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: Free Response Questions (FRQ)

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
- Question 1 0:15
- Part A: I
- Part A: II
- Part A: III
- Part B
- Part C
- Part D: I
- Part D: II
- Question 2 5:46
- Part A: I
- Part A: II
- Part B: I
- Part B: II
- Part B: III
- Part B: IV
- Part B: V
- Question 3 13:30
- Part A: I
- Part A: II
- Part A: III
- Part A: IV
- Part A: V
- Part B
- Part C
- Part D: I
- Part D: II

### 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: Free Response Questions (FRQ)

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

*In our final lesson of Mechanics, we are going to go over the free response portion of the AP Physics C Mechanics practice exam.*0003

*Take a minute, print that out, give it a try, and come back here and we will see how things look.*0011

*Taking a look at question number 1 A, it says determine the average speed of glider A for those following time intervals, we want to know first from 0.1 to 0.3 s.*0017

*In that interval, our average velocity is change in position divided by time.*0029

*Our change in position from 0.1 to 0.3 s is 0.2 m and the time interval is 0.2 s.*0035

*The velocity there is 1 m/s.*0043

*For the next part, we are asked to find the average velocity between 0.9 and 1.1 s.*0048

*We used the same formula but now the displacement from 0.9 to 1.1 s, it looks like that is 0.12 m in the same time interval 0.2 s which is just going to be 0.6 m/s.*0057

*And A3, find the average speed from 1.7 to 1.9 s.*0075

*Same formula again, Δ x / Δ t, but now our displacement is 0.04 m in 0.2 s which is just going to be 0.2 m/s.*0083

*And that covers part A.*0098

*The let us move on to part B, we are asked to make a graph.*0100

*Here in part B, we are asked to sketch the speed of the glider as a function of time.*0106

*Our graph will look roughly like this, take your time, make sure you have everything plotted very neatly.*0112

*We will go from 1 to 2 s, so there is 1.5 to 0.5, 0.*0126

*Here is our velocity in meters per second from 0.5, 1, 1.5.*0134

*Our graph should look something like this.*0142

*It should go straight over for a spell then we are going to have a drop down to our final 0.2, make sure you move that over a little bit.*0146

*That looks like that is a little bit light, over to somewhere around there and off like that should give it a shape your graph if you plot it very carefully.*0158

*And for C, use the data to calculate the speed of glider B immediately after it separates from the spring.*0171

*That to me looks like a conservation of momentum sort of question.*0177

*Let us do this one with a momentum table.*0182

*We will have the momentum before, momentum after for A, B, and their total.*0185

*The momentum of A before hand is 1 × 0.9 so that will be 1 × 0.9 = 0.9.*0200

*B is at rest so that is 0.*0207

*The total before is 0.9, and after the collision we have the same mass 0.9 × 0.2 is going to be our .18.*0209

*We also have for B, its mass 0.6 × unknown velocity VB, so 0.9 is going to be equal to 0.18 + 0.6 VB.*0221

*We can solve that for VB, 0.9 -0.18 is going to be 0.72 = 0.6 VB or VB = 0.72/0.6 is going to be 1.2 m/s.*0231

*There is part C, C1.*0249

*C2, asks us to plot the speed of the glider B as a function of time.*0252

*We can even, for our purposes, I'm just going to plot that in blue on our same graph because we have the same axis.*0258

*It is going to come down like that and it is going to then come up here, into roughly the same rate and same points, up to 1.2 m/s at its highest point.*0265

*There is our graph for glider B compared to A.*0279

*Let us give ourselves a little bit more room for part D.*0287

*It shows us the total kinetic energy as a function of time as the collision is elastic.*0295

*The kinetic energy before the collision = the kinetic energy after, therefore it is elastic, yes.*0300

*Justify your answer.*0309

*The kinetic energy before = the kinetic energy after, state that in word somehow.*0310

*For part 2, why is their minimum and kinetic energy at 1 s?*0315

*The kinetic energy is being stored as elastic potential energy at that point while the spring is compressed.*0320

*I would explain that in words too.*0328

*During that deep end kinetic energy, some of the kinetic energy is being transferred to stored potential elastic energy,*0330

*then the spring decompresses again and will be back kinetic energy.*0337

*That covers number 1, moving on to Mechanics question 2.*0344

*Let us see here, a space shuttle astronaut playing around.*0353

*We have 2 masses connected by rigid rod of length L and negligible mass and the device is a small lump of clay of mass M at some speed V knot.*0358

*Determine the total kinetic energy of the system after the collision.*0370

*For M2 part A, we have MV initial must be equal to the total mass × the final velocity which will be 3 M, let us call that V final.*0374

*Complies that V final is just going to be equal to MV knot / 3 M which is V0 / 3.*0387

*If we want the final kinetic energy after the collision, the final kinetic energy is going to be ½ × 3 M the total mass × our final velocity V knot /3²,*0396

*which will be 3 M /2 × V knot² /9, which is MV knot² /6.*0408

*There is A, for part 2, determine the change in kinetic energy as a result of the collision.*0420

*A2, initial kinetic energy is 1/2 M V initial² which is just M V knot² /2.*0427

*Our change in kinetic energy is going to be the final kinetic energy - the initial kinetic energy*0440

*which is MV knot² /6 from up there - MV knot² /2 from right there.*0448

*That is MV knot² / 6 -3 MV knot² / 6 which is just going to be - MV knot² / 3.*0457

*Moving on 2 part B, as we look at B part 1, the assembly is brought to rest, the lump of clay is pulled off,*0472

*we hit again but this time we had it stick to one of the spheres of the assembly over on the side.*0483

*Determine where the center of mass is going to be after the collision.*0489

*Our position vector to the center of mass is just the sum for all of our different masses of MIRI divided by the total mass M so that is going to be,*0493

*if we count the left side as 0, we have got M × 0 + 2 M and the distance L /3 M our total mass which is just going to be 2 L /3.*0505

*Alright and for part 2 B2, on the figure above, indicate the direction of the motion of the center of mass immediately after the collision.*0522

*That is straightforward, if the clay that is coming toward is going up, afterwards we must have the exact same thing so up would be the correct direction there.*0532

*In part 3 B3, determine the speed of the center of mass immediately after the collision.*0544

*The final velocity is just going to be the initial velocity divided by 3, conservation of momentum.*0553

*Moving on to B4, determine the angular speed of the system immediately after the collision.*0563

*B4, the way I would do that is look at the angular momentum, initial = the final angular momentum*0574

*which implies that MVR sin θ = I ω or M, V knot L /3 r sin θ, sin 90 is 1 = Iω.*0585

*We can solve for ω which is just going to be, we will have MV knot L /3 × the moment of inertia.*0604

*To figure out the moment of inertia, if I want to go any further with this, the moment of inertia is just the sum of our Mr²*0617

*which is going to be M × 2 L/3² + 2 M × its distance from the center of mass L /3² which is going to be M × 4 L² /9 + 2 ML² /9, which will give us 6 ML²/9 or 2 ML² /3.*0626

*I can use that in there for my moment of inertia.*0653

*Ω = M V knot L /3 divided by our moment of inertia which is 2 ML² /3*0658

*3 makes a ratio of 1, our masses will cancel out, we will lose one of our L's which implies then that ω is just going to be equal to V knot /2 L.*0672

*B5, determine the change in kinetic energy as a result of the collision.*0688

*B5, our initial kinetic energy was 1/2 M V initial².*0694

*Our final kinetic energy, now we have translational and rotational components.*0701

*We have to add those up, that is going to be ½ × our total new mass 3 M × V knot /3² + our rotational ½ × I*0707

*which we determined up here was 2 ML² /3 × √angular velocity which was V knot /2 L².*0721

*This implies then that our final kinetic energy is going to be equal to, this left hand side we are going to have / 3,*0740

*we are going to have MV knot² /6 + 2 ML² / 6 × V knot² /4 L² is going to be MV knot² / 6 + MV knot² /12, which is going to be MV knot² /4.*0750

*Our change in kinetic energy which is final - initial is going to be MV knot² /4 for right there, - our initial 1/2 MV knot², which is just - MV knot² /4.*0781

*That finishes up question 2, as we move to our third question, this is a pretty complex situation.*0806

*However, they try and make it simpler by walking through the different pieces step by step using there are*0819

*a lot of little pieces we are going to take bit by bit in order to put the whole situation together.*0823

*For M3 A1, that first asks us to draw vector on the block and determine the magnitude of the force.*0829

*We are looking for the normal for on M1.*0841

*There is M1, normal force on it going up and M1 = M1 G.*0845

*For part 2, a frictional force exerted on the block 1 by block 2.*0858

*It is at rest, we have got M1 here, the frictional force is going to be 0.*0865

*A3, find the force T exerted on block 2 by the spring.*0873

*Will here is our M2, by the string there is our tension.*0883

*Tension must be equal to MG.*0890

*For A4, now we are looking at the normal force N2 inserted on block 2 by the tabletop.*0896

*Here is M2, we are looking for the normal force exerted on it.*0907

*There is N2, to figure this out I will probably get to the point where I'm drawing free body diagram just help me out.*0911

*We have the force N2 up, we have M2 G down, and M1 G down.*0919

*Those are all balance so I can write here that N2 = M1 G + N2 G.*0924

*That will cover A4, let us go to the next page for A5.*0932

*The frictional force F2 exerted on block 2 by the tabletop.*0942

*Block 2, there it is the frictional force F2 opposing motion.*0946

*To figure our free body diagram, we would have F2 and over here T which we know is MG.*0954

*I can write then that F2 = MG.*0962

*We have got all our free body diagrams.*0969

*For part B, determine the largest value of M for which the blocks can remain at rest.*0971

*When this happens, our force of friction is going to be μ S2 × N2 which has to equal MG.*0977

*But we just previously determined that N2 was equal to M1 G + M2 G so we can rewrite this as μ S2 × M1 G + M2 G = MG.*0987

*Or solving for M, that is going to be μ S2 × (M1 + M2).*1007

*Moving on to C, now suppose M is large enough that the hanging blocked descends when the blocks are released.*1021

*Assume blocks 1 and 2 are moving easy without slipping, find the acceleration.*1028

*A little bit trickier situation, let us draw our free body diagram for our hanging block.*1035

*We have T and MG.*1040

*MG - T = MA calling down our positive y direction here and T, therefore, = MG – MA.*1043

*And if we go look at our block situation, we have the normal force up, we have their combined weight, M1 + M2 down,*1054

*we have the tension pulling them to the right, and we have our force of friction.*1066

*In this case, as we look in the direction of the motion T - F2 is going to be equal to M1 + M2 A.*1073

*But we know a little bit more here as well, F2 is μ K2 × the normal force, which means that F2 = μ K2 × M1 + M2 G.*1084

*We can write that as T - μ K2 × (M1 + M2 G) = M1 + M2 × A.*1101

*We can combine this equation and that equation, to write that our tension MG – MA.*1116

*We still have our – μ K2 × M1 + M2 G = M1 + M2 × A or with a little bit of rearrangement, trying to get all my G on one side and all of my A on the other.*1128

*We have G × M – μ K2 × (M1 + M2) = a × M1 + M2 + M or solving for A by dividing both sides by (M1 + M2 + M),*1147

*I find out that a = M – μ K2 × (M1 + M2) ÷ the sum of our masses M1 + M2 + M all of that multiplied by G.*1168

*And that should cover us for part C, looking at part D, now suppose M is large enough that the hanging block ascends block 1 is slipping on block 2.*1188

*Find the magnitude A1 of the acceleration the block 1.*1204

*D part 1, our M1 block we have normal force 1 up on it, we have the frictional force 1 to the right, and M1 G down.*1209

*The net force in the x direction, in the direction of motion is F1 = M1 a, which implies then that A1 = F1 / M1.*1223

*We know that F1 = μ K1 M1 G so then A1 = μ K1 M1 G / M1 or making a ratio of 1 from M1/M1.*1238

*A1 = μ K1 × G.*1261

*Part D2, find the magnitude A2 of the acceleration the block 2.*1270

*Our free body diagram for block 2, there it is, we have got N2 up, we have tension to the right, we have friction 1 to the left,*1278

*we have friction 2 to the left, we have M1 G down, and we have M2 G down.*1292

*For our hanging mass M, we have tension up, and MG.*1301

*As we look at this, starting with the one on the right MG - T = MA 2, which implies that T = MG - MA 2.*1310

*And if we look at our left most free body diagram over here, we have T - F1 - F2 = M2A2 which implies then that the T we can replace with MG - MA2.*1326

*This becomes MG - MA2 - F1 - F2 = M2A2 which implies then that MG - F1 - F2 = we will have M2A2 + MA2,*1345

*which implies that A2 is going to be equal to MG - F1 - F2 ÷ M2 + M.*1372

*We know a couple more things about friction.*1385

*We know our frictional force 1 = μ K1 M1 G and we know there frictional force 2 = μ K2 × (M1 + M2 G).*1388

*Therefore, we can write this as A2 B2 = MG - F1 which is μ K1 M1 G - F2, which is μ K2 M1 + M2 × G / M2 + M.*1406

*A little bit of simplification, A2 = M – μ KM 1 - μ K2 × (M1 + M2) ÷ M 2 + M, that whole thing × G.*1433

*Pretty involved there but the previous questions help walk you through that.*1458

*Alright, that ends our free response test.*1463

*Hopefully, you get a good feel for where you are strong and areas you need to work a little bit more.*1466

*Thank you so much for joining us at www.educator.com.*1471

*Good luck and make it a great day everyone.*1473

6 answers

Last reply by: Sohan Mugi

Fri Apr 15, 2016 8:46 AM

Post by Sohan Mugi on March 23, 2016

Hey Professor Fullerton. I just had a quick question about the 2016 AP Physics C: Mechanics Exam. How are these questions exactly scored? What is the process that collegeboard uses and will most likely use this year to grade our exams to determine our AP Score(3,4,5)?

0 answers

Post by Professor Dan Fullerton on March 27, 2015

http://apcentral.collegeboard.com/apc/public/courses/211624.html