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 $2\times {10}^2 m/s$. The bullet gets embedded within the bob. Obtain the height to which the bob rises before swinging back.

Applying principle of conservation of linear momentum

$\mathrm{mu}=(\mathrm{M}+\mathrm{m})\mathrm{v}\Rightarrow {10}^{-2}\times \left(2\times {10}^2\right)=(1+.01)\mathrm{v}\Rightarrow \mathrm{v}=\frac{2}{1.01}\mathrm{m}/\mathrm{s}$

Initial $KE$ of the block with bullet in it, is fully converted into $PE$ as it rises through a height $h$, given by

$\frac{1}{2}(\mathrm{M}+\mathrm{m}){\mathrm{v}}^2=(\mathrm{M}+\mathrm{m})\mathrm{gh}\Rightarrow {\mathrm{v}}^2=2\mathrm{gh}\Rightarrow \mathrm{h}=\frac{{\mathrm{v}}^2}{2\mathrm{g}}={\left(\frac{2}{1.01}\right)}^2\times \frac{1}{2\times 9.8}=0.2\mathrm{m}$