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For more information, please see full course syllabus of AP Physics C: Mechanics
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Lecture Comments (14)

1 answer

Last reply by: Professor Dan Fullerton
Sat May 6, 2017 5:28 PM

Post by Woong Ryeol Yoo on May 6, 2017

Hi mr. Fullerton.
Is the equation sheet given for the Multiple Choice on both the AP physics c exams?
I know the past exams only gave table of information for the MCs. Does the current exam give equation sheet and table of information for the MCs?

1 answer

Last reply by: Professor Dan Fullerton
Thu Mar 23, 2017 5:17 AM

Post by Woong Ryeol Yoo on March 22, 2017

Hi Mr. Fullerton.
So I only missed two questions in the 1998 released exam. But I was wondering if this exam is outdated in terms of the problem style and contents. Is the current c mech exam way different than the 1998 exam? Is it more difficult now?

1 answer

Last reply by: Professor Dan Fullerton
Tue Dec 20, 2016 6:14 AM

Post by Simon Fei on December 19, 2016

How did you get the formula |a|=omega^2*A for number 29?

1 answer

Last reply by: Professor Dan Fullerton
Sun Sep 18, 2016 7:08 AM

Post by Shive Gowda on September 18, 2016

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

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 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

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 because0413

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 / 20531

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 displacement0574

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 to0704

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 derivative0713

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 force1030

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 acceleration1492

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 velocity1536

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