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In a system of two particles of masses 'mi' and 'm2', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged,
the second particle will have to be moved by a distance
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the second particle will have to be moved by a distance
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Verified Answer
The correct answer is:
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}$, towards the centre of mass.
the 2 masses are miand m2
let $x$ and $y$ be the distance of miand $m 2$ from the centre of mass respectively...
now, $m_{1} x=m_{2} y$
the mass $m 1$ is moved by a distance $d$,
let the mass $\mathrm{m} 2$ be moved by a distance $\mathrm{D}$
therefore, $m_{1}(x-d)=m_{2}(y-D)$
$\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{D}$
by eqn (i) $m_{1} x=m_{2} y$
$=>-\mathrm{m}_{1} \mathrm{~d}=-\mathrm{m}_{2} \mathrm{D}$
$=\mathrm{D}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m} 2}$
let $x$ and $y$ be the distance of miand $m 2$ from the centre of mass respectively...
now, $m_{1} x=m_{2} y$
the mass $m 1$ is moved by a distance $d$,
let the mass $\mathrm{m} 2$ be moved by a distance $\mathrm{D}$
therefore, $m_{1}(x-d)=m_{2}(y-D)$
$\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{D}$
by eqn (i) $m_{1} x=m_{2} y$
$=>-\mathrm{m}_{1} \mathrm{~d}=-\mathrm{m}_{2} \mathrm{D}$
$=\mathrm{D}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m} 2}$
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