Question

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


Please anybody solve my problem​

Answer 1

Explanation:

For Gun -

Mass of gun ( M ) = 3 kg

Initial velocity ( U ) = 0 m/s

Let, final or Recoil velocity ( V ) = V m/s

For bullet -

Mass of bullet ( m ) = 30 g or 0.03 kg

Initial velocity ( u ) = 0 m/s

final or Recoil velocity ( v ) = 100 m/s

time taken = 0.003 second

(i) LAW OF CONSERVATION OF MOMENTUM

initial momentum of object before collision = final momentum after collision

(MU)+(mu) = (MV)+(mv)

(3×0)+(0.03×0) = (3×V)+(0.03×100)

0 = 3V + 3

V = -3/3

Recoil velocity of gun = -1 m/s

Note - ( minus sign show gun Recoil in opposite directions with 1 m/s )

(i) F = ma

acceleration = ( v - u )/t

= ( 100-0 )/0.003

= 100/0.003

= 100000/3 m/s²

F = 0.03×100000/3

= 300000/300

= 1000 N