Question

A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod in horizontal direction. Speed of particle P at a distance l/6 from the centre towards A of the rod after time $t=\frac{\pi ml}{12J}$ is

• A. $2\frac{J}{m}$
• B. $\frac{J}{\sqrt{2}m}$
• C. $\frac{J}{m}$
• D. $\sqrt{2}\frac{J}{m}$

Answer 1

The correct option is D $\sqrt{2}\frac{J}{m}$

Let V and w be the linear and angular velocity of the rod after applying an impulse so that.

Impulse j = change in momentum

or J = (mv - 0) = V = J/m

By conservation of linear momentum.

$IW=y\left(l/2\right)$ about center

$m\frac{l^2}{12}w=J\frac{1}{2}$ or, $w=\frac{6J}{ml}$

after the given time t, rod will rotate an angle = wt.

$=\frac{6J}{ml}\times \frac{\pi ml}{12J}=\pi /2$

$\therefore V=\frac{1}{6}w=J/m$

$V_p=\sqrt{2}V=\frac{\sqrt{2}J}{m}$