Question

A 70 g bullet moving east 400m/s strikes a 1.3 kg pendulum block on the right and become embedded on it. How high will the block-bullet rise above its original position?

Answer 1

Given:

$m=0.07\:\rm kg$

$u=400\:\rm m/s$

$M=1.3\:\rm kg$

The law of conservation of momentum gives

$$mu=(m+M)v$$

The law of conservation of energy says

$$(m+M)v^2/2=(m+M)gh$$

So

$$h=v^2/2g=(mu/(m+M))^2/2g\\=(0.07*400/(0.07+1.3))^2/(2*9.8)=21\:\rm m$$