Question

A body is projected up with a speed $v_0$, along the line of greatest slope of an inclined plane of angle of inclination $\beta$. If the body collides elastically perpendicular to the inclined plane, find the time after which the body passes through its point of projection.

Answer 1

On striking the plane at B, the velocity along the x-axis should be zero. So using $v_x=u_x+a_{x}t$, we get

$0=v_{0}cos\alpha -g\sin \beta T$, where T is the time of flight

$T=\frac{v_0\cos \alpha }{g\sin \beta }$ ....(i)

Also $S_y=0$ from $A$ to $B$. So using $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$, we get

$0=\left(v_0\sin \alpha \right)T-\frac{1}{2}\left(gcos\beta \right)T^2\Rightarrow T=\frac{2v_0\sin \alpha }{g\cos \beta }$ ....(ii)

From (i) and (ii), $\frac{v_{0}cos\alpha }{g\cos \beta }=\frac{2v_0\sin \alpha }{g\cos \beta }\Rightarrow \tan \alpha =\frac{1}{2}cot\beta$

$\Rightarrow \cos \alpha =\frac{2}{\sqrt{4+{\cot }^2\beta }}=\frac{2\sin \beta }{\sqrt{1+3{\sin }^2\beta }}$

Put the value of cosa in (i) to get $T=\frac{2v_0}{g\sqrt{1+3{\sin }^2\beta }}$

Since the body collides elastically, it will rebound perpendicular to the inclined plane with the same speed $v$. In consequence, it retraces its path elapsing equal time $T$ for backward journey to the point of projection. Hence, the required time of motion is

$T^{\prime }=2T=\frac{4v_0}{g\sqrt{1+3{\sin }^2\beta }}$