Question

For the shown collision of the ball with a wall, the coefficient of restitution $e$ is found to be $\frac{1}{N}$. Then, the value of $N$ is

Answer 1

Let the velocity of the ball before and after collision be $v$ and $v^{\prime }$ respectively.


As there is no external force along $X-$ direction, velocity of ball will not change in $X-$direction..
$\Rightarrow v^{\prime }\sin {\theta }^{\prime }=v\sin \theta ...\left(1\right)$
Along $Y-$ direction, normal force by wall will act on the ball, hence velocity of ball will change along $Y-$ direction.
Applying equation for $e$ along $Y-$ direction,
$e=\frac{\text{speed of separation}}{\text{speed of approach}}$
$\Rightarrow e=\frac{v^{\prime }\cos {\theta }^{\prime }}{v\cos \theta }$
$\Rightarrow v^{\prime }\cos {\theta }^{\prime }=ev\cos \theta ...\left(2\right)$
Dividing Eq. $\left(1\right)$ by $\left(2\right)$, we get
$\tan {\theta }^{\prime }=\frac{\tan \theta }{e}$
$\Rightarrow e=\frac{\tan \theta }{\tan {\theta }^{\prime }}=\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}$
[$\theta ={30}^{\circ }$ and ${\theta }^{\prime }={60}^{\circ }$, given]
$\because \tan {30}^{\circ }=\frac{1}{\sqrt{3}},\tan {60}^{\circ }=\sqrt{3}$
$\Rightarrow e=\frac{1}{3}$
$\&e=\frac{1}{N}$ according to question.
On comparing both values, we get $N=3$