Question

A ball of mass $m$ strikes and fixed inclined plane after falling through a height $h$. If it rebounds without losing energy, find the impulse on the ball.

Answer 1

N'N is perpendicular to the inclined face. N'N is therefore the line of impact. $\overrightarrow{PQ}$ is direction of initial velocity of the ball
$\overrightarrow{QR}$ is the direction of final velocity of the ball
Because this is an elastic collision
$\angle PQ{N}^{\prime }=\angle N^{\prime }QR=\theta$
Impuse is measured along the impact line. The change in velocity of the ball along $N{N}^{\prime }=m\Delta v=m(-2v\cos \theta )=-2mv\cos \theta$
Since the ball falls through a height $h$ just before collision $v=\sqrt{2gh}$
Impulse $=-2m\cos \theta \sqrt{2gh}$