AP Physics C: Mechanics 1998 AP Practice Exam: Free Response Questions (FRQ)

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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 energy0440

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 momentum0574

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 Mr²0617

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 /30658

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 × I0707

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 are0819

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