Question

a machine gun of mass 10kg, fires 10 gm bullets at the rate of two every second each bullet comes at with a velocity of 200m/s. the velocity of recoil of gun at the end of the fourth second after firing starts is

Answer 1

Mass of the bullet (m)=10gm =10/1000kg =0.01kg
Velocity of the bullet (v)=200m/s

Momentum of the bullet= m × v
=0.01kg × 200m/s
=2kg.m/s
According to the conservation of momentum,
Momentum of bullet = Momentum of Gun
=>2kg.m/s = mass of gun × velocity of gun
=>2kg.m/s = 10kg × V
=> 2/10 = V
=> 0.2m/s = V