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.

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.

Related Articles:

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

m1 = 4kg

m2 = 10kg

v1 = ?

v2 = 5m.s-1

We know from law of conservation of momentum that,

Pinitial = 0, as the cars are at rest

Pfinal = p1 + p2

Pfinal = m1.v1 + m2.v2

= 4kg.v1 + 10kg.5m.s-1

Pi = Pf

0=4kg.v1+50kg.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 1.5kg. The recoil velocity of pistol is 1.5m.s-1.
Ans: Given,

Mass of bullet, m1 = 5 gram = 0.005kg

Mass of pistol, m2 = 1.5kg

Velocity of bullet, v1 = ?

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

Using law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

Here, Initial velocity of bullet, u1 = 0

Initial recoil velocity of pistol, u2 = 0

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

0 = (0.005kg)(v1)+(2.25kg.m.s-1)

v1=-450m.s-1

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

1 Comment

Leave a Comment

Your email address will not be published. Required fields are marked *