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:
(i) The velocity with which the gun recoils.
(ii) The force exerted on gunman due to recoil of the gun
(i) For gun,
m - 1=3 kg
v1=?
For bullet,
m2 = 30 g = 0.03kg
v2 = 110m/s
By law of conservation of momentum
M1×v1 = m2×v2
3 × v1 = 0.03 × 110
V1 = 0.03 × 110/3
V1 = 1.1m/s
Hence, the velocity of recoil of gun is 1.1m/s
(ii) Acceleration of bullet, a = v-u/t
110-0/0.03 = 3666.6m/s2
Force, F = ma
= 0.03 × 3666.6
= 109.9N