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 FAB be the force exerted by the 1st particle on 2nd particle. Then according to Newton’s third law of motion, the 2nd particle also exerts equal and opposite force on 1st particle FBA.
So, $$F_{AB} = – F_{BA}$$
Now, $$F_{AB} = \frac{\text{Change in momentum of a body of mass}(m_1)}{t}$$

$$or, F_{AB} = \frac{{\text{Final momentum of} (m_2)} – {\text{Initial momentum of }(m_1)}}{t}$$

$$or, F_{AB} = \frac{{m_2V_2} – {m_2U_2}}{t}…….(ii)$$


$$F_{BA} = \frac{{m_1V_1} – {m_1U_1}}{t}…….(iii)$$
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.


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