AP AP Success Physics

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4th edition

John W. Dooley, Ph.D.Matthew G. Sexton, M.S.Steven O. Nelson, M.S.Gabriel Lombardi, Ph.D.

Jay Streib, Ph.D.


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About The Thomson Corporation and Petersons

The Thomson Corporation, with 2002 revenues of US$7.8 billion, is a global leader in providing integratedinformation solutions to business and professional customers. The Corporations common shares are listedon the Toronto and New York stock exchanges (TSX: TOC; NYSE: TOC). Its learning businesses and brandsserve the needs of individuals, learning institutions, corporations, and government agencies with productsand services for both traditional and distributed learning. Petersons (www.petersons.com) is a leadingprovider of education information and advice, with books and online resources focusing on education search,test preparation, and financial aid. Its Web site offers searchable databases and interactive tools forcontacting educational institutions, online practice tests and instruction, and planning tools for securingfinancial aid. Petersons serves 110 million education consumers annually.

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ISBN: 0-7689-1265-2

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Fourth Edition


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CONTENTS

INTRODUCTION ......................... v

RED ALERT: STUDY PLAN ............. 1Diagnostic Test ............................................................................ 11

General Physics ................................................................... 11

Mechanics ............................................................................ 16

Electricity and Magnetism ................................................... 21

Answers and Explanations .........................................................26

AP PHYSICS REVIEW

Unit 1 Newtonian Mechanics

Chapter 1. Kinematics.......................................................... 41

Chapter 2. Newtons Laws of Motion..................................55

Chapter 3. Work, Energy, Power ......................................... 59

Chapter 4. System of Particles, Linear Momentum..............63

Chapter 5. Circular Motion and Rotation ............................ 67

Chapter 6. Oscillations and Gravitation .............................. 73

Unit 2 Thermal Physics

Chapter 7. Temperature and Heat ........................................81

Chapter 8. Kinetic Theory and Thermodynamics ................89

Unit 3 Electricity and Magnetism

Chapter 9. Electrostatics...................................................... 97

Chapter 10. Conductors, Capacitors, Dielectrics ..............105

Chapter 11. Electric Circuits ............................................. 109

Chapter 12. Magnetostatics................................................ 117

Chapter 13. Electromagnetism ........................................... 121Unit 4 Waves and Optics

Chapter 14. Wave Motion ..................................................131

Chapter 15. Physical Optics .............................................. 137

Chapter 16. Geometric Optics ........................................... 141


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AP Success: Physics B/C

CONTENTS

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Unit 5 Atomic Physics and Quantum Effects

Chapter 17. Atomic Physics and Quantum Effects .............149

Chapter 18. Nuclear Physics..............................................153

PRACTICE TESTSPhysics B, Practice Test 1 .......................................................... 159

Answers and Explanations ..................................................180

Physics B, Practice Test 2 .......................................................... 199


Physics C, Practice Test 1 .......................................................... 239


Physics C, Practice Test 2 .......................................................... 293


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CONTENTS

vAP Success: Physics B/C

INTRODUCTION

ABOUT THIS BOOK

The AP Physics exam may appear daunting at first, but if youve prepared forthe test throughout the year and take the time to use this book properly, itshould not be that difficult. We have tried to make this a workable book. Inother words, the book is set up so that you will be able to find the materialthat is necessary to study and be fully prepared when its time to take theactual test.

The book begins with a Physics Diagnostic Test. The purpose of the Diag-nostic Test is to help you get a handle on what you know and what needsmore work. We have included material from the General Physics section, aswell as questions from both the Physics B and Physics C exams. Take thisexam (and all of the tests) under simulated exam conditions, if you can. What

this means is that you should find a quiet place in which to work set up a time or a clock take the test without stopping

When you are finished, take a break and then go back and check youranswers. Always reread those questions you got wrong, since sometimeserrors come from merely misreading the question. Again, double-check youranswers, and if theyre still not clear, read the review material. We alsosuggest that you answer all of the questions, regardless of the version of the

exam you plan to take.Once youve completed the Diagnostic Test, its time to move on to thephysics review. Study the material carefully, but feel free to skim portions ofthe review section that are easy for you. There are eighteen chapters in all.In fact, before you begin any of this work, it would be helpful to consult thesuggested study plans that follow this introduction.

Then, take the actual practice tests. There are two practice exams for PhysicsB and two practice exams for Physics C. These tests are designed to give youan idea of the types of questions you will encounter on the exam. While theseare not actual exams, the questions themselves are the same types of ques-tions you will find the on the actual AP Physics tests.

As you complete each exam, take some time to review your answers. Wethink youll find a marked improvement in your scores from the time you takethe Diagnostic Test to the time you complete all of the full-length practicetests. As you go through the tests, circle those answers that you are not sureof, so you can go back and review them. Always take the time to check thereview section for clarification, and if you still dont understand the material,go to your teacher for help.


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INTRODUCTION

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ABOUT THE TESTS

The Physics B exam is 3 hours long. The first section contains 70 multiple-choice questions, and the second is a free-response section that contains 6 to8 questions. You will have 1 hours to take each section of this test.

The Physics C exam also consists of two parts, each 1 hours long. Onepart covers mechanics and the other part covers electricity and magnetism.You may take either part, or you may take both parts, and you will getseparate grades for each section. There are 35 multiple-choice questions ineach section, and each part has a free response section as well. There areusually three questions in each free response section.

TAKING THE TEST

1. Since you will have 3 hours in which to complete the exam, it is important

to pace yourself. You should also make sure you are thoroughly familiarwith the directions for the tests so that you dont have to waste time tryingto understand them once youve opened your test booklet.

2. Work through the easy questions first. The faster you complete thosequestions, the more time youll have for those that are more difficult. Youmay use your test book for scratch paper, but keep your answer sheetclean, since they are machine-readable, and any stray marks might beconstrued as an answer.

3. The multiple-choice questions have five lettered choices. As with anymultiple-choice question, you should approach each one by first trying to

select the correct answer. If the answer is clear to you, select it at once. Ifyoure unsure, the first technique is the process of elimination. Try tocross off any answers that dont seem to make sense or that you know arecompletely wrong. This improves your odds of guessing the correctanswer. If, for example, you can eliminate three choices, you have a 50/50chance of guessing the correct answer. Otherwise, if you cant eliminateany choices, you have only a 20 percent, rather than a 50 percent, chanceof getting the answer correct. The penalty for an incorrect answer is one-quarter point, so it may be advisable to guess.

With diligent studying, careful preparation, and a positive attitude, you canhelp yourself succeed.

Good luck!


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RED ALERT1

AP PHYSICS STUDY PLAN

When you begin to study for your AP Physics exam, the most important thingyou should have is a plan. The first thing you should do is to estimate howmuch time you have before exam day. The more time, the better. If, how-ever, youre somewhat short on time, this study plan will be extremelyvaluable for you. We offer these different study plans to help maximize yourtime and studying. The first is a 12-Week Plan, which involves concentratedstudying and a focus on the sample test results. The second is the more

leisurely 24-Week Plan, the one thats favored by schools. Finally, if time isrunning short, you should use the Panic Plan. We dont want you to reallypanicthis plan is supposed to help you conquer that panic and help youorganize your studying so that you can get the most out of your review workand still be as prepared as possible.

These plans are supposed to be flexible and are only suggestions. Feel free tomodify them to suit your needs and your own study habits. But start immedi-ately. The more you study and review the questions, the better your resultswill be.

THE 12-WEEK PLAN2 LESSONS PER WEEK

WEEK1Diagnostic Test. The AP Physics Diagnostic Test is designed to help youdetermine what you need to know and where to focus your study. Take thistest under simulated test conditions in a quiet room and keep track of the timeit takes to complete the test. The test consists of three sections: GeneralPhysics,Mechanics, andElectricity and Magnetism. Each section consistsof fourteen multiple choice questions and one free-response question. Re-

gardless of which specific test you intend to take, you should answer all of thequestions on this test to get an idea of your weakest areas.

Diagnostic TestAnswers. Once you have completed the test, carefullycheck all of your answers, and read through the explanations. This may takequite a bit of time, as will all of the tests, but it will enable you to select those

Lesson 1Lesson 1Lesson 1Lesson 1Lesson 1


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subject areas that you should focus on and spend the most amount of timestudying. With this information, you can start reviewing the chapters in therest of the book. If youre short on study time, use the results of this test tofocus your study efforts on the specific chapters in the review section that willbetter help you understand the material that you missed on the test.

WEEK2Chapter One: Kinematics. The review section of this book consists ofeighteen chapters. Its an enormous amount of work, so youll have to beextremely diligent about reviewing this material. These chapters fall underfive major sections:Newtonian Mechanics, Thermal Physics,Electricity andMagnetism, Waves and Optics, andAtomic and Nuclear Physics.

Kinematics is the first chapter under the Newtonian Mechanics section, and50 percent of the C-Level test consists of Newtonian Mechanics questions.Take your time to read through the first chapter. Underline or use a marker to

highlight those areas that are unclear to you.

Chapter Two: Newtons Law of Motion. Again, read through this chapter,mark whatever is unclear, and go back and read the material again, if neces-sary.

WEEK3Chapter Three: Work, Energy, Power. As you continue your lessons, try tostudy in a quiet room, uninterrupted by others in your household or the TV,radio, or any outside noises.

Chapter Four: System of Particles, Linear Momentum. Again, readthrough this chapter, mark whatever is unclear, and go back and read thematerial again, if necessary.

WEEK4Chapter Five: Circular Motion and Rotation. You can, of course, breakthese lessons into sections. Work on half the chapter in the morning and theother half in the afternoon.

Chapter Six: Oscillations and Gravitations. Read through this chapter,mark whatever is unclear, and then go back and read the material again, if

necessary. You can always ask your teacher for additional information ifyoure having difficulty.

If you are taking the C-Level exam, you might want to spend the next weekreviewing chapters one through six.








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RED ALERT

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WEEK5Chapter Seven: Temperature and Heat. This is the first of two chaptersunder the Thermal Physics section.

Chapter Eight: Kinetic Theory and Thermodynamics. If you find thatyouve finished reading and reviewing Temperature and Heat with time tospare, you can double up and complete Chapter 8. You will then have moretime to reread these two chapters and go to the next lesson.

WEEK6Chapter Nine: Electrostatics. The third unit isElectricity and Magnetism,and this material in the next five chapters represents 50 percent of the C-Levelexam, so it pays to focus heavily on these chapters.

Chapter Ten: Conductors, Capacitors, Dielectrics. This is the secondchapter in this unit. Take your time to make sure you fully understand all ofthe material.

WEEK7Chapter Eleven: Electric Circuits. This is the midway point of this unit ifyoure preparing only for the C-Level exam. These questions onElectricityand Magnetism also represent at least 25 percent of the B-Level exam, so itsimportant to understand what youre studying.

Chapter Twelve: Magnostatics.Again, if you find yourself finished with asection faster than you anticipated, or the pace of this study plan is too slow,feel free to add additional reading to your lessons.

WEEK8Chapter Thirteen: Electromagnetism. This is the last chapter of this unit,and if youre taking the C-Level exam, you have several choices. You caneither reread the material in the two major units that are covered on yourexam, you can skip to the final tests given at the end of this book, or you cancontinue to read through the rest of the chapters to make sure you have acomplete understanding of AP Physics.

Chapter Fourteen: Wave Motion. This chapter is part of the Waves andOptics unit that consists of three chapters.

WEEK9Chapter Fifteen: Physical Optics. If you find these chapters difficult, youmight want to take a break from your reading. Give yourself a day or two tojust relaxassuming you have the time.











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Chapter Sixteen: Geometric Optics. This has been a very concentratedperiod of study, and youre almost done. Make sure to keep highlightinganything you dont fully understand.

WEEK10Chapter Seventeen: Atomic Physics and Quantum Effects. This chapter isthe first in theAtomic and Nuclear Physics unit. These last two chaptersrepresent about 15 percent of the B-Level test.

Chapter Eighteen: Nuclear Physics. The end of a long study road. This isthe last review chapter. If you have time to spare when youve completed allof these chapters, you might want to go back and check any topics orquestions that you didnt understand, and make an appointment with yourteacher to go over these topics.

WEEK11AP Physics Practice Test 1, Level B. Take this test and answer all of thequestions you can, and then guess at those you dont know. Circle the ques-tions that you guessed at so that you can zero in on those specific answers.Its important to evaluate what you know. Check all of your answers.

AP Physics Practice Test 2, Level B. Take this test and answer all of thequestions you can, and then guess at those you dont know. Circle thosequestions that you guessed at so that you can zero in on those specific an-swers. Check all of your answers.

WEEK12AP Physics Practice Test 1, Level C. Take this test and answer all of the

questions you can, and then guess at those you dont know. Circle thosequestions that you guessed at so that you can zero in on those specific an-swers. Its important to evaluate what you know. Check all of your answers.

AP Physics Practice Test 2, Level C. Take this test and answer all of thequestions you can and then guess at those you dont know. Circle thosequestions that you guessed at so that you can zero in on those specific an-swers. Check all of your answers.









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RED ALERT

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THE 24-WEEK PLAN1 LESSON PER WEEK

If youre lucky enough to have the extra time, the 24-Week Plan will enableyou to better utilize your study time. You will now be able to spread out yourplan into one lesson a week. This plan is ideal because you are not under anypressure, and you can take more time to review the material in each of the

chapters. You will also have enough time to double-check the answers tothose questions that might have given you problems. Keep in mind that thebasis for all test success is practice, practice, practice. If youre taking theC-Level exam, you can either shorten your study time or take the extra fewweeks to reread everything.

THE PANIC PLAN

While we hope you dont fall into this category, not everyone has the luxury

of extra time to prepare for the AP Physics test. Perhaps, however, we canoffer you a few helpful hints to get you through this period.

Read through the official AP Physics bulletin and thisAP Success: PhysicsB/C book and memorize the directions. One way of saving time on this, orany, test, is to be familiar with the directions in order to maximize the timeyou have to work on the questions. On this test, the directions are prettysimple. You may also want to take the time to look at additional material onthe Internet. You can find more information about the AP Physics test on theInternet at www.collegeboard.org/ap/physics/.

Read the introduction to this book. It will be helpful in preparing for the testand give you an understanding of what you can expect on the exam and howmuch time you will have to complete both sections of the test. Take theDiagnostic Test as well as the practice tests. Focus whatever time you haveleft on those specific areas of the test that gave you the most difficulty whenyou took the practice tests. Whatever time you have before the exam, keep inmind that the more you practice, the better you will do on the final exam.


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Physics Formulas

TABLE OF INFORMATION

Constants and Conversion Factors

1 unified atomic mass unit 1 u = 1.66 1027kg = 931 MeV / c2

Proton mass mp= 1.67 1027kg

Neutron mass mn= 1.67 1027kgElectron mass m

e= 9.11 1031kg

Magnitude of electron charge e = 1.60 1019 CAvogadros number N

0= 6.02 1023mol1

Universal gas constant R = 8.31 J / (mol K)Boltzmanns constant k

B= 1.38 1023J/K

Speed of light c = 3.00 108m/sPlancks constant h = 6.63 1034J s = 4.14 1015eV s

Hc = 1.99 1025J m = 1.24 103eV nmVacuum permittivity e

0= 8.85 1012C2/N m2

Coulombs law constant k = 1/4e0= 9.0 109N m2/C2

Vacuum permeability m0= 4 107(T m)/A

Magnetic constant k= m0/4 107(T m)/A

Universal gravitational constant G = 6.67 1011m3/kg s2Acceleration due to gravityat the Earths surface g = 9.8 m/s2

1 atmosphere pressure 1 atm = 1.0 105N/m2 = 1.0 105Pa1 electron volt 1 eV = 1.60 1019J1 angstrom 1 = 1 1010m

UnitsName Symbolmeter m

kilogram kgsecond sampere Akelvin Kmole molhertz Hz

newton Npascal Pajoule Jwatt W

coulomb Cvolt Vohm henry Hfarad Fweber Wbtesla T

degree Celsius Celectron-volt eV

Prefixes

Factor Prefix Symbol109 giga G

106 mega M103 kilo k102 centi c103 milli m106 micro m109 nano n1012 pico p

Values of Trigonometric FunctionsFor Common Angles

Angle Sin Cos Tan

0 0 1 030 1 / 2 2/3 3/337 3 / 5 4 / 5 3 / 445 2/2 2/2 1

53 4 / 5 3 / 5 4 / 360 2/3 1 / 2 390 1 0

Newtonian Mechanics

a= acceleration F= forcef = frequency h= heightJ= impulse K= kinetic energyk= spring constant l= lengthm= mass N = normal forceP = power p = momentumr = radius or distance s = displacementT = period t = timeU = potential energy v = velocity or speedW = work x = position


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ANSWER SHEET FORANSWER SHEET FORANSWER SHEET FORANSWER SHEET FORANSWER SHEET FOR

DIADIADIADIADIAGNOSTICGNOSTICGNOSTICGNOSTICGNOSTICTESTTESTTESTTESTTEST

General Physics

1 A B C D E

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Mechanics

Electricity and Magnetism


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Diagnostic Test

Directions: Each question listed below has five possible choices. Select the best answer given theinformation in each problem, and mark the corresponding oval on the answer sheet.(You may assume g= 10 m/s2).

1. In Olde English measure, 4 fingers equal onepalm, 2 spans equal one cubit, 3 feet equal oneell, 2 cubits equal one ell, and 3 palms equal onespan. How many inches are there in one finger?

(A) 43

(B) 3

4

(C) 16

(D) 1

16

(E) 2

2. A calorimeter contains 200 g of ice at 20o

C

.Heat is added to the system at the rate of100 calories/s. In these units, the specific heatsof ice, water, and steam may be taken as0.5 cal/gCo, 1.0 cal/gCo, and 0.5 cal/gCo,respectively. The heat of fusion of ice is80 cal/g, and the heat of vaporization of water is540 cal/g. Neglecting the specific heat of thecalorimeter, describe quantitatively the state ofthe system at 920 s.

(A) All steam

(B) All water(C) All ice(D) 100g ice, 100g water(E) 100g water, 100g steam

3. Two polarizing sheets have their transmissiondirections arranged so that no light gets through.A third sheet is inserted between the two so thatits transmission direction makes a 30oangle withthe transmission direction of the first sheet.

Unpolarized light of intensityIois incident on thefirst sheet. Find the intensity transmittedthrough the last sheet. Note that

1sin 30 = cos 60 = sin 60

2

3= cos 30 =

2

and

(A) None

(B) I

o

8

(C) 3

32

Io

(D) Io16

(E) log104

Io

SECTION IGENERAL PHYSICS


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4. A string of lengthLand linear mass density isheld taut by a force Fexerted at either end.Find the time required for a transverse pulse totravel from one end of the string to the other.

(A) 2LF

(B) FL2

(C) FL

(D) F

L

(E) LF

5. A box of massMstarts from a state of rest on atable. The coefficient of kinetic frictionbetween the box and the table is (where < 1). A cord attached to the side of the boxpasses over a pulley at the edge of the table andis connected to an equal massMthat hangs adistance,L,above the floor. If static friction issufficiently small that the system starts tomove, how long will it take the hanging mass tohit the ground?

(A) 21

L

g( )

(B) L

g

(C) 4 2

2 2

L M

g

(D) 2L

g

(E) M L

g

2 2

2

6. A pith ball of mass mhas a positive charge of qon its surface. The ball is thrown verticallyupward with an initial speed of voin a uniformvertically downward electric field withmagnitudeE. How high does the ball go?Neglect air resistance, but do not neglect gravity.

(A) v

g

o

2

2

(B) qE

vo

(C) v

qmE

o

(D) mv

qE mg

o

2

2( )+

(E) 3mEqv

g

o

7. A +2Ccharge is at point (6m, 0), and a 8Ccharge is at point (2m, 0). Find the coordinatesof a point (not at infinity) where the electric fieldis zero.

(A) (10m, 0)

(B) 143

0m,

(C) 26

50m,

(D) (3m, 0)(E) (4m, 2m)


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DIAGNOSTIC TEST


8. A string is passed over a frictionless pulleysuspended from the ceiling with a 3kg masssuspended from one end of the string and a 2kgmass at the other end. The 2kg mass starts outat floor level, and the 3kg mass starts somedistance above the floor. If the system is

released from rest, the 3kg mass hits the floorwith a speed of 6 m/s. Find the initial distanceof the 3kg mass from the floor. For simplicity,assume that g = 10 m/s2.

(A) 2m(B) 5m(C) 9m(D) 20m(E) 32.5m

9. Suppose that in a given location on the earthssurface, the earths magnetic field has adownward component of 5 105T, a northwardcomponent of 1 105T, and no east-westcomponent. What are the magnitude anddirection of the force on a 4m length of wirethat carries 200Ahorizontally, from S to N?

(A) 0.008N, North

(B) 0.04N, West

(C) .008 26N , East

(D) 0.008N, South(E) 0.04N, East

10. A merry-go-round of radiusRrotates at constantspeed with period T. What is the minimumcoefficient of static friction between the merry-go-round and a box of mass, m,placed at itsedge that will enable the box to remain on thesurface without sliding?

(A) 2RT

(B) 4 2

2

RgT

(C) gT

R

2

2

(D) Zero

(E) mgTR3

11. Two trains are following one another at slightlydifferent speedsthe one in front at speed v1and the one trailing behind at speed v2, wherev1 > v2. Both sound their horns at the samefrequency,f. Take the speed of sound in air asv, and calculate the beat frequency of thecombined sounds for a passenger in the trailingtrain.

(A) f v v

v v

1

2

(B) f v v

v v

++

2

1

(C) f v v

v v

1 2

1 2

+

(D) f v vv v

1 2

1

+

(E) f v

v v

2

1 2+


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12. The radius of a newly discovered planet is halfthe radius of the earth, and its mass is one tenthof the mass of the earth. If an object weighs200Non the surface of the earth, what will itweigh on the surface of the other planet?

(A) 20N(B) 5N(C) 2000N(D) 10N(E) 80N

13. In the circuit shown below, the current throughthe 80resistor is 2A, the voltage across the180resistor is 240V, and the battery has an

EMF of 440V and an unknown internalresistance, r. Find the value of the resistance,X.

(A) 180(B) 220(C) 360(D) 440(E) 60

14. A 0.25kg object is connected to a spring ofspring constant k= 25N/mand is set intooscillation with an initial spring potential energyof 12J and an initial kinetic energy of 4J. Atwhat displacement are the kinetic and potentialenergies equal?

(A) 0.8m(B) 1.0m(C) 1.25m(D) 2.0m(E) 16m


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Directions:Answer the following free-response question. Each question is designed to take approximately15 minutes to answer. Note that each part within a question may not have equal weight. Show all work to

obtain full credit, and avoid leaving important work on the green insert. Assume g= 10 m/s2

.

15. A cord is used to vertically lower a block of

mass,M,a distance, d,at a constant downward

acceleration of g

3. In terms ofM, g,and d,

(a) find the tension in the cord.

(b) find the work done by the cord on theblock.

(c) find the work done by gravity.(d) what is the change in kinetic energy of

the block?

SECTION IIFREE RESPONSE


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SECTION IMECHANICS

1. A moving body of mass, m,makes a perfectlyinelastic collision with a second body of twiceits mass, initially at rest. Find the fraction ofthe initial kinetic energy that is lost in the

collision.

(A) None(B) One third(C) One half (D) Two thirds(E) All

2. An object of mass 2kg makes an elasticcollision with another object at rest andcontinues to move in the original direction butwith one fourth of its original speed. What isthe mass of the struck object?

(A) 0.5 kg(B) 1.2 kg(C) 2 kg(D) 3.4 kg(E) 8 kg

3. A boy throws a ball into the air as hard as he canand then bicycles as fast as he can, alwaysstaying under the ball in order to catch it.Assume the throw is from ground level. If theinitial speed of the ball is 20m/s and the boyscycling speed is 10m/s, find the time of flight.

Take g as 10m/s2

, and note that sin30o

=cos60o = 0.5 and sin60o = cos30o = 0.866.

(A) 3.46s(B) 5.00s(C) 6.28s(D) 7.50s(E) 8.87s

4. Assume that NASA wishes to send a mannedspaceship to explore a large asteroid at a distance

of 1 1010m from the earth. To approximatelysimulate earth gravity, NASA intends toaccelerate the spaceship (from rest) at 10 m/s2

for the first half of the trip, then give the ship anequal negative acceleration for the remainder.However, the astronauts forget to reverse theengines until they cover 80% of the distance.Assuming that the astronauts arrive at theasteroid with zero velocity and that the reversedengines gave the ship a constant negativeacceleration, what is the total elapsed time for

the trip?

(A) 10,000 s(B) 20,000 s(C) 30,000 s(D) 40,000 s(E) 50,000 s


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DIAGNOSTIC TEST


5. The three blocks in the figure below are released

from rest and accelerate at the rate of 2m/s10g ,

where gis the acceleration due to gravity. What

is the coefficient of friction between the table

and the horizontally moving block?

(A) 0.25(B) 0.2(C) 0.2MG(D) 0.5(E) 2

6. Suppose that the earth was to somehow expandto become a sphere four times its presentradius, but the total mass of the earth stayed

constant. What is the new escape velocity of arocket from the surface of the earth, in terms ofthe old escape velocity vo?

(A) No change

(B) 2vo

(C) vo2

(D) vo

4(E) 2vo

7. A flywheel with diameterDis pivoted on ahorizontal axis. A rope is wrapped around theoutside of the flywheel, and a steady force, F, isexerted on the rope. It is found that (startingfrom rest)Lmeters of rope are unwound intseconds. What is the moment of inertia of the

flywheel?

(A) LtF

D2

(B) 4 2 2

FL t

D

(C) DLtF

2

4

(D) 3 3

2L FDt

(E) FD t

L

2 2

8

8. A 2kg block traveling at 10 m/s on a horizontal,frictionless table strikes and becomes fastened tothe end of a spring without a loss of energy.The other end of the spring is fixed. If thespring is compressed 100 cm before the blockmomentarily stops, what is the period of theresulting simple harmonic motion?

(A) 2 s(B) 2s

(C) 5 s

(D) 1 s

(E) 2 s


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DIAGNOSTIC TEST


9. A particle of mass mslides down a frictionlesscircular track of radiusR, starting from rest froma position horizontally across from the center ofthe circle, as shown in the figure. Find themagnitude of the force exerted by the track on mat point B.

(A) 0

(B) mg

4

(C) mg

2

(D) mg(E) 1.5mg

10. An Atwoods machine consists of a pulleyhanging from a ceiling with unequal massesM1andM2 (M1 >M2) attached to opposite ends ofa rope hanging over the pulley. Assume that anAtwoods machine is set up on the surface of aplanet. The pulley is a uniform solid disk (I =

MR2

/2) of mass M kg and radiusR, rotating ona frictionless axle. It is found that massM1descendsLmeters in tseconds, starting fromrest. No slippage occurs between the rope andthe pulley. Find the acceleration due to gravityon the surface of the planet.

(A) ( )

( )

M M M L

M M t

+ +

2 21 2

1 2

2

(B) ( )M M M R

Lt

+ +1 2

2

2

(C) ( )2 1 22

2

M M M L

Rt

+ +

(D) ( )M M M R

t

1 224

(E) 22

L

t

11. A stone is tied to a string of lengthRand whirled

around in a vertical circle. Assuming that theenergy remains constant, find the minimumspeed it must have at the bottom of the circle inorder for the string to remain taut at the top ofthe circle.

(A) 2 gR

(B) 5gR

(C) gR

(D) 2 gR

(E) 1

gR


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DIAGNOSTIC TEST


12. CornersA, B, and Cof a right triangle areoccupied by masses of 3, 8, and 6 kgrespectively. SideAC is 3m,BCis 4m, andABis 5m. If the magnitude of the net gravitationalforce exerted on the 6 kgmass by the other twomasses is represented by F, and Gis the

universal gravitational constant, find the valueof F.

(A)2

25

m

kgG

(B)2

256

m

kgG

(C)2

213

m

kgG

(D)2

217m

kgG

(E) 0

13. A small block of mass, m,slides down thefrictionless loop-the-loop shown below, the circlehaving a radiusR. The speed of the block as itpasses a height, h,is v. Find the magnitude ofthe force that the track exerts on the block as itpasses pointAat the top of the loop.

(A) 2

5

mgh

R mg

(B)m

v gh

R

R

2 2 2+

(C) mv

R

2

(D) m v gh gRR

( )2 2 5+

(E) mg

14. A dog running at a constant velocity of 11 m/s is18 m behind its owner when the owner startsfrom a state of rest on a motor bike with aconstant acceleration of 2 m/s2. For a period oftime, the dog will find itself ahead of its owner.How long is that time interval?

(A) 2 s(B) 3 s(C) 6 s(D) 7 s(E) 9 s


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DIAGNOSTIC TEST


15. The position of an object as a function of time is

given by 18242 3 = ttx , where tis inseconds andxis in meters. Displacementsmeasured to the right are positive.

(a) What is the average velocity of theobject between t= 1sand t= 3s?(b) What is its velocity at t= 3s?(c) At what time or times does the object

stop?(d) What is its acceleration at each of the

times in part (c)?



.



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DIAGNOSTIC TEST


SECTION IELECTRICITY& MAGNETISM

1. Five equal positive charges, Q,are equallyspaced on a semicircle of radiusRas shown inthe following diagram. What is the magnitudeof the Coulomb force on a positive charge qlocated at the center of the semicircle?

(A) Zero

(B) kQq

R2

(C) 52

kQqR

(D) 32

kQq

R

(E) 1 22

+( )kQqR

2. A positive point charge of magnitude Qis placedat the origin, and an unknown charge is placed atpoint (a, 0). It is found that the electric field iszero at (2a, 0). Find the field at (3a, 0).

(A) kQ

a12 2

(B) 7

144 2kQ

a

(C) Zero

(D) 3

22 2kQ

a

(E) 22

kQ

a

3. A thin rod stretches along thex-axis from(2m, 0) to (6m, 0) and has a uniform chargeper unit length of 3 C/mdistributed along itslength. Find the potential at the origin.Use k = 9 109N m2/C2.

(A) 6.75 103V(B) 2.7 104ln(3) V(C) 9 103V

(D) Zero(E) 0.025 V


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DIAGNOSTIC TEST


4. The charge density on the surface of aconducting sphere of radiusRis . A positivecharge, q,of mass mis released from rest at apoint outside the sphere at distance, a, fromthe center of the sphere. Find its speed at theinstant it is at distance, b, from the center.

(A) 8 2

k R q b a

mab

( )

(B) 2kq Ra

mb

(C) Zero

(D) qmk aR b

ln

3( )

(E) 4 2ke

abR

5. When a voltage, V,is placed across a resistor ofresistanceR, the power generated is 20W. Theresistor is then snipped into three equalpiecestwo of the pieces are combined inparallel and the third in series with thatcombination. If the same voltage, V,is placedacross this combination, what will be the total

power generated?

(A) 10W(B) 26.7W(C) 30W(D) 40W(E) 60W

6. The current in a 5resistor varies with timeaccording to the relation i= 3t2 4,wherei is inamperes and tis in seconds. Consider a timeinterval from t= 1 s to t = 5 s. What constantcurrent would transport the same charge in thattime interval as the time-varying current?

(A) 4A(B) 9A(C) 8 ln(2)A(D) 15A(E) 26A

7. A circular loop of wire of radius 2mandresistance 8, located in the plane of the paper,is placed in a magnetic field perpendicular to thearea of the wire. The magnetic field through the

loop varies with time according toB = 6t t2,where t= time in seconds and the positivedirection is into the paper. Find the magnitudeand direction of the current flow in the loop at t= 1 s.

(A) Zero, no direction(B) 2A, clockwise(C) 2A, counterclockwise(D) 4A, clockwise

(E) 4A, counterclockwise

8. A set of axes is laid out with the positivey-axispointing toward the north and the positivex-axispointing east. A long, straight wire carries acurrent of 12 amp along they-axis toward thenorth. Find the magnitude and direction of theforce on a +3Ccharge moving due north with avelocity of 500 m/s, at the instant it passesthrough the point (6m, 2m). The magnetic

constant is k o= =

4107N/A2.

(A) 6 1010N, west(B) 4 1010N, north(C) 6 1010N, vertically up(D) 4 1010N, west(E) 1 109N, vertically down


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DIAGNOSTIC TEST


9. A positive point charge +75Cis located on they-axis at the point (0m, 4m), and a negativepoint charge 50Cis located on they-axis at(0m, 4m). Find theycoordinates of alllocations on they-axis (not including ) atwhich the potential is zero.

(A) 4 3 2

3 2

( )+( )

monly

(B) 0.8monly(C) 20monly(D) 0.8mand 20monly(E) All points between the two charges

10. A 20,000resistor is connected in series with acapacitor, and a 40Vpotential is suddenlyapplied to the combination. The charge on thecapacitor rises to 25% of its final value in 2s.Find the capacitance of the capacitor.

(A) 2 108F(B) 16F

(C)

1 10

4

3

10

ln

F

(D) 6 3e F

(E) 4pF

11. A hollow rubber ball of inner radius aand outerradius bhas a uniform charge density distributed through the rubber. Find the electricfield at a distance rfrom the center, wherea < r < b.

(A) 40

2r

(B)

( )b a

r

2 2

0

2

(C) rb a( )

(D)

lnb

a

(E)

( )r a

ro

3 3

23

12. For the circuit given below, calculate the powerdissipated in the 4resistor.

(A) 81(B) 36(C) 29.16(D) 64(E) 3.25


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DIAGNOSTIC TEST


13. A charge, q, of mass, m,moves in a circularorbit of radius,R,perpendicular to a magneticfield of magnitudeB. Find its kinetic energy.

(A) q B R

m

2 2 2

2

(B) mv

R

2

(C) qvxB

(D) mv

qB

(E) mB

qR4 2

14. A long, hollow cylindrical shell of radius a,carrying a uniform negative surface chargedensity , is surrounded by a coaxialcylindrical shell of radius b, carrying a surfacecharge of the same density but opposite sign.Find the electric field in terms of at a

distance rfrom the axis of the cylinder, wherer > b .

(A) Zero

(B)

( )b a

ro

(C)

b a

r

2 2

0

2

( )

(D)

4 2or

(E)

0


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DIAGNOSTIC TEST


15. In each of the following cases, determine thepotential difference between pointsxandy,andstate which point is at the higher potential.[Parts (a), (b), and (d) each show only a portionof the complete circuit.]

(a)

(b)

(c)

(d)



.



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26

Diagnostic TestANSWERS AND EXPLANATIONS

SECTION IGENERAL PHYSICS

1. The correct answer is (B). Unit conversions should be done by

cancellation:

(1finger)(1 palm/4 fingers)(1 span/3 palms)(1 cubit/2 spans)

(1 ell/2 cubits)(3 feet/1 ell) (12inches/1 foot) =3

4inch

2. The correct answer is (E). For a change in temperature, the heat suppliedis given by Q = mcT. To heat the ice to 0oC, Q = (200 g)(0.5 cal/gCo)(20Co) = 2,000 cal. At the rate of 100 cal/s, this will take 20 s. To melt theice requires Q = mL,whereLis the heat of fusion. Then Q =(200 g)(80 cal/g), requiring 16,000 cal or 160 s. To bring the water formed

up to 100o

C requires Q = (200 g)(1 cal/gCo

)(100Co

) = 20,000 cal, oranother 200 s. The elapsed time so far is 380 s, leaving 920 380 = 540 sfor boiling. This will supply 54,000 cal. At a heat of vaporization of540 cal/g, this is sufficient to boil 100 g of water.

3. The correct answer is (C). Starting with unpolarized light, the lighttransmitted through the first polarizer will be linearly polarized with

intensityI

o

2. The succeeding intensities are determined by Brewsters

Law: Ifinal

=Iinitial

cos2. The intensity through the second polarizer is then

( )020 30cos2

I

83 0I= , since 2330cos =

o . The final intensity is then

I I I

o o=

=3

860

3

32

2cos , since cos60 1

2 = .

1. B 3. C 5. A 7. A 9. B 11. D 13. C

2. E 4. E 6. D 8. C 10. B 12. E 14. A

QUICK-SCORE ANSWERS


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27

ANSWERS AND EXPLANATIONS

www.petersons.comAP Success: Physics B/C

4. The correct answer is (E). The velocity of a transverse pulse traveling

down a string is given by =Fv . The time required to travel a distance

Lis thenF

LT

= .

5. The correct answer is (A).Considering the hanging mass and takingdownward as positive,Mg T = Ma,where Tis the tension in the string.Considering the mass on the table,f = N = Mg. Then summing thehorizontal forces, T Mg = Ma. Combining the first and last equations

and solving for the acceleration,g( )1

2

. Then, since L at=

2

2, the time is

t = 21

L

g( ) .

6. The correct answer is (D). The forces on the pith ball are qEand mg,both downward. Taking upward as positive and using

F = ma,(qE + mg) = ma, yielding a = m

mgqe )( +.Using

v v ayo

2 2 2= + (or by using energy conservation) and solving foryafter

substituting the value for acceleration, y=)(2

2

mgqE

mvo

+ .

7. The correct answer is (A). Since electric fields are directed away frompositive charges and toward negative charges, the point in question mustbe outside of the positive charges, closer to the smaller charge, and on theline joining them. Let that point have coordinates (x, 0). Then,

k C

x m

( )

( )

2

6 2

= k C

x m

( )

( )

8

2 2

.

Canceling like factors, taking the square root of both sides, and solving forx, x = 10m. The coordinates of the point are then (10m, 0).

8. The correct answer is (C). The problem may be solved either by dynam-

ics or by energy conservation. By the latter method, since the initialkinetic energy of the system is zero,(3kg)(10 m/s2)(x) = 0.5(3kg)(6m/s)2+ 0.5(2kg)(6 m/s)2+ (2kg)(10 m/s2)(x),wherexis the desired distance. Solving,x= 9m.


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28

DIAGNOSTIC TEST


9. The correct answer is (B). The northward component of the earths fieldwill have no effect on a current directed northward. The force exerted bythe downward component is given byF = ILBsin= (200A)(4m)(5 105T)sin90o= 0.04N. The direction of theforce, as determined by the right-hand rule, is toward the west.

10. The correct answer is (B). The coefficient of friction is = =fN fmg .

Since f mv

R=

2

, we have v2 =gR. But v R

T=

2. Substituting and

solving for ,

=4

2

2R

gT.

11. The correct answer is (D). The wavelength behind the first train is thespeed of sound relative to that train divided by the frequency sounded bythat train, or

= +v v

f

1 . The frequency heard by the second train is the velocity of

sound relative to that train divided by the wavelength, or

f f v v

v v2

2

1

= +

+

.

Since v1> v

2, f

2< f. The beat frequency is then the difference, f f

2.

Inserting the expression above forf2and simplifying, Beat frequency =

f

v v

v v

1 2

1

+

.

12. The correct answer is (E). The gravitational force exerted by a planet ona mass located on its surface is directly proportional to the planets massand inversely proportional to the square of its radius. Using subscripts Pfor the planet andEfor the earth,

F M

M

R

RF N N

P

P

E

E

P

E=

=

=2

21

102 200 80( ) ( ) .


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29



13. The correct answer is (C). The current through the 180resistor is

240

180

4

3

VA

=

. Since the current through the 80resistor is 2A, this

leaves2

3

A to go through the unknown resistor. The voltage across this

resistor is 240V, so the resistance is given by=

360

32

240

A

V

.

14. The correct answer is (A). The total energy is 16J, which is conserved.At the point where the kinetic and potential energies are equal, each will be

8J. The potential energy is given by U= 0.5kx2= 0.5(25m

N)x2= 8J from

whichx= 0.8m.


15. (a) Taking upward as positive, and using F = Ma,

T Mg =

g

3or T = 2

3

Mg .

(b) W= Tdcos180o= 23

Mgd

(c) W =Mgdcos0o =Mgd

(d) DK= Wtotal

=Mgd

3


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30

DIAGNOSTIC TEST


SECTION IMECHANICS

1. The correct answer is (D). From momentum conservation,

mv = 3mV, so Vv

=3

, where v= original velocity and V= final velocity.

Then63

)3(21 2

2mvv

mKf =

= and Klost =Ki Kf =mv

2

3.

The fraction lost is K

K

lost

i

=23

.

2. The correct answer is (B). From momentum conservation,

2 24

v v

mV=

+ , yielding V

v

m=

3

2. From kinetic energy conservation,

22

2

21

4)2(

21

)2(21

mVv

v +

= . Substituting expression (1) for Vand

simplifying, m= 1.2 kg.

3. The correct answer is (A). Thex-component of the balls velocity mustmatch the boys velocity of 10 m/s. The initial velocity triangle is then a30o 60o 90otriangle, yielding an initial vertical velocity of 17.3 m/s.

Taking vertically upward as positive and using 20 21

attvy += , with

a = 10 m/s2, andy = 0, since the ball returns to ground level, t may befactored out, yielding t= 3.46 s.

4. The correct answer is (E). Using 20 2

1attvx += for the first leg,

t= 4 105s. Using v = vo+ at, the velocity at the end of the first leg is

4 105 m/s. This is the initial velocity for the second leg. Using

0and2

=+

= finalfinalo

vtvv

x for that leg, t= 1 104 s. The total time is

then 5 104s.

1. D 3. A 5. A 7. E 9. E 11. B 13. D

2. B 4. E 6. C 8. C 10. A 12. C 14. D

QUICK-SCORE ANSWERS


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5. The correct answer is (A). For the block of massM, T Mg Mg

110

= ,

where T1

is the tension in the right-hand rope and the positive direction is

upward. This yields T

Mg

1

11

10

= . For the other hanging block, taking

downward as positive, 25

2Mg T

Mg = . This yields T

Mg2

9

5= . For

the block on the table, taking the positive direction to the left,

T T f Mg

2 1

2

10 = . Substituting the tensions yields f

Mg=

2. But

f =N = (2Mg). Thus, = 0.25.

6. The correct answer is (C). Conserving energy, KI+ U

I= K

f+ U

f.

1

202mv

GMm

Resc

E

= , whereM= mass of Earth and m= mass of rocket.

Then v GM

Resc

E

= 2

, and, thus, if the radius of the earth is quadrupled,

the escape velocity will become half of its former value.

7. The correct answer is (E). Since = I , the moment arm is R D=

2,

and = a

R, we have

FD Ia

D2

2

=

. Also, L at=

2

2. Eliminating afrom

the two equations, I FD tL

= 2 2

8.

8. The correct answer is (C). From energy conservation,1

2

1

2

2 2kx mv= ,

from which m/200 mN

k= . For a spring, T m

k=2 , yielding T s=5 .

9. The correct answer is (E).Taking pointBas the zero of potential energy

and conserving energy betweenAandB, mgR mv

2 2

2

= , so v2 = gR.


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32

DIAGNOSTIC TEST


A radius drawn fromB makes a 30oangle with the horizontal. The twoforces on mare the normal forceNand the weight mg. If the weight isbroken into radial and tangential components, then in the radial direction,

F N mg mv

R= =sin30

2

. Using v2 = gR, we obtainN= 1.5mg.

10. The correct answer is (A). Applying F = matoM1, where T

1is the

tension in the rope connected toM1, and taking downward as positive,

M1g T

1= M

1a.

Similarly forM2, taking upward as positive, T

2 M

2g = M

2a.

Using =I, where = a

R, T R T R

MR a

R1 2

2

2 =

.

Simplifying and adding the three equations yieldsM1g M

2g =

(0.5M + M1+ M

2)a

ButL = 0.5at2from kinematics. Substituting and solving for g,

g= ( )

( )

M M M L

M M t

+ +

2 21 2

1 2

2.

11. The correct answer is (B). If vis the speed at the bottom and Vis the

speed at the top, then from energy conservation,2

)2(2

22MV

RMgMv

+= .

The centripetal force at the top is supplied by gravity only: Mg MV

R=

2

.

Eliminating Vbetween the two equations, v gR= 5 .

12. The correct answer is (C). The force Fexerted by one mass on another is

given by F GMm

r=

2. Then, the force exerted on the 6kgmass by the 3kg

mass is2

22

m

Gkg , and the force exerted on the 6kgmass by the 8kgmass is

2

23

m

Gkg . These forces are at right angles, so the net force is given by the

Pythagorean theorem as2

213

m

Gkg .


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13. The correct answer is (D). By conserving energy between height handpointA, and using Vas the velocity atA,

1

2

1

222 2mv mgh mV mg R+ = + ( ) .

Using Newtons second Law at pointA, N mg mV

R+ =

2

, whereNis thenormal force exerted by the loop. Eliminating Vbetween the two equa-

tions, Nm v gh gR

R=

+ ( )2 2 5 .

14. The correct answer is (D). From kinematics, 221

attvxx oo += .

For the dog,x=(11 m/s)t.For the owner,( )

2/2

1822

tsmmx = . Eliminating

xand solving the factorable quadratic equation that results, t= 2s or 9s.Consequently, there is a 7-second interval between the two times that thedog and the owner were at the same location.


15. Sincex= 2t3 24t 18, the position of the object is at 36m at t = 3 s andat 40m at t= 1 s. The average velocity is the displacement over the timeinterval, or 4m/2 s = 2 m/s.

(a) The instantaneous velocity is the derivative with respect to time of theposition, yielding v= 6t2 24. At t= 3 s, v= 30 m/s.

(b), (c) The velocity is zero when 6t2 24 = 0, yielding t= 2 s. (Bothanswers are meaningful. The negative time simply means 2 s before theclock was started.)

(d) The acceleration is the time derivative of the instantaneous velocity,yielding a= 12t. At t= 2 s, a= 24 m/s2. At t = 2 s, a= 24 m/s2.


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34

DIAGNOSTIC TEST


SECTION IELECTRICITY AND MAGNETISM

1. The correct answer is (E). The forces exerted by the charges at the 90o and270opositions are equal and opposite and cancel each other. The force

exerted by the charge at 180ois kQq

R2

toward the right. The forces exerted

by the other two charges, each of magnitudekQq

R2, are at right angles to

each other and can be combined by the Pythagorean theorem into a single

force2

2

kQq

Rto the right. (Alternatively, this can be accomplished by

breaking the forces into components.) The resultant force is then of magni-

tude( )

2

21

R

qQk+.

2. The correct answer is (B). If the unknown charge is called q, the field at

(2a, 0) is given by( )

02 22

=+=a

kq

a

kQE , yielding q

Q=

4

. The field at

(3a, 0) is then

222 144

7

)2(4

)3( a

kQ

a

Qk

a

kQE =

+= .

3. The correct answer is (B). The potential dVdue to a small piece of charge

dqat distancexfrom the origin isx

dxk

x

kdqdV

== , where = 3 106C/m.

Integrating fromx = 2mtox = 6m, and recalling that ln(6) ln(2) = ln (6/2)= ln(3), we have V = 2.7 104ln3 Volts.

1. E 3. B 5. D 7. C 9. D 11. E 13. A

2. B 4. A 6. E 8. A 10. C 12. C 14. B

QUICK-SCORE ANSWERS


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35



4. The correct answer is (A). The amount of charge on the surface of thesphere is Q =(4R2). At a point outside the sphere, a distance rfrom the

center, the charge on the sphere creates a potential given by V kQ

r= . The

potential energy of a point charge, q,is then U = qV. Conserving energy,

Ki+ UI= Kf+ Uf , where iandfstand for initial and final, respectively.Then,

0 4 1

2

42 22

+ = +k R q

amv

k R q

b

. Solving for v,

v = 8 2

k R q b a

mab

( ) .

5. The correct answer is (D). Since P V

R=

2

, the relationship between V

andRis given by 202

W V

R= . The three resistors will each have resis-

tanceR

3. The parallel combination will then have resistance

R

6, and the

series combination will have a total resistanceR

2. The power across that

combination is then given by WWR

V

R

VP 40)20(2

2

2

22

==== .

6. The correct answer is (E). Current is given by i dq

dt= ,

so Cdttq 104)43(5

1

2 = = . The constant current would then be

As

C26

4104

= .

7. The correct answer is (C). The flux at any instant is given by

= =

B A t t( ) ( )6 22 2 Wb, and the inducedEMFis

= ( )

d

dtt

24 8 . At t = 1 s, theEMFis then 16Voltsand the


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36

DIAGNOSTIC TEST


current is I=

= 16

2

V

8A

. The current (as indicated by the negative

sign or by application of Lenzs Law) will be counterclockwise, so as tocreate a magnetic field out of the paper to counteract the increasing fluxinto the paper.

8. The correct answer is (A). The field at a distance rfrom a long straight

wire is B I

r

o=

2. With the current going north, the field at (6m, 2m) will

be vertically down. The force on a charge moving north will be directed

west and will have magnituder

IqvqvBF o

==2

, where in this instance r=

6. Substituting, F = 6 1010N.

9. The correct answer is (D). At a point between the two charges withcoordinates (0,y), the total potential is given by

V k C

m y

k C

y m=

++

=( ) ( )50

4

75

40

.

Dividing through by 25k Cand solving,y = 0.8m.At a point with coordi-nates (0,y) located above the upper charge,

V k C

y m

k C

y m=

++

=( ) ( )50

4

75

40

. Solving in a similar fashion, y = 20m.

There is no point below the lower charge at which the potential is zero.

10. The correct answer is (C). The charge on a capacitor being charged is

given by

Q Q em

t

RC=

( )1 , where Qmis the maximum charge after a long period

of charging. In this instance, Q Q

m=4

, so that43

=

RC

t

e , yielding

=

34

n1R

tC

.

Substituting the given values, C = 1 1043

10

lnF.


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37



11. The correct answer is (E). Gauss Law applies to a Gaussian sphere of

radius r(where a < r < b) yields

E dA q

o

= .

E r

r a

o

4

4

3

4

32

3 3

=

Solving forE,E =

( )r a

ro

3 3

23

.

12. The correct answer is (C). The parallel combination of 1and 2gives

a combined resistance of2

3

, which combines in series with the other

resistors to give an equivalent resistance of 203

. Using V = IR, the

current through the equivalent circuit is AV

=

1027

32018

, which is then

the current through the 4resistor. The power loss through that resistor is

P = I2R, yielding =

= 16.291002916

P .

13. The correct answer is (A). The force on the particle is f qvB mv

R= =

2

,

from whichm

qBRv= . Since

2

2mv

KE= , we obtainm

RBqKE

2

222

= .

14. The correct answer is (B). Taking a Gaussian cylinder of radius randlengthL, and applying Gauss Law,

E dA qo

=

E rL bL aL

o

2 2 2

=

( ), from which E =

( )b a

ro

.


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38

DIAGNOSTIC TEST



15. In diagram 1, Vxy

= (2A)(3) 10V + (2A)(1) = 2V,indicating that pointyis at the higher potential.

In diagram 2, Vxy

= (2A)(1) 6V + (2A)(2) = 0V,indicating that pointxis

at the higher potential.

In diagram 3, no current goes through the capacitor branch. The 2and3resistors are then in series, and the current through the left-hand loopisI = V/R= 10V/5= 2Aclockwise.

Then Vxy

= 6V (2A)(3) = 12V,indicating that pointyis at the higherpotential.

In diagram 4, the total resistance of the top branch is 2+ 4 = 6. Theequivalent resistance of the parallel combination is then obtained from

+

=

21

611

R, so that R = 1.5.

The voltage across the combination is then V = (8A)(1.5) = 12V, fromwhich the current in the top branch is 12V/6= 2A. The voltage V

xyis

then Vxy

= (2A)(2) = 4V, indicating that pointxis at the higher potential.


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UNIT 1Newtonian Mechanics


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41

Chapter 1KINEMAKINEMAKINEMAKINEMAKINEMATICSTICSTICSTICSTICS

MOTION IN ONE DIMENSIONMotion in one dimension is exemplified by motion along a straight line.

DEFINITIONSCoordinate SystemMotion is a change of position, so in order to discuss

motion, we must first discuss position. To discuss position, we must choose

an origin and a reference direction. These choices are arbitrary and must be

made before the position can be defined.

For example, the origin can be in the middle of a horizontal line, andthe reference direction can be to the right. We will call the reference direc-

tion the + direction. In one dimension, only one other direction is pos-

sible. In this example, that direction is toward the left. We call that direction

the direction.

When we have identified a reference direction and origin, we say that

we have defined the (one dimensional) coordinate system that we will be

using.

PositionThe position of an object in one dimension is specified by stating

two things: the distance from the origin and the direction from the origin tothe object. Thus, an object with position 6 meters is 6 meters to the left of

the origin. If it moves to a position of 3 meters, it is now 3 meters to the

left of the origin.

DisplacementThe change in position is calculated by subtracting the

initial position from the final position. In the subtraction, the + and

direction signs are treated as algebra signs. An object that moves from 6

meters to 3 meters has a change of position of +3 meters.

The algebraic representation of this calculation usesxas the symbol

for position and x as the symbol for change in position. We write:

m3)m6(m3 +=== INITIALFINAL xxxx is called the displacement of the object.


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CHAPTER 1


Velocity can be positive or negative, depending on the direction of the

displacement. The velocity can be zero while the position is not zero. The

velocity can be large at a time when the position is zero.

AccelerationAcceleration is defined as the rate of change of velocity.

Average acceleration, < >a , is defined as the change in velocity, v ,

divided by the time required to make the change, t: < > =a v

t

VelocityVelocity is defined as the rate of change of position. Average

velocity, < >v , is defined as the change in position, x , divided by the time

required to make the change, t: < > =v x

t

Instantaneous velocity is the limit of the average velocity as the time

interval (and thus also the change in position) approaches zero. It is the timederivative of the position.

v x

t

dx

dtt= =

lim

0

Instantaneous acceleration is the limit of the average acceleration as the

time interval (and thus also the change in velocity) approaches zero. It is the

time derivative of the position.

a v

t

dv

dtt= =

lim

0

Acceleration can be positive or negative, depending on the direction of

the change in velocity. The acceleration can be zero while the velocity is notzero. The acceleration can be large at a time when the velocity is zero.

MOTIONWITHCONSTANTACCELERATIONEQUATIONS

Position as a Function of TimeFor one dimensional motion with constantacceleration, a, the position,x, as a function of time, t, is given by

x at v t x= + +1

2

2

0 0

wherex0

is the position at time t= 0 (the initial position) and v0

is the velocityat time t= 0 (the initial velocity). The position is the definite integral of thevelocity from t = 0 to t.

Velocity as a Function of TimeFor one dimensional motion with constantacceleration, a, the velocity, v, as a function of time, t, is given by


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43

KINEMATICS


v= at+ v0, where v

0is the velocity at time t= 0 (the initial velocity). The

velocity is the definite integral of the (constant) acceleration from t= 0 to t.

It is also the time derivative of the position given above.

Velocity as a Function of PositionThe two equations give the complete

solution for the motion of an object (for the case of constant acceleration). Apartial solution can be created by combining those two so as to eliminate the

time variable. The result gives the velocity as a function of position:

v v a x x2

0

2

02 = ( )This equation does not allow determination of the direction of the

velocity since a velocity of either sign will satisfy it. This equation is re-

placed later, for any acceleration, by the work-energy theorem.

FREE FALL

It has been determined experimentally that any object falling withoutresistance near the surface of the earth has a downward acceleration of

9.8 m/sec2. This acceleration is said to be due to the earths gravity.

An object moving vertically and subject to only the earths gravity is an

ideal example of one dimensional motion with constant acceleration. It is

important to note that the acceleration is the same whether the object is

a. moving upward (with decreasing magnitude of velocity),

b. moving downward (with increasing magnitude of velocity), or

c. standing still at the top of the path (zero velocity).

Problem solutions are typically set up with the origin at the lowest point in

the problem and with t= 0 when the object begins its flight.

In many physics exercises, cars are understood to accelerate forward at a

constant rate when the gas pedal is pressed. They are understood to acceler-

ate backward at a constant rate when the brake pedal is pressed.

GRAPHS

If the position of a particle is plotted versus time, the slope of theposition graph is the velocity of the particle.

The dashed line is tangent to the position curve at 1 second in the first graph

on the next page. The slope of that line is about 2 m/sec.Calculus users, note that the slope of the graph is just the derivative of

the position function with respect to time.


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CHAPTER 1


If the velocity of a particle is plotted versus time, the slope of thevelocity graph is the acceleration of the particle. In the graph, the slopeof the velocity graph is constant, as is the acceleration.

Calculus users, note that the slope of the graph is just the derivative of the

velocity function with respect to time.

Calculus users, note that the definite integral of velocity from one time toanother is the change in position of the particle represented by the areaunder the velocity curve on the graph. Also, the definite integral of


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45

KINEMATICS


acceleration from one time to another is the change in velocity of theparticle, represented by the area under the acceleration curve on thegraph.

The area under the acceleration curve between .5 and 1.0 seconds is equal tothe change in velocity during that time interval. The area under the velocitycurve between 1.5 and 2.0 seconds is the change in position during that time.


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CHAPTER 1


As an exercise, see how the slopes (derivatives) of position and velocity agreewith velocity and acceleration in the graphs below.

Calculus users, note that the formula used for the position in the graphs

above is x t t= + 2 3 12cos( ) .


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47

KINEMATICS


MOTION IN TWO DIMENSIONS, INCLUDINGPROJECTILE MOTION

COORDINATESYSTEMANDVECTORS

Motion is a change of position, so to discuss motion, we must first discuss

position. To discuss position, we must choose an origin and a referencedirection. These choices are arbitrary and must be made before the positioncan be defined.

DISTANCE/ANGLE METHOD

As in one dimension, we may describe the position of an object by itsdistance from the origin and the direction in which it is displaced from theorigin.

We choose a location for the origin and a reference direction. Tradi-tionally, the reference direction points to the right along a horizontal straightline.

We draw an arrow from the origin to the object. The length of thearrow is a distance and is called the magnitude of the position vector.

For us, a vectoris a quantity that has both magnitude and

direction.

The angle that the arrow line makes with the reference direction is takenas the directionof the positionvector.

COMPONENT METHOD

For this description, we add a second reference direction. Calling theoriginal direction (traditionally to the right) thex direction,our seconddirection is called they direction.

Theydirection is by definition perpendicular to thex direction. Tradi-

tionally, this is taken to be upward on the page. Wheny points up (instead ofdown), withxto the right, we say that we have a right handedcoordinatesystem.

The component description of a vector tells how much of the vector isalong thexdirection and how much is along theydirection. The figuresbelow show two different vectors and their components.


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48

CHAPTER 1


VECTOR ALGEBRA

Vector Addition and SubtractionVectors are added following rules thatwork for adding steps in a journey on foot.

A B+ is determined by placing

the tail of the

B vector on the tip of the

A vector. The resultant vector,

A B+ is the vector from the tail of

A to the tip of

B , as in the figure

below.The vector difference

A B is as the sum of the vector

A , with thereverse of the vector

B , called

B . As in the figure,

B points opposite

to

B .


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KINEMATICS


Using the component method as sketched above, the magnitude of the

vector sum of

A Band is

A B A B A B A B A B

x x y y x y+ = + + + = + + +[ ] [ ] [ ] [ ]2 2 2 2 .

The angle, , between

A B+ and the reference direction (x=axis) is given

by tan[ ]

[ ]

[ ]

[ ]

A B

y y

x x

y

x

A B

A B

A B

A B+

= +

+ =

+

+.

The results for the difference between BA

and can be found by simply

placing a sign in front of each B

above.

Scalar Multiplication of a VectorWhen a vector is multiplied by a scalar

(a number with no direction), the magnitude (length) of the vector is multi-

plied by that number, and the direction is unchanged.

Scalar (dot) Product of Two VectorsThe scalar product,

A B AB A B A Bx x y y

= = +cos( ) [ ] [ ] is the product of the magnitudes times

the cosine of the angle between the two vectors. This product can be positive

or negative, depending on the sign of the cosine.

Vector (cross) Product of Two VectorsThe vector product of twovectors has magnitude / / sin( )

A B AB

= .

The vector product also has direction. It is perpendicular to the planedefined by the

A Band vectors. There are two directions that satisfy that

condition. The ambiguity is resolved with a right-hand rule.If

A Band lie in the page, and if

A must turn clockwise to become

parallel to

B , then the direction of

A B is into the page.

Note that

A B B A = .

DisplacementThe change in position is calculated by subtracting theinitial position vector from the final position vector.

The algebraic representation of this calculation uses

ras the symbol forposition and

r as the symbol for change in position. We write:

r r rFINAL INITIAL

= The arrows above remphasize the vector nature of the position by

reminding us that it has a direction as well as a magnitude.

r is called the displacement of the object.


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CHAPTER 1


VelocityVelocity is defined as the rate of change of position. Average

velocity vector, , is defined as the change in position, r , divided by

the time required to make the change, t:

< > =

v r

t

The instantaneous velocity vector is the limit of the average velocity asthe time interval (and thus also the change in position) approaches zero. It isthe time derivative of the position.

v r

t

dr

dtt= =

lim

0

AccelerationAcceleration is defined as the rate of change of velocity.

Average acceleration, , is defined as the change in velocity, v , divided

by the time required to make the change, t:

< > =a v

t

Instantaneous acceleration is the limit of the average acceleration asthe time interval (and thus also the change in velocity) approaches zero. It isthe time derivative of the velocity.

a v

t

dv

dtt= =

lim

0

THE EXPERIMENTAL SITUATION

Near the surface of the earth, it is found that allfreely fallingobjects have

the same acceleration, a

m

secDOWNWARD g=

9 8

2. , .

Freely falling means that no force except gravity acts on the object. Inparticular, we ignore wind resistance. When an object is launched into theair with some initial velocity, it is freely falling after launch, even though itmight not be moving downward.

The motion of such an object near the surface of the earth is thesimplest example of projectile motion.

THE EQUATIONS OF MOTION AS A FUNCTION OF TIME

If we choose ourx-axis to be horizontal and oury-axis to be vertical (thetraditional choices) then the acceleration in thexdirection is zero, while the

acceleration in theydirection is 9 82

. m

sec.


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KINEMATICS


In these coordinates, the projectile motion separates; thex motion isindependent of they motion (as long as the object is free falling).The equation of motion for they component of the motion is

y m

sect v t y

y=

+ +

1

29 8

2

2

0 0.

where y0 is theycomponent of the initial position, and v y0 is theycompo-nent of the initial velocity.

(Thexandycomponents of initial velocity are found from the magni-tude and direction of the initial velocity. One uses the same trigonometricmethod as was used to find thexandycomponents of a position vector.)Theycomponent of the velocity can be obtained by differentiation:

v m

sect v

y y=

+9 8

2 0.

Thexcomponent of the motion is more simple, since there is no accelerationin thex direction:

x v t xx

= +0 0

and thexcomponent of the velocity is constant:

v vx x

= 0

THE TRAJECTORY

The trajectory is a plot of theycomponent of the motion versus thexcomponent of the motion. Each point on the trajectory represents theposition at a particular time. It can be found by solving thexequation for tand replacing t in they equation:

y m

sec

x x

vv

x x

vy

x

y

x

=

+

+

1

29 8

2

0

0

2

00

0

0.

This simplifies if we choose our origin so that x y0 00 0= =and :

y m

sec

x

vv

x

vx

y

x

=

+

4 9 2

0

2

0

0

.

In either case, the trajectory is a segment of a parabola, curvingdownward. We may ask, for example, for thexcomponent of the positionwhen the particle has returned to its initial height. The answer is often

called the range of the projectile.


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CHAPTER 1


CIRCULARMOTIONTWO DIMENSIONAL VECTOR DESCRIPTION

For an object moving in a circular path, we choose the origin of the coordi-nate system to be at the center of the circle. The position vector has constantmagnitude rand a direction that varies as the object moves around the circle.

The velocity vector,

v dr

dt= , is non-zero solely because of the chang-

ing direction of r. The velocity vector is tangent to the circular path andalways perpendicular to

r.

The acceleration vector,

a dv

dt= , has two componentsa radial

component and a tangential component.The radial component is called the centripetal acceleration. It is

directed toward the center of the circle and has magnitudea

v

rC

=2

.

The tangential component of acceleration is zero if the magnitude ofthe velocity is constant. Otherwise, it causes changes in the velocity magni-tude.

If the velocity is constant,v

r

T=

2,where Tis the period of the

motionthe time to make one round trip. For constant velocity,

a

r

TC

=4 2

2

, and the tangential acceleration is zero.

REDUCTION TO A ONE DIMENSIONAL DESCRIPTIONIf we narrow down our focus to the circular path, like a driver on a circularrace track, circular motion can be described as one dimensional motion (on acurved path).

Position is measured from a reference position on the curve (tradition-ally the intersection of thex-axis with the curve, with thex-axis passingthrough the center of the circle). Traditionally, counter-clockwise is taken tobe the + direction around the circle.

We name the position along the circle s. sis a one-dimensional vector.

The velocity is also one-dimensional: v ds

dt= . v is tangent to the circle.

When we use this one dimensional model, we ignore the centripetalacceleration. The only acceleration is the component tangent to the circle,

a dv

dt= . If the object moves at constant speed, v=constant , the one-

dimensional acceleration (tangential acceleration) is zero.


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KINEMATICS


ANGULAR DESCRIPTION

Continue to use anx-axis that passes through the center of the circle as areference. The position vector, r, makes an angle with thex-axis. Tradi-tionally, is positive when it opens in the counterclockwise direction.

Taking the magnitude of ras given, measures the angular positionof the object.

We define angular velocity,

=d

dt, and angular acceleration,

= d

dt. For constant angular acceleration, we have the usual equations

for one dimensional motion with constant acceleration:

= + +1

2

2

0 0t t and = +t 0 , where the initial angular velocity

and initial angular position are 0 0and , respectively.

The relation between the angular and the linear description follows:s r= v v r= =

a rTANGENTIAL

=

a rCENTRIPETAL

= 2

TWO DIMENSIONAL COMPONENT DESCRIPTION

Taking the usualx-andy-axesxhorizontal andyvertical, with the origin atthe center of the circle, and using as defined above, we can write the vector

components of the position of an object that follows a circular path:x r y r= =cos sin and

We limit ourselves to the special case of zero angular acceleration.

Then, = +0 0t and = 0 , so = +t 0 .

We have x r t y r t= +( ) = +( )cos sin 0 0 and . Then the velocity

components are v r t v r t x y

= +( ) = +( ) sin cos0 0and .

The acceleration components are

a r t a r t x y

= +( ) = +( ) 20

2

0

cos sinand .


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Chapter 2NEWTNEWTNEWTNEWTNEWTONS LAONS LAONS LAONS LAONS LAWS OF MOWS OF MOWS OF MOWS OF MOWS OF MOTIONTIONTIONTIONTION

EQUILIBRIUM (FIRST LAW)

If you give an object a position and then arrange things so that it is left

alone, it keeps the position that you gave it.

If you give an object a velocity and then arrange things so that it is left

alone, it keeps the velocity (both magnitude and direction) that you gave it.

If you give an object an acceleration and then arrange things so that it

is left alone, the acceleration drops to zero the moment that you release it.Sir Isaac Newton wrote, Every body continues in its state of rest, or of

uniform motion in a right [straight] line, unless it is compelled to

change that state by forces impressed upon it. Principia, Mott

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