To explain conservation of momentum, let us take the following example. Consider two balls A and B having masses *m** _{1} *and

*m*

_{2}, respectively. Let the initial velocity of ball A be

*u*

_{1}, and that of ball B be

*u*

_{2}(u

_{1}> u

_{2}). Their collision takes place for a very short interval of time

*t*and after that A and B start moving with velocities

*v*

_{1}and

*v*

_{2}(now

*v*

_{1}<

*v*

_{2}) respectively as shown in Figure.

The momentum of ball A before and after the collision is *m*_{1}*u*_{1} and *m*_{1}*v*_{1} respectively. If there are no external forces acting on the body, then the rate of change of momentum of ball A, during the collision will be

= {𝑚_1(𝑣_1− 𝑢_1) \over 𝑡}

and, similarly the rate of change in momentum of ball B

= {𝑚_2(𝑣_2− 𝑢_2) \over 𝑡}

Let *F*_{12} be the force exerted by ball A on B and *F*_{21} be the force exerted by ball B on A.

Then,

according to Newton’s second law of motion

*F*_{12} = {𝑚_1(𝑣_1− 𝑢_1) \over 𝑡} and *F*_{21} = {𝑚_2(𝑣_2− 𝑢_2)\over 𝑡}

According to Newton’s third law of motion, we have

*F*_{12} = – *F*_{21}

Or, {𝑚_1(𝑣_1− 𝑢_1)\over 𝑡} = – {𝑚_2(𝑣_2− 𝑢_2)\over 𝑡}

Or, *m*_{1}*v*_{1} – *m*_{1}*u*_{1} = – *m*_{2}*v*_{2} + *m*_{2}*u*_{2}

Or, *m*_{1}*u*_{1} + *m*_{2}*u*_{2} = *m*_{1}*v*_{1} + *m*_{2}*v*_{2}

*i.e., *Total momentum before collision = Total momentum after collision

Thus, we find that in a collision between the two balls the total momentum before and after the collision remains unchanged or conserved provided no net force acts on the system. This result is law of conservation of momentum.