Question

A homogeneous 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 ${\overline{v}}_0$ perpendicular to the rod. If the velocity of the farther end of the rod just after impact is in the opposite direction of ${\overline{v}}_0$ then.

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

Answer 1

The correct option is D $\eta <6$

(I) Linear momentum conservation

$m{v}_0=(M+m)v_{CM}$

(II) Angular momentum conservation about point $)$:

$m{v}_{0}x=I^{\prime }w$ $I^{\prime }=\frac{(m+M)l^2}{12}\$Momentofinertiaofarodaboutcenter\$$

$m{v}_{0}x=\frac{(m+M)l^2}{12}w$

using (I)

$(M+m)v_{CM}x=\frac{(m+M)l^2}{12}w$

$\to [v_{CM}x=\frac{{\eta }^2{x}^2}{12}w$ $\{l=\eta x\}$ given

$\to v_{CM}=\frac{{\eta }^2\times w}{12}$

given, after collision velocity of top most point is opposite to $v_0$ i.e.

$\frac{wL}{2}-v_{CM}>0$

$\to \frac{w\eta x}{2}-\frac{{\eta }^2\times w}{12}>0\Rightarrow \eta <6$