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.