Question

A marble starts falling from rest on a smooth inclined plane forming an angle $\alpha$ with horizontal. After covering distance $h$ the ball rebound off the plane. The distance from the impact point where the ball rebounds for second time is

• A. $8hsin\alpha$
• B. $8hcos\alpha$
• C. $hsin\alpha$
• D. $hcos\alpha$

Answer 1

The correct option is A $8hsin\alpha$

The balls fall from $A$ on to incline and bounces off to $B$

$B{B}^{{}^{\prime }}$ is the normal to the incline at $B$.

Since it is smooth plane and collision l bouncing is elastic, the ball bounces at the same angle with the normal $B{B}^{{}^{\prime }}$

The velocity remains same

Hence the angle of projection of the ball w.r.t. horizontal$=90°2\alpha$

Velocity of the ball at $B=\nu =\sqrt{2gh}$

Let $B=(0,0)$

The equation of the trajectory of the ball (parabolic path)

$y=x\tan (90°-2\alpha )-\frac{g{x}^2}{2{\nu }^2}{\sec }^2(90°-2\alpha )$

Equations of the inclined plane: $y=-x\tan \alpha$

Ball hits the plane at

$-x\tan \alpha =x\cot 2\alpha -\frac{g{x}^2}{4gh}{\csc }^2\left(2\alpha \right)$

$\therefore x=0$ or $x=(\cot 2\alpha +\tan \alpha )4h{\sin }^{2}2\alpha$

or $c=\frac{4h(1-{\tan }^2\alpha +2{\tan }^2\alpha ){\sin }^{2}2\alpha }{2\tan \alpha }$

$x=4h\sin 2\alpha$

$y=-4h\sin 2\alpha \cdot \tan \alpha =-8h{\sin }^{2}2\alpha$

Distance$=\sqrt{x^2+y^2}=8h\sin \alpha$

The distance on inclined plane from $B$ (point of first bounce to point of second bounce)$=\frac{x}{\cos \alpha }=8h\sin \alpha$