Question

A marble starts falling from rest on a smooth inclined plane of inclination $\alpha$. After covering distance $h$ the ball rebounds off the plane. The distance from the impact point where the ball rebounds for the second time is :

• A. $8h\cos \alpha$
• B. $8h\sin \alpha$
• C. $2h\tan \alpha$
• D. $4h\sin \alpha$

Answer 1

Answer: (B)

$8h\sin \alpha$

Using the energy balance we get the velocity of the marble near the incline as
$\frac{m{v}_o^2}{2}=mgh$
or
$v_o=\sqrt{2gh}$
The distance traveled in the direction of the incline is given as
$S=v_{o}sin\alpha t+gsin\alpha \frac{t^2}{2}$
and
$o=v_{o}cos\alpha t-gcos\alpha \frac{t^2}{2}$
or
$t=\frac{2v_o}{g}$
Substituting this in the equation for $S$ we get
$S=\frac{4v_0^2}{g}sin\alpha$
Substituting $v_o^2=2gh$
we get $S=8hsin\alpha$