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

b) the force exerted on the gunman due to the recoil of the gun.

Question

Philoid · Accepted Answer

Let m₁ = 3kg

m₂ = 0.03kg

u₁ = 0m/s

u₂ = 0m/s

v₁ = ?

v₂ = 100m/s

a) According to law of conservation of momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

3kg (0m/s) + 0.03kg (0m/s) = 3kg (v₁) + 0.03kg (100m/s)

0 = 3v₁ + 3

–1m/s = v₁

NOTE: Negative sign indicates that gun recoils opposite to that of bullet.

b) Force = dp/dt

Initial momentum= p₁= 0 kgm/s

Final momentum =p₂ = 0.03 × 100

= 3kgm/s

Time = 0.003s

Force = dp/dt

= (3)/0.003

= 1000N

A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 seconds to move through the barrel of the gun and acquires a velocity of 100 m/s.Calculate a) the velocity at which the gun recoilsb) the force exerted on the gunman due to the recoil of the gun.