1. Unit B Changes in Motion
Chapter 1
Describing Motion
Science 20 Unit 2

2. Average Speed
Average speed is equal to the total distance traveled divided by the total time.
Average speed = total distance
elapsed time
v = d t
Pg 169 # 1-3
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3. Uniform Motion
Uniform motion is motion in a straight line at a constant speed.
Uniform motion is rare
Non-uniform motion is when there is a change in speed (speeding up or slowing down) or a change in direction.
Instantaneous speed is the speed at any one point in time
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4. Scalar Quantity
Scalar quantities consist of magnitude only and no indication of direction.
Speed, time and volume are scalar quantities
Pg #6-8
Pg 173 # 2-4
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5. Velocity
Position is a vector quantity describing the location of a point relative to a reference point
Vector quantity is a quantity consisting of magnitude and direction
Sign convention – north is positive, to the right is positive – south is negative and to the left is negative
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6. Displacement is a vector quantity describing the length and direction in a straight line from the starting position to the final position.
Average velocity is a vector quantity describing the change in position over a specified time
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7. Example
A high school athlete runs 100 m south in s. What is the velocity in m/s and km/hr?
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8. Scale Diagrams
Resultant displacement is the vector sum of individual displacements.
Head-to-tail method: a method where the tail of a succeeding vector arrow begins at the head of the preceding vector arrow.
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9. Draw a vector diagram to represent the following
Draw a vector diagram to represent the following. A person walks 300 m south and then turns around and walks 150 m north. What is the persons displacement?
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10. Pg 181 #15
Pg 184 # 17
Pg 185 # 2-5
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11. Lab Activity
Using a ticker tape timer pull a dynamics cart at constant speed.
Mark ‘0’time when uniform motion starts – dots are equally spaced.
Count every five dots and mark ticker tape
Measure and chart data from the ticker tape
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12. Graph a Position vs Time Graph, calculate the slope of the line.
Graph a Distance vs Time Graph
Graph a Velocity vs Time Graph, calculate displacement at time = to 5 tocks
Hand in Chart and three graphs
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13. Work
Pg 193 #3&4
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14. Acceleration
Acceleration is a change in velocity during a time interval(speeding up or slowing down)
Acceleration is a vector quantity
A force is required to change motion in some way
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15. Acceleration cont’d Units for acceleration – m/s2 Formula is
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16. vi = initial speed (m/s)
v = vf – vi
vf = final speed (m/s)
vi = initial speed (m/s)
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17. Example
A car traveling a 50 m/s speeds up to 95 m/s over 6 s. What is the car’s acceleration?
vi = 50 m/s
vf = 95 m/s
t = 6 s
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18. Examples
The velocity of a car increases from 2 m/s at 1.0s to 16 m/s at 4.5 s. What is the car’s average acceleration?
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19. Rearrange Formula vf = vi + at
Use this formula to find final speed when an object is accelerating
Example – If a car with a velocity of 2.0 m/s at time zero, accelerates at a rate of +4.0 m/s for 2.5 s, what is its velocity at the end of its acceleration?
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20. Work
Pg 200 #25
Pg 203 #26 & 28
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21. Acceleration Lab
Using the ticker tape timer drop an object from the top of the stair well.
Chart data from ticker tape
1. Time
2. Position
3. Velocity
4. Acceleration
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22. Lab Continued Graphs to be completed 1. Position vs Time graph
2. Velocity vs Time graph – calculate slope
3. Acceleration vs Time graph
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23. Displacement Equation
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24. Acceleration Due to Gravity
Acceleration due to gravity is 9.81 m/s2
Gravitational acceleration will act on any object moving up or down in the atmosphere
Example jumping, throwing a ball up, falling off a building etc
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25. Example
A rock is thrown straight up in the air. It reaches a height of 18.6 m in 2.1 s. Calculate the initial velocity
t = 2.1 s
d = 18.6 m
a =
Vf = 0
Vi = ?
Vi = vf - at
= 0 – (-9.81)(2.1)
Vi = 20.6 m/s
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26. Work
Pg 199 # 24
Pg 203 # 27
Pg 208 # 32
Pg 209 # 33
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27. Another Distance (Displacement) Equation
When final velocity is not given in the original data and acceleration is given use the following formula
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28. Example 1
A boy leaves the surface of a trampoline with an initial velocity of 11.8 m/s, straight up. Determine the displacement after 0.8 s.
vi = 11.8 m/s
t = 0.8 s
a = m/s2
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29. Example 2
A diver steps off the ledge of a platform and enters the water 5.0 m below. If the initial velocity of the diver was zero, determine the time it took the diver to reach the water.
d = -5.0 m
a = m/s2
vi = 0
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30. Reaction Distance
Reaction time is critical in the stopping of a vehicle.
Includes – the time it takes the drivers brain to recognize there is a need to stop and the time it takes the driver’s foot to move from the gas pedal to the brake pedal.
Reaction time varies from person to person
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31. Stopping Distance = Reaction Distance + Braking Distance
Reaction distance is the distance the vehicle travels while the driver is reacting.
Braking distance is the distance a vehicle travels from the moment the brakes are first applied to the time the vehicle stops.
Stopping Distance = Reaction Distance + Braking Distance
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32. Example
The typical reaction time for most drivers is considered to be about 1.50 s. This includes the time required to identify the danger (0.75 s) and the time required to react to the danger (0.75 s) The ability of vehicles to decelerate varies greatly however. Traffic safety engineers often use a deceleration value of 5.85m/s2 to calculate the minimum stopping distance for a vehicle on smooth, dry pavement.
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33. Determine the distance traveled while reacting, the distance traveled while braking and the minimum stopping distance of a vehicle traveling 110 km/h.
While reacting
d = vt = (31)(1.50) = 46.5 m
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34. While stopping (braking): Vi = 31 m/s Vf = 0 a = - 5.85 m/s
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35. Science 20 Unit 2

36. Work
P. 213 #3, 4
Pg 216 # 38
Pg 218 # 39-40
Pg 220 #1 & 4
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37. Braking
Force of friction is contact between two surfaces that acts to oppose the motion of one surface past the other.
Friction is a force
All forces are a push or pull.
Forces are measured in Newtons – N.
Brakes – particularly brake pads and rotors – are designed to produce additional friction between the rotating wheels and the fixed frame of the vehicle
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38. Net force – is the vector sum of all forces acting on an object.
In the case of braking the net force includes
1. Force of air resistance
2. Force of road resistance
3. Force applied by the braking system
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39. Larger trucks require a much greater stopping distance
Another factor that affects the rate of deceleration is the mass of the vehicle.
Larger trucks require a much greater stopping distance
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40. Newton’s Second Law of Motion
Newton’s Second Law of Motion states that an object will accelerate in the direction of the net force applied.
Fnet = ma
Units are N = kg m/s2
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41. Example
A vehicle with a mass of 1250 kg is traveling 45 km/h east, when the driver engages the brakes to stop at an intersection.If the net force on the vehicle is 7000 N west, determine the magnitude and direction of the deceleration of the vehicle while the net force is applied.
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42. Determine the length of time the net force must be applied to stop the vehicle.
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43. Work Pg. 222 #43 Pg 226 # 45 Pg 227 # 1-4 (copy questions)
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44. Speeding Up
Newton’s second law also explains what happens when a vehicle increases its velocity.
The additional force required to make the vehicle move faster is called the applied force.
The net force results from the vector sum of the applied force and the force of friction
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45. Science 20 Unit 2

46. Newton’s First Law of Motion (Inertia)
This law states that in the absence of a net force, an object in motion will tend to maintain its velocity, and an object at rest will tend to remain at rest.
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47. Example - One
The engine of a motorcycle supplies an applied force of 1880 N, west, to overcome frictional forces of 520 N, east. The motorcycle and rider have a combined mass of 245 kg. Determine the acceleration.
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48. Science 20 Unit 2

49. Example - Two
A car with a mass of 1075 kg is traveling on a highway, The engine of the supplies an applied force of 4800 N, west, to overcome frictional forces of 4800 N, east. Determine the acceleration.
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50. Science 20 Unit 2

51. Inertia
Inertia is the property of an object to resist changes in its state of motion.
Example an object at rest will remain at rest or an object in motion will remain in motion
The amount of inertia an object has depends upon its mass. The greater the mass the greater the inertia.
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52. Try to push a small car stopped on the road or a transport truck.
Work Pg 233 # 47-49
Work Pg 235 # 1-13, 16-20
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53. Momentum (p)
Momentum is the product of an objects mass and its velocity.
p = mv p – momentum(kg m/s)
m – mass (kg)
v – velocity (m/s)
Momentum is a vector quantity.
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54. Rearranged formulas:
m = p
v v = p m
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55. Example 1
Determine the momentum of a vehicle with a mass of 2100 kg moving at a velocity of 22 m/s (E).
m = 2100 kg
v = 22 m/s
p = ? p = mv = (2100)(22)
p = 46,200 kg m/s
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56. Example 2
An airplane has a momentum of 8.7 x 107 kg m/s . If the airplane is flying at a velocity of 990 km/h, determine its mass.
p = 8.7 x 107 kg
v = 990 km/h = 275 m/s
m = ? m = p/v
m = (8.7 x 107)/(275)
m = 3.2 x 105 kg
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57. Work Read Pg 244 – 245 P. 244 #1-3 Pg 245 # 2 – 6 (copy questions)
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58. Newton vs Momentum
Newton’s second law can show that a net force will cause a mass to accelerate in the direction of the applied force.
F = ma
a = vf – vi t
Therefore F = m( vf – vi)
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59. F = change in momentum Or
time
Ft = m( vf – vi)
Ft = Impulse
m( vf – vi) = Change in momentum
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60. Example
A 2.1 kg barn owl flying at a velocity of 15 m/s (E) strikes head-on with the windshield of a car traveling 30 km/h (W).
If the time interval for the impact was 6.7 x 10-3 s, determine the force that acted on the owl.
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61. m = 2.1 kg vi = 15 m/s vf = - 30 km/h = - 8.3 m/s t = 6.7 x 10-3 s
F = ? F = m(vf – vi) t
F = 2.1(-8.3 – 15)
6.7 x 10-3
F = N
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62. Read Pg 249 – 250 Discuss results on Pg 250 Do questions Pg 247 # 4 –6
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63. Impulse
Impulse is the product of the net force applied to an object and the time interval during which the force is applied.
Impulse does not have its own symbol.
It is represented by FΔt and has the units kgm/s
When it comes to roadside safety some types of barriers are less damaging to vehicles and their occupants.
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64. Remember that FΔt = Δp Example
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.
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65. Example 2
A raw egg with a mass of kg falls to the floor. At the moment the egg strikes the floor, it is travelling 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.
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66. Work Pg 254 # 9-12 Read Pg 254 Pg 255 # 13-14 Pg 256 # 1,3,5,7,8,10-12
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67. Collisions There are 3 classes of collisions
1. Primary collision – the vehicle colliding with another object, such as another vehicle
2. Secondary collision – the occupant colliding with the interior of the vehicle
3. Tertiary collision – the occupant’s internal organs colliding within the occupant’s body
Pg 257
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68. Newton’s Third Law
Newton’s Third Law states that whenever one object exerts a force on a second object, the second object exerts an equal but opposite force on the first object.
F1on 2 = F 2on1
Forces occur in pairs- action vs reaction
Forces are the same magnitude but push in opposite directions
Forces push on different objects.
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69. Computer Work Pg 260 – 261 Pg 262 # 1-5 Computer Lab Pg 265-267
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70. Momentum is Conserved
Collisions observed in the activity were one of three kinds: hit and stick, hit and rebound, or explosion.
In all cases the momentum of one object was transferred to another such that the total momentum always remained the same.
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71. Law of Conservation of momentum – if the net force acting on a system is zero, the sum of the momentum before an interaction equals the sum of the momentum after the interaction.
Σ p before = Σ p after
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72. Example 1
A kg freight car traveling west with a velocity of 1.5 m/s collides with a
kg freight car at rest. After the collision, the freight cars stick together. Determine the velocity of the fright cars after the collision.
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73. Example 2
A 3.0 kg ball rolling east with a velocity of 1.5 m/s collides with another 6.0 kg ball at rest. After the collision, the first ball rebounds and is traveling at a velocity of 0.50 m/s west.
A) Determine the velocity of the second ball after the collision
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74. B) Determine the momentum values of the balls before and after the collision
C) Use scale diagrams of momentum vectors to demonstrate that momentum is conserved in this collision.
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75. Example 3
A kg firecracker at rest explodes into two pieces. If a kg piece flies off to the right at a velocity of 3.00 m/s, determine the velocity of the other kg piece.
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76. Work
Pg 270 # 17-19
Pg 271 # 20
Pg 271 # 1-8
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77. Designing a Helmet
You are to design a helmet for an egg. An egg is much like your head, it has a thin hard shell protecting a soft inside. Your skull is a hard shell covering soft brain tissue.
Your design must address three characteristics of helmets
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78. 2. The egg test dummy must have no covering over its eyes and ears.
1. The egg test dummy must be protected from a frontal collision against a rigid barrier
2. The egg test dummy must have no covering over its eyes and ears.
3. The egg test dummy’s helmet must continue to provide protection after a number of severe frontal impacts.
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79. Canadian Standards Association (CSA) tests all helmets to ensure that they will remain on the head during an impact and will provide sufficient protection.
To test your helmet the egg test dummy with its helmet on will be placed in a plastic bag and suspended at the end of a long string and then swung till it hits a solid barrier.
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80. When the egg is pulled back prior to release the work done to change the position of the egg is equal to the gravitational potential energy of the egg.
Ep(grav) = W
= FΔd
= mgh
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81. Then according to the Law of Conservation of Energy, the gravitational potential energy should be converted to kinetic energy as it swings toward the barrier
Ek = ½ mv2
v =
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82. Example
The mass of the egg, plastic bag, helmet, and paper clip is kg. If they are pulled back such that they are now 0.40 m higher than they were at rest and then released to swing forward and hit a solid barrier.
A) Calculate the gravitational potential energy of the egg when it is 0.40 m above the resting point.
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83. Eg = mgh = (0.085 kg)(9.81m/s2)(0.40 m) = 0.33 J
B) Determine the kinetic energy of the egg just before it hits the barrier
Ek = Eg
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84. C) Calculate the speed of the egg just before impact. Ek = ½ mv2 v =
0.085kg
= 2.8 m/s
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85. The forward motion will be positive p = mv = (0.085 kg)(2.8 m/s)
D) Calculate the magnitude and direction of the momentum of the egg just before impact.
The forward motion will be positive
p = mv
= (0.085 kg)(2.8 m/s)
= 0.24 kgm/s
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86. E) Assuming the egg stops immediately upon impact, calculate the change in momentum upon impact.
Δp = pf – pi (pf is zero)
= - pi
= kgm/s
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87. F) Calculate the impulse required to stop the egg during impact.
impulse = FΔt
= Δp
= kgm/s
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88. G) If the impact lasts for 0
G) If the impact lasts for s, determine the force that acted upon the egg during the collision using the equation for impulse.
FΔt = Δp
F = Δp Δt
= kgm/s = N
0.040 s
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89. H) Determine the acceleration of the egg over the 0.040 s a = vf – vi
= 2.8 m/s
0.040 s
= - 70 m/s2
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90. I) Use Newton’s second law to confirm your answer to question g by calculating the force that acted on the egg.
F = ma
= (0.085 kg)(- 70 m/s2)
= N
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91. Practice
Pg 277 # 23
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92. Do Lab Pg
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93. Work
Pg 281 # 2-6
Pg # 1,2,5,7,9,11,13,15,19,20
Pg # 1,3 – 7, 17, 19
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94. Name Symbol Unit Formula Momentum p kgm/s p=mv Δmomentum Δp p1 - p2
Impulse Ns
FΔt = pf-pi
Newtons
F1on2 =F2on1 N
F1on2 = F2on1
Conservatio
Σp = Σp
Σpbefore = Σpafter
Work W J
W = Fd
Kinetic Energy KE
E = 1/2mv2
Gravitational Eg
E = mgh
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95. Science 20 Unit 2

96. Science 20 Unit 2

97. Science 20 Unit 2