Question

A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion of gun powder. What is the average force (in N) exerted on a 0.0200 kg bullet to accelerate it to a speed of 575 m/s in a time of 2.70 ms (milliseconds)? (Enter the magnitude.)

Answer 1

Answer:4.259 kN

Step-by-step explanation:

Given

mass of bullet
`m=0.02 kg`

final velocity of bullet
`v=575 m/s`

time taken
`=2.7 ms\approx 2.7* 10^(-3) s`

We know

`Impulse=F_(avg)\cdot t=change\ in\ momentum`

thus

Initial Momentum
`(P_i)=m* 0`

Final Momentum
`(P_f)=m* v=0.02* 575`

`P_f=11.5 kg-m/s`

`F_(avg)* 2.7* 10^(-3)=P_f-P_i=11.5`

`F_(avg)=4.259* 10^3`

`F_(avg)=4.259 kN`