Question

A body is projected up with a speed $V$ of an inclined plane of angle of inclination β. If the body collides elastically perpendicular to the inclined plane, find the time after which it crosses through its point of projection after collision ?

• A. $\frac{2V}{g\left(\sin \beta \right)}$
• B. $\frac{2V}{g\left(\tan \beta \right)}$
• C. $\frac{2V}{g\left(\sqrt{1+3{\sin }^2\beta }\right)}$
• D. $\frac{2V}{g\left(\sqrt{1+3{\tan }^2\beta }\right)}$

Answer 1

The correct option is C $\frac{2V}{g\left(\sqrt{1+3{\sin }^2\beta }\right)}$

Taking x axis along the inclined plane and y axis perpendicular to plane and resolving gravity in that direction -

On striking the plane perpendicularly, the velocity along x axis should be zero. So using $v_x=u_{x+}{a}_{x}t$
We get
$0=V\cos \alpha -g\sin \beta t\Rightarrow t=\frac{V\cos \alpha }{g\sin \beta }$. Also $s_y=0$from A to B.
So using
$s_y=u_{y}t+\frac{a_y{t}^2}{2}$

We get,
$0=\frac{2V\left(\sin \alpha \right)t-g\left(\cos \beta \right)t^2}{2}\Rightarrow t=\frac{2V\sin \alpha }{g\cos \beta }$

Equating $t$ from above two equation, we get :
$\frac{V\cos \alpha }{g\sin \beta }=\frac{2V\sin \alpha }{g\cos \beta }\Rightarrow \tan \alpha =\frac{cot\beta }{2}$
$\Rightarrow \cos \alpha =\frac{2}{\sqrt{4+co{t}^2\beta }}=\frac{2\sin \beta }{\sqrt{1+3{\sin }^2\beta }}$
Putting value of $\cos \alpha$in $t$ we get $t=\frac{2V}{g\sqrt{1+3{\sin }^2\beta }}$ .
Since collision is elastic at perpendicular to inclined plane, so it will trace back same path and will take same time t as mentioned above to get back to point of projection.