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A girl of mass ‘M’ stands on the rim of a friction less merry-go-round, of radius ‘R’ and rotational inertia ‘I’, that is not moving. She throws a rock of mass ‘m’ horizontally in a direction that is tangent to the outer edge of the merry-go-round. The speed of the rock, relative to the ground is ‘v’. Afterwards, the linear speed of the girl is

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$\frac{m v R^{2}}{I + (M + m) R^{2}}$

$\frac{m v R^{2}}{I + M R^{2}}$

$\frac{(m + M) v R^{2}}{I + M R^{2}}$

$\frac{m v R^{2}}{I + (M - m) R^{2}}$

answer is A.

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Detailed Solution

$(I + M R^{2}) ω = m R v W = \frac{m R v}{I + M R^{2}} V = R W V = \frac{m R^{2} v}{I + M R^{2}}$

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