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

Explanation:

Given

mass of bullet $m=0.02 kg$

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

time taken $=2.7 ms\approx 2.7\times 10^{-3} s$

We know

$Impulse=F_{avg}\cdot t=change\ in\ momentum$

thus

Initial Momentum$(P_i)=m\times 0$

Final Momentum$(P_f)=m\times v=0.02\times 575$

$P_f=11.5 kg-m/s$

$F_{avg}\times 2.7\times 10^{-3}=P_f-P_i=11.5$

$F_{avg}=4.259\times 10^3$

$F_{avg}=4.259 kN$