Need Help?

Get in touch with us

searchclose
bannerAd

Law of Conservation of Momentum

Grade 10
Aug 29, 2022
link

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 towards each other and when one object is at rest. 

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 

parallel

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

parallel

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

Conservation of momentum 

Application of Law 

  1. When moving towards each other: 
Collision of objects

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 = 0 m/s 

We know the formula of momentum, 

So, Initial momentum (A) = ma ua 

& Initial momentum (B) = -mb ub = 0 m/s 

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 mb vb 

According to law of conservation,  

Magnitude of the final momenta = magnitude of the initial momenta 

Such that the equation becomes, 

-ma va + mb  vb = ma ua 

  1. When one object is stationary and other collides with it 
Collision of objects 

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 = 0 m/s 

We know the formula of momentum, 

So, Initial momentum (A) = ma ua 

& Initial momentum (B) = mb ub = 0 m/s 

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 mb vb 

According to law of conservation,  

Magnitude of the final momenta = magnitude of the initial momenta 

Such that the equation becomes, 

-ma va + mb  vb = ma ua 

Questions and answers 

Question 1: There are two balls of masses 3 kg and 4 kg respectively moving towards each other. 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 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. 

Answer:

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) = 6 kgm/s 

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

Equating both, 

3v + 20 = 6 

3v = -14 

v =

−143−143

= -3.66 m/s 

Velocity of ball after collision is -3.66 m/s. 

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

Answer:

Mass of car A = 1500 kg 

Mass of car B = 3000 kg 

Before collision, 

Initial velocity , uA = 30 m/s 

                               uB = 0 m/s 

Initial momentum, = ma ua + mb ub    

                                    = 1500 (30) + 3000 (0)  

                                    = 45,000 kgm/s 

After collision, 

Final velocity , vA = -40 m/s 

                             vB = +v 

Final momentum,  = mA vA + mB vB    

                                    = 1500 (-40) + 3000 (+v)  

                                    = -60,000 + 3000 v 

According to law of conversation of momentum. 

Initial momentum = Final momentum  

                      45000 = -60000 + 3000 v 

                 = 3000 v = 105000 

v = 105000/3000 = 35 m/s 

Velocity of car B after collision is 35 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

Comments:

Related topics

Types of Waves

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 […]

Read More >>
Dispersion of Light

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 […]

Read More >>
Force

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 […]

Read More >>
magnets

Magnets: Uses, Materials, and Their Interactions

Introduction: Nowadays magnets are widely used for many applications. In this session, we will discuss the basics of magnets and their properties, and the way they were and are used. Explanation: Magnets: Magnetic and Non-magnetic Materials: Poles of a Magnet: Fig No. 1.2: Poles of a magnet Compass: Interaction Between Magnets: The north pole of […]

Read More >>

Other topics