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(D) Let the center of mass be at the origin. Initially,the positions of the particles are $-x_1$ and $x_2$ such that $m_1(-x_1) + m_2(x_2) = 0$,which

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$\frac{m_1}{m_2} d$

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(D) Let the center of mass be at the origin. Initially,the positions of the particles are $-x_1$ and $x_2$ such that $m_1(-x_1) + m_2(x_2) = 0$,which implies $m_1 x_1 = m_2 x_2$ (Equation $1$).When the first particle is moved by distance $d$ towards the center of mass,its new position becomes $-(x_1 - d)$. Let the second particle be moved by distance $d'$ towards the center of mass,so its new position becomes $(x_2 - d')$.To keep the center of mass at the origin,the new condition is:$m_1(-(x_1 - d)) + m_2(x_2 - d') = 0$$-m_1 x_1 + m_1 d + m_2 x_2 - m_2 d' = 0$Since $m_1 x_1 = m_2 x_2$ from Equation $1$,the terms $m_1 x_1$ and $m_2 x_2$ cancel out:$m_1 d - m_2 d' = 0$$m_2 d' = m_1 d$$d' = \frac{m_1}{m_2} d$

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

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