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A ball of mass 3M rolls from rest on a smooth flat table towards another ball of mass 2M at rest with a constant velocity 5m/s. When the two balls are separated by a distance of 6m, the position of centre of mass from the mass 3M is

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2.4m

2.0m

answer is B.

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Detailed Solution

m₁ = 3M, m₂ = 2M, d = 6m
$\large {x_1} = \frac{{{m_2}d}}{{{m_1} + {m_2}}} = \frac{{2M \times 6}}{{3M + 2M}} = \frac{{12M}}{{5M}} = 2.4m$

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