Linear momentum of a body is defined by the product of the mass of that body and the linear velocity of that body. If \( m \) be the mass of the body and \( \vec{v} \) be the linear velocity of that body then the linear momentum is given by,

\( \vec{p}=m\vec{v} \)

If \( \vec{r} \) be the displacement vector,

then \( \displaystyle{\vec{v}=\frac{d\vec{r}}{dt}} \)

Now, \( \displaystyle{\vec{p}=m\frac{d\vec{r}}{dt}} \)

\( or,\ \displaystyle{\frac{d\vec{p}}{dt}=m\frac{d\vec{r}}{dt^2}} \)

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

Where, \( \displaystyle{\vec{F}=m\frac{d^2{\vec{r}}}{dt^2}} \) is the force acting on the particle.

So the time rate of change of linear momentum is equal to the force acting on the particle.

If \( \vec{F}=0 \) then \( \displaystyle{\frac{d\vec{p}}{dt}=0} \), or, \( \vec{p}=constant \).

The Principle of conservation of linear momentum states that, the total linear momentum of a system of particles remains unchanged when there is no external force acting on it and is subjected only to their mutual interaction.

To understand this, let us consider a bullet of mass \( m \) is fired from a gun of mass \( M \). The bullet moves forward with velocity \( \vec{v} \) but after firing the bullet the gun moves backward with velocity \( \vec{V} \).

So the momentum of the bullet is \( \vec{p_1}=m\vec{v} \)

and the momentum of the bullet is \( \vec{p_2}=-M\vec{V} \).

Here the negative sign indicates that the direction of the movement of the bullet and the gun are in opposite directions.

So according to the conservation of linear momentum, we can write,

\( \vec{p_1}=\vec{p_2} \)

\( or,\ m\vec{v}=-M\vec{V} \)

\( Or,\ m\vec{v}+M\vec{V}=0 \)

This shows that the total linear momentum of the gun and the bullet before firing the bullet is zero. So the linear momentum is conserved.