Physical Sciences Grade 11 Quanta by Impaq

Quanta

Miemie Pretorius

• Contents according to the

Curriculum and Assessment Policy Statement (CAPS)

• Contents divided into small bite-sized “quanta” – each “quantum” contains a crisp and to the point summary of a specific topic, enriched with hints, followed by an exercise containing numerous questions. • Step- by step answers to all the questions in the memorandum in the back of the book.

Physical Sciences Grade 11 Quanta

Physical Sciences Grade 11

Physical Sciences Grade 11 Miemie Quanta Pretorius

Miemie Pretorius

ISBN 0-620-35266-3

9 780620 352666

Cover GR11 ENG.indd 1

CAPS 29/01/2013 02:43:20 PM

Physical Sciences Grade 11 Quanta Miemie Pretorius

ALL RIGHTS RESERVED ©Copyright Copyright subsists in this work. No part of this work may be reproduced in any form or by any means without the author’s written permission. Any unauthorised reproduction of this work will constitute a copyright infringement and render the doer liable under both civil and criminal law. Whilst every effort has been made to ensure that the information published in this work is accurate, the editors, publishers, printers and author take no responsibility for any loss or damage suffered by any person as a result of the reliance upon the information contained therein. All the answers are the original work of the author and have not been taken from the official memoranda of the Departments of Education. ISBN 978-0-620-35266-3 Cover page and graphic design: Maximum Exposure, Somerset West. DVD recordings: University of Stellenbosch, Telematic Services Animations: Egbert Westra from Enjoythecompany If you have any query, recommendation or discussion on the contents of this book, please feel free to contact the author at: www.quantabooks.co.za 021 852 5598 or 082 852 1826 or info@quantabooks.co.za or mimi.pretorius@gmail.com or 086 6596 220 (fax) Other books in this series by the author: Fisiese Wetenskappe Graad 12 – Kwanta (KABV) Physical Sciences Grade 12 – Quanta (CAPS) Fisiese Wetenskappe Graad 11 – Kwanta (KABV) Physical Sciences Grade 10 – Quanta (CAPS) Fisiese Wetenskappe Graad 10 – Kwanta (KABV)

CONTENTS MECHANICS Resultant Parallelogram (Tail-to-Tail Method) Triangle or Polygon (Head-to-Tail Method) Resultant is Zero if Object is Stationary or Moving at Constant Velocity The Angle Between Two Vectors Force Diagrams and Free-Body Diagrams Resolving a Vector into its Perpendicular Components Body on an Inclined Plane Contact Forces Newton’s First Law of Motion Newton’s Second Law of Motion: Law and Experiment Newton’s Second Law of Motion: Applications Newton’s Second Law of Motion: Velocity/Time Graphs Newton’s Second Law of Motion, Apparent Weight and “Lifts” Newton’s Second Law of Motion and Systems Newton’s Third Law of Motion Newton’s Universal Law of Gravity Newton’s Law of Gravity and Newton’s Third Law of Motion: Newton’s Law of Gravity on a Planet Terminal Velocity MATTER AND MATERIALS Atomic Combinations: Molecular Structure: Forces, Bond Energy, Bond Length Atomic Combinations: Molecular Structure: Lewis Diagrams Atomic Combinations: Molecular Structure: Molecular Shape Atomic Combinations: Molecular Structure: Electronegativity Kinds of Particles and Intermolecular Forces Intermolecular Forces and the Physical State of Substances The Chemistry of Water

1 1 1 3 4 5 7 10 11 15 17 22 28 29 33 36 40 42 42 44 46 49 51 53 56 59 65

WAVES, SOUND AND LIGHT Background Knowledge Geometrical Optics: Refraction Critical Angle and Total Internal Reflection Applications of Reflection and Total Internal Reflection 2D and 3D Wave Fronts: Diffraction

67 68 70 72 73

MATTER AND MATERIALS (Continued) Ideal Gases: The Kinetic Theory of Gases Ideal Gases: Definition Ideal Gases: Relationship between p and V (Boyle’s Law) Ideal Gases: Relationship between V and T (Charles’s Law) Ideal Gases: Relationship Between p and T (Guy Lussac’s Law) Ideal Gases: General Gas Law: p1V1/T1 = p2V2/T2 Ideal Gases: General Gas Law: PV = nRT Ideal Gases: Deviations from Ideal Gas Behaviour

78 79 80 82 83 85 87 93

CHEMICAL CHANGE Quantitative Aspects of Chemical Change: Basic Concepts Quantitative Aspects of Chemical Change: Concentration Quantitative Aspects of Chemical Change: Mol Calculations Quantitative Aspects of Chemical Change: Determination of Composition Quantitative Aspects of Chemical Change: Stoichiometric Calculations

96 97 100 102 104

ELECTRICITY AND MAGNETISM Electrostatics: Conservation of Charge Electrostatics: Electric Fields Electrostatics: Coulomb’s Law Electrostatics: Electric Filed Strength Electrostatics: Force on a Charge in an Electric Field Electromagnetism: Right Hand Rule Electromagnetism: Current Induced by Changing Magnetic Field Current Electricity: Background Knowledge Current Electricity: Ohm’s Law and Resistance Current Electricity: Applications of Ohm’s Law Current Electricity: Potential Difference between Two Sets of Parallel Resistors Current Electricity: Special Power Problems Current Electricity: Resistors in Parallel Current Electricity: A Special Type of Parallel Resistor Problem

110 111 113 118 122 126 127 130 133 137 147 149 154 155

ENERGY AND CHEMICAL CHANGE Energy change during Chemical Reactions Catalysts

161 167

TYPES OF REACTION Acid-Base Reactions: Background Knowledge Acid-Base Reactions: Definitions Acid-Base Reactions: Conjugate Acid-Base Pairs Acid-Base Reactions: Ampholytes (Amphiprotic Substances) Acid-Base Reactions: Titration, Neutralisation Redox Reactions: Oxidation and Reduction Writing Equations for Redox Reactions

171 173 177 179 181 182 189

CHEMICAL SYSTEMS Exploiting the Lithosphere: Key Terms Exploiting the Lithosphere: Mining and Mineral Processing Exploiting the Lithosphere: Mining of Gold Exploiting the Lithosphere: Mining of Iron Exploiting the Lithosphere: Calcium Carbonate Exploiting the Lithosphere: Environmental Impact of Mining and Mineral Processing Exploiting the Lithosphere: Burning of Fossil Fuels

193 193 193 194 194 194 195

Grade 11 Paper Paper 1 (Physics) Paper 2 (Chemistry)

Content Mechanics Waves, Sound and Light Electricity and Magnetism Chemical Change Chemical Systems Matter and Materials

Grade 12 Examination Physics from grade 11 Chemistry from grade 11 Newton’s Laws & forces Intermolecular forces Electrostatics Mol and Stoichiometry Electric Circuits Energy and Change

Marks 68 32 50 70 20 60

Total 150 150

CONTENTS: MEMORANDUM MECHANICS Exercise 1: Resultant, Parallelogram, Triangle or Polygon Exercise 2: Resultant is Zero if Object is Stationary or Moving at Constant Velocity Exercise 3: The Angle Between Two Vectors Exercise 4: Force Diagrams and Free-Body Diagrams Exercise 5: Resolving a Vector into its Perpendicular Components Exercise 6: Body on an Inclined Plane Exercise 7: Contact Forces Exercise 8: Newton’s First Law of Motion Exercise 9: Newton’s Second Law of Motion: Law and Experiment Exercise 10: Newton’s Second Law of Motion: Applications Exercise 11: Newton’s Second Law of Motion: Velocity/Time Graphs Exercise 12: Newton’s Second Law of Motion, Apparent Weight and “Lifts” Exercise 13: Newton’s Second Law of Motion and Systems Exercise 14: Newton’s Third Law of Motion Exercise 15: Newton’s Universal Law of Gravity Exercise 16: Newton’s Law of Gravity and Newton’s Third Law of Motion: Exercise 17: Newton’s Law of Gravity on a Planet Exercise 18: Terminal Velocity

199 199 200 201 201 203 204 207 208 210 213 214 216 218 220 221 221 222

MATTER AND MATERIALS Exercise 19: Atomic Combinations: Forces, Bond Energy, Bond Length Exercise 20: Atomic Combinations: Molecular Structure: Lewis Diagrams Exercise 21: Atomic Combinations: Molecular Structure: Molecular Shape Exercise 22: Atomic Combinations: Molecular Structure: Electronegativity Exercise 23: Kinds of Particles and Intermolecular Forces Exercise 24: Intermolecular Forces and the Physical State of Substances Exercise 25: The Chemistry of Water

222 223 223 224 225 225 227

WAVES, SOUND AND LIGHT Exercise 26: Background Knowledge Exercise 27: Geometrical Optics: Refraction Exercise 28: Critical Angle and Total Internal Reflection Exercise 29: Applications of Reflection and Total Internal Reflection Exercise 30: 2D and 3D Wave Fronts: Diffraction

227 227 229 229 230

MATTER AND MATERIALS Exercise 31: Ideal Gases: The Kinetic Theory of Gases Exercise 32: Ideal Gases: Definition Exercise 33: Ideal Gases: Relationship between p and V (Boyle’s Law) Exercise 34: Ideal Gases: Relationship between V and T (Charles’s Law) Exercise 35: Ideal Gases: Relationship Between p and T (Guy Lussac’s Law) Exercise 36: Ideal Gases: General Gas Law: p1V1/T1 = p2V2/T2 Exercise 37: Ideal Gases: General Gas Law: PV = nRT Exercise 38: Ideal Gases: Deviations from Ideal Gas Behaviour

231 231 232 232 233 234 236 240

CHEMICAL CHANGE Exercise 39: Quantitative Aspects of Chemical Change: Basic Concepts Exercise 40: Quantitative Aspects of Chemical Change: Concentration Exercise 41: Quantitative Aspects of Chemical Change: Mol Calculations Exercise 42: Quantitative Aspects of Chemical Change: Determination of Composition Exercise 43: Quantitative Aspects of Chemical Change: Stoichiometric Calculations

241 242 243 245 246

ELECTRICITY AND MAGNETISM Exercise 44: Electrostatics: Conservation of Charge Exercise 45: Electrostatics: Electric Fields Exercise 46: Electrostatics: Coulomb’s Law Exercise 47: Electrostatics: Electric Filed Strength Exercise 48: Electrostatics: Force on a Charge in an Electric Field Exercise 49: Electromagnetism: Right Hand Rule Exercise 50: Electromagnetism: Current Induced by Changing Magnetic Field Exercise 51: Current Electricity: Background Knowledge Exercise 52: Current Electricity: Ohm’s Law and Resistance Exercise 53: Current Electricity: Applications of Ohm’s Law Exercise 54: Potential Difference between Two Sets of Parallel Resistors Exercise 55: Current Electricity: Special Power Problems Exercise 56: Current Electricity: Resistors in Parallel Exercise 57: Current Electricity: A Special Type of Parallel Resistor Problem

250 250 251 254 255 257 257 258 260 261 267 268 270 271

ENERGY AND CHEMICAL CHANGE Exercise 58: Energy change during Chemical Reactions Exercise 59: Catalysts

274 275

TYPES OF REACTION Exercise 60: Acid-Base Reactions: Background Knowledge Exercise 61: Acid-Base Reactions: Definitions Exercise 62: Acid-Base Reactions: Conjugate Acid-Base Pairs Exercise 63: Acid-Base Reactions: Ampholytes (Amphiprotic Substances) Exercise 64: Acid-Base Reactions: Titration, Neutralisation Exercise 65: Redox Reactions: Oxidation and Reduction Exercise 66: Writing Equations for Redox Reactions

276 277 278 279 279 280 282

CHEMICAL SYSTEMS Exercise 67: Exploiting the Lithosphere

283

SKILLS FOR PHYSICAL SCIENCES LEARNERS

What is ‘Rate’?

Scientific Notation

Rate is the change of a quantity per second. E.g. power is rate at which work is done or the amount of work done per second, i.e. (work done)/second. Current is rate at which charge flows or the amount of charge flowing past a point per second, i.e. (amount of charge)/second.

Very large or very small numbers are preferably written in the scientific notation, e.g. a charge of 0,0000000326 C can be written as 3,26 x 10–8 C or a pressure of 350 625 Pa can be written as 3,51 x 105 Pa. Notice that in the scientific notation the number is written as a product of two numbers – the first number is between 1 and 10 and the second as 10n. The first number is usually rounded off to two decimal places. Please ensure that you know how to multiply and divide by numbers with exponents and how to use your calculator when dealing with exponents.

Conversion of Units The 7 basic SI units: Mass (m) is measured in kilogram (kg) Time (t) is measured in second (s) Length ( ) is measured in metre (m) Electric current (I) is measured in ampere (A) Amount of substance (n) is measured in mole (mol) Temperature (T) is measured in kelvin (K) Luminous intensity is measured in candela (cd) All answers should be given in SI units, unless otherwise specified. Mass is often given in gram. In order to convert it to kg, it is useful to know that ‘kilo’ means ‘1 000’ or 103. Therefore 1 kg = 103 g and 1 g = 10–3 kg. Time is often given in hours (h). Remember that 1 h is 60 minutes and 1 minute is 60 seconds. Hence 1 h = 3 600 s or 3,6 x 103 s and 1 s = 1/(3,6 x 103) h Length is sometimes given in millimetre (mm) or centimetre (cm) or decimetre (dm) or kilometre (km). In order to convert from one to the other, it is useful to understand the following diagram: For each space that you move to the right, you x10–1 x10–1 x10–1 x10–1 x10–1 x10–1 10 or x10–1, –1 –2 e.g. 5 mm = 5 x 10 cm = 5 x 10 dm mm cm dm m Dm Hm km = 5 x 10–3 m = 5 x 10–6 km, etc. For each space that you move to the left, you x10 x10 x10 x10 x10 x10 x10, e.g. 5 m = 5 x 10 dm = 5 x 102 cm = 5 x 103 mm, etc. If the conversion is for area, which is x10–2 x10–2 x10–2 x10–2 x10–2 x10–2 measured in m2, then for each space that you move to the right, you 102 or x10–2, e.g. mm2 cm2 dm2 m2 Dm2 Hm2 km2 5 cm2 = 5 x 10–2 dm2 = 5 x 10–4 m2, etc. For each space you move to the left, you x102 x102 x102 x102 x102 x102 x102, e.g. 6 km2 = 6 x 106 m2, = 6 x 108 dm2 = 6 x 1012 mm2, etc. x10–3 x10–3 x10–3 x10–3 x10–3 x10–3 If the conversion is for volume, which is measured in m3, then for each space that you mm3 cm3 dm3 m3 Dm3 Hm3 km3 move to the right, you 103 or x10–3, e.g. 250 cm3 = 250 x 10–3 dm3 = 250 10–6 m3, etc. x103 x103 x103 x103 x103 x103 For each space you move to the left, you x103, e.g. 1 km3 = 1 x 109 m3 = 1 x 1015 cm3, etc. Electric current is sometimes indicated as micro ampere ( A) or milli ampere (mA). ‘Micro’ means one millionth, or 10–6 and ‘milli’ means one thousandth, or 10–3. The scientific temperature scale that is used in Science is the kelvin scale, but the celsius scale is popular for everyday use. In order to convert from celsius to kelvin and vice versa, the following equation can be used: T = t + 273, where T is the temperature in kelvin and t is the temperature in celsius.

Directly and Inversely Proportional If we say that quantity A is directly proportional to quantity B, it means that if A doubles, B also doubles; if A increases 10 times, B also increases 10 times, if A is halved, B is halved, etc. This is written as A B, which we read as “A is directly proportional to B”. To change this into a mathematical equation, we write it as A = kB, where k is a constant. Example: This means that it we measure the distance covered by different people for the same time, say 10 s, then the greater the speed of the person, the greater the distance covered by the person. D v, which can be written as D = kv, where the constant k is the time – the equation becomes The graph of two quantities that are directly proportional, is always a straight line passing through the origin. If we say that P is inversely proportional to Q, it means that if P increases 2 times, Q decreases 2 times; if P increases 10 times, Q decreases 10 times, etc. This is written as P 1/Q, which we read as “P is inversely proportional to Q”. To change this into a mathematical equation, we write it as P = k/Q, where k is a constant and PQ = k. The graph for two quantities that are inversely proportional is a hyperbola. Graph of A and B, which are Graph of P and Q, which are directly proportional inversely proportional Q

Ratios In Physical Sciences we are often required to divide a quantity into ratios. Suppose three workers, X, Y and Z divide the profit of a business in a ratio of 1:3:5. This means that for every R1 that X earns, Y earns R3 and Z earns R5, i.e. every R(1 + 3 + 5) = R9 is split up into R1, R3 and R5. Hence X earns 1/9 of the profit, Y earns 3/9 of the profit and Z earns 5/9 of the profit. Now it easy to calculate each one’s share for any profit, e.g. if the profit is R1 800, X earns 1/9(R1 800) = R200; Y earns 3/9(R1 800) = R600 and Z earns 5/9(R1 800) = R1 000 Example: A potential difference of 24 V divides in a ratio of 2:2:4 across resistors P, Q and R. Share of potential difference for P: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V Share of potential difference for Q: 2/(2 + 2 + 4) = 2/8 2/8(24 V) = 6 V Share of potential difference for R: 4/(2 + 2 + 4) = 4/8 4/8(24 V) = 12 V

Practical Investigations A practical investigation involves the relationship between two quantities. All other quantities have to be kept constant. The investigative question should always relate the two quantities investigated, e.g. suppose we investigate the relationship between temperature and volume, then the question can be “What is the relationship between temperature and volume?” or “Does volume increase with temperature?”, etc. The hypothesis should be an ‘answer’ to the investigative question, e.g. “Volume is directly proportional to temperature” or “Volume increases as temperature increases”, etc. The hypothesis may even be a wrong statement, but it has to ‘answer’ the investigative question. The results should be recorded in a table with suitable headings that contain units, e.g. the column that contains the temperature should have the heading “Temperature (°C)” and the column containing the volume should have the heading “Volume (cm3)”. The independent variable is the quantity that you change – suppose you decide you are going to take the volume at 0 °C, 10 °C, 20 °C, then temperature is the independent variable. The dependent variable is the quantity that changes because of the change in the one that you control – in this case, the volume. The constant variables are those quantities that you will have to keep the same in order to have a fair investigation, i.e. you cannot have anything that will influence your readings except your two variables. E.g. you will have to work with the same substance – you cannot use hydrogen in one measurement and then iron in the next. You will also have to keep the mass of the substance constant – you cannot take 10 g of hydrogen in one measurement and 50 g of hydrogen in the next. You will also have to keep the pressure constant – one measurement cannot be taken at sea level and the other at high altitudes, etc. The graph should always have the two axes labelled properly, similar to the table, including units and then also a suitable scale. The independent variable should always be on the horizontal or X-axis and the dependent variable on the vertical or Y-axis. The conclusion should state the relationship between the two variables. Precautions have to be taken to exclude dangers and factors which could cause the investigation to fail. A quantitative analysis involves measurements, e.g. when you determine the relationship between the volume of a gas and temperature, you have to measure temperature and volume. A qualitative analysis involves a study, e.g. when you determine which materials conduct electricity; you only change the substances and observe in which cases the light bulb glows – no measurement required.

Graphs: Gradient and Area In both Chemistry and Physics you often will come across graphs which you will have to interpret. At this stage there are two important skills to master that come in very useful: Look at the gradient of the graph and the area included by the graph and the axes. In any graph the . This means, the gradient will be v whatever is on the Y-axis divided by whatever is on the X-axis. Next look for an equation on the data sheet that contains these two quantities and rewrite the equation in such a way that the expression on the right hand side is the same as that of the gradient. Example: In a velocity versus time graph, t Now look for an equation that contains the two variables, v and t, and see if you can rewrite the equation in such a way that v and t are the subject – the rest of the equation then represents the gradient. In our example we find a = ient represents a. Whatever the gradient does, acceleration a does the same – the first part of the graph has a constant and negative gradient, hence the acceleration a will be constant and negative. In the second part of the graph the gradient is constant and positive; hence the acceleration is constant and positive. In any graph the area included by the graph and the axes is calculated as length times width. In our example it is v x t. Again see if you can find an equation that you can rearrange in such a way

Models in Science Models are used to represent something. It tries to make a difficult concept understandable. Most models do have some shortcomings, but as new discoveries are made, models keep on improving.

How to Solve Problems in Physical Sciences As you read through the question, list all the quantities that are given, and very important, the quantity which is asked. Look for an equation under the relevant section on your data sheet, for an equation containing the quantities given and asked. Substitute the given values into the equation, but first ensure that your units are correct. Usually you have to use SI units, but in some instances you have to use other units, e.g. in n = m/M, the mass has to be in gram and not in the SI unit of kilogram!

MECHANICS

Resultant The resultant of vectors is that single vector which has the same effect as all the vectors together. The resultant of two vectors that are not in a straight line can be determined in ONE of three ways: 1 Using a parallelogram/rectangle (tail-to-tail method), discussed below; 2 Using a triangle (head-to-tail method), discussed below; 3 Resolving the vectors into their perpendicular components, discussed on page 7. NEVER MIX THE THREE METHODS – USE JUST ONE AT A TIME!! A vector may be moved around until it is tail–to–tail or head–to–tail with another vector, provided its direction and magnitude remain the same. Any one of the three methods can be used to determine the resultant and in all of these methods either a scale drawing or calculation can be used. If the question specifically asks for a scale drawing, you are not allowed to use a calculation. If the question states that you have to do a calculation, you are not allowed to do a scale drawing. If the question only requires you to determine the answer, without specifying a method, you may use a calculation or a scale drawing.

A B C D 4

If a parallelogram is used, the two vectors are drawn as two adjacent sides of a parallelogram, starting at the same point. The diagonal, drawn from the same point, represents the resultant. Whenever this method is used, it is important to note that the two vectors, as well as the resultant, are all drawn starting from the same point. The parallelogram method can be used for only two vectors at a time.

accelerates towards the south-east. accelerates towards the south-west. accelerates towards the north-east. does not accelerate.

N 45º 90º

(2)

In which ONE of the following vector diagrams is the resultant of the three vectors zero? A

(2)

According to the vector diagram A

F1 + F2 + F3 = 0

F1 + F2 = F3

F1 + F3 = F2

F2 + F3 = F1

F1 F2 (2)

Parallelogram (Tail–to–Tail Method)

Three horizontal forces of equal magnitude act on a box initially at rest as shown. The box

Vectors p, q and r form a vector diagram as indicated. Which statement regarding the relationship between the vectors is true? A C

Resultant

p+q+r =0 p +r=q

B D

p+q=r r+q=p

(2) In which ONE of the following vector diagrams is the resultant of the three vectors zero? A B C D

(2)

Define the resultant of two vectors.

(3)

Johannesburg is 1 500 km from Cape Town on a bearing of 045º (N 45º E). Durban is 500 km from Johannesburg on a bearing of 150º (E 60º S). An aircraft on a test flight starts from Cape Town and flies to Johannesburg and then to Durban. Find the magnitude of the resultant displacement of the aircraft for the total flight by means of an accurate scale vector diagram using a scale: 1 cm = 250 km. Label the vectors.

Triangle (or Polygon) (Head–to–Tail Method) If a triangle is used, the two vectors are placed head–to–tail, i.e. where Resultant the one arrow ends, the next one starts. It does not matter in what order you take the vectors. The resultant is represented by the arrow drawn from the starting point of the first to the end point of the second vector, i.e. the arrow closing the triangle. If we want to find the resultant of more than two vectors, the vectors are placed head-to-tail, in any order. The resultant is then the arrow drawn from the start of the first to the end of the last vector, i.e. the arrow closing the polygon. In a vector diagram the resultant is the vector closing the polygon. In a vector diagram the resultant can be recognised as the vector that is not head-to-tail (i.e. it is lying ‘the wrong way round’ – head-to-head and tail-to-tail). If the vector diagram is a closed polygon, with all vectors lying head-to-tail, the resultant is zero. In a right-angled triangle, the resultant can be calculated using the Law of Pythagoras (r2 = x2 + y2) and trigonometry .

Exercise 1: 1

A student walks 4 km due north in one hour and 3 km due west in the next hour. What is his resultant displacement after two hours? A C

7,0 km 2,5 km, 37º

B D

5,0 km, 323º 3,5 km, 323º

3 km

(2)

N 4 km 37º

An aircraft is flying directly north at constant airspeed in still air. As the flight progresses, a wind arises, blowing from the west. What will be the effect on the speed and direction of the aircraft relative to the ground? A B C D

No effect. Speed unchanged but direction changes. Speed increases but direction unchanged. Speed increases and direction changes.

(2)

Johannesburg 1 500 km

500 km Durban

Cape Town (8)

A fisherman in a boat first sails 3 km due south from the harbour, then he sails 4 km due east and then 2 km in a 45º N of E direction. Use a scale drawing (in which 10 mm represents 1 km) to answer the following questions:

10.1 After these movements, how far will the fisherman be from the harbour? 10.2 In what direction should he now sail to get back to the harbour in a straight line? 11

A scale drawing in which vectors are used to represent four forces which act on an object forms a closed quadrangle as shown by the following diagram: What can be deduced by this vector diagram? Explain your answer. (4)

Use a scale drawing to represent the following vector equation to show its meaning properly: (Let 20 mm represent 1 vector unit.) 3+4=4

(9) (2)

(4)

The Resultant is Zero if the Object is Stationary or Moving at Constant Velocity When an object is at rest, the object is in equilibrium, i.e. the resultant is zero. When an object is moving at constant velocity, the body is also in equilibrium, i.e. resultant is zero. It is extremely important to immediately realise that whenever you see the expression constant velocity, you should know this implies the forces acting on the object have a resultant of zero. This knowledge is tested very frequently and is often skilfully camouflaged even in other sections of Physics. We will learn more about this in Newton’s First Law of Motion. When an object is in equilibrium under the influence of three (or four) forces, the three (or four) forces can be represented in magnitude and direction by the three (or four) sides of a closed triangle (or quadrangle), placed head-to-tail. To draw this triangle (or quadrangle), you can start with any force. Next any of the remaining forces can be moved around (keeping its direction) until it lies head-to-tail with the first force. Now move the third force until it lies head-to-tail with the second force, closing the triangle. Any unknown force or angle can now be calculated using trigonometry and Pythagoras if the triangle is a right-angled triangle or by a scale drawing. Remember, an object is in equilibrium when it is at rest or moving at constant velocity. Please note that if you come across a closed vector diagram, with all the sides lying head-to-tail, it means the resultant is zero and the body is in equilibrium, i.e. at rest or moving at constant velocity.

6.3 6.4

falling freely from rest. moving in a circular path.

B D

A car moves along a horizontal road to the right at a constant velocity of 80 km h-1. Which ONE of the following diagrams best represents the forces acting on the car? (2)

accelerating. moving with constant velocity. A

7.1 7.2

Strut What is the magnitude of the downward force acting on the wall? (3) Draw a labelled force diagram which shows all the forces acting on point P. (5) 30º Explain whether the forces acting at P are in equilibrium. (2) Rod Calculate the magnitude of the force exerted by the horizontal rod on P. (4) If the angle between the rods were 45º instead of 30º, would the magnitude of force exerted by the horizontal rod at P be smaller, greater or the same? (2) A car engine with a mass of 200 kg is lifted from DIAGRAM A DIAGRAM B its mountings by means of a chain and pulley system. In sketch A, the engine is suspended by the chain, hanging stationary. In sketch B, the engine is pulled sideways by a mechanic, using a chain chain rope. In the questions that follow, the masses of the chain and the rope can be ignored.

7.3 7.4 7.5 8

(2)

8.1

8.2 8.3

A body moving at a CONSTANT VELOCITY on a horizontal plane, has a number of unequal forces acting on it. Which ONE of the following statements is TRUE? A B C D

6.1 6.2

At least two of the forces must be acting in the same direction. The resultant of the forces is zero. Friction between the body and the plane causes a resultant force. The vector sum of the forces causes a resultant force which acts in the direction of motion. (2) The diagram shows an object of weight W on a string. A horizontal force F is applied on the object so that the string makes an angle with the vertical when the object is at rest. The force exerted by the string is T. Which ONE of the following expressions is wrong? F A F +T+ w=0 B w = T cos C T2 = w2 + F2 D tan = F/T (2) A boy pushes a lawnmower of mass 40 kg with a constant velocity over a horizontal lawn. He exerts a push on the handle of 200 N at an angle of 20º to the horizontal. What is the resultant force acting on the mower? (2) A mass m of weight W, is suspended from a light rope by means of a strong hook which is fixed at point X. The rope is strung between vertical poles as shown in the figure. Say why a vector diagram of the forces acting at point X, plotted tail to head, will be a triangle. (3) Draw a rough diagram of this triangle (in 6.1) of forces. Label each force and indicate the size of each angle. (6)

20º

9.1 9.2

B 90º X

A heavy crate, mass M, is lifted by means of a rope S and a pulley, so that it can be taken through a window to the first floor of a building. A second rope T, pulls the crate horizontally to the left so that it does not scratch the wall. At lunchtime the ropes are fixed and the crate is left to hang in equilibrium as shown in the diagram. The tension in the rope S is 6 500 N.

m W

S T

40º

M Now calculate 9.1.1 the mass of the crate. (4) 9.1.2 the tension in the rope T. (3) Will the tension in the rope S increase if the crate is pulled further to the left? Explain your answer. (4)

The Angle Between Two Vectors Whenever we talk about the angle between two vectors we refer to the angle measured between their tails and not between the head of one and the tail of the other. The resultant of two vectors is a maximum when the angle between them is 0º and a minimum when the angle is 180º.

right

wrong

3N This resultant is 7 N, because the angle between the vectors is 0º.

4N 50º

engine What is the magnitude of the tension in the chain rope (i.e. force exerted by the chain) in sketch A? (3) engine Does the tension in the chain increase or decrease by pulling the engine sideways? Explain your answer by referring to a properly labelled force diagram. (Do not use calculations.) (6) Can the tension in the rope ever exceed that in the chain? (2)

E.g. 40º

(7)

A bell of mass 5 kg is supported by two rods bolted together at point P and attached to the wall as shown in the sketch. Ignore the weight of the rods in the questions which follow:

The resultant of all the forces acting on a body, is zero if the body is A C

(2)

Exercise 2: 1

In which part of the rope is the force greater, in A or in B? The rope will break if the force in the rope exceeds 500 N. Calculate the maximum mass which the above arrangement can support. (Correct to one decimal place).

This resultant is 1 N because the angle between the vectors is 180º.

Any value between the minimum and maximum is possible, i.e. as the angle between them increases the resultant decreases.

Exercise 3: 1

Which ONE of the following values of the angle rise to the smallest resultant? A

90º

120º (2)

10 N

(2)

Increases Remains the same

B D

between them

Decreases Increases and then decreases

(2)

An object is suspended by means of a light string. The sketch shows a horizontal force F which pulls the object from the vertical position until it reaches an equilibrium position as shown. Which ONE of the following vector diagrams best represents all the forces acting on the object? (2) A B C D

(2)

13 N

(2)

Two forces of magnitude 5 N and 7 N act on an object. Which of the following cannot be a resultant of the two forces? A

A car moves along a horizontal road to the right at a constant velocity of 80 km h-1. Which ONE of the following diagrams best represents the forces acting on the car? A B C D

An object experiences two successive displacements of 3 m and 5 m. Which ONE cannot be the magnitude of the resultant displacement (in m)? A

60º

1 Z

Two forces of constant magnitude act at the same point, as shown. If the angle increases from 10º to 80º, how will the magnitude of the resultant force change? A C

between forces X and Z, gives

Two forces, X and Y, can be replaced by a single force of magnitude 5 N. If the magnitude of X is 2 N, which ONE of the following can be the magnitude of force Y? A

0º

Exercise 4:

11 N

A bullet is fired by a rifle as indicated in the sketch. If the bullet experiences air resistance, which diagram correctly indicates the forces acting on the bullet at the moment when it is in the position as indicated? (The forces are not drawn to scale.) (2) A B C

Two forces of 5 N and 7 N act on an object. They can be replaced by a single force which has the same effect as the other two. This force must have A C

a magnitude of 2 N. B a magnitude between 5 N and 7 N. D

a magnitude of 12 N. a magnitude between 2 N and 12 N. 20 N

7 varies between 0º and 180º, the magnitude of the resultant force Fres … A C

is always less than 20 N. varies between 40 N and 80 N.

B D

is always more than 60 N. varies between 20 N and 60 N.

(2)

A child sits on a swing. Her mother pulls the seat of the swing horizontally to the right as shown in the sketch. Which ONE of the following diagrams best represents the forces acting on the seat while the mother holds it? (2) A B C D

A boy pushes a lawnmower of mass 40 kg with a constant velocity over a horizontal lawn. He exerts a push on the handle of 200 N at an angle of 20º to the horizontal. On a copy of the sketch indicate (using arrows) and name all the forces acting on the mower while it is being pushed. (6)

60 N

(2)

Force Diagrams and Free-Body Diagrams Normal force of The force diagram for an object includes the object with all floor on crate the forces acting on the object only, drawn as arrows indicating their directions. Note that a pushing force acts Force by boy towards the object and a pulling force acts away from the on crate Frictional object. Gravitational force force by floor Example: A boy pushes a crate across a rough floor. by earth on crate on crate In the free-body diagram for an object the object is indicated as a dot and all the forces are moved, whilst keeping their respective directions, until all the starting points start on the dot, Normal force of floor on crate i.e. all the forces act away from the dot. Force by boy Label each force in a full sentence and specify where possible on crate the nature of the force – e.g. frictional force exerted by the plane Frictional on the object or gravitational force by the earth on the object, force by floor Gravitational force etc. by earth on crate on crate Only forces exerted on the object should be mentioned, therefore make certain each label reads: Force exerted by ….. on the object. Should you mention forces acting on any other objects, you lose marks! In all situations the gravitational force of the earth on the object (weight) will be one of the forces. The relative magnitudes (not exactly according to scale) of the forces must also be indicated by the lengths of the arrows.

•

S 40º

O M

An elephant is dragging a tree trunk along a rough horizontal surface by means of an attached rope which makes an angle of 30º with the horizontal. The force in the rope is 4 000 N. Draw a force diagram of the log and on it indicate and label all the forces (other than the 4 000 N force) acting on the log. (6)

199

200

Exercise 1: 1

B. = + r = 5 km Direction measured clockwise from base line 360º – 37º = 323º

Zero, because constant velocity implies equilibrium, no resultant.

6.1

The object is at rest, therefore in equilibrium (no resultant) and therefore these three forces will produce a closed head-to-tail triangle. 6.3 A. Opposite largest angle in a triangle is the longest side, which represents the largest force. 6.4 Because A experiences the largest force, it will reach 500 N first, while B 40º A will still be less than 500 N. We therefore ignore B and take A as 500 N. W cos 40º = A/W = 500/W, hence W = 500/cos 40º = 652,7 N 90º Mass is then, from W = mg, 50º B m = W/9,8 = 652,7/9,8 = 66,6 kg

6.2

Resultant velocity (diagonal) > velocity of aircraft only and therefore resultant speed > speed of aircraft only. Direction changes as indicated in the diagram.

B. If the three vectors are placed head-to-tail, the fourth vector, closing the figure, is the resultant, which is south-west.

B. Resultant is vector which lies ‘the wrong way round’. This is a closed triangle with all vectors head-to-tail, therefore no resultant.

A. Resultant is vector lying the ‘wrong way round’ – in this case they are all head-to-tail, therefore no resultant or resultant = 0.

D. p is the vector which lies ‘the wrong way round’, hence p is the resultant.

A. All lie head-to-tail, i.e. no resultant or resultant is zero.

It is the single vector that has the same effect as the two vectors together.

45º

60º

1 500 km

45º

10.1

75º 500 km

7.5

8.1

8.2

10.2

56 mm

74º 30 mm 40 mm

16º

20 mm He is 5,6 km from the harbour.

74º 8.3

16º N of W or 74º W of N

The four forces are all lying head-to-tail and the figure is closed, therefore the body in equilibrium. The resultant is zero.

68º 3

60mm

80mm 4

68º 80mm

44º

7.1 7.2

7.3 7.4

Construct a triangle of which one side is 6 cm (represents 1 500 km) and another side is 2 cm (representing 500 km) with an angle of 75º between the two sides. Complete the triangle and measure the third side in cm and convert back to km by x250. Answer should be about r = 1 453,2 km

9.1

Draw a triangle of which the three sides are 3 x 20 mm, 4 x 20 mm and 4 x 20 mm. The direction of the arrows should be such that the 3 and 4 units are head-to-tail and the other 4 unit (resultant) should be head-to-head and tail-to-tail. 9.2

Exercise 2:

A Weight of bell only W = mg = 5 x 9,8 = 49 N W A is force exerted by strut on P. B is force exerted by rod on P. B 30º W is force exerted by bell on P due to its weight. Hint: To determine whether a force acts towards or away from an object, you can imagine what the effect would be if that force were removed, i.e. whether it was pushing or pulling. A string cannot push, only pull. Yes, it is in equilibrium, because point P is stationary. Because P is in equilibrium, the three forces can be represented by a triangle: tan W/B B = W/tan Smaller. (From triangle it is evident if W remains the same and the angle increases, B decreases.) Because it is stationary, it is in Diagram B equilibrium, forces balance; hence force Diagram A in chain upwards is equal in magnitude Force exerted by Force exerted by to gravitational force (weight) down. rope chain Force in chain = W = mg = 200(9,8) Weight of = 1 960 N engine Weight remains the same and from the Force exerted by Weight of diagrams it is obvious that the force chain engine exerted by the chain in diagram A < force in diagram B. It has therefore increased. No. (Force by chain is hypotenuse, which is always > the weight.) Force of T on O 9.1.1 cos 40º = (weight of M)/6 500 weight of M = 6 500 cos 40º Force of = 4 979,3 N M on O Force of S W = Mg M = W/g (Weight of M) on O 40º = 4 979,3/9,8 = 508,1 kg (6 500 N) 9.1.2 Tension in rope T = force of T on O 2 2 therefore T = S – W 2 = (6 500)2 – (4 979,3)2 T = 4 178,1 N Yes. According to the vector diagram in 9.1, the force in S (hypotenuse) will increase if the force in T increases while the weight remains the same.

D. NNB! Equilibrium means object is at rest or moving at constant velocity.

D. Constant velocity implies equilibrium, i.e. resultant force is zero, which is the case in D only.

D. The larger the angle between the vectors, the smaller the resultant.

B. Constant velocity implies equilibrium, i.e. resultant force is zero.

D. Tan = F/W and not F/T T A: Because object is at rest, resultant is zero. B: In triangle cos W/T, hence W F W = T cos C: It is a right-angled triangle, therefore according to Pythagoras T2 = W 2 + F2.

C. When the two forces act in the same direction, with a resultant of 5 N, and X = 2 N, then Y = 3 N. If they act in opposite directions, with a resultant of 5 N, and X = 2 N, then Y = 7 N. Therefore Y can be anything from 3 N to 7 N, which excludes all but C.

B. The larger the angle between the vectors, the smaller the resultant

Exercise 3:

T F