In a system of two particles of masses m_{1} | vedclass

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(A) Let the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass be $r_{1}$ and $r_{2}$ respectively. By the definition of the centre of ma

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$\frac{m_{2}}{m_{1}} d$,towards the centre of mass.

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(A) Let the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass be $r_{1}$ and $r_{2}$ respectively. By the definition of the centre of mass,$m_{1}r_{1} = m_{2}r_{2}$.If the second particle $(m_{2})$ is moved by a distance $d$ towards the centre of mass,its new distance becomes $(r_{2} - d)$.To keep the centre of mass unchanged,let the first particle $(m_{1})$ be moved by a distance $d'$ towards the centre of mass,so its new distance becomes $(r_{1} - d')$.The new condition for the centre of mass is $m_{1}(r_{1} - d') = m_{2}(r_{2} - d)$.Expanding this,we get $m_{1}r_{1} - m_{1}d' = m_{2}r_{2} - m_{2}d$.Since $m_{1}r_{1} = m_{2}r_{2}$,these terms cancel out,leaving $-m_{1}d' = -m_{2}d$.Thus,$d' = \frac{m_{2}}{m_{1}} d$.Since the displacement is towards the centre of mass for both to maintain the balance,the first particle must move by $\frac{m_{2}}{m_{1}} d$ towards the centre of mass.

In a system of two particles of masses $m_{1}$ and $m_{2}$,the second particle is moved by a distance $d$ towards the centre of mass. To keep the centre of mass unchanged,the first particle will have to be moved by a distance:

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