Question

A body is projected up with a speed $V$
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 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 .