Question

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun and acquires a velocity of 100 m/s. Calculate:

1. the velocity with which the gin recoils.
2. the force exerted on gunman due to recoil of the gun

Answer 1

Given Data:

• Mass of the gun (M) = 3 kg
• Mass of the bullet (m) = 30 g = 0.03 kg
• Velocity of the bullet (v) = 100 m/s
• Time taken by the bullet to move through the barrel (t) = 0.003 s

1. Velocity with which the gun recoils:

From the law of conservation of momentum:

Total initial momentum = Total final momentum

Initially, both the gun and bullet are at rest, so total momentum is zero. After firing:

M⋅V + m⋅v = 0

Where V is the recoil velocity of the gun. Rearrange the equation to find V:

`V = -(m.v)/M`

`V = - (0.03 xx 100)/3`

−1 m/s

2. Force exerted on the gunman due to recoil:

The force exerted on the gunman can be calculated using Newton's Second Law:

F = M × a

`a = (ΔV)/t`

Here, ΔV = 1 m/s and t = 0.003 s

`a = 1/0.003` = 333.33 m/s²

F = 3 × 333.33 = 1000 N

Final Answers:

• Recoil velocity of the gun = 1 m/s
• Force exerted on the gunman = 1000 N