Question 85 of 130

A simple pendulum of length $1m$ has a wooden bob of mass $1kg$. It is struck by a bullet of mass ${10}^{-2}kg$ moving with a speed of $2\times {10}^{2}m{s}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $g=10m{s}^{-2}$)

A$0.30m$B$0.20m$C$0.35m$D$0.40m$

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Answer: (B) $0.20m$

Linear momentum will be conserved, therefore
$mu=\left(M+m\right)V$
$\Rightarrow {10}^{-2}\times 2\times {10}^{2}\cong 1\times V$
$\Rightarrow V\cong 2m{s}^{-1}$
Again from energy conservation, we get height achieved, $h=\dfrac{{V}^{2}}{2g}=0.2m$.