Question

A gun of mass 4 kg fires 10 g with a speed of 100 m/s in the forward direction,find

a] the total initial momenntum

b] recoil velocityof the gun

Answer 1

Answer:

Mass of gun, M = 4 kg

Mass of bullet, m = 10 g = 0.01 kg

Speed of bullet, u = 100 m/s

(i)

Initial momentum of the system = 0 kg m/s

Final Momentum of bullet = m × u = 1 kg m/s

(ii)

Let the recoil velocity of the revolver be ‘v’

Mv + mu = 0

=> 4v + 1 = 0

=> v = -0.25 m/s

The negative sign implies the revolver recoils in the direction opposite to that of the bullet.

Explanation:

Answer 2

a) The total initial momentum= (4 + 0.01) x 0 = 0

Total momentum after the bullet is fired = (4xv) + (0.01 x 100) = 4v +1

By law of conservation of momentum , 4v +1 = 0

b) Therefore, recoil velocity (v) = - 1/4 = -.25 m/s

The recoil velocity of the gun is less than the velocity of the bullet because the mass of the gun is more than the mass of the bullet.