A simple pendulum of length `1m` has a wooden bob of mass `1kg`. It is struck by a bullet of mass `10g` moving with a speed of `200m//s`. The bullet gets embedded into the bob. Obtain the height to which the bob rises before swinging back.
Take `g=10m//s^(2)`.

Explanation

Applying principles of conservation of linear momentum `mu = (M+m)v rArr 10^(-2) xx (2xx 10^(2)) = (1+0.01)v`
`rArr v=(2/1.01)`

`KE_(1)` of the block with bullet in it, is converted into PE as it rises through a height h
`1/2(M+m)v^(2) = (M +m)gh`
`rArr h=(v^(2)/2g) xx (1)/(2 xx 9.8) = 0.2m`