Question

A simple pendulum of length 1 m has a wooden bob of mass 1 kg. It is struck by a bullet of mass `10^(-2)` kg moving with a speed of `2 xx 10^(2) m s^(-1)`. The height to which the bob rises before swinging back is (Take `g = 10 m s^(-2)`)
A. 0.2 m
B. 0.6 m
C. 8 m
D. 1 m

Answer 1

Correct Answer - A
Momentum of bullet =`10^(-2)xx2xx10^2 = 2 kg m s^(-1)`
Let the combined velocity of the bob + bullet =v
Momentum of bob + bullet =`(10^(-2) + 1)v=1.01 v`
By conservation of momentum, 1.01 v =`2 kg m s^(-1)`
or `v=2/1.01 = 1.98 m s^(-1)`
By conservation of energy
`1/2(M+m)v^2 =(M+m)gh`
or `h=v^2/(2g)=((1.98)^2)/(2xx10)=0.196 ~=0.2` m