Question

A gun of mass 3 kg fires a bullet of mass 30 grams. The bullet takes 0.003 seconds to move through the barrel of the gun and acquire a velocity of 100 meters per second. Calculate the velocity with which the gun recoils and the force exerted on the gunman, due to recoil of the gun.

Answer 1

For gun m1=3kg
For bullet m2=30g=0.03kg,v2= 100m/s
By law of conservation of momentum m1*v1= m2 *v2 3*v1 =0.03*100 v1 =0.03*100/3
(i) The velocity with which the gun recoils=v1=1 m/s Acceleration of bullet=v-u/t 100-0/0.03=3333.33m/s2
(ii)The force exerted on gunman due to recoil of the gun=F=ma=0.03*3333.33=100 N