Conservation Of Momentum

SUBJECT

TOPITOPICC

Group 2

Team Members

Au Yeung Sum Yee Licia (99197960)

Chung On Wing (99163700)

Lee Chun Kit (99197390)

Leung Siu Fai (99269080)

Mak Ka Man (99163620)

Responsibility

InformationSearch

Content GraphicDesign

VisualDesign

Arrangement

Au YeungSum YeeChung

On WingLee

Chun KitLeungSiu Fai

MakKa Man

About the PackageAbout the Package

• Target Audience Band 3 Form 4 Students

• Function of this Package

•as an auxiliary teaching aids

Previous KnowledgePrevious Knowledge

• Newton's Second Law

• Momentum

• Impulse

ObjectivesObjectives

• Recall 3 kinds of collisions and momentum

• Define the law of conservation of momentum

• Apply the law of conservation of momentum to solve problems

Shooting gunShooting gun

• Why does the gun recoil(move backwards) after firing?

Table of contentTable of content Collision

Elastic collision Inelastic collision Partially elastic collision

Momentum Conservation of momentum Example

MC Exercise Explosion Demonstration of water rocket Problem solving strategy of firing a gun

CollisionCollision

• There are three kinds of collision

– Elastic collision

– Inelastic collision

– Partially elastic collision

Elastic CollisionElastic Collision

The two identical balls hit each other and bounce back to the same level

Partially Elastic CollisionPartially Elastic Collision

The balls bounce back to a lower level

Inelastic CollisionInelastic Collision

The balls do not bouncei.e. they stick

together

• Momentum = mass (kg) velocity (ms-1) p = m v

• the quantity of motion• how much stuff is moving (mass)• how fast the stuff is moving(velocity)

Animation of a football and a tennis move at same velocitysame velocity

Which one has greater momentum?

Football / Tennis ballTennis ball

Animation of two identical cars move at different velocity

Which one has greater momentum?

Car A Car BCar B

Car ACar A

Car BCar B

/

Car A

Car B

Conservation of Momentum

What is conservation of momentum?

In any collision, the total momentum before collision is equal to the total momentum after collision, provided that there is no external force acting.

If Fext = 0

P before = P after

When Fext = 0,

=m1u1 + m2u2

Before collision

m1v1 + m2v2

After collision

A 20 g marble travels to the right at 0.4 ms-1 on a smooth, level surface. It collides head-on with a 60 g marble moving to the left at 0.2 ms-1. After collision, the 20 g marble rebounds at 0.1 ms-1. Find the velocity of the 60 g marble.

SolutionStep 1

Make a sketch showing the direction, masses and velocities of each object before collision.

Before collision:

0.02kg0.06kg

0.4ms-1

0.2ms-1

Step 2

Assign one direction as the positive direction.

Assume that the 0.06 kg marble continues to move to the left after the collision at a velocity v. Take the direction to the left as positive.

+ ve

Step 3

Make a sketch showing the direction, masses and velocities of each object after collision.

0.02kg0.06kg

v0.1ms-1

After collisionBefore collision:

0.02kg0.06kg

0.4ms-1

0.2ms-1

+ ve+ ve

Step 4

Write down the equation for conservation of momentum and substitute the known values of each object after collision.

As no external force exists during the collision,

by the law of conservation of momentum m1u1 + m2u2 = m1v1 + m2v2

(0.06 kg * 0.2 ms-1) + (0.02 kg*-0.4ms-1)

= (0.06 kg * v) + (0.02 kg*0.1ms-1)

v = 0.033 ms-1

Step 5

Solve the equation to find out the unknown value.

The velocity of the 60 g marble is 0.033 ms-1 to the left.

m2u2m1u1

m1v1 m2v2

+

+=

15 cm s-110 cm s-1 20 cm s-1

25 cm s-1 30 cm s-1

In the following figure, two particles of masses 1kg and 2kg are moving in the same direction at speed of 30 cms-1 and 15 cms-1 respectively, If they stick together after collision, the final speed of the particles is

30 cms-115 cms-1

1 kg 2 kg

QUESTION 1

3 m s-1 towards the left

Two objects A and B of masses 2 kg and 1 kg respectively move in opposite directions. They collide head on. After the collision, the velocity of A becomes 1 m s-1 towards the left. What would be the velocity of B ?

2 m s-1 towards the right

4 m s-1 towards the left 3 m s-1 towards the right

4 m s-1 towards the right

2 ms-1 4 ms-1

A B

QUESTION 2

1 m s-1

A trolley of mass 1 kg travelling at 3 m s-1 collides with a stationary trolley of mass 2 kg. If the two trolleys remain together after collision, their combined speed immediately after collision, is

10 m s-1

1.5 m s-1 2 m s-1

3 m s-1

QUESTION 3

6 reversed

Trolley A towards a stationary trolley B ahead. After collision, it is found that trolley B moves at a speed of 18 m s-1. The final velocity of trolley A is

6 same as before

12 same as before

12 reversed

21 reversed

Speed Direction

QUESTION 4

moving at a constant speed of 4 m s-1

Stationary

moving at a constant speed less than 4m s-1

decelerating from a speed of 4m s-1

decelerating from a speed less than 4 m s-1

QUESTION 5

After collision, both trolleys sticked together.They are

Rocket gains momentum in the up direction

The hot gases gain momentum in the down direction

Shooting gun

• Why does the gun recoil(move backwards) after firing?

Remember the question asked at the beginning of this lesson?

By the law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

Before firing, both the gun and the bullet have zero momentum

i.e. m1u1 + m2u2 = 0

So, the gun must have a backward momentum

After firing, the bullet moves forward, and have a forward momentum

Sorry, you are wrongSorry, you are wrong

Don’t disappointedDon’t disappointed

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