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A bullet fired into a fixed target loses half of its velocity after penetrating $3 \mathrm{~cm}$. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?
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The correct answer is:
$1.0 \mathrm{~cm}$
$1.0 \mathrm{~cm}$
$F .3=\frac{1}{2} m v^2-\frac{1}{2} m \frac{v^2}{4}$
$F(3+x)=\frac{1}{2} m v^2$
$x=1 \mathrm{~cm}$
$F(3+x)=\frac{1}{2} m v^2$
$x=1 \mathrm{~cm}$
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