Infinite bouncing in finite time? 
Is it possible to calculate the time in which a bouncing ball with restitution coefficient $k$ which first hits the ball with velocity $v$ will be considered stopped? Note that each collision lasts $\tau$ seconds.

I know how to calculate the answer if $\tau \to 0$. Let's see! Let's denote $t = \frac{kv}{g}$ the time after the ball fill first reach velocity $0$ at a height $h$. So, between the first collisions, we have a total time $t_1 = 2t = \frac{2kv}{g}$. We can prove easily (I think we can even try mathematical inductory reasoning) that $t_n = \frac{2k^nv}{g}$. Let's compute the infinite sum/series:
$$T = \sum_{i = 1}^\infty t_i = \frac{2v}{g} \sum_{i = 1}^\infty k^n$$
As $k < 1$ (restitution coefficient), we can enter Geometric series:
$$T = \frac{2v}{g} \cdot \frac{k}{1 - k}$$
Howerevr, if $\tau \gg 0$, the times are $t_n = \frac{2k^nv}{g} + \tau$. So the series is:
$$T = \frac{2v}{g} \sum_{i = 1}^\infty k^n + \sum_{i = 1}^\infty \tau$$
Calculating the Geometric series and the divergent sum:
$$T = \frac{2v}{g} \cdot \frac{k}{1 - k} + \infty = \infty$$
So, the time looks infinite... Is this true? Just by taking into account a collision time/delay, the process should be considered infinite (in time)? Or, where are my mistakes and what should I do to avoid this situation? Thanks a lot!
 A: Your approach looks reasonable. Since the ball always keeps some proportion of its kinetic energy, it will bounce an infinite number of times. You might expect the time for each bounce to approach 0 as the bounces get lower and faster, but that's apparently not the case, as the problem states that each rebound requires a fixed amount of time (which isn't terribly realistic). Doing anything that takes a finite amount of time an infinite number of times will of course take an infinite amount of time.
A: It is sure that with an infinite number of bounces and a finite duration for each bounce, the total time will become infinite. So we can conclude that the model is not good, at least to predict the total duration !
To have a finite total duration, you could introduce a rebound duration which decreases according to the collision speed with the ground: it is a sphere which deforms all the less as the impact speed is low.
But in practice, if you want a model which take into account the last bounces, when the height of rebound is of the order of the deformation, it is surely necessary to return to a much more complicated model.
See for exemple this link : Hertz contact theory
