Question

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)`.

Answer 1

Applying principle of conservation of linear momentum, we get`m upsilon=(M+m)V`
`10^(-2)xx(2xx10^(2))=(1+0.01)V`
`V=(2)/(1.01)`
`K.E.` of the block with the bullet in it, is converted into `P.E.` as it rises through a height `h`.
`:. (1)/(2)(M+m)V^(2)=(M+m)gh`
`V^(2)=2gh` or `h=(V^(2))/(2h)`
`h=((2)/(1.01))^(2)xx(1)/(2xx9.8)=0.2m`