# 5.2: Bouncing Balls

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When a ball is dropped to the ground, one of four things may happen:

1. It may rebound with exactly the same speed as the speed at which it hit the ground. This is an elastic collision.
2. It may come to a complete rest, for example if it were a ball of soft putty. I shall call this a completely inelastic collision.
3. It may bounce back, but with a reduced speed. For want of a better term I shall refer to this as a somewhat inelastic collision.
4. If there happens to be a little heap of gunpowder lying on the table where the ball hits it, it may bounce back with a faster speed than it had immediately before collision. That would be a superelastic collision.

The ratio

$\dfrac{\text{speed after collision}}{\text{speed before collision}} \nonumber$

is called the coefficient of restitution, for which I shall use the speed before collision symbol $$e$$. The coefficient is 1 for an elastic collision, less than 1 for an inelastic collision, zero for a completely inelastic collision, and greater than 1 for a superelastic collision. The ratio of kinetic energy (after) to kinetic energy (before) is evidently, in this situation, $$e^{2}$$.

If a ball falls on to a table from a height $$h_{0}$$, it will take a time $$t_{0} = \sqrt{2H_{0}lg}$$ to fall. If the collision is somewhat inelastic it will then rise to a height $$h_{1}=e^{2}h_{0}$$ and it will take a time $$et$$ to reach height $$h_{1}$$. Then it will fall again, and bounce again, this time to a lesser height. And, if the coefficient of restitution remains the same, it will continue to do this for an infinite number of bounces. After a billion bounces, there is still an infinite number of bounces yet to come. The total distance travelled is

$h = h_{0} +2h_{0}(e^{2}+e^{4}+e^{6}+...) \tag{5.2.1}\label{eq:5.2.1}$

and the time taken is

$t = t_{0} +2t_{0}(e + e^{2}+e^{3}+...). \tag{5.2.2}\label{eq:5.2.2}$

These are geometric series, and their sums are

$h = h_{0} \left(\frac{1+e^{2}}{1-e^{2}}\right), \tag{5.2.3}\label{eq:5.2.3}$

which is independent of g (i.e. of the planet on which this experiment is performed), and

$t = t_{0} \left(\frac{1+e}{1-e} \right) \tag{5.2.4}\label{eq:5.2.4}$

For example, suppose $$h_{0}$$ = 1 m, $$e$$ = 0.5, $$g$$ = 9.8 m s−2, then the ball comes to rest in 1.36 s after having travelled 1.67 m after an infinite number of bounces.

##### Discuss

Does the ball ever stop bouncing, given that, after every bounce, there is still an infinite number yet to come; yet after 1.36 seconds it is no longer bouncing...?

This page titled 5.2: Bouncing Balls is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Jeremy Tatum via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.