Question

Consider a system of two particles having masses $2\text{kg}$ and $4\text{kg}$. If the particle of mass $2\text{kg}$ is pushed towards the centre of mass of the particles through a distance $2\text{m}$, by what distance would the particle of mass $4\text{kg}$ move, so as to keep the centre of mass at the original position?

• A. $\frac{2}{3}\text{m}$
• B. $1\text{m}$
• C. $2\text{m}$
• D. $4\text{m}$

Answer 1

The correct option is B $1\text{m}$


Assuming,
Distance of $m_1$ from origin = $x_1$
Distance of $m_2$ from oxigin =$x_2$
$x_{\text{com}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=\frac{2x_1+4x_2}{2+4}=\left(\frac{x_1+2x_2}{3}\right)---\left(1\right)$

Now
$x_1\to (x_1+2)$
Let $x_2\to (x_2+d)$
Then $x_{com}=\frac{2(x_1+2)+4(x_2+d)}{2+4}$
$=\frac{2x_1+4x_2+4+4d}{6}$
$=\frac{x_1+2x_2+2+2d}{3}--\left(2\right)$

$\because$ Position of Centre of mass is not changing, $\left(1\right)=\left(2\right)$
$\Rightarrow \frac{x_1+2x_2}{3}=\frac{x_1+2x_2+2+2d}{3}$
$\Rightarrow 2+2d=0$
$\Rightarrow d=-1\text{m}$
i.e $m_2$ should be displaced $1\text{m}$ towards COM.