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A uniform thin rod of length $L$, mass $m$ is lying on a smooth horizontal table. A horizontal impulse $P$ is suddenly applied perpendicular to the rod at one end. The total energy of the rod after the impulse is
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$\frac{2 P^{2}}{M}$
$\mathrm{P}=(\mathrm{mv}-0)$
angular impulse $\mathrm{J}=\mathrm{P} \times \frac{\mathrm{L}}{2}=\mathrm{L}_{\mathrm{f}}-0$
$\mathrm{LE}_{\mathrm{f}}=\frac{\mathrm{PL}}{2}$
$=\frac{\mathrm{K}_{\mathrm{r}}}{2 \mathrm{~m}}+\mathrm{K}_{\mathrm{t}}=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}+\frac{\mathrm{L}_{\mathrm{f}}^{2}}{2 \mathrm{I}}$
$=\frac{(\mathrm{PL} / 2)^{2}}{2 \times \frac{\mathrm{mL}^{2}}{12}}$
$=\frac{\mathrm{P}^{2}}{2 \mathrm{~m}}+\frac{\mathrm{P}^{2} \mathrm{~L}^{2}}{4 \times \frac{\mathrm{mL}^{2}}{6}}=\frac{\mathrm{P}}{2 \mathrm{~m}}+\frac{3 \mathrm{P}^{2}}{2 \mathrm{~m}}=\frac{2 \mathrm{P}^{2}}{\mathrm{~m}}$
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