Troublesome Cylinder!

I guess I overstated the case.

Suppose we understand "perfectly elastic" to mean "energy-conserving". Suppose that the cylinder has horizontal velocity $v_x$ away from the wall, and vertical velocity $v_y$ , and also has angular velocity $\omega$ . Conservation of angular momentum (about the point of contact of the cylinder with the wall) tells us that $r\omega + 2v_y \; = \; v_0$ while, assuming conservation of energy, we deduce that $r^2\omega^2 + 2v_x^2 + 2v_y^2 \; = \; 3v_0^2$ Since the impulsive normal reaction is four times larger than the impulsive friction force, we also deduce that $x_v + v_0 \; = \; 4v_y$ Putting these three equations together we deduce that $(r\omega - v_0)(19r\omega + v_0) \; = \; 0$ and hence (either $r\omega = v_0$ , which is what happens if no collision occurs, or) $r\omega = -\tfrac{1}{19}v_0$ , and so the cylinder has, in fact, a nonzero angular velocity after the collision.

If we take "perfectly elastic" to mean "the cylinder moves away from the wall with the same speed as it approached the wall (so the coefficient of restitution is $1$ ", then we would have $v_x = v_0$ and $v_y = \tfrac12v_0$ , which would imply that $\omega = 0$ . But then energy is not conserved.

Elastic collisions are more complicated when the colliding objects are no longer point particles!