In a system of two particles of masses ' $\mathrm{m}_{1}$ ' and ' $\mathrm{m}_{2}$ ', the second particle is…
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
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}}$ d, towards the centre of mass.
$\frac{\mathrm{m}_{2}}{\mathrm{~m}} \mathrm{~d}$, away from the the centre of mass. $\mathrm{m}$
$\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \mathrm{~d}$, towards the centre of mass.
$\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}$, away from the centre of mass.
Solution
In system of two particles of masses ' \(m_1\) ' and ' \(m_2\) ', 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
\(\frac{m_1}{m_2} \mathrm{~d}\), towards the centre of mass.
Explanation:
Let \(x_1\) and \(x_2\) be the position of masses \(m\), and \(m_2\), respectively.
The position of centre of mass is
\(\mathrm{x}_{\mathrm{CM}}=\frac{\mathrm{x}_1 \mathrm{~m}_1+\mathrm{x}_2 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\)
If \(\Delta x_1\) and \(\Delta x_2\) be the changes in positions, then change in the position of centre of mass,
\(\Delta \mathrm{x}_{\mathrm{CM}}=\frac{\Delta \mathrm{x}_1 \mathrm{~m}_1+\Delta \mathrm{x}_2 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\)
Given that, the centre of mass remains unchanged i.e., \(\Delta \mathrm{X}_{\mathrm{CM}}=0\) and \(\Delta \mathrm{x}_1=\mathrm{d}\).
\(\Rightarrow 0=\frac{\mathrm{dm}_1+\mathrm{m}_2 \Delta x_2}{\mathrm{~m}_1+\mathrm{m}_2}\)
or \(\Delta x_2=-\frac{m_1}{m_2} \mathrm{~d}\)
Here, negative sign shows that the second particle should be moved towards the centre of mass.