Question

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

• A. $\frac{m}{M}v$
• B. $\frac{m}{M+m}v$
• C. $\frac{Mv}{M+m}$
• D. $\frac{M}{m}v$

Answer 1

The correct option is B

$\frac{m}{M+m}v$

No external force acts on the system (gun + bullet) during the impact (till the bullet leaves the gun). Therefore the momentum of the system remains constant. Before the impact the system (gun + bullet) was at rest. Hence its initial momentum is zero. Therefore just after the impact, its momentum of the system (gun + bullet) will be zero.

$\Rightarrow m{\overrightarrow{v}}_b+M{\overrightarrow{v}}_g=0\Rightarrow m[{\overrightarrow{v}}_{bg}+{\overrightarrow{v}}_g]+M{\overrightarrow{v}}_g=0$

$\Rightarrow {\overrightarrow{v}}_g=-\frac{m{\overrightarrow{v}}_{bg}}{M+m},where|{\overrightarrow{v}}_{bg}|$ = Velocity of bullet relative gun = v

$\Rightarrow V_g=\frac{mv}{M+m}$ Opposite to the direction of v