Question

Consider a two particle system with particles having masses m, and m2. 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 $\frac{m_{1} \vec{v_{1}}+m_{2} \vec{v_{2}}}{m_{1}+m_{2}}$ (where $\vec{v_{1}}$ and $\vec{v_{2}}$ are velocities of particles 1 and 2 respectively.] $\Rightarrow\, \quad m_{1} \frac{d \vec{r_{1}}}{d t}+m_{2} \frac{d \vec{r_{2}}}{d t}=0$ $\left[\because\vec{v_{1}} =\frac{d \vec{r_{1}}}{d t} \& \vec{v_{2}} =\frac{d \vec{r_{2}}}{dt}\right]$ $\Rightarrow m, d \vec{r_{1}}+m_{2}d \vec{r_{2}} =0 $[d $\vec{r_{1}}$ dnd d $\vec{r_{2}}$ represent the change in displacement of particles] Let 2nd particle has been displaced by distance x. $\Rightarrow\, \quad m_{1} \left(d\right)+m_{2} \left(x\right)=0 \Rightarrow x=\frac{m_{1}d}{m_{2}}$ -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.