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A bullet of mass $m=50\,\text{g}$ strikes a sandbag of mass $M=5\,\text{kg}$ hanging from a fixed point, with a horizontal velocity $v$. If the bullet sticks to the sandbag, then the ratio of final to initial kinetic energy of the bullet is approximately:

1. $10^{-2}$
2. $10^{-3}$
3. $10^{-6}$
4. $10^{-4}$

Correct answer: $10^{-4}$

Solution

Since the bullet sticks to the sandbag, the collision is perfectly inelastic and momentum is conserved. The common speed becomes $V=\frac{mv}{m+M}$, so the final kinetic energy of the bullet-sandbag system is much smaller than the bullet’s initial kinetic energy. The ratio is approximately $\frac{m}{m+M}\approx\frac{0.05}{5.05}\approx10^{-2}$ for speed, and for kinetic energy it scales as the square, giving about $10^{-4}$.

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