Question

A ball is released from position $A$ and travels $5\text{m}$ before striking the smooth fixed inclined plane as shown. If the coefficient of restitution in the impact is $e=\frac{1}{2}$, the time taken by the ball to strike the plane again is

Answer 1

Let $u_x,u_y$ be components of velocity parallel and perpendicular to the inclined plane before collision and
$v_x,v_y$ be components of velocity parallel and perpendicular to the inclined plane after collision.
Just before first collision,
Speed $u=\sqrt{2g\left(5\right)}=10\text{m/s}$
So,
$u_y=-10\cos {30}^{\circ }=-5\sqrt{3}$
$u_x=10\sin {30}^{\circ }=5\text{m/s}$

Impulse imparted due to collision is along common normal, as surface is smooth.
Along the plane there will no change in speed during collision
$v_x=u_x=5\text{m/s}$

Applying Newton's law of collision along common normal
$v_y=-e{u}_y=\frac{5}{2}\sqrt{3}$

The ball after collision acts similar to projectile motion from top of the incline.
It strikes the plane again at time,
$t=\frac{2v_y}{g\cos {30}^{\circ }}=\frac{5\sqrt{3}}{5\sqrt{3}}=1\text{sec}$