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

Answer 1

Answer: (C)

To keep the COM at the same position, velocity of COM is zero, so
$\frac{m_1{\vec{v}}_1+m_2\overrightarrow{v_2}}{m_1+m_2}=0$
(where ${\vec{v}}_1$ and ${\vec{v}}_1$ are velocities of particles 1 and 2 respectively.]
$\Rightarrow m_1\frac{d{\vec{r}}_1}{dt}+m_2\frac{d{\vec{r}}_2}{dt}=0$
$\stackrel{⃗}{\left[\because {\vec{v}}_1=\frac{d\overrightarrow{{\stackrel{⃗}{r}}_1}}{dt}\&{\vec{v}}_2=\frac{d{\vec{r}}_2}{dt}\right]}$
$\Rightarrow md{\vec{r}}_1+m_{2}d{\vec{r}}_2=0$ [$d{\vec{r}}_1$ and $d{\vec{r}}_1$ represent the change in displacement of particles]
Let 2nd particle has been displaced by distance $x$.
$\Rightarrow m_1\left(d\right)+m_2\left(x\right)=0\Rightarrow x=-\frac{m_{1}d}{m_2}$
-ve sign shows that both the particles have to move in opposite directions.
So, $\frac{m_{1}d}{m_2}$ is the distance moved by 2nd particle to keep COM at the same position.