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(A) The position of the centre of mass $(R_{cm})$ is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.For the centre of mass to remain stationa

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$d' = \frac{m_1}{m_2} d$

Vedclass · Answer

(A) The position of the centre of mass $(R_{cm})$ is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.For the centre of mass to remain stationary,the change in the position of the centre of mass must be zero: $\Delta R_{cm} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$.This implies $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.Given that mass $m_1$ is moved towards the centre of mass by distance $d$,we have $\Delta x_1 = d$ (taking the direction towards the centre of mass as positive).To keep the centre of mass at the same position,mass $m_2$ must move in the opposite direction by distance $d'$,so $\Delta x_2 = -d'$.Substituting these into the equation: $m_1(d) + m_2(-d') = 0$.$m_1 d = m_2 d'$.Therefore,$d' = \frac{m_1}{m_2} d$.

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