Q. A bullet of mass $m$ leaves a gun of mass $M$ kept on a smooth horizontal surface. If the speed of the bullet relative to the gun is $v$, the recoil speed of the gun will be

A) $\frac{m}{M}v$
B) $\frac{m}{M + m},v$
C) $\frac{m}{M + m}v$
D) $\frac{M}{m}v$
Explanation

(b) Speed of the bullet relative to ground ${\overrightarrow{v}}_{b} = \overrightarrow{v} + {\overrightarrow{v}}_{r},$ where $v_{r}$ is recoil velocity of gun. Now for gun-bullet system applying the conservation law of momentum, we get
$m{\overrightarrow{v}}_{b} + M{\overrightarrow{v}}_{r} = 0$ or $m\left( \overrightarrow{v} + {\overrightarrow{v}}_{r} \right) + M{\overrightarrow{v}}_{r} = 0$
$\Rightarrow \ \ {\overrightarrow{v}}_{r} = - \frac{m\overrightarrow{v}}{m + M}$ or $v_{r} = \frac{mv}{m + M}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.