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(B) The position of the centre of mass $(R_{cm})$ for a two-particle system is given by the formula $R_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$.For t

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Towards $M$

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(B) The position of the centre of mass $(R_{cm})$ for a two-particle system is given by the formula $R_{cm} = \frac{m_1r_1 + m_2r_2}{m_1 + m_2}$.For two masses $m$ and $M$ separated by a distance $d$,the centre of mass lies on the line joining them.Specifically,the distance of the centre of mass from the mass $M$ is $r_M = \frac{m \cdot d}{M + m}$ and from the mass $m$ is $r_m = \frac{M \cdot d}{M + m}$.Since $M > m$,it follows that $r_M Therefore,the centre of mass is always located closer to the heavier mass,which is $M$.

Where will the centre of mass be located when combining two masses $m$ and $M$ $(M > m)$?

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