State the law of conservation of momentum. Derive it by using Newton's third law of motion.

(N/A) The law of conservation of momentum states that if no external force acts on a system of particles,the total linear momentum of the system remains conserved.
Consider two particles $A$ and $B$ with masses $m_{1}$ and $m_{2}$ moving in a straight line. Let their initial velocities be $u_{1}$ and $u_{2}$ respectively,where $u_{1} > u_{2}$. They collide for a time $t$. After the collision,their final velocities become $v_{1}$ and $v_{2}$ respectively.
Initial momentum of particle $A = m_{1}u_{1}$
Initial momentum of particle $B = m_{2}u_{2}$
Final momentum of particle $A = m_{1}v_{1}$
Final momentum of particle $B = m_{2}v_{2}$
Rate of change of momentum of particle $A$ (Force exerted by $B$ on $A$,$F_{AB}$) = $\frac{m_{1}(v_{1} - u_{1})}{t}$
Rate of change of momentum of particle $B$ (Force exerted by $A$ on $B$,$F_{BA}$) = $\frac{m_{2}(v_{2} - u_{2})}{t}$
According to Newton's third law of motion,the force exerted by $A$ on $B$ is equal and opposite to the force exerted by $B$ on $A$:
$F_{AB} = -F_{BA}$
Substituting the values:
$\frac{m_{1}(v_{1} - u_{1})}{t} = -\frac{m_{2}(v_{2} - u_{2})}{t}$
Multiplying both sides by $t$:
$m_{1}v_{1} - m_{1}u_{1} = -(m_{2}v_{2} - m_{2}u_{2})$
$m_{1}v_{1} - m_{1}u_{1} = -m_{2}v_{2} + m_{2}u_{2}$
Rearranging the terms:
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
This shows that the total initial momentum of the system is equal to the total final momentum of the system,which proves the law of conservation of momentum.