COLLISION

1. 1
Grade 9
Students!

2. PRAYER

3. MOMENTUM
Which is the product of
Mass Velocity
Can be changed by applying
Impulse
Which is the product of
Force Time

4. TELL THE STANDARD FORMULA OF
THE FOLLOWING:
MOMENTUM IMPULSE
MOMENTUM –
IMPULSE THEOREM
p = mv I = Ft Ft = mv

5. Who is Sir Isaac Newton?
What are the 3 laws of motion?
Law of Inertia
Law of
Acceleration
Law of
Interaction
He was the one who
proposed the LAWS OF
MOTION
MOMENTUM IMPULSE ???

6. WHAT ARE WE GOING TO EXPLORE TODAY?
• When is the conservation of momentum
applied in your daily life.
• To understand the outcomes of collisions that
may be predicted using some conservation
laws.
• To know how the study of collision and
conservation laws came up with the idea of
the Third Law of Motion-Interaction.
LEARNING OUTCOMES:

7. Law of CONSERVATION
OF MOMENTUM
Newton came up with the idea of the Third Law of motion through
the study of Collision and Law of Conservation of momentum.

8. What do you mean
by the word
Conserve?
Conserve means “to
keep"
MOMENTUM
Since Momentum means the quantity of
motion of any object that has mass, ”to keep
the motion of a moving object after applying
force means
CONSERVATION

9. Law of CONSERVATION OF MOMENTUM
✗ The momentum of an object can't change unless an external force acts
on the object.
✗ Based on this definition, if the total force is zero then the initial and
final momentums must be the same,. This is momentum
conservation.
9

10. 10
Internal
forces, such
as a push on
the
handlebars
exerted by a
bicycle rider,
act between
objects
within a
system.
External
forces, such
as the force
the road
exerts on a
rear bicycle
tire, are
exerted on
the system
by
something
outside the
system
The figure below shows both the internal and external
forces acting on a rider and bicycle.

11. ✗ Summarizing:
✗ Internal forces have no effect on the
total
momentum of a system.
✗ If the total external force acting on a
system is
zero, then the system's total
momentum is conserved.
11

12. 12
What is the
difference of
Internal
forces and
External
forces?
Internal forces are the
forces that particles of a
system exert on one
another.
External forces are
applied on any part of the
system by other objects
outside the system.

13. EXAMPLES OF CONSERVATION
OF MOMENTUM
1

16. COLLISION

17. COLLISION
✗ A collision
occurs when
two or
more bodies
come in
contact
with each
other.
17
DEFINITION

18. COLLISION
✗ A collision occurs
when two objects
free from
external forces
strike one
another.
18
DEFINITION

19. What happen when
there is collision?
During a collision, the
interacting bodies hit
and exert a force on
each other. In general,
when two or more
bodies collide, linear
momentum is always
conserved.
19

20. 20
TYPES OF
COLLISION
1. ELASTIC COLLISION
This happens when
two objects collide and
separate to move with
different velocities.
NEWTON’S CRADLE

21. OTHER EXAMPLES OF ELASTIC COLLISION
21

22. 22
TYPES OF
COLLISION
2. INELASTIC COLLISION
This doesn’t make
the colliding body
separate. They
couple or stick
together and
move with one
velocity.
TWO CLAY BALLS

23. OTHER EXAMPLES OF inELASTIC
COLLISION
23

24. DIFFERENCES OF ELASTIC AND INELASTIC COLLISION
✗ Bodies separate after
collision
✗ Bodies cling to each other
after collision moving
common velocities
24
ELASTIC COLLISION INELASTIC COLLISION
✗ Kinetic energy is conserved. ✗ Kinetic energy is not
conserved.
✗ Formula is:
m1v1 + m2v2 = m1v1 + m2v2
✗ Formula is:
m1v1 + m2v2 = (m1+m2) v

25. HOW TO COMPUTE for elastic collision
25
STANDARD FORMULA FOR
ELASTIC COLLISION
• With elastic collision, two objects are observed.
m1v1 + m2v2 = m1v1’ + m2v2’
Where:
m1 = mass of the first object
v1 = velocity of the first object
m2 = mass of the second object
v2 = velocity of the second object
TRY TO OBSERVE THE FORMULA:
m1v1 + m2v2 = m1v1’ + m2v2’
The left side of the
equation is the
“BEFORE
COLLISION”
The right side of the
equation is the
“AFTER COLLISION”

26. SAMPLE COMPUTATION #1
26
A 2,500 kg car moving at 25.0 m/s hits the rear of a 2,000 kg car moving
slower on a straight hi-way. If after the collision, the heavier car moves at
20.0 m/s while the other at 24.0 m/s in the same direction along the
highway, with what speed is the lighter car moving before the collision?
GIVEN:
2,500 kg
25.0 m/s
BEFORE AFTER
m1
v1
m2
v2
2,500 kg
2,000 kg 2,000 kg
20.0 m/s
24.0 m/s
=???
STANDARD FORMULA
m1v1 + m2v2 = m1v1’ + m2v2’
m2 m2
m1v1’ + m2v2’- m1v1
v2=
m2
Final formula:
m1v1’ + m2v2’- m1v1
v2=
m2

27. Continuation for sample #1
27
SOLUTION
GIVEN:
2,500 kg
25.0 m/s
BEFORE AFTER
m1
v1
m2
v2
2,500 kg
2,000 kg 2,000 kg
20.0 m/s
24.0 m/s
=???
m1v1’ + m2v2’- m1v1
v2=
m2
FORMULA
v2= (2,500 kg)(20.0 m/s) + (2,000 kg)(24.0 m/s) - (2,500 kg)(25.0 m/s)
2,000 kg
v2=
50,000 kg●m/s + 48,000 kg●m/s - 62,500 kg●m/s
2,000 kg
v2= 98,000 kg●m/s - 62,500 kg●m/s
2,000 kg
v2= 35,500 kg●m/s
2,000 kg
FINAL ANSWER:
v2= 17.75 m/s

28. SAMPLE COMPUTATION #2
28
A 1,650 kg car moving at 15.0 m/s hits the rear of a 1,000 kg car moving
at 4.0 m/s on a straight hi-way. If after the collision, the heavier car moves
at 10.0 m/s while the other in the same direction along the highway, with
what speed is the lighter car moving after the collision?
GIVEN:
1,650 kg
15.0 m/s
BEFORE AFTER
m1
v1
m2
v2
1,650 kg
1,000 kg 1,000 kg
10.0 m/s
4.0 m/s =???
STANDARD FORMULA
m1v1 + m2v2 = m1v1’ - m2v2’
m2 m2
m1v1 + m2v2- m1v1’
V2’=
m2
Final formula:
m1v1 + m2v2- m1v1’
v2’=
m2

29. Continuation for sample #2
29
SOLUTION
GIVEN:
1,650 kg
15.0 m/s
BEFORE AFTER
m1
v1
m2
v2
1,650 kg
1,000 kg 1,000 kg
10.0 m/s
4.0 m/s =???
m1v1 + m2v2 - m1v1’
v2’=
m2
FORMULA
v2'= (1,650 kg)(15.0 m/s) + (1,000 kg)(4.0 m/s) - (1,650 kg)(10.0 m/s)
1,000 kg
v2'=
24,750 kg●m/s + 4,000 kg●m/s - 16,500 kg●m/s
1,000 kg
v2'= 28,750 kg●m/s
1,000 kg
FINAL ANSWER:
v2'= 12.25 m/s
- 16,500 kg●m/s
v2'= 12,250 kg●m/s
1,000 kg

30. 30
HOW TO COMPUTE for INelastic collision
✗ STANDARD FORMULA FOR INELASTIC COLLISION:
✗ m1v1 + m2v2 = (m₁+m₂)V’
Where:
m1 = mass of the first object
v1 = velocity of the first object
m2 = mass of the second object
v2 = velocity of the second object
V’= prime velocity

31. 31
SAMPLE COMPUTATION #1
✗ A .010 kg pebble moving at 5.00 m/s collides head-on
with a 0.500 kg clay moving at 2.00 m/s. How fast will
the clay-pebble combination move?
GIVEN:
m1
v1
m2
v2
0.10 kg
5.00 m/s
0.500 kg
2.00 m/s
STANDARD FORMULA
m1v1 + m2v2 = (m₁+m₂) V’
(m₁+m₂) (m₁+m₂)
V’= m1v1 + m2v2
m1+m2
Final formula:
V’= m1v1 + m2v2
m1+m2
V’= ?

32. 32
Continuation for sample #1
GIVEN:
m1
v1
m2
v2
0.10 kg
5.00 m/s
0.500 kg
2.00 m/s
FORMULA
V’= m1v1 + m2v2
m1+m2
SOLUTION
v'= (0.10 kg)(5.00 m/s)+ (0.500 kg) (2.00 m/s)
(0.10 kg + 0.500 kg)
v'= 0.5 kg●m/s + 1 kg●m/s
(0.6 kg)
v'= 1.5 kg●m/s
(0.6 kg)
FINAL ANSWER:
v'= 2.5 m/s

33. seatwork
33
A 2.5 kg mass is moving to the right at 10.0 m/s while a
3.0 kg mass is moving to the left at 14 m/s. They collide
head-on and stick together. Find their velocity after collision.