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

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(D) Let $x_{1}$ and $x_{2}$ be the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass,respectively.By the definition of the centre of mas

Vedclass · Accepted Answer

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

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(D) Let $x_{1}$ and $x_{2}$ be the distances of masses $m_{1}$ and $m_{2}$ from the centre of mass,respectively.By the definition of the centre of mass,$m_{1}x_{1} = m_{2}x_{2}$.When the first particle is moved by a distance $d$ towards the centre of mass,its new distance becomes $(x_{1} - d)$.Let the second particle be moved by a distance $D$ to keep the centre of mass unchanged.For the centre of mass to remain at the same position,the new distances must satisfy: $m_{1}(x_{1} - d) = m_{2}(x_{2} - D)$.Expanding this,we get $m_{1}x_{1} - m_{1}d = m_{2}x_{2} - m_{2}D$.Since $m_{1}x_{1} = m_{2}x_{2}$,the equation simplifies to $-m_{1}d = -m_{2}D$.Thus,$D = \frac{m_{1}}{m_{2}} d$.Since the first particle moved towards the centre of mass,the second particle must move away from the centre of mass to maintain the balance.

In a 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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