A rifle man, who together with his rifle has a mass of `100 kg,` stands on a smooth surface and fires `10` shots horizontally. Each bullet. has a mass `10 g` and a muzzle velocity of `800 m//s`,
a. What velocity does the rifle man acquire at the end of `10` shots?
b. If the shots are fired in `10 s`, what will he the average force exerted on him?
c. Compare his kinetic energy with that of `10` bullets

Explanation

a. Let `m_(1)` and `m_(2)` be the masses of bullet and the rifleman, `v_(1)` and `v_(2)` their respective velocities after the first shot.
Initially, the rifleman and bullet are at rest, therefore initial momentum of the system `=0`.
As external force is zero, momentum of the system is constant i.e., initial momentum `=` final momentum `0= m_(1)v_(1)=m_(2)v_(2)`
or `v_(2)=-(m_(1)v_(1))/(m_(2))=-((10xx10^(-3)kg)(800m//s))/(100kg)=-0.08m//s`
As mass of the bullet is negligible as compared to (rifle `+` man),
`:.` Velocity acquired after `10` shots `=10v_(2)=0.8m//s`
i.e., the velocity of rifleman is `0.8m//s `in a direction opposite to that of the bullet.
b. The momentum is acquired by the rifleman is
`P=m_(2)v_(2)=100xx0.8kgm//s=80kgm//s`
This momentum is acquired in `10s`, therefore the average force exeed `=(/_\p)/(/_\t)=80/10=8N`
c. Kinetic energy of the rifleman,
`E_(1)=1/2m_(2)v_(2)^(2)=1/2xx100(0.8)^(2)=32J`
Kinetic energyy of `10` bullet
`E_(2)=10[1/2xx10xx10^(-3)]=(800)^(2)=32000J`
Kinetic energy of bullet is `1000` times the energy of rifleman.