A stick of length L and mass M lies on a fric | vedclass

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Question

Applying conservation of angular momentum

$\frac{\mathrm{mvL}}{2}=\frac{\mathrm{ML}^{2}}{12} \omega$

$\omega=\frac{6 \mathrm{mv}}{\mathrm{ML}}$&

Vedclass · Accepted Answer

$m=M/4$

Vedclass · Answer

Applying conservation of angular momentum

$\frac{\mathrm{mvL}}{2}=\frac{\mathrm{ML}^{2}}{12} \omega$

$\omega=\frac{6 \mathrm{mv}}{\mathrm{ML}}$         $...(1)$

conservation of linear momentum

$\mathrm{mv}=\mathrm{Mv}_{2}$

$\mathrm{v}_{2}=\frac{\mathrm{mv}}{\mathrm{M}}$          $...(2)$

Applying conservation of kinetic energy (elastic collision)

$\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} \mathrm{Mv}_{2}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}$          $...(3)$

solving eq. $(1),(2) \&(3)$

$\mathrm{m}=\frac{\mathrm{M}}{4}$

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