Inelastic but the disc comes back

The best way to solve this problem is by using angular momentum conservation in the following way

Just After colloision the disc rebounds with a velocity $e . V$ towards the left with no change in angular velocity.

Now after some time the disc will again start pure rolling now what we want is that the disc to move towards the right when it again starts pure rolling .At that moment the angular momentum about any point on the ground would be tending to 0 and into the plane (let's denote it by -ve sign).

Now since the angular momentum remains conserve about that point so

Angular momentum just After colloision= final angular momentum

$\Rightarrow M.e.V.R- \omega I_{com} \leq 0$

$\Rightarrow M.e.V.R \leq \dfrac{MR^{2}}{2} \omega$

$\Rightarrow e \leq \dfrac{1}{2}$

After the collision (and until rolling resumes), let the disc have speed $V$ towards the wall, and angular velocity $\omega$ . The disc experiences a friction force $F = \mu mg$ towards the wall. Then Newton's Laws state that $m\dot{V} \; = \; F \hspace{2cm} \tfrac12mr^2 \dot{\omega} \; = \; -Fr$ so that $\dot{V} = \mu g$ and $\dot\omega = -\frac{2\mu g}{r}$ . Thus we deduce that (with time $t=0$ representing the moment of collision) $V \; = \; \mu g t - ev \hspace{2cm} \omega \; = \; \frac{v}{r} - \frac{2\mu g t}{r}$ until rolling resumes, and this occurs when $\mu g t - ev \; = \; V \; = \; r\omega \; =\; v - 2\mu gt$ namely when $t = \frac{(1+e)v}{3\mu g}$ . At this time we have $V \; = \; \tfrac13(1 - 2e)v$ and so rolling resumes with the disc still heading towards the wall (and so a second collision occurs) provided that $e < \tfrac12$ . The supremum of possible values of $e$ is thus $\boxed{\tfrac12}$ .

Note that, if $e=\tfrac12$ , the cylinder stops skidding precisely when $V=0$ and $\omega = 0$ . Thus the cylinder stops skidding, coming to rest at the same time. This means that the cylinder will not return to collide with the wall a second time in this case. The set of values of $e$ for which a second collision is possible does not have a maximum.

It is interesting to note that, during the skidding process, the distance of the cylinder from the wall is $evt - \tfrac12\mu gt^2$ Thus, if $\frac{2ev}{\mu g}<\frac{(1+e)v}{3\mu g}$ , which occurs if $e<\tfrac15$ , the cylinder strikes the wall a second time before the skidding process stops.