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
TRY AGAINTRY AGAIN