Bouncing Ball in a Cavity II

I thought today I'd review the bouncing ball in a cavity problem I looked at a couple days ago. In the original post, I had calculated the number of bounces that it would take the ball to hit the ground. I was thinking today that I'd just finish off the problem by calculating the trajectory. Nothing really fancy here, I'm afraid. I put everything into a normalized coordinate system, with normalized X given by $X^\prime = \frac{x}{L}$, $Y^\prime = \frac{y}{\phi^2}$ and $T^\prime = \frac{t}{\frac{L}{V_0}}$. I then just worked out the positions vs time by brute force--because X'(T') is not differentiable, there isn't really an elegant way to make these plots. I only plotted for $\alpha = 0.9$ and $\phi = 6$, which gives $N = 4.84$, so four bounces total. Here are the results.

X'(T') is pretty much what you'd expect. Linear and non-differentiable as I said. You'll notice that the time between each direction change is growing as you'd expect in this situation.

Y'(T') is perfectly parabolic, since in the original problem I had the loss in velocity only along X'. The proper way to do this is probably to have the loss be a loss in kinetic energy rather than X velocity, but this complicates matters substantially, since the x and y components of velocity change at different rates. I don't know that there's a really nice way to do this.

And here's the trajectory of the bouncing ball. Y' depends on $X^'2$, of course, although looking at the plot it is a bit hard to tell. I expect if I had chosen a different value of $\phi$ and $\alpha$, then this would probably be more pronounced.

Bouncing Ball in a Cavity

Here's a quick, fun problem I just thought up. Suppose you have a rectangular cavity like the one I've drawn below. A ball is fired vertically across the cavity with initial velocity $V_0$. When it hits the wall, its direction is reversed and it bounces back the other way. Suppose that at each bounce, its x-velocity is decreased by $\alpha$. How many bounces will the ball make during the trip?

I decided to do this problem with only the x-velocity changing on bounces since it gets pretty complicated if both x and y (or kinetic energy, say) is changing, but it should be doable in principle either way. To start off, let's calculate the time it takes the ball to hit the ground.

$h = \frac{1}{2}gt^2$

$t = \sqrt{\frac{2h}{g}$

Easy enough. Now what about the velocity in the x direction? Well, for the first pass, we have:

$L = V_0 t_0$

But the next pass takes longer...

$L = V_0 \alpha t_1$

If we add up all of the times that it travels along x, we should get the total y time...

$\sqrt{\frac{2h}{g}} = \frac{L}{V_0} + \frac{L}{V_0\alpha} + \frac{L}{V_0\alpha^2} + \ldots$

Let's define $\phi = \frac{\sqrt{\frac{2h}{g}}}{\frac{L}{V_0}}$. This is the ratio of the total travel time to the single pass time. If $\alpha$ were equal to zero, this value alone would give us the number of passes N. As it is, we've still got some more work to do:

$\phi = \Sigma_{i=0}^{N-1} \alpha^{-i}$

This is a geometric series and can be computed analytically:

$\phi = \frac{\alpha^{-N} - 1}{\alpha^{-1}-1}$

Rearrange a bit:

$\phi (\alpha^{-1}-1) + 1 = \alpha^{-N}$

And finally

$N = \frac{\ln \big( \phi (\alpha^{-1} - 1) \big)}{\ln (\alpha^{-1})}$

Here's some graphs of N vs $\phi$ for different values of $\alpha$. Intuitively, this is pretty much what we might expect: as $\phi$ (that is, the ratio between the total time and the single-pass time) grows, we get progressively more bounces, but the effect is moderated significantly by the energy lost per bounce. For relatively small $\alpha$ like the pink graph, we get many bounces, although we still end up with only about half as many bounces as the $\alpha = 1$ case where no energy is lost.

I'll look at the trajectories in part II