## Bouncy, bouncy

Posted by Chris on May 27, 2012 – 4:54 pm

A ball is dropped from a height of 1 m on to a hard surface. Each time it bounces, the ball loses half of it’s energy. How long does the ball bounce for? Is there a paradox?

May 28th, 2012 at 1:50 am

If both the ball and the surface are infinitely hard, i.e. zero deformation at each bounce, then the ball would “bounce” for ever. However, these bounces would soon become virtually undetectable.

However, in the real world outside of ToM-land, no material is infinitely hard. So the height of each bounce would soon fall within the range of deformation of the ball and the surface. All that could then be detected would be a slight vibration at each “bounce”. That, too, would soon become virtually undetectable.

May 28th, 2012 at 1:55 am

im pretty sure he means the amount of time the ball takes to stop bouncing

May 28th, 2012 at 6:32 am

Hi danv – that’s right.

Hi Wiz. You have partially misunderstood. Even if the ball and surface are undeformable, it is stipulated that half the energy is lost at each bounce, so the height of each bounce is smaller than the previous one.

For the ideal case, there is a definite time for all the bounces, even though the ball bounces infinitely many times >:~

Otherwise, what you wrote is pretty good – it contributes greatly toward resolving the paradox that I hinted at.

May 28th, 2012 at 2:17 pm

so it would take

1=.5*9.8t^2

t = root 2/9.8 before the first bounce

.5 = .5*9.8t^2

t = root 1/9.8 on the way up and way down after bounce 1

.25 = .5*9.8t^2

t = root 1/9.8 on the way up and way down after bounce 2

so t(1) + 2t(2) + 2t(3)…

which is like 2.633 unless messed up

May 28th, 2012 at 4:29 pm

2.633 seconds is what I get too. As Wiz pointed out, that can’t really be achieved. At some point, the deformation of the ball will be higher than the bounce height, and then the ball will vibrate and dissipate it’s remaining energy.

Part of the calculation involved the series

x = 1 + 1/√2 + 1/√4 + 1/√8 + … = 1 + 1/√2 (1 + 1/√2 + 1/√4 + …)

= 1 + x/√2 => x = 2 + √2 ≈ 3.414

May 31st, 2012 at 5:30 am

Yes there is a paradox involved. This particular paradox is called the dichotomy paradox. This is one of Zeno’s (an ancient Greek philosopher’s) famous paradoxes where for example; Suppose Homer wants to catch a stationary bus. Before he can get there, he must get halfway there. Before he can get halfway there, he must get a quarter of the way there. Before traveling a quarter, he must travel one-eighth; before an eighth, one-sixteenth; and so on. This description requires one to complete an infinite number of tasks, which Zeno maintains is an impossibility.

Therefore, according to the paradox the ball will bounce for an infinite amount of time/ not stop.

May 31st, 2012 at 5:57 am

Hi Loz. Thanks for posting that. I conclude that Zeno was wrong (yet again).

June 3rd, 2012 at 11:54 pm

Just out of stipulation of what i can stir up here, but can anyone prove Zeno was wrong? Whose’s to say the ball “stops moving?” Nothing is ever really at a standstill, as in the case of this ball. As you will find if you observe very closely, the ball will continue bouncing, each bounce smaller than the last, with each bounce occurring at an increasing frequency. This means that the ball’s height, on a graph, will look like a parabola, reaching an infinite number of bounces per second, each at an infinitely smaller height.

So please correct my logic if i am wrong, but this seems to make sense, as it avoids the “flexible floor problem”

June 5th, 2012 at 4:31 pm

Hi Spencer. The ball stops bouncing when the bounce height is on a par with the deformation that would exist due to the ball being at rest. It will then vibrate, and dissipate its energy. The vibrations will eventually become indistinguishable from thermal vibrations.

If the ball continued to lose 1/√2 of it’s amplitude for each vibration, the whole process would still be over in 2.633 seconds. I’d venture that such a ball cannot exist.