# Law of Conservation of Momentum Derivation

Law of conservation of momentum is defined for two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

Law of conservation of momentum is an important consequence of Newton’s third law of motion.

## Derivation of conservation of momentum

Consider two colliding particles A and B whose masses are m1 and m2 with initial and final velocities as u1 and v1 of A and u2 and v2 of B. The time of contact between two particles is given as t.

$A=m_{1}(v_{1}-u_{1})$ (change in momentum of particle A)

$B=m_{2}(v_{2}-u_{2})$ (change in momentum of particle B)

$F_{BA}=-F_{AB}$ (from third law of motion)

$F_{BA}=m_{2}*a_{2}=\frac{m_{2}(v_{2}-u_{2})}{t}$

$F_{AB}=m_{1}*a_{1}=\frac{m_{1}(v_{1}-u_{1})}{t}$

$\frac{m_{2}(v_{2}-u_{2})}{t}=\frac{-m_{1}(v_{1}-u_{1})}{t}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

Therefore, above is the equation of law of conservation of momentum where, $m_{1}u_{1}+m_{2}u_{2}$ is the representation of total momentum of particles A and B before collision and $m_{1}v_{1}+m_{2}v_{2}$ is the representation of total momentum of particles A and B after collision.

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#### Practise This Question

The velocity - time graph of a particle moving in a straight line is shown in figure. The mass of the particle is 2kg.Work done by all the forces acting on the particle in time interval between t = 0 to t = 10 s is