1. VCE Physics Unit 2 Topic 1 Movement View physics as a system of thinking about the world rather than information that can be dumped into your brain without integrating it into your own belief systems.

2. Unit Outline To achieve this outcome the student should use scientific methods, data, theories and knowledge to: Describe non-uniform and uniform motion along a straight line graphically; Analyse motion along a straight line graphically, numerically and algebraically; Describe how changes in movement are caused by the actions of forces; Model forces as external actions through the centre of mass point of each body; Explain movement in terms of the Newtonian model and some of its assumptions, including Newton’s 3 laws of motion, forces act on point particles, and the ideal, frictionless world. Compare the accounts of the action of forces by Aristotle, Galileo and Newton. Apply the vector model of forces including vector addition, vector subtraction and components to readily observable forces including weight, friction and reaction forces; Model mathematically work as force multiplied by distance for a constant force and as area under the force versus distance graph. Interpret energy transfers and transformations using an energy conservation model applied to ideas of work, energy and power, including transfers between –kinetic energy and gravitational potential energy close to the Earth’s surface; –potential energy and kinetic energy in springs;

3. Chapter 1 Introduction

4. 1.0 An Ideal World In the ideal world the laws of motion apply exactly, eg. objects which are moving will continue to move with the same speed unless or until something occurs to change this. To make life easier for Physics students situations or events which require mathematical analysis are often described as occuring in an ideal, frictionless world. In the ideal world an object under the influence of Earth’s gravity will accelerate at 9.8 ms -2 throughout its journey never reaching a terminal velocity. In the ideal world energy transformations are always 100% efficient, so that the potential energy of a pendulum at the top of its swing is all converted to Kinetic Energy (motion energy) at the bottom. In the ideal world perpetual motion machines are commonplace.

5. 1.1 The S.I. System In 1960, the “General Conference of Weights and Measures”, a Paris based international organisation, agreed that one set of units would be adopted world wide for the measurement of physical quantities. This system is called the Systeme Internationale d’Units, or more simply the S.I. System. The system is used and recognised worldwide and defines 7 fundamental units. Physical Quantity S.I. Unit Symbol Length metre m Mass kilogram kg Time second s Electric Current ampere A Temperature kelvin K Luminous Intensity candela cd Amount of Substance mole mol A derived unit is the force unit, the Newton, which is found from mass x length x 1/(time) 2 Thus the Newton has dimensions kg x m x s -2 All other units are derived from these 7 fundamentals.

6. 1.2 S.I. Definitions Length; metre [m] It is the distance light travels, in a vacuum, in 1/299,792,458th of a second. Luminous Intensity; candela [cd] It is the intensity of a source of light of a specified frequency, which gives a specified amount of power in a given direction. Amount of Substance; mole [mol] It is the amount of substance that contains as many elementary units as there are atoms in 0.012 kg of 12 C Temperature; kelvin [K] It is 1/273.16th of the thermodynamic temperature of the triple point of water. Current; ampere [A] It is that current which produces a force of 2 x 10 -7 N between two parallel wires which are 1 metre apart in a vacuum. Time; second [s] It is the length of time taken for 9,192,631,770 periods of vibration of the caesium-133 atom to occur. Questions Mass; kilogram [kg] It is the mass of a platinum-iridium cylinder kept at Sevres in France. It is now the only basic unit still defined in terms of a material object.

7. Fundamentals QUESTIONS 1. Which of the following quantities have fundamental units and which have derived ? Quantity (Unit)FundamentalDerived Power (Watts) Distance (metre) Time (second) Force (Newton) Energy (Joule) Mass (kilogram) Electrical Resistance (ohms) Temperature (kelvin) Electric Current (amperes) √ √ √ √ √ √ √ √ √

8. Fundamentals 2. From which of the fundamental units do the following derive their units ? Quantity (Unit)Fundamental Units e.g Force (Newton)Mass (kg), length (m), time (s) Acceleration (ms -2 ) Momentum (kgms -1 ) Impulse (Newton.second) Velocity (ms -1 ) Work (Joule) Note: W = F.d Length (m), time (s) Mass (kg), length (m), time (s) Length (m), time (s) Mass (kg), length (m), time (s)

9. Fundamentals 3. Show that 1 ms -1 = 3.6 kmh -1 1m x 1km x 3600s s 1000m 1h Two relevant conversion factors are: 1 km = 1000 m, 1 h = 3600 s so 1 ms -1 = 3.6 kmh -1 Which ones to use ? Easy, you want to end up with km on the top line and h on the bottom These can be written as 1km 1000m or 1000m 1km and 1h 3600 s 1h or 3.6

10. 1.3 Position In order to specify the position of an object we first need to define an ORIGIN or starting point from which measurements can be taken. For example, on the number line, the point 0 is taken as the origin and all measurements are related to that point. -40 -35 -30 -25 -20 -15 -10 -5 0 5 10 15 20 25 30 35 40 Numbers to the right of zero are labelled positive Numbers to the left of zero are labelled negative A number 40 is 40 units to the right of 0 A number -25 is 25 units to the left of 0 Questions

11. Position 4. What needs to be defined before the position of any object can be specified ? A zero point needs to be defined before the position of an object can be defined 5. (a) What distance has been covered when an object moves from position +150 m to position + 275 m ? Change in position = final position – initial position = +275 – (+150) = + 125 m. Just writing 125 m is OK (b) What distance has been covered when an object moves from position + 10 m to position -133.5 m ? Change in position = final position – initial position = -133.5 – (+10) = - 143.5 m. Negative sign IS required

12. Chapter 2 Vectors & Scalars

13. 2.0 Scalars and Vectors Before proceeding further we need to define two new quantities: SCALAR QUANTITIES These are completely defined by A Number andA Number and A UnitA Unit Examples of scalars are: Temperature 17 0, Mass 1.5 kg VECTOR QUANTITIES These are completely defined by A NumberA Number A Unit andA Unit and A DirectionA Direction Examples of vectors are: Displacement 25 km West, Force 14 Newtons South Vectors are usually represented by an ARROW, with the length of the arrow indicating the size of the quantity and the direction of the arrow the direction of the quantity. This vector represents a Force of 4 N, acting North West N Questions

14. Vectors and Scalars 6. Which of the following quantities are scalars and which vectors ? QuantityUnitScalarVector Distancemetre Momentumkgms -1 East Kinetic Energyjoule Accelerationms -2 N45 o E Gravitational Field StrengthNkg -1 downwards Displacementmetre sideways Ageyears Velocityms -1 West Temperature oCoC √ √ √ √ √ √ √ √ √

15. 2.1 Vector Addition & Subtraction Vectors can be at any angle to one another and still be added. This can be done in two ways: Draw accurate, scale vectors on graph paper and measure the size and direction of the result of the addition, called the “resultant vector” Draw sketch vectors and use trig and algebraic methods to calculate the size and direction of the resultant. ADDITION The tail of the second adds to the head of the first The resultant is drawn from the tail of the first to the head of the second SUBTRACTION To subtract, reverse the direction of the negative vector then add. 7.1 units East 5.0 units NE NE + 5.0 units SE = NE NE _ 5.0 units SE = = NE NE 5.0 units NW + 7.1 units North

16. 2.2 Vector Components A single vector can be broken up into two or more parts called COMPONENTS. This process is useful when, for example, trying to find the vertical and horizontal parts of a force which is accelerating a mass through the Earth’s atmosphere. F = 5 x 10 6 N 30 o F = 5 x 10 6 N FHFHFHFH FVFVFVFV The Horizontal component of the force (F H ) can be found using trig methods: F H = F cos 30 o = (5 x 10 6 ) ( 0.866) = (5 x 10 6 ) ( 0.866) = 4.3 x 10 6 N = 4.3 x 10 6 N Similarly for the Vertical component (F V ), F V = F sin 30 o = (5 x10 6 )(0.5) = (5 x10 6 )(0.5) = 2.5 x 10 6 N = 2.5 x 10 6 N At present, the total force is directed at 30 o above the horizontal F H and F V are the COMPONENTS of the force F. 30 0 Questions

17. Vector Addition 7. What is the resultant force when 2 forces (6.0 N west and 4.0 N south) act on an object at the same time ? θ = tan -1 4/6 = 33.7 o θ Resultant Force = √ (6) 2 + (4) 2 = 7.2 N 6 N west 4 N south

18. Vector Subtraction 8. Calculate the change in velocity of an object initially travelling at 8.5 ms -1 East whose final velocity was 8.5 ms -1 West. (remember Change in Velocity = Final Velocity – Initial Velocity) - = 8.5 ms -1 East 8.5 ms -1 West + = 17 ms -1 West

19. Vector Components 9. An boy fires a stone from slingshot. The stone leaves with a velocity of 27 ms -1 at an angle 32 0 above the horizontal. Calculate the vertical and horizontal components of the stone’s velocity. VHVH VvVv θ = 32 o V = 27 ms -1 V v = 27 sin 32 o = 14.3 ms -1 V H = 27 Cos 32 o = 22.9 ms -1

20. Vector addition/subtraction 10. Calculate the acceleration of a car whose velocity changes from 16 ms -1 west to 21 ms -1 north in 1.5 seconds (acceleration = change in velocity/change in time) Acceleration is a vector quantity so a vector calculation is required to calculate it. 16 ms -1 West Initial Velocity Final Velocity 21 ms -1 North Change in Velocity = V F – V i - = + = Resultant Velocity = √ (21) 2 + (16) 2 = 26.4 ms -1 θ θ = tan -1 16/21 = 37.3 o Acceleration = change in velocity/change in time = 26.4/1.5 = 17.6 ms -2 at N37.3 o W

21. Chapter 3 Kinematics

22. 3.0 Distance & Displacement Distance is a SCALAR quantity. It has a Unit (metres) but no Direction. Distance is best defined as “How far you have travelled in your journey” Displacement is a VECTOR quantity Having both a Unit (metres) and a Direction. Displacement is best defined as “How far from your starting point you are at the end of your journey” The difference between these two quantities is easily illustrated with a simple example. You are sent on a message from home to tell the butcher his meat is off. 2 km At the end of the journey, Distance travelled = 2 + 2 = 4 km Positive Direction while Displacement = +2 + (-2) = 0 km At this point in the journey, Distance travelled = 2 km and Displacement = + 2 km

23. 3.1 Speed & Velocity These two terms are used interchangeably in the community but strictly speaking they are different: Speed is the time rate of change of distance, i.e., Speed = Distance Time Time Speed is a SCALAR QUANTITY, having a unit (ms -1 ), but no direction. Thus a speed would be: 100 kmh -1 or, 27 ms -1 Velocity is the time rate of change of displacement, i.e., Velocity = Displacement Time Time Velocity is a VECTOR QUANTITY, having a unit (ms -1 ) AND a direction. Thus a velocity would be: 100 kmh -1 South or - 27 ms -1 - 27 ms -1

24. 3.2 Acceleration Acceleration is defined as the time rate of change of velocity, i.e., Acceleration = Velocity Time Acceleration is a VECTOR QUANTITY having both a unit (ms -2 ) and a direction. There is no scalar measurement of acceleration, so acceleration MUST always be quoted with a direction. Typically, Acceleration means an increase in velocity over time, while Deceleration means a decrease in velocity over time. av When v and a are in the same direction, the car accelerates and its velocity will increase over time. v a When v and a are in the opposite direction, the car decelerates and its velocity will decrease over time.

25. 3.3 Instantaneous & Average Velocity The term velocity can be misleading, depending upon whether you are concerned with an Instantaneous or an Average value. The best way to illustrate the difference between the two is with an example. You take a car journey out of a city to your gran’s place in a country town 90 km away. The journey takes you a total of 2 hours. The average velocity for this journey, v AV = Total Displacement = 90 = 45 kmh -1 Total Time 2 Total Time 2 However, your instantaneous velocity measured at a particular time during the journey would have varied between 0 kmh -1 when stopped at traffic lights, to, say 120 kmh -1 when speeding along the freeway. Average and Instantaneous velocities are rarely the same. Unless otherwise stated, all the problems you do in this section of the course require you to use Instantaneous Velocities. Questions

26. Kinematics 11. A runner completes a 400 m race (once around the track) in 21 seconds what is: (a) her distance travelled (in km), (b) her displacement (in km), (c) her speed (in ms -1 ) and (d) her velocity (in ms -1 ) ? (a) Distance = 0.4 km (b) Displacement = 0 km (c) Speed = distance/time = 400/21 = 19ms -1 (d) Velocity = displacement/time = 0/21 = 0 ms -1

27. Acceleration 12. A roller coaster, at the end of its journey, changes it’s velocity from 36 ms -1 to 0 ms -1 in 2.5 sec. Calculate the roller coaster’s acceleration. a = change in velocity/change in time = (0 – 36)/2.5 = - 14.4 ms -2

28. Chapter 4 Motion by Graphs

29. 4.0 Graphical Relationships It is often useful and convenient to represent information about things like position, velocity, acceleration etc., using graphs. Graphs “tell you a story”. You need to develop the skills and abilities to “read the story”. There are two basic types of graphs used in Physics: (a)Sketch Graphs – give a “broad brush” picture of the general relationship between the two quantities graphed. (b) Numerical Graphs – give the exact mathematical relationship between the two quantities graphed and may be used to calculate or deduce numerical values.

30. 4.1 Sketch Graphs Distance Time Sketch graphs have labelled axes but no numerical values, they give a general broad brush relation between the quantities. The Story: As time passes, the distance of the object from its starting point does not change. This is the graph of a stationary object The Story: The object begins its journey at the origin at t = 0. As time passes its displacement increases at a constant rate (slope is constant). So time rate of change of displacement which equals velocity is constant. This is a graph of an object travelling at constant velocity Displacement Time Velocity Time The Story: As time passes the velocity remains constant. This is a graph of an object travelling at constant velocity The Story: As time passes its displacement gets larger at an increasing rate. This is the graph of an object moving with constant acceleration Displacement Time Questions

31. Sketch Graphs Distance versus time graph. As time passes displacement remains the same. This is the graph of a stationary object Displacement versus time graph. As time passes its displacement is increasing in a uniform manner. This is a graph of an object travelling at constant velocity. Distance Time 13 (a) Displacement Time (b)

32. Sketch Graphs Velocity Time (c) Velocity versus time graph. As time passes the velocity of the object remains the same. This is a graph of an object travelling at constant velocity. Displacement versus time graph. As time passes its displacement gets larger at an increasing rate. This is a graph of an accelerating object. (if the shape is parabolic the object is increasing its speed in a uniform fashion i.e. it has a constant acceleration) Displacement Time (d)

33. 4.2 Exact Graphical Relationships The graphs you are required to interpret mathematically are those where distance or displacement, speed or velocity or acceleration are plotted against time. The information available from these graphs are summarised in the table given below. Learn this table off by heart. Put it on any cheat sheet you are allowed to use. Graph Type Distance or DisplacementversusTime Speed or Velocity versusTime AccelerationversusTime Read directly from the Graph Obtain from Slope of the Graph Obtain from Area Under the Graph Distance or Displacement Speed or Velocity Acceleration Velocity Acceleration No Useful Information Information Velocity Distance or Displacement Questions

34. Graphical Interpretation 14. Given below is the Distance vs Time graph for a cyclist riding along a straight path. 0 10 203040 50 60 ABCD Time (s) 10 20 Distance (a) In which section (A,B,C or D) is the cyclist stationary ? (b) In which section is the cyclist travelling at her slowest (but not zero) speed ? (c) What is her speed in part (b) above ? (d) What distance did she cover in the first 40 seconds of her journey ? (e) In which section(s) of the graph is her speed the greatest ? (f) What is her displacement from her starting point at t = 50 sec ? (a) Stationary in section C (b) Section B (c) Travels 10 m in 20 s  speed = 10/20 = 0.5 ms -1 (d) 20 m (read directly from graph) (e) Section D (travels 20 m in 10 s) speed = 2 ms -1 (f) Displacement at t = 50 s is 0 m (i.e., back at starting point)

35. Graphical Interpretation 15. Shown below is the Velocity vs Time graph for a motorist travelling along a straight section of road. 123456 78 91011 12 1314151617181920 0 2 4 6 8 10 -2 -4 -6 -8 Time(s) -10 Velocity (ms -1 ) (a) What is the motorist's displacement after 4.0 sec ? (b) What is the motorists acceleration during this 4.0 sec period ? (c) What distance has the motorist covered in the 20.0 sec of his journey ? (d) What is the motorist's displacement at t = 20.0 sec (e) What happens to the motorists velocity at t = 20.0 sec? Is this realistic ? (f) Sketch an acceleration vs time graph for this journey. (a) Displacement = area under velocity time graph. Between t = 0 and t = 4 s. Area = ½ (10 x 4) = 20 m (b) Acceleration = slope of velocity time graph = (10 – 0)/(4 – 0) = 2.5 ms -2 (c) Distance = area under graph (disregarding signs) Total area = ½(10 x 4) + (6 x 10) + ½(10 x 2) + ½(9 x 2) + (6 x 9) = 20 + 60 + 10 + 9 + 54 = 153 m (d) Displacement = area under graph (taking signs into account) = ½(10 x 4) + (6 x 10) + ½(10 x 2) - ½(9 x 2) - (6 x 9) = 20 + 60 + 10 - 9 – 54 = 27 m (e) Velocity falls from 9 ms -1 to zero in no time – no realistic, as it would require an infinite deceleration to achieve this.

36. Graphical Interpretation 15, continued -4.5 4 10 1214 20 To Infinity a -5 2.5 (e) t

37. Graphical Interpretation 16. An object is fired vertically upward on a DISTANT PLANET. Shown below is the Velocity vs Time graph for the object. The time commences the instant the object leaves the launcher Velocity (ms -1 ) Time (s) 30 -30 24681012 0 (a) What is the acceleration of the object ? (b) What is the maximum height attained by the object ? (c) How long does the object take to stop ? (d) How far above the ground is the object at time t = 10.0 sec ? (a) Acceleration = slope of velocity time graph. Slope = (30 – 0)/(0 – 6) = -5.0 ms -2 (b) Displacement = area under velocity time graph = ½ (6 x 30) = 90 m (c) Stops at t = 6.0 sec (d) The rocket has risen to a height of 90 m in 6 sec. It then falls a distance of ½ (4 x 20) = 40 m, so it will be 90 – 40 = 50 m above the ground at t = 10 s

38. Chapter 5 The Equations of Motion

39. 5.0 The Equations of Motion The Equations of Motion are a set of equations linking displacement, velocity, acceleration and time. They allow calculations of these quantities without the need for graphical representations. THESE EQUATION CAN ONLY BE USED IF THE ACCELERATION IS CONSTANT THESE EQUATION CAN ONLY BE USED IF THE ACCELERATION IS CONSTANT The 3 main equations are: v = u + at v = u + at v 2 = u 2 + 2as s = ut + ½at 2 Where, u = initial velocity (ms -1 ) v = final velocity (ms -1 ) a = acceleration (ms -2 ) s = displacement (m) t = time (s) When using the equations, always list out the information given and note what you need to find, then choose the most appropriate equation. In some cases you also need to define a positive direction, up or down for vertical motion, left or right for horizontal motion questions u = v = a = s = t = +ve

40. 5.1 Motion Under Gravity Objects (close to the surface) falling through the Earth’s gravitational field are subject to a constant acceleration of 9.8 ms -2. Since the acceleration is constant this motion can be analysed by the equations of motion. You need to go through the same process of listing information u = v = a = s = t = +ve The acceleration in this case is ALWAYS directed downward. Objects thrown or fired directly upwards would thus have their velocity and acceleration in opposite directions. v = u + at v = u + at v 2 = u 2 + 2as s = ut + ½at 2 The calculations using the equations of motion always ignore the effects of friction and air resistance and deciding on a positive direction Questions

41. Equations of Motion 17. A truck travels from rest for 10.0 sec with an acceleration of 3.0 ms -2. Calculate the truck's final velocity and total distance travelled. List information: u = 0, v = ?, a = 3.0 ms -2, s = ?, t = 10 s firstly find v, use v = u + at  v = 0 + (3.0)(10) = 30 ms -1 then find x use x = ut + ½at 2  (0)(10) +½(3.0)(10) 2 = 150 m. 18. A ball rolling down an inclined plane from rest travels a distance of 20.0 m in 4.00 sec. Calculate its acceleration and its final speed List information: u = 0, v = ?, a = ?, x = 20.0 m, t = 4.0 s Firstly find a, use x = ut + ½at 2  20.0 = (0)(4.0) + ½a(4.0) 2  a = 1.25 ms -2 The find v, use v = u + at  v = 0 + (1.25)(4.0)  5.0 ms -1

42. Equations of Motion 19. The speed of a freewheeling skateboard travelling on a level surface falls from 10.0 ms -1 to 5.00 ms -1 in moving a distance of 30.0 m. If the rate of slowdown is constant, how much further will the skateboard travel before coming to rest ? List information u = 10 ms -1, v = 5.0 ms -1,a = ?, x = 30 m, t = ? Cannot get to answer in 1 step. First find acceleration Use v 2 = u 2 + 2ax  a = (v 2 – u 2 )/2x  a = - 1.25 ms -2 Now new information u = 5.0 ms -1, v = 0, a = -1.25ms -2, x = ?, t = ? Use v 2 = u 2 + 2ax  x = (v 2 – u 2 )/2a  x = 10 m 20. A bullet leaves the barrel of a gun aimed vertically upwards at 140 ms -1. How long will it take to reach its maximum height ? (Ignore air resistance and use g = 10 ms -2 ). List information (up is +ve) u = 140 ms -1 v = 0, a = -10 ms -2, x = ? t = ? Use v = u + at  t = (v – u)/a = (0 – 140)/-10 = 14 s

43. Chapter 6 Forced Change

44. 6.0 What is a Force ? First, a force is an "interaction". You can compare a force to another common interaction - a conversation. A conversation is an interaction between 2 people involving the exchange of words (and ideas). Some things to notice about a conversation (or any interaction) are: To have a conversation, you need two people. One person can't have a conversation A conversation is something that happens between two people. It is not an independently existing "thing" (object), in the sense that a chair is an independently existing "thing". In the definition, "(material) objects" means that both objects have to be made out of matter - atoms and molecules. They both have to be "things", in the sense that a chair is a "thing". A force is something that happens between 2 objects. It is not an independently existing "thing" (object) in the sense that a chair is an independently existing "thing". "A force is an interaction between two material objects involving a push or a pull." How is this different from the usual textbook definition of a Force simply being a “push or a pull” ? Forces are like conversations in that: To have a force, you have to have 2 objects - one object pushes, the other gets pushed. Questions

45. Force 21. A force is an interaction between 2 objects. Therefore a force can be likened to A: Loving chocolate B: Fear of flying C: Hatred of cigarettes D: Having an argument with your partner 22. Between which pair can a force NOT exist ? A: A book and a table B: A person and a ghost C: A bicycle and a footpath D: A bug and a windscreen

46. 6.1 What Kinds of Forces Exist ? 2. Field Forces are forces in which the two interacting objects are not in contact with each other, yet are able to exert a push or pull despite a physical separation. Examples of field forces include Gravitational Forces, Electrostatic Forces and Magnetic Forces For simplicity sake, all forces (interactions) between objects can be placed into two broad categories:  1. Contact forces are types of forces in which the two interacting objects are physically contacting each other.  Examples of contact forces include frictional forces, tensional forces, normal forces, air resistance forces, and applied forces. Force is a quantity which is measured using the derived metric unit known as the Newton. One Newton (N) is the amount of force required to give a 1 kg mass an acceleration of 1 ms -2. So 1N = 1 kgms -2 Force is a vector quantity, you must describe both the magnitude (size) and the direction. Questions

47. Contact or Field Forces 23. Classify the following as examples of either Contact or Field forces in action (or maybe both acting at the same time). EXAMPLECONTACT FORCE FIELD FORCE (a) A punch in the nose (b) A parachutist free falling (c) Bouncing a ball on the ground (d) A magnet attracting a nail (e) Two positive charges repelling each other (f) Friction when dragging a refrigerator across the floor (g) A shotput after leaving the thrower’s hand √ √√ √ √ √ √ √ √

48. 6.2 What Do Forces Do ? Forces affect motion. They can: Begin motion Change motion Stop motion Have no effect BEGINNING MOTION: A constant force (in the same direction as the motion) produces an ever increasing velocity. STOPPING MOTION: A constant force (in the opposite direction to the motion) produces an ever decreasing velocity. CHANGING MOTION: A constant force (at right angles to the motion) produces an ever changing direction of velocity. NO EFFECT: A total applied force smaller than friction will not move the mass FRFR

49. 6.3 Where Forces Act Forces acting on objects must have a point of application, a place where the force acts. For Contact Forces the point of application is simply the point at which the force initiator contacts the object. For Field Forces, the only one applicable in movement being gravity, will act through the centre of mass of the object Force of finger on carton Force of carton on finger Gravitational Force C of M Questions

50. Net Force 24. A body is at rest. Does this necessarily mean that it has no force acting on it ? Justify your answer. NO – A body will remain at rest if the NET FORCE acting is zero – it could have any number of forces acting on it. So long as these forces add to zero it will remain at rest. 25. Calculate the net force acting on the object in each of the situations shown. 1200 N N 900 N 75 N 95 N 250 N 150 N 450 N (a) (b) (c) (d) 300 N Left 20 N Left 0 N 300 N Down

51. 6.4 Forces in 2 Dimensions Tim Tom Tam Tim, Tom and Tam, the triplets, are fighting over a teddy bear. Each exerts a different force. What will be the net force on the bear ? F TIM = 26 N F TOM = 25 N F TAM = 18 N Add the vectors head to tail. Tim Tom Tam F RESULTANT The resultant force is the vector joining the starting point to the finishing point A force diagram shows each boy’s contribution The bear will then accelerate in the direction of the resultant force Forces can act in any direction and the total or resultant force is the vector sum of all the forces acting. Questions

52. Net Force 26. Tim, Tom and Tam, the triplets, are fighting over a teddy bear. Each exerts a different force. A force diagram shows each boy’s contribution. What will be the net force on the bear ? F TIM = 35 N F TOM = 26 N F TAM = 12 N Resolve F TIM and F TOM to give F = 9 N left Then resolve this force with F TAM 12 N 9 N X θ X = √ (9 2 + 12 2 ) = 15 N θ = sin -1 (9/12) = 48.6 0 Net force is 15 N directed at S48.6 o W

53. 6.5 Weight Weight is the outcome of a gravitational field acting on a mass Weight is a FORCE and is measured in Newtons. Its direction is along the line joining the centres of the two bodies which, between them, generate the Gravitational Field. Near the surface of the Earth, each kilogram of mass is attracted toward the centre of the earth by a force of 9.8 N. (Of course each kilogram of Earth is also attracted to the mass by the same force, Newton 3) So, the Gravitational Field Strength near the Earth’s surface = 9.8 Nkg -1 Weight and mass are NOT the same, but they are related through the formula: W = mg Where: W = Weight (N) m = mass (kg) g = Grav. Field Strength (Nkg -1 ) Strength (Nkg -1 ) 1 kg 9.8 N Questions

54. Mass & Weight 27. Fill in the blank spaces in the table based on a person whose mass on earth is 56 kg PlanetMass on planet (kg) Grav Field Strength (Nkg -1 ) Weight on planet (N) Earth569.81 Mercury0.36 Venus0.88 Jupiter26.04 Saturn11.19 Uranus10.49 56 549.4 20.2 1458.2 626.6 587.4

55. 6.6 Reaction Force All objects on and near the Earth’s surface are subject to the gravitational force. Any object subject to a net or resultant force will accelerate in the direction of that force (Newton 2). Why then do objects placed on a table on the Earth’s surface remain stationary ? The Reaction Force only exists as a result of the action of the weight of the vase acting on the table top and as such the reaction force does not exist as an isolated force in its own right. Because there is no net or resultant force on the vase, it remains stationary on the table Remove the table, the reaction force disappears and the vase accelerates under the action of W, until it encounters the floor and probably smashes. W R W There must be a force equal in size and opposite in direction to cancel out the gravitational force. There is such a force. It is called the REACTION or NORMAL FORCE. Note: W and R are NOT an action reaction pair. Why? Because when R disappears W does not.

56. Chapter 7 Centre of Mass

57. 7.0 Centre of Mass In order to deal with large objects it is useful to think of all the object’s mass being concentrated at one point, this point being the Centre of Mass of the object. Centre of Mass For odd shaped objects such as a boomerang, the Centre of Mass may fall outside the perimeter of the object. C of M For regularly shaped objects eg. squares or rectangles, cubes or spheres the Centre of Mass of the object is in the geometric centre of the object The C of M is the point around which the object it will spin if a torque or turning force is applied to the object.

58. 7.1 Translation and Rotation When a Force acts C through the Centre of Mass (of M) of an object or structure, it causes Translational Motion, ie. The object moves in the direction of the applied force according to Newton’s 2nd Law. (see slide 6.3) A Force acting through the Centre of Mass causes Translational Motion only When the force is applied to another part of the object or structure, a TORQUE or TWISTING FORCE or TURNING MOMENT is applied and Rotational as well as Translational motion occurs A Force acting at a point other than the C of M will cause BOTH Translational AND Rotational Motion.

59. Chapter 8 Newton’s Laws

60. 8.0 Aristotle to Newton Isaac Newton (1642 – 1727) was the first to realise there WAS a universal set of laws which could describe the motion of ALL bodies, BUT these laws had to be modified for use within the friction riddled confines of the Earth and its atmosphere. Attempts to explain the “causes of motion” (a field of study called dynamics), were first recorded in the time of the ancient Greek philosopher Aristotle (384 – 322 BC). However there were problems with the theories which could not, for example, explain why falling objects tended to increase their speed in the absence of any visible force or why heavenly bodies behaved differently than those on earth. This seemed logical as everyone could see that a horse needed to apply a constant pull to haul a cart at constant speed. It was Galileo (1564 – 1642) who was the first to define the property of matter we call INERTIA, (matter’s tendency to resist changes in its motion), with his law which said “when no force exists a body will stay at rest or move with constant speed It was believed that constant speed required a constant force. Philosophers prior to Newton believed a set of laws covering motion on earth could be developed, but they needed to be modified to explain the motions of heavenly bodies. Galileo Newton Aristotle Questions

61. Aristotle to Newton ScientistStatement NewtonConstant speed requires constant force AristotleDefined the property of matter called inertia GalileoA universal set of laws applicable everywhere but must be modified for use on earth

62. 8.1 Newton’s Laws Newton developed 3 laws which cover all aspects of motion (provided objects travel at speeds are well below the speed of light). Law 1 (The Law of Inertia) A body will remain at rest, or in a state of uniform motion, unless acted upon by a net external force. Law 2 The acceleration of a body is directly proportional to net force applied and inversely proportional to its mass. Mathematically, a = F/m more commonly written as F = ma Law 3(Action Reaction Law) For every action there is an equal and opposite reaction. Motion at or near the speed of light is explained by Albert Einstein’s Theory of Special Relativity. Newton, at age 26

63. 8.2 Newton’s 1 st Law Newton 1 deals with non accelerated motion. It does not distinguish between the states of “rest” and “uniform motion” (constant velocity). As far as the law is concerned these are the same thing (state). There is no experiment that can be performed in an isolated windowless room which can show whether the room is stationary or moving at constant velocity. Objects want to keep on doing what they are doing Newton’s 1 st Law states: A body will remain at rest, or in a state of uniform motion, unless acted upon by a net external force. If NO net external force exists No Net Force means No Acceleration It requires an unbalanced force to change the velocity of an object Another way of saying this is: Most importantly: Force is NOT needed to keep an object in motion Is this how you understand the world works ?

64. 8.3 Newton’s 2 nd Law Newton’s 2 nd Law states: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force F NET, in the same direction as the net force, and inversely proportional to the mass of the object. Mathematically, a = F NET /m more commonly written as F NET = ma The Net Force on an object equals the rate of change of its momentum Using the formula F NET = ma is only valid for situations where the mass remains constant Newton actually expressed his 2nd law in terms of momentum. So, F NET = change in momentum = Δp = mΔv = ma change in time Δt Δt Momentum (p) = mass x velocity F NET is the VECTOR SUM of all the forces acting on an object. The acceleration and F NET are ALWAYS in the same direction. Newton 2 deals with accelerated motion.

65. 8.4 Newton’s 3 rd Law Newton's 1 st and 2 nd Laws tell you what forces do. Newton's 3 rd Law tells you what forces are. "opposite" means that the two forces always act in opposite directions - exactly 180 o apart. This statement is correct, but terse and confusing. You need to understand that it means: "action...reaction" means that forces always occur in pairs. Single, isolated forces never happen. For example, first Suzie annoys Johnnie (action) then Johnny says "Mommy! Suzie’s annoying me!" (reaction). This is NOT an This is NOT an example what is going on here! The action and reaction forces exist at the same time. "action " and "reaction " are unfortunate names for a couple of reasons : For every action there is an equal and opposite reaction 1. Either force in an interaction can be the "action" force or the "reaction" force. "equal" means : Both forces are equal in magnitude. Both forces exist at exactly the same time. They both start at exactly the same instant, and they both stop at exactly the same instant. They are equal in time. 2. People associate action/reaction with "first an action, then a reaction” Questions

66. Newton’s Laws 29. At what speeds are Newton’s Laws applicable ? At speeds way below the speed of light 30. Newton’s First Law: A: Does not distinguish between accelerated motion and constant velocity motion B: Does not distinguish between stationary objects and those moving with constant acceleration C: Does not distinguish between stationary objects and those moving with constant velocity D: None of the above 31. Newton’s Second Law: A: Implies that for a given force, large masses will accelerate faster than small masses B: Implies that for a given force, larger masses will accelerate slower than smaller masses C: Implies that for a given force, the acceleration produced is independent of mass D: Implies that for a given force, no acceleration is produced irrespective of the mass.

67. Newton’s laws 32. Newton’s Third Law: A: Does not distinguish which force of a pair is the “action” force and which is the “reaction” force. B: Implies that both action and reaction forces begin and end at the same instant C: Implies that forces always exist in pairs D: All of the above. 33. Which of Newton’s Laws require that the vector sum of all the forces acting is needed before a calculation of acceleration can be made ? A: Newton’s 1st Law B: Newton’s 2nd Law C: Newton’s 3rd Law D: Newtons 1st and 2nd Laws

68. Newton’s 2 nd Law 34. A car of mass 1250 kg is travelling at a constant speed of 78 kmh -1 (21.7 ms -1 ). It suffers a constant retarding force (from air resistance, friction etc) of 12,000 N (a)What is the net force on the car when travelling at its constant speed of 78 kmh -1 ? At constant velocity, acc = 0 thus ΣF = 0 (b) What driving force is supplied by the car’s engine when travelling at 78 kmh -1 ? (c) If the car took 14.6 sec to reach 78 kmh -1 from rest, what was its acceleration (assumed constant) ? At constant velocity ΣF = 0, so driving force = retarding force = 12,000 N Use eqns of motion u = 0 ms -1, v = 21.7 ms -1, a = ?, x = ?, t = 14.6 s use v = u + at -> 21.7 = 0 + 14.6(a) -> a = 1.49 ms -2

69. 8.5 The Horse and Cart Problem If the horse and cart exert equal and opposite forces on each other, how come the combination can move ? An explanation hinges on a couple of simple points: (Lets assume no friction) 1. An object accelerates (or not) because of the forces that push or pull on it. (Newton 2) 2. Only the forces that act on an object can cancel. Forces that act on different objects don't cancel - after all, they affect the motion of different objects! Why does the cart accelerate? Looking at the cart alone, just one force is exerted on it, (F HC ) - the force that the horse exerts on it. The cart accelerates because the horse pulls on it! The cart’s acceleration equals the net force on it divided by its mass F CH F HC F HR F RH The cart pulls the horse backwards (F CH ), and the road pushes the horse forward (F RH ). The net force is the vector sum of these two forces. The horse’s acceleration equals the net force on it divided by the its mass. There are 2 pairs of Newton 3 forces in this situation: F HC and F CH F HR and F RH If F NET on the horse is zero, what happens ? The obvious answer is the horse and cart are at rest. BUT, they could also be moving at constant speed ! Newton 1 Why does the horse accelerate? There are 2 forces acting on the horse. Questions

70. Newton’s 3 rd Law 35. Explain why, if a cart exerts an equal an opposite force on a horse as the horse exerts on the cart, the combination is able to move forward. It is the forces that act on the individual components (i.e., on the horse or cart individually) that decide whether each will move. If the friction of the road on the horse’s feet is larger than the force of the cart on the horse then the horse will accelerate. If the force of the horse on the cart is greater than the frictional force acting on the cart then the cart will accelerate. 36. A car mass 1500 kg is towing a trailer of mass 750 kg. The car/trailer combination accelerate at 3.4 ms -2. The trailer suffers a constant retarding force of 500 N, while the car suffers a constant retarding force of 1000 N. Calculate the net force acting on the trailer. Calculate the driving force supplied by the car’s engine. The net force is that force that provides the acceleration. From Newton 2, ΣF = ma ΣF = (750)(3.4) = 2550 N (2.55 x 10 3 N) The driving force must (i) overcome friction and (ii) provide extra force to accelerate the combination ΣF = (1000 + 500) + ma = 1500 + (1500 + 750)(3.4) = 1500 + 7650 = 9150 N (9.15 x 10 3 N)

71. 8.6 Momentum and Impulse Newton described Momentum as the “quality of motion”, a measure of the ease or difficulty of changing the motion of an object. Momentum is a vector quantity having both magnitude and direction. Mathematically, p = mv Where, p = momentum (kgms -1 ) m = mass (kg) v = velocity (ms -1 ) In order to change the momentum of an object a mechanism for that change is required. This mechanism of change is called Impulse. Mathematically, I = Ft Where, I = Impulse (N.s) F = Force (N) t = Time (s) The relationship between momentum and impulse can be derived from Newton’s 2 nd Law: F = ma and a = v/t, so F = mv/t Rearranging we get: Ft = mv Ft = mv ie. Impulse = Momentum

72. 8.7 Conservation of Momentum The concept of Momentum is particularly useful in analysing collisions. This is because of the Law of Conservation of Momentum which states: IN AN ISOLATED SYSTEM, TOTAL MOMENTUM IS CONSERVED. The term “isolated system” means no external forces are acting in the situation under investigation. In a crash situation, where the vehicle comes to a halt after, say, hitting a tree, both its velocity and momentum fall to zero. The apparently “lost” momentum, has, in fact, been transferred via the tree to the Earth. Since the Earth has a huge mass (6 x 10 24 kg). The change in its velocity is so small as to be negligible. In the crash mentioned, the momentum change is a fixed quantity so the Impulse (the product of F and t) is also a fixed quantity. However the individual values of F and t can vary as long as the multiply to give that fixed value. If t, the time during which the crash occurs, can be lengthened, then the force which needs to be absorbed by the car and its occupants is reduced. Modern vehicles use this concept in crumple zones and air bags as both are designed to extend the time and so reduce the force. Questions

73. Momentum 37. A car (and its occupants) is of total mass of 2250 kg and is travelling at 50 kmh -1. Approaching, head on, is a motorcycle (and rider) of total mass 350kg travelling at 180 kmh -1 (a)Which vehicle (car or bike) has the greater momentum ? (b) They collide head on and stick together. What velocity will the “wreck” have immediately after collision ? Firstly need to convert speeds to ms -1 50 kmh -1 = 13.9 ms -1 : 180 kmh -1 = 50.0 ms -1 p 4wd = mv = (2250)(13.9) = 31275 kgms -1 p cycle = mv = (350)(50) = 17500 kgms -1 So 4WD has the greater momentum. Assume 4WD is travelling to the right and motorcycle to the left. So Σp = p 4wd - p cycle = 31275 – 17500 = 13775 kgms -1 to the right. If the vehicles stick together total mass = 2250 + 350 = 2600 kg So Σp = m total v  v = Σp/m total = 13775/2600 = 5.3 ms -1 to the right

74. Momentum and Impulse 38. While talking on a mobile phone a truck driver loses concentration and runs off the road and hits a tree. His speed goes from 20 ms -1 to 0 ms -1 in 0.7 sec. his truck has a mass of 42 tonnes (1 tonne = 1000 kg) (a)Calculate his change in momentum (b) Calculate the Impulse during the collision (c) Calculate the force he will experience during the collision Change in momentum = final mom – initial mom = mv final - mv initial = m (v final – v initial ) = 4.2 x 10 4 (0 – 20) = - 8.4 x 10 5 kgms -1 (negative sign can be omitted from answer as it only indicates direction of momentum change) Since change in momentum = impulse. Impulse = 8.4 x 10 5 Ns I = Ft  F = I/t = (8.4 x 10 5 )/0.7 = 1.2 x 10 6 N

75. Airbags/Crumple Zones 39. Explain why, in a modern car equipped with seat belts and an air bag, he would likely survive the collision whereas in the past, with no such safety devices, he would most likely have been killed. The change in momentum in any collision is a fixed value thus impulse is also fixed, but the individual values of F and t can vary as long as their product is the that fixed value. In modern vehicles seat belts and crumple zones are designed to increase to time it takes to stop thus necessarily reducing the force needed to be absorbed by the driver because Impulse = Ft. This reduced force will lead to reduced injuries. In the old days the driver would have been “stopped” be some hard object like a metal dashboard and his time to stop would have been much shorter and thus the force experienced would have been larger leading to more severe injury and likely death.

76. Chapter 9 Work, Energy & Power

77. 9.0 Work In Physics, the term WORK is very strictly defined. When a force moves an object through a distance, work has been done. Mathematically: W = F x d Where, W = Work (Joules) F = Force (N) d = distance (m) If a force is applied and the object does not move, NO WORK has been done. If the force applied is constant, the work done can be calculated from the formula, W = F x d But, if the force varies during the course of doing the work, as in compressing a spring, the work must be calculated from the area under the force versus distance graph ForceDistance Area = Work done Work is a SCALAR quantity, meaning it has a magnitude but no direction. Questions

78. Work 40. Calculate the work done on a refrigerator when a net force of 125 N acts over a distance of 4.5 m Work = Fxd = (125) x (4.5) = 562.5 J (5.63 x 10 2 J) 41. The graph shows the force required to compress a spring Distance (cm) Force (kN) 1.02.03.0 4.0 1500 3000 4500 6000 (a) Calculate the work done in compressing the spring by 3.0 cm. (b) Calculate the further work required to compress the spring from 3.0 cm to 4.0 cm Work done = area under graph = ½ base x height = ½ (3.0 x 10 -2 )(4.5 x 10 6 ) = 1.35 x 10 5 J Further work to compress from 3.0 cm to 4.0 cm = area under graph between these two distances = area of trapesium = ½ (height(1) + height(2)) x base = ½ (4.5 x 10 6 + 6.0 x 10 6 )(1.0 x 10 -2 ) = 5.25 x 10 4 J

79. 9.1 Work and Energy The concept of WORK was developed BY PHYSICISTS as a means of quantifying and measuring ENERGY. It is very easy to say what energy can do, but very difficult to define exactly what energy is. The relation between work and energy is summarised by one simple but powerful statement: WORK DONE = ENERGY TRANSFERRED If work has been done on an object, the amount of energy it has MUST have increased. By how much ? By exactly the amount of work done on the object. If an object has done some work, the amount of energy it has MUST have decreased. By how much ? By exactly the amount of work done by the object. Questions

80. Work & Energy 42. How much energy is stored in the spring in question 41 when it has been compressed by 2.0 cm Work Done = ½ base x height = ½ (2.0 x 10 -2 )(3.0 x 10 6 ) = 3.0 x 10 4 J Since Work Done = Energy Transferred, the energy stored in the spring = 3.0 x 10 4 J

81. 9.2 Kinetic Energy Kinetic Energy is the energy possessed by moving objects. It is called the “Energy Of Motion”. Kinetic Energy is a SCALER quantity. Mathematically: K.E. = ½mv 2 Where: K.E. = Kinetic Energy (Joule) m = mass (kg) m = mass (kg) v = speed (ms -1 ) v = speed (ms -1 ) Horse has K.E. due to its movement Arrow has K.E. due to its motion Gears have K.E. due to their rotation

82. 9.3 Gravitational Potential Energy Gravitational Potential Energy, often just called Potential Energy, is the energy possessed by an object due to its position. It is called the “Energy of Position” Potential Energy is a SCALAR quantity. Mathematically: P.E. = mgh Where: P.E. = Potential Energy (Joules) m = mass (kg) m = mass (kg) g = Grav. Field Strength (Nkg -1 ) g = Grav. Field Strength (Nkg -1 ) h = height (m) h = height (m) Potential Energy needs a zero point for the measurement of the height, h. The zero point is usually, but not always, the surface of the Earth. The zero point needs to be known for the calculation to have meaning. The man has P.E. due to his height above the ground height above the ground Skier has P.E. while he is on the slope + KE due to his speed Questions

83. Kinetic & Potential Energy 43. A cyclist is riding her bike along a flat road. She and her bike have a mass of 105 kg. she is travelling at a constant speed of 15 ms -1. (a) Calculate her Kinetic Energy She accidentally rides over a 15 m cliff. (b) What is her potential energy at the top of the cliff ? (take g = 10 ms -2 ) KE = ½ mv 2 = ½ (105)(15) 2 = 1.18 x 10 4 J (c) If all the PE she had at the top of the cliff is converted to KE at the bottom, calculate her vertical speed just before she hits the ground. PE = mgh = (105)(10)(15) = 1.58 x 10 4 J PE TOP = KE BOTTOM  1.58 x 10 4 = ½ (105) v 2  v = 17.3 ms -1

84. 9.4 Hooke’s Law Developed by English scientist Robert Hooke in 1676, the law states that the Restoring Force in an elastic material is directly proportional to its extension. Mathematically: F = - kx Where: F = Restoring Force (N) k = Spring Constant (Nm -1 ) x = Extension (m) The spring constant (k) is a measure of the nature or quality of the elastic material. The higher its value the greater is the restoring force for a given extension. The negative sign in the equation indicates that the restoring force and the extension are in opposite directions. Questions

85. Hooke’s Law 44. A spring of length 100 cm and spring constant 2.5 x 102 Nm -1 hangs vertically from a retort stand. A total mass of 15.6 kg is hung from the spring. Calculate the extent of the spring’s extension under this load. (Take g = 10 Nkg -1 ) F = -kx: Force produced by hanging mass = mg = (15.6) (10) = 156 N. So Restoring Force = 156 N x = -F/k = -156/(2.5 x 10 2 ) = -0.62 m (could do this calculation without negative sign) by using applied force in Hooke’s Law equation) 45. Calculate the amount of elastic potential energy stored in the extended spring in question 44. EPE = ½ kx 2 = ½ (2.5 x 10 2 ) (0.62) 2 = 48.1 J

86. 9.5 Energy Transfers ENERGY CANNOT BE CREATED OR DESTROYED BUT ONLY TRANSFERRED FROM ONE FORM TO ANOTHER. When doing problems concerning energy and energy transfers, it is assumed that the transfers are 100% efficient, meaning no energy losses occur. For instance, a roller coaster will have a large Gravitational Potential Energy component at its highest point most of which will have been converted to Kinetic Energy at its lowest point and in calculations a direct equality can be made between P.E. at the top and K.E at the bottom In real life, these types of conversion processes are never 100% efficient. Friction and the production of heat and sound mean significant losses occur. The efficiency of energy transfer processes can be calculated from: % Eff = Energy Out x 100 Energy In 1 Energy In 1 The Law of Conservation of Energy says:

87. 9.6 Elastic Potential Energy Energy stored in springs is called elastic potential energy Elastic materials store energy when they are deformed and release that energy when they return to their original condition.Elastic materials store energy when they are deformed and release that energy when they return to their original condition. The amount of energy stored can be found from the Elastic Potential Energy Formula: E S = ½kx 2 whereThe amount of energy stored can be found from the Elastic Potential Energy Formula: E S = ½kx 2 where E S = Elastic P. E. (J ) k = Spring Constant (Nm -1 ) x = extension and or compression (m) Area = ½Fx = ½kx 2 = Energy stored = Energy stored up to extension x up to extension x Force Extension Slope = Spring constant (k) x F “REGULAR” ELASTIC BEHAVIOUR The stored energy can be used to increase the kinetic energy of the “arrows” such as used in cross bows Questions

88. 9.7 Energy Transfers & Heat The end result of the action of these frictional effects is the production of what is called “low grade, unrecoverable heat” Large sources of this kind of heat are the exhausts from fossil fuelled transport such as cars, trucks and trains and from the electricity generation industry. This means the heat energy cannot be harnessed to do any useful work, such as boil water or run a pump. In our friction riddled world, no energy transfer process is 100% efficient.

89. 9.8 Power Power is the time rate of doing work, and since work and energy are equivalent is also the time rate of energy transfer. Mathematically: P = W/t = E/t Where: P = Power (W, Watts) P = Power (W, Watts) W = Work (J) E = Energy (J) E = Energy (J) t = time (s) t = time (s) Since P = W/t and W = F.d, we can say P = F.d/t but d/t = v so P = F.v So, the power for a body moving at constant velocity can be found in a one step calculation. Questions

90. Power 46. A train of mass 1.5 x 10 4 kg is travelling at a constant speed of 70 kmh -1. If the engine is providing a driving force of 3.0 x 10 7 N, at what power is the engine operating ? 70 kmh -1 = 19.4 ms -1 Power = F.v = (3.0 x 10 7 )(19.4) = 5.82 x 10 8 W A B C D E F 47. A roller coaster moves through its journey from A to F. The coaster has no motor and is not powered after it leaves point A. Its total mass is 650 kg. The heights above ground of each portion of the track are given below. Positio n Height above Ground (m) A25.0 B5.0 C12.5 D E15.0 F7.5

91. Energy Transformations (a) Between which points is (i) the Kinetic Energy increasing, (ii) the Potential Energy falling, (iii) the coaster accelerating (d) Assuming no frictional losses, calculate the Kinetic Energy at point F (i) AB, EF (ii) BC, DE (iii) All except CD (b) At which point is the force exerted on the coaster by the track at its greatest ? Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms -2 ) B (At this point the track must supply the reaction force to counteract the weight of the coaster AND the centripetal force to make the coaster go round the ‘corner” PE = mgh = (650)(10)(25) = 1.63 x 10 5 J (c) Calculate the Gravitational Potential Energy at Point A. (take g = 10 ms -2 ) (e) Calculate the speed of the coaster at point F Loss in PE between A and F = gain in KE between A and F PE LOSS = mgh 1 – mgh 2 = (650)(10)(25 – 7.5) = 1.14 x 10 5 J At point F, KE = 1.14 x 10 5 = ½ mv 2  v = 18.7 ms -1

92. Ollie Leitl 2008