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A ball of mass $3 M$ rolls from rest on a smooth flat table towards another ball of mass $M$ at rest with a constant velocity $5 m / s .$ When the two balls are separated by a distance 6 m, the position of centre of mass from the mass $3 M$ is

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$1.5 m$

$3 m$

$2.4 m$

$2.0 m$

answer is D.

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Detailed Solution

$x c m = \frac{m_{2} d}{m_{1} + m_{2}}$ $= \frac{m \times 6}{3 m + m} = \frac{6}{4} = \frac{3}{2} = 1.5$

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