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Problem 90 Easy Difficulty

Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an clastic collision
with the floor and then rebounds to make an elastic collision with the still-descending small ball. Just before the collision between the two balls, the large ball is moving upward with velocity $\overrightarrow{\boldsymbol{v}}$ and the small ball has velocity $-\overrightarrow{\boldsymbol{v}}$ . (Do you see why? Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large hall? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?


(a) $\overrightarrow{\mathbf{v}}_{A 2}^{\prime}=\left(\frac{m_{B}-m_{A}}{m_{A}+m_{B}}\right) \vec{v}$
(b) $h_{2}=\frac{9 v^{2}}{2 g}=9 h_{1}$


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Video Transcript

{'transcript': "So in this exercise we have a classic, uh, one D elastic collision situation on which we have a ball off mess and one and a smaller bowl off mess and two on which we know that the mass off the ball one is much greater than the mass of Paul Chu. So we take these two balls and dropped them from the same height age. And then when the first ball reaches the ground in this case, the ball with the greater mass, Uh, we know that it will collide with the floor in a less a collision. So we know that we will have a final velocity minus V, and right after the collision, the velocity will change. Directions will be the same magnitude but pointing upwards. So just after this collision, the both the smaller bowl to will continue to fall. And in this case, we have that the ball one will have a motion pointing upwards. Where is the ball? Chu will have a motion pointing outwards so this trip balls will collide and when to collide. Uh, it is in a lasted collision, so the kinetic energy and the linear momentum will be conserved so here is just before the collision. And just after the collision will have that that the small bowl to will gain an extra velocity. That all that I called v two f. So let me race this, and we don't know what will be the velocity off the ball. One. Okay, so in question A in this exercise, you want to calculate what will be the velocity off the ball chew after it collides with the ball One. Okay, so let me write it here. So question A, um So for the conservation off the linear momentum from this scenario here to this scenario here, So right before and right after the collision off the two balls we have that s o the mass off the ball one times the final velocity of ball one that I'll call the one F uh, plus the mass of ball two times the final velocity off the ball to which is the quantity we want to find has to be equal to the mass of ball one times the initial velocity off the ball one. So the so the philosophy Well, well, before it collides with the ball chew mine, plus the mass off the ball, Chu and the velocity off the ball right before it collides with the ball one so vey chew. I okay, notice that I didn't replace any off this off. The losses with the concept is that we know because there is a general expression we cannot detain from this. So let me rewrite. So pass everything that hasn't want to the left side of the equation and everything that has any to to the right side of the equation so that we can put someone in evidence on the left side. And we have that. And one times we won f uh, sorry. I write the I f So V one f minus V one. I has to be equal to m two V, uh, to I minus V two f. Okay, so let's keep this expression that we found from the conservation off linear momentum. So expression one. And now we know that the collision is elastic. So we also have conservation off mechanical energy, so conservation off the kinetic energy. So we have that from the conservation off kinetic energy will have that m one view one f squared over two plus m two v two f squared over two has to be able to m one V one. I squared over two plus and to be too I squared over two so we can divide both sides by two. So sorry multiply both sides by two. And so the same thing we did with the leader Momentum. So everything with and we'll keep to this side of the equation and everything that has into to the right side of the equation. So we have on one times we one f squared minus V one. I have to be able to, um, to V two I squared minus V two f squared. Okay, sorry, there's a squared here. So remember that there is also one way we can rewrite the this terms inside off the parenthesis so we can rewrite it as we won f algae just for this term. But it's the same thing for this other one. So it is. We want f minus. We won, I times we won f plus v. Well, I Okay, so we have this expression from the conservation of kinetic energy. So if we divide expression from the kinetic energy by the expression from the conservation of linear momentum. We can get rid off the masses and we have a relation between the velocities on Lee. So we find something like this. So dividing, too by one expression to buy expression one, we find that we won f plus V one. I has to be equal to V two. I plus D two f now is to be stood by the values that we know. So uh, no. Sorry. Uh, sorry. Don't be sued anything yet. Eso Let's just isolate Veach u f in this equation and we'll find that we can rewrite B two F as V one f plus V one I minus V two I Okay, the whole call this expression three and also be stood this back into the mo mentum conservation. Okay, So sorry. So, I'll Yeah, I also instituted this back here and we'll have something that looks like this. So putting three in two and one. Sorry, we have, um that the left side remains the same, So m one times V one, Uh, minus V one. I is equal shoe and to V to I minus V one f minus V one. I plus me too. I Okay, notice that if we divide both sides by M one so you can get rid of this. And we have this Okay, But in the exercise, when we have that and one is much greater than M two. So this division is approximately zero. So all of this becomes zero as well. And we have that for the ball with great amass. The final velocity has to be equal in magnitude and in direction to the initial velocity. And we know that the initial velocity city so we can go back to our diagram. This is a culture of E. Okay. Uh, okay. So no. In this, we can be stewed. Uh, V one f in here. So let's go back to this expression. So rewrite it here. The expression we found by dividing too by one. So we had that V one f close V one. I is equal to V two. I close the two f so we know that the initial and final the losses of the well worn are equal in magnitude and in direction. So this will be good to TV and we know that initial velocity off the ball. So the velocity off the smaller ball before the collision is negatively because it's pointing downwards so is equal to minus V plus V two f. And by this we find that the final velocity of culture you will be able to three times the velocity, the initial velocity. Okay, three times Be now, question Be asked us to find what will be the maximum height that about two we reach after it gains this velocity off tree V. So after the collision, the smaller ball will will go upwards with this velocity. Uh, three v. So let me put it here. Three v and it will reach a point off heights that I'll call. Uh, h f. Okay, so you want to find what is H f in terms off V. Okay. So we can use conservation off energy. So we have that, uh, in the beginning, because a bishop energy for the ball one. We have that. Sorry. My program stopped a little bit. Um, so you have that the energy, the total energy off dull too, just after the collision. So the reputational potential, plus the kinetic energy has to be equal to the gravitational potential. Sorry. And the kind of energy when it reaches the top off the its trajectory. Okay, so you should have plus, Kate U f okay, if I said the zero off the, um, off the gravitational potential to be here. So in the beginning, we'll have that this term, ventures. And we also know that in the highest point of the trajectory, the velocity off the ball zero. So, Kate, um, goes to zero, and we have the following quality. So I am too. V two f squared over two has to be equal to m two j the height, so h f okay, so we can divide both sides by m two. We can pass this factor of two to the right on the side, and we also can be stewed feature f by the value that we found in the previous exercise, which is three v. So we have nine. The squared is equal to to times g h f. Okay. And we have that, um h f will be equal to nine. We squared over two G. But if we want to find instead the height, the final height in terms off the initial height from which the ball was dropped. So in terms off this age year. Yeah, we can use kind of medics equation to express V in terms off the initial height. So how do we do this? So we can take the ball one for instance. And we have that, uh, the ball one is being accelerated downwards due to the force of gravity F G. So from Newton's second law, we have that in one G has to be equal to, um 18 So the acceleration off the ball one will be put to the acceleration of gravity. And from this, we can find a relation between the acceleration and the velocity, which is equal to so having that the particle was initially so having that the ball one was initially at rest. So we have that be the final velocity is equal to GT, and we can find t by the equation for the displacement. So we have that the height that the displacement in the Y direction, which is equal to the height, will be equal to one half g t squared. Okay. So we can isolate uh, t in here. And we have that t I'm sorry. You have that. Um sorry. So we can isolate t in this expression and so be suited back into the expression for the velocity V And we have that V will be go to the square, root off two times j eight. So if we should be sued this back in the expression in terms off the velocity we have that the final velocity off the final height. Sorry. So we see this in the expression off the high to have that the height off dull to will be equal to nine times we squared, which is, uh, two times G h over to G. We can cancel out to G with two G and we have that the final height off the ball one off the ball to will be equal to nine times the initial height."}