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

The Correct Option is C

To keep the COM at the same position, velocity of COM is zero, so $\frac{m_{1}\,\vec{v}_{1}+m_{2}\,\vec{v_{2}}}{m_{1}+m_{2}}=0$ (where $\vec{v}_{1}$ and $\vec{v}_{1}$ are velocities of particles 1 and 2 respectively.] $\Rightarrow m_{1} \frac{d \vec{r}_{1}}{dt}+m_{2} \frac{d \vec{r}_{2}}{dt}=0$ $\vec{\left[\because \vec{v}_{1}=\frac{d \vec{\vec{r}_{1}}}{dt} \,\&\, \vec{v}_{2}=\frac{d \vec{r}_{2}}{dt} \right]}$ $\Rightarrow m\,d\,\vec{r}_{1}+m_{2}d\,\vec{r}_{2}=0$ [$d\,\vec{r}_{1}$ 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.