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

Answer 1

Answer: (C)

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


$\text{v sin 60}=u\sqrt{2}cos45^{\text{ο}}$
$\left(u\sqrt{2}\right)\frac{1}{\sqrt{2}}=\frac{v\sqrt{3}}{2}$
$v=\frac{2u}{\sqrt{3}}m{s}^{-1}$