A ball of mass $m$ strikes the fixed inclined plane after falling through a height $h$. If it rebounds elastically, the impulse imparted on the ball is

The correct option is A $2m\sqrt{2gh}\cos \theta$
Speed of ball just before the collision $u=\sqrt{2gh}$
Collision is elastic hence, $e=1$
$e={\left(\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}\right)}_{\text{along line of impact}}$


$\Rightarrow e=\frac{v}{u\cos \theta }$
$\Rightarrow v=eu\cos \theta =u\cos \theta \because e=1$

Here, $v$ is the velocity component of ball just after collision along line of impact.
Impulse $J=\Delta P=mv-(-mucos\theta )$
$\Rightarrow \Delta P=mu\cos \theta +mu\cos \theta$
$\Rightarrow J=\Delta P=2m\cos \theta \sqrt{2gh}$