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 the second particle be moved so as to keep the centre of mass at the same position?

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

Answer 1

Answer: (A)

To keep the centre of mass at the same position,velocity of centre of mass is zero, so
$\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}=0$
(where, $v_1$ and $v_2$ are the velocities of particles $1$ and $2$ respectively)
$\Rightarrow m_1\frac{d{r}_1}{dt}+m_2\frac{d{r}_2}{dt}=0$
$[\because v_1=\frac{d{r}_1}{dt}$ and $v_2=\frac{d{r}_2}{dt}$ and $m_1{m}_2\ne 0]$
$\Rightarrow m_{1}d{r}_1+m_{2}d{r}_2=0$
($d{r}_1$ and $d{r}_2$ represent the small changes 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
$m_1(d)+m_2(X)=0$
$X=-\frac{m_{1}d}{m_2}$
Negative sign shows that both the particle have to move in opposite directions
So, $\frac{m_1}{m_2}d$ is the distance moved by $2nd$ particle to keep position of centre of mass unchanged.