Question

A particle projected from point $A$ with velocity $u\sqrt{2}$ at an angle of 45^o with the horizontal, strikes a plane BC making an angle $60^{0}$ as shown in the figure. The velocity of the particle at the time of colission is

• A. $\frac{\sqrt{3 u}}{2}$
• B. $\frac{u}{2}$
• C. $\frac{2 u}{\sqrt{3}}$
• D. $u$

Answer 1

Answer: (C)

Let $v$ be the velocity at the time of collision


$\textit{v sin 60} = u \sqrt{2} cos 45^{\text{ο}}$
$\left(u \sqrt{2}\right) \frac{1}{\sqrt{2}} = \frac{v \sqrt{3}}{2}$
$v = \frac{2 u}{\sqrt{3}} ms^{- 1}$