Let us consider \( \vec{F_i} \) be the resultant external force acting on the ith particle of a system of n particles and \( \vec{F_{ij}} \) be the internal force on the ith particle due to the jth particle.

So the resultant internal force on the ith particle due to all other particle is \( \displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}} \), where \( i\neq{j} \).

If \( \vec{p_i} \) be the linear momentum of the ith particle of mass \( m_i \) moving with velocity \( \vec{v_i} \) and \( \vec{r_i} \) be the position vector of ith particle then the total force on the ith particle is given by,

\( \vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\displaystyle{\frac{d}{dt}\vec{p_i}}=\frac{d}{dt}(m_i\vec{v_i}) \)

or, \( \vec{F_i}+\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\frac{d^2}{dt^2}(m_i\vec{r_i})\tag{1} \)

Summing over \( i \) in equation (1) we get,

\( \displaystyle{\sum_{i=1}^{n}}\vec{F_i}+\displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=\frac{d^2}{dt^2}\left[\displaystyle{\sum_{i=1}^{n}}(m_i\vec{r_i})\right] \)

Now according to Newton’s 3rd law, \( \vec{F_{ij}}=-\vec{F_{ji}} \), so \( \displaystyle{\sum_{i=1}^{n}}\displaystyle{\sum_{j=1}^{n}\vec{F_{ij}}}=0 \)

Therefore,

\( \displaystyle{\sum_{i=1}^{n}}\vec{F_i}=\frac{d^2}{dt^2}\left[\displaystyle{\sum_{i=1}^{n}}(m_i\vec{r_i})\right] \)

or, \( \vec{F}=\frac{d^2}{dt^2}(M\vec{R}) \)

where, \( \displaystyle{\vec{R}=\frac{\displaystyle{\sum_{i=1}^{n}}m_i\vec{r_i}}{M}} \) is the position vector of centre of mass of the system and \( \displaystyle{(M=\sum_{i=1}^{n}m_i)} \) is the total mass of the system.

or, \( \displaystyle{\vec{F}=M\frac{d^2\vec{R}}{dt^2}} \)

\( \vec{F} \) is the total force acting on the system of particles.