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 the figure. Initially the centre of mass \[{{r}_{CM}}=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]                           (i) When mass w, moves towards centre of mass by a distance d, then let mass \[{{m}_{2}}\]move a distance d away from CM to keep the CM in its initial position. So,       \[{{r}_{CM}}=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]               (ii) Equating Eqs. (i) and (ii), we get \[\frac{{{m}_{1}}{{r}_{1}}+{{m}_{1}}{{r}_{2}}}{{{m}_{1}}{{m}_{2}}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d')}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \]\[-{{m}_{1}}d+{{m}_{2}}d'=0\] \[\Rightarrow \]\[d'=\frac{{{m}_{1}}d}{{{m}_{2}}}\]