Law of Conservation of Momentum Derivation

What is Law of Conservation of Momentum?

Law of conservation of momentum states that

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.

The 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 the collision and \(m_{1}v_{1}+m_{2}v_{2}\) is the representation of total momentum of particles A and B after the collision.

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Law of Conservation of Momentum Examples

Following are the examples of law of conversation of momentum:

  • Air-filled balloons
  • System of gun and bullet
  • Motion of rockets

Law of Conservation of Momentum Problems

Q1. There are cars with masses 4 kg and 10 kg respectively that are at rest. A car having the mass 10 kg moves towards the east with a velocity of 5 m.s-1. Find the velocity of the car with mass 4 kg with respect to ground.
Ans: Given,

m1 = 4 kg

m2 = 10 kg

v1 = ?

v2 = 5 m.s-1

We know from the law of conservation of momentum that,

Pinitial = 0, as the cars are at rest

Pfinal = p1 + p2

Pfinal = m1.v1 + m2.v2

= 4 kg.v1 + 10 kg.5 m.s-1

Pi = Pf

0=4 kg.v1+50 kg.m.s-1

v1 = 12.5 m.s-1

Q2. Find the velocity of bullet of mass 5 gram which is fired from a pistol of mass 1.5 kg. The recoil velocity of pistol is 1.5 m.s-1.
Ans: Given,

Mass of bullet, m1 = 5 gram = 0.005 kg

Mass of pistol, m2 = 1.5 kg

The velocity of a bullet, v1 = ?

Recoil velocity of pistol, v2 = 1.5 m.s-1

Using law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

Here, Initial velocity of the bullet, u1 = 0

Initial recoil velocity of a pistol, u2 = 0

∴ (0.005 kg)(0) + (1.5 kg)(0) = (0.005 kg)(v1) + (1.5 kg)(1.5 m.s-1)

0 = (0.005 kg)(v1)+(2.25 kg.m.s-1)

v1=-450 m.s-1

Hence, the recoil velocity of pistol is 450 m.s-1.

7 Comments

  1. Thanks so much for the derivation

  2. Thank you

  3. Very Nice notes and explain

  4. thank you

  5. Jhilam Chattopadhyay

    thank you very much

  6. Very well explained and thanks for giving Questions. This makes it easier 2 understand the concept

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