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In a system of two particles of masses ' $\mathrm{m}_{1}$ ' and ' $\mathrm{m}_{2}$ ', the second particle is moved
by a distance 'd' towards the centre of mass. To keep the centre of mass
unchanged, the first particle will have to be moved by a distance
Options:
by a distance 'd' towards the centre of mass. To keep the centre of mass
unchanged, the first particle will have to be moved by a distance
Solution:
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Verified Answer
The correct answer is:
$\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \mathrm{~d}$, towards the centre of mass.
Correct option is E)
given masses are $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$
let $x$ and $y$ be the distance of $m_{1}$ and $m_{2}$ from the centre of mass respectively, then
$\mathrm{m}_{1} \mathrm{x}=\mathrm{m}_{2} \mathrm{y}$
If the mass $m_{1}$ is moved by a distance $d_{1}$, and mass $m_{2}$ be moved by a distance $\mathrm{d}_{2}$, then
$\mathrm{m}_{1}\left(\mathrm{x}-\mathrm{d}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{y}-\mathrm{d}_{2}\right)$
$\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{~d}_{2}$
$\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{~d}_{2}$
$\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}_{1} \quad(\because$ constants $)$
$\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d} \quad\left(\because \mathrm{d}_{1}=\mathrm{d}\right)$
given masses are $\mathrm{m}_{1}$ and $\mathrm{m}_{2}$
let $x$ and $y$ be the distance of $m_{1}$ and $m_{2}$ from the centre of mass respectively, then
$\mathrm{m}_{1} \mathrm{x}=\mathrm{m}_{2} \mathrm{y}$
If the mass $m_{1}$ is moved by a distance $d_{1}$, and mass $m_{2}$ be moved by a distance $\mathrm{d}_{2}$, then
$\mathrm{m}_{1}\left(\mathrm{x}-\mathrm{d}_{1}\right)=\mathrm{m}_{2}\left(\mathrm{y}-\mathrm{d}_{2}\right)$
$\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{~d}_{2}$
$\mathrm{m}_{1} \mathrm{~d}_{1}=\mathrm{m}_{2} \mathrm{~d}_{2}$
$\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}_{1} \quad(\because$ constants $)$
$\mathrm{d}_{2}=\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d} \quad\left(\because \mathrm{d}_{1}=\mathrm{d}\right)$
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