Getting stuck

In the first collision the angle of incidence of the ball, measured relative to the plane of the wall, is $\alpha_1=\alpha_0$ . After the collision the angle will be $\alpha'_1 = \tan^{-1} ( e\tan (\alpha_1))$ . Simple geometry tell us that in the next collision the angle of incidence is $\alpha_2= \alpha'_1+\alpha_0$ . Subsequently, we get $\alpha_3= \alpha'_2+\alpha_0$ , and in genera $\alpha_n= \alpha'_{n-1}+\alpha_0$ , where $\alpha'_{n-1} = \tan^{-1} ( e\tan (\alpha_{n-1}))$ .

The ball gets stuck, if the angle of incidence between the subsequent collisions does not change. For that we have to solve the transcendent equation

$\alpha= \tan^{-1} ( e\tan (\alpha))+\alpha_0$

This can be solved numerically.

In the three graphs below I plotted two quantities. On the left side I show how the angle evolves with the number of collisions. On the right side I plotted $\tan^{-1} ( e\tan (\alpha))+\alpha_0- \alpha$ , so that the zero crossing of this function gives the solution for $\alpha$ .

For $\alpha_0=9^{\circ}$ there are two zero crossings in the right-side graph. The solution clearly exists. Indeed, on the left side we see that the angle converges to a finite value. For $\alpha_0=10.2^{\circ}$ we are at the critical angle, because the curve just touches the zero line and the convergence of the angle is very slow. For $\alpha_0=11^{\circ}$ there is no solution, and the angle as a function of the number of collisions does not converge. That means the ball will come out of the trap. The solution is $\alpha_{critical}=10^{\circ}$ (rounded to the nearest integer).

Note: Here we did not consider what happens with the absolute value of the velocity. It is clear that this velocity decreases with every collision. Therefore when we predict that the ball will come out, but the the number of collisions is large, it will take very long time the see the ball back.