Two particles have masses m_1 and m_2. If the | vedclass

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

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$\frac{m_1}{m_2} d$

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(D) The position of the center of mass $R_{cm}$ is given by $R_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$.For the center of mass to remain at the same position,the change in the position of the center of mass must be zero,i.e.,$\Delta R_{cm} = 0$.Therefore,$m_1 \Delta r_1 + m_2 \Delta r_2 = 0$.Let the first particle $(m_1)$ be moved by a distance $d$ towards the center of mass,so $\Delta r_1 = d$.Let the second particle $(m_2)$ be moved by a distance $x$ in the opposite direction,so $\Delta r_2 = -x$.Substituting these values into the equation: $m_1 d + m_2 (-x) = 0$.$m_1 d = m_2 x$.Solving for $x$,we get $x = \frac{m_1}{m_2} d$.

Two particles have masses $m_1$ and $m_2$. If the first particle $(m_1)$ is moved by a distance $d$ towards the center of mass,by what distance should the second particle $(m_2)$ be moved so that the center of mass of the system remains at the same position?

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