Principle of Conservation of Linear Momentum

It states that,”In the absence of external force, the total momentum of a system remains conserved.”
It can be verified from Newton’s Law of Motion.
For single particle system:
Let us consider a particle of mass m moving with velocity v. Now, linear momentum of the particle is given by:
P = mv………(i)
Now, according to Newton’s second law of motion,
Force acting on a particle is equal to the rate of change in momentum.
$$F = \frac{dP}{dT}$$
$$F = \frac{d}{dT}(mv)….(ii)$$
If no external force is acted on a particle then F = 0, Now:
$$\frac{d}{dT}(mv) = 0$$
$$or, mv = 0$$
Integrating both sides,
$$\int mv = \int 0$$
$$\therefore \text{mv = Constant}$$
For two particle system:

Conservation of linear momentum

Let us consider two particle A and B of masses m₁ and m₂ moving with velocity U₁ and U₂. The particle with greater mass has more velocity (V₁>V₂) in the same direction. After while they collide with each other then move with the velocity V₁ and V₂. The momentum is transferred from the particle having more mass to the particle having smaller mass. So, the velocity of the particle with smaller mass becomes more than that of particle with greater mass.

Now, during collision, Let F_AB be the force exerted by the 1^st particle on 2^nd particle. Then according to Newton’s third law of motion, the 2^nd particle also exerts equal and opposite force on 1^st particle F_BA.
So, $$F_{AB} = – F_{BA}$$
Now, $$F_{AB} = \frac{\text{Change in momentum of a body of mass}(m_1)}{t}$$
$$or, F_{AB} = \frac{{\text{Final momentum of} (m_2)} – {\text{Initial momentum of }(m_1)}}{t}$$
$$or, F_{AB} = \frac{{m_2V_2} – {m_2U_2}}{t}…….(ii)$$
Similarly,
$$F_{BA} = \frac{{m_1V_1} – {m_1U_1}}{t}…….(iii)$$
Now, From equation (i),
$$F_{AB} = -F_{BA}$$
$$\frac{{m_2 V_2} – {m_2U_2}}{t} = -\frac{{m_1V_1} – {m_1U_1}}{t}$$
$$\therefore m_1U_1 + m_2U_2 = m_1V_1 + m_2V_2$$
Here, momentum before collision = momentum after collision
Hence, the law of conservation of linear momentum is verified.