Question

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 magnitude of 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. $\frac{M}{m}v$

Answer 1

Answer: (B)

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