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Suppose we have two solid spheres with masses $m$ and $M$, respectively, and that $m$ is significantly less than $M$. The lighter sphere is placed directly on top of the heavier one, and the two are dropped together onto the ground from a height $h$.

We are told that all subsequent collisions involving the balls and floor are elastic, and are asked to determine the maximum height to which the small sphere will rise on the rebound.

The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation, perhaps something like: $$(M + m)gh_i = mgh_f$$

Edit—4/30: I reviewed the original wording of the problem, which did mention that we could assume the presence of a small gap between the balls. I do realize that, in the absence of this assumption, the behavior of the balls would vary depending on their material characteristics.

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  • $\begingroup$ @Floris Its pretty obvious from context but for clarity you might want to say "not going to get you very far" instead of "t going to get you very far." $\endgroup$ – Reid Erdwien Apr 18 '15 at 15:35
  • $\begingroup$ @ReidErdwien - it was a typo (darn phone keyboards). I will repost the comment as you can't edit them... $\endgroup$ – Floris Apr 18 '15 at 15:37
  • $\begingroup$ A single equation with two unknowns isn't going to get you very far. The problem is the the bottom ball will also rise after the bounce so you have two unknowns. $\endgroup$ – Floris Apr 18 '15 at 15:37
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    $\begingroup$ Both are conserved in elastic collisions. Not sure what you mean by "choose" - you simply use whatever's important to find out whatever you want to find out. A general elastic collision with known masses and initial velocities is usually solved with both. $\endgroup$ – Abhimanyu Pallavi Sudhir Apr 25 '15 at 10:27
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    $\begingroup$ I know. I originally thought that it might be possible to solve for the final height using a single cons.-of-energy equation. I wanted to know how to choose between using said approach and using a cons.-of-energy + cons.-of-momentum approach—precisely because I knew that both forms of conservation were occurring in the scenario. But at this point, my mistakes have been identified and my question answered. $\endgroup$ – Rations Apr 26 '15 at 16:45
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.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound.

The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear momentum calculations altogether and use a single conservation of energy

It is a contradiction because an elastic collision is defined as one in which both KE and momentum are conserved. Only one equation is not enough to define the outcome of the collision.

Knowing the radii doesn't help, by the way, to solve any such problem, since the balls are usually considered of uniform density and the center of mass coincides with the center of gravity and with the point of application of all the vectors.

If you know the formulae, put any value you like, and solve for a particular example, unless they are asking for a general formula.

Since this is a homework question, nobody is allowed to give you more details.

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  • $\begingroup$ Thanks. I didn't mean to imply that I could just disregard the concept of cons. of momentum altogether. I thought that knowing the radii would affect the heights in any cons. of energy equations I might use (though finding the center of mass might be needlessly laborious when a cons. of momentum approach is available). Anyway, it looks like I incorrectly assumed that the bottom sphere would not bounce (see other comment thread). (Also, to be transparent, the question is one my college friend sent to me for fun, and I'm interested in better understanding the concepts it relates.) $\endgroup$ – Rations Apr 18 '15 at 16:10
  • $\begingroup$ I have added a clarification to my original post, and believe that I understand both interpretations of this problem now. Sorry about the confusion. $\endgroup$ – Rations Apr 30 '15 at 21:43
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In an elastic collision involving two objects both momentum and energy are conserved. You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other. Since energy is the square of the momentum divided by twice the mass, this means that the total kinetic energy is proportional the square of the momentum and this must thus be the same before and after the collision. the magnitude of the momentum therefore cannot change, we can thus conclude that all that can happen is that the momentum changes direction. In a one dimensional setting like in this problem, this means that the momenta of both object just changes sign.

So, using both energy and momentum conservation we've reduced all one dimensional collision problems to a mere triviality: Transform to the center of mass frame, change the sign of the momenta and then transform back to the original frame. Then in the case that one of the masses happens to be much larger than the other mass, the center of mass frame is the rest frame of that mass which makes it even easier to solve the problem.

Let's see if we can tackle this problem using the method I derived above. Here you need to consider that when the ball collides with the ground that's a collision of the big ball with the Earth obviously you should then take the big ball to the light mass. Also when that collision happens the small ball will initially keep on moving toward the ground and just a fraction later will it feel the impact due to the big ball having changed direction.

So, if the ball hits the ground with velocity v then this velocity will reverse sign as we're working in the rest frame of the Earth which is the relevant center of mass frame for that collision with the ground. But then what happens is that the small ball which is still moving at velocity v toward the ground hits the big ball that is moving upward with velocity v. In the rest frame of the big ball, the small ball is thus moving toward it with velocity 2 v. Since this frame is to a good approximation the relevant center of mass frame for the collision that is about to happen, it follows that after the collision the velocity will be 2 v relative to the big ball. Then transforming back the the original frame, we see that after the collision, the small ball will have a velocity of 3 v relative to the ground.

This means that the kinetic energy of the small ball after the collision will be 9 times what is was before the collision and that then implies that it will reach 9 times the height from which it was dropped.

Note: The outcome does not depend on the assumption that there be a small gap between the two balls

Some commentators and some other answers wrongly claim that the answer given here is wrong, because the Hyperphysics website claims that without there being a small gap between the balls, the outcome will be different. Now, apart from the fact the Hyperphysics websites make no such claim at all, the apparent dependence on the gap is an artifact of the way one treats the simultaneous collisions. Of course, one may argue that things are really a bit different when the collision happen simultaneously, but all the assumptions made should be made explicit, which they didn't do.

Simply put, if with a gap of zero the small ball doesn't rise as high, then what happened to the energy that would have gone into the small ball if the gap wasn't there? So, the hidden assumptions made leads to energy being dissipated in the form of vibrations in he the big ball. But as I show below, this is not at all obvious, you are not led to this conclusion if you just analyze the problem in more detail.

So, let me explain here why the gap doesn't matter in the spirit of this problem where we make the assumption of elastic collisions, where the two balls are certainly not glued to each other. The moment the bottom ball collides with the ground, the top ball together with the contact point are moving toward the ground with velocity v while the bottom part of the ball has come to a stop. The compression of the ball at the contact point at the ground thus happens first, the shock wave needs to propagate to the other end of the ball before the contact point with the other ball will come to a stop and the top ball will start to compress it there. When that happens the contact point with the top ball will end up being moved into the ball with the relative velocity of 2 v. This means that when that elastic motion rebounds this relative velocity of 2 v will reverse sign. After that point the contact point will slow down, creating a gap between the top ball (it's not glued to the bottom ball so it will move away). So, we again arrive at the same result.

Of course, one may argue that a more detailed treatment of the elastic motions of the balls is necessary and that the 2 v relative velocity isn't accurate. One has to note here that internal motion of the elastic balls is the process by which kinetic energy gets dissipated into heat, so this actually points to the impossibility of having a perfect elastic collisions to begin with.

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    $\begingroup$ Well...the smaller ball is right on the top of the bigger one. You can not treat the velocity changes as if it was first collided with earth and then with the smaller ball. Because it collides with both of them simultaneously. $\endgroup$ – Dvij Mankad Apr 19 '15 at 18:38
  • $\begingroup$ Either you treat them as separate collisions or you must use a realistic model consistent with only elastic interactions. In the latter case, you cannot impose that the velocities stay equal, the moment the big ball collides with the ground it is only the part in contact with the ground that stops moving the other side still keeps moving, it will rebound. The end result will only depend on the conservation laws, so it actually doesn't matter if you replace the true situation with one where the same interactions take place but without the unnecessary complications. $\endgroup$ – Count Iblis Apr 19 '15 at 18:44
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    $\begingroup$ Your answer is true only if there is a little gap between the balls. If they are just touching each other than you can not treat the two collisions to be taking place one after the other. You can super impose the effect of the two collisions on the bigger ball.But can not treat that once this collision is done and then the other is taking place. (if they are just touching each other ) $\endgroup$ – Dvij Mankad Apr 19 '15 at 18:55
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    $\begingroup$ exo.net/~pauld/activities/physics/bouncingballs.html $\endgroup$ – Dvij Mankad Apr 19 '15 at 18:55
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    $\begingroup$ Count Iblis: "Some commentators and another answer wrongly claim that the outcome depends on whether or not there is a small gap between the balls. The apparent dependence on the gap is an artifact ..." That is wrong, that claim is made by Hyperphysics and my quote is from the quoted link. If the balls are not separated the topball will never reach 3v and 9h. $\endgroup$ – user78409 Apr 29 '15 at 8:42
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The problem does not mention any radii, but if we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether

The accepted answer has confused you:

You can simplify things by considering these conservation laws in the center of mass frame. There the total momentum is zero, therefore one object will have the opposite momentum of the other.

You can find a clear explanation of the problem at hyperphysics:

enter image description here if $m_B>m_b$ enter image description hereif masses $m_B = m_b$enter image description here

Changing the frame of reference here is useless and misleading, since in this case the Lab frame coincides with the Center of mass frame. (the floor-Earth is 10^24 times greater than B). In the link they show the more interesting case when the frame of reference of the bigger ball B is considered.

Another aspect to be underlined is that: - the two balls B, b must not touch each other in the first place: "..The two slightly separated balls dropped from the same height are seen by a ground observer to approach the surface with velocity v..." If the upper ball hits the bottom ball before it has bounced, then the outcome is not the same.

The problem is quite simple if you consider the two separate elastic collisions:

  • B hits the floor and rebounds with (almost) same speed -v
  • B collides with oncoming b with speed +v and rebounds with speed +2v with reference to +v

I mistakenly assumed that the bottom ball would halt completely and transfer all its kinetic energy into the upper ball. Two follow-up questions: (1) Because M is so great in comparison to m, the second collision should not really affect the rebound height of M, right? (2) If we conducted this experiment using two identical balls, should both return to the original height after "bouncing"? – Rations

The ratio between the mass of a basketball and a tennis ball is roughly 1:11, if M/m decreases, the outcome is always different:

  • if the two balls have equal mass, in the second collision they exchange their velocities , the bottom ball halts completely as you say, but then gets the top ball's velocity.

Therefore the upper ball rebounds at -v and the bottom ball at +v and consequently has a third collision with the floor where it rebounds for the second time at roughly -v

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  • $\begingroup$ "Changing the frame of reference here is useless and misleading" It is not because the person is learning physics here. $\endgroup$ – Count Iblis Apr 26 '15 at 15:45
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    $\begingroup$ The lab frame coincides with the CM frame, saying that one must change to the CM frame is illogical and therefore useless and confusing, especially for a learner. As to the separation of the ball the quote and the claim is from hyperphysics $\endgroup$ – user78409 Apr 26 '15 at 16:55
  • $\begingroup$ It only coincides with the CM in case of the collision with the ground. The fact that the lab frame is already the CM frame in case of the first collision is something trivial that you would hope the student will notice. If not, then any confusion that causes is a good thing. $\endgroup$ – Count Iblis Apr 26 '15 at 17:01
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If we did know the radius of each sphere, would it be possible to skip conservation-of–linear momentum calculations altogether and use a single conservation-of-energy equation...

The radius of a ball is always irrelevant to the outcome of a collision, what counts is the ratio of the masses. Two equations are always necessary to determine the outcome of a n impact (m, M): if m = 1 and v = 10, KE = 50. In an elastic collision that value is conserved, but there are many ways in which it will and that depends on the masses: 50 + 0, 25+25, 12.5+37.5 etc.

This link is very useful and has a calculator where anyone can check his calcs, learn the formulae and check all the statements in this answer.

If h = 0.8 m, $M>>m : M/m =99$, and the balls are separated so that they impact after M has rebounded and regained roughly its speed in the opposite direction (v = 4 m/s), m will rebound with a speed of 11.84 m/s and M will continue with a slightly reduced speed (v = 3.84 m/s). KE (800) is conserved (70+730) an also momentum (392 = 11.84+380.16)

enter image description here,in the case quoted by Dvij:enter image description here

of a tennis+basket-ball M/m = 10, the rebound of the tennis ball will be only 2.5v (= 10.5 m/s).

Lastly, the bottom ball will stop dead only when M/m = 3 and m will rebound exactly at 8 m/s

Bottom Ball stops dead enter image description here

Since OP hasn't got yeat a clear picture and has accepted a wrong answer, it is necessary to clarify that, if the ball are touching when they reach the ground, there is only one collision and the Lab-frame is exactly theCM-frame

So, assuming that both balls are of "perfect" hardness (i.e., non-deforming), would we simply treat them as one unified system and then conclude that they would both rebound to the original height?

No Gapenter image description here Hard Top-Ballenter image description here

The outcome of the collision with the floor depends not only on the ratio between the masses but also on on the hardness of the balls and on the time theey take to recover their original shape and velocity: a golf ball, even on its own, is completely flattened when it impacts a wall, if the top ball is hard it will mearly impact the floor, if it is soft it will get squashed too. Depending on these factors the final velocity of m varies between v and 2 v.

It is unlikely the 2 balls will ever rebound to the same height.

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    $\begingroup$ I posted an answer since I can neither downvote nor post a comment on any other post $\endgroup$ – user78660 Apr 29 '15 at 13:01
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    $\begingroup$ Thanks. I think I am being unclear in my comments. I looked at the text of the problem again, and it mentioned that we should assume that there is a small gap between the balls. (I understand what you guys are saying about the no-gap scenario, though). I also understand that the behavior of the balls depends on the ratio of their masses. What I did assume was that $M >> m$ to the point where $M$ would lose a negligible amount of speed following its collision with $m$ (and would thus essentially rebound to its original height). $\endgroup$ – Rations Apr 30 '15 at 21:28
  • $\begingroup$ "Depending on these factors the final velocity of m varies between v and 2 v." That's not true, the range will be from v and 3 v. $\endgroup$ – Count Iblis Apr 30 '15 at 22:13

protected by Qmechanic Apr 19 '15 at 17:41

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