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## AP Physics 1 free response questions

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# Question 1c: 2015 AP Physics 1 free response

## Video transcript

- [Voiceover] Let's now tackle part C. So they tell us block 3 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 3 and the table. Alright, indicate whether the magnitude of the acceleration of block 2 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 the 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 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 this though the difference in the weights between the two blocks but now
that difference in those, inbetween the weights of the two blocks is moving more mass and we know that force is equal to mass times acceleration or acceleration is equal to force divided by mass and our
force, 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 plus m2 plus m3 and so you're going to have
a smaller acceleration. And so what you could
write is acceleration, acceleration smaller because same difference, difference in weights, in weights, between m1 and m2 is now accelerating more mass, accelerating more mass. And that's the intuitive
explanation for it and 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 1 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, we'll do that in a different color, so you're going to have, whoops, let me do it, alright so you're going to have this tension, let's call that T1, you're now 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 I'm going to do the same magnitude, T1. Now since block 2 is a larger weight than block 1 because it has a larger mass, we know that the whole system is going to accelerate, is 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. And 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, alright so that 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, and then finally we have the weight of the block, we have the weight of block 2, 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 on an upwards, it's exerting an upwards force on it so of the same magnitude
offsetting its weight. So that's 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 accleration depending on where we are on the table and so we can just use Newton's second law like we've used before,
saying the net forces in a given direction are equal to the mass times the magnitude of the accleration in that given direction, so the magnitude on that force is equal to mass times the magnitude of the acceleration. And so we can do that first with block 1, so block 1, actually I'm just going to do this with specific, so block 1 I'll do it with this orange color. So block 1, 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 3? Well block 3 we're
accelerating to the right, we're going to have T2, we're going to do that
in a different color, block 3 we are going to have T2 minus T1, minus T1 is equal to m 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, they're all, I can denote
them with this lower-case a. And then finally we can
think about block 3. We could say that the net forces, well that's m2g minus T2, that's going against m2g, is equal to m2 times its acceleration and now if we want to
solve for acceleration, and 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 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. And so what are you going to get? So if you add up all of this, this T1 is going to cancel out
with the 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 m1a plus m3a plus m2a. Well we could of course factor the a out and so let me just write this as that's equal to a times m1 plus m2 plus m3, and then we could divide both sides by m1 plus m2 plus m3. So let's just do that. So m1 plus m2 plus m3, m1 plus m2 plus m3, these cancel out and so this is your, the magnitude of your acceleration. And 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, so more mass to accelerate, more mass, more mass to accelerate
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 and so if you're taking the same net force and you're dividing it by more mass you're going to have a smaller, smaller acceleration. Hopefully that all made sense to you.

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