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## VCE Physics Unit 2 Topic 1

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**VCE PhysicsUnit 2 Topic 1**Motion 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.**Unit Outline**• Identify parameters of motion as scalars or vectors • Analyse straight line motion under constant acceleration graphically, numerically and algebraically; • Analyse graphically non-uniform in a straight line; • Compare the effects of a force as defined by Aristotle, Galileo and Newton. • Describe the change in motion that result from the application of a force; • Model weight, w, as the force of gravity acting at the centre of mass point (approximated as the geometric centre) of a body, w = mg • Model forces as vectors acting at the point of application (with magnitude and direction) labelling these forces using the convention “force of … on…” • Apply Newton’s 3 laws of motion to a body on which a resultant vector force acts • Model forces as external actions through the centre of mass point of each body; • Apply the vector model of forces including vector addition, vector subtraction and components to readily observable forces including weight, friction and reaction forces; • 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. • Apply the concept of work done by a constant force work done = force multiplied by distance in the direction of the force work done = area under the force versus distance graph. • Analyse Hooke’s Law for an ideal spring, F = -kΔx • Analyse energy transfers and transformations using an energy conservation model, including transfers between • gravitational potential energy near the Earth’s surface, mgΔh ; and kinetic energy ½mv2 • potential energy in ideal springs ½ kx2 and kinetic energy ½mv2; • Apply rate of energy transfer, power P = E/t • Apply the concept of momentum, p = mv; • Describe how action of a net force causes changes in momentum • Analyse impulse (momentum transfer) in an isolated system, for elastic collisions between objects moving in a straight line. • Apply graphical, numerical and algebraic models to primary data collected during practical investigations of motion, and to secondary data. • Identify and apply safe and responsible practices when investigating motion**Chapter 1**Introduction**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. 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.**Physical QuantityS.I. Unit Symbol**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. 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 All other units are derived from these 7 fundamentals. 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**Length; metre [m]**• It is the distance light travels, in a vacuum, in 1/299,792,458th of a second. • 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. • Temperature; kelvin [K] • It is 1/273.16th of the thermodynamic temperature of the triple point of water. • 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 12C 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. • 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. • 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. 1.2 S.I. Definitions Questions**Fundamentals**QUESTIONS 1. Which of the following quantities have fundamental units and which have derived ? √ √ √ √ √ √ √ √ √**Fundamentals**2. From which of the fundamental units do the following derive their units ? Length (m), time (s) Mass (kg), length (m), time (s) Mass (kg), length (m), time (s) Length (m), time (s) Mass (kg), length (m), time (s)**3.6**Fundamentals 3. Show that 1 ms-1 = 3.6 kmh-1 Two relevant conversion factors are: 1 km = 1000 m, 1 h = 3600 s 1km 1000m 1000m 1km 1h 3600 s 3600s 1h These can be written as or and or Which ones to use ? Easy, you want to end up with km on the top line and h on the bottom 1m x 1km x 3600s s 1000m 1h so 1 ms-1 = 3.6 kmh-1**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. 0 5 10 15 20 25 30 35 40 -40 -35 -30 -25 -20 -15 -10 -5 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**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**Chapter 2**Vectors & Scalars**2.0 Scalars and Vectors**N This vector represents a Force of 4 N, acting North West Before proceeding further we need to define two new quantities: • SCALAR QUANTITIES • These are completely defined by • A Number and • A Unit • Examples of scalars are: • Temperature 170, Mass 1.5 kg • VECTOR QUANTITIES • These are completely defined by • A Number • A Unit and • A 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. Questions**Vectors and Scalars**6. Which of the following quantities are scalars and which vectors ? √ √ √ √ √ √ √ √ √**=**+ 5.0 units SE 5.0 units NE 7.1 units East = = _ + 7.1 units North 5.0 units SE 5.0 units NW 5.0 units NE 5.0 units NE 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. The tail of the second adds to the head of the first ADDITION 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.**2.2 Vector Components**F = 5 x 106 N F = 5 x 106 N FV 300 30o FH 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. FH and FV are the COMPONENTS of the force F. At present, the total force is directed at 30o above the horizontal The Horizontal component of the force (FH) can be found using trig methods: FH = F cos 30o = (5 x 106) ( 0.866) = 4.3 x 106 N Similarly for the Vertical component (FV), FV = F sin 30o = (5 x106)(0.5) = 2.5 x 106 N Questions**6 N west**θ θ = tan-1 4/6 = 33.7o 4 N south Resultant Force = √(6)2 + (4)2 = 7.2 N 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 ?**8.5 ms-1 East**8.5 ms-1 West 8.5 ms-1 West 8.5 ms-1 West = - = + 17 ms-1 West 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)**V = 27 ms-1**Vv θ = 32o VH Vector Components 9. An boy fires a stone from slingshot. The stone leaves with a velocity of 27 ms-1 at an angle 320 above the horizontal. Calculate the vertical and horizontal components of the stone’s velocity. VH = 27 Cos 32o = 22.9 ms-1 Vv = 27 sin 32o = 14.3 ms-1**θ**θ = tan-1 16/21 = 37.3o Resultant Velocity = √(21)2 + (16)2 = 26.4 ms-1 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. Initial Velocity Final Velocity 16 ms-1 West 21 ms-1 North Change in Velocity = VF – Vi + - = = Acceleration = change in velocity/change in time = 26.4/1.5 = 17.6 ms-2 at N37.3oW**Chapter 3**Kinematics**3.0 Distance & Displacement**Positive Direction At this point in the journey , Distance travelled = 2 km and Displacement = + 2 km 2 km Distance is a SCALAR quantity. It has a Unit (metres) but no Direction. 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” Distance is best defined as “How far you have travelled in 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. At the end of the journey, Distance travelled = 2 + 2 = 4 km while Displacement = +2 + (-2) = 0 km**3.1 Speed & Velocity**These two terms are used interchangeably in the community but strictly speaking they are different: Velocity is the time rate of change of displacement, i.e., Velocity = Displacement Time Speed is the time rate of change of distance, i.e., Speed = Distance 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 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**3.2 Acceleration**v v a a There is no scalar measurement of acceleration, so acceleration MUST always be quoted with a direction. Acceleration is defined as the time rate of change of velocity, i.e., Acceleration = Velocity Time Typically, Acceleration means an increase in velocity over time, while Deceleration means a decrease in velocity over time. Acceleration is a VECTOR QUANTITY having both a unit (ms-2) and a direction. When v and a are in the opposite direction, the car decelerates and its velocity will decrease over time. When v and a are in the same direction, the car accelerates and its velocity will increase over time.**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, vAV = Total Displacement = 90 = 45 kmh-1 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**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**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**Chapter 4**Motion by Graphs**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: • 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.**4.1 Sketch Graphs**Distance Displacement Velocity Displacement 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 Time Time Time Time Sketch graphs have labelled axes but no numerical values, they give a general broad brush relation between the quantities. 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 The Story: As time passes its displacement gets larger at an increasing rate. This is the graph of an object moving with constant acceleration The Story: As time passes the velocity remains constant. This is a graph of an object travelling at constant velocity Questions**Distance**13 (a) Time Displacement (b) Time 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.**Velocity**(c) Time Displacement (d) Time Sketch Graphs 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)**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. Obtain from Slope of the Graph Obtain from Area Under the Graph Read directly from the Graph Graph Type Distance or Displacement versus Time Distance or Displacement Speed or Velocity No Useful Information Speed or Velocity versus Time Speed or Velocity Distance or Displacement Acceleration Acceleration versus Time No Useful Information Velocity Acceleration Learn this table off by heart. Put it on any cheat sheet you are allowed to use. Questions**Distance**20 10 A B C D Time (s) 0 50 60 20 30 40 10 Graphical Interpretation 14. Given below is the Distance vs Time graph for a cyclist riding along a straight path. (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)**Velocity (ms-1)**10 8 6 4 Time(s) 2 0 7 8 12 1 2 3 4 5 6 9 10 11 13 14 15 16 17 18 19 20 -2 -4 -6 -8 -10 Graphical Interpretation 15. Shown below is the Velocity vs Time graph for a motorist travelling along a straight section of road. (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.**(e)**a To Infinity 2.5 t 12 14 10 4 20 -4.5 -5 Graphical Interpretation 15, continued**Velocity (ms**-1) 30 Time (s) 0 2 4 6 8 10 12 -30 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 (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 ? • 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**Chapter 5**The Equations of Motion**5.0 The Equations of Motion**+ve 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. The 3 main equations are: v = u + at v2 = u2 + 2as s = ut + ½at2 Where, u = initial velocity (ms-1) v = final velocity (ms-1) a = acceleration (ms-2) s = displacement (m) t = time (s) THESE EQUATION CAN ONLY BE USED IF THE ACCELERATION IS CONSTANT u = v = a = s = t = 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**5.1 Motion Under Gravity**+ve 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. 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. The calculations using the equations of motion always ignore the effects of friction and air resistance v = u + at v2 = u2 + 2as s = ut + ½at2 u = v = a = s = t = You need to go through the same process of listing information and deciding on a positive direction Questions**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 s = ut + ½at2 (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 = ?, s = 20.0 m, t = 4.0 s Firstly find a, use s = ut + ½at2 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**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 = ?, s = 30 m, t = ? Cannot get to answer in 1 step. First find acceleration Use v2 = u2 + 2as a = (v2 – u2)/2s a = - 1.25 ms-2 Now new information u = 5.0 ms-1, v = 0, a = -1.25ms-2, s = ?, t = ? Use v2 = u2 + 2as s = (v2 – u2)/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, s = ? t = ? Use v = u + at t = (v – u)/a = (0 – 140)/-10 = 14 s**Chapter 6**Forced Change**6.0 What is a Force ?**"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. 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". Questions**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**6.1 What Kinds of Forces Exist ?**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 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 Force is a vector quantity, you must describe both the magnitude (size) and the direction. Questions**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). √ √ √ √ √ √ √ √ √**FR**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. NO EFFECT: A total applied force smaller than friction will not move the mass 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.**Force of carton on finger**Force of finger on carton C of M Gravitational Force 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 Questions**(a)**(b) 900 N 1200 N 75 N 95 N 150 N 250 N (c) (d) 250 N 450 N 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. 300 N Left 20 N Left 300 N Down 0 N