A) \[\frac{{{m}_{2}}}{{{m}_{1}}}d\]
B) \[\frac{{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}}d\]
C) \[\frac{{{m}_{1}}}{{{m}_{2}}}d\]
D) \[d\]
Correct Answer: C
Solution :
To keep the centre of mass at the same position, velocity of centre of mass is zero, so \[\frac{{{m}_{1}}{{{\vec{v}}}_{1}}+{{m}_{2}}\,{{{\vec{v}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}\,=0\] (where, \[{{\vec{v}}_{1}}\] and \[{{\vec{v}}_{2}}\] are velocities of particles 1 and 2 respectively) \[\Rightarrow \] \[{{m}_{1}}\frac{d{{{\vec{r}}}_{1}}}{dt}+{{m}_{2}}\frac{d{{{\vec{r}}}_{2}}}{dt}=0\] \[\left( \because {{v}_{1}}=\frac{d{{r}_{1}}}{dt}and\,{{v}_{2}}=\frac{d{{r}_{2}}}{dt}and\,{{m}_{1}},{{m}_{2}}\ne 0 \right)\] \[\Rightarrow \]\[md{{r}_{1}}+{{m}_{2}}d{{r}_{1}}=0\](\[d{{r}_{1}}\]and\[d{{r}_{2}}\]represent the small change in displacement so that\[d{{r}_{2}}\to 0\] and\[d{{r}_{2}}\to 0\]of particles) Let 2nd particle has been displaced by distance \[x,\]then \[{{m}_{1}}(d)+{{m}_{2}}(x)=0\] \[\Rightarrow \]\[x=-\frac{{{m}_{1}}d}{{{m}_{2}}}\] Negative sign shows that both the particles have to move in opposite directions. So,\[\frac{{{m}_{1}}d}{{{m}_{2}}}\]is the distance moved by 2nd particle to keep position of centre of mass unchanged.
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