WordPress parses by typing “$lat.ex $”, (without the period in the middle of “latex”. I just put that there so it would display) in case you want it for future comments.

I’d take the same approach to that problem as you outlined. Another problem is that if the collisions are perfectly elastic, then *no* speed is sufficient to crack your skull, since cracking your skull takes energy.

I’m not sure whether the balls at the tail ends of the distribution should be expected to have had more impacts. If a ball is on the tail end, then impacts are more likely to bring it back towards the center of the distribution. Off the top of my head it’s a hard question.

When the collisions are inelastic, I suppose we could try to estimate the number of collisions necessary to come to equilibrium, and then estimate what the remaining energy would be. I might think about it a little more. You could calculate, for example, the distribution in energies after a single collision, and see how wide that is.

]]>If you dropped 1000 super-bouncy-balls en masse off the roof of a tall building (say 100m high) what is the probability that, following the ensuing series of collisions at ground level, one ball shoots out fast enough to crack someone’s skull? I guess we’d have to define “fast enough to crack someone’s skull,” for this to make sense, so let’s just assume that corresponds to some kinetic energy threshold.

The simplest starting point seems to be to treat the balls like atoms of an ideal gas and assume that just before hitting the ground they all have the same kinetic energy $E\left( h\right)$ (determined by the height $h$ of the building), which is then distributed among them by numerous collisions (assumed to be completely elastic) until the distribution of ball speeds corresponds to a Maxwell-Boltzmann distribution with average kinetic energy $\left\langle E\left(h\right)\right\rangle$. Then the number of balls dropped simply influences the probability that at least one ends up above the kinetic energy threshold for cracking someone’s skull.

However, one realistic issue immediately makes the problem more complex: the collisions aren’t perfectly elastic. If I recall correctly (from Mr. Wizard, no less), the very best rubber balls return to about 90% of their initial height after bouncing off a hard surface, meaning 10% of their energy is dissipated inelastically. But a 10% loss (or even a 1% one) becomes significant as soon as you need to have many collisions take place in order to go from a narrow kinetic energy distribution to a broad Maxwell-Boltzmann-esque one. Plus I imagine the outliers at the tails of the distribution have probably undergone (on average) a greater number of collisions than the median.

Anyway, that’s as far as I’ve ever gone in my thinking. Perhaps you’ll get further.

]]>