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Question

Physics Question on work, energy and power

A simple pendulum of length 1 m has a wooden bob of mass 1kg1 \,kg . It is struck by a bullet of mass 102kg10^{-2}\, kg moving with a speed of 2×102ms12 \times 10^2\, ms^{-1} . The height to which the bob rises before swinging back is (Take g=10ms2g = 10 \,ms^{-2} )

A

0.2m0.2\, m

B

0.6m0.6\, m

C

8m8\, m

D

1m1\, m

Answer

0.2m0.2\, m

Explanation

Solution

Momentum of bullet =102×2×102=2kgms1= 10^{-2} \times 2 \times 10^2 = 2\, kg\, ms^{-1} . Let the combined velocity of the bob ++ bullet =v= v . Momentum of bob ++ bullet =(102+1)v=1.01v= (10^{-2} + 1)v = 1.01\, v . By conservation of momentum, 1.01v=2kgms11.01\, v = 2\, kg\, ms^{-1} or v=21.01=1.98ms1v = \frac{2}{1.01}=1.98\,ms^{-1} . By conservation of energy 12(M+m)v2=(M+m)gh\frac{1}{2}\left(M + m\right)v^{2} = \left(M + m\right)gh or =h=v22g=(1.98)22×10= h = \frac{v^{2}}{2\,g} = \frac{\left(1.98\right)^{2}}{2 \times 10} =0.190.2m= 0.19 \simeq 0.2\,m