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A meter stick, pivoted about horizontal axis through its centre, has a body of mass $2 k g$ attached to one end and a body of mass $1 k g$ attached to the other end. The system released from rest with the stick horizontal, the velocity of either mass as the stick swing through vertical position :

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$2.2 m / s$

$1.5 m / s$

$0.75 m / s$

$1.8 m / s$

answer is B.

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Detailed Solution

$0 = - 2 g \times 0.5 + 1 \times g \times 0.5$

$+ \frac{1}{2} \times 2 \times v^{2} + \frac{1}{2} \times 1 \times v^{2}$

$∴ v = 1.8 m / s$

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