In a closed system, the total linear momentum of the system remains constant or conserved.

Proof: Consider two bodies A and B of masses m₁ and m₂ moving in the same direction with uniform velocities u₁ and u₂ respectively. After the collision let their uniform velocities be v₁ and v_2. Let ‘t’ be the time of impact.

Change in momentum of A

= m₁v₁ – m₁u₁

Rate of change of momentum of A

= \(\frac{ m_1v_1-m_1μ_1}{t}\)

Change in momentum of B

= m₂v₂ – m₂u₂

Rate of change of momentum of B

= \(\frac{ m_2v_2-m_2μ_2}{t}\)

If F₁ is the force exerted by A on B then according to second law, 

F₁ = \(\frac{ m_2v_2-m_2μ_2}{t}\)(action)

If F₂ is the force exerted by B on A then

F₂ = \(\frac{ m_1v_1-m_1μ_1}{t}\) (reaction)

According to Newton’s third law, action and reaction are equal and opposite i.e.

F₁ = – F₂

 \([\frac{ m_2v_2-m_2μ_2}{t}]\) = - \(\frac{ m_1v_1-m_1μ_1}{t}\)

m₂v₂ – m₂u₂ 

= – m₁v₁ + m₁u₁

OR

m₁u₁ + m₂u₂ 

= m₁v₁ + m₂v₂

i.e., Total momentum before collision = Total momentum after collision. Hence the momentum is conserved.