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. What is the final angular velocity of the rod ?

• A. $\omega =\frac{v}{9L}$
• B. $\omega =\frac{2v}{9L}$
• C. $\omega =\frac{3v}{9L}$
• D. $\omega =\frac{5v}{9L}$

Answer 1

The correct option is B $\omega =\frac{2v}{9L}$

Angular momentum about $O$ remains constant just before and just after collision
$\therefore L_i=L_f$ or $\left(\frac{m}{6}v\frac{L}{2}\right)=I\omega$
$=[\frac{m{L}^2}{3}+\frac{m}{6}\frac{L^2}{4}]\omega$
Solving, we get

$\frac{mvL}{12}=\frac{9m{L}^2}{24}$
$\omega =\frac{2v}{9L}$