Question

A bullet of mass 100g is fired from a gun of mass 20kg with a velocity of 100m/s.then velocity of recoil of the gun is(in m/s)

Answer 1

As we know that we need to find conservation of momentum.

We know

Momentum = mass X velocity

Thus, we can say:

Momentum before gun is fired = momentum of gun after firing + momentum of bullet after firing

If we consider recoil velocity of gun as V m/s. and units will be 100g = 0.1 kg

By conservation of momentum:

 (20kg + 0.1 kg) X 0 m/s = (20 kg X V m/s) + (0.1 kg X 100 m/s)

 =>0 = 20V + 10

 =>V = -0.5 m/s

 So the velocity of recoil is -0.5 m/s and the minus sign states that the gun recoils in the opposite direction to the direction of the bullet.

Answer 2

Answer:

mass of bullet, m = 100g=100/1000kg=1/10kg

velocity of bullet, v= 100 m/s^-1

mass of gun, M= 20 kg

let recoil velocity of gun=V

Step-1

before firing , the system (gun+bullet) is at rest, therefore initial momentum of the system =0

final momentum of the system = momentum of bullet + momentum of gun

=mv+MV=1/10×100+20V=10+20V

Step-2

apply the law of conservation of momentum

final momentum = initial momentum

that is, 10+20V=0

20V= -10

or

V=-10/20=-0.5m/s^-1