Question

A gun of mass 4 kg fires a bullet of mass 20 g with a velocity of 50 m/s, recoil velocity of the gun is

• A. 25 m/s
• B. 0.25 m/s
• C. 15 m/s
• D. 0.15 m/s

Answer 1

The correct option is B 0.25 m/s

Recoil of the gun is the result of the conservation of momentum. As the bullet is fired, the gun recoils in order to conserve the momentum.
Given,
Mass of bullet, m = 20 g = 0.02 kg
Mass of gun, M = 4 kg
Velocity of bullet after firing, v = 50 m/s
Let the recoil velocity of the gun be V.

Since both the gun and the bullet are at rest before firing,
Initial momentum, P1 = 0

After firing:
Final momentum, P2 = mv + MV

As per law of conservation of momentum,
P1 = P2
0 = mv + MV
0.02 × 50 + 4 × V = 0
1 + 4V = 0
This gives,
V = -¼ m/s = -0.25 m/s

Therefore, the recoil velocity of the gun is 0.25 m/s. The negative sign indicates that the direction of recoil is opposite to the direction of the motion of the bullet.