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## AP Physics 1 free response questions

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# Question 1b: 2015 AP Physics 1 free response

## Video transcript

- Alright, 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 1, 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
ways you can 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 gonna accelerate downwards on
the right-hand side and upwards on the left-hand side. The other way to 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're gonna accelerate upwards on the left-hand side, accelerate
downwards on the right-hand side and 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
the acceleration, here, has a magnitude a, it's gonna
go in the downward direction the acceleration on the left-hand
side is gonna be the same magnitude, but it's gonna
go in the upwards direction. So that just gives us a sense of things. They say, derive the magnitude
of acceleration of block 2 Alright, so this is, let me leave the labels up there, so this is block 2 up here. And we know, from Newton's 2nd Law, that if we pick a direction,
and the direction that matters here, is the vertical direction. All the forces are acting in either the upwards or downwards direction. So, the magnitude of our net forces, we care about the vertical dimension here, is going to be equal to the
mass times the acceleration, in that, in that dimension,
in that direction, I guess you could say. And so let's just think about block 2. Block, Block 2. And since the acceleration,
we know is downwards, and we wanna figure out what a is, let's just assume that positive, positive magnitude specifies downwards. So what are the net forces? Well, the net forces are going
to be the force of weight minus the tension, and
that's going to be positive. If we think about it in
the downward direction. The downward direction being positive. So we're gonna have m2g, the weight, minus the tension, tension is going against the weight, minus the tension is going
to be equal to the mass. Is going to be equal to m2
times, times our acceleration, and we need to figure out
what that acceleration is going to be. So, we do that same blue color. Times the acceleration. Now, we could divide both sides by m2, but that's not going to
help us too much, just yet. Because then, we would've
solved for acceleration in terms of 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. So, somehow, we have to get rid of this T. And what we can do to get rid of the T, is set up a similar equation for block 1. Block 1, Block 1 Here, since we're concerned
with magnitude and especially the magnitude
of acceleration, and here the acceleration is
going in the upward direction, we could say that the upward direction is the positive direction. And so, we could say that T minus, we know that the tension is larger than the weight, T minus m1g is going to be equal to is equal to m1, is going to be equal to
m1 times the magnitude of the acceleration. And to be clear, these
magnitudes are the same. And we already know that the
magnitudes of the tension are the same. And now we have two equations with two unknowns, and so, if we can eliminate the tension, we can solve for acceleration. And we can acutally 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 learned this probably
first in algebra 1. If, 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 gonna
get two things that are equal to each other. So when you add the left-hand sides, what are you going to get? So, you're gonna get m2g, m2g minus m1g, minus m1g, and then you're gonna get T minus T. These two are going to cancel out. So, let me just cross them out. So that was convenient. Is going to be equal to m2a, is going to be equal to m2a plus m1a, plus m1 times a. And now, we just need to solve for a. And how do we do that? Well, we can factor out an a
out of this right-hand side here, so this going to be m2g minus m1g is equal to let's factor out an a, a times, times m2 plus m1, and now to solve for a, we just divide both sides by m2 plus m1. m2 plus m1, m2 plus m1, and there you have it, we get a is equal to a is equal to this. And, notice, we have solved it, we have solved for a in terms of we have solved for a in terms of m1, m2, and g. And this is the magnitude of
the acceleration of either block 1 or block 2. Now, some of you might be thinking, wait, there might be an
easier way to think about this problem, and I went straight
from the free-body diagrams which is, ya know, it's
implied that this is the way to tackle it, using the tension. But, another way to tackle it, you could've said, well, this would be analogous, it's not the exact same thing, but
it would be analogous to imagine, two, these two
blocks floating in space. So, this is m2, here, and I'm not gonna use pulleys here, so that's m2. And it's connected by a massless string, to m1, which has a smaller mass, m1. And let's say that you
are pulling on, pulling in the rightward direction, now we're just drifting in space. With a force of m2g, we're not, we just care about the magnitude here, and I know you might be saying, wait, okay, is this is 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,
that has a magnitude of m2g, and let's say you're
pulling in this direction, with a force that has
a magnitude of m1, m1g. Now, this isn't exactly the
same as our, as where we started where we started, we saw what
I have just kind of drawn, but I have it in the presence
of a gravitational field, and I have it wrapped
around those pulleys, and then the gravitational
field is providing these forces. But let's just assume,
just for simplicity that that you're drifting in space and you're pulling on m2 to the
right with a force that's equivalent to m2g, and you
are pulling to the left on m1, with a force of m1g. Well, you could just view this as one big, you could just view the 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, alright, well from that one combined mass, I am, whoops, that one combined mass, I am I am pulling to the right
with a magnitude of m2g, and I am pulling to the left let me make that a
slightly different color so you can see it. And I am pulling to the
left with a magnitude of m1g and now, this becomes a
pretty straight-forward thing. What would be the acceleration here? Well, you could say the net, 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, m2g minus m1g minus m1g, and then, so
that's the net force, in the right direction, the magnitude of the net force
in the rightward direction and then you divide it by its mass, and you're gonna get acceleration. Force divided by mass
gives you acceleration. So, you divide that by our
mass, which is going to be 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 they, notice, either of them give
you the exact same answer. That's one of the fun
things about science, as long as you do logical things, you get to the same point.

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