2015 AP Physics 1 free response 1b

All right, let's tackle part B now. Derive the magnitude of the acceleration of block 2. Express your answer in terms of m1, m2, and g. And like always, try to pause the video and see if you can work through it yourself.

We already worked through part one, or part A, I should say, based on this diagram above, and there's a previous video. So now we're ready to do part B, and we've already drawn the free body diagrams which will help us determine the acceleration of block 2.

Let's just think about what the acceleration is first. We know it's going to accelerate downwards because there's a couple of ways you could think about it. The weight of block 2 is larger than the weight of block 1. They're connected by the string, so we know we're going to accelerate downwards on the right-hand side and upwards on the left-hand side.

The other way you think about it is the weight of block 2 is larger than the upward force of tension, and the weight of block 1 is less than the tension pulling upwards. So, you can accelerate upwards on the left-hand side, accelerate downwards on the right-hand side. A key realization is the magnitude of the acceleration is going to be the same because they're connected by that string.

So the acceleration, I'll just draw it a little bit away from the actual dot. So if the acceleration here has a magnitude 'a', it's going to go in the downward direction. The acceleration on the left-hand side is going to be the same magnitude; that's going to go in the upward direction. So that just gives us a sense of things.

They say derive the magnitude of acceleration of block 2. All right, so let me leave the labels up there. This is block two up here, and we know from Newton's second law that if we pick a direction, the direction that matters here is the vertical direction. All the forces are acting either in the upwards or downwards direction.

So the magnitude of our net forces—and we care about the vertical dimension here—is going to be equal to the mass times the acceleration in that direction.

Let's just think about block two. Since the acceleration we know is downwards and we want to figure out what 'a' is, let's just assume that the positive magnitude specifies downwards. So, what are the net forces? The net forces are going to be the force of weight minus the tension. That's going to be positive if we think about it from the downward direction, the downward direction being positive. So we're going to have m2g, the weight, minus the tension.

The tension is going against the weight. So, m2g minus the tension is going to be equal to the mass m2 times our acceleration. We need to figure out what that acceleration is going to be.

Now, we could divide both sides by m2, but that's not going to help us too much just yet because then we would have solved for acceleration in terms of mg, m2g, and t. We don't have any m1s here, so we're not solving in terms of m1, m2, and g; we're solving in terms of m2, t, and g.

Somehow, we have to get rid of this t. What we can do to get rid of the t is set up a similar equation for block one. Here, since we're concerned with magnitude, and especially the magnitude of acceleration, we could say that the upward direction is the positive direction.

So we could say that t minus m1g is going to be equal to m1 times the magnitude of the acceleration. To be clear, these magnitudes are the same, and we already know that the magnitudes of the tensions are the same.

Now we have two equations with two unknowns. If we can eliminate the tension, we could solve for acceleration. We can actually do that by just adding the left-hand side to the left-hand side and the right-hand side to the right-hand side.

You learn this probably first in Algebra I. If this is equal to that and that is equal to that, if we add the left side to the left side and the right side to the right side, well, we're still going to get two things that are equal to each other.

So when you add the left-hand sides, you're going to get m2g minus m1g, and then you're going to get t minus t. These two are going to cancel out, so let me just cross them out. So that was convenient.

This is going to be equal to m2a plus m1a. Now we just need to solve for a, and how do we do that? Well, we could factor out an a from the right-hand side.

So this is going to be m2g minus m1g is equal to a times (m2 + m1). Now, to solve for a, we just divide both sides by (m2 + m1).

And there you have it; we get a is equal to (m2g - m1g) / (m2 + m1). Notice we have solved for a in terms of m1, m2, and g, and this is the magnitude of the acceleration of either block one or block two.

Now, some of you might be thinking, "Wait, there might be an easier way to think about this problem." I went straight from the free body diagrams, which is, you know, it's implied that this is the way to tackle it using the tension.

But another way to tackle it, you could have said, well, this would be analogous—it’s not the exact same thing, but it would be analogous to imagine these two blocks floating in space. So this is m2 here, and I'm not going to use pulleys.

So that's m2, and it's connected by a massless string to m1, which has a smaller mass. Let's say that you are pulling in the rightward direction. Now we're just drifting in space with a force of m2g.

We're not, you know, we just care about the magnitude here. I know you might be saying, "Wait, okay, is this the gravitational field or whatever else?" But I'm saying let’s just say you're pulling in this direction with a force that happens to be equal to m2g, and you're pulling in the left direction with a force that has a magnitude of m1g.

Now, this isn't exactly the same as where we started, but I have it in the presence of a gravitational field, and I have it wrapped around those pulleys. The gravitational field is providing these forces. Let’s just assume, for simplicity, that you're drifting in space, and you're pulling on m2 to the right with a force that's equivalent to m2g.

You're pulling to the left on m1 with a force of m1g. Well, you could just view this as one big—you could view m1, the string, and m2 as just one combined mass. You could just view this as one combined mass of m1 plus m2, and you could say, "All right, well, from that one combined mass, I am pulling to the right with a magnitude of m2g."

And I'm pulling to the left with the magnitude of m1g. Now this becomes a pretty straightforward thing. What would be the acceleration here? Well, you could say the net magnitude of the force or the magnitude of the net force would be m2g.

We'll take the rightward direction as the positive direction, so it would be m2g minus m1g. Then, you divide by its mass, and you're going to get acceleration. Force divided by mass gives you acceleration, so you divide that by our mass, which is going to be m1 plus m2.

That's going to give you your acceleration. So you could view this as a simpler way of thinking about it. But notice either of them gives you the exact same answer, and that's one of the fun things about science. As long as you do logical things, you get to the same point.