Question

A bullet of mass 10 g strikes the wall with velocity 300 m/s and rebounds with same velocity. If the contact period between bullet and wall is 1 ms, then the impulse exerted by bullet on the wall is __ n-s

Answer 1

Hii,
take a look to the solution below
may you get that☺


we know,
impulse= force × time
(and force= change in momentum)
=>impulse= change in momentum×time
= Delta p × t
as we know when it rebounds with same speed the Delta p = 2mv
=>impulse =2mv×t
=2×(10/1000) ×300 ×1
=6n-s