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\[{{r}_{GM}}=\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}\].(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 \[\frac{{{m}_{1}}{{r}_{1}}+{{m}_{2}}{{r}_{2}}}{{{m}_{1}}+{{m}_{2}}}=\frac{{{m}_{1}}({{r}_{1}}-d)+{{m}_{2}}({{r}_{2}}+d')}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \]\[-{{m}_{1}}d+{{m}_{2}}d'=0\Rightarrow d'=\frac{{{m}_{1}}}{{{m}_{2}}}d\]You need to login to perform this action.
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