Two identical blocks are resting on supports as shown. Two identical bullets travelling with the same speed hit the blocks & get embedded in them in very short time. The blocks leave the contact immediately after bullet strikes them. In case (i), bullet's velocity is toward centre of mass and in case (ii), bullet's velocity is off centre. The kinetic energies after collision are $K_{1}$ & $K_{2}$ and maximum height reached by the centre of mass are $h_{1}$ & $h_{2}$ respectively. Then

K_{1} < K_{2}

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K_{1} = K_{2}

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h_{1} > h_{2}

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h_{1} = h_{2}

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Solution

The correct options are
A $K_{1} < K_{2}$
D $h_{1} = h_{2}$
By conservation of linear momentum both will get same velocity of centre of mass
$∴ \frac{1}{2} m v_{c m}^{2} = m g h$
Therefore CoM of both blocks will achieve equal heights.
$h_{1} = h_{2}$
But by conservation of angular momentum about centre of mass, system 2 will have angular velocity also hence its kinetic energy will be due to both linear and angular velocity while system 1 has only linear velocity. So

$∴ K_{1} < K_{2}$
[Where $K = \frac{1}{2} m v_{c m}^{2} + \frac{1}{2} I ω^{2}$ ]

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