Download presentation

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

7
Example A high school athlete runs 100 m south in s. What is the velocity in m/s and km/hr? Science 20 Unit 2

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. Science 20 Unit 2

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? Science 20 Unit 2

10
Pg 181 #15 Pg 184 # 17 Pg 185 # 2-5 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

13
Work Pg 193 #3&4 Science 20 Unit 2

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 Science 20 Unit 2

15
**Acceleration cont’d Units for acceleration – m/s2 Formula is**

Science 20 Unit 2

16
**vi = initial speed (m/s)**

v = vf – vi vf = final speed (m/s) vi = initial speed (m/s) Science 20 Unit 2

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 Science 20 Unit 2

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? Science 20 Unit 2

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? Science 20 Unit 2

20
Work Pg 200 #25 Pg 203 #26 & 28 Science 20 Unit 2

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 Science 20 Unit 2

22
**Lab Continued Graphs to be completed 1. Position vs Time graph**

2. Velocity vs Time graph – calculate slope 3. Acceleration vs Time graph Science 20 Unit 2

23
**Displacement Equation**

Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

26
Work Pg 199 # 24 Pg 203 # 27 Pg 208 # 32 Pg 209 # 33 Science 20 Unit 2

27
**Another Distance (Displacement) Equation**

When final velocity is not given in the original data and acceleration is given use the following formula Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

34
**While stopping (braking): Vi = 31 m/s Vf = 0 a = - 5.85 m/s**

Science 20 Unit 2

35
Science 20 Unit 2

36
Work P. 213 #3, 4 Pg 216 # 38 Pg 218 # 39-40 Pg 220 #1 & 4 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

42
**Determine the length of time the net force must be applied to stop the vehicle.**

Science 20 Unit 2

43
**Work Pg. 222 #43 Pg 226 # 45 Pg 227 # 1-4 (copy questions)**

Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

54
Rearranged formulas: m = p v v = p m Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

57
**Work Read Pg 244 – 245 P. 244 #1-3 Pg 245 # 2 – 6 (copy questions)**

Science 20 Unit 2

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) Science 20 Unit 2

59
**F = change in momentum Or**

time Ft = m( vf – vi) Ft = Impulse m( vf – vi) = Change in momentum Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

62
**Read Pg 249 – 250 Discuss results on Pg 250 Do questions Pg 247 # 4 –6**

Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

66
**Work Pg 254 # 9-12 Read Pg 254 Pg 255 # 13-14 Pg 256 # 1,3,5,7,8,10-12**

Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

69
**Computer Work Pg 260 – 261 Pg 262 # 1-5 Computer Lab Pg 265-267**

Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

76
Work Pg 270 # 17-19 Pg 271 # 20 Pg 271 # 1-8 Science 20 Unit 2

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 Science 20 Unit 2

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. Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

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 = Science 20 Unit 2

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. Science 20 Unit 2

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 Science 20 Unit 2

84
**C) Calculate the speed of the egg just before impact. Ek = ½ mv2 v = **

0.085kg = 2.8 m/s Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

87
**F) Calculate the impulse required to stop the egg during impact. **

impulse = FΔt = Δp = kgm/s Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

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 Science 20 Unit 2

91
Practice Pg 277 # 23 Science 20 Unit 2

92
Do Lab Pg Science 20 Unit 2

93
Work Pg 281 # 2-6 Pg # 1,2,5,7,9,11,13,15,19,20 Pg # 1,3 – 7, 17, 19 Science 20 Unit 2

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 Science 20 Unit 2

95
Science 20 Unit 2

96
Science 20 Unit 2

97
Science 20 Unit 2

Similar presentations

Presentation is loading. Please wait....

OK

Unit 4 FORCES AND THE LAWS OF MOTION

Unit 4 FORCES AND THE LAWS OF MOTION

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google