Law of Conservation of Momentum Derivation: Class 11 Physics Notes, Definition, Working Principle, Formula & Examples

The law of conservation of momentum states that the total momentum of a system remains constant if no external force acts on the system. For example, when two balls collide, the momentum of the balls before and after the collision is the same.

 

The Principle of the Conservation of Momentum is an essential aspect of modern physics. It says that overall momentum before and after a collision is identical (provided that no external forces, for example, friction acting on the system). 

This is immensely beneficial since scientists can foresee an event’s results long before it takes place. 

Imagine a truck of mass ‘m1’ moves with a velocity of ‘u1,’ and a car of mass ‘m2’ moves at a given velocity ‘u2.’ Therefore, net momentum at this point = m1u1 + m2u2. 

If the vehicle collides with the truck for a brief period, the velocity can shift. Due to the truck’s velocity and vehicle study, the variables v₁ and v₂ are obtained. However, their mass does not alter. The average momentum is (m₁v₁ + m₂v₂).

Acceleration of car = (V₂–U₂)/t. 

F = ma. 

F = force exerted on the car by the truck. 

F₁ = m₂v₂−u₂/t. 

Acceleration of truck = (v₁–u₁)/t. 

If F₂ = m₁ (v₁–u₁), then F₁ = –F₂ 

m₁ (v₁– u₁)/t = –m₂(v₂– u₂)/t 

m₂v₂-m₂u₂ = -m₁v₁+m₁u₁. 

or m₁u₁+m₂u₂ = m₂v₂+m₁v₁

Consider an example of a balloon. The gas particles travel quickly colliding with one another and the balloon’s walls. Even if the gas particles move quicker and slower as they clash, the particles’ overall velocity remains the same. 

Law of Conservation of Momentum Derivation in Class 9

This concept is taught in the chapter ‘Force and Law of Motion.’ You will learn about the basics of momentum and its formula. The weightage of this chapter is eight marks in the final exam.

1. A shell is shot from a pistol flying at 300 meters per second, creating an angle of 60 degrees with the horizontal. The arrow collides into two parts before it hits its peak stage. The mass ratio of the two components is 1:3. If the collision wholly halts the smaller piece, find the velocity of the other piece.

Sol: Velocity at the highest point = 360 x cos 60

= 150 m/s

Using momentum conservation,

150 x m = 3m/4 x v

v = 200 m/s.

2. A 10-ton wagon travelling at 12 kmph collides an 8-ton wagon in the same direction at a speed of 10 kmph. If the first vehicle’s speed reduces to 8 kmph, determine the other vehicle’s pace after the collision.

Sol: Conserving momentum,

10×12+8×10=10×8+8×v

200−80=8×v

v =15 kmph.

3. Two rigid rubber balls A and B of similar mass are pushed around in different directions at a velocity of 0.3 m/second. After the collision, the two balls come to rest with B possessing the maximum velocity of?

Sol: Let the velocity of 2nd ball be v m/s

M_A = 0.2 kg;

V_A=0.3 m/s;

M_B = 0.4 kg;

V_B = v

Applying Momentum Conservation:

Initial Momentum= Final Momentum

M_AV_A +M_BV_B =0 (final momentum)

0.2×0.3+0.4×v=0

−0.06/0.4=v

v = −0.15 m/s.

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