In Kibble-Berkshire Classical Mechanics there is a problem which says:

A ball is dropped from height $h$ and bounces. The coefficient of restitution at each bounce is $e$. Find the velocity immediately after the first bounce, and immediately after the $n$th bounce. Show that the ball finally comes to rest after a time $$\frac{1+e}{1-e}\sqrt{\frac{2h}{g}}.$$

Even though I did not face any problem proving what I was asked to, the whole problem seems to me irrational. The velocity after the $n$th collision equals $u_n=e^nu_0$. Since the ball never stops how can we find that the ball comes to rest after a certain amount of time?. Any help to my misunderstanding is really appreciated.

(1st Edit: If we could reproduce such an experiment in lab exactly as the conditions of the problem predict, supposing that from the equations above the ball has to stop at time $t_1$ then at time $t_2$: $t_2>t_1$ what would we see? Would we find the ball being between $k$th and $(k+1)$th collision? The maths however of this collision problem would predict that both $k$th and $(k+1)$th collision happened before $t_1$.)

(2nd Edit: Even if we argued that the idealization of physical reality (neglecting the deformation of the ball, neglecting the time interval for which the ball touches the ground, etc) is responsible for the paradox, we would observe the same paradox if we could run a PC simulation with a virtual ball hitting a virtual ground changing its velocity in the way suggested by the excercise.The total time that we would predict mathematicaly that the ball is moving would be finite but an observer of such a simulation would see the ball moving infinitely)

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    $\begingroup$ Show the duration time of the n'th bounce, and sum all these times. What do you find? You know lots of infinite series summing to a finite value. $\endgroup$ May 19, 2023 at 19:28
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    $\begingroup$ The ball "never" stops in the same sense Achilles "never" catches the tortoise: it doesn't happen after finitely many iterations, but does happen after a finite time in which the infinitely many iterations occur. $\endgroup$
    – J.G.
    May 20, 2023 at 8:23
  • $\begingroup$ +1 for bringing up a question I have wondered about, but never got about solving it. $\endgroup$
    – fraxinus
    May 20, 2023 at 15:24
  • $\begingroup$ Is there a missing minus sign, in your velocity expression? $\endgroup$
    – Brondahl
    May 20, 2023 at 18:08
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    $\begingroup$ Does this answer your question? Finite distance, Infinite time? $\endgroup$ May 21, 2023 at 16:32

2 Answers 2


Since the ball never stops...

Using that model, the ball bounces an infinite number of times in a finite amount of time, then it stops.

But what is really happening here is that the model is only partially accurate*. That model is for an infinitely rigid ball bouncing off of an infinitely rigid surface, that takes zero time to actually bounce, yet loses a finite amount of energy with each bounce.

Basically, that model ignores the fact that the ball and surface are 3D objects made of real elastic material. Because they are elastic, the bounce itself (i.e., the time the ball is in contact with the surface) takes finite time, and during that time the ball and the surface will deform.

As long as the mechanics of the ball bouncing are not significantly affected by the details of the dynamics of the materials of the ball and the surface it's bouncing upon, the model is accurate. Once the bounces get small enough that the period of time during which the ball stays in contact with the surface it's bouncing on significantly affects the outcome, then that model can't be used accurately.

What really happens, for any real materials, is that at some time- and distance-scales, the ball touches the surface, then both the ball and the surface deform, turning the kinetic energy of the ball into potential energy in the deformation, as well as heat and sound (and maybe EM radiation or other, weirder things). Then both the ball and the surface rebound -- generating even more heat and sound -- and the ball comes out of contact with the surface.

Eventually there will be a "last bounce" where the ball will hit the surface, there won't be enough potential energy in the deformation to make it lose contact, and the ball and surface will transition to decaying vibratory motion -- at least until the "ball and surface are uniform elastic material" model is no longer valid and one has to resort to either quantum mechanics, or the observation that the vibration of the ball is smaller than random thermal motion so who cares anyway?

That "quivers without bouncing" part is explained here: What happens when a ball stops bouncing?

* All models are only partially accurate. There is never One Model to Rule them All -- this is the challenge to modeling physical phenomenon. You need to choose the right model for the problem at hand -- and that may not be the same model as you used last week, or the model that your boss wants you to use.

Choose a model without the right details, and your results are too inaccurate to be useful.

Choose a model with too many extraneous details, and your computation time goes up, as well as your opportunities for numerical errors or outright bugs in the model to corrupt your results.

  • $\begingroup$ The model cited in the question does take into account elasticity. However, in an idealized elastic collision, e=1, and the equation doesn't have a finite solution. The point is that heat is dissipated through inelastic mechanisms. en.wikipedia.org/wiki/Coefficient_of_restitution $\endgroup$
    – Joooeey
    May 22, 2023 at 9:47
  • $\begingroup$ The point -- at least my point -- is that any model is wrong if you depend on it outside of its underlying assumptions. I think it's obvious from the framework of the answer that I'm not talking about perfectly elastic materials, and even if you were to do so, a perfectly elastic ball with non-zero radius and a perfectly elastic surface, and a starting point at a finite height, would not behave the same as a perfectly elastic point bouncing off the same surface. $\endgroup$
    – TimWescott
    May 22, 2023 at 15:47
  • $\begingroup$ Your answer is good overall but the talk about "infinitely rigid bodies" is misleading. That would be the special case e=0 where all energy would be dissipated at once. What is true is that the equation assumes that the size of the ball is insignificant but that doesn't make it rigid. $\endgroup$
    – Joooeey
    May 22, 2023 at 16:18
  • $\begingroup$ @TimWescott I have started doubting that your answer is correct. I've been thinking that the ideal conditions that were supposed is not the problem. If we could imagine a PC simulation of a ball falling in the exact way suggested in my question, changing its velocity every time it hits the virtual ground in the way supposed by the excercise, an observer of this PC simulation would still see a paradox: Total time calculated mathematicaly=finite , Time for which the ball is observed moving=infinite $\endgroup$
    – Kani Pen
    May 23, 2023 at 1:21
  • $\begingroup$ @KaniPen: by definition, if the simulation is correct then it will stop after a finite amount of simulated time. The fact that the simulation itself, if accurate, would take an infinite amount of time (and would require infinite precision) illustrates why simulation is not always the correct approach to finding answers to questions. $\endgroup$
    – TimWescott
    May 23, 2023 at 17:29

After infinite number of bounces the ball attains a fixed position and what we are seeing is the value which time approaches after infinite number of collisions.$$\Sigma_{n=0}^{\infty} T_n=1+1/2+1/4...$$ After infinite collisons,Time will approach a finite value. In case of real ball there is certain loss of energy due to deformations or loss of energy to air resistance,this also causes the ball to attain a fixed time span. The above case has coefficient of restitution e<1 which confirms the loss of energy.

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    $\begingroup$ "Theoretically ball never comes to rest." is a bit misleading IMO; as you say the ball comes to rest after a finite amount of time (in which it happens to bounces infinitely often). This is a bit like the Zeno paradox – infinitely many steps only take finite time. (Of course a real ball won't bounce infinitely often, but that's another topic.). +1 nevertheless. $\endgroup$ May 19, 2023 at 19:17
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    $\begingroup$ @sebastian Riese what i wrote might look a bit misleading but if we look through the perspective of calc then i might look a bit easier. Thank you for +1. $\endgroup$
    – Alv
    May 19, 2023 at 19:22
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    $\begingroup$ No ... "looking through the perspective of calculus" doesn't change the fact that your first sentence is just completely wrong (and thus obviously misleading). Even in the fictional perfect scenario, the ball does come to a rest, at a finite time, but after an infinite sequence of increasing fast bounces. $\endgroup$
    – Brondahl
    May 20, 2023 at 18:06
  • $\begingroup$ @Brondahl Thanks for the suggestion. Infinite collisions does leads to ball attaining a fixed position and this is the apt perspective for such phenomenon. $\endgroup$
    – Alv
    May 20, 2023 at 19:44

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