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 ?

  • $\begingroup$ Inelastic is a general name for every case energy isn't conserved. This about the case that only a tiny fraction of the energy goes to waste, meaning energy is approximately conserved and the angles are almost the same. If that is the case, then surely it can't be that the general formula is $\alpha+\beta=\pi / 2$. $\endgroup$ Commented Dec 10, 2021 at 19:23
  • $\begingroup$ @OfekGillon yes, but I've seen examples when we had $\beta=\pi/6$ and we had to consider $\alpha=\pi/3$ just to proceed with the sum. It was a problem where we had been given the coefficient of restitution and the angle of deflection, and had to calculate percentage change in kinetic energy. $\endgroup$
    – RayPalmer
    Commented Dec 10, 2021 at 19:28
  • $\begingroup$ if you were given the angle of deflection, then what seems to be the problem? This was a specific case that you couldn't know the angle a priori without extra knowledge about "what happened during the collision" $\endgroup$ Commented Dec 10, 2021 at 19:36
  • $\begingroup$ What source states alpha + beta = pi/2? If the ball is sticky putty it will stick to the wall. $\endgroup$
    – John Darby
    Commented Dec 10, 2021 at 19:40
  • $\begingroup$ @JohnDarby is it true then that when two balls collide, but not along the line joining their COM, they are deflected in different directions, and the sum of angles of deflection is $\pi/2$ ? $\endgroup$
    – RayPalmer
    Commented Dec 10, 2021 at 19:44

2 Answers 2


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 $\alpha+\beta\ne \pi/2$.

  • $\begingroup$ with \begin{align*} &\tan(\beta)=\frac{1}{\epsilon}\,\tan(\alpha)\\ &\alpha+\beta=\frac{\pi}{2}\\ \end{align*} you obtain unique solution for $~\alpha~,\beta~$ so what is wrong with this solution ?? $\endgroup$
    – Eli
    Commented Dec 11, 2021 at 22:41
  • $\begingroup$ @Eli Because $\alpha$ is an arbitrary angle between 0 and $\pi/2$. It's determined by how you throw the ball at the wall. $\endgroup$
    – mike stone
    Commented Dec 11, 2021 at 22:56
  • $\begingroup$ O.k I see , but if I know the coefficient of restitution i can choose $\alpha$ ? and obtain $\alpha+\beta=\pi\2$ $\endgroup$
    – Eli
    Commented Dec 12, 2021 at 8:31

For an elastic collision between balls of equal mass $\alpha + \beta = {\pi \over 2}$. This is not true in general. See a physics mechanics test, such as Mechanics by Symon for the general case.


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