In a two-particle system with particle masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$, the first parti

Question

In a two-particle system with particle masses $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$, the first particle is pushed towards the centre of mass through a distance $d$, the distance through which second particle must be moved to keep the centre of mass at the same position is\

Professor PC Thomas Classes · Answer

D)\ \ $	heta=\frac{m_{1}\left(-x_{1}ight)+m_{2} x_{2}}{m_{1}+m_{2}}$\ $\Rightarrow \mathrm{m}_{1} \mathrm{x}_{1}=\mathrm{m}_{2} \mathrm{x}_{2}$\  Finally,\ $0=\frac{-\mathrm{m}_{1}\left(\mathrm{x}_{1}-\mathrm{d}ight)+\mathrm{m}_{2}\left(\mathrm{x}_{2}-\mathrm{d}ight)}{\mathrm{m}_{1}+\mathrm{m}_{2}}$\ $0=\mathrm{m}_{1}\left(\mathrm{~d}-\mathrm{x}_{1}ight)+\mathrm{m}_{2}\left(\mathrm{x}_{2}-\mathrm{d}^{\prime}ight)$\ $\Rightarrow 0=\mathrm{m}_{1} \mathrm{~d}-\mathrm{m}_{1} \mathrm{x}_{1}+\mathrm{m}_{2} \mathrm{x}_{2}-\mathrm{m}_{2} \mathrm{~d}^{\prime}$\ $\Rightarrow \mathrm{d}^{\prime}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}$\ (using (i))\