Question

A ball collides with smooth and fixed inclined plane of inclination $\theta$ after falling vertically through a distance $h\text{m}$. If it moves horizontally just after the impact, then the coefficient of restitution is

• A. ${\tan }^2\theta$
• B. ${\cot }^2\theta$
• C. $\tan \theta$
• D. $\cot \theta$

Answer 1

The correct option is A ${\tan }^2\theta$

Let the situation be as shown in the below figure.


Speed of ball just before collision $u=\sqrt{2gh}$
Collision is perfectly elastic hence, $(e=1)$ along the line of impact
$\Rightarrow e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$
$\Rightarrow e=\frac{v\sin \theta }{u\cos \theta }$ ...(i)

As we know, perpendicular to the $\text{LOI}$, momentum will be conserved for individual particle.
Thus, $u\sin \theta =v\cos \theta \Rightarrow v=u\tan \theta$ ...(ii)
So, from eq. (i) and (ii) we have
$e=\frac{u\tan \theta \cdot \sin \theta }{u\cos \theta }$
$\Rightarrow e={\tan }^2\theta$
Hence, the coefficient of restitution is $e={\tan }^2\theta$