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A uniform thin rod of length $2 \mathrm{~L}$ and mass $\mathrm{m}$ lies on a horizontal table. A horizontal inpulse $\mathrm{J}$ is given to the rod at one end. There is no friction. The total K.E. of the rod just after the impulse will be :
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The correct answer is:
$2 \mathrm{~J}^{2} / \mathrm{m}$
$\mathrm{J}=\mathrm{mv} \quad \mathrm{J} . \mathrm{L}=\frac{\mathrm{M}(2 \mathrm{~L})^{2}}{12} \omega$
$v^{v}=\mathrm{J} / \mathrm{M}$
$\mathrm{Cm}$
$J . L=\frac{M L^{2}}{3} \omega$
$\omega=\frac{3 J}{M L}$
$\mathrm{ke}=\frac{\mathrm{J}^{2}}{2 \mathrm{~m}}+\frac{3 \mathrm{~J}^{2}}{2 \mathrm{~m}}=2 \mathrm{~J}^{2} / \mathrm{m}$
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