potentials).

CONCEPTS TO REMEMBERCharges are either positive or negative. The fundamental unit of charge is equal to the

charge of a single proton or a single electron. Protons have positive charge, whileelectrons have negative charge.

Coulomb’s law gives the the force between two charges that are separated by somedistance. The force vector lies along the straight line that connects the two charges.Like charges exert repulsive forces, while unlike charges exert attractive forces.

Coulomb’s law is analogous to the gravitational force equation, except that it has amuch larger constant and uses charge rather than mass. Electrostatic andgravitational forces are inversely proportional to the square of the distance betweenthe two charges or masses, respectively.

Every charge sets up an electric field, which surrounds it and by which it exerts forceson other charges. The electric field is defined as the ratio of force that is exerted (orwould be exerted) on a test charge that is (or could be) at some point in space within



the electric field. Electric field vectors can be represented as field lines that radiateoutward from positive source charges and radiate inward from negative sourcecharges.

Positive test charges will move in the direction of the field lines; negative test chargeswill move in the direction opposite of the field lines.

The electric potential energy of a charge is the amount of work it took to put thecharge in a given position. It is analogous to gravitational potential energy. When likecharges move away from each other, their electrical potential energy decreases.When unlike charges move toward each other, their electrical potential energydecreases.

Electric potential is the electric potential energy per unit charge. Different points in thespace of an electric field surrounding a source charge will have different electricpotential values. Test charges will move spontaneously in whichever direction resultsin a decrease in their electrical potential energy. Positive test charges will movespontaneously from high potential to low potential. Negative test charges will movespontaneously from low potential to high potential.

Equipotential lines are concentric lines (or concentric spheres in 3-D space) of pointsin an electric field that have the same electric potential. Work will be done whencharge is moved from one equipotential line to another, and the work is independentof the pathway taken between them. No work is done when a charge moves from apoint on one equipotential line to another point on the same equipotential line.

Two charges of opposite sign separated by a distance d generate an electric dipole ofmagnitude p = qd.

In an external electric field, an electric dipole will experience a net torque until it isaligned with the electric field vector. However, an electric field will not induce anytranslational motion, regardless of its orientation with respect to the electric fieldvector.


F = qoE

τ = (qd)Esin θ Practice Questions

1.



In the figure above, the magnitude of the electric force on R due to S isA. F/2B. F C. 2F D. 4F

2. If the distance between the centers of the spheres above is halved, the magnitude ofthe force on S due to R will be

A. F/2B. F/4C. 2F D. 4F

3. If an electron were placed midway between R and S above, the resultant electric forceon the electron would be

A. toward R.B. toward S.C. up.D. down.

4. If the electric field at a distance r away from charge Q is 36 N/C, what is the ratio ofthe electric fields at r, 2r and 3r?

A. 36:9:4B. 9:3:1C. 36:18:12D. 36:18:9

5.

A positive charge of +Q is fixed at point R a distance d away from another positivecharge of +2Q fixed at point S. Point A is located midway between the charges, andpoint B is a distance d/2 from +2Q. In which direction will a positive charge move ifplaced at point A?

A. Toward the +Q chargeB. Toward the +2Q chargeC. Upward and diagonally toward the +2Q chargeD. It will remain stationary.

6. Two parallel conducting plates, one carrying a charge +Q and the other a charge -Q,are separated by a distance d. The voltage between the plates is 12 V If a +2 ?Ccharge is released from rest at the positive plate, how much kinetic energy does ithave when it reaches the negative plate?



A. 2.4 × 10-6 JB. 2.4 × 10-5 JC. 4.8 × 10-5 JD. 4.8 × 10-6 J

7. The negative charge (-1 ?C) in the figure below goes from y = -5 to y = +5 and ismade to follow the dashed line in the vicinity of two equal positive charges (= +5 C).What is the work required to move the negative charge along the dashed line?

A. -5 JB. -10 JC. 10 JD. No net work is done.

8. If the electric field at 3 m away from a charge is 8 N/C, what is the electric potential ata point 6 m away from the charge?

A. 2 VB. 6 VC. 12 VD. 24 V

9. Given an electric dipole, the electric potential is zeroA. only at the midpoint of the dipole axis.B. anywhere on any perpendicular bisector of the dipole axis and at infinity.C. anywhere on the dipole axis.D. only for points at infinity.

10. The electric potential applied to a certain electron is increased by a factor of 4. Thevelocity of the electron will increase by a factor of

A. 16B. 8C. 4D. 2

11. Which of the following accurately depicts the field lines around a proton that ismoving toward the right of this page?

A.



B.

C.

D.

12. A certain 9-volt battery is used as a power source to move a 2 C charge. How muchwork is done by the battery?

A. JB. 9 JC. 18 JD. 36 J

13. A proton and an alpha particle (a helium nucleus) repel each other with a force of Fwhile they are 20 nm apart. If each particle combines with three electrons, what is thenew force between them?

A. 9F B. 3F C. F D. F/9



14. The diagram below represents the field lines for two charges of equal magnitude.What are the signs on each charge?

A. q1 is positive; q2 is negative.B. q1 is negative; q2 is positive.C. q1 and q2 are both positive.D. q1 and q2 are both negative.

Small Group Questions1. Three point charges are placed at the corners of a square so that the net electric field

at one corner equals zero. Do these charges all possess the same sign? Magnitude?2. A proton is held in place. An electron is released from rest and allowed to collide

with the proton. The experiment is repeated with the roles of the proton and electronreversed. Which particle travels faster upon collision?

Explanations to Practice Questions 1. BAccording to Newton’s third law, if R exerts a force on S, then S exerts an equal butoppositely directed force back on R. Therefore, the magnitude of the force on Rdue to S is F. 2. DThe force is inversely proportional to r2. Thus,

F 1/r2

If r2 equals r1/2, thenFold 1/r1

2

andF new 1/r2

2

Fnew 1/(r1/2)2

Fnew 1/(r12/4)

Fnew 4/r12 = 4Fold

So the new force is 4 times greater. 3. B



If an electron with its negative charge (-e) were to be placed midway between R andS, the electron would be attracted by S (unlike charges attract) and repelled by R(like charges repel). Thus, the resultant force would be toward S (and away from R). 4. AThe first step in answering this question is to remember that the electric field isinversely proportional to the square of the radius:

E = kq/r2

E 1/r2

Therefore, if the electric field at radius r, Er, is 36, then the electric field at radius 2rwill be

E2r 1/(2r)2

E2r ¼r2

E2r = Er/4 = 36/4 = 9 N/CSimilarly, the electric field at radius 3r, E3r, is equal to

E3r 1/(3r)2

E3r 1/9r2

E3r = Er/9 = 36/9 = 4 N/CLastly, the ratio of Er:E2r:E3r is 36:9:4. Choice (A) matches with our response and istherefore the correct answer. 5. AA positive charge placed at A will experience two forces, a force to the left due to+2Q and a force to the right due to +Q. Since point A is the same distance from +Qand +2Q, the force due to +2Q will be larger than that due to +Q, and there will be anet force to the left (towards +Q). 6. BRecall that the change in potential energy, ΔU, and the change in potential, ΔV, arerelated by ΔU = qΔV. So ΔU= (2 × 10-6 C) × 12 V = 2. 4 × 10-5 J. To move apositive charge away from negative charges and towards positive charges, we mustexert a force to counteract the attractive force of the negative charges on the negativeplate and the repulsive force of the positive charges on the positive plate. Thus,positive work must be done. Conservation of energy applies to charges as it does tomasses; thus, the loss in electric potential energy as the charge moves from thepositive plate to the negative plate must equal the gain in the kinetic energy of thecharge. The loss in potential energy is just the amount of potential energy gainedoriginally when going from the negative to the positive plate (i.e., 2.4 × 10-5 J). So thegain in kinetic energy is just 2.4 × 10-5 J, which matches with choice (B). 7. DThere will be work done in moving the negative charge from its initial position to theorigin. However, in moving the negative charge from the origin to the final position,the same amount of work is done but with the opposite sign. This is because theforce changes direction as the electron crosses y = 0. Therefore, the two quantitiesof work cancel each other out. This argument depends crucially on the symmetry of



the initial and final positions. 8. CLet’s start by defining the relationship between the electric field and the electricpotential of a charge:E = kQ/r2

V = kQ/rSimply put, while the electric field is a vector that is proportional to 1/r2, the electricpotential is a scalar that is proportional to 1/r. Since E is 8 N/C at r = 3 m, then at adistance twice as far away (r = 6 m), E would be four times smaller; that is, 2 N/C.To determine the electric potential, given the electric field, we can multiply theelectric field by r:

E = kQ/r2

kQ/r2 × r = kQ/r =VTherefore, V at r = 6 is 2 N/C × 6 m = 12 V Choice (C) matches our result and isthus the correct answer. 9. BPotential is a scalar quantity. The total potential is the sum of the potentials of thepositive and negative charges. V+ = kq+/r+ and V- = kq-/r-. In a dipole,-q+ = q-, so thepotential will be zero wherever r+ = r-. This will be at any point along anyperpendicular bisector of the dipole axis. (Note the potential is also zero for allpoints at infinity since kq/∞ = 0, but choice(D) says only points at infinity. Clearly many points not infinitely far away also havezero potential.) 10. DThe electric potential (V) is equal to the amount of work done (W) divided by thetest charge (q0), according to the equation V = W/q0. This means that the potential isdirectly proportional to the amount of work done, which is equal to the amount ofenergy exerted by the particle; therefore, the overall amount of energy increases by afactor of 4. Since energy is directly proportional to the square of the velocity(according to E = (½)mv2), the velocity must increase by a factor of 2. 11. DYou should know that the field lines for a positively charged particle will alwayspoint away from the particle in a radial pattern, regardless of the direction in whichthe particle is moving. Choice (A) represents the field lines of a negatively chargedparticle. Choices (B) and (C) do not represent any particular scheme. 12. CElectric potential (V) is equal to the quotient of the amount of work done (W)divided by the charge of the particle on which the work is done (q0), according to theequation V = W/q0. Since the potential equals 9 V and the charge equals 2 C, thework done must equal 9 V × 2 C = 18 J.



13. CThe electrostatic force is given by the equation F = kq1q2/r. Since the distance doesnot change during the interaction in the question, the value of r is irrelevant to theanswer. Currently, q1 and q2 are equal to +1 e and +2 e, respectively; the addition ofthree electrons (each of which carries a charge of -e) will change the charges to -2 eand -1 e. Therefore, the product q1q2 before the interaction is equal to the product q1

q2 after the interaction (q1q2 = +2 e2). Because k and r remain constant in this system,the value of kq1q2/r does not change. 14. DA field line shows the direction of the electrostatic force that would act on apositively charged test particle. The field lines in this diagram are pointed toward q1

and q2, meaning that a positive particle would be accelerated in the direction of q1

and q2. This can only happen if both charges are negative, choice (D). The lines from q1 can never intersect with the lines from q2, since a positively charged test particleat any given point will always move in one specific direction; if the lines intersected,then a test particle at the intersection point would have to move in both directions atonce.



7Magnetism

Picture it: A class of 30 high school students brought into the middle of a woodedarea, each supplied only with a compass, a topographical map, a canteen of water,and a candy bar, and told that they have three hours to find their way back to thedesignated meeting point along one of the edges of the woods. No, this is not theopening scene for a new low-budget teen slasher film—we’re quite certain thecreative reserve for that cinematographic genre has been depleted long ago. Neitheris this some sick revenge fantasy of high school teachers who’ve “had it up to herewith these hooligans.” Actually, this is all just a fun and educational challenge in theskills of orienteering—a cross-country race in which competitors use maps andcompasses to navigate their way through unfamiliar territory. While orienteeringusually takes place in woods and other undeveloped terrain, one could certainlyemploy the same skills of compass navigation and topographical map reading in anurban environment as well, although it has to be said that the grid system of roads inmany major cities makes compasses and topographical maps rather obsolete. Thismight serve well in the urban jungle of LA but probably not in neatly gridded NewYork or Chicago.



A map and a compass are the only essential tools for the orienteer. The map willprovide helpful information about rock formations, elevations, rivers, and otherlandmarks, which help orienteers recognize the surrounding area. The compass helpsguide the orienteer by providing a continuous reference to the earth’s magneticpoles, thereby allowing the orienteer to determine the direction of her movementsthrough the unfamiliar territory.



A compass, one of the simplest tools developed by humans, takes advantage of theearth’s natural magnetic field and the tendency of certain elements to be attracted tomagnetic fields. The type of compass that most of us are familiar with, the magneticcompass, consists of a magnetized needle (a very small, light magnet) that isbalanced on a pivot so that it is free to turn, encased within some kind of clearhousing like glass or plastic. Usually there is fluid filling the housing to stabilize theneedle. The magnetic needle will always be aligned with the earth’s magnetic poles.The north-seeking end of the compass needle is usually painted or colored red, andit will always point to the north magnetic pole (which, by the way, is not in exactlythe same location as the geographic north pole, and to complicate things slightly, thenorth magnetic pole is actually the “south” pole of the imaginary bar magnet thatexists inside the earth). Using a magnetic compass, you can be certain where “northis,” and you can use the compass to guide you in whatever direction you need to go.

Real WorldThe magnetic field of the earth is created by the movement of molten metals underthe surface of the earth.This chapter reviews the subject of magnetism. Unlike electrostatics, in whichelectric charges create electric fields that exert forces on other electric charges,magnetism has no fundamental magnetic charges. Instead, magnetic fields arecreated by moving charges, sometimes in the form of currents in wires, andpermanent magnets. These magnetic fields, in turn, exert magnetic forces on othermoving charges, current-carrying wires, and magnets. The two necessary conditions,then, for the generation of magnetic fields are charge and movement of that charge.Likewise, the two necessary conditions for the generation of magnetic forces arecharge and movement of that charge.

BridgeHave you heard the term electromagnetism before? Two words are combined into asingle concept because a changing magnetic field produces an electric field and achanging electric field produces a magnetic field. It is because of thisinterconnectedness that both are considered part of a single fundamental force.The first half of this chapter will discuss the generation of magnetic fields bypermanent magnets and moving charge. We will focus on the two most commonconfigurations of current: the straight wire and the loop of wire. In the second half ofthe chapter, we will review the calculations involved in determining the magneticforce exerted by an external magnetic field on a moving charge or current.



ConclusionThis chapter continued our investigation and review of charge that began in Chapter6, Electrostatics. Here, we discussed the magnetic field created by moving charge(s)and magnets and the forces that are exerted by magnetic fields when charges movethrough them. We reviewed the different types of magnet materials: diamagnetic,paramagnetic, and ferromagnetic. We also demonstrated the use of the variousequations to calculate the strength of magnetic fields at different distances from astraight, current-carrrying wire and at the center of a circular loop of current-carryingwire, as well as two equations for calculating the magnitude of force acting on eithera point charge moving through a magnetic field or a current-carrying wire placed in amagnetic field. From magnetic compasses and kitchen magnets to MRI machines and maglev(magnetic levitation) transportation systems of the future, magnetic fields andmagnetic forces are integral to our lives. They record our history (magnetic strip on acredit card), reveal our present (MRI), and will carry us into the future (ShanghaiMaglev Train).

CONCEPTS TO REMEMBERMagnetic fields are created either by magnets or moving charge. The SI unit for

magnetic field is the tesla; smaller magnetic fields are measured in gauss.Materials are classified as diamagnetic, in which all electrons are paired and these

materials are weakly antimagnetic; paramagnetic, in which some electrons areunpaired and these materials become weakly magnetic in an external magnetic field;and ferromagnetic, in which some electrons are unpaired, atoms are organized intomagnetic domains, and these materials are strongly magnetic. Ferromagneticmaterials become permanently magnetized at temperatures below their respectiveCurie temperatures.

Straight, current-carrying wires create magnetic fields that are perpendicular concentriccircles surrounding the wire. The strength of the field decreases as the distance fromthe wire increases. Use the first right-hand rule to determine the direction of themagnetic field vector.

Circular loops of current-carrying wires create magnetic fields that are alsoperpendicular, concentric circles surrounding the wire. We can calculate themagnitude of the magnetic field at the center of the loop of wire, which decreases asthe radius of the loop increases. Use the first right-hand rule to determine thedirection of the magnetic field vector.

External magnetic fields exert forces on charges only if they are moving with a velocitythat has a component perpendicular to the magnetic field vector.

Point charges moving into a magnetic field perpendicular to the magnetic field vectorwill assume uniform circular motion for which the centripetal force is the magneticforce acting on the point charge. Use the second right-hand rule to determine thedirection of the magnetic force acting on the moving charge.



A current-carrying wire placed in an external magnetic field will experience a magneticforce on it as long as the current has a directional component that is perpendicular tothe magnetic field vector. Use the third right-hand rule to determine the direction ofthe magnetic force on the current-carrying wire.


(for straight current-carrying wire)

(for circular loop of current-carrying wire)F = qvB sin θ (for point charge moving through magnetic field)F = iLB sin θ (for current-carrying wire in magnetic field)

Practice Questions1. A negative charge placed in an external magnetic field circulates counterclockwise in

the plane of the paper as shown. In which direction is the magnetic field pointing?

A. Into the pageB. Out of the pageC. Toward the center of the circleD. Tangent to the circle

2. If, as in the accompanying figure, an electron is traveling straight out of the pagecrossing a magnetic field that is directed from left to right, in which direction will theelectron be deflected?

A. DownwardB. UpwardC. To the rightD. To the left

3. Which of the following is NOT a necessary condition for having a magnetic force on aparticle?

A. The particle must have a charge other than zero.B. The particle must move in a direction that is neither parallel nor antiparallel to the

direction of the magnetic field.C. There must be a current.D. The particle must have a certain length.



4. If the magnetic field a distance r away from a current carrying wire is 10 T, what willbe the total magnetic field at r if another wire is placed a distance 2r from the originalwire and has a current twice as strong flowing in the opposite direction? Themagnetic field generated by a current carrying wire is B = μoi/(2πr).

A. 0 TB. 15 TC. 20 TD. 30 T

5. A long, straight wire carries a current of 5 × 10-4 A as shown in the figure. A chargedparticle moving parallel to the wire experiences a force of 1 × 10-6 N at point P.Assuming the same charge and same velocity, what is the magnetic force on thecharge at point R and at point S ?

A. FR=1 × 10-6 N; Fs = 5 × 10-7 NB. FR = 0.5 × 10-6 N; FS = 0.5 × 10-6 NC. FR = 1 × 10-6 N; FS= 5 × 10-6 ND. FR = 1 × 10-6 N; Fs = 1 × 10-6 N

6. A positively charged particle enters a region with a uniform electric field, as shown,and is initially moving perpendicular to the field. In what direction must a uniformmagnetic field be oriented so that the net force on the particle is zero (assuming thatthe strength of the magnetic field can be appropriately adjusted)?

A. Into the pageB. Out of the pageC. Top of the pageD. Bottom of the page

7. A circular loop of wire of radius 20 cm carries a current of 5 A. A positively chargedparticle with velocity perpendicular to the plane of the loop passes directly throughthe center of the loop. What is the force on the charged particle the instant it passesthrough the center? (The magnetic field at the center of a circular loop is B = μoi/(2r)and μo = 4π × 10-7 T • m/A.)

A. 0 NB. π NC. π × 10-6 ND. 0.5π × 10-5 N

8. A certain urban hospital wants to expand its radio-oncology department and will needa source of short-lived radioisotopes. A 6.5 MeV proton cyclotron is to be built, butit must fit into an existing 10 m × 10 m room in the basement of the hospital.Assuming that the maximum magnetic field attainable is 1 T, can the hospital goahead with its plans? (The mass of a proton is 1.67 × 10-27 kg, and the charge of a



proton is + 1.6 × 10-19 C.)A. Yes, because the magnetic field needed is smaller than the maximum magnetic field

attainable.B. No, because the cyclotron would big too big for the room in the basement.C. It depends on the number of short-lived radioisotopes needed.D. Cannot be determined.

9. Two positive charges are traveling in opposite directions parallel to the uniformmagnetic field. It can be inferred that the magnetic force on both charges

A. is equal in magnitude and opposite in direction.B. has the same direction but different magnitudes.C. is zero.D. is different in both magnitude and direction.

10. A negatively charged particle of 2 C and 0.005 g is spinning in a uniform magneticfield along the circle with a radius of 8 cm. Knowing that the strength of magneticfield is 5 T, calculate the speed of the particle.

A. 1.3 × 10-2 m/sB. 160 m/sC. 1.6 × 105 m/sD. 1.3 × 10-5 m/s

11. A power cable, stretched 8 m above the ground, is carrying 50 A of current.Determine the magnetic field produced by the current at the ground level under thewire if μo, the permeability of free space, is 1.26 × 10-6 Tm/A.

A. 1.26 × 10-6 TB. 4.5 × 10-6 TC. 33 × 10-6 TD. 0.5 × 10-6 T

12. A researcher is interested in creating a particle accelerator that can spin particles inthe uniform magnetic field at the highest possible speed. This can be achieved by allof the following EXCEPT

A. increasing magnetic field strength.B. increasing mass of the particles.C. increasing orbital radius.D. increasing charge of the particle.

13. A proton and an electron are traveling in the uniform magnetic field with identicalvelocities. If the movement of both particles is perpendicular to the magnetic fieldlines, which of the following is/are NOT true?

I. The acceleration of the proton is greater than that of the electron.II. The proton will experience a greater kinetic energy change than the electron.III. The magnetic force on both particles is zero.A. III onlyB. I and III onlyC. II and III onlyD. I, II, and III

14. A velocity filter detects particles of particular speed at the point when electric forceF = Eq and magnetic force F = qvB produced by the filter are equal. It can be



reasonably assumed that, in order to select a particle with a higher speed, one shouldincrease

A. the electric field..B. the charge of the particle.C. the magnetic field.D. both electric and magnetic fields.

Small Group Questions1. Magnetic field lines, like electric field lines, never intersect. Suppose two magnetic

field lines were allowed to intersect. What would this suggest about a charge movingthrough such a point?

2. Explain how a compass works.Explanations to Practice Questions

1. BThis problem is an application of the right-hand rule. The velocity vector v is alwaystangent to the circle. Because you are dealing with a negative charge, qv is alsotangent to the circle but going clockwise the other way. The magnetic force is thecentripetal force and points toward the center of the circle. Knowing the directionsof qv and F, use the right-hand rule to find the direction of B. Pick a convenientpoint along the circular path. Using your right hand, put your thumb in the clockwisedirection. Now the direction of B is unknown, but you know that F points towardthe center of the circle. With your thumb pointing in the direction of qv and the palmfacing the center of the circle, your remaining fingers should be pointing out of thepage. Therefore, the direction of the magnetic field is out of the page. (B) is thecorrect answer. 2. AThe problem asks only for the direction of deflection, so it will only be necessary toapply the right-hand rule to find the direction of the force. An electron travels straightout of the page, so v is out of the page and qv is into the page (the electron isnegatively charged). Holding the thumb of the right hand into the page and theremaining fingers pointing toward the right, your palm, and therefore the force,points downward. The electron therefore will be deflected downward, just as (A)states. 3. DThree main conditions must be satisfied in order for a magnetic force to be exertedon a particle in a magnetic field. The fastest way for determining these conditions isto look at the two formulas for the magnetic force:FB = qvB sin θFB = iLB sin θwhere q is the charge of the particle, v is the velocity, B is the magnetic field, θ is theangle between v and B, i is the current, and L is the length of the wire through whichcurrent runs. The first formula describes the force on a particle, while the secondformula describes the magnetic force on a current-carrying wire. Thus, we can infer



that for FB to exist, the particle must be charged; that is, it cannot be neutral. Crossout (A). Furthermore, the velocity cannot be zero; that is, the particle must bemoving. The flow of charge is called electric current. (C) is also out. Finally, for sinθ to have a nonzero value, θ must be different from 180°, 360°, or any multiple ofthese numbers; in other words, the direction of the velocity vector with respect tothe magnetic field vector has to be other than parallel or antiparallel. Because (B) is atrue statement, you are left with (D) as the correct answer. Indeed, the length L thatwe see in the second formula refers to the length of the current-carrying wire, not thelength of the particle. 4. DThe safest way to answer this question is to quickly draw a diagram:

Notice right away that in between the two wires, the direction of the magnetic field isthe same, into the page. Therefore, since the vector direction is the same, we canfocus only on the magnitude of the two magnetic fields. We know that B1 = 10 T at adistance r. B2 can be calculated using this formula:

Given that the current in the second wire is twice as strong as the one in the firstwire, at a distance r from each wire, B2 is equal to

Thus, overall, the magnitude of the magnetic field at a distance r away from eachwire is 10 T + 20 T = 30 T, making (D) the correct answer. 5. AThe magnetic force on a charge moving in the vicinity of the given wire is F = qvBsin θ where B is given by the magnetic field of a long, straight wire and is equal to B= µ0i/(2πr). Thus,

In other words, F is inversely proportional to the distance between the charge andthe wire. Because point R is the same distance from the wire as point P, the force at



R is the same as at P. Point S is given to be twice as far from the wire as point P, sothe magnetic force at S will be one half the force at P. Plugging in numbers,determine the following values for the force at R and S, respectively:FR= 1 × 10-6 NFs = 0.5 × 10-6 N = 5 × 10-7 N(A) matches our result and is thus the correct answer. 6. BThe electric field lines always point from positive to negative; in the diagram, thepositive end is therefore on the bottom of the page, and the negative end is towardthe top of the page, in the direction of the field lines. The positively charged particlewill experience a force in the direction of the electric field, pushing it toward the topof the page (i.e., toward the negative end). Using the right-hand rule for magneticforces on charged particles, you can see that a magnetic field coming out of the pagewill give a force toward the bottom of the page. Assuming that the strength of themagnetic field is appropriately set (the assumption stated in the question), the netforce on the charged particle due to both the electric and magnetic fields will bezero. Thus, (B), out of the page, correctly describes the direction of the magneticfield. 7. AThe magnetic field at the center of the loop is directly perpendicular to the plane ofthe loop. Therefore, the particle’s velocity is parallel to the magnetic field, whichmeans that the field doesn’t exert any force on the particle. Remember that FB = iLBsin θ, and if the velocity (or current) is parallel (θ= 360°) or antiparallel (θ= 180°),sin θ= 0. (A) is the correct answer. 8. AThe problem is asking whether the proposed cyclotron will fit in the room with thegiven dimensions. You could assume the maximum value of the magnetic field andthen calculate the cyclotron orbit radius to see if it will fit in the room. Alternatively,you could assume that the protons in the cyclotron will orbit such that they are justcontained by the room, then calculate the necessary field and see if it is feasible.Because the cost of providing a strong magnetic field is high, take the last approach.For circular motion, the centripetal acceleration a = v2/r. The centripetal force is themagnetic force F = qvB. From F = ma, we obtain the following:

F = ma = mv2/r = qvBNow solve for B:

B = mv/(qr)Let the energy be denoted by E. E = 6.5 MeV = 6.5 × 106 eV = (6.5 × 106) · (1.6 ×10-19) = 1 × 10-12 J. We can find mv by using the definition of kinetic energy:



The orbit diameter must fit into the 10 × 10 m room, so the orbit radius must beabout 5 m (actually less than this to allow room for the walls of the cyclotron andvarious wires, etc.). At any rate, combining the above expressions yields an upperlimit to the required field:

So if the maximum field needed is only 0.08 T and if the materials are available tomake a field up to 1 T, then clearly it will be possible to go ahead with the project.(A) mirrors this decision and is thus the correct answer. 9. CThe magnetic force experienced by both charges is zero, as one charge is travelingparallel and another charge antiparallel to the direction of magnetic field. That makesthe angle between v and B = 0°. sin 0 = 0; thus, F = qvB sin 0 = 0. 10. CA charged particle moving perpendicular to the direction of a uniform magnetic fieldis rotating. A centripetal force associated with circular motion is the magnetic forceacting on the particle. Thus, mv2/r = qvB. Eliminating v on both sides yields mv/r =qB, and from there, v = qBr/m. For the question, convert mass to kg (0.005 g = 5 ×10-6 kg) and circle radius to meters (8 cm = 8 × 10-2 m). Plugging the numbers intothe equation, we have v = 2 • 5 • 8 × 10-2/ (5 × 10-6) = 1.6 × 105 m/s. 11. AThe magnetic field produced by the current at particular distance from the wire canbe calculated by applying the formulaB = µoi/(2πr), where µo is the permeability of free space and r is a distance awayfrom the wire, in this case 8 m. Plugging values into the formula, we can estimate ouranswer:

This matches most closely with (A).



12. BA charged particle moving perpendicular to the direction of a uniform magnetic fieldis rotating. A centripetal force associated with circular motion is the magnetic forceacting on the particle. Thus, mv2/r = qvB, and eliminating v on both sides, we havemv/r = qB. From there, you get v = qBr/m. From this formula, to obtain the highestvalue for v, the charge of the particle, (D); the magnetic field strength, (A); andorbital radius, (C), can be increased, yet increasing the mass of the particles, (B),would decrease the speed of rotation. 13. DAll of the statements are false. The magnetic force on both particles is not zero, asthe particles move perpendicular to the magnetic field lines and not parallel (III).Acceleration of the proton would be less than that of an electron, as the net force F= ma, according to Newton’s second law, and Fnet is the same for both proton andelectron. Thus, as the proton’s mass is greater, its acceleration would be less (I).The magnetic force does not do work, so it cannot change the kinetic energy ofeither particle (II). 14. AAccording to the question, the speed of the particle detected occurs when Felectric

equals Fmagnetic or Eq = qvB; from here, v = E/B. To increase the velocity of a charge,you would need to increase the electric field strength, (A), or decrease the magneticfield strength. As the particle charge, (B), is not in the equation, increasing it wouldnot bring the result needed. Increasing both electric and magnetic fields, (D), wouldcause the effects to oppose each other, resulting in no change in velocity.



8DC and AC Circuits

In this age of portable electronic devices and sophisticated long-life and rechargeablebatteries, it’s good to remember that whenever you find yourself in a pinch with littlebattery life left and no way to recharge, if you happen to be carrying around a sackof potatoes or a few dozen lemons (and really, who doesn’t carry this amount ofproduce with them?), you can make a battery sufficiently powerful to give yourdying electronic a little more life. Back in your grade school science class, you mighthave created a battery out of nothing more than a potato, a penny, a galvanized nail,and a little copper wire. It only takes a few of these put together to create sufficientvoltage and current to run a clock or light up an LED.



What’s happening inside that potato? Is veggie-power the answer to our energycrisis? (Actually, the answer to that question might be “In part, yes,” but that’s atopic far too complex and political for us to be addressing here.) The first questionis perfectly appropriate, and you may wish to review general chemistry, specificallyelectrochemistry. For now, it will suffice to note that the galvanized nail, which iscoated with zinc, serves as the anode and the penny, which contains a small amountof copper, serves as the cathode. The copper wire provides the conductive pathwaythrough which the flow of electrons can move from the galvanized nail to the penny.And the potato? Well, the potato is sufficiently juicy to provide an electrolytesolution that actually contains a low concentration of phosphoric acid (H3PO4). Thezinc is oxidized, losing electrons. These electrons travel through the copper wire,creating a current, to the penny. There H+ ions from the potato’s phosphoric acid arereduced to hydrogen gas (H2). Replace the potato with a lemon, and you’ve createda veggie battery that can generate even more current because the electrolyte in lemon(citric acid) is more concentrated.



We’re not suggesting that you go around everywhere with a sack of potatoes orlemons in case you have a sudden need to make a battery (although, that is what wejust suggested), but we do want to make the point that batteries and electric circuitsand electrical equipment pervade our everyday world. Think of any piece ofequipment, tool, or toy that has a battery or a power cord, and you’ve just identifiedan object whose very function depends upon the movement of electrons and thedelivery of electric potential energy. Turn on a light, ring a buzzer, or toast bread,and you are witnessing moving electrons at work (literally).



This chapter reviews the essentials of DC (direct current) circuits and brieflyconsiders AC (alternating current) circuits, a topic that is tested in a very basicmanner on the MCAT. Within this broader consideration of circuits, we will reviewthe many topics of circuit theory: emf, resistance, power, Kirchhoff’s laws,resistors, capacitors, and series and parallel arrangement of circuit components. Aword of encouragement: The MCAT approaches the topic of circuits with a greateremphasis on the concepts than on the math. Certainly, you will be expected tocalculate, say, the resultant resistance for resistors in series and/or in parallel, but thecircuits you encounter on Test Day will, on the whole, be simpler than what you maybe used to from your university physics class.



ConclusionThis chapter covered a lot of material. We’re exhausted, so we can only imaginehow you feel: You did all the thinking! Let’s review what we discussed so that youcan go take your well-deserved break. This chapter began with a review of current,taking special note of the conventional definition of current as the movement ofpositive charge (when in fact, the electrons are moving). We considered the basiclaws of electricity and circuits: Kirchhoff’s laws, which are expressions ofconservation of charge and energy, and Ohm’s law, which relates voltage, current,and resistance. We defined resistance and determined the relationship betweenresistance and length (directly proportional) and resistance and cross-sectional area(inversely proportional). We also defined capacitance as the ability to store charge atsome voltage, thereby storing energy. We stressed the importance of themathematical treatment of resistors and capacitors in series and in parallel as a majortesting topic on the MCAT. Finally, we briefly reviewed the key differences betweenDC and AC circuits and defined the rms values of current and voltage for ACcircuits. Now, throw the switch in your brain to the off position for a little while so that youcan rest and relax, and then, when you are refreshed, tackle some MCAT practiceproblems to reinforce these concepts.

CONCEPTS TO REMEMBERCurrent is the movement charge between two points that exist at different electric

potentials. Current, by convention, is defined as the movement of positive chargefrom the higher potential (positive) end of a voltage source to the lower potential(negative) end. In actuality, negative charge (electrons) is moving from the lowerpotential end to the higher potential end. The SI unit of current is the ampere(coulombs per second).

Kirchhoff’s laws are expressions of conservation of charge and energy. The first lawstates that the sum of currents directed into a point within a circuit equals the sum ofthe currents directed away from that point. The second law states that the sum of thevoltage sources is equal to the sum of the voltage drops around a closed circuitloop.

Resistance is the opposition to the movement of electrons through a material.Materials that have low resistance are called conductors. Conductive materials thathave moderate resistance are called resistors. Materials that have very high resistanceare called insulators. Resistance is related to resistivity and is proportional to lengthof the resistor and inversely proportional to the cross-sectional area of the resistor.

Ohm’s law states that for a given resistance, the voltage drop across a resistor isproportional to the magnitude of the current through the resistor.

Resistors in series are additive to give a resultant resistance that is the sum of all theindividual resistances. Resistors in parallel cause a decrease in overall resistance.

For resistors in parallel, the magnitude of current through each circuit division will beinversely proportional to the magnitude of the individual resistances of each circuit



division.Capacitors in series cause a decrease in overall capacitance. Capacitors in parallel are

additive to give a resultant capacitance that is the sum of all the individualcapacitances.

Dielectric materials act as insulators and increase the capacitance by a factor equal tothe material’s dielectric constant K.

Direct current (DC) is in one direction only; alternating current (AC) reverses directionperiodically.

For AC circuits, use the rms values of current and voltage for Ohm’s lawapplications.


V= iR

Rs=R1 +R2+R3+...+Rn

C’= KC

Cp=C1+C2+C3 +...+Cn

Vs=V1 + V2+ V3+...+ Vn

Vp=V1=V2=V3=...=Vn

i = Imaxsin(2πft) = Imaxsin(ωt)

Practice Questions1. If a defibrillator passes 15 A of current through a patient’s body for 0.1 seconds, how

much charge goes through the patient’s skin?A. 0.15 CB. 1.5 C



C. 15 CD. 150 C

2. A charge of 2 μC flows from the positive terminal of a 6 V battery, through a 100 Ωresistor, and back through the battery to the positive terminal. What is the totalvoltage drop experienced by the charge?

A. 0 VB. 0.002 VC. 0.2 VD. 6 V

3. If the resistance of two conductors of equal cross-sectional area and equal lengths arecompared, they are found to be in the ratio 1:2. The resistivities of the materials fromwhich they are constructed must therefore be in what ratio?

A. 1:1B. 1:2C. 2:1D. 1:3

4. If a voltaic cell provides a current of 0.5 A, the resistor in the circuit has a resistanceof 3 Ω, and the internal resistance of the battery is 0.1 Q, what is the voltage acrossthe terminals of the battery when there is no current flowing?

A. 0.05 VB. 1.5 VC. 1.505 VD. 1.55 V

5. A transformer is a device that takes an input voltage and produces an output voltagethat can be either larger or smaller than the input voltage, depending on thetransformer design. Although the voltage is changed by the transformer, energy isnot, so the input power equals the output power. A particular transformer producesan output voltage that is 300 percent larger than the input voltage. What is the ratio ofthe output current to the input current?

A. 1:3B. 3:1C. 1:300D. 300:1

6.



Given that R1 = 20 Ω, R2 = 4 Ω, R3 = R4 = 32 Ω, R5 = 15 Ω, and R6 = 5 Ω, what is thetotal resistance in the circuit shown?

A. 0.15 ΩB. 6.67 ΩC. 16.7 ΩD. 60 Ω

7. How many moles of charge pass over a period of 10 seconds through a circuit with abattery of 100 V and a resistance of 2 Ω? (F = 9.65 × 104 C/mol.)

A. 500 molesB. 5.2 × 106 molesC. 5.18 × 103 molesD. 5.18 × 10-3 moles

8.

In the above circuit, what is the voltage drop across the 2/3 Ω resistor?A. 1/2 VB. 2/3 VC. 5 VD. 10 V

9. If the area of a capacitor’s plates is doubled while the distance between them ishalved, how will the final capacitance (Cf) compared to the original capacitance (Ci)?

A. Cf = Ci

B. Cf = (1/2)Ci

C. Cf = 2Ci

D. Cf= 4Ci

10. The energy stored in a fully charged capacitor is given by E= (1/2)CV2. In a typicalcardiac defibrillator, a capacitor charged to 7,500 V has a stored energy of 400 W•s.If the charge Q and voltage V on a capacitor are related by Q = CV, what is thecharge on the capacitor in the cardiac defibrillator?

A. 1.1 × 10-5 CB. 5 × 10-2 CC. 1.1 × 10-1 C



D. 3.1 × 106 C11. A 10 Ω resistor carries a current that varies as a function of time as shown. What

energy has been dissipated by the resistor after 5 s?

A. 40 JB. 50 JC. 80 JD. 120 J

12. In the figure, six charges meet at point P. What is the magnitude and direction of thecurrent between points P and x?

A. 2 A, toward x B. 2 A, toward PC. 10 A, toward xD. 10 A, toward P

13. Which of the following will most likely increase the electric field between the platesof a parallel plate capacitor?

A. Adding a resistor that is connected to the capacitor in seriesB. Adding a resistor that is connected to the capacitor in parallelC. Increasing the distance between the platesD. Adding an extra battery to the system

14. Each of the resistors shown carries an individual resistance of 4 Ω. Assumingnegligible resistance in the wire, what is the overall resistance of the circuit?



A. 16 ΩB. 8 ΩC. 4 ΩD. 3 Ω

Small Group Questions1. How many possible ways can four identical resistors be connected to achieve an

overall resistance of 4 ohms?2. Several neighbors are competing for a prize for the best holiday decorations.

However, the rules state that each household can only use one electrical outlet.Which family will have the brightest lights, the one that connects their lights in seriesor in parallel? Why?

Explanations to Practice Question 1. BThis is a straightforward question that tests your knowledge about the current in acircuit. Electric current is defined as charge flow, or in mathematical terms, chargeover time:

i = q/ΔtA 15 A current that acts for 0.1 s means that q = iΔt = 15 A × 0.1 s = 1.5 C flowedthrough the patient’s body. (B) matches your answer. 2. AKirchhoff’s second law states that the total voltage drop around a complete circuitloop is zero. Another way of saying this is that the voltage lost through the resistorsis gained in going through the battery, so the net change in voltage is zero. Becausethe voltage is electric potential energy per unit charge, this is also equivalent to sayingthat the amount of potential energy lost by the charge in going through the resistor isgained back when the charge goes through the battery. 3. BThe resistance of a resistive material is given by this formula:

For two different materials,



From the information given in the question stem, A1 = A2, L1 = L2 and R1:R2 = 1:2.Because the resistance is directly proportional to resistivity, ρ1: ρ2 = 1:2, making (B)the correct answer. 4. DThis question tests our understanding of batteries in a circuit. The voltage across theterminals of the battery when there is no current flowing is referred to as theelectromotive force (emf or ε of the battery). However, when a current is flowingthrough the circuit, the voltage across the terminals of the battery is decreased by anamount equal to the current multiplied by the internal resistance of the battery. Inother words,

V = ε - irint

To determine the emf of the battery, first calculate the voltage across the batterywhen the current is flowing. For this, we can use Ohm’s law:V = iRV= 0.5 A × 3 ΩV= 1.5 VBecause you know the internal resistance of the battery, the current and the voltage,calculate the emf:ε = V + irint

ε = 1.5 V + 0.5 A x0.1 Ωε = 1.5 V + 0.05ε = 1.55 VThe answer makes sense in the context of a real battery, because its internalresistance is supposed to be very small so that the voltage provided to the circuit isas close as possible to the emf of the cell when there is no current running. (B)matches the result and is thus the correct answer. 5. AYou are told that transformers conserve energy so that the output power equals theinput power. Thus Pout = Pin, where Pout = ioutVout and Pin = iinVin. So = ioutVout = iinVin,which means that iout/iin =Vin/Vout. You are told that the output voltage is 300 percentlarger than the input voltage, which means that Vout/Vin = 300% = 300/100 = 3/1. Thismeans that Vin/Vout = 1/3 and iout/iin = 1/3 making (A) the correct answer. 6. BThe fastest way to tackle these kinds of questions is to simplify the circuit as muchas possible. For example, notice that R3 and R4 are in parallel and in series with R2;similarly, R5 and R6 are in series. If you determine the total resistance on each branch,you will be left with three branches in parallel, each with a different resistance. Tostart with, find the total resistance in the middle branch:



1/R3,4 = 1/R3 + 1/R4

1/R3,4 = 1/32 + 1/32 = 2/32 = 1/16R3,4 =16 ΩR2,3,4 = 4 Ω + 16 Ω = 20 ΩNext, take a look at the total resistance in the bottom branch:R5,6 = R5 + R6

R5,6 = 15 Ω + 5 Ω = 20 ΩThe circuit can now be viewed as three resistors in parallel, each providing aresistance of 20 Ω. The total resistance in the circuit is thus

Therefore, Rtot = 20/3 ≈ 7 Ω. (B) most closely matches our result and is thus thecorrect answer. 7. DThis question is testing your understanding of circuits and charge. To determine themoles of charge that pass through the circuit over a period of 10 s, first calculate theamount of charge using Faraday’s constant. Obtain the amount of charge from thecurrent in the circuit, which can be calculated using Ohm’s law:i = V/Ri = (100 V)/(2 Ω)i = 50 AYour result implies that every second, 50 A worth of current pass through thecircuit. Determine the amount of charge for 10 seconds by using the followingcalculation:i = q/Δtq = iΔtq = (50 A)(10 s)q = 500 CLastly, calculate the number of moles of charge that this represents by using theFaraday constant and approximating F as 105 C. # moles of charge = (500 C)(1 mol)/(105 C) = 5 × 10-3 mol (D) matches our result and is thus the correct answer. 8. CTo determine the voltage drop across the 2/3 Ω resistor, simply calculate the voltagedrop across both parallel resistors. Because they are arranged in parallel, the voltagedrop will be the same. To begin with, calculate the total resistance in the circuit. Forthe resistors in parallel, the total resistance is 1/Rtot = 1/2 + 3/2 = 4/2 = 2, so Rtot = 1/2Ω. The total resistance in the circuit is the sum of the two main resistances: Rtot = 1/2Ω + 1/2 Ω = 1 Ω. The current in the circuit is therefore i = V/Rtot = (10 V)/(1 Ω) = 10A. Lastly, determine the voltage drop across the two parallel resistors. Theirresistance added together is 1/2 Ω. Given that the current that goes through both of



them together is 10 A, the voltage drop is V = iR = 10 A × 1/2 Ω = 5 V This makessense because the voltage drop across the 1/2 Ω resistor must also be 5 V, whichadds up to the total voltage of the battery. The voltage drop across the 2/3 Ωresistor is thus 5 V, making (C) the correct answer. 9. DTo answer this question, you have to make use of our knowledge of capacitance.Because C = ε0A/d, where ε0 is the permittivity of free space, A is the area of theplates, and d is the distance between the plates, you can infer that doubling the areawill double the capacitance, while halving the distance will also double thecapacitance. Therefore, the new capacitance is four times larger than the initial one,making (D) the correct answer. 10. CBecause the question is asking you to calculate the charge on the capacitor, use theQ = CV formula. You are given V = 7,500 V and can calculate C from the formulafor energy, E= (1/2)CV2.E = (1/2)CV2

C = 2E/V2

Lastly, to calculate the charge, simply plug in the equation for capacitance into the Q= CV formula:Q = V × 2E/(V2)Q = 2E/VQ = 2 × 400 W.s/(7,500 V)Q ≈ 800/8000 = 0.1 CYour result best matches with (C), the correct answer. 11. DPower is energy dissipated per unit time; therefore, the energy dissipated is

E = PtThe power dissipated in a resistor R carrying a current I is

P = i2RTherefore, the energy dissipated in the first 2 s isE = i2RtE = 22 A2 × 10 Ω × 2sE = 80 JThe energy dissipated in the next 2 s is zero, because there is no current, andtherefore no power is dissipated. The energy dissipated during the one secondinterval from t = 4 s to t = 5 s isE = i2RtE = 22A2 × 10Ω × 1 sE = 40 JThe total energy dissipated is therefore

80 J + 40 J = 120 J(D) is thus the correct answer.



12. AKirchhoff’s current law states that the sum of all currents directed into a point isalways equal to the sum of all currents directed out of the point. The currentsdirected into point P are equal to 8 A, 2 A, and 3 A, so the sum is 13 A. Thecurrents directed out of point P are equal to 5 A and 6 A, so the total is 11 A.Because the two numbers must always be equal, an additional current of 2 A mustbe directed away from point P, which is (A). 13. DThe electric field between two plates of a parallel plate capacitor is related to thepotential difference between the plates of the capacitor and the distance between theplates (according to the formula E = V/d). The addition of another battery willincrease the total voltage applied to the circuit, which consequently is likely toincrease the electric field. (D) is correct. The addition of a resistor, (A) and (B),whether in series or parallel, will increase the resistance and decrease the voltageapplied to the capacitor. Increasing the distance between the plates, (C), would notwork because electric field is inversely proportional to the distance between theplates. 14. DThe resistance of the three resistors wired in series is equal to the sum of theindividual resistances (12 Ω). This means that the circuit essentially contains a 12 Ωresistor and a 4 Ω resistor. To determine the overall resistance of this system, usethe formula 1/R = 1/R1 + 1/R2 + ... where R is equal to the overall resistance when R1

and R2 are wired in parallel. Based on this formula, the resistance is equal to 3 Ω. (D)is therefore the correct answer.



9Periodic Motion, Waves, and Sound

In the epic and ongoing battle between convenience store owners and maraudinggroups of bored teenagers, there’s a new weapon that may ultimately prove to bringvictory to merchants in their quest against loitering. No, it doesn’t involve parents ortruant officers. It will not cause permanent injury to either body or self-esteem. It’snontoxic, safe, and effective, and best of all, anyone over the age of 25 is completelyunaffected by it. What is it? It’s a noise machine, but not just any noise machine.This special device emits a sound whose frequency is so high, it can only be heardby young people—and they find it really, really annoying. As people get older, theylose the ability to hear these high-frequency sounds, but teenagers and even somepeople in their early 20s can still hear them. The device goes by the name of aparticularly annoying blood-sucking insect, and the company that distributes thedevice throughout North America has a name that would be fitting for a pest controlservice.



The sound that is emitted is a modulated tone at 17.5 KHz and 18.5 KHz and is setto be automatically louder than background noise by about 5-8 dB. At its loudest, itis about as loud as a lawnmower, though there is a model for police use that is asloud as a jet engine. Those who can hear it describe the sound as being likefingernails continuously scraped across a chalkboard, and most can’t tolerate it formore than a few minutes.



There is some opposition to its use from civil liberties groups and children’sadvocacy agencies who say that the device is needlessly cruel and smacks ofageism, pointing out that if such a device were designed to drive away the elderly ora particular ethnic group, there would be greater resistance to its use. We will leavethat for the courts to decide but will take this opportunity to marvel at the uniqueway in which the wave character of sound has been employed.



This chapter lays the foundation for understanding wave phenomena by reviewingthe subject of simple harmonic motion. The general properties of waves will then beintroduced, including the concepts of wavelength, frequency, wave speed, amplitude, and resonance. We will review the interaction of waves meeting at a point in space,resulting in constructive or destructive interference as a result of their superposition;examine the mathematics of standing waves; and recognize that standing waveformation is the means by which all musical instruments produce their characteristicsounds. The subject of sound is reviewed as a specific example of the longitudinalwaveform, with a focus on certain wave phenomena, such as the Doppler effect, thatcan be observed through our sense of hearing.



ConclusionPeriodic motion was the topic of discussion in this chapter. Periodic motion isrepetitive motion about an equilibrium position. We began with a review of simpleharmonic motion demonstrated by spring-mass systems and simple pendulums.These conservative systems (ignoring friction) experience restoring forces that areproportional to their displacements. Restoring forces are always directed opposite todisplacement in the direction of the equilibrium position. At maximum displacement,the system has maximal potential energy, which is converted to kinetic energy as itaccelerates toward its equilibrium position. Maximum kinetic energy is reached as thesystem reaches equilibrium. We then reviewed the general characteristics of waves,including the phenomena of interference and resonance, and analyzed thecharacteristics and behaviors of sound as an example of a longitudinal waveform.Sound is the mechanical disturbance of particles creating oscillating regions ofcompression and rarefaction along the direction of movement of the wave. Theintensity of a sound wave is perceived as the loudness of the sound and is measuredin decibels, which is a logarithmic scale used to describe the ratio of a sound’sintensity to a reference intensity (usually the intensity of the threshold of hearing).The Doppler effect is a phenomenon of a sound’s frequency, perceived as pitch,that occurs when the source of a sound and the detector of the sound are movingrelative to each other. The perceived frequency is either higher or lower than theactual frequency, depending on whether the source and detector are moving towardeach other or away from each other, respectively. Finally, we reviewed themathematics governing the formation of standing waves, important in the formationof musical sounds in strings, open pipes, and closed pipes. Continue to review these important topics for the MCAT, taking time to apply whatyou’ve learned to MCAT-style passages and questions. Practice is the best way toensure that you’ll be ready to tackle any problem on Test Day!

CONCEPTS TO REMEMBERPeriodic motion is repetitive motion about an equilibrium position. The simple

harmonic motion demonstrated by springs and pendulums is an important type ofperiodic motion.

Springs are governed by Hooke’s law, which states that the restoring force isproportional to the displacement and always directed toward the equilibriumposition. Springs experience maximum restoring force, maximum acceleration,maximum potential energy, and minimum kinetic energy at maximum displacement.They experience minimum restoring force, minimum acceleration, minimum potentialenergy, and maximum kinetic energy at minimum displacement. The angularfrequency of springs is a function of the spring constant and the mass attached tothe spring but not its displacement.

Pendulums are governed by a restoring force that is a function of gravity. Theequilibrium position of a pendulum is the alignment with the gravitational accelerationvector (that is, the vertical position). Pendulums experience maximum restoring



force, maximum acceleration, maximum potential energy, and minimum kineticenergy at maximum angular displacement. Pendulums experience minimum restoringforce, minimum acceleration, minimum potential energy, and maximum kinetic energyat minimum angular displacement. The angular frequency of pendulums is a functionof the acceleration of gravity and the length of the pendulum arm.

Waves can be transverse (oscillation of wave particles perpendicular to the directionof propagation of the wave) or longitudinal (oscillation of the wave particles alongthe direction of propagation of the wave). Water waves and EM waves aretransverse; sound waves are longitudinal.

Waves meeting in space in phase result in constructive interference; waves meeting inspace out of phase result in destructive interference.

Sound waves are longitudinal waves of oscillating mechanical disturbances of amaterial. The pitch of a sound is related to its frequency; the loudness of a sound isrelated to its intensity (related to its amplitude).

The Doppler effect is the shift in the perceived frequency of a sound compared to theactual frequency of the emitted sound when the source of the sound and the detectorare moving relative to each other. When they are moving toward each other, theapparent frequency will be higher than the actual frequency; when they are movingaway from each other, the apparent frequency will be lower than the actualfrequency.

Standing waves are produced by the constructive and destructive interference of twowaves of the same frequency traveling in opposite direction in the same space. Theresultant wave appears to be standing still; the only oscillation is that of the amplitudeat points called antinodes. Points where there is no oscillation of amplitude are callednodes.

String secured at both ends and open pipes (open at both ends) support standingwaves, and the length of the string or pipe is equal to integer multiples of thehalf-wavelength.

Closed pipes (closed at one end) support standing waves and the length of the pipe isequal to odd integer multiples of the quarter-wavelength.

EQUATIONS TO REMEMBERF=-kx ω=2πf=√(k/m)

F = -mg sin θ

ν=fλ



fbeat = |f1 - f2|

(for strings and open pipes; where n is a positive non-zero integer)

(for closed pipes; where n is an odd integer)Practice Questions

1. If the length of a pendulum is increased from 2 m to 8 m, how will the period ofoscillation be affected?

A. It will double.B. It will be halved.C. It will quadruple.D. It will decrease by one-fourth.

2. A child is practicing the second harmonic on his clarinet. If his brother covers one endof the clarinet for a brief second, how will the sound change?

A. The pitch of the sound will be higher.B. The pitch of the sound will be lower.C. The pitch of the sound will not change.D. The pitch of the sound depends on the song the boy is playing.

3. A mass is attached to a horizontal spring and allowed to move horizontally on africtionless surface. The mass is displaced from equilibrium and released. At whatpoint does the mass experience the minimum force from the spring and its maximumacceleration, respectively?

A. At the equilibrium point; at the equilibrium pointB. At maximum compression or expansion; at the equilibrium pointC. At the equilibrium point; at maximum compression or expansionD. At maximum compression or expansion; at maximum compression or expansion

4. How far away from equilibrium will the kinetic energy be equal to the potential energyof a spring that has a spring constant k = 0.1 N/m, a speed of 3 m/s, and a 0 kgmass attached?

A. 3 mB. 6 mC. 12 mD. 18 m

5. At what frequency will a spring with an attached mass resonate?



6. If the speed of a wave is 3 m/s and its wavelength is 10 cm, what is the period?A. 0.01 sB. 0.03 sC. 0.1 sD. 0.3 s

7. What is the angular frequency of a sound in the third harmonic that comes out of apipe with one end closed? (The length of the pipe L = 0.6 m, and speed of thesound is 340 m/s.)

A. 200B. 400πC. 400D. 800π

8. A certain sound level is increased by an additional 20 dB. By how much does itsintensity increase?

A. 2B. 20C. 100D. log 2

9. At the end of a show, the audience is clapping enthusiastically. A physics student whois in the audience measures that the lower level is clapping with a frequency of 5claps/second, while the first and second balconies clap with a frequency of 4claps/second and 2 claps/second, respectively. What is the beat frequency at whichthe big round of applause of all three seating levels can be heard?

A. -1 clap/secondB. 1 clap/secondC. 0 claps/secondD. 2 claps/second

10. If two waves are 180° out of phase, what is the amplitude of the resultant wave if theamplitudes of the original waves are 5 cm and 3 cm?

A. 2 cmB. 3 cmC. 5 cmD. 8 cm

11. A student is standing on the side of a road, facing east, and measuring soundfrequencies. For which of the following situations would the student determine thatthe difference between the perceived frequency and the actual emitted frequency iszero?

A. A plane flying directly above him from east to west



B. A police car chasing a driver offenderC. A person playing piano right behind the studentD. A dog barking in a car that moves from north to south

12. In which of the following media does sound travel the fastest?A. VacuumB. AirC. WaterD. Glass

13. A mass on a pendulum oscillates under simple harmonic motion. A student wants todouble the period of the system. She can do this by which of the following?

I. Increasing the massII. Dropping the mass from a higher heightIII. Increasing the mass length of the stringA. I onlyB. III onlyC. II and III onlyD. I and III only

14. As an officer approaches a student who is studying with his radio playing loudlybeside him, he experiences the Doppler effect. Which of the following statementsremains true as the officer moves closer to the student?

I. The apparent frequency of the music increases.II. The same effect will be produced if the officer is stationary and the student

approaches him.III. The apparent velocity of the wave increases.A. I onlyB. II onlyC. I and III onlyD. I, II, and III

15. If the frequency of a pendulum is four times greater on an unknown planet than it ison earth, then the gravitational constant on that planet is

A. 16 times greater.B. 4 times greater.C. 4 times lower.D. 16 times lower.

Small Group Questions1. Can dribbling a basketball be considered simple harmonic motion if the ball returns to

its original height each time it is bounced?2. Suppose a grandfather clock (a simple pendulum) is moving too fast. Should you

shorten or lengthen the pendulum to make sure it keeps the correct time?Explanations to Practice Questions

1. ATo answer this question, first determine the relationship between the length of apendulum and its period of oscillation. You know that the angular frequency, ω, of apendulum is given by



Since f= 1/T, rewrite the first formula asω =2π/T

Equating these two equations, you obtain the following equality:

From here, you see that the period is equal to

Finally, you can see that quadrupling the length of the pendulum will double theperiod, making (A) the correct answer. 2. BThis question is basically testing your understanding of pipes open at one or bothends. To begin with, establish the relationship between the sound change and thetype of pipe. A change in sound can best be characterized by a change in itsfrequency; low-frequency sounds have a low pitch (e.g., bass), while high-frequencysounds have a high pitch (e.g., soprano). Your task is to determine how thefrequency of a second harmonic differs between a pipe that is open at both endsfrom one that is open at only one end. For a pipe open at both ends and of length L,the wavelength for the second harmonic is equal to L. In contrast, for a pipe open atone end and closed at the other, the wavelength is equal to 4L/3. In other words,when the brother covers one end of the clarinet, the wavelength increases by 4/3.Given that the wavelength and the frequency of a sound are inversely proportional,an increase of 4/3 in wavelength corresponds to a decrease of 3/4 in frequency. Inother words, when the brother covers one end of the clarinet, the sound producedby the instrument will be slightly lower than the original sound. The explanationmatches (B), the correct answer. 3. CHooke’s law gives the force, F, of the spring on the mass: F = -kx. The minimumforce is actually zero and occurs where x = 0. By definition, x = 0 is the equilibriumpoint (i.e., the position of the mass so that the spring is neither stretched norcompressed). Based on this, eliminate (B) and (D). Newton’s second law tells youthat F = ma and therefore a = F/m, so the point of maximum acceleration is thepoint at which the mass experiences the greatest force. From Hooke’s law, thegreatest force occurs at the point of the largest displacement from equilibrium.Because the displacement at maximum expansion equals the displacement atmaximum compression, the largest force and therefore largest acceleration occurs ateither maximum expansion or maximum compression. (C) matches the explanationand is thus the correct answer. 4. B



This question is testing your understanding of springs and how they conserve

energy. The kinetic energy of a spring is given by KE = mv2, while the potential

energy is equal to . Your task is to determine at what speed anddistance from the equilibrium point the two energies will be equal. Equate the twoequations:

(B) is therefore the correct answer. 5. AThe resonance frequency of a mass on a spring is just the natural frequency:

At this frequency and any multiple of it, the spring will be in resonance. (A) is thusthe correct answer. 6. BThis question is testing your understanding of traveling waves. To find the period ofthe wave, manipulate the calculation for frequency. Since one knows the speed andwavelength, the frequency and period can be calculated using the following equation:v = λff = v/λ = 1/TT = λlvT = (0.1 m)/(3 m/s) ≈ 0.03 s(B) is therefore the correct answer. 7. DYou know that the angular frequency is related to the frequency of a wave throughthe following formula: ω = 2πf. Thus, your initial task is to calculate the frequency ofthe wave. Knowing its speed, determine the frequency by first calculating itswavelength. For the third harmonic of a standing wave in a pipe with one end closed,the wavelength is

λ = 4L/nwhere n is the harmonic. In this case, n = 3. The wavelength is

λ = 4 • 0.6 m/3



λ = 0.8 mThe frequency of the wave is therefore

v = λff = v/λ

f= (340 m/s)/(0.8 m)f≈ 400 s-1

Finally, obtain the angular frequency simply by multiplying the frequency of the waveby 2π:

ω = 2πfω = 800π

(D) matches your result and is thus the correct answer. 8. CLet Ia be the intensity before the increase and Ib the intensity after the increase. Usingthe equation that relates decibels to intensity, obtain the ratio of Ib to Ia.

Δβ = 10 log Ib/Ia

log Ib/Ia = Δβ/10log Ib/Ia = 20/10

Ib/Ia = 102

Ib/Ia = 100Ib = 100Ia

The intensity is increased by a factor of 100, making (C) the correct answer. 9. BIn spite of the complicated wording of this question, the answer is quitestraightforward. You are asked to calculate the beat frequency of a certain eventgiven three individual frequencies. The beat frequency is the absolute value of thedifference between all the individual frequencies:

fbeat = |f1 - f2 - f3|fbeat = |5 - 4 -2|

fbeat = 1 clap/second(B) is therefore correct answer. 10. AWhen two waves are out of phase by 180°, the resultant amplitude is the differencebetween the two separate amplitudes. Because this is destructive interference, expectthat the amplitude of the resulting wave will be less than the amplitude of eachindividual wave. In this case, the resulting wave will have an amplitude of 5 - 3 = 2cm, making (A) the correct answer. 11. CThis question is testing us on your understanding of the Doppler effect. A differenceof zero between the perceived and the emitted frequencies implies that the source ofthe sound is not moving. From the given choices, the only source of sound that isnot moving is the person playing the piano right behind the student. (C) is thereforethe correct answer.



12. DSound is a mechanical disturbance propagated through a deformable medium; it istransmitted by the oscillation of particles along the direction of motion of the soundwave. As such, sound needs a medium to travel through, and the more tightlypacked the particles are within that medium, the faster the propagation of vibration ofthe sound waves. Therefore, sound cannot be transmitted in a vacuum, and its speedof propagation increases from gas to liquid to solid. Sound will thus travel fastestthrough glass, making (D) the correct answer. 13. BRecall the formula for the period of a pendulum, T = 2π It is a commonmisconception that mass has an effect on the frequency of a pendulum. Rememberthat this is a pendulum and not a spring! In a pendulum system, the only factors thataffect the motion are the ones in the formula: L, the length of the string, and g, thegravitational acceleration (usually 9.8 m/s2, unless the pendulum is on a differentplanet or under special conditions). Only item III is correct. 14. AHere, an observer is moving closer to a stationary source. The formula needed is f‘= f[(v + vD) /v], where v is the velocity of the sound. Because the numerator is greaterthan the denominator, f’ will be greater than f; therefore, item I is correct. (A) is thecorrect answer. The scenario described in item II will produce a similar, but notidentical, effect. The frequency formula here will be f’ = f [v/(v - vs)]. The listener’sfrequency will increase, but the increase will not be exactly the same. Because themedium through which the wave travels does not change, the velocity of the wavedoes not change, either; therefore, item III is also incorrect. 15. A

The frequency of a pendulum is defined as f= . Because g is under a squareroot, the gravitational constant has to be f2 as large, or 16 in this case. (A) is thecorrect answer. (B) and (C) assume that the relationship between frequency andgravitational acceleration is 1:1. (C) and (D) may be obtained if you mistakenly wroteg in the denominator instead of the numerator when solving the equation.



10Light and Optics

The next time you’re browsing the aisles of your local convenience store (presumingyou are not one of the rascally youngsters driven away by the antiloitering high-pitchnoise machines), look for the security mirrors. There are usually at least a couple ofthese round mirrors placed strategically throughout the store, often near the entrance,the cash register, and the beer cooler. If you are able, stand in front of the securitymirror and notice your image and the images of your surroundings; observe yourappearance and the way everything else looks in the mirror. Everything seems a littledistorted—smaller than you would expect and maybe even a little “curvy.” Noticeyour field of view in that mirror. It’s wider than you might expect, isn’t it?



Security mirrors, the kind you find in convenience stores, are a type of sphericalmirror called convex, because the mirror surface is on the outer curvature—just likea circular portion of a disco ball. Convex mirrors are classified as diverging opticalsystems, because parallel light rays that strike them are reflected in such a way thatthe reflected light rays diverge from each other.



Convex mirrors make ideal security mirrors because they always reflect an image thathas the same physical orientation as the object that produces the image. In otherwords, when you are standing in front of that security mirror at the conveniencestore, no matter how close to or far away from the mirror you are standing, yourimage will be “standing” upright just like you are; it won’t be upside down. As youmight imagine, store owners who are using security mirrors to protect from theft orvandalism find it much easier to view images of the store environment that areoriented in the same direction as the people and objects that are producing theimages. Furthermore, the convex shape of the mirror, which causes light rays todiverge, allows for a much greater field of view compared to that available in a plane(flat) mirror. Again, this makes sense given the security purposes that the mirrorsserve. These advantages do come at a small price, however. The field of view isexpanded at the expense of image size: Convex mirrors always produce images thatare reduced in size. This gives the appearance that the objects producing the imagesare farther away than they actually are.



Does that last line ring a bell, at all? For most, if not all of you, it probably does,even if you can’t place it. If you have access to a car, go to it and read the fine-printwarning on the driver’s side-view mirror. It will say something to the effect of“Warning: Objects in mirror are closer than they appear.” Based on what we’ve justdiscussed, it seems that automobile side-view mirrors just might be convex mirrors.In fact, they are. The benefits provided by the convex mirror in the conveniencestore are the same as those provided by the convex mirrors attached to the sides ofour cars—and thank goodness for that! Can you imagine your confusion if side-viewmirrors projected upside-down images of the cars behind you, or images that startedout upside down, momentarily disappeared, and then became right side up as the carfollowing you got closer and closer (as would happen with a concave mirror)?Driving is dangerous enough as it is!



This chapter focuses on light and optics. Our initial discussion will be a continuationof a topic from Chapter 9, Periodic Motion, Waves, and Sound: the transversewaveform of visible light and other electromagnetic (EM) waves. We will then moveinto a consideration of the behavior of electromagnetic waves—reflection, refraction,and diffraction—taking the waves of visible spectrum as representative of the wholeEM spectrum. In consideration of reflection and refraction, we will review the fouroptical systems tested on the MCAT: concave and convex mirrors, which produceimages by reflection, and concave and convex lenses, which produce images byrefraction.



ConclusionThis chapter illuminated (!) information about the behaviors and characteristics oflight and optical systems. First, we described the nature of the electromagnetic wave,noting the full spectrum of EM waves, of which visible light is the only segment thatwe can perceive visually. We then focused (!) on geometrical optics to consider thereflective and refractive behaviors of light, noting the ways in which mirrors reflectlight to produce images and lenses refract light to produce images. Weacknowledged the fact that light doesn’t always travel in straight-line pathways butcan bend and spread out through diffraction. Finally, we examined the pattern ofinterference that occurs when light passes through a double slit, as demonstrated inYoung’s double-slit experiment.

CONCEPTS TO REMEMBERElectromagnetic waves are transverse waves consisting of an oscillating electric field

and oscillating magnetic field. The two fields are perpendicular to each other and tothe propagation of the wave. The EM spectrum includes, from low to high energy,radio waves, microwaves, infrared, visible, ultraviolet, x-ray, and gamma-rayradiation Reflection.

is the rebounding of incident light waves at the boundary of a medium. The law ofreflectionstates that the angle of incidence will equal the angle of reflection, wheneach is measured from the normal.

Mirrors reflect light to form images of objects. Spherical mirrors have centers andradii of curvature, as well as focal points. Concave mirrors are converging mirrorsand will produce real, inverted or virtual, upright images. Convex mirrors arediverging mirrors and will only produce virtual, upright images. Plane mirrors may bethought of as spherical mirrors with infinite radii of curvature. Plane mirrors produceonly virtual, upright images.

Refraction is the bending of light as it passes from one medium to another andchanges speed. Refraction depends on the wavelength(causing dispersion through aprism). The law of refraction is Snell’s law, which states an inverse relationshipbetween the index of refraction and the sine of the angle when measured from thenormal.

Total internal reflection occurs when the angle of incidence is greater than the criticalangle for light leaving a medium with a higher index of refraction and entering amedium with a lower index of refraction. The internally reflected light does notactually leave the original medium; rather, it is totally reflected back into it.

Lenses refract light to form images of objects. Thin, bilaterally symmetrical lenseshave equal focal points on each side. Convex lenses are converging lenses and willproduce real, inverted or virtual, upright images. Concave lenses are diverging lensesand will only produce virtual, upright images.

Diffraction is the bending and spreading out of light waves as they pass through anarrow slit.

Interference demonstrates the wave/particle duality of light. Young’s double-slit



experiment shows the constructive and destructive interference of waves as lightpasses through a double slit to produce, respectively, maxima and minima ofintensity.

Plane- polarized light has been passed through a polarizer, and the electric fields of allthe waves are oriented in the same direction.

EQUATIONS TO REMEMBERc = fλ θ1 = θ2 (law of reflection)

(for mirrors)

(index of refraction)n1 sin θ1 = n2 sin θ2 (law of refraction: Snell’s law)

(determining critical angle)

(for lenses)

(for mirrors and lenses)

(for lenses on non-negligible thickness)a sin θ = nλ (n = 1, 2, 3, ... ) (for determining location of dark fringes of diffraction)d sin θ = mλ (for maxima of interface m = 0, 1, 2, 3 ...)d sin θ = (m + ½)λ (for minima of interface m = 0, 1, 2, 3 ...)

Practice Questions1. If a light ray has a of frequency 5.0 × 1014 Hz, in which region of the electromagnetic

spectrum can it be located?A. X-rayB. UVC. VisibleD. Infrared

2. A standing child has a plane mirror 5 m away from his left arm and another planemirror 7 m away from his right arm. How far apart are the two images produced bythe mirrors if the child has an arm span of 0.5 m?

A. 2 mB. 12 mC. 12.5 m



D. 24.5 m3. An object is placed at the center of curvature of a concave mirror. Which of the

following is true about the image?A. It is real and inverted.B. It is virtual and inverted.C. It is virtual and upright.D. It is real and upright.

4. When monochromatic light is refracted as it passes from air to glass, which of thefollowing does not remain the same? (Assume that the wave is fully transmitted.)

A.WavelengthB.FrequencyC.AmplitudeD.Period

5. A ray of light (f= 5 × 1014 Hz) travels from air into crystal into chromium. If the indicesof refraction of air, crystal, and chromium are 1, 2, and 3, respectively, and the angleof incidence is 30°, which of the following describes the frequency and the angle ofrefraction in the chromium? (The speed of the light ray is 3 × 108 m/s in air, 2.0 × 104

m/s in crystal, and 1.0 × 103 m/s in chromium.)A. 5 × 1014 Hz; 9.6°B. 5 × 1014 Hz; 57°C. 1.0 × 1010 Hz; 9.6°D. 1.0 × 1010 Hz; 57°

6. A source of light (f= 3.0 × 108 Hz) passes through three polarizers. The first twopolarizers are in the same direction, while the third is rotated 90° with respect to thesecond polarizer. What is the frequency of the light that comes out of the thirdpolarizer?

A. Light will not pass through the third polarizer.B. 3.0 × 108 HzC. 6.0 × 108 HzD. 9.0 × 108 Hz

7. Which phenomenon would cause monochromatic light entering the prism along pathAB to leave along path CD?

A.Reflection B.Refraction C.Diffraction D. Polarization

8. Which of the following describes the image formed by an object placed in front of aconvex lens at a distance smaller than the focal length?

A. Virtual and invertedB. Virtual and upright



C. Real and uprightD. Real and inverted

9. A submarine is inspecting the surface of the water with a laser that points from thesubmarine to the surface of the water and through the air. At what angle will the lasernot penetrate the surface of the water but rather reflect itself entirely back into thewater? (Assume nwater = 2 and nair = 1.)

A. 15°B. 25°C. 30°D. 35°

10. A student is analyzing the behavior of a light ray that is passed through a smallopening and a lens and allowed to project on a screen a distance away. Whathappens to the central maximum (the brightest spot on the screen) when the slitbecomes narrower?

A. The central maximum remains the same.B. The central maximum becomes narrower.C. The central maximum becomes wider.D. The central maximum divides into smaller light fringes.

11. Which of the following are able to produce a virtual image?I. Convex lensII. Concave lensIII. Plane mirrorA. I onlyB. III onlyC. II and III onlyD. I, II, and III

12. Monochromatic red light is allowed to pass between two different media. If the angleof incidence in medium 1 is 30° and the angle of incidence in medium 2 is 45°, whatis the relationship between the speed of the light in medium 2 compared to that inmedium 1?

A. v2 = √2 v1

B. √2 v2 = v1

C. v2 = √3 v1

D. √3 v2 = v1

13. The near point, or nearest point at which an object can be seen clearly, of one ofyour eyes is 100 cm. You wish to see your friend’s face clearly when she stands 50cm in front of you. If you use a contact lens to adjust your eyesight, what must bethe radius of curvature and power of the contact lens?

A. r = 33 cm; P = 3 dioptersB. r = 67 cm; P = 3 dioptersC. r = 100 cm; P = 2 dioptersD. r = 200 cm; P = 1 diopter

14. Imagine that a beam of monochromatic light originates in air and is allowed to shineupon the flat surface of a piece of glass at an angle of 60° with the horizontal. Thereflected and refracted beams are perpendicular to each other. What is the index of



refraction of the glass?

A. B. 1C. √3D. This scenario is not possible.

Small Group Questions1. Children often play with mirrors to ignite leaves with sunlight. Which kind of spherical

mirror, concave or convex, should be used? How far from the mirror should theleaves be placed?

2. When you look at the backside of a shiny teaspoon, your image appears upright.When you look at the other side of the spoon, your image appears upside down.Why does this occur?

Explanations to Practice Questions 1. CTo determine the region of the spectrum in which this light ray is located, calculateits wavelength:c = λfλ = c/fλ = (3 × 108 m/s)/(5.0 × 1014 1/S)λ = 6.0 × 10-7 mλ = 600 nmThis wavelength falls within the visible spectrum (400-700 nm) and it has anorange-red color. (C) correctly identifies this light ray. 2. DTo answer this question, you must understand that in plane mirrors, the image is asfar away from the mirror as the object. In other words, the image produced by theleft mirror is 5 m away from the mirror because the child is standing 5 m away fromthe mirror. Similarly, the right mirror produces an image that is 7 m away from thecenter of the mirror. To calculate how far away the two images are, you have to takeinto consideration not only the image distance but also the distance of the object (thechild) away from the mirrors and the child’s arm span of 0.5 m. Therefore, theimages are 5 + 5 + 0.5 + 7 + 7 = 24.5 m apart, making (D) the correct answer. 3. AFirst draw a diagram:

With the mirror equation and what we’re given, there should be no problem:



We know o = r, and since 1/f= 2/r, we can substitute into the equation:

Because i is positive, the image is real, eliminating choices (B) and (C). Themagnification will then tell us the image’s orientation:

The negative sign tells us the image is inverted, which means the correct choice is(A). 4. AIf you did not know right away that (A) was the correct answer, you could havesolved this question by process of elimination.Frequency is related to period (f = 1/T), so if either of these quantities changes, the other would have to change as well.Because they cannot both be correct, (B) and (D) can both be eliminated. Further,because the wave is fully transmitted, there is no absorption or reflection, and theamplitude should not change. Therefore, (C) is eliminated. Looking at the equation c= f λ, you might be tempted to eliminate wavelength, too. However, when light isrefracted, its speed changes; therefore; although the frequency cannot change, thewavelength will. 5. AThis question contains two parts—you have to determine the frequencyand theangle of refraction of the light ray. The first part, however, is straightforwardbecause the frequency of a light ray traveling from one medium to another does notchange. Because the frequency must be 5 × 1014 Hz, you can cross out (C) and (D).



For the angle of refraction, you can either calculate it or determine it using logic.First, the light ray goes from air into crystal; that is, from a low index of refraction toa higher one. According to Snell’s law, the angle of refraction will be lower (i.e.,closer to the perpendicular). When the light ray moves from crystal to chronium, itagain goes from a lower index of refraction into a higher one, thus making the angleof angle of refraction even smaller. Between (A) and (B), the only one that illustratesthis situation is (A), the correct answer. This question could also be answered bycalculation using Snell’s law, but the calculations are time consuming andunnecessary. 6. APlane-polarized light is light in which the electric fields of all the waves are oriented inthe same direction. Light passing through the first two polarizers will change so thatall the electric field vectors point in the same direction. When it reaches the thirdpolarizer, however, the light will not be able to pass through because all the light rayswill be oriented in the direction dictated by the first and second polarizer. Becauseno light will exit the third polarizer, the answer matches (A). 7. BEven though the light is traveling through a prism, the change in the light’s directionis caused by refraction, not dispersion. Dispersion involves the breaking up ofpolychromatic light into its component wavelengths because of the index ofrefraction’s dependence on wavelength. You are told that the incident light ismonochromatic or, in other words, of only one wavelength; therefore, light will notbe dispersed. 8. BThe image produced by a convex lens can be either real or virtual: It is real if theobject is placed at a distance greater than the focal point, virtual if the object isplaced at a distance less than the focal point(between the focal point and the lens). Itis also important to keep in mind that an image that is real must be inverted and onethat is virtual must be upright. In this question, the object is placed in front of thefocal point, so the image must be virtual and, therefore, upright. (B) matches yourprediction. 9 DThis question is testing your understanding of total internal reflection. As the laserbeam travels from water to air—that is, from a higher to a lower index of refraction—the angle of refraction increases. However, there is a limit to this increase. At acertain angle of incidence ( θi), the angle of refraction becomes 90°; at this point, therefracted ray is parallel to the surface of the water. When the angle of incidence isgreater than the critical angle, all the light is reflected back into the water. Thequestion asks you to determine the angle at which the laser beam will reflect itselfentirely into the water. To find this, first calculate the critical angle of the beam usingSnell’s law:nwater sin θi= nair sin θr



nwater sin θi = nair sin 90°nwater sin θi = nair

sin θi = nair/nwater

This is the sine of the critical angle, and you can memorize this last equation so thatyou don’t have to derive it each time.sin θc = nair/nwater

sin θc = 1/2θc = 30°What this tells you is that at exactly 30°, the refracted laser beam will be parallel tothe surface of the water. After 30°, the laser beam will reflect itself entirely back intothe water, an event known as total internal reflection. (D) is therefore the correctanswer, because it is greater than 30°. 10. CThis question is testing your understanding of diffraction. When light passes througha narrow opening, the light waves spread out; as the slit narrows, the light wavesspread out even more. When a lens is placed between the narrow slit and the screen,a pattern consisting of alternating bright and dark fringes can be observed on thescreen. As the slit becomes narrower, the central maximum (the brightest and mostcentral fringe) becomes wider. Imagine doing the experiment yourself. If you let theopening be as wide as a door, you wouldn’t even notice bright and dark fringesbecause each fringe is extremely small. As you make the opening smaller andsmaller, you are allowing only certain rays to penetrate, and the bright and darkfringes become wider. (C) matches your explanation and is thus the correct answer. 11. DAll images produced by plane mirrors will be virtual (III). The same goes for mirrorsthat are convex. Lenses are a bit different; all concave lenses (II) will produce virtualimages, but this is not necessarily true for convex lenses. Virtual images will nottypically be produced unless an object is placed between the center and the focalpoint of the lens, but it is still possible (I). 12. AFirst, the color of the light is irrelevant here; the ratio would be the same even if thespecific color were not mentioned. Second, recall the equation: n1 sin θ1 = n2 sin θ2.Although you don’t know the value of n for either medium, you do know the simplerelationship n = c/v. Replacing n in the first equation, canceling out c, andrearranging, you can ultimately get sin θ2/sin θ1 = v2/v1. You’re asked for the ratiobetween 2 and 1, so rearrange accordingly:



Thus, v2 = √2 v1. Having the relationship the other way around would give you (B). 13. DA basic optics principle is tested here, but you need to know how to discernbetween and assign values to variables, as well as choose the correct equation. Here,your choice of equation should not be very difficult. The question points to 1/o + 1/i= 1/f = 2/r. You know that the goal of the lens is to create a virtual image of yourfriend’s face at the near point of your eye. Consider this when assigning amounts tovariables: 1/50 cm + 1/(-100) cm = 2/r; therefore, r = 200 cm. If you fail to realizethat the question asks about r rather than f, you might get (A) or (C). (B) would beyour result had you used a positive value, getting a value of 67 cm for r. Payattention to signs! Finally, to get power, you know that P = 1/f and 1/f = 2/r. Thismeans that P = 1.0 diopters. Remember, this equation works only if f and r aremeasured in meters. 14. DDrawing a diagram is best here. Because the angle given is with respect to thehorizontal, you know that our angle in question, θ1 must equal 30°. You know thatthe reflected beam will produce a mirror image of the incident beam angles on theother side of the plane of symmetry. Therefore, the reflected beam will make anangle of 60° with the horizontal. Because you’re given that the reflected andrefracted beams are perpendicular to each other, the refracted beam will make a 30°angle with the horizontal. Your θ2 must be with respect to the plane of symmetry, soit equals 60°. Be careful in this analysis: Confusing 30° and 60° throughout will leadyou to an incorrect n2 = √3. Using n1 sin θ1 = n2 1 sin θ2, you have 1 sin 30° = n2 sin

60°, and n2 = , which matches (A). However, the index of refraction can never beless than 1 because that would imply that v is larger than c in n = c/v, which isimpossible, so (D) is correct.



11Atomic Phenomena

The great 45-carat Hope Diamond, currently housed in the Smithsonian Institute, is adiamond of incredible history and intrigue. Mined in the 1600s, most likely from theKollur mine in Golconda, India, the diamond was first purchased by Jean BaptisteTavernier and at the time had a weight of over 112 carats. Tavernier sold thediamond to King Louis XIV in 1668, who had it recut, and the diamond becameknown as the “Blue Diamond of the Crown” or the “French Blue.” It was passed toLouis the XV and then to Louis XVI and Marie Antoinette (she of the dubiouslyascribed phrase, “Let them eat cake.”), but when they tried to escape town in 1791,the royal jewels were turned over to the government. Within a year, the jewels werelooted and the diamond was stolen. The diamond didn’t reappear until 1812 inLondon, having been recut (to its present shape) at some point. It passed throughthe hands of King George IV of England to Henry Philip Hope (whose name thediamond now bears) through his nephew and that nephew’s grandson, eventuallyending up in New York City. It was acquired by Pierre Cartier in Paris, who reset itfor Mrs. Evalyn Walsh McLean, of Washington D.C., who owned and famouslywore it from 1911 through her death in 1947. Harry Winston purchased McLean’sentire collection, including the Hope Diamond, and after a decade of displaying it atexhibits and charitable events, he donated the diamond to the Smithsonian Institutionin 1958, where it has been showcased since.



The diamond carries a mystique and a mystery that is befitting of its history. Manyclaim that the diamond is cursed, and many apocryphal stories have been told of theunhappy fates that have befallen its possessors. The legend begins with the curseagainst Tavernier who, according to the story, did not rightfully purchase thediamond but stole it from the eye of a statue of the Hindu goddess, Sita. For his sin,as the legend goes, he was torn apart by wild dogs while on a trip to Russia, after hehad sold the diamond. His death was the first of many tragic endings attributed tothe curse. The diamond has been blamed for the beheading of Louis XVI and MarieAntoinette and the eventual destitution of the Hope family fortunes (more likelycause: gambling). Even Evalyn Walsh McLean, who believed the diamond to be agood luck charm—in fact, so much so, there’s a story that she nearly refused toremove it from her neck for a goiter surgery—seems not to have escaped its curse:Her first-born son was killed in a car crash when he was 9 years old, her daughtercommitted suicide at the age of 25, and her husband was declared insane andconfined to a mental institution until his death in 1941.



Part of the modern allure of the diamond is its behavior under UV light. A smallpercentage of diamonds fluoresce under UV light and emit, usually, a bluish glowthat ranges from very faint to deep. The Hope Diamond, in keeping with its stature,is extraordinary. Not only does it fluoresce, it fluoresces a faint red, an exceedinglyrare phenomenon for diamonds, and then glows a brilliant red for a few secondsafter the UV light has been removed. This phenomenon of “afterglow” is calledphosphorescence. Phosphorescence is similar to fluorescence and is the result ofelectrons, which have been excited by high-frequency radiation, returning to theirground state in multiple steps that correspond to lower-frequency radiation, typicallyin the visible range. The difference is that the electrons get “stuck” in unusual energystates and can exit from them only by classically “forbidden” transitions; althoughpossible, these are kinetically unfavorable, so the return of the excited electrons totheir ground state is delayed. The slower return to the ground state results inprolonged release of photons whose frequencies correspond to the visible range.The object glows, even after the removal of the UV light, for anywhere from a fewseconds to many hours.



Fluorescence and phosphorescence are atomic phenomena that we now understandthrough the theory of quantum mechanics. From the end of the 19th century throughtoday, research has shown that different sets of laws take effect at short distancesdue to the dual (wave and particle) nature of discrete bits of matter. The theory ofquantum mechanics that has been developed over the last century has helped toexplain phenomena that seem to go against predictions based on classical theories.This chapter will primarily cover particular applications of quantum mechanical ideasto atomic physics but will not cover the formal theory of quantum mechanics itself.Our first topics of discussion are blackbody radiation and the photoelectric effect,which will provide a first look at the quantum, or discrete, aspects of nature at theatomic level, particularly the discrete or particle nature of light. We will then reviewthe structure of the hydrogen atom as understood by the quantum theory. Bohr’smodel is adequate for describing the hydrogen atom and other one-electron systems,and we will also rely on Bohr to understand the interaction of electromagnetic quanta(photons) with atoms. Finally, we will review briefly the phenomenon offluorescence.



ConclusionQuantum mechanical theory helps us understand phenomena that occur at themolecular level, phenomena that would otherwise seem counter to our understandingbased on classical physics. This chapter examined a few behaviors andcharacteristics of substances that are best understood through the quantummechanical theory. We first discussed blackbody radiation, noting the significanceof the relationships between the peak wavelength and absolute temperature andintensity of radiation and temperature for blackbody objects. We then analyzed thephotoelectric effect, paying particular attention to the fact that quantum theoryexplains why ejection is a function of frequency of the photon, not its intensity. (Themagnitude of the current is a function of the intensity, once the threshold frequencyhas been exceeded.) We then re-examined Bohr’s model of the hydrogen atom,which he developed by incorporating the work of Rutherford and Planck. Althoughhis model is inadequate for describing more complex atoms, the Bohr model workswell to predict the discrete energy levels that the single electron may have in thehydrogen atom. Finally, we concluded our discussion of atomic phenomena with thetopic of fluorescence. When a fluorescent substance is excited by high-energyradiation, typically UV radiation, the substance’s excited electrons will return to theirground state through multiple steps and, in the process, will emit photons of lowerenergy whose frequencies correspond to the visible light range. Can you feel the excitement? Only one more chapter to go! Don’t stop now. Youare well on your way to achieving great success on the MCAT. Keep up the hardwork, the focus, and the energy. We know this is a tough road to Test Day, but youare mere steps away from reaching a major milestone!

CONCEPTS TO REMEMBERBlackbody radiators are also ideal absorbers: They will radiate an amount of energy

equal to the amount they absorb but at different peak wavelengths depending on thetemperature.

The peak wavelength for a blackbody radiator is the wavelength at which the objectradiates the greatest amount of energy. The peak wavelength is inversely proportionalto the absolute temperature.

The intensity of energy being radiated by a blackbody is proportional to the fourthpower of the absolute temperature.

The photoelectric effect is the ejection of an electron from the surface of a metal in avacuum due to an incident photon whose frequency (energy) is at least as great asthe threshold frequency (energy known as the work function) necessary to eject theelectron from that particular metal. The greater the energy of the incident photonabove the threshold energy, the more kinetic energy the ejected electron will possess.

Once the threshold frequency has been exceeded, the magnitude of the current of theejected electrons will be proportional to the intensity of the beam of photons.

The Bohr model of the hydrogen atom proposes a dense nucleus consisting of a



single proton surrounded by a single electron traveling in orbits of discrete energyvalues. For the electron to jump from a lower-energy orbital to one of higher energy,the electron must absorb an amount of energy exactly equal to the differencebetween the two energy levels. For the electron to jump from a higher-energy orbitalto one of lower energy, the electron must emit an amount of energy exactly equal tothe difference between the two energy levels.

Fluorescence is the phenomenon that is observed when a substance glows with visiblelight upon being excited by higher-energy light, typically in the UV range. Theelectrons jump to an excited state by absorbing the high-frequency photons and thenreturn to their ground state in multiple steps, releasing lower-frequency photons witheach step. These lower-frequency photons are in the visible range.

EQUATIONS TO REMEMBER( peak(T) = constantET = σT4

E=hƒ

K=hƒ-W W = hƒT

(energy in joules)

Practice Questions 1. The Stefan-Boltzman law says that E α T4, where E is the power per unit arearadiated by a body of absolute temperature T. A given body is heated to 380 K andthen subsequently further heated to 760 K. What is the ratio of the power per unitarea radiated at the higher temperature to that at the lower temperature?

A. 2B. 16C. 24D. 64

2. An infrared thermogram can detect a breast cancer tumor even if its temperature isonly 1°C above the rest of the breast. Assuming a skin temperature of 25°C (notethat this is significantly below the 37°C temperature of the inner body), how muchmore radiant energy is emitted by the skin over the carcinoma than by the skin overnormal tissue? (Assume for ease of calculation that the skin surface emits blackbodyradiation).

A. 1% brighterB. 15% brighterC. 1% darkerD. 15% darker

If the work function of a metal is 2,130 J and a ray of light with a frequency of 1.0 ×1037 Hz is incident on the metal, what will be the speed of the electrons ejected? (h =



6.63 × 10-34 J × s and melectron = 9.1 × 10-31 kg.)A. 4,500 m/sB. 3 × 10-29 m/sC. 1017 m/sD. 1034 mls

If the work function of a certain metal is 4.14 eV and light of frequency 2.42 × 1014

Hz is incident on this metal, will there be an ejection of electrons? What is thethreshold frequency of the metal? (Use h = 4.14 × 10-15 eV•s.)

A. Electrons will not be ejected; ƒT= 1.0 × 1015 HzB. Electrons will not be ejected; ƒT= 1.0 × 10-15 HzC. Electrons will be ejected;ƒT= 10 × 1015 HzD. Electrons will be ejected;ƒT= 10 × 10-15 Hz

5. A light of what wavelength must be incident on a metal for the ejected electrons tohave a kinetic energy of 50 J? (The work function of the metal is 16 J, and h = 6.63 ×10-34 J × s.)

A. 3 × 10-27mB. 3 × 10-26

m

C. 1.0 × 10-34mD. 1.0 × 1035 m

6. In the hydrogen atom, how many electron states fall in the energy range -10.2 eVto -1.4 eV?

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

7. Which of the following statements is NOT consistent with Bohr’s set ofpostulates regarding the hydrogen atom model with regard to the emission andabsorption of light?

A. Energy levels of the electron are stable and discrete.B. An electron emits or absorbs radiation only when making a transition from one energy

level to another.C. To jump from a lower energy to a higher energy, an electron must absorb a photon of

precisely the right frequency such that the photon’s energy equals the energydifference between the two orbits.

D. To jump from a higher energy to a lower energy, an electron absorbs a photon of afrequency such the photon’s energy is exactly the energy difference between the twoorbits.8. What is the difference in ionization energy for a hydrogen atom with its electron inthe ground state and a hydrogen atom with its electron in the n = 4 state?

A. 0 eVB. 0.85 eVC. 12.75 eVD. 13.6 eV

When a hydrogen atom electron falls to the ground state from the n = 2 state, 10.2eV of energy is emitted. What is the wavelength of this radiation? (Use 1 eV = 1.60 ×10-19 J, h = 6.63 × 10-34 J × s.)



A. 1.22 × 10-7mB. 3.45 × 10-7 mC. 5.76 × 10-9 mD. 2.5 × 1015 m

10. Radiation is emitted from a small window in a large furnace. When thetemperature of the furnace is doubled, the peak emitted energy

A. remains constant.B. increases by a factor of 2.C. increases by a factor of 4.D. increases by a factor of 16.

11. The figure shown here illustrates an electron with initial energy of-10 eV movingfrom point A to point B. What change accompanies the movement of the electron?

A. Absorption of a photonB. Emission of a photonC. Decrease in the atom’s work functionD. Increase in the atom’s total energy

12. Ultraviolet light is more likely to cause a photoelectric effect than visible light.This is because photons of ultraviolet light

A. have a longer wavelength.B. have a higher velocity.C. are not visible.D. have a higher energy.

13. All of the following are characteristics of the photoelectric effect EXCEPTA. The intensity of the light beam does not affect the photocurrent.B. The kinetic energies of the emitted electrons do not depend on the light intensity.C. A weak beam of light of frequency greater than the threshold frequency yields more

current than an intense beam of light of frequency lower than the threshold.D. For light of a given frequency, the kinetic energy of emitted electrons increases as the

value of the work function decreases.Small Group Questions

1. Why does the amount of energy required to eject an electron from a metal need to bequantized?

2. Why are the energy levels of hydrogen described with negative numbers?Explanations to Practice Questions



1. BAlthough this question may seem complicated, it is actually rather straightforward.The ratio of the power per unit area radiated at the higher temperature to that at thelower temperature is: T4

higher /T4lower. Since Thigher = 760 K and Tlower = 380 K, you can

obtain the ratio (760/380)4 = 24 = 16, making (B) the correct answer. 2. AAssume that the skin over the tumor is also 1°C hotter than the skin over thenoncancerous tissue. Recall the Stefan-Boltzmann law for radiant energy, which saysthat radiant energy is proportional to the fourth power of the temperature (this law istrue only for absolute temperatures, so use the Kelvin scale). The skin over thetumor will therefore be (299 K)4/(298 K)4 = (299/298)4 14 1 times more radiant,or 1% “brighter.” (A) is therefore the correct answer. 3. CTo determine the speed of the electrons ejected, you must first calculate their kineticenergy:KE = hƒ-WKE = (6.63 × 10-34 J × s) × (1.0 × 1037 Hz) - 2,130 JKE = 6,630 J - 2, 130 JKE = 4,500 JUsing the formula for the kinetic energy, we can now calculate the speed of theejected electrons:

(C) matches the result and is thus the correct answer. 4. AFirst determine if electrons will be ejected. Each photon has an energy given byE hƒE (4 × 10-15eV•s)(2.5 × 1014s-1)E 1 eVTherefore, the photons do not have enough energy to allow an electron in the metalto overcome the 4.14 eV barrier, and there will be no ejection of electrons. You caneliminate (C) and (D). Next, from the relation between the work function and thethreshold frequency of a metal, W = hƒT

, we can solve for the threshold frequency:W = hƒT

ƒT = W/h



ƒT= 4.14 eV/(4.14 × 10-15 eV•s)ƒT= 1.0 × 1015 Hz(A) matches the result and is thus the correct answer. 5. ATo determine the wavelength of the light ray, first calculate its frequency from thefollowing equation:KE = hƒ - Wƒ = (KE + W)/hƒ = (50 J + 16 J)/(6.63 × 10-34 J × s)ƒ = 66 J/6.63 × 10-34 J × sƒ = 1035 HzNext, determine the wavelength of the incident ray of light by relating the frequencyto the speed of light:c = f

= c/ƒ= (3 × 108 m/s)/(1035 1/s)= 3 × 10-27 m

(A) best matches your result and is thus the correct answer. 6. BThe energy states of an electron in hydrogen are given by this equation:

En = -13.6 eV/n2nwhere n = 1, 2, 3,... For n = 1, E1 = -13.6 eV. For n = 2, E2 = -13.6 /4 = -3.4 eV.For n = 3, E3 =-13.6/9 = -1.5 eV. For n = 4, E4 = -13.6/16 = -0.85. Therefore, thereare two energy states in the range, -10.2 eV. to -1.4 eV, at n = 2 and n = 3. (B) istherefore the correct answer. 7. DBohr put forth a set of postulates that forms the basis of the model of the hydrogenatom with regard to the emission and absorption of light. First, energy levels of theelectron are stable and discrete; they correspond to specific orbits. Next, an electronemits or absorbs radiation only when making a transition from one energy level toanother. To jump from a lower to a higher energy orbital, an electron must absorb aphoton of precisely the right frequency such that the photon’s energy equals theenergy difference between the two orbitals. To jump from a higher to a lower energyorbital, an electron emits a photon of a frequency such the photon’s energy isexactly the energy difference between the two orbitals. From the given answerchoices, all of them are consistent with the Bohr theory except for (D), the correctanswer. 8. CThe energy required to ionize an atom is the energy needed to bring the electronenergy up to 0 eV. Using the equation for the energy states of an electron in ahydrogen atom, En = -13.6 eV/n2, calculate the ionization energies for the ground



state and the n = 4 state. For a hydrogen atom with its electron in the ground state (E1 = -13.6 eV), the ionization energy is 13.6 eV. For a hydrogen atom with its electronin the n = 4 state (E4 = -0.85 eV), the ionization energy is 0.85 eV. The difference inionization energies is (13.6 - 0.85) eV = +12.75 eV. (C) is therefore the correctanswer. 9. ATo solve this question correctly, you must first make sure to be consistent with theunits. As such, we have to convert ΔE = 10.2 eV to joules: ΔE = 10.2 eV (1.60 × 10-19 J/eV) = 1.6 × 10-18 J. Next, to determine the wavelength of the radiation, first findthe frequency using the following equation:ΔE = hfƒ =ΔE/hƒ = (1.5 × 10-18 J)/(6.5 × 10-34 J × s)ƒ ≈ 2.5 × 1015 HzLastly, from the wave equation c = ƒ, you can calculate the wavelength of theradiation:

= clƒ= (3 × 108 m/s)/(2.5 × 10” 1/s)≈ 1 × 10-7 m

(A) most closely matches your result and is thus the correct answer. 10. DA window in a furnace is an approximate replica of a blackbody system. In such asystem, the total emitted energy is proportional to the fourth power of thetemperature (according to the equation ET = σT4). If the temperature is doubled, thenthe total energy must increase by a factor of 16. Therefore, (D) is the correct answer. 11. BThe electron moves from a higher energy level to a lower energy level; this can onlyoccur if the extra energy is dissipated through the emission of a photon. (A)suggests the opposite. (C) is incorrect because the work function is equal to theamount of energy required to eject the electron, which increases as the electronbecomes more dif ficult to separate from the nucleus. (D) is incorrect because theenergy of the atom will decrease after it emits a photon. Therefore, (B) is the correctanswer. 12. DThe photoelectric effect occurs when a photon of sufficiently high energy strikes anatom with a sufficiently low work function. This means that a photon with higherenergy is more likely to produce the effect. Because ultraviolet light has a higherfrequency and lower wavelength than visible light, it also carries more energyaccording to the formula E = hƒ.



13. AThe greater the intensity, the greater the number of incident photons and, therefore,the greater the number of photoelectrons that will be ejected from the metal surface(provided that the frequency of the light is above the threshold). This means a largercurrent. (A) is not a characteristic of the photoelectric effect and is therefore theanswer to the question. All of the other choices are characteristics of thephotoelectric effect.



12Nuclear Phenomena

So here we are: the last chapter of this review of physics for the MCAT. We knowthat it has not been easy. We know that through the time spent with us in review ofthese important concepts for Test Day, there may have been tears shed, hair pulledout, foreheads banged against hard surfaces, screams and curses uttered, moretears, frustration, and exhaustion. Nevertheless, we are confident that, in the end, youhave learned what you need to understand for success on the physics problems ofthe MCAT, and we are hopeful that you have come to understand that physics istested because it is relevant to your future practice as a health care professional. Justas much, we are hopeful that you have enjoyed the learning and remembering journeywith us. Let us repeat: We know that it hasn’t been easy, but just because it hasn’tbeen easy doesn’t mean that you should have been miserable. We hope that ourattempts at bringing humor and real-life relevancy to the discussion have helped tolessen the anxiety, discomfort, fear, or boredom that some or many of you may havefelt as you opened this book for the first time. We hope that you will take from thesepages not only an appreciation for the ways in which the MCAT will test yourunderstanding of these basic physics concepts but also a true love and appreciationfor the complexity and beauty of the physical world, both seen and “unseen,” and ajoyful curiosity in the playfulness of the world that surrounds us.



This last chapter reviews the organization of the atomic nucleus and describesnuclear phenomena. We will begin with a review of some of the standardterminology used in nuclear physics, then discuss the concept of binding energy andthe related concept of the mass defect. We will investigate what Einstein’s famousdictum, E = mc2, really means. The remainder of the chapter focuses on nuclearreactions, fission and fusion, and radioactive decay, which will be presented in twoparts. The first part deals with the four types of radioactive decay and a discussionof the reaction equations that describe them. The second part covers the generalproblem of determining the half-life of a decay process and the associatedcalculations of the number of nuclei that remain after a period of time.



ConclusionOur final chapter examined the characteristics and behavior of the atomic nucleus.We began by defining some of the important terms that are used to describe thenucleus and the atom, including atomic number, which is the number of protons inthe nucleus and is a unique identifier for each element; atomic mass, which is thetotal number of protons and neutrons in the nucleus; isotopes, which are atoms ofthe same element that differ in the number of neutrons and therefore have the sameatomic number but different mass numbers; and finally, atomic mass and atomicweight, which are defined as the mass of a single atom (in amu) and the mass of amole of atoms (in grams) of a given element, respectively. We reviewed the relationbetween the binding energy and the mass defect and used Einstein’s equation E =mc2 to understand that the energy released when the nucleons are bound together bythe strong nuclear force is from the conversion of a small amount of mass of theconstituent nucleons. We reviewed the different types of radioactive decay in whichthe nucleus decays and emits a specific type of particle, such as an alpha particle,electron, positron, or gamma photon. Finally, we concluded our discussion ofnuclear decay with a brief consideration of radioactive decay half-life and calculationof the remaining radioactive nuclei as a function of time. Congratulations! You did it! You have accomplished so much in completing thisreview of physics for the MCAT. We know that you’ve worked hard and investedmuch time, effort, and attention to the concepts reviewed in this book. We know thatyour hard work will pay off in points on Test Day. You’ve done much, but there’smore to do. Continue to review the concepts discussed in these chapters andpractice applying them to MCAT passages and questions. Success on Test Day isbased not just on your ability to recall important facts, figures, equations, or laws,but also—and even more importantly—on your ability to think critically, analyze newinformation, and synthesize your funds of knowledge with the passage and questionpresentation.

CONCEPTS TO REMEMBERAtomic number is the number of protons in the nucleus; all the atoms of a given

element have the same number of protons, and no two elements have the sameatomic number.

Atomic mass is the number of protons and neutrons in the nucleus. Atoms of thesame element may have different numbers of neutrons and, therefore, will have thesame atomic number but different mass numbers. Atoms of a given element thathave different mass numbers are called isotopes.

The atomic mass is the mass of a single atom of a given element in atomic mass units(amu). The average mass of a single proton or neutron is one amu, so the atomicmass of an atom is equal to the mass number of that atom. Atomic weight is themass of one mole of atoms of a given element in grams. One mole of an atom or acompound is the number of atoms or molecules equal to 6.022 × 1023 (Avogadro’snumber).



Nuclear binding energy is the amount of energy that is released when the nucleons(protons and neutrons) bind together through the strong nuclear force. The morebinding energy per nucleon released, the more stable the nucleus. The mass defect isthe difference between the mass of the unbound constituents and the mass of thebound constituents in the nucleus. The unbound constituents have more energy and,therefore, more mass than the bound constituents. The mass defect is the amount ofmass converted to energy through the nuclear reactions of fusion or fission.

Fusion occurs when small nuclei combine into larger nuclei. Fission occurs when alarge nucleus splits into smaller nuclei. Energy is released in both processes since thenuclei formed in both processes are more stable.

Radioactive decay includes alpha decay (emission of an alpha particle), beta decay(decay of a neutron and emission of electron), positron decay (decay of proton andemission of positron), gamma decay (emission of gamma photon), and electroncapture (capture of electron which combines with proton to form neutron).

Radioactive decay half-life is the amount of time required for half a sample ofradioactive nuclei to decay.

The rate at which radioactive nuclei decay is proportional to the number of nuclei thatremain.

EQUATIONS TO REMEMBERn = noe-λt (exponential decay)

Practice Questions1. Which of the following correctly identifies the following process?

A. β- decayB. α decayC. e- captureD. γ decay

2. Consider the following fission reaction.

The masses of the species involved are given in atomic mass units below eachspecies, and 1 amu = 932 MeV What is the energy liberated due to transformation ofmass into energy?

A. 0.003 MeVB. 1.4 MeVC. 2.8 MeVD. 5.6 MeV

3. Calculate the binding energy of the argon-40 isotope in MeV using the followinginformation: 1 proton = 1.0073 amu, 1 neutron = 1.0087 amu, 40Ar = 39.9132 amu, c2

= 932 MeV/amu.A. 381.7 MeVB. 643.8 MeVC. 0.4096 MeVD. 40.3228 MeV



4. An α particle and a β+ particle (positron) possess the same kinetic energy. What is theratio of the velocity of the β+ particle to that of the α particle? (Assume that theneutron mass is equal to the proton mass and neglect binding energy.)

5. A student is trying to determine the type of nuclear decay for a reaction. Which of thefollowing would be an indication that the reaction he is observing is a gamma decay?

A. Zdaughter Zparent + 1; Adaughter =Aparent

B. Zdaughter = Zparent - 1; Adaughter = Aparent

C. Zdaughter =Zparent -2; Adaughter=Aparent-4 D. Zdaughter =Zparent; Adaughter =Aparent

6. Element X is radioactive and decays via α decay with a half-life of four days. If 12.5percent of an original sample of element X remains after N days, what is the value ofN?

A. 4B. 8C. 12D. 16

7. A patient undergoing treatment for thyroid cancer receives a dose of radioactive iodine(131I), which has a half-life of 8.05 days. If the original dose contained 12 mg of 131I,what mass of 131I remains after 16.1 days?

A. 3 mgB. 6 mgC. 9 mgD. 12 mg

8. In an exponential decay, if the natural logarithm of the ratio of intact nuclei (n) at time tto the intact nuclei at time t = 0 (no) is plotted against time, what does the slope ofthe graph correspond to?

A. λB. -λC. e-λt

D. n/no

9. The mass of a proton is about 1.007 amu, and the mass of a neutron is about 1.009amu. The mass of a helium nucleus is

A. less than 4.032 amu.B. exactly 4.032 amu.C. greater than 4.032 amu.D. Cannot be determined from the information given.

10. A certain carbon nucleus dissociates completely into α particles. How many particles



are formed?A. 1B. 2C. 3D. 4

11. The reaction here is an example of

A. α decay.B. β decay.C. γ decay.D. x-ray decay.

12. The half-life of 14C is approximately 5,730 years, while the half-life of 12C is essentiallyinfinite. If the ratio of 14C to 12C in a certain sample is 25% less than the normal ratioin nature, how old is the sample?

A. Less than 5,730 yearsB. Approximately 5,730 yearsC. Significantly greater than 5,730 years, but less than 11,460 yearsD. Approximately 11,460 years

13. A nuclide undergoes 2 alpha decays, 2 positron decays, and 2 gamma decays. Whatis the difference between the atomic number of the parent nuclide and the atomicnumber of the daughter nuclide?

A. 0B. 2C. 4D. 6

14. A helium nucleus fuses with a hydrogen nucleus and then captures an electron. Whatis the identity of the daughter nuclide?

A. 5HeB. 5He+C. 5LiD. 5Li+

Small Group Questions1. Creating bonds releases energy, while holding nucleons together requires energy

(binding energy). Why do these two processes follow different paths?2. Two nuclei have the same radius. Does this mean they have the same number of

protons and neutrons? Why or why not?Explanations to Practice Questions

1. CThis process can be described as electron capture. Certain unstable radionuclidesare capable of capturing an inner electron that combines with a proton to form aneutron. The atomic number becomes one less than the original, but the massnumber remains the same. Electron capture is a relatively rare process and can bethought of as inverse β- decay. Notice that the equation is similar to that of β+ decay



but not identical. (C) is therefore the correct answer. 2. CThis problem presents a reaction and asks for the energy liberated due totransformation of mass into energy. To convert mass into energy, we are given theconversion factor 1 amu = 932 MeV (Note that this is actually the c2 from Einstein’sequation E = Δmc2). All we need to do now is calculate how much mass, in amu, islost in the reaction. Because we are given the atomic mass for each of the elements inthe reaction, this is simply a matter of balancing the equation:

This is the amount of mass that has been converted into kinetic energy. To obtainenergy from mass, we have to multiply by the conversion factor (1 amu = 932 MeV):

KE = 0.003 × 932 ≈ 3 MeV(C) best matches our prediction and is thus the correct answer. 3. AGlancing at the periodic table, you can see that argon has an atomic number of Z=18, so we can determine the number of neutrons: A - Z= 40 - 18 = 22. Therefore, themasses of the constituent nucleons of 40Ar add up as follows:

Therefore, if the actual mass of 40Ar is 39.9132 amu, then the mass defect isΔm = 40.3228 - 39.9132 = 0.4096 amu

Convert this value into binding energy in MeV:E= Δmc2

E= (0.4096)(932) = 381.7 MeV(A) matches our prediction and is thus the correct answer. 4. BAn α particle is a helium nucleus and therefore contains 2 protons and 2 neutrons.Assuming the neutron mass equals the proton mass, and neglecting binding energy,the mass of an α particle is 4mproton, where mproton is the mass of a proton. A β+ particleis a positron (Recall that a positron has the same mass but opposite charge to that ofan electron). Thus, the mass of a β+ particle is melectron, where melectron is the mass of anelectron. Given that the particles have equal kinetic energy, you have

Solving for νβ /να gives



(B) is therefore the correct answer. 5. DGamma decay is the emission of a γ particle, which is a high-energy photon. In thisprocess, the mass number and the atomic number remain the same. Therefore, (D) isthe correct answer. 6. CBecause the half-life of element X is four days, 50 percent of an original sampleremains after four days, 25 percent remains after eight days, and 12.5 percentremains after 12 days. Therefore, N = 12 days. Another approach is to set (½)n =0.125, where n is the number of half-lives that have elapsed. Solving for n gives n =3. Thus, 3 half-lives have elapsed, and because each half-life is four days, we knowthat N = 12 days, making (C) the correct answer. 7. AGiven that the half-life of 131I is 8.05 days, you know that 2 half-lives have elapsedafter 16.1 days, which means that 25 percent, or one-fourth, of the original amountof 131I is still present. Therefore, only 25 percent of the original number of 131I nucleiremains, which means that only 25 percent of the original mass of 131I remains.Because the original dose contained 12 mg of 131I, only 3 mg remain after 16.1 days.(A) is therefore the correct answer. 8. BThe expression n = noe-λt is equivalent to nlno = e-λt. Taking the natural logarithm ofboth sides of the latter expression, you find

ln(n/no) = -λtFrom this expression, it is clear that plotting ln(n/no) versus t will give a straight linewith a slope of -λ. (B) is therefore the correct answer. 9. AThe mass of the protons and neutrons is (1.007 amu/proton) (2 protons) + (1.009amu/neutron)(2 neutrons) = 4.032 amu. Some of this mass is converted to energy inorder to overcome the binding energy of the nucleus, so the overall mass must beless than 4.032 amu. Therefore, (A) is the correct answer. 10. CA typical carbon nucleus contains 6 protons and 6 neutrons. An α particle contains 2protons and 2 neutrons. Therefore, one carbon nucleus can dissociate into 6/2, or 3α particles, making choice (C) the correct answer. 11. BP decay occurs when an electron or a positron is released by the nucleus. Thismeans that a proton is converted to a neutron or vice versa. Therefore, a β particle is



emitted in any reaction in which the atomic number of the parent atom increases ordecreases by 1 while the mass number stays the same. 12. ABecause the half-life of 12C is essentially infinite, a 25 percent decrease in the ratio of14C to 12C means the same as a 25 percent decrease in the amount of 14C. If less thanhalf of the 14C has deteriorated, then less than one half-life has elapsed. Therefore,the sample is less than 5,730 years old, making (A) the correct answer. 13. DIn alpha decay, an element loses two protons. In positron decay, a proton isconverted to a neutron. Gamma decay, meanwhile, has no impact on the atomicnumber of the nuclide. Therefore, two alpha decays and two positron decays willyield a daughter nuclide with six less protons than the parent. 14. AThe fusion of a hydrogen nucleus (one proton) and a helium nucleus (two protons,two neutrons) will produce 5Li. If the 5Li nucleus captures an electron, a proton willbe converted to a neutron, producing 5He. Although the atom will initially carry apositive charge in most cases, the positive charge is a property of the atom and notof the nuclide.



13High-Yield Problem Solving Guide for Physics

High-Yield MCAT ReviewThis is a High-Yield Questions section. These questions tackle the most frequentlytested topics found on the MCAT. For each type of problem, you will be providedwith a stepwise technique for solving the question and key directional points on howto solve for the MCAT specifically.



At the end of each topic you will find a “Takeaways” box, which gives a concisesummary of the problem-solving approach, and a “Things to Watch Out For” box,which points out any caveats to the approach discussed above that usually lead towrong answer choices. Finally, there is a “Similar Questions” box at the end so youcan test your ability to apply the stepwise technique to analogous questions.



We’re confident that this guide can help you achieve your goals of MCAT successand admission into medical school!



Good luck!



Part IIPractice Sections

INSTRUCTIONS FOR TAKING THE PRACTICE SECTIONS



Before taking each Practice Section, find a quiet place where you can uninterrupted.Take a maximum of 70 minutes per section (52 questions) to get accustomed to thelength and scope.



Keep in mind that the actual MCAT will not feature a section made up of Physicsquestions alone, but rather a Physical Sciences section made up of both Physics andGeneral Chemistry questions. Use the following three sections to hone your Physicsskills.



Good luck!



Practice Section 1Time—70 minutes



Answers and Explanations



GlossaryAbsolute pressureThe actual pressure.AccelerationA vector quantity; the time-rate of change in velocity.AdiabaticA process in which heat flows neither into nor out of a system.Alternating currentA current that changes directions periodically, often in a sinusoidal fashion.Alpha (α) particleA helium nucleus.AmplitudeThe maximum displacement from an equilibrium position; the magnitude of themaximum disturbance in a wave.AntinodePoints of maximum amplitude in a standing wave.Archimedes’ principleA body immersed in fluid experiences an upward buoyant force equal to the weightof the displaced fluid.Atomic massThe mass of an atom of an element in amu (atomic mass units).Atomic mass unit (amu)

One-twelfth (1/12) of the mass of the carbon-12 atom.Atomic numberThe number of protons in a nucleus; this number characterizes each element.Atomic weightThe average mass of an element’s atoms—a weighted average of the differentisotopes, weighted according to each isotope’s naturally occurring abundance.BeatsThe sound produced by the alternating constructive and destructive interferencebetween sound waves of slightly different frequencies.Beta (β) particleAn electron; usually refers to one emitted in radioactive beta decay.Binding energyThe energy required to separate an electron from an atom or to completely separatethe protons and neutrons in a nucleus.Buoyant forceThe upward force felt by a body partially or wholly submerged in a fluid that is equalto the weight of the fluid that the submerged body has displaced (see alsoArchimedes’ principle).CapacitanceA measure of the ability of a capacitor to store charge; the absolute value of themagnitude of the charge on one plate divided by the potential difference between theplates.CapacitorTwo conducting surfaces that store equal and opposite charges when connected to avoltage source.



Center of gravityA point such that the entire force of gravity can be thought of as acting at that point.If the acceleration of gravity is constant, then the center of gravity and the center ofmass are at the same point.Center of massThe point that acts as if the entire mass was concentrated at that point.Centripetal accelerationThe acceleration of an object that travels in a circle at a constant speed. Themagnitude of the acceleration is equal to the velocity squared divided by the radiusof the circle. It is always directed towards the center of the circle. (If the velocity isnot constant, there is a second component of acceleration tangent to the circle.)Centripetal forceThe force responsible for the centripetal acceleration. Its magnitude is equal to theproduct of the mass and square of the velocity divided by the radius of the circle.Like centripetal acceleration, it is always directed towards the center of the circle,provided that the velocity of the object is constant.ConductorA material, such as metal, in which electrons can move relatively freely.Conservative forceA force is conservative if the work done on a particle in any round trip is zero, or ifthe amount of work done by moving a particle from one position to another isindependent of the path taken.ConvectionThe transfer of heat through the bulk motion of the heated material.Critical angleThe critical angle is the angle of incidence such that the refracted angle is 90° whenlight is going from a medium having a higher index of refraction into a mediumhaving a lower index of refraction.Decay constantThe proportionality constant between the rate at which radioactive nuclei decay andthe number of radioactive nuclei remaining.DecibelA unit of sound level.DensityMass per unit volume.DielectricA nonconducting material often placed between plates of a capacitor to increase thecapacitance.DiffractionThe bending of waves as a result of passing through a slit; the bending of wavesaround an obstacle.DiopterA measure of a lens’s power, defined as the inverse of the focal length in meters.Dipole momentA vector whose magnitude is the product of the charge of the dipole and thedistance separating the two charges. For physicists, the vector points from thenegative charge towards the positive charge (chemists sometimes use the reverse



direction).DispersionThe variation of wave-speed with frequency in a medium; the separation of visiblelight into its constituent colors (for example, by a prism) as a result of this variation.DisplacementA vector quantity; the straight-line distance and direction going from some initialposition to some final position.Doppler effectThe change in frequency of a wave as a result of the motion of the source and/orobserver along the line joining them.Electric currentThe net charge per unit time passing through a given cross section; by convention,the direction is that in which positive charge would flow.Electric dipoleTwo equal and opposite electric charges separated by a small distance.Electric fieldThe electrical force on a stationary positive test charge divided by the charge.Electric potentialThe work needed to move a positive test charge from infinity to a given point in anelectric field divided by the charge. The electric potential can also be considered theelectric potential energy per unit charge.Electromagnetic waveA transverse wave of changing electric and magnetic fields.Electromagnetic spectrumThe full range in frequency and wavelength of all electromagnetic waves.Electromotive forceThe voltage across the terminals of a cell or battery when no current is flowing.EnergyThe ability to do work.EntropyThe measure of a system’s disorder.EquilibriumSee Rotational equilibrium, Translational equilibrium.Excited state An atom in which an electron occupies an energy state above the minimum energystate or ground state; a nucleus that is in an energy level above its ground stateenergy level.Field lineA technique used to better visualize electric or magnetic field patterns. The tangent toa field line at any point is the direction of the field itself at that point. The more fieldlines per unit area, the greater the magnitude of the field.FissionThe splitting of a heavy nucleus into two or more lighter nuclei accompanied by therelease of energy.FluorescenceA process in which certain substances emit visible light when excited by otherradiation, usually ultraviolet radiation.



Focal lengthThe distance from a mirror or lens to the focal point.Focal pointThe point at which rays of light parallel to the optic axis converge, or appear todiverge from, when reflected from a mirror or refracted by a lens.FrequencyThe number of cycles per second; the number of wavelengths of a travelingsinusoidal wave passing a fixed point per second.FrictionThe force that two surfaces in contact exert on each other in a direction parallel totheir surfaces and opposite to their motion.Fundamental frequencyThe lowest frequency at which a standing wave can be produced on or in a body.FusionThe combining of lighter nuclei into a heavier nucleus accompanied by the release ofenergy.Gamma (γ) radiationHigh-energy photons, often emitted in radioactive decay.Gauge pressureThe difference between the absolute pressure and atmospheric pressure.GravityA fundamental force of attraction between all matter. Its magnitude is directlyproportional to the product of the masses and inversely proportional to the square ofthe distance between their centers.Ground stateThe lowest energy state of an atom or nucleus.Half-lifeThe time in which one-half of the radioactive nuclei that were originally presentdecay.HeatThe energy that is transferred between two objects as a result of a difference intemperature.Heat conductionThe transfer of heat energy through a body without bulk motion of the materialwithin the body.ImpulseA vector quantity; the force acting on an object multiplied by the time the force acts;also the change of momentum of an object.Index of refractionThe ratio of the speed of light in a vacuum to the speed of light in a medium.InertiaAn object’s resistance to a change in its motion when a force is applied.InsulatorA material in which electrons do not move freely.IntensityThe average rate per unit area of energy transported by a wave across aperpendicular surface.Interference



The combined effects of two waves—constructive when the waves are in phase,destructive when they are out of phase.IsobaricA process in which the pressure of a system remains constant.IsothermalA process in which the temperature of a system remains constant.IsotopesAtoms of a given element whose nuclei have the same number of protons but adifferent number of neutrons and, therefore, have the same atomic number butdifferent mass numbers.Kinetic energyThe energy a body has as a result of its motion.Line of forceSee Field line.Longitudinal waveA wave in which the oscillation is parallel to the direction of propagation.Magnetic fieldA magnetic property associated with a point; the maximum magnetic force on amoving positive test charge divided by its charge and its velocity.MassA measure of a body’s inertia.Mass defectThe difference between the sum of the masses of neutrons and protons forming anucleus and the mass of that nucleus.Mass numberThe total number of protons and neutrons in the nucleus of an atom.MomentumA vector quantity; mass times velocity.NodePoints where the displacement of a standing wave remains zero at all times.NormalIn optics, the normal is a line drawn perpendicular to the boundary between twomedia.Normal forceThe force that two surfaces in contact exert on each other in a directionperpendicular (normal) to the area of contact.NucleonA member of the nucleus, either proton or neutron.Pascal’s principleThe pressure applied to an enclosed fluid is transmitted undiminished to everyportion of the fluid and to the walls of the vessel containing the fluid.PeriodThe time necessary to complete one cycle.Polarized lightLight that has all of its electric field vectors parallel.PositronAntiparticle of an electron; it has the same mass as an electron and a charge that isequal to the charge of an electron in magnitude but opposite in sign (i.e., positive).



Potential differenceThe difference of electrical potential between two distinct points; the work needed tomove a unit positive charge between two points.Potential energyThe energy that a body has as a result of its position when a conservative force isacting on it.PowerThe time-rate at which work is done or energy is transferred.PressureForce per unit area.QuantumA discrete bundle of energy, such as a photon (particle of light).RadiationThe transfer of energy by electromagnetic waves.ResonanceThe phenomena in which an oscillatory system absorbs, with extremely highefficiency, the energy transmitted by an external oscillatory force having a frequencyequal to one of the system’s natural frequencies.ReflectionThe change in direction of a wave at the boundary of a medium when the waveremains in the medium.RefractionThe change in direction of a wave as it passes obliquely through a boundary fromone medium to another.ResistanceA characteristic property of a conductor, which measures the opposition to currentflowing through it; the ratio of the voltage applied to a conductor and the current thatresults.ResistivityA measure of the intrinsic resistance of a material independent of its shape or size.rms currentThe maximum current divided by the square root of 2; used to measure the currentin an AC circuit in lieu of the average current, which is zero.RotationThe turning of an extended body about an axis or center; change in orientation.Rotational equilibriumThe condition where the sum of the torques acting on a body is zero; also called thesecond condition of equilibrium.ScalarA quantity that has a magnitude but not a direction; an ordinary number.Simple harmonic motionPeriodic motion about an equilibrium position resulting from a linear restoring force.Sound levelTen times the logarithmic ratio of the intensity of a given sound to the intensity of thefaintest sound that can be heard by humans.Specific gravityThe ratio of the density of a substance to the density of water.Specific heat



The number of calories needed to raise the temperature of one gram of a substanceby 1°C.SpeedA scalar quantity; the instantaneous rate at which distance is being covered by amoving object.Standing waveA wave with a displacement that appears fixed in space and time. It can be producedwhen two waves of the same frequency, amplitude, and speed travel in oppositedirections.Superposition principleThe resulting displacement of a medium at a point where two or more waves meet isthe algebraic sum of the individual displacements of each wave at that point.Thermal expansionThe change in an object’s size with a change in temperature. In general, objectsexpand as the temperature increases.TorqueThe magnitude of the force acting on a body times the perpendicular distancebetween the direction of the force and the point of rotation.TranslationMotion through space without a change in orientation.Translational equilibriumThe condition in which the sum of the forces acting on a body is zero, also calledthe first condition of equilibrium.Transverse waveA wave in which the oscillation is perpendicular to the direction of propagation.VectorA quantity that has both magnitude and direction.VelocityA vector quantity whose magnitude is speed and whose direction is the direction ofmotion.ViscosityA fluid’s internal resistance to flow.WavelengthThe distance between two corresponding points in successive cycles of a sinusoidalwave; the distance from one crest to the next.WeightThe force of gravity on an object, not to be confused with mass.WorkA scalar quantity; the force acting on an object times the distance the object movestimes cos θ, where θ is the angle between the force and the direction of motion.



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