Question

A homogenous rod of length $l=\eta x$ and mass $M$ is lying on a smooth horizontal floor. A bullet of mass m hits the rod at a distance x from the middle of the rod at a velocity $v_0$ perpendicular to the rod and comes to rest after collision. If the velocity of the farther end of the rod just after the impact is in the opposite direction of $v_0$, then:

• A. $\eta >3$
• B. $\eta <3$
• C. $\eta >6$
• D. $\eta <6$

Answer 1

The correct option is D $\eta <6$

Let after collision velocity of rod be V and angular velocity be $\omega$, then
$m{v}_0=mV+MV$
and
$m{v}_{0}x=mVx+\frac{M{l}^2}{12}\omega$
$m{v}_0=mV+\frac{M\omega {\eta }^{2}x}{12}$
From above equation, $\frac{M\omega {\eta }^{2}x}{12}=MV\Rightarrow \omega {\eta }^{2}x=12V$
Velocity of farther end of the rod $=V-\frac{l\omega }{2}$
$=\frac{\omega {\eta }^{2}x}{12}-\frac{\omega \eta x}{2}=\frac{\omega \eta x}{2}(\frac{\eta }{6}-1)$
If this velocity is opposite to $v_0$, then,
$\frac{\eta }{6}-1<0\Rightarrow \eta <6$