Question

A rod AB of mass $M$ and length $L$ is kept on a smooth horizontal surface. A particle of mass $m$ moving with speed $v_0$ collides with the stationary rod (shown in figure) at point C, at $\frac{L}{4}$ distance from the centre of the rod. After collision, the particle becomes stationary, but the rod translates with speed $v$ and rotates anticlockwise with angular speed $\omega$ about the centre of the rod O. Following conservation of angular momentum, which of the following statements is correct for the (rod + particle) system?

• A. About point O, $m{v}_0\frac{L}{4}=\frac{M{L}^2}{12}.\omega$
• B. About point C, $Mv.\frac{L}{4}=\frac{M{L}^2}{12}.\omega$
• C. About point A, $m{v}_0.\frac{3L}{4}=Mv.\frac{L}{2}+\frac{M{L}^2}{12}.\omega$
• D. About point B, $m{v}_0.\frac{3L}{4}=Mv.L+\frac{M{L}^2}{12}.\omega$

Answer 1

Answer: (A), (B), (C)

About point A, $m{v}_0.\frac{3L}{4}=Mv.\frac{L}{2}+\frac{M{L}^2}{12}.\omega$

There is no external torque acting on the rod + particle system.
Using conservation of angular momentum about different points,


About O - $m{v}_0\times \frac{L}{4}=\frac{M}{12}{L}^2\times \omega$
About C - $Mv\times \frac{L}{4}=\frac{M{L}^2}{12}\times \omega$
About A - $m{v}_0\times \frac{3L}{4}=Mv\times \frac{L}{2}+\frac{M{L}^2}{12}\times \omega$
About B $m{v}_0\times \frac{L}{4}=Mv\times \frac{L}{2}+\frac{M{L}^2}{12}\times \omega$

Option (a), (b), (c) are correct.