Question

A solid ball of radius $0.2$ $m$ and mass $1$ $kg$ lying at rest on a smooth horizontal surface is given an instantaneous impulse of $50$ $Ns$ at point $P$ as shown. The number of rotations made by the ball about its diameter before hitting the ground is

• A. $\frac{625\sqrt{3}}{2\pi }$
• B. $\frac{2500\sqrt{3}}{2\pi }$
• C. $\frac{3125\sqrt{3}}{2\pi }$
• D. $\frac{1250\sqrt{3}}{2\pi }$

Answer 1

Answer: (C)

Impulse = change in momentum
Along vertical
$V_V=\frac{50\cos 30^{\circ }}{1}\frac{m}{s}=25\sqrt{3}m{s}^{-1}$
Along horizontal
$V_H=\frac{50cos60^{\circ }}{1}\frac{m}{s}=25$ $m{s}^{-1}$
Time of flight $T=\frac{2V_V}{g}=5\sqrt{3}s$
Also $\omega =\frac{50\times 0.2}{\frac{2}{5}m{r}^2}$ $rad{s}^{-1}$
$=625$ $rad{s}^{-1}$
$2\pi n=\omega T$
$\Rightarrow n=\frac{\omega T}{2\pi }=\frac{3125\sqrt{3}}{2\pi }$