Question

A uniform rod of length $L$ rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed $v$ strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. The ratio of the kinetic energy of the system before the collision to the kinetic energy of the bullet after the collision is $\frac{1}{x}$. Find the value of $x$.

Answer 1

Angular momentum about $O$ remains constant just before and just after collision

$\therefore L_i=L_f$ or $\left(\frac{m}{6}v\frac{L}{2}\right)=1\omega$

$=[\frac{m{L}^2}{3}+\frac{m}{6}\frac{L^2}{4}]\omega$

Solving, we get

$\omega =\frac{2v}{9L}$

$\frac{K_f}{K_i}=\frac{\frac{1}{2}I{\omega }^2}{\frac{1}{2}\left(\frac{m}{6}\right)v^2}$

$=\frac{6I{\omega }^2}{v^2}$

$=\frac{6[\frac{m{L}^2}{3}+\frac{m{L}^2}{24}]{\left(\frac{2v}{9L}\right)}^2}{v^2}$

$=6\times \frac{9}{24}\times \frac{4}{81}=\frac{1}{9}$