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AP Practice Exam: Free Response, Part 1

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 0:23
  • Question 2 8:55

Transcription: AP Practice Exam: Free Response, Part 1

Hi folks and welcome back to Educator.com. 0000

Today we are going to continue as we tackle the 1998 AP Physics B released examination. 0003

What we are going to start today are the free response problems and we are going to take them in batches of two problems each because there are eight problems in the test, so we will do 1 and 2 now, then we will take a break and come back to 3 and 4. 0010

For now, why not dive in and let us get right in to problem Number 1. 0020

What we have to begin with are a couple of blocks with one on a table and the other hanging off the table attached by a light string. 0025

To begin with, it asks us to determine the acceleration of block (B), the hanging block as it descends. 0032

First thing I am going to do in a problem like this is I am going to draw some free-body diagrams (FBD), so for number 1 here in section A, if I were to draw the FBD of block (A), I will have my object, we have its weight with the force of gravity down, the normal force up, and the force of tension to the right. 0039

I am going to neglect friction, because it talks about it being a smooth table top. 0062

Now, for block (B), we will do the same thing. 0068

For its FBD, we have gravity pulling it down and tension pulling it up. 0072

In this case, we know the tensions of these are going to have to be equal; it is the same string. 0079

Let us start and write Newton's Second Law equation, but as we do this, since it is going to go over the table and down, let us call that the positive y-direction. 0084

We are going to wrap our axis around the corner of the table. 0094

The Newton's Second Law in the y-direction for (A) is going to have (t) going in that direction and that is just going to be equal to mass (A), which is (m) times its acceleration (a). 0097

Let us do the same thing for (B). 0114

Net force in the y-direction for (B) -- well (mg) is down in the direction of positive y, so mg - t = ma.0116

With those two: t = ma and mg - t = ma, I can combine those since they are both equal to (ma), I could say t = mg - t. 0128

Let us do that: t = mg - t and then if I solve this, I can find that 2t = mg or t = mg/2, but wait, I also know that t = ma...0139

...so I am going to replace (t) here with (ma) to say that ma = mg/2 and if I divide an (m) out of both sides, I found the acceleration (a = g/2). 0159

Excellent. Good start. 0175

Now for part B -- It says (B) strikes the floor, so determine the time at which it strikes the floor. 0177

Well, that is going to be a kinematics equation, where δy is going to be our initial velocity in the y-direction times (t) plus 1/2 ay(t)2. 0186

Since the initial velocity of (B) in the y-direction is 0, that term goes away, therefore, I could write this as t = 2δy/ay (square root), where δy is just going to be the distance that (B) falls, which is (h) and ay = g/2. 0200

So square root there and that is going to be square root of 4h/g or pull out the square root of 4 and we have two square root, so h/g must be the time. 0226

All right, so there is B. 0243

Now for Part C, it says describe the motion of block (A) from the initial time until (B) strikes the floor. 0247

Well, during that time, block (A) is just going to accelerate horizontally across the table, so let us just write that. 0258

Block (A) accelerates horizontally across the table. Terrific!0268

Now on to Part D -- Describe the motion of (A) from the time (B) strikes the floor until (A) leaves the table. 0288

Well, in this case, there is no longer anything causing an acceleration, but we also do not have any friction, so it is just going to keep moving at a constant velocity...0295

...so I would write block (A) continues horizontally at constant speed. 0303

We can do that because we already assumed it was frictionless because we said the table is smooth. 0322

Part E -- Determine the distance between the landing points of the two blocks. 0329

If we draw our table here, we know that (B) just went straight down and (A) is going to fly off at some velocity, so we need to know this distance, which is really how far (A) goes as a projectile. 0336

To do this, the first thing I am going to do is I am going to find out what (B)'s velocity was right before it hit the ground because that will be the same as the initial velocity of (A) as it leaves the table. 0351

To do that, I will say that the final velocity of (B) in the y-direction is going to be V-initialB + ay × t. 0362

And again the initial velocity of (B) was 0, so that is just going to be our acceleration (g/2) times our time, which we already determined was 2 square root (h/g), so that is going to be (g) square root (h/g) or just the square root of h × g. 0377

If that is the velocity of (B) right before it hits the ground that has to be the velocity of (A) just when (B) hits the ground and when it leaves the table because there is no friction...0401

...so we can say that the initial velocity of (A) in the x-direction, which is going to be its final velocity in the x-direction too if nothing is going to speed it up or slow it down while it is in the air horizontally, must be square root of h × g. 0410

Now we have to find the time that (A) is in the air. 0427

To do that, let us go with δy for (A), its vertical displacement is going to be V-initialA in the y-direction, which is going to be 0 times (t) plus 1/2ay time for (a) squared. 0431

Its initial velocity vertically was 0, so then as we solve for (t), the time for (A) is going to be 2δyA/ay (square root) or 2 times...0449

...well δy for (A) is the entire height of the table or 2h and ay = g, the acceleration due to gravity because it is a free fall projectile here, so square root or that is going to be square root 4h/g. 0468

Now finally to figure out how far it went by putting all of that together. 0489

Well, δx for block (A) is going to be vx for (A) times the time for (A) or...0493

...well we have square root of (hg) multiplied by the time (square root 4h/g), which is going to be the square root of 4h2 or just 2h. 0503

All right, that finishes up problem Number 1. Let us move on to problem 2. 0527

In free response question 2, we have a wall that has a negative charge distribution creating an electric field and we have a small plastic ball of some mass that is charged that is suspended by a non-conducting thread and because of that electric field from the wall, it is actually held out away from the wall. 0536

Part A is on the diagram and it wants us to label the forces acting on the ball or draw a FBD. 0555

If there is our ball, of course we have gravity acting down and it is connected by a string, so we will call that force the tension and not that that has to be on the diagram, but that angle is what we will call θ right there, and we have an electric force over to the right, otherwise it would not hang suspended there. 0563

That must be in equilibrium. 0585

Now for Part B, it wants us to calculate the magnitude of the electric field at the ball's location due to the wall. 0588

Well the electric field is just the electric force divided by the charge and it gives us the force as 0.032 N and it even gives us the charge as 80 × 10-6 C, so that is going to be 400 N/C. 0597

How about its direction? Well, it is a negatively charged ball and we know that we have a force going to the right. 0614

If the force is to the right and it is a negative charge, then we must have the electric field pointing to the left, which only makes sense because we said there is a negative charge distribution in the wall. 0623

If it is negatively charged, the electric field points in to negative charges. 0636

Let us check out Part C as we get a little more involved here. 0644

Determine the perpendicular distance from the wall to the center of the ball. 0649

I will start this by drawing my pseudo FBD. 0654

If there is my object -- I still have electric force to the right, (mg) down, but I am going to break up that tension into its components. 0657

The component up here is going to be the cosine based on where I define that angle, T cos(θ), which means over here to the left, I must have T sin(θ) and now I can start applying Newton's Second Law. 0667

The net force in the x-direction is going to be the electric force minus T sin(θ) and that must be equal to 0, because it is in equilibrium, therefore I could say that the electric force must equal T sin(θ). 0683

We will do the same thing in the y-direction and net force in the y-direction equals T cos(θ) minus (mg), so that must be equal to 0, and therefore we could say then that mg = T cos(θ). 0701

Now to find a couple of more things. Let us figure out that angle. 0721

I notice I have T sin(θ) and T cos(θ). 0726

That should be pretty easy then to say that T sin(θ) divided by T cos(θ) must be equal to electric force divided by (mg). 0729

Why did I do that? Well the T's cancel out and sin(θ)/cos(θ) is tan(θ), so tan(θ) = Fe/mg or θ equals the inverse tangent of Fe/mg...0742

...where Fe = 0.032 and mg = 0.01 × 10, so I come up with an angle θ of about 17.7 degrees. 0756

Now if I want to know the perpendicular distance from the wall, well let us draw a diagram here. 0772

There is our wall, here is our string of length (.3 m) and we know our angle here, which we just determined is 17.7 degrees, I can use Trig now to figure out the length of this side, our opposite side, which is the distance from the wall. 0777

Sin(θ) -- sine is opposite over hypotenuse, therefore the opposite side that we are trying to find is the hypotenuse times the sine of theta...0800

...or the opposite side is going to be our hypotenuse (0.3 m) times the sin(17.7 degrees), which is 0.091 m. 0814

All right that was a little bit more involved. 0830

Going on to Part D -- We are going to cut the string. 0834

Find the magnitude of the resulting acceleration of the ball and its direction. 0844

Well when we cut the string, we no longer have any tension, therefore, we are going to have the electric force to the right, which is 0.032 N and we are going to have the force of gravity (mg), which is 0.1 N...0848

...so the total force is going to be the sum of those two -- so there is our total force and we will try and find that angle θ while we are at it. 0864

To find the magnitude, we can use the Pythagorean Theorem and that is going to be the square root of 0.12 + 0.0322, which gives us a total force magnitude of 0.105 N. 0876

As far as the angle goes -- well θ is going to be the inverse tangent of the opposite side over the adjacent side, which is the inverse tangent, our opposite side is 0.1, and our adjacent side is 0.032, or 72.3 degrees. 0901

Now to get the acceleration -- acceleration is the net force divided by the mass, so that is going to be 0.105 N/.01 kg (mass) or 10.5 N/kg or 10.5 m/s2. 0925

As far as its direction goes -- well, if we draw our axis here, we already determined that that is going to be an angle of 72.3 degrees below the horizontal. 0945

Moving on to Part D2 -- Describe the resulting path of the ball. 0960

Well we are going to have a force from the electric field, we are going to have a force for gravity, so we are going to have a net force, and therefore an acceleration basically that way -- a straight line down and to the right. 0971

Those were the first two free response questions. 0996

If you had trouble with these, I recommend going back to the video lectures previously on those specific topics, checking them out and then seeing if you can come back to them and do better. 0998

In our next segment, we will do free response questions 3 and 4. 1006

Thanks everyone and make it a great day!1011