For more information, please see full course syllabus of AP Physics C: Mechanics

For more information, please see full course syllabus of AP Physics C: Mechanics

### 1998 AP Practice Exam: Multiple Choice

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Problem 1
- Problem 2
- Problem 3
- Problem 4
- Problem 5
- Problem 6
- Problem 7
- Problem 8
- Problem 9
- Problem 10
- Problem 11
- Problem 12
- Problem 13
- Problem 14
- Problem 15
- Problem 16
- Problem 17
- Problem 18
- Problem 19
- Problem 20
- Problem 21
- Problem 22
- Problem 23
- Problem 24
- Problem 25
- Problem 26
- Problem 27
- Problem 28
- Problem 29
- Problem 30
- Problem 31
- Problem 32

- Intro 0:00
- Problem 1 0:30
- Problem 2 0:51
- Problem 3 1:25
- Problem 4 2:00
- Problem 5 3:05
- Problem 6 4:19
- Problem 7 4:48
- Problem 8 5:18
- Problem 9 5:38
- Problem 10 6:26
- Problem 11 7:21
- Problem 12 8:08
- Problem 13 8:35
- Problem 14 9:20
- Problem 15 10:09
- Problem 16 10:25
- Problem 17 11:30
- Problem 18 12:27
- Problem 19 13:00
- Problem 20 14:40
- Problem 21 15:44
- Problem 22 16:42
- Problem 23 17:35
- Problem 24 17:54
- Problem 25 18:32
- Problem 26 19:08
- Problem 27 20:56
- Problem 28 22:19
- Problem 29 22:36
- Problem 30 23:18
- Problem 31 24:06
- Problem 32 24:40

### AP Physics C: Mechanics Online Course

I. Introduction | ||
---|---|---|

What is Physics? | 7:12 | |

Math Review | 1:00:51 | |

II. Kinematics | ||

Describing Motion I | 23:47 | |

Describing Motion II | 36:47 | |

Projectile Motion | 30:34 | |

Circular & Relative Motion | 30:24 | |

III. Dynamics | ||

Newton's First Law & Free Body Diagrams | 23:57 | |

Newton's Second & Third Laws of Motion | 23:57 | |

Friction | 20:41 | |

Retarding & Drag Forces | 32:10 | |

Ramps & Inclines | 20:31 | |

Atwood Machines | 24:58 | |

IV. Work, Energy, & Power | ||

Work | 37:34 | |

Energy & Conservative Forces | 28:04 | |

Conservation of Energy | 54:56 | |

Power | 16:44 | |

V. Momentum | ||

Momentum & Impulse | 13:09 | |

Conservation of Linear Momentum | 46:30 | |

Center of Mass | 28:26 | |

VI. Uniform Circular Motion | ||

Uniform Circular Motion | 21:36 | |

VII. Rotational Motion | ||

Rotational Kinematics | 32:52 | |

Moment of Inertia | 24:00 | |

Torque | 26:09 | |

Rotational Dynamics | 56:58 | |

Angular Momentum | 33:02 | |

VIII. Oscillations | ||

Oscillations | 1:01:12 | |

IX. Gravity & Orbits | ||

Gravity & Orbits | 34:59 | |

X. Sample AP Exam | ||

1998 AP Practice Exam: Multiple Choice | 28:11 | |

1998 AP Practice Exam: Free Response Questions (FRQ) | 28:11 |

### Transcription: 1998 AP Practice Exam: Multiple Choice

*Hello, everyone, and welcome back to www.educator.com.*0000

*I’m Dan Fullerton and in this lesson we are going to take an old AP practice exam and walk through it step by step.*0004

*We are going to start with a multiple choice portion of the test.*0010

*You can find it with the www.google.com search or from the AP Physics site.*0014

*The links are down below as well.*0018

*We are just going through the multiple choices one by one.*0020

*Take a minute, give it a try, and come back here and we will see how you did as we walk through each of the problems.*0023

*Starting with the number 1, we have got a force exerted by a broom handle on a mass at some angle, the work done is what?*0031

*This is just the definition of work, force × displacement × cos of the angle between them and that looks like answer B.*0040

*Number 2, we have got a projectile launched with the horizontal component and vertical component of velocity, no air resistance, *0053

*and when it is at its highest point what do we know about the vertical velocity, horizontal velocity, and vertical acceleration?*0060

*Its highest point, the vertical velocity is going to be 0.*0067

*We know the horizontal velocity is not going to change, it is going to remain constant.*0072

*Our vertical acceleration, the entire time it is in the air is just g.*0076

*It looks like our answer there must be E.*0081

*Taking a look at number 3, here we have a graph showing the velocity as a function of time for an object moving in a straight line, *0087

*find the corresponding graph that shows displacement as a function of time.*0095

*The trick here is realizing that displacement is the area under the VT graph or if we look at the position time graph, the slope of that should give you velocity.*0101

*To me, the only graph that it looks like it comes even close there is going to be D.*0112

*Number 4, position of a toy locomotive is given by x = T³ -6 T² + 9 T and we want to find the net force.*0121

*In right away, I’m thinking if we want net force, let us find acceleration and multiply that by the mass.*0137

*If we are given position, velocity which is X prime, the first derivative that is going to be 3 T² -12 T + 9, *0143

*so acceleration which is the second derivative of velocity or the second derivative of position or the first derivative of velocity is just going to be 6T-12.*0152

*We are looking the net force on locomotive is equal to 0 when T is equal to, for that to be the case, our acceleration has to be 0, that means 6T-12 equal 0.*0164

*6T must equal 12 or T = 2s, so 2s is the answer D.*0178

*Number 5, we have a system of wheels fix to each other free to rotate about a frictionless axis, four forces are exerted on the rims of the wheels, find the net torque.*0189

*The net torque, we just add those forces × their displacements, making sure we take into account their direction.*0200

*It looks like on all of these, the sign of the angle sin θ that is going to be 90° so 1.*0207

*Our net torque, we have from top to bottom, we have -2 F × 3 R, negative because it is causing a clockwise rotation.*0213

*We have then + F × 3 R /on the left, we have + F × 2 R and we have + F × R 3.*0224

*When I put that all together, I have -6 FR + 8 FR, for a net torque of 2 FR and our answer is C.*0244

*Looking at number 6, the wheel of mass M and radius r, rolls on a level surface without slipping.*0260

*If the angular velocity of the wheel is ω what is its linear momentum?*0268

*If V = ω r and we know p=MV, that is just going to be M ω r.*0272

*Now that is easy, the answer is A.*0283

*Let us take a look at number 7, 7 and 8 refer to the same situation.*0288

*A ball tossed straight up from the surface of a small spherical asteroid with no atmosphere.*0294

*It rises to a height equal to the asteroid’s radius and then fall straight down towards the surface of the earth.*0301

*What forces act on the ball on its way up?*0305

*The only force it is going to act on it is going to be gravity pulling it down and that is going to decrease the higher the ball gets, so the answer has to be A.*0309

*And number 8, the acceleration of the ball at the top of its path is?*0320

*The acceleration is going to be ¼ what it was when it was on the surface due to the inverse square law?*0325

*That is going to have to be answer is D, it looks like.*0332

*There is 7 and 8, let us move on and take a look at number 9.*0336

*The equation of motion of the simple harmonica oscillator is d² x / dt² = -9x, find the period of oscillation.*0343

*Let us write this first in our standard differential equations form, d² x / dt² + 9x = 0.*0354

*And remember, that value right there is what we call ω².*0364

*If ω² = 9 then ω must equal 3 and period is 2π / ω which is going to be 2π / 3.*0369

*We will see if that is one of our answers, yes that is D.*0381

*There we go, taking a look at number 10.*0385

*A pendulum with a period of 1s on earth with the acceleration due to gravity is G, is taken to another planet where its period is 2 s.*0391

*The acceleration due to gravity on the other planet is most nearly, on earth we know the period on any planet at 2 π √L /G.*0399

*As we look at that, on the new planet its period goes to 2s which means we must have 1/4 gravity because *0413

*we have this inverse square root relationship because it is proportional to √L /G must be ¼ G in order to double the period.*0422

*The answer has to be a, G/4.*0432

*You can work that out mathematically as well with all the details and numbers if you want to.*0437

*Number 11, a satellite of mass M moves in the circular orbit of radius r with constant speed V, which of these statements are true?*0444

*Its angular speed is V /r, angular speed is V/r, that is one of our definitions, one is true.*0454

*Its tangential acceleration is 0, of course it is moving at a constant speed so 2 is true.*0463

*Let us see, the magnitude of it centripetal accelerations is constant, AC = V² /R, V is not changing, r is not changing, *0471

*so that has to be constant, so 3 is true.*0481

*All 3 of those, E must be the right answer.*0484

*Number 12, we have a graph of force vs. Time and for the time interval from 0 to 4 s, the total change of momentum is?*0491

*For number 12, change in momentum impulse is the area under the force time graph and our total area under the graph there is 0.*0502

*Therefore, the answer has to be C.*0511

*Taking a look at 13, we have a disk of mass M moving horizontally to the right with some speed V.*0516

*It is going to collide with a disk of mass 2 M moving with b/2, find the speed after the collision.*0524

*That looks like a conservation momentum problem where initial momentum we have MV + 2 M × V / 2 *0531

*which is going to be equal to our total combined mass 3 M times some unknown velocity V prime.*0540

*MV + MV 2 MV = 3 MV prime or V prime is just going to be equal to 2 V /3 and that is the answer C.*0548

*On the 14, 14 and 15 both refer to this object attached to a spring, moving in a circle, what is the centripetal force on the disk?*0562

*That we can get from Hooke’s law, that force is going to be jx which is 100 N/m × the displacement *0574

*from its equilibrium our happy position, 0.03 m is just going to be 3 N, answer B.*0582

*What is the work done on the disk by the spring during one full circle?*0593

*There is a trick question, the force is being applied towards the center of the circle but the displacement is always tangent to that, perpendicular to that at 90°.*0599

*The work done in this case is going to be 0.*0610

*For 15, we can state that A must be the right answer.*0613

*Moving on to 16 here, 16 and 17 refer to this graph.*0621

*If a particle was released from rest, the position are 0, its peak position toward 0 is most nearly.*0629

*That looks like a conservation of energy problem where our initial potential + our initial kinetic energy must equal our final potential + final kinetic energy.*0635

*And our initial potential energy when we are at our 0 is 3u 0 must equal, at 2r 0 our potential is u 0 + whatever energies left must be our final kinetic energy.*0647

*Therefore, final kinetic energy must be 2 u0 and kinetic energy is ½ MV².*0663

*Solving this, MV² must equal 4u 0 and therefore, V² must equal 4u 0 /M or V = √4u 0/M and that looks like answer is C.*0670

*Let us take a look at 17, if the potential energy function is given by that equation which of the following is an expression for the force?*0693

*Force is the opposite of the derivative of the function along that path r which is going to be - the derivative with respect to *0704

*r of our function VR to the -3/2 + C which is going to be, we can pull the B out of there is a constant that will be -2 B × our derivative *0713

*which will be -3/2 r⁻⁵/2 which implies then that our force is going to be equal to, our negative will cancel out and I will have 3 B /2 r⁻⁵/2, *0726

*which looks like that is going to be answer A.*0743

*Moving on to 18, we have got a frictionless pendulum of length 3m swinging with an amplitude of 10° so we can use that small angle approximation it is under about 15.*0749

*Its maximum displacement, the potential energy is 10 joules, what is the kinetic energy when its potential energy is 5 joules?*0761

*Energy is conserved that is going to be 5 joules, answer B, 5 + 5 = 10.*0772

*Alright 19, we have got a descending elevator, a 1000 kg, uniformly accelerate to rest over a distance of 8 m and tell us the tension in the cable.*0782

*The speed Vi of the elevator at the beginning is most nearly what?*0792

*We can start by finding the acceleration by using a free body diagram.*0796

*Tension up it tells us is 11,000 N we will call down the positive y direction and we have the weight of the elevator, MG which is going to be 10,000 N.*0802

*The net force in the y direction must be -1000 N and that must be equal to our mass × our acceleration.*0813

*Our mass is 1000, therefore, the acceleration must be just -1 m /s².*0822

*It becomes a kinetics problem, we are trying to find V initial, V final = 0, Δ y is 8m, and the acceleration in the y is -1 m /s².*0829

*We could use V final² = V initial² + 2a Δ y or V initial² = V final² -2a Δ y, substitute in for our values, VF² that will be 0² -2 × -1 × 8m.*0842

*Vi² = 6 T m² / s², Vi is square root of that which is 4 m /s, which is the answer A, so there is 19.*0863

*Taking a look here at 20, two identical stars at fix distance D apart, revolve at the circle about their center of mass.*0881

*Each star has a mass M, speed V, which is a cracked relationship among those quantities?*0890

*As I look at 20 here, if they are revolving around each other, they are moving in the circular path, the net centripetal force is MV² /r, *0897

*which implies then that G × the mass of the first × the mass of the second divided by the square of the distance between them must equal MV² /r is D /2, *0907

*the circular path which implies with a little manipulation that V² = GM × M × D /D² × 2 M.*0922

*Or solving for V² that is GM /2 D which is answer was B.*0936

*On 21, block of mass M is accelerated across a rough surface by a force of magnitude F exerted an angle 5 with a horizontal.*0946

*Frictional force has magnitude F, find the acceleration of the block.*0959

*Let us start with a free body diagram here.*0963

*We have our normal force up, we have the weight down, we have some applied force F at some angle of I, and we have a frictional force.*0966

*The net force in the x direction is going to be F cos I, the component of F along the x - F is equal the MA, *0979

*which implies that the acceleration must be F cos i - the frictional force divided by the mass which looks like that is answer D.*0990

*And 22, what is the coefficient of friction between the block and the surface?*1005

*To do that, let us start looking in the y direction.*1009

*Net force in the y direction is going to be M + F sin i - MG all equal to 0 because it is not accelerating in the y direction, *1012

*which implies that the normal force = MG - F sin i.*1023

*The frictional force, friction is fun so that is μ × the normal force which implies that the coefficient of friction μ is our frictional force / normal force *1030

*is going to be our frictional force / our normal force we said was MG – F sin i.*1041

*It looks like that answer E.*1050

*Alright, 23, was a problem that they skipped on the exam.*1058

*There is either a problem with the question or the solution, something did not come out right.*1062

*We will skip that one and move on to 24.*1066

*Alright, 2 people are initially standing still on frictionless ice, they push on each other so that one person of mass 120 kg moves to left at 2 m /s.*1077

*The other person mass 80 kg moves to the right at 3 m /s, what is the velocity of the center of mass?*1086

*Another trick question, the initial velocity of the center of mass is 0 and there no external forces.*1093

*The final velocity of the center of mass must be 0, the answer there must be A.*1099

*Moving on to 25, for 25 here, we have a figure of a dancer on a music box moving counterclockwise at constant speed, *1108

*which of those graphs best represents the magnitude of the dancer’s acceleration as a function of time during one trip around beginning at point P?*1121

*It begins from point P, from P to Q there is no acceleration and from R to S there is no acceleration.*1131

*And from Q to R, during those current parts the acceleration is constant AC = V² / r.*1137

*I think we are looking for something that looks like B.*1144

*26, at target T lies flat on the ground 3 m from the side of a building that is 10 m tall.*1150

*Student rolls a ball off the horizontal roof and the direction of the target.*1158

*The horizontal speed of which the ball must leave the roof distract the target is most nearly what?*1162

*Let us first figure out how long it is going to take in order to hit the ground.*1168

*We have done that many times now, that is a vertical problem, where Vi vertically is 0 or call down positive y.*1173

*V final we do not know, Δ y is 10 m, AY 10 m /s² and T is what we are trying to find.*1182

*Δ y = V initial T + ½ AYT² and since V initial is 0, that term goes away.*1192

*Δ y = ½ AYT² or T = 2 Δ y / √ay which is going to be 2 × 10 m / 10 m /s² square root which is just going to be √2s.*1204

*The horizontal analysis, we know that Δ x needs to be 3 m.*1225

*We know our time is √2s, so the velocity in the x direction is just Δ x / t which is 3m /√2s, 3m /√ 2s, for number 26 looks like the answer must be C.*1232

*And 27, to stretch a certain nonlinear spring by an amount, x requires that force, what is the change in potential energy when it stretch 2 m from its equilibrium position?*1257

*We need to find the amount of work done in stretching the spring from 0 to 2 m and the work done will be the potential energy stored in that spring.*1268

*Our potential energy will be the integral from x = 0 to 2 m of FX DX which is the integral from 0 to 2.*1279

*Our force is 40x-6 x² Dx, so that is going to be 40x² / 2 -6x³/ 3 all evaluated from 0 to 2, which is going to be, *1290

*let us simplify this first, 20 x² -2x³ evaluated from 0 to 2, which will be 20 × 2² -2 × 2³ -0 -0 is going to be 20 × 2² that is going to be 80 - 16 or 64 joules.*1306

*27 must be D.*1335

*Alright 28, when the block slide a certain distance down an incline, the work done by gravity is 300 joules.*1339

*What is the work done by gravity if the block slides the same distance up the incline?*1347

*That is going to be -300 joules, the answer is C.*1351

*29, particle moves in the xy plane with coordinates given by x = a cos ω T and Y = a sin ω T, where A is 1½ m and ω is 2 radians /s.*1358

*What is the particles acceleration?*1372

*Remember, the magnitude of the acceleration is ω² A which is going to be 2 radians /s² × A 1.5 m is just going to be 6 m/s², answer must be E.*1375

*Let us take a look here at number 30, for the wheel and axel system shown, *1398

*which in the following expresses the condition required for the system to be in static equilibrium?*1405

*For 30, we are looking for all of our torques and our forces has to be balanced.*1411

*Right away, I can say that M1 G, the force of gravity on block 1 × its distance from our axle A must be equal in magnitude, 2M2 G B.*1421

*Therefore, AM1 = BM 2, that is going to be answer B.*1435

*Taking a look here at 31, an object having an initial momentum is represented by that vector above, *1448

*which in the following sets of vectors represent the momentum of the 2 objects after the collision?*1456

*The sum of those 2 vectors has to be the exact same of what you have above because you do not have any external forces here .*1461

*If conservation of linear momentum, that looks like the only vectors that we could add together to get what you have up above there is E.*1468

*32, a wheel with some rotational inertia mounted on a fix frictionless axle with the angular speed ω is increased from 0 to ω final at time T.*1481

*What is the net torque required?*1490

*Net torque is moment of inertia × angular acceleration which is going to be moment of inertia × angular acceleration *1492

*is the change in angular velocity with respect to time, which is going to be I ω final / T, since ω initial was 0.*1502

*At 32, it looks like our answer must be E.*1514

*33, what is the average power input to the wheel during this time interval?*1522

*Remember, power is force × velocity in a linear world.*1527

*Rotationally, we can do that same analog.*1531

*Instead of force, we are going to have torque.*1533

*Instead of average velocity, we are going to have average angular velocity so that is going to be I ω final / T our torque × average rotational velocity *1536

*is going to be halfway between the initial and final values because it is a constant angular acceleration or ω final /2, which will give us I ω final² / 2 T, which is answer B.*1550

*To 34, an object is released from rest to time T = 0 and falls with an acceleration given by a = G – BV, where B is the object’s speed.*1571

*V is the object’s speed, B is a constant.*1583

*Our drag force, retarding force question.*1584

*Which of the following is a possible expression for the speed of the object as an explicit function of time?*1588

*A = G - BV and we know we can draw the graph to begin with, knowing that our acceleration start at some value G.*1593

*We are going to have an exponential BK as a function of time.*1602

*We just have to find which of those functions fits that shape.*1607

*Initially, a if t is 0, we are going to have V = G × 1 -0 / B, so that is going to be G × 1 ÷ B.*1611

*We are going to start at the high point and as it gets bigger, we are going to get lower and lower.*1625

*Right away, A fix that shape.*1629

*And 35, ideal mass of spring fix to the wall, block of mass M oscillate with amplitude A and maximum speed VM.*1636

*Find the force constant of the spring.*1646

*Looking at conservation of energy, ½ KA² the maximum potential energy in the spring must equal ½ MV² its maximum kinetic energy, *1650

*which implies then that K is going to be equal 2 MV² × 2/2 A² which is just MVM² / a².*1662

*The correct answer is D.*1675

*Hopefully that gets you a good feel for where you are strong on the concepts for this test and areas that need a little bit more improvement.*1679

*Thank you so much for watching www.educator.com.*1686

*I look forward to seeing you again soon and make it a great day everybody.*1688

1 answer

Last reply by: Professor Dan Fullerton

Sun Sep 18, 2016 7:08 AM

Post by Shive Gowda on September 18 at 12:32:27 AM

I could not find the book.

1 answer

Last reply by: Professor Dan Fullerton

Sat Apr 25, 2015 5:04 PM

Post by Micheal Bingham on April 24, 2015

For number 7 why isn't the gravitational force constant on the asteroid, like on Earth?

3 answers

Last reply by: Professor Dan Fullerton

Thu Mar 17, 2016 5:55 AM

Post by Dianfan Zhang on March 27, 2015

Where are the links for this 1998 practice exam? The links from google seem to be different exams. Thanks