Question

A particle is projected from a point A with velocity u$\sqrt{2}$ at an angle of 45º with horizontal as shown in figure. It strikes the plane BC at right angles. The velocity of the particle at the time of collision is :

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

Answer 1

Answer: (C)

$\frac{2u}{\sqrt{3}}$

Answer 2

Let v be the velocity at the time of collision.

Then, u$\sqrt{2}$cos 45º = v sin 60º

(u$\sqrt{2}$)$\left( \frac{1}{\sqrt{2}} \right)$=$\frac{\sqrt{3}v}{2}$

\ v =$\frac{2}{\sqrt{3}}$u