In the system of two particles of masses m_1 | vedclass

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(C) Let $x_1$ and $x_2$ be the initial distances of particles $m_1$ and $m_2$ from the centre of mass $C$ respectively. For the centre of mass to rema

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$\frac{m_1}{m_2} d$,towards the centre of mass.

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(C) Let $x_1$ and $x_2$ be the initial distances of particles $m_1$ and $m_2$ from the centre of mass $C$ respectively. For the centre of mass to remain at $C$,the condition is $m_1 x_1 = m_2 x_2$ ... $(i)$When particle $m_1$ is moved by distance $d$ towards the centre of mass,its new distance from $C$ becomes $(x_1 - d)$.Let the second particle $m_2$ be moved by a distance $d'$ to keep the centre of mass unchanged. Its new distance from $C$ will be $(x_2 - d')$.For the centre of mass to remain unchanged,the new condition is $m_1(x_1 - d) = m_2(x_2 - d')$ ... (ii)Expanding equation (ii): $m_1 x_1 - m_1 d = m_2 x_2 - m_2 d'$Since $m_1 x_1 = m_2 x_2$ from equation $(i)$,we can substitute this into the expanded equation:$m_2 x_2 - m_1 d = m_2 x_2 - m_2 d'$$-m_1 d = -m_2 d'$$d' = \frac{m_1}{m_2} d$Since the displacement $d'$ is positive in the direction towards the centre of mass,the second particle must also be moved towards the centre of mass.

In the system of two particles of masses $m_1$ and $m_2$,the first particle is moved by a distance $d$ towards the centre of mass. To keep the centre of mass unchanged,the second particle will have to be moved by a distance

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