Question

A particle is projected with velocity $u$ horizontally from the top of a smooth sphere of radius $a$ so that it slides down the outside of the sphere. If the particle leaves the sphere when it has fallen a height $\frac{a}{4},$ the value of $u$ is

• A. $\sqrt{ag}$
• B. $\frac{\sqrt{ag}}{4}$
• C. $\frac{\sqrt{ag}}{2}$
• D. $\frac{\sqrt{ag}}{3}$

Answer 1

Answer: (C)

$\frac{\sqrt{ag}}{2}$

Let the particle leaves the sphere at point P, so $N=0$ at P.

From geometry, $cos\theta =\frac{3}{4}$

Using work-energy theorem, $mg\frac{a}{4}=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

Circular motion equation at P, $\frac{m{v}^2}{a}=mgcos\theta$

Eliminating $v$, we get $mg\frac{a}{4}=\frac{3}{8}mga-\frac{1}{2}m{u}^2$

$⟹u=\frac{\sqrt{ag}}{2}$