Question

A bullet of mass 60 gram moving with velocity of 500 m per second is brought to rest in 0.01 seconds its impulse will be

Answer 1

The impulse will be

V=u+at

0=500+a(0.01)

a= -50000

F=ma

F=0.06(-50000)

F=-300N

Impulse ,J=Ft

J=3000*0.01

J= 30Ns