(III) The two masses shown in Fig. 4–57 are each initially 1.8...

Question

(III) The two masses shown in Fig. 4–57 are each initially 1.8 m above the ground, and the massless frictionless pulley is 4.8 m above the ground. What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its 'launch' speed. Assume the mass doesn't hit the pulley or the ceiling. Ignore the mass of the cord.]
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Accepted Answer

Welcome back. Everyone. In this problem. A student hangs two balls by a string from a pulley connected with the roof of his house. As shown in the figure below. In the beginning, both the two balls are stationary at a height of 1.2 m from the ground. While the frictionless pulley of negligible mass is 5.2 m above the ground. Calculate the maximum height the lighter ball reaches if the system moves freely, hint evaluating the lighter ball's acceleration, calculates its speed when the other ball touches the ground. Think of it as its launch speed and presume the balls don't collide with anything. And the mass of this string is negligible for our answer choices. A says that it's 0.38 m. B 1.6 m C 2.8 m and D 5.2 m. No. Before we start to solve for the maximum height, the lighter ball reaches, let's make a note of all the information we have here. So let's see that the mass of our first ball based on a diagram, the mass of the heavier ball M one is 2.5 kg. OK? And the mass of our lighter ball. Let's call that M two, which is 1.3 kg. Now, we're trying to figure out the maximum height the lighter ball reaches if the system moves freely. Now, if we think about it that height, ok? When, when our pulley starts, when our pulley system is activated, the ball is going to increase in height, but we don't know to what point. Ok? But that height plus the height, it's already off the ground. 1.2 m is going to be the maximum height of the lighter ball. OK. So let's call that height that is already off the ground at ground as S one, which means then that we are looking for max the maximum height which is, or change in height plus or height that's already off the ground. S one. No, we know what S one is from our diagram here, but we don't know what the change in height is for our pully. So let's try to figure out what that height is. First, let's make sure we understand all the forces that are acting here. OK. So let's let the initial magnitude of the acceleration of the two balls be a. Now, if we were to draw a free body diagram here, so we have our smaller ball or pulley, then our larger ball M one K and for our system here, then the rope is going to have a tension tee. OK? For both balls going upward, we have M one as a larger ball, M two as a smaller ball. And for both of these, then the weight of M one will be acting downwards as M one G while the weight of M two will also be doing the same thing. Now, if we think about the motion of our pulley here, OK. So our larger ball will be going down, which tells us then that for M two, we can take up as the positive direction. That's the direction the system is moving in. While for M one, we can take downwards as our positive direction. Again, that's the direction our system is moving in because M one is the heavier mass. So it's going to come down and the lighter mass M two would go up. So we can tell them that the tension T is less than M one G and greater than M two G. This is a connected system. So let's think about the forces acting on both masses. No recall that when it comes to motion, Newton's second law tells us that the sum of the forces is directly proportional to the acceleration by the formula that says the sum of the forces equals the mass multiplied by the acceleration. So applying it here for both our masses, then that means for M one, then the, some of the forces acting where we have uh t going upwards. So that would be negative direction because we are considering downwards as the positive direction. So negative T, all right, let me write it beside it here, negative T loss M one GM one G is going in the direction of the system. The weight of our mass M one is going to be equal to or mass M one multiplied by A. Remember we said we're considering both the two balls that have an initial acceleration of A. No, let's call that equation one next for M two K. Then considering the forces we have T going upwards positive in this case minus the weight of M two. So that's M two G that was going downward. So it would be negative and that would be equal to M two A. Now we can take it a step further here to make a connection for both of our formula. Because if T minus M two G equals M two A, that means T equals M two G plus M two A, which if we factor out M two is going to be M two multiplied by G plus A. So let's call that equation two. So we have a formula for T that we can put into our equation for equation one. So that means then that T we can eliminate T because T is not going to go into negative T. So that's negative M two multiplied by G plus A. OK. Substituting equation two into equation one plus M one G is going to be equal to M one A. Let me just make a note here. That's what we did. So we substitute equation two into equation one. Since that's the case, then we can solve for acceleration in terms of the masses and the G, the acceleration due to gravity. Because now this tells us if we, if we expand here, OK, then we're going to have negative M two G minus M two A being equal to M sorry plus M one G equals M one A. We can group our like terms here in terms of A. So M one A plus M two A is going to be equal to M one G minus M two G. I know we can group based on or, or we can factorize based on a here. So that's a multiplied by M one plus M two on the other side, that's M one minus M two are multiplied by G. We factored over G. So now that tells us A equals M one minus M two G divided by M one plus M two. So we have a, in terms of the masses and G. OK? No, how is that going to help us? Well, let's start to think about or motion here for our system. OK. So now that we have our free body diagram to understand the forces, let's get into the actual movement. So let's imagine that no, we have our ball that has moved by that height edge that we specified before. So again, we have our system, ok. We have our pulley and now we have the heavier ball on the ground. Ok. And now just to show that initially, it's moving from sorry M two moved from its initial position. Ok. Oops, let me add that back M to move from its initial position in our dotted line here. And likewise, M one would have also moved from its initial position. Yeah. And again, let me just make sure I have both of these looking pretty good here. Just to make sure, remember M one is the smaller mass. I think I needed to say that for me, not for you, but no, we have M one. It has moved by a height of edge, a changing edge and we have sorry, we have M two that's moved by a height of h my apologies. And then we have M one that has moved by our height that we mentioned earlier. S one. OK. So that's our initial height, our initial distance that M one has moved. Now we're talking about distances and we, and for those distances, we are considering their acceleration. OK. Now, do we know any kinematic equation that relates distance to acceleration? Well, again, recall that we know that the difference between our final velocity and the square of our final velocity and the square of our initial velocity, our speed is going to be equal to two A S where is the acceleration? And S is the distance that they've moved. So now we can think about both of our distances. Let's start with the distance covered for M one. Ok. Ok. No four M one. We're talking about the distance of S one where initially, OK, initially was at rest. So let's call that initial speed V not which is 0 m per second. And then when it touches the ground, let's call that a final speed V. OK. Final speed V that we, we, we don't necessarily know yet. OK. So I plan this formula here or M one or from, let's say from N one from M one, then V squared or final velocity minus V not squared or initial velocity is going to be equal to two multiplied by a multiplied by our distance covered, which is S one like we said, we know our initial speed here was zero. So if we take that O that means V squared equals two a multiplied by R by delta S one. OK. Now what about our other mass? What about M two or mass that moves a height H because remember this is the height that we really want. What do we know? Well from M two? Yeah, we can think of V as our launch speed. OK? We can think of V as the launch speed of the mass M two. So at that land speed where we have V then when M two gets, when M two is raised by a height of H it's going to come to a stop. So that final height, we can think of that height as V prime. OK. That's the maximum height for our, our height when it comes to rest. Ok? And now at that maximum height, the speed V prime is going to be equal to 0 m per second. Now again, after M two has been launched, its acceleration is solely due to the gravitational acceleration G which acts in the downward motion. So for our formula, we can take a to be negative G. Now let's use all of those ideas because no, that means V prime, OK. Squared, sorry minus V squared or launch speed is going to be equal to two multiplied by negative G or negative two G delta H or change in our height H like we said, we know that the speed at the maximum height is 0 m per second. So we can cancel that and that's going to leave us with negative V squared being equal to negative two G delta H, we can cancel the negative signs from both sides. And thus we can solve for delta H because remember that's what we really want. That note tells us delta H equals V squared divided by two G. But guess what? We already have an expression for V squared, we know what V squared is V squared equals two A delta S. So since V squared equals to A delta S, then we can substitute it into our formula for delta H because now that tells us delta H is going to be equal to two A delta S divided by two G. We can cancel our tools here. And that leaves us with a delta S divided by G. But again, we are also in for some good news because remember earlier, we expressed a in terms of our masses and G so we can substitute it here to then say that delta H is going to be equal to am one. Me get that here. M one minus M two divided by M one plus M two G. OK? Multiplied by delta S divided by G. Notice we can cancel G and no delta H will be M one minus M two divided by M one M two multiplied by delta S which we already know is 1.2 m since we know these values, let's substitute. So we know M one is 2.5 kg. M two is 1.3 kg. So it's gonna be 2.5 kg minus 1.3 kg divided by 2.5 kg plus 1.3 kg, all multiplied by 1.2 m. Now we can put this expression into our calculator. OK? And when we do that, we get delta H to be equal to 0.38 m. Now let's just take a quick check of this what this means for us. Let's go back to our diagram. OK. Now when we go back to our diagram, we just solved for delta H. In other words, we just figured out the height that our smaller ball is going to move by the maximum height. Remember that? Sorry. Yeah, the maximum height is going to move for here for that distance from its original height. My apologies. Now, remember our maximum height, the height it is overall from the ground equals delta H plus delta S. So now we can find that maximum height. Ok. So since we have delta H, OK. Therefore, that means our maximum height is going to be the sum of delta H 0.38 m. And the height it was originally off the ground 1.2 m. And when we add both of those, we get 1.58 m. Recall that all of our answers were written into two significant figures. So when we do the same to this answer, we get it to be approximately equal to 1.6 m. Therefore, that means the maximum height the lighter ball reaches if the system moves freely is 1.6 m. If we go back to our answer choices that tells us B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Newton's Second Law of Motion

Conservation of Energy

Kinematics of Projectile Motion

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