In a ballistics demonstration a police officer fires a bullet of mass $50.0 \;g$ with speed $200 \;m s ^{-1}$ on soft plywood of thickness $2.00 \;cm .$ The bullet emerges with only $10 \%$ of its initial kinetic energy. What is the emergent speed of the bullet?

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Answer The initial kinetic energy of the bullet is $m v^{2} / 2=1000 J .$ It has a final kinetic energy of $0.1 \times 1000=100 J$. If $v_{f}$ is the emergent speed of the bullet.

$\frac{1}{2} m v_{f}^{2}=100 J$

$v_{f}=\sqrt{\frac{2 \times 100 J }{0.05 kg }}$

$=63.2 m s ^{-1}$

The speed is reduced by approximately $68 \%$ (not $90 \%)$

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