A bullet of mass `2 g` travelling at a speed of `500 m//s` is fired into a ballistic pendulum of mass `1.0 kg` suspended from a cord `1.0 m` long. The bullet penetrates the pendulum and emerges with a velocity of `100 m//s`. Through what vertical height will the pendulum rise?

Explanation

Let `m=2xx10^(-3k), M=1.0kg`

`u=500m//sv_(1)=100m//s`
` v_(2)=` speed of the pendulum after the impact
`mu=mv_(1)+Mv_(2)` (coservation of momentum)
`v_(2)=(m(u-v_(1)))/=2/1000(500-100)=0.8m//s`
The block swings and its kinetic energy gets converted into potential energy.
`1/2Mv_(2)^(2)=Mgh`
`implies h=v_(2)^(2)/(2g)=(0.8xx0.8)/(2xx10)=0.032m`