Question

A uniform rod AB of mass $m$ and length $l$ is are on a smooth horizontal surface. An impulse $J$ is applied to the end $B$, perpendicular to the rod in the horizontal direction. Speed of particle $P$ at a distance $\frac{l}{6}$ 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 $\sqrt{2}\frac{J}{m}$

As the rod is hit, its Center of Mass will start moving with a velocity calculate as

From Impulse momentum theorem, we know that Impulse is equal to change in momentum

$\Rightarrow J=m\times V_{COM}$

$V_{COM}=\frac{J}{m}$

Also, at the same time the rod also starts rotating. By applying the principle of conservation of angular momentum, we have

If the angular speed of the rod after the impact is $\omega$, then

$\frac{l}{2}J=I_{COM}\omega$

$\Rightarrow \omega =\frac{\frac{lJ}{2}}{\frac{m{l}^2}{12}}$

$\Rightarrow \omega =\frac{6J}{ml}$

Now, we need to find the speed of particle $P$ after time $t=\frac{\pi ml}{12J}$

Using the formula,

$\theta =\omega t$, we have

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

This means that at $t=\frac{\pi ml}{12J}$, the rod is turned by $\frac{\pi }{2}$ angle.

The horizontal velocity of point P is same as center of mass while the vertical component is due to rotation and is given by

$v_t=r\times \omega$

$r=\frac{l}{6}$

$\Rightarrow v_t=\omega \times \frac{l}{6}=\frac{6J}{ml}\times \frac{l}{6}=\frac{J}{m}=v$

$\therefore v_{resultant}=\sqrt{{v_t}^2+{v_h}^2}=\sqrt{v^2+v^2}=\sqrt{2}v$

Hence, the velocity of point $P$ is $\sqrt{2}v=\sqrt{2}\frac{J}{m}$.

Hence, the correct answer is OPTION D.