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(B) The position of the center of mass is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.For the center of mass to remain unchanged,the chang

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$\frac{m_1}{m_2} d$

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(B) The position of the center of mass is given by $R_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$.For the center of mass to remain unchanged,the change in the position of the center of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta x_1 + m_2 \Delta x_2 = 0$.Here,mass $m_1$ is moved by a distance $d$ towards the center of mass. Let the displacement be $\Delta x_1 = d$.Substituting this into the equation: $m_1 d + m_2 \Delta x_2 = 0$.Solving for $\Delta x_2$,we get $\Delta x_2 = -\frac{m_1}{m_2} d$.The negative sign indicates that mass $m_2$ must be moved in the opposite direction to $m_1$ (i.e.,towards the center of mass) by a distance of $\frac{m_1}{m_2} d$.

Consider a system of two particles of masses $m_1$ and $m_2$. If mass $m_1$ is pushed towards the center of mass of the system by a distance $d$,by what distance should the mass $m_2$ be displaced so that the center of mass of the system remains unchanged?

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