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 = 2 xx 10^(-3) kg, M = 1.0 kg`
` v = 500 m/s, v_(1) = 100 m/s`,
` v_(2)` = speed of the pendulum after impact

`mu = mv_(1) + Mv_(2)` (conservation of momentum)
`Rightarrow v_(2) = (m (u - v_(1)))/M = 2/1000(500-100) = 0.8 m/s`
The block swings and its kinetic energy gets converted into potential energy.
` 1/2 Mv_(2)^(2) = Mgh Rightarrow h = (v_(2)^(2))/(2g) = (0.8 xx 0.8)/(2 xx 9.8) = 8/245 m`
` = 0.033 m`