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(D) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the sam

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$\left( \frac{m_1}{m_2} \right) d$

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(D) The position of the centre of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the centre of mass to remain at the same position,the change in the position of the centre of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.Given that the first particle is moved by a distance $d$ towards the centre of mass,let $\Delta r_1 = d$.Substituting this into the equation: $m_1 d + m_2 \Delta r_2 = 0$.Solving for $\Delta r_2$: $\Delta r_2 = -\left( \frac{m_1}{m_2} \right) d$.The negative sign indicates that the second particle must be moved in the opposite direction to the first particle,also towards the centre of mass,by a distance of $\left( \frac{m_1}{m_2} \right) d$.

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$,by what distance should the second particle be moved,so as to keep the centre of mass at the same position?

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