A girl of mass M stands on the rim of a frictionless merrygo- 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-goround. The speed of the rock, relative to the ground, is v. Afterward, the linear speed of the girl is

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$\frac{(m + M) {vR}^{2}}{I + {MR}^{2}}$

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

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

$\frac{{mvR}^{2}}{I + {MR}^{2}}$

answer is A.

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Detailed Solution

The initial angular momentum of the system is zero. The final angular momentum of the girl-plus-merrygo- round is $(I + {MR}^{2}) ω$ which we will take to be positive. The final angular momentum of the system = $(I + m R^{2}) ω + (- m v R)$
Angular momentum conservation leads to
$0 = (I + {MR}^{2}) ω - mRv \Rightarrow ω = \frac{mRv}{I + {MR}^{2}}$
The girl's linear speed is $Rω = \frac{{mvR}^{2}}{I + {MR}^{2}}$