Question

A bullet of mass $0.01$ kg is fired from a gun of mass $5$ kg with velocity of $250$ m/s, calculate the speed with which the gun recoils.

• A. 0.50 m/s
• B. 0.25 m/s
• C. 0.05 m/s
• D. 0.025 m/s

Answer 1

Answer: (A)

0.50 m/s

As the net external force on the gun bullet system is zero, the sum of the momentum of the gun and bullet system will be constant.

Now as the gun+bullet system was initially at rest, the initial momentum of the system =-0.

Hence, according to the law of conservation of momentum,

(the final momentum of the bullet fired) + (final momentum of the gun that recoils (opposite direction)) $=0$

Momentum of the bullet =$\left(0.01\right)\times 250=2.5kgm/s$

Let $v$ be the velocity of the recoil of the gun.

Momentum of the recoiling gun =$\left(5\right)\times v=5vkgm/s$

Equating we have $2.5+5v=0$

$\Rightarrow v=-0.5m/s$

The speed with which the gun recoils is $0.50m/s$ in the opposite direction of speed of bullet.