Law of Conservation of Momentum Derivation

The law of conservation of momentum is one of the most prominent laws in physics. The conservation of momentum law principle tells us that the total momentum of a system is always conserved for an isolated system. Let us learn more about the conservation of momentum along with derivation and solved problems.

Momentum Conservation Principle

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.

Derivation of Conservation of Momentum

Newton’s third law states that for a force applied by an object A on object B, object B exerts back an equal force in magnitude, but opposite in direction. This idea was used by Newton to derive the law 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.

Examples of Law of Conservation of Momentum

Following are the examples of law of conservation of momentum:

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

Solved Problems on Law of Conservation of Momentum

Q1. There are cars with masses 4 kg and 10 kg respectively that are at rest. The 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

P_initial = 0, as the cars are at rest

P_final = p₁ + p₂

P_final = m₁.v₁ + m₂.v₂

= (4 kg).(v₁) + (10 kg).(5 m.s^-1)

We know from the law of conservation of momentum that,

P_initial = P_final

0=4 kg.v₁+50 kg.m.s^-1

v₁ = 12.5 m.s^-1

Q2. Find the velocity of a bullet of mass 5 grams which is fired from a pistol of mass 1.5 kg. The recoil velocity of the 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 the pistol is 450 m.s^-1.