I'm facing a lot of confusion regarding the conservation of momentum when a ball deflects from an infinitely massive wall.
Suppose, the ball makes an angle $\alpha$ with the normal, before the deflection. After it, it makes the angle $\beta$ against the normal.
Now I know, in a perfectly elastic collision, the coefficient of restitution $e$ is equal to $1$. In that case, it turns out that the angle of incidence is equal to the angle of reflection, such that $\alpha=\beta$.
My confusion comes, when we consider the non-elastic general cases. I want to know, what is the relation between $\alpha$ and $\beta$ when the collision is not perfectly elastic.
According to some sources, we have : $$\alpha+\beta=\frac{\pi}{2}$$
In some other examples, this is not true. In those examples, $\alpha+\beta$ can take any value.
I'm inclined to believe the latter, that the sum of the angles can take any random value depending on the coefficient of restitution and so on.
However, what is the special case, when the sum of these angles is $\pi/2$, and how can I prove it ?
Best Answer
Suppose that the wall is parallel to the $y$ axis, and that the incoming velocity vector is $$ {\bf v}_{\rm in}= (v_x,v_y) $$ giving $\alpha= \tan^{-1}(v_y/v_x)$. With coefficient of restitution $c$, the outgoing velocity is $$ {\bf v}_{\rm out}= (-cv_x, v_y) $$ making $\beta = \tan^{-1}(v_y/cv_x)$ Then $$ \tan(\alpha+\beta)=\frac{\tan \alpha +\tan\beta}{1-\tan\alpha\tan\beta}, $$ is not going to be infinity, so $\alpha+\beta\ne \pi/2$.