According to Newton’s 1^st law of motion, everybody persists in a state of rest or of uniform motion along a straight line (i.e. with constant velocity) unless and until it is acted upon by an external force to change that state.

Again according to Newton’s 2^nd law of motion, the rate of change of momentum is directly proportional to the impressed force and is in the same direction along which the force acts.

Let us consider a particle of mass \( m \) is moving with velocity \( \vec{v} \). So the linear momentum of the particle is \( \vec{p}=m\vec{v} \).

Now, when an external force \( \vec{F} \) is applied on the particle, then the rate of change of linear momentum is \( \displaystyle{\frac{d\vec{p}}{dt}} \).

Therefore, \( \displaystyle{\vec{F}=\frac{d\vec{p}}{dt}} \).

\( or,\ \displaystyle{\vec{F}=\frac{d}{dt}(m\vec{v})} \)

When there is no external force acting on the particle, i.e. \( \vec{F}=0 \) then \( \displaystyle{\frac{d\vec{p}}{dt}=0} \), this means that the linear momentum \( \vec{p} \) is constant, i.e. \( m\vec{v}=constant \). This indicates that a particle continues to move with constant velocity when there is no net external force acting on the particle. So we can say that, Newton’s 1^st law of motion is only a special case of Newton’s 2^nd law of motion.

Now let us consider a system of two particles (particle 1 and particle 2) are taken in such a way that there is no external force acting on the system. They are acted only by the force of interaction, which may be the coulomb’s force or gravitational force.

Let \( \vec{F}_{12} \) be the force exerted on the particle 2 due to the particle 1.

Let \( \vec{F}_{21} \) be the force exerted on the particle 1 due to the particle 2.

\( \vec{F}_{12} \) is known as action and \( \vec{F}_{21} \) is known as reaction.

According to Newton’s 3^rd law of motion, action and reaction forces are equal in magnitude and opposite in direction. So we can write, \( \vec{F}_{12}=- \vec{F}_{21} \).

Let us consider, particle 2 of mass \( m_2 \) is moving with velocity \( \vec{v}_2 \). So the linear momentum of particle 2 is \( \vec{p}_2=m_2\vec{v}_2 \).

According to Newton’s 2^nd law of motion,

\( \displaystyle{\vec{F}_{12}=\frac{d\vec{p}_2}{dt}=\frac{d}{dt}( m_2\vec{v}_2)=m_2\frac{d \vec{v}_2}{dt}} \)
where mass \( m_2 \) is constant.

Similarly, let us consider, particle 1 of mass \( m_1 \) is moving with velocity \( \vec{v}_1 \). So the linear momentum of particle 1 is \( \vec{p}_1=m_1\vec{v}_1 \).

According to Newton’s 2^nd law of motion,

\( \displaystyle{\vec{F}_{21}=\frac{d\vec{p}_1}{dt}=\frac{d}{dt}( m_1\vec{v}_1)=m_1\frac{d \vec{v}_1}{dt}} \)
where mass \( m_1 \) is constant

Now, \( \vec{F}_{12}=- \vec{F}_{21} \)

\( or,\ \vec{F}_{12}+ \vec{F}_{21}=0 \)

\( or,\ \displaystyle{\frac{d\vec{p}_1}{dt}+\frac{d\vec{p}_2}{dt}=0} \)

\( or,\ \displaystyle{\frac{d}{dt}(\vec{p}_1+\vec{p}_2)=0} \)

Integrating both side with respect to time t, we get

\( \vec{p}_1+\vec{p}_2=constant \)

\( or,\ m_1\vec{v}_1 +m_2\vec{v}_2=constant \)

So the total linear momentum of the system of two particles remains constant when Newton’s 2^nd and 3^rd laws of motion hold good.

If we extend this two particles system to three or more particles system then we can write, If there is no external force acting on the particles consisting of a system, then the total linear momentum of the system remains constant, here the particles are subjected only to their mutual interaction.