Question

Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance d, by what distance should be second particle be moved, so as to keep the centre of mass at the same position?

• A. $\frac{m_1}{m_2}d$
• B. d
• C. $\frac{m_2}{m_1}d$
• D. $\frac{m_1}{m_2+m_1}d$

Answer 1

The correct option is A $\frac{m_1}{m_2}d$

If $x_1$ and $x_2$ are the positions of masses $m_1$ and $m_2$, the position of the centre of mass is given by $x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
If $x_1$ changes by $\Delta x_1$ and $x_2$ changes by $\Delta x_2$, the change in $x_{cm}$ will be
$\Delta x_{cm}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}$ .........(I)
Given $\Delta x_{cm}=0$ and $\Delta x_1=d$.
Using these values in eq.(i), we get $m_{1}d+m_2\Delta x_2=0$
or $\Delta x_2=-\frac{m_{1}d}{m_2}$
$\therefore$ Distance moved by $m_2=\frac{m_{1}d}{m_2}$, which is choice (a).