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A toy train of mass $M$ is moving on a circular track of radius $' R'$ with constant speed $v$ . The length of the train is half of the perimeter of the track. The linear momentum of the train will be

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zero

$\frac{2 M V}{π}$

$M V$

$M V R$

answer is B.

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Detailed Solution

$p_{x} = \int_{0}^{π} (\frac{M}{πR} R d θ) V \sin θ$

$p_{y} = \int_{0}^{π} (\frac{M}{πR} R d θ) V \cos θ$

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