# Elastic Collision

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```					SUBJECT

TOPIC
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
Information   Content   Graphic   Visual   Arrangement
Search               Design    Design
Au Yeung
Sum Yee
Chung
On Wing
Lee
Chun Kit
Leung
Siu Fai
Mak
Ka Man
• Target Audience
 Band 3
 Form 4 Students
• Function of this Package
•as an auxiliary teaching aids
Previous Knowledge

• Newton's Second Law
• Momentum
• Impulse
Objectives

• 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 gun

• Why does the gun recoil(move backwards)
after firing?
Table 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
Collision
• There are three kinds of collision
– Elastic collision

– Inelastic collision

– Partially elastic collision
Elastic Collision

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

The balls
bounce back
to a lower
level
Inelastic Collision

The balls do
not bounce
i.e. they stick
together
• the quantity of motion
• how much stuff is moving (mass)
• how fast the stuff is moving(velocity)

• Momentum = mass (kg)  velocity (ms-1)
p = mv
Animation of a football and a
tennis move at same velocity

Which one has greater momentum?

Football / Tennis ball
Animation of two identical cars
move at different velocity

Car A                                Car A

Car B                                Car B

Which one has greater momentum?
Car A / Car B
Conservation of Momentum
What is conservation of momentum?

total momentum before
In any collision, theFext = 0
If
collision is equal to the total momentum
P before P after
after collision, provided=that there is no
external force acting.
Before
When Fext = 0,   collision

m1u1 + m2u2 = 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.
Solution
Step 1                     Before collision:
0.2ms-1
Make a sketch
0.4ms-1
showing the direction,                   + ve
masses and velocities
of each object before
0.06kg
collision.                     0.02kg

Step 2                     Assume that the 0.06 kg marble
continues to move to the left after
Assign one direction as    the collision at a velocity v. Take the
the positive direction.    direction to the left as positive.
After collision
Before collision:
Step 3                                          0.2ms-1
0.4ms-1
Make a sketch                0.1ms-1     + ve    v
showing the direction,
masses and velocities
0.06kg
of each object after           0.02kg
collision.                                      0.06kg
0.02kg
Step 4                     As no external force exists
Write down the             during the collision,
equation for               by the law of conservation of
conservation of            momentum
momentum and
substitute the known        m1u1 + m2u2 = m1v1 + m2v2
values of each object                              mkg*-0.4ms-1)
(0.06 kg *u1 ms-1) + (0.02 2u2
m1 0.2      +
after collision.
m1 kg           m kg*0.1ms-1)
= (0.06v1 * v)+ (0.02 2v2
+
Step 5
Solve the equation to find
The velocity
v = 0.033 ms-1 of the 60 g marble
is 0.033 ms-1 to the left.
out the unknown value.
QUESTION 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-1                               15 cms-1

1 kg              2 kg

10 cm s-1          15 cm s-1      20 cm s-1

25 cm s-1         30 cm s-1
QUESTION 2

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 ms-1      4 ms-1

A                        B

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

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

4 m s-1 towards the right
QUESTION 3

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 m s-1

1.5 m s-1                  2 m s-1

3 m s-1
QUESTION 4

Trolley A towards a
stationary trolley B
is found that trolley B
moves at a speed of 18
m s-1. The final velocity
of trolley A is
Speed         Direction
6          same as before
6          reversed
12         same as before
12         reversed
21         reversed
QUESTION 5

After collision, both trolleys sticked together.
They are

Stationary
moving at a constant speed of 4 m s-1
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
Rocket gains
momentum in
the up
direction

The hot gases
gain
momentum in
the down
direction
Shooting gun

Remember the question asked at the
beginning of this lesson?

• Why does the gun recoil(move backwards)
after firing?
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
After firing, the bullet moves forward, and have a
forward momentum

So, the gun must have a backward momentum
Sorry, you are wrong

Don’t disappointed
TRY AGAIN

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