1. Momentum and Collisions
Chapter 6

2. International regulations specify the mass of official soccer balls.
How does the mass of a ball affect the way it behaves when kicked?
How does the velocity of the player’s foot affect the final velocity of the ball?

3. Knowledge to Expect
Energy cannot be created or destroyed, but only changed from one form to another.
An object that is not being subjected to a force will continue to move at constant speed in a straight line.

4. Knowledge to Review
A force on an object is a push or pull that tends to cause a change in motion. Forces can be field or contact forces.
Newton's laws of motion describe the effects of forces on objects and the idea that forces always exist in pairs.
Energy of motion, called kinetic energy, depends on the mass and speed: KE=1/2mv2
Energy is neither created nor destroyed, but it can be converted from one form to another.

5. Momentum
A vector quantity defined as the product of an object's mass and velocity.

6. Equation
“p” is the letter used to denote momentum. It comes from Leibniz’s use of the term progress, defined as the “the quantity of motion which which a body proceeds in a certain direction.”
Momentum = mass x velocity
p=mv

7. Who was Leibniz?
Leibniz was one of the great polymaths of the modern world. 
As an engineer he worked on calculating machines, clocks and even mining machinery. 
As a librarian he more or less invented the modern idea of cataloguing. 
As a mathematician he came up with the calculus independently of (though a few years later) than Newton, and his notation has become the standard. 
As a physicist he made advances in mechanics, specifically the theory of momentum. 

8. Momentum is a vector quantity. It has magnitude and direction
Momentum Unit
kg·m/s
Momentum is a vector quantity.
It has magnitude and direction

9. Practice
A 2250 kg pickup truck has a velocity of 25 m/s to the east. What is the momentum of the truck?
M=2250 kg v=25 m/s p=?
P=2250 kg x 25 m/s
P= 5.6 x 104 kg·m/s

10. Momentum is closely related to force.
A change in momentum takes force and time.
A change in momentum is caused by an impulse.

11. What is Impulse?
Impulse is the product of the force and the time over which it acts on an object for a constant force.
Impulse (J) is force x the change in time.
J=Ft

12. Impulse –Momentum Theorem
Impulse is equal to the change in momentum.
J= p v= Vf - Vi
p= mvf - mvi
Ft= mv

13. Practice Ft= mvf - mvi F=-7.0 x 104 N to the east vi= 15 m/s F=?
A 1400 kg car moving westward with a velocity of 15 m/s collides with a utility pole and is brought to rest in 0.30 s. Find the magnitude of the force exerted on the car during the collision.
M= 1400 kg t= 0.30 s vf= 0 m/s
vi= 15 m/s F=?
Ft= mvf - mvi
F= 1400 kg(0m/s) – 1400 kg (15)
0.30s
F=-7.0 x 104 N to the east

14. Impulse-Momentum Theory
A change in momentum over a longer time requires less force.
This relationship is used to design safety equipment
Trampolines
Nets
Safety belts

15. Impulse-Momentum Theory
Ft Large force with small impact time….much damage
fTSmall force with large impact time….less damage

16. Ft = P
Ft = M1Vf - M1Vi
Ft = M2Vf – M2Vi

17. The Law of Conservation of Momentum
When two objects interact the sum of each of their individual momentums BEFORE the interaction will EQUAL the sum of their individual momentums AFTER the interaction.

18. When is momentum conserved?
Elastic Collisions
Objects pushing away from each other.
PA + PB = PA’ + PB’
TOTAL INITIAL MOMENTUM = TOTAL FINAL MOMENTUM

19. PRACTICE
A 76 KG BOATER, INITIALLY AT REST IN A STATIONARY 45 KG BOAT, STEPS OUT OF THE BOAT AND ONTO THE DOCK. IF THE BOATER MOVES OUT OF THE BOAT WITH A VELEOCTIY OF 2.5 M/S TO THE RIGHT, WHAT IS THE FINAL VELOCITY OF THE BOAT?
PA + PB = PA’ + PB’

20. SOLUTION M1 = 76 KG M2 = 45 KG V1 = 0 V2 = 0 V1’ = 2.5 M/S V2’ = ?
PA + PB = PA’ + PB’
M1V1 + M2V2 = M1’V1’ + M2’V2’
76 X X 0 = 76 X X V2’
0 = V2’
-4.2 = V2’

21. Two Types of Collisions
Elastic – mechanical energy is conserved
M1V1 + M2V2 = M1’V1’ + M2’V2’
Inelastic – some mechanical energy is transformed into other forms of energy there fore mechanical energy is not conserved.
M1V1 + M2V2 = (M1’ + M2’)Vf’

22. COLLISION FORCES ARE NOT CONSTANT
FORCES VARY WITH TIME THROUGHOUT THE COLLISION.
FORCES ARE ALWAYS EQUAL AND OPPOSITE DURING THE COLLISION.

23. PERFECTLY INELASTIC COLLISION
WHEN TWO OBJECTS COLLIDE AND MOVE TOGETHER AS ONE MASS
A NEARLY PERFECT INELASTIC COLLISION IS WHEN ONE OBJECT COLLIDES WITH ANOTHER OBJECT AND STOPS WITHIN THE OBJECT IT STRUCK.
M1V1 + M2V2 = (M1’ + M2’)Vf’

24. PRACTICE
A 1850 kg LUXURY SEDAN STOPPED AT A TRAFFIC LIGHT IS STRUCK FROM THE REAR BY A COMPACT CAR WITH A MASS OF 975 KG. THE TWO CARS BECOME ENTANGLED AS A RESULT OF THE COLLISION. IF THE COMPACT CAR WAS MOVING AT A VELOCITY OF 22.0 M/S TO THE NORTH BEFORE THE COLLISION, WHAT IS THE VELOCITY OF THE ENTANGLED MASS AFTER THE COLLISION?

25. SOLUTION M1 = 1850 KG M2 = 975 KG V1 = 0 V2 = 22 M/S V’ = ?
M1V1 + M2V2 = (M1’ + M2’)Vf’
1850 X X 22 = ( ) V’
7.59 M/S TO THE NORTH = V’

26. ENERGY KINETIC ENERGY IS NOT CONSTANT IN INELASTIC COLLISIONS
KE = ½ MV2
KEi = KE1i + KE2i
KEf = KE1f + KE2f
∆KE = KEf – KEi

27. PRACTICE
TWO CLAY BALLS COLLIDE HEAD-ON IN A PERFECTLY INELASTIC COLLISION. THE FIRST BALL HAS A MASS OF .500 KG AND AN INTIAL VELCITY OF 4.00 M.S TO THE RIGHT. THE MASS OF THE SECOND BALL IS .250 KG, AND IT HAS AN INITIAL VELCOTIY OF 3.00 M/S TO THE LEFT. WHAT IS THE FINAL VELOCITY OF THE COMPOSITE BALL OF CLAY AFTER THE COLLISISION?

28. SOLUTION M1 = .500KG M2 = .250 KG V1 = +4.00 M/S V2 = -3.00 M/S
V’ = ? ∆KE = ?
M1V1 + M2V2 = (M1’ + M2’)Vf’
.500 X X = ( ) Vf’
1.67 M/S = Vf’

29. WHAT IS THE DECREASE IN KINETIC ENERGY DURING THE COLLISION?
KEi = KE1i + KE2i
KEi = 1/2(.500)(4.OO)2 + 1/2(.250)(-3.00)2
KEi = 5.12J
KEf = KE1f + KE2f
KEf = 1/2( )(1.67)2
KEf = 1.05 J
∆KE = 1.05J – 5.12J
∆KE = -4.07

30. KINETIC ENERGY IS CONSERVED IN ELASTIC COLLISIONS
MOMENTUM AND KINETIC ENERGY REMAINS CONSTANT IN AN ELASTIC COLLISION.
M1V1i + M2V2i = M1V1’ + M2V2’
½ M1V1i2 + ½ M2V2i2 = ½ M1V1’2 + ½ M2V2’2

31. PRACTICE
A .015 KG MARBLE MOVING TO THE RIGHT AT .225 M/S MAKES AN ELASTIC HEAD ON COLLION WITH A .030 KG SHOOTER MARBLE MOVING TO THE LEFT AT .180 M/S. AFTER THE COLLISION, THE SMALLER MARBLE MOVES TO THE LEFT AT .315 M/S. ASSUME THAT NEITHER MARBLE ROTATES BEFORE OR AFTER THE COLLISION AND THAT BOTH MARBLES ARE MOVING ON A FRICITONLESS SURFACE. WHAT IS THE VELOCITY OF THE .030 KG MARBLE AFTER THE COLLISION?

32. SOLUTION M1V1 + M2V2 = M1’V1’ + M2’V2’
MAKE SURE YOU GIVE DIRECTIONS BY POSITIVE AND NEGATIVE.
V2; = .09 M/S TO THE RIGHT