State the law of conservation of momentum and derive related expression.

Law of conservation of momentum states that when two objects collide with each other , the sum of their linear momentum always remains same or we can say conserved and is not effected by any action, reaction only in case is no external unbalanced force is applied on the bodies.

Let,

$m_A$ = Mass of ball A

$m_B$= Mass of ball B

$u_A$= initial velocity of ball A

$u_B$= initial velocity of ball B

$v_A$= Velocity after the collision of ball A

$v_B$= Velocity after the collision of ball B

$F_{ab}$= Force exerted by A on B

$F_{ba}$= Force exerted by B on A

Now,

Change in the momentum of A= momentum of A after the collision - the momentum of A before the collision

= $m_A{v}_A-m_A{u}_A$

Rate of change of momentum A= Change in momentum of A/ time taken

= $\frac{m_A{v}_A-m_A{u}_A}{t}$

Force exerted by B on A ($F_{ba}$);

$F_{ba}=\frac{m_A{v}_A-m_A{u}_A}{t}$........ [i]

In the same way,

Rate of change of momentum of B=

$\frac{m_b{v}_B-m_B{u}_B}{t}$

Force exerted by A on B ($F_{ab}$)=

$F_{ab}=\frac{m_B{v}_B-m_B{u}_B}{t}$.......... [ii]

Newton's third law of motion states that every action has an equal and opposite reaction, then,

$F_{a}b=-F_{b}a$ [ ' -- ' sign is used to indicate that 1 object is moving in opposite direction after collision]

Using [i] and [ii] , we have

$\frac{m_B{v}_B-m_B{u}_B}{t}=-\frac{m_A{v}_A-m_A{u}_A}{t}$

$m_B{v}_B-m_B{u}_B=-m_A{v}_A+m_A{u}_A$

Finally we get,

$m_B{v}_B+m_A{v}_A=m_B{u}_B+m_A{u}_A$

This is the derivation of conservation of linear momentum.