#### Need Help?

Get in touch with us

# Law of Conservation of Momentum & Derivation

Aug 27, 2022

## Key Concepts

• Conservation of momentum

## Introduction:

In our daily life we make many observations such as a fast bowler taking a run up before bowling, a tennis player moving her racket backwards before hitting the tennis ball and a batsman moving his bat backwards before hitting the cricket ball. All these activities are performed in order to make the ball move with a great speed when hit/thrown. The balls would rather move slowly if these activities are not done. But did you know in all of these the momentum is conserved. In this session we are going to discuss about law of conservation of momentum in objects moving in the same direction.

### Explanation:

Momentum

Momentum is a vector quantity, as velocity, in its formula, is a vector quantity. It is directed along the velocity of the moving object.

Momentum, p α m and v

This means,  p = mv

The S I unit of mass is ‘kg’ and of velocity is ‘m/s’.

Therefore, the S I unit of p = S I unit of mass x S I unit of velocity.

=  kg * m/s

=  kg.m/s

Thus, the S I unit of momentum is “kg.m/s”

Law of Conservation of Momentum

The sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them.

This is known as the law of conservation of momentum

Derivation of the Law

Consider an object A has mass ‘ ma,’ and object B has mass ‘ mb.’

The initial velocity of an object A is ua

and initial velocity of an object B is ub

We know the formula of momentum,

So, Initial momentum (A) = maua

& Initial momentum (B) = mbub

Both collide such that,

The final velocity of an object A is va

and final velocity of an object B is vb

Final momentum after the collision:

Final momentum of an object A is ma va

Final momentum of an object B is ma vb

The force on A due to B is applied and thus an equal and opposite force will act on B due to A according to Newton’s third law of motion. Thus,

Fab = -Fba

Since F = mass x acceleration

Thus, mbab = -maaa

Since a = (v – u )/ t, so

mb ( vb – ub ) = -ma (va – ua

mbvb – mb ub = -ma va + ma ua

mbvb + ma va = ma ua+ mb ub

Such that,

Magnitude of the final momenta = magnitude of the initial momenta

Question 1: There are two balls of masses 3 kg and 4 kg respectively moving in the same direction. 3kg ball is moving with a speed of 10 m/s and 4 kg ball is moving with a speed of 6 m/s.  After collision of the balls the 2nd ball of mass 4 kg moves with velocity 5 m/s . Find the velocity of the ball at mass 3 kg with respect to ground.

Mass of ball , m1 = 3 kg

m2 = 4 kg

Initial speed of the ball, u1 = 10 m/s

u2 = 6 m/s

Final speed of the ball, v1 = ?

v2 = 5 m/s

According to law of conversation of momentum.

Initial momentum = 3 (10) + 4 (6) = 54 kgm/s

Final momentum = 3 (u1) + 4 (5) = (3u1 + 20) kgm/s

Equating both,

3 u1 + 20 = 54

u1 =

343343

= 11.3 m/s

Velocity of ball after collision is 11.3 m/s

Question 2: There are two cars, A and B. Car A of mass 1500 kg, traveling at 30 m/s collides with car B of mass 3000 kg traveling at 20 m/s in the same direction. After collision, the velocity of car A becomes 40 m/s. Calculate the velocity of car B after the collision?

Mass of car A = 1500 kg

Mass of car B = 3000 kg

Before collision,

Initial velocity , uA = 30 m/s

uB = 20 m/s

Initial momentum, = ma ua + mb ub

= 1500 (30) + 3000 (20)

= 105000 kgm/s

After collision,

Final velocity , vA = 40 m/s

vB = v

Final momentum, = ma va + mb vb

= 1500 (40) + 3000 (v)

= 60000 + 30000 v

According to law of conversation of momentum.

Initial momentum = Final momentum

105000 = 60000 + 30000 v

⇒ 30000 v = 45000

v  =

45000300004500030000

= 1.5 m/s.

Velocity of car B after collision is 1.5 m/s.

### Summary

• Law of conservation of momentum- “The sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them.
• ” m1u1+ m2u2 = m1v1+ m2V2

#### Different Types of Waves and Their Examples

Introduction: We can’t directly observe many waves like light waves and sound waves. The mechanical waves on a rope, waves on the surface of the water, and a slinky are visible to us. So, these mechanical waves can serve as a model to understand the wave phenomenon. Explanation: Types of Waves: Fig:1 Types of waves […]

#### Dispersion of Light and the Formation of Rainbow

Introduction: Visible Light: Visible light from the Sun comes to Earth as white light traveling through space in the form of waves. Visible light contains a mixture of wavelengths that the human eye can detect. Visible light has wavelengths between 0.7 and 0.4 millionths of a meter. The different colors you see are electromagnetic waves […]

#### Force: Balanced and Unbalanced Forces

Introduction: In a tug of war, the one applying more force wins the game. In this session, we will calculate this force that makes one team win and one team lose. We will learn about it in terms of balanced force and unbalanced force. Explanation: Force Force is an external effort that may move a […]