1. Chapter B2 – Impulse and Momentum
Unit B - Physics

2. B 2.1-Momentum

3. Objects in motion tend to stay in motion…
Recall Newton’s first law of motion: Objects at rest tend to stay at rest, and objects in motion tend to stay in motion
since 𝐹 = m 𝑎 , the amount of effort required to change an object’s motion depends on two factors:
the mass of the object
the degree of change in speed (ie - the acceleration)

4. Objects in motion tend to stay in motion…
A car going 100 km/h has more of a tendency to stay in motion than a tennis ball going 100 km/h.
The more mass and the more velocity an object has, the harder it will be to stop that object.

5. Practice problem #1
Suggest two reasons why it is harder to stop a semi truck traveling at 100 km/h than a small bird traveling at 45 km/h
the semi is heavier (greater mass)
the semi is traveling faster (greater velocity)

6. Momentum This “quantity of motion” is referred to as momentum. 𝑝 = m 𝑣
Momentum is measured in kg•m/s and is denoted by the letter 𝑝 .
The formula for momentum is:
𝑝 = m 𝑣
where
𝑝 : momentum (kg•m/s )
m : mass (kg)
𝑣 : velocity (m/s)

7. Momentum m = 𝑝 𝑣 𝑣 = 𝑝 𝑚 𝑝 = m 𝑣 The formula for momentum is:
Rearrange the formula to solve for both mass and velocity.
m = 𝑝 𝑣 𝑣 = 𝑝 𝑚

8. Practice Problem #2
An airplane has a momentum of 8.3 x 107 kg•m/s [N]. If the airplane is flying at a velocity of
230 m/s [N], determine its mass.
m = 𝒑 𝒗
m = 8.3 x 107 kg•m/s
230 m/s
m = 3.6 x 105 kg
𝑝 = 8.3 x 107 kg•m/s [N]
𝑣 = 230 m/s [N]
m =?

9. Try These!!!
Practice Problems
1-3 (pg 244)

10. Protective sporting equipment
Picture the type of equipment that is worn by athletes in three different sports: soccer, hockey and tennis

11. Protective sporting equipment
a soccer ball has about 3 times the mass as a hockey puck
a hockey puck can reach velocities of over 160km/h, but a tennis ball can reach velocities of over 200km/h
You would expect that soccer and tennis players would require more protective equipment than a hockey player then…

12. Protective sporting equipment
what makes hockey the most dangerous of the three sports is the combination of velocity ( 𝑣 ) and mass (m) of the puck; it has the greatest momentum

13. Practice Problem #3 “Characteristics of an Object in Motion” Lab
Purpose: You will identify the most significant characteristics of an object in motion.
Materials: Set up the materials as shown in the following diagram.
Read the procedure and complete the following questions in your notes.

14. Assignment: Complete the Practice Problem #3 in your note package
2.1 Section Questions
2-5 & 8 (pg 245)

15. B2.2 – Change in Momentum
Chapter B2

16. The formula for momentum is 𝑝 = m 𝑣
For change to occur in momentum, either the mass or the velocity of an object must change.
because the object in question does change its mass mid motion, change in momentum is usually a result of a change in the object’s velocity
If velocity is changing, changes in momentum must be caused by acceleration
We can amend the formula for change in momentum to:
Δ 𝑝 = m•Δ 𝑣 or
Δ 𝑝 = m ( 𝑣 f - 𝑣 i )

17. Changing momentum: m = Δ 𝑝 Δ 𝑣 𝑎 = Δ𝑣 Δ𝑡 𝐹 = Δ 𝑝 Δ𝑣 • Δ𝑣 Δ𝑡
To change an object’s motion also requires a force:
𝐹 = m 𝑎
Since…
So the formula for “force need to change momentum” can be re-written as:
𝐹 = Δ 𝑝 Δ𝑣 • Δ𝑣 Δ𝑡
m = Δ 𝑝 Δ 𝑣
𝑎 = Δ𝑣 Δ𝑡

18. Changing momentum: Δ𝑝 : change in momentum (kg•m/s) ∆t: time(s)
𝐹 = Δ𝑝 Δ𝑣 • Δ𝑣 Δ𝑡 𝐹 = 𝑝 Δ𝑣 • Δ𝑣 Δ𝑡
Simplified: 𝐹 = Δ 𝑝 Δ𝑡
Where
𝐹 : Force (N)
Δ𝑝 : change in momentum (kg•m/s)
∆t: time(s)

19. Changing momentum: Δ 𝑝 = m ( 𝑣 f - 𝑣 i ) Δ𝑝 = 𝐹 ∆t
So there are two formulas to find change in momentum, depending on the information given in the question:
If mass and BOTH final and initial velocity is known use:
Δ 𝑝 = m ( 𝑣 f - 𝑣 i )
If force and the time interval is known use :
Δ𝑝 = 𝐹 ∆t

20. Changing momentum
During a collision between two objects, we use the formula 𝐹 = Δ 𝑝 Δ𝑡 to calculate the force exerted by one object onto another,
based on the time it takes for a change in momentum to occur, as both objects are involved in the collision (applying force to one another) for the same length of time.

21. Practice Problem #1
A 2.1 kg barn owl flying at a velocity of 15 m/s [E] strikes head-on with a windshield of a car traveling 30 m/s [W].
Calculate the change in momentum of the owl.

22. Practice problem #1
A 2.1 kg barn owl flying at a velocity of 15 m/s [E] strikes head-on with a windshield of a car traveling 30 m/s [W].
Calculate the change in momentum of the owl.
m = 2.1kg
𝑣 i= +15m/s [N]
𝑣 f = – 30 m/s [W]
Δ 𝑝 = ?
Δ 𝑝 = m ( 𝑣 f - 𝑣 𝑖 )
= 2.1kg (-30 m/s [W]– 15 m/s [N])
= -95 kg•m/s [W]

23. Practice Problem #1
A 2.1 kg barn owl flying at a velocity of 15 m/s [E] strikes head-on with a windshield of a car traveling 30 m/s [W].
b) If the time interval for the impact was 6.7 x s determine the force that acted on the owl

24. Practice problem #1
If the time interval for the impact was 6.7 x 10-3 s determine the force that acted on the owl
Δ 𝑝 = - 95 kg•m/s [W]
Δt = 6.7 x 10-3 s
𝐹 = ?
𝐹 = Δ 𝑝 Δ𝑡
= - 95 kg•m/s [W]
6.7 x 10-3s
= -1.4 x 104 N [W]

25. Practice Problem #1
A 2.1 kg barn owl flying at a velocity of 15 m/s [E] strikes head-on with a windshield of a car traveling 30 m/s [W].
c) Predict the effects of the collision on the owl and on the
car.

26. Practice problem #1
Predict the effects of the collision on the owl and on the car.
The owl will be killed, while the car will only have surface damage.
This is b/c the owl had a smaller velocity and thus a lower momentum.
The owl would apply an insufficient force to the car to change its large momentum

27. Factors affecting change in momentum
By rearranging the force equation, we get
Δ𝑝 = 𝐹 ∆t
The factors that affect change in momentum are
the force exerted on an object,
the time it takes for that change to occur
In the illustration above, because the truck has mass and velocity, it has momentum. The damage of the collision is dependent on the amount of force and the time of the impact.

28. Try These!!!
Practice Problems
5 & 6 (pg 247)
7 & 8 (pg 251)

29. Protective sports equipment
Think back to the sports example:
what made hockey the most dangerous of the three sports was the large change in momentum of the hockey puck, compared to a smaller change in momentum for either the soccer ball or the tennis ball.

30. Protective sports equipment
a collision refers to the impact of one object with another (in this example, the collision of the puck with the hockey player)
the greater the change in momentum, the more serious the damage of the collision

31. Vehicular accident applications
Since change in momentum is affected by force and time, one of those two things has to be reduced in order to make cars better able to handle collisions
changing the force:
the only way to reduce force in a collision is for the driver to travel at a slower velocity – this is not in the manufacturer’s control
𝐹 = m 𝑎
∴ if 𝑎 (change in 𝑣 ) goes down, 𝐹 goes down proportionally.
changing the time:
automotive manufacturers have created ‘crumple zones’ in vehicles.
crumple zones increase the time it takes for a car to stop when it hits another object, therefore reducing the force and minimizing the damage

32. Assignment:
2.2 Section Questions
2-6 (pg 251)

33. B2.3 Impulse
Chapter B2

34. Impulse Therefore I = 𝐹 ∆t too
Change in momentum is such an important concept, it receives its own name: impulse (I)
I = Δ𝑝
Therefore I = 𝐹 ∆t too
where:
I = impulse (N•s )
𝐹 = Force ( N)
∆t = time ( s)
Notice, impulse and the change in momentum are exactly the same formula, but they have different units!
𝛥𝑝 = 𝐹 ∆t

35. Practice example #1
A raw egg drops to the floor. If the floor exerts a force of 9.0 N over a time interval of s, determine the impulse required to change the egg’s momentum.
I = 𝐹 ∆t
= (9.0N)(0.030s)
= 0.27 N•s
F = 9.0 N
∆t = s
I = ?

36. Impulse- I = ∆ 𝒑 I = ∆ 𝒑 and I = 𝐹 ∆t…
BUT change in momentum can also be found using: Δ 𝑝 = mΔ 𝑣 which means I = mΔ 𝑣
This can also be expanded out as: I = m 𝑣 f - m 𝑣 i
where:
I = impulse (N•s )
m = mass ( kg)
𝑣 = velocity (final & initial) ( m/s)
So even if you are not given a force and time, you can still
calculate the impulse of a collision by using the mass and velocity,
Always collect your variables first, then use them to figure out which formula best suits the situation.

37. Practice example #2
A raw egg with a mass of kg falls to the floor. At the moment the egg strikes the floor, it is traveling 4.2 m/s. Assuming that the final velocity of the egg is zero after impact, determine the impulse required to change the momentum of the egg.
I = m 𝑣 f - m 𝑣 i
= (0.065 kg)(0 m/s) - (0.065 kg)(4.2 m/s)
= – 0.27 Ns
m= kg
𝑣 i= 4.2m/s
𝑣 f= 0 m/s
I = ?

38. Large force vs. large time interval
Think about throwing a baseball straight up into the air, and then catching it in your hands.
If you hold your hands ‘rigid’ when the ball hits your hand it will stop almost instantly (small Δt), but will strike with greater force (large F ), hurting your hand.
If you let your hands ‘give’ when the ball hits your hands it will take stop over a greater time period (large Δt), but will strike with smaller force, and will not hurt your hand.
The impulse to stop the baseball in both cases is the same, however by increasing the time period of the change in momentum, we therefore decrease this force.

39. Large force vs. large time interval
Recall the illustration of the truck.
In both illustrations below, the truck’s momentum is the same.
This means that if one of the factors (force or time) is increased, the other factor has to be decreased.
Observe the consequences on the vehicle.
Large force,
small time interval
Small force,
large time interval

40. Try These!!!
Practice
Problems
9 & 11 (pg 254)
13 (pg 255)

41. Assignment:
2.3 Section Questions 2-5, 7, 8 (pg 256)

42. Homework Check-up
Mini-Formative Quiz

43. Impulse Lab

44. Applications to car accidents
Making a car have ‘give’ (i.e. a crumple zone),can increase the time it takes for a car to stop, thus decreasing the force exerted on both the car its passengers.

45. Part 1: How crumple zones work (in comp. lab)
Purpose: The objective of this lab is to calculate the force exerted on a car during a collision, and compare it to a car that is not designed with a crumple zone.
Part 2: Breaking an egg (outside)
Purpose: compare the force required to break an egg against a hard surface with no “give” and a surface with more “give”.

46. B2.4 -Newton’s Third Law and Collisions
Chapter B2

47. BBC Horizon - Surviving a Car Crash

48. Collisions
Collisions involving a vehicle include three classes of collisions:
Primary Collision: Vehicle collides with another object
Secondary Collision: Occupant collides with interior of vehicle
Tertiary Collision: Organs of occupant collide within the body

49. Practice Problem #1
The technologies identified in the following table are designed to reduce injury in a motor vehicle collision. Identify the class for which each technology is designed.
Safety Technology
Class of collision
Shock-absorbing bumpers
Crumple zones
Padded dashboard, steering wheel, etc.
Seat belts
Air bags
1st
2nd

50. Collisions
Think about what would happen if a car driving down the street loses control and hits a pedestrian.
The car will apply an enormous force on the pedestrian, causing injury
The person will also apply a force on the car, causing some minor front end damage to the vehicle (denting the hood or windshield)

51. Conservation of momentum
This concept is summarized in Newton’s third law of motion:
whenever one object exerts a force on another object, the second object exerts an equal but opposite force on the first object.
You may have heard this before: for every action, there is an equal and opposite reaction
This can be shown by the equation:
𝐹 1 on 2 = – 𝐹 2 on 1
* Note the negative. This indicates that although the 𝑭 values are the same, they are in the OPPOSITE direction from one another.

52. Practice Problem #2
Two people on skates, hold two ends of a rope. Identify the action and reaction forces when the person on the right pulls on the rope.
Freaction
Faction
Direction of motion

53. Practice Problem #3
Identify the action force and reaction force in the following situations:
Ex.
Person flies forward off of skateboard
Skateboard flies backwards out from under person

54. a passenger colliding with a deployed air bag
Action
Reaction
Force of airbag on person.
Force of person on airbag.

55. a car bumper colliding with a concrete barrier
Action
Reaction
Force of bumper on concrete.
Force of concrete barrier on bumper

56. a hiker pushes against the ground
Action
Reaction
Force of hiker pushing against the ground
Force of ground holding on hiker.

57. while moving forward, a car tire pushes against the ground
Action
Reaction
Force of tire on the ground.
Force of ground on tire.

58. Try These!!!
Complete Practice Problem #4
in your Notes

59. Practice Problem #4
In an interaction between a large vehicle and a smaller vehicle, state whether or not the following quantities must be the same for each vehicle.
Circle the correct response, and then justify your answers.
Mass: Same Different
Velocity: Same Different
Acceleration: Same Different
Time: Same Different
Change in Momentum: Same Different
Force: Same Different
Impulse: Same Different

60. Practice Problem #4
In an interaction between a large vehicle and a smaller vehicle, state whether or not the following quantities must be the same for each vehicle.
Mass: Same Different
B/c the vehicles are different, they will have differing masses
Velocity: Same Different
B/c each vehicle is travelling independently, their 𝒗 will not be the same.
Acceleration: Same Different
B/c each vehicle has a different 𝒗 𝐢 and 𝒗 f their acceleration will be different.
Time: Same Different
time must be the same because the forces can only act when the vehicles are touching, thus making the time interval equal.

61. Practice Problem #4
In an interaction between a large vehicle and a smaller vehicle, state whether or not the following quantities must be the same for each vehicle.
Force: Same Different
B/c 𝑭 1 on 2 = – 𝑭 2 on 1
Change in Momentum: Same Different
B/c if 𝐹 (force) and Δt (time), then change in momentum must be the same b/c 𝜟𝒑 = 𝑭 ∆t
Impulse: Same Different
B/c I = 𝜟𝒑 , and if change in momentum is equal, so is impulse.

62. B2.5- Conservation of Momentum
Chapter B2

63. Law of Conservation of Momentum
“The total momentum gained by one object is equal to the total momentum lost by the other”
Just like the “Law of Conservation of Matter” OR “Law of Conservation of Energy”
Represented by the formula:
Σp(before) = Σp(after)

64. Momentum is conserved Σ = “sum of” (to add up)
Σ 𝒑 (before) = Σ 𝒑 (after)
Where:
Σ = “sum of” (to add up)
𝑝 = momentum ( 𝑝 = m 𝑣 )
So the equation can be expanded out to:
m1 𝒗 1 + m2 𝒗 𝟐= m1 𝒗 ’ 1+ m2 𝒗 ’2
m = mass (you can assume the mass of the vehicles doesn’t change, so there’s no need to distinguish between m and m’)
𝒗 = velocity (we use 𝒗 1 and 𝒗 2 to represent the speed of each vehicle before the collision)
the prime (‘) symbol indicates conditions AFTER the collisions; 𝒗 ’1 and 𝒗 ’2 represent the speed of the vehicles after the collision)

65. There are three types of collisions:
Hit and Rebound
Hit and Stick
Explosions

66. Hit and Rebound
the two vehicles collide and bounce back, often in the opposite direction
in these questions you will be given three of the four velocities (v1, v2, v1’, and v2’) and you will be asked to find the fourth, using this formula:
m1 𝒗 1 + m2 𝒗 2 = m1 𝒗 1’ + m2 𝒗 2’

67. Hit and Rebound – Example #1
A 3.0kg ball strikes a stationary ball with a mass of 6.0kg. The first ball is moves at 1.5m/s until the collision, but bounces back going -0.50m/s. Determine the velocity of the second ball after the collision.
m1 =
m2 =
𝒗 1 =
𝒗 2 =
𝒗 1’ =
𝒗 2’ =
3.0 kg
6.0 kg
1.5 m/s
0 m/s (stationary)
-0.50 m/s ?
mass of moving ball
mass of stationary ball
speed of 3.0kg ball before collision
speed of 6.0kg ball before collision
speed of 3.0kg ball after collision
speed of 6.0kg ball after collision

68. Hit and Rebound – Example #1
A 3.0kg ball strikes a stationary ball with a mass of 6.0kg. The first ball is moves at 1.5m/s until the collision, but bounces back going -0.50m/s. Determine the velocity of the second ball after the collision.
Σp = Σp’
m1 𝒗 1 +m2 𝒗 2 =
[m1 𝒗 1 +m2 𝒗 2 ] - (m1 𝒗 1’ ) = 𝒗 2’
[m1 𝒗 1 +m2 𝒗 2 ] - (m1 𝒗 1’ ) = 𝑣 2’ m2
- (m1 𝒗 1’ )
m1 𝒗 1’ +
m2 𝒗 2’
- (m1 𝒗 1’ ) m2 m2 m2

69. Hit and Rebound – Example #1
A 3.0kg ball strikes a stationary ball with a mass of 6.0kg. The first ball is moves at 1.5m/s until the collision, but bounces back going -0.50m/s. Determine the velocity of the second ball after the collision.
[m1 𝒗 1 +m2 𝒗 2 ] - (m1 𝒗 1’ ) = 𝑣 2’ m2
[(3.0kg)(1.5m/s) + (6.0kg)(0m/s)] – (3.0kg)(-0.50m/s) = 𝒗 2’
6.0kg
(4.5 kg•m/s) –(1.5 kg•m/s) = 𝒗 2’
1.0m/s = 𝒗 2’

70. Hit and Stick m1 𝒗 1 + m2 𝒗 2 = (m1 + m2) 𝒗 ’
the two vehicles collide, then travel together as one combined mass
because the two vehicles are traveling as one unit after the collision, v1’ and v2’ can be simplified to v’
the formula for hit and stick collisions is therefore:
m1 𝒗 1 + m2 𝒗 2 = (m1 + m2) 𝒗 ’

71. Hit and Stick – Example #2
A 1000-kg car is traveling at 12m/s when it strikes a 2000-kg truck that is stopped at a red light. After the collision, the two vehicles stick together. What is the speed of the two vehicles after the collision?
m1 =
m2 =
𝒗 1 =
𝒗 2 =
𝒗 ’ =
mass of car
mass of truck
speed of car before collision
speed of truck before collision
speed of both vehicles together after collision
1000 kg
2000 kg
12 m/s
0 m/s (stopped) ?

72. Hit and Stick – Example #2
A 1000-kg car is traveling at 12m/s when it strikes a 2000-kg truck that is stopped at a red light. After the collision, the two vehicles stick together. What is the speed of the two vehicles after the collision?
Σp = Σp’
m1 𝑣 1 +m2 𝑣 2 = 𝒗 ’
m1 𝑣 1 +m2 𝑣 2 = 𝑣 ’
(m1+m2)
(m1+m2)
(m1+m2)
(m1+m2)

73. Hit and Stick – Example #2
A 1000-kg car is traveling at 12m/s when it strikes a 2000-kg truck that is stopped at a red light. After the collision, the two vehicles stick together. What is the speed of the two vehicles after the collision?
m1 𝑣 1 +m2 𝑣 2 = 𝑣 ’
(m1+m2)
(1000kg)(12 m/s) + (2000kg)(0m/s) = 𝑣 ’
(3000kg)
4.0m/s = 𝑣 ’

74. (m1and2 )( 𝒗 1and2)= m1 𝒗 1’ + m2 𝒗 2’
Explosion
in an explosion, the mass starts as one contained object (m1 + m2) at rest ( 𝒗 1 and 2 = 0 m/s)
after the explosion, the solid object has broken up into two separate masses (m1 and m2), each with their own separate momentum
the formula for explosions is
(m1and2 )( 𝒗 1and2)= m1 𝒗 1’ + m2 𝒗 2’

75. Explosion – Example #3
A 50-g firecracker at rest explodes into two pieces, a 15-g piece which flies to the right at a velocity of 3.50m/s, and a 35-g piece. Determine the velocity and direction of the 35-g piece.
m1and2 =
m1 =
m2 =
𝒗 1and2 =
𝒗 1’ =
𝒗 2’ =
50 g *you can leave in grams as long as all your masses are in grams
15 g
35 g
0 m/s (at rest)
3.50 m/s
? *will be a negative number, meaning it flies to the left
mass of firecracker
mass of first piece
mass of second piece
speed of firecracker before explosion
speed of first piece after collision

76. Explosion – Example #3 Σp = Σp’ (m1and2 ) = + m2 𝒗 2’
A 50-g firecracker at rest explodes into two pieces, a 15-g piece which flies to the right at a velocity of 3.50m/s, and a 35-g piece. Determine the velocity and direction of the 35-g piece.
Σp = Σp’
(m1and2 ) = m2 𝒗 2’
-(m1 𝒗 1’) = 𝒗 2’
-(m1 𝒗 1’) = 𝒗 2 m2
- m1 𝒗 1’
0 m/s
( 𝒗 1and2)
m1 𝒗 1’
- m1 𝒗 1’ m2 m2 m2

77. Explosion – Example #3
A 50-g firecracker at rest explodes into two pieces, a 15-g piece which flies to the right at a velocity of 3.50m/s, and a 35-g piece. Determine the velocity and direction of the 35-g piece.
-(m1 𝒗 1’) = 𝒗 2’ m2
- (15 g)(3.50m/s) = 𝒗 2’
35 g
- 1.5 m/s [left] = 𝒗 2’

78. Assignment:
Complete the Ch 2.4 & 2.5 Assignment: Newton’s Third Law & Collisions

79. B2.6-Work and Energy
Chapter B2

80. Recall from Science 10: Condition #1:
Work is the energy transferred to an object, when a force is applied, that causes the object to move
in order for work to be done, there are three conditions that must be satisfied:
Condition #1:
there must be a force applied
a push or pull must be applied to the object
an object that is coasting is not having work done it

81. Recall from Science 10: Condition #2:
Work is the energy transferred to an object, when a force is applied, that causes the object to move
in order for work to be done, there are three conditions that must be satisfied:
Condition #2:
there must be movement
if the force is not large enough to change the object’s motion, there is no work being done

82. Recall from Science 10: Condition #3:
Work is the energy transferred to an object, when a force is applied, that causes the object to move
in order for work to be done, there are three conditions that must be satisfied:
Condition #3:
the direction of the force and the direction of movement must be the same
the movement has to be as a result of the force, so they must be in the same direction

83. Energy Energy is the ability to do work Types of Energy:
kinetic (movement)
thermal (heat)
solar / light
electrical (movement of electrons)
potential (energy stored in readiness)
chemical (stored in chemical bonds)
gravitational (stored in an object that can fall)
elastic (as in a stretched elastic or compressed spring)
Energy is the ability to do work
when a certain amount of work is done on an object, that object “gains” that much energy

84. Mechanical Energy Em = Ep + Ek
Ep(grav) = Gravitational Potential Energy
if an object is lifted up, the energy acquired is gravitational potential energy
energy due to the position of an object above the Earth’s surface
gravitational potential energy is energy stored in an object that has the potential to fall:
Ep (grav) = mgh
where
Ep (grav) = Gravitational potential energy (J)
m = mass (kg)
g = acceleration due to gravity (-9.81m/s2)
h = height (m)
Gravitational potential energy is affected proportionally, by mass and height.

85. Mechanical energy Em = Ep + Ek
Ek= Kinetic Energy
if an object is moved, the energy acquired, due to the work done, is kinetic energy.
Kinetic energy is energy of movement
Ek = ½ mv2
where
EK = Kintic energy (J)
m = mass (kg)
v = velocity (m/s)
Kinetic energy is affected proportionally by mass and exponentially by the speed at which it is traveling

86. Conservation of Energy
in a closed system; where no work or energy is added/ subtracted, the Law of Conservation of Energy applies.
this means that when energy changes from one form to another, the total energy of the system remains constant
in the case of an object that is rising straight up in the air, or falling, there is a transfer between potential and kinetic energy

87. Conservation of Energy
When the object is at maximum height A), it has maximum potential energy
When the object is just about to hit the ground, (or just as it leaves the ground) C), it has maximum kinetic energy
Anywhere in the middle of the journey there is a combination of potential and kinetic energy
The sum of the energies at any point in the falling objects journey will remain equal.

88. Example:
At the TOP of the diving board, the diver has maximum gravitational potential energy, but as he is not falling yet, NO kinetic energy
As the diver FALLS, his height decreases, so as his potential energy decreases
BUT as he falls, his velocity increases (because gravity causes objects to accelerate), so his kinetic energy increases.
The moment the diver strikes the water’s surface, he is at “ground level” (h = zero) so he has NO potential energy. However, he has reached his maximum velocity, so he has maximum kinetic energy.

89. Questions involving a falling or rising object
Because energy is conserved, in a question involving a falling or rising object, the two formulas (Ep and Ek) can be set equal to each other
Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = ½ v2
 notice mass is the same, as it is the same object, and is present on both sides of the equation, so it will cancel out

90. Practice Problem #1 2(9.81m/s2)(10m) = v
If the diving platform is 10m high, what will be the diver’s velocity when he reaches the water?
h = 10 m
g = 9.81m/s2
v = ?
Ep(top) = Ek(bottom)
mgh = ½ mv2
gh = v2
2gh = v2
 2gh = v
2(9.81m/s2)(10m) = v
* we can make g positive because the height refers to a negative change in position (down) when the object falls, so the two negatives cancel out
10 m
(2x)
(x2)  
14m/s = v

91. Pendulums another example of a conversion between Ep and Ek
at position A, the pendulum is not moving, but is at a maximum height  maximum Ep, minimum Ek
at position B, the pendulum is moving at a maximum speed, but is the closest to the ground  minimum Ep, maximum Ek
at position C, the pendulum is at the same position as position A

92. Assignment: Prepare for Unit Exam
Complete Practice Problem 23 (pg 277) Read
Use Example Problem 2.13 (pg ) as a guide compare the answers you get to the answers for Example Problem 2.13, explain the condition that resulted in a difference in the values.
2.5 Section Questions 2-6 (pg 281)
Prepare for
Unit Exam