question_answer

Consider a system of two particles having masses\[{{m}_{1}}\]and\[{{m}_{2}}\]. If the particle of mass\[{{m}_{1}}\]is pushed towards the mass centre  of particles through a distance d, by what distance would the particle of mass\[{{m}_{2}}\]move so as to keep the mass centre of particles at the original position?

A)  \[\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}d\]

B)  \[\frac{{{m}_{1}}}{{{m}_{2}}}d\]

C)  \[d\]

D)  \[\frac{{{m}_{2}}}{{{m}_{1}}}d\]

Correct Answer: B

Solution :

 The system of two given particles of masses\[{{m}_{1}}\]and\[{{m}_{2}}\]are shown in figure. Initially the centre of mass \[{{r}_{CM}}=\frac{{{m}_{1}}{{r}_{}}_{1}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]         ...(i) When mass\[{{m}_{1}}\]moves towards centre of mass by a distance d, then let mass\[{{m}_{2}}\]moves a distance d away from CM to keep the.CM in its initial position. So, \[{{r}_{CM}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d)}{{{m}_{1}}+{{m}_{2}}}\] ?(ii) Equating Eqs. (i) and (ii), we get \[\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d)}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \] \[-{{m}_{1}}d+{{m}_{2}}d=0\] \[\Rightarrow \] \[d=\frac{{{m}_{1}}}{{{m}_{2}}}d\] NOTE: If. both the masses are equal ie, \[{{m}_{1}}={{m}_{2}},\] then second mass will move a distance equal to the distance at which first mass is being displaced.