2015 AP Physics 1 free response 1c

Let's now tackle part C. They tell us block three of mass m sub 3, so that's right over here, is added to the system, as shown below. There is no friction between block three and the table. All right, indicate whether the magnitude of the acceleration of block two is now larger, smaller, or the same as in the original two-block system. Explain how you arrived at your answer.

So let's just think about the intuition here. If you think about the net forces on the system itself, they're the same as we had before. Now their internal forces are going to be different. You're actually going to have two different tensions now, now that you have two different strings. But the net forces, the ones that are causing this thing to accelerate, to in the upward direction on the left-hand side, to the right on the top, and then downwards on the right-hand side, it's still the difference in the weight between the two blocks. But now that difference in those between the weights of the two blocks is moving more mass.

We know that force is equal to mass times acceleration, or acceleration is equal to force divided by mass. Our net force is being is the differential between the weights or the difference between the weights of the block, but now we're going to be moving more aggregate mass. This is going to be M1 + M2 + M3. So you're going to have a small acceleration. So what you could write is acceleration is smaller because the same difference in weights between M1 and M2 is now accelerating more mass.

That's the intuitive explanation for it. If you wanted to dig a little bit deeper, you could actually set up free body diagrams for all of these blocks over here, and you would come to that same conclusion. So let's just do that, just to feel good about ourselves.

So what are on mass one? What are going to be the forces? Well, you're going to have the force of gravity, which is M1g. Then you're going to have the upward tension pulling upwards, and it's going to be larger than the force of gravity. Let me do that in a different color. So you're going to have, whoops, let me do it all right. You're going to have this tension, let's call that T1. You're not going to have two different tensions here because you have two different strings now. The tension there is T1, the tension over here is also going to be T1, so let me do the same magnitude, T1.

Now, since block two has a larger weight than block one because it has a larger mass, we know that the whole system is going to accelerate. It's going to accelerate on the right-hand side, it's going to accelerate down on the left-hand side, it's going to accelerate up, and on top, it's going to accelerate to the right. So if the top is accelerating to the right, then the tension in this second string is going to be larger than the tension in the first string.

So we do that in another color. I'm having trouble drawing straight lines. All right, so we could call T2, and if that is T2, then the tension through—so then this is going to be T2 as well, because the tension through—the magnitude of the tension through the entire string is going to be the same. Finally, we have the weight of block two, which is going to be larger than this tension, so that is m2g.

Now I’ve just drawn all of the forces that are relevant to the magnitude of the acceleration. If I wanted to make a complete, I guess you could say, free body diagram, where I'm focusing on M1, M3, and M2, there are some more forces acting on M3. M3, in the vertical direction, you have its weight, which we could call M3g, but it's not accelerating downwards because the table is exerting force on it, exerting an upwards force on it of the same magnitude, offsetting its weight.

So if you wanted to do a more complete free body diagram for it, but we care about the things that are moving in the direction of the acceleration depending on where we are on the table. So we can just use Newton's Second Law, like we've used before, saying that the net forces in a given direction are equal to the mass times the magnitude of the acceleration in that given direction.

So the magnitude of net force is equal to mass times the magnitude of the acceleration. We can do that first with block one. So with block one—actually, let me just do this with specifics. So block one, I’ll do it with this orange color.

So block one, what's the net forces? Well, it is T1 minus M1g. That's going to be equal to mass times acceleration, so it's going to be M1 times the acceleration. Now, what about block three? Well, block three, we're accelerating to the right. We're going to have T2—let me do it in that different color. Block three, we are going to have T2 minus T1, and T1 is equal to M3, and the magnitude of the acceleration is going to be the same.

Here, we're accelerating to the right; here, we're accelerating up; here, we're accelerating down, but the magnitudes are going to be the same. I can denote them with this lowercase a. Then finally, we could think about block three. We could say that the net forces—well, that's M2g minus T2. That's going against M2g, is equal to M2 * its acceleration.

Now, if we want to solve for acceleration, this will be quite convenient. We can just add up all of the left-hand sides to get a new left-hand side and add up all the right-hand sides to get a new right-hand side. We can do that algebraically because they’re all equal. This is equal to that; that is equal to that; that is equal to that. So if you add up these and then you add up those, well then the sums are going to be equal to each other.

So what are you going to get? If you add up all of this, this T1 is going to cancel out with subtracting the T1. This T2 is going to cancel out with the subtracting the T2, and you're just going to be left with an M2g. M2g minus M1g minus M1g; M2g minus M1g is equal to—and just for—well let me just write it out—is equal to M1 a + M3 a + M2 a. But we can, of course, factor the a out, so let me just write this as that's equal to a * (M1 + M2 + M3).

Then we could divide both sides by M1 + M2 + M3, so let's just do that. M1 + M2 + M3, M1 + M2 + M3; these cancel out. So this is the magnitude of your acceleration. Notice you have the same difference in weights that's providing the net force on the system, but is now accelerating more mass.

So you could even say, hey, look, I have more mass here; more mass to accelerate. More mass, while I have the same net force acting on the system—we're not talking about the internal forces; those all canceled out when I added these equations. So if you're taking the same net force and you're dividing it by more mass, you're going to have a smaller acceleration.

Hopefully, that all made sense to you.