A solid cube of wood of side 2a and mass M is | vedclass

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(D) Since there is no external torque on the system of the bullet and the block about the axis $AB$,the angular momentum about $AB$ remains conserved.

Vedclass · Accepted Answer

$mv/2Ma$

Vedclass · Answer

(D) Since there is no external torque on the system of the bullet and the block about the axis $AB$,the angular momentum about $AB$ remains conserved.First,calculate the moment of inertia of the block about the axis $AB$ using the parallel axis theorem:$I_{AB} = I_{CM} + Mh^2$Given $I_{CM} = 2Ma^2/3$ and the distance $h$ from the centre of mass to the axis $AB$ is $\sqrt{a^2 + a^2} = \sqrt{2}a$.$I_{AB} = \frac{2}{3}Ma^2 + M(\sqrt{2}a)^2 = \frac{2}{3}Ma^2 + 2Ma^2 = \frac{8}{3}Ma^2$.Now,apply the conservation of angular momentum about the axis $AB$:$L_{initial} = L_{final}$$mvr = I_{AB}\omega$where $r = 4a/3$ is the perpendicular distance of the bullet's path from the axis $AB$.$m v (\frac{4a}{3}) = (\frac{8}{3}Ma^2) \omega$Solving for $\omega$:$\omega = \frac{mv(4a/3)}{8Ma^2/3} = \frac{4mva}{8Ma^2} = \frac{mv}{2Ma}$.

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