Question

A gun of mass 3 kg fires A bullet of mass 30 gram the bullet in 0.003 s to move through the barrel of the gun and acquires a velocity of 100
metre per second calculate the velocity with which the gun recoils the force exerted on Gunman due to recoil of the gun Please ans fast I will mark him as Brainliest ​

Answer 1

Answer:

Explanation:

Sorry for the figure I have just drawn for the reference purpose only.

Answer 2

The velocity of recoil of the gun = 1 m/s

The force exerted on the gunman due to recoil of the gun = 10 N

M - mass of the gun = 3 kg

m - mass of the bullet = 0.03 kg

v - recoil velocity

Initial momentum of the (gun+bullet) = (M+m)v

= (3+0.03)v

= 3.03v kg m/s

Final momentum = mv' = 0.03 × 100 = 3

According to Law of conservation of momentum - initial momentum = final momentum

=> 3.03v = 3

=> v = 300/303 = 0.99 m/s = 1 m/s

Force exerted = change in momentum / time interval

= (3.03 - 3)/0.003

= 0.03/0.003

= 10 N