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 \[{{r}_{CM}}=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\]???..         (1) 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,\[{{r}_{CM}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d')}{{{m}_{1}}+{{m}_{2}}}\] 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 i.e., \[{{m}_{1}}={{m}_{2}},\]then second mass will move a distance equal to the distance at which first mass is being displaced.