# A bullet is fired from a gun. The force on the bullet is given by f= 600-[2*10^5]t. The force on bullet become zero as given by equation as soon as it leaves the barrel. The impulse imparted is?

Vikas TU
14149 Points
10 years ago

600 N

SOURAV MISHRA
37 Points
10 years ago

first we shall find the time taken by the bullet to come out of te barrel.

this will help us calculate the impulse on the bullet

the required time is 3*10-3 s.

now we can calculate the impulse by two ways.

the first way is integration which is the most common method.

J = ∫F(t)dt = ∫{600-(2*105t)}dt

the limits shall be from t = 0 to t = 3*10-3 s.

the other way is very simple. it is the method of average.

when a quantity has a linear variation we can take the average of the initial and final quantiites to evaluate such an integral

that is if F is a linear function of t then

0TF(t)dt = <F(t)>T = {F(t = 0) + F(t = T)}/2.

both these ways give the same answer as 300 N-s.

shashi
11 Points
7 years ago
A small amendment in sourav mishra’s ans to correct the answer...
300 in the last what we got is avg force and not the impulse imparted on the bullet.
impulse=F*T
=300*0.003=0.9 N-s
hope it helps...thumbs up plz