Write the expression of centre of mass of a system of $'n'$ particles and derive the formula of force acting on its centre of mass. 

The centre of mass of a system of ' $n$ ' particle

$\vec{v}_{\mathrm{cm}} \text { or } \overrightarrow{\mathrm{V}}=\frac{m_{1} \overrightarrow{v_{1}}+m_{2} \overrightarrow{v_{2}}+\ldots m_{n} \overrightarrow{v_{n}}}{m_{1}+m_{2}+\ldots m_{n}}$

$\therefore \quad \mathrm{MV}=m_{1} \overrightarrow{v_{1}}+m_{2} \overrightarrow{v_{2}}+\ldots m_{n} \overrightarrow{v_{n}} \quad \ldots$

Where $\mathrm{M}=m_{1}+m_{2}+\ldots, m_{n}$ total mass of a system.

Assume that the value of masses does not change with time, then differentiate equation $(1)$ w.r.t. time,

$\mathrm{M} \frac{d \overrightarrow{\mathrm{V}}}{d t}=m_{1} \frac{d \overrightarrow{v_{1}}}{d t}+m_{2} \frac{d \overrightarrow{v_{2}}}{d t}+\ldots m_{n} \frac{d \overrightarrow{v_{n}}}{d t}$

but $\frac{d \overrightarrow{\mathrm{V}}}{d t}=\overrightarrow{\mathrm{A}}$ acceleration of centre of mass

$\frac{d \overrightarrow{v_{1}}}{d t}=\overrightarrow{a_{1}}$ acceleration of first particle

$\frac{d \overrightarrow{v_{2}}}{d t}=\overrightarrow{a_{2}}$ acceleration of second particle and

$\frac{d \overrightarrow{v_{n}}}{d t}=\overrightarrow{a_{n}}$ acceleration of ' $n$ ' particle

$\therefore \quad \mathrm{MA}=m_{1} \vec{a}_{1}+m_{2} \overrightarrow{a_{2}}+\ldots m_{n} \overrightarrow{a_{n}} \ldots$

OR

$\vec{a}_{\mathrm{cm}} \text { or } \overrightarrow{\mathrm{A}}=\frac{m_{1} \overrightarrow{a_{1}}+m_{2} \overrightarrow{a_{2}}+\ldots m_{n} \overrightarrow{a_{n}}}{m_{1}+m_{2}+\ldots m_{n}}$

is a formula for acceleration of centre of mass.

From Newton's second law, $\overrightarrow{\mathrm{F}}_{i}=m_{i} \vec{a}_{i}$ but $\mathrm{MA}=\overrightarrow{\mathrm{F}}_{1}+\overrightarrow{\mathrm{F}}_{2}+\ldots \overrightarrow{\mathrm{F}}_{n}$

Hence, the vector sum of all the forces acting on the all particles of a system is equal to the product of total mass of system and the acceleration of its centre of mass.