1. Tardigrade
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  4. 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 × 102 ms-1 . The height to which the bob rises before swinging back is (Take g = 10 ms-2 )

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Q. 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 \times 10^2\, ms^{-1} $ . The height to which the bob rises before swinging back is
(Take $ g = 10 \,ms^{-2} $ )

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Solution:

Momentum of bullet $ = 10^{-2} \times 2 \times 10^2 = 2\, kg\, ms^{-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\, ms^{-1} $
or $ v = \frac{2}{1.01}=1.98\,ms^{-1} $ .
By conservation of energy
$ \frac{1}{2}\left(M + m\right)v^{2} = \left(M + m\right)gh $
or $ = h = \frac{v^{2}}{2\,g} = \frac{\left(1.98\right)^{2}}{2 \times 10} $
$ = 0.19 \simeq 0.2\,m $