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A ball of mass $4 M$ rolls from rest on a smooth flat table towards another ball of mass $1 M$ at rest with a constant velocity $5 m / s$ . When the two balls are separated by a distance $3 m$ ,the position of centre of mass from the mass $4 M$ is

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$0.6 m$

$1.6 m$

$0.3 m$

$0.62 m$

answer is B.

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Detailed Solution

$X_{1} = \frac{m_{2} d}{m_{1} + m_{2}} = \frac{1 \times 3}{4 + 1} = 0.6 m$

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