InterviewSolution
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InterviewSolution
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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 mB. 0.6 mC. 8 mD. 1 m |
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Answer» 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 |
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