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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 to 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 two 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 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 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 the tension pulling upwards so you can 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 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 it's going to go in the upwards direction so that just gives us a sense of things and so say they say derive the magnitude of acceleration of block 2 all right so this is let me leave the labels up there so this is block 2 up here and we know from Newton's second law that if we pick a direction and the direction that matters here is the vertical direction all the forces are acting in the either 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 in that dimension in that direction I guess you could say and so let's just think about block to block block - and since the acceleration we know is downwards and we want to 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 - the tension and that's going to be positive if we think about it from in the downward direction the downward Direction being positive so we're going to have m2 G the weight - the tension the tension is going against the weight - 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 let me do that same blue color times the acceleration now we could divide both sides by M - but that's not going to help us too much just yet because then we would have solved for acceleration in terms of M jus m2 G and T we don't have any M ones here we don't so we're not solving in terms of M 1 M 2 and G we're solving in terms of M 2 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 - we know that the tension is larger than the weight t minus m1g is going to be equal to is equal to M 1 is going to be equal to M 1 times the magnitude of the acceleration and it 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 could solve for acceleration and we can actually do that by just adding the left hand side to the left hand side on the right hand side to the right hand side you learn this probably first in in algebra 1 if if this is equal to that and that is equal to that if we add the the left side to the left side and the right side to the right side well we're still going to get to things that are equal to each other so when you add the left the left-hand sides what are you going to get so you're going to get M m2 G m2 G minus m1 G minus m1 G 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 is going to be equal to m2 a is going to be equal to m2 a plus em 1 a plus m1 times a and now we just need to solve for a and how do we do that well we could factor out an a out of this right hand side here so this is going to be m2 G minus m1 G 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 except the magnitudes acceleration of either block 1 or block 2 now some of you might be thinking away there there might be an easier way to think about this problem and I went straight from the Freebody diagrams which is you know it's 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 two these two blocks floating in space so this is m2 here and I'm not going to 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 m2 G we're not necessary just care about the magnitude here and I know you might be saying wait okay this is the gravitational field or whatever else but I'm just let's just say you're pulling in this direction with a force that happens to be equal to M that has a magnitude of M 2g and let's say you're pulling in this direction with a force that has a magnitude of m1 m1 G now this isn't exactly the same as where we started where we started we saw what I've 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 m2 G and you're pulling to the left on m1 with a force of m1 G 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 we'll find that one combined mass I am whoops on that one combined mass I am I am pulling to the right with a magnitude of m2 G and I am pulling to the left let me make that in a slightly different color so you can see it and I am pulling to the left with a magnitude of m1 G and now this becomes a pretty straightforward 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 it would be m2 G we'll take the rightward direction is the positive direction m2 G 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 going to 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 and that's one of the fun things about science as long as you do two logical things you get to the same point
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