(N/A) Consider a system of two particles of masses $m_1$ and $m_2$ located at positions $x_1$ and $x_2$ respectively from the origin $O$ on the $X$-axis. The centre of mass $C$ of this system is located at a position $X$ given by:
$X = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$
Here,$X$ (also denoted as $r_{cm}$) is the mass-weighted mean of the positions. If the two particles have equal masses $(m_1 = m_2 = m)$,then:
$X = \frac{m x_1 + m x_2}{m + m} = \frac{x_1 + x_2}{2}$
This shows that for two particles of equal mass,the centre of mass lies exactly at the midpoint. For a system of $n$ particles with masses $m_1, m_2, \dots, m_n$ at positions $x_1, x_2, \dots, x_n$ respectively,the position of the centre of mass $X$ is given by the weighted average:
$X = \frac{\sum_{i=1}^{n} m_i x_i}{\sum_{i=1}^{n} m_i}$
Defining the total mass of the system as $M = \sum_{i=1}^{n} m_i$,the expression becomes:
$X = \frac{1}{M} \sum_{i=1}^{n} m_i x_i$