A machine gun fires bullets at the rate of 180 shots per minute. Each bullet is of mass 20 g and moves with a velocity of 1 km `cdot s^(-1)` . After colliding perdicularly with a steel plate, the bullets rebound with half the incident speed. What will be the force required to keep the steel plate in position?

Explanation

Velocity of each bullet before impact
= 1 km`cdot s^(-1)` = 1000 m `cdot s^(-1)`
Velocity of each bullet after impact
= -`(1)/(2) km cdot s^(-1) = - (1)/(2) xx 1000 m cdot s^(-1) = - 500 m cdot s^(-1)`
Negative sign indicates, this velocity is oppositely directed.
`therefore` Change in velocity of each bullet after impact
= 1000 -(-500) = 1500 m`cdot s^(-1)`
Number of bullets incident per second on the steel plate
`(180)/(60)` = 3
`therefore` Rate of change of momentum of the three bullets
= `3xx (20)/(1000) xx 1500 = 90` N = force exerted on the steel plate.
Hence, the force required to hold the steel plate in position = 90 N.