Derive the formula of conservation of …

According  to  the  law  of  conservation  of  momentum, the  sum  of  momenta  of two objects  before collision is equal to the sum of momenta after collision, provided there is no external unbalanced force acting on the objects.

Suppose two balls A and B, of masses m_A and m_B are travelling in the same direction along a straight line at different velocities u_A and u_B. And there are no other external unbalanced forces acting on them. Let u_A > u_B and the two balls collide with each other. During collision which lasts for a time t, the ball A exerts a force F_AB on ball B and the ball B exerts a force F_BA  on ball A.

Suppose v_A  and v_B are the velocities of the two balls A and B after the collision, respectively.

The momenta of ball A before and after the collision are m_A u_A and m_A v_A. The rate of change of its momentum (or F_AB, action) during the collision will be

m_A (v_A - u_A)/t

Similarly, the rate of change of momentum of ball B (= F_BA or reaction) during the collision will be

m_B (v_B - u_B)/t.

According to the third law of motion, the force F_AB  exerted by ball A on ball B (action) and the force F_BA  exerted by the ball B on ball A (reaction) must be equal and opposite to each other. Therefore,

F_AB = – F_BA

or

m_A (v_A - u_A)/t = -m_B (v_B - u_B)/t. This gives,

m_A u_A + m_B u_B = m_A u_A + m_B v_B

where m_A and m_B are masses of the bodies, u₁ and u₂ are initial velocities of the bodies and v₁ and v₂ are final velocities of the bodies.