Question

top mass 0.005 kg moving with a speed of 200 m per second enters heavy wooden block and is stopped after a distance of 50cm what is the average force exerted by the block on the bullet ?Also find impulse

Answer 1

From third law of motion
V^2-u^2=2as
200×200-0=2a×0.05
a=40000 m/s^2
Force exerted by block
F=ma
=0.005×40000
=200N