MTG Physics Eng

i e n c e Pa r t 1 Grade Physical Sc Physics 12 Study Guide

Colour Diagrams: Emission and Absorption Spectra* Colour Spectrum and White Light Emission Spectra Absorption Spectrum (Source: Wikimedia Commons) * Refer to pages 144, 145 and 146

© Department of Basic Education 2015 This content may not be sold or used for commercial purposes. Curriculum and Assessment Policy Statement (CAPS) Grade 12 Mind the Gap study guide for Physical Science Part 1: Physics ISBN 978-1-4315-1937-8 This publication has a Creative Commons Attribution NonCommercial Sharealike license. You can use, modify, upload, download, and share content, but you must acknowledge the Department of Basic Education, the authors and contributors. If you make any changes to the content you must send the changes to the Department of Basic Education. This content may not be sold or used for commercial purposes. For more information about the terms of the license please see: http://creativecommons.org/licenses/by-nc-sa/3.0/. Copyright © Department of Basic Education 2015 222 Struben Street, Pretoria, South Africa Contact person: Dr Patricia Watson Email: watson.p@dbe.gov.za Tel: (012) 357 4502 http://www.education.gov.za Call Centre: 0800202933 The first edition, published in 2012, for the Revised National Curriculum Statement (RNCS) Grade 12 Mind the Gap study guides for Accounting, Economics, Geography and Life Sciences; the second edition, published in 2014 , aligned these titles to the Curriculum and Assessment Policy Statement (CAPS) and added more titles to the series in 2015, including the CAPS Grade 12 Mind the Gap study guide for Physical Science Part 1: Physics. ISBN 978-1-4315-1937-8 Mind the Gap team Series managing editor: Dr Patricia Watson Production co-ordinators: Lisa Treffry-Goatley and Radha Pillay Production assistants: Nomathamsanqa Hlatshwayo and Motshabi Mondlane Authors: Physical Science: Mariaan Bester, Nkosinathi Michael Khathi, Karen Yvonne Kornet, Jenny Koster, Mosala Daniel Leburu, Karen Reynecke, Lerato Mabogatsu Mphachoe, Thabiso Precious Ndaba, Gilberto Izquierdo Rodriguez, Meshack Jabu Sithole, Kumar Kuttan Pillai Suresh, Sehubedu Jerry Tladi Expert readers: Morongwa Masemolo, Prof Mundslamo Fathuwani, Dr T.D.T Sedumedi, Veena Maharaj, John Ostrowick and Karen van Niekerk Proofreaders: John Ostrowick, Sehubedu and Angela Thomas Designers: Sonja McGonigle and Erika van Rooyen Illustrators: Michele Dean, Vusi Malindi, Khosi Pholosa, John Ostrowick, Kenneth Tekane Cover illustration: Alastair Findlay Afrikaans Translation: Johan Steenkamp Onsite writers’ workshop IT support: Wayne Cussons Special thanks to Winning Teams, CEO Denzil Hollis, for the organisation’s subject expertise and workshop support. Winning Teams board game facilitators: Mantse Khoza and Sue Jobson

© Department of Basic Education 2015 Ministerial foreword The Department of Basic Education (DBE) has pleasure in releasing the second edition of the Mind the Gap study guides for Grade 12 learners. These study guides continue the innovative and committed attempt by the DBE to improve the academic performance of Grade 12 candidates in the National Senior Certificate (NSC) examination. The study guides have been written by teams of exerts comprising teachers, examiners, moderators, subject advisors and coordinators. Research, which began in 2012, has shown that the Mind the Gap series has, without doubt, had a positive impact on grades. It is my fervent wish that the Mind the Gap study guides take us all closer to ensuring that no learner is left behind, especially as we celebrate 20 years of democracy. Matsie Angelina Motshekga, MP The second edition of Mind the Gap is aligned to the 2014 Curriculum and Minister of Basic Education Assessment Policy Statement (CAPS). This means that the writers have considered the National Policy pertaining to the programme, promotion requirements and protocols for assessment of the National Curriculum Statement for Grade 12 in 2014. The CAPS aligned Mind the Gap study guides take their brief in part from the 2013 National Diagnostic report on learner performance and draw on the Grade 12 Examination Guidelines. Each of the Mind the Gap study guides defines key terminology and offers simple explanations and examples of the types of questions learners can expect to be asked in an exam. Marking memoranda are included to assist learners to build their understanding. Learners are also referred to specific questions from past national exam papers and examination memos that are available on the Department’s website – www.education.gov.za. The CAPS editions include Accounting, Economics, Geography, Life Sciences, Mathematics, Mathematical Literacy and Physical Sciences Part 1: Physics and Part 2: Chemistry. The series is produced in both English and Afrikaans. There are also nine English First Additional Language (EFAL) study guides. These include EFAL Paper 1 (Language in Context); EFAL Paper 3 (Writing) and a guide for each of the Grade 12 prescribed literature set works included in Paper 2. These are Short Stories, Poetry, To Kill a Mockingbird, A Grain of Wheat, Lord of the Flies, Nothing but the Truth and Romeo and Juliet. (Please remember when preparing for EFAL Paper 2 that you need only study the set works you did in your EFAL class at school.) The study guides have been designed to assist those learners who have been underperforming due to a lack of exposure to the content requirements of the curriculum and aim to mind-the-gap between failing and passing, by bridging the gap in learners’ understanding of commonly tested concepts, thus helping candidates to pass. All that is now required is for our Grade 12 learners to put in the hours required to prepare for the examinations. Learners, make us proud – study hard. We wish each and every one of you good luck for your Grade 12 examinations. __________________________________ Matsie Angelina Motshekga, MP Minister of Basic Education 2015

© Department of Basic Education 2015

© Department of Basic Education 2015 Table of contents Dear Grade 12 learner .... ...........................................................vii How to use this study guide.......................................................viii Top 10 study tips...........................................................................ix Mnemonics. ....................................................................................x Mind maps. ....................................................................................xi On the day of the exam...............................................................xii Question words to help you answer questions. ...................... xiii Vocabulary.................................................................................... xiv General terms.............................................................................. xiv Technical terms.......................................................................... xvii The maths you need. .............................................................. xxxiii Resources Sheets . ................................................................... xliii • SI multipliers ..................................................................... xliv • Constants .......................................................................... xliv • Formulas . .......................................................................... xlv Unit 1: Mechanics: Force and Newton’s Laws. ......................... 1 1.1 Revision: Vectors.......................................................................... 2 1.2 What is force?............................................................................... 3 1.3 Different types of forces. ............................................................. 3 1.4 Force diagrams and free body diagrams.................................... 6 1.5 Resultant (net) force. ................................................................. 11 1.6 Newton’s First Law of Motion (Law of Inertia).......................... 14 1.7 Velocity and acceleration: Revision........................................... 16 1.8 Newton’s Second Law of Motion in terms of acceleration...... 16 1.9 Newton’s Third Law of Motion................................................... 27 1.10 Newton’s Law of Universal Gravitation. .................................... 28 1.11 The difference between mass and weight. .............................. 30 Unit 2: Momentum and impulse...............................................33 2.1 Momentum. ................................................................................ 34 2.2 Change in momentum. .............................................................. 35 2.3 Newton’s second law of motion in terms of momentum......... 37 2.4 Impulse. ...................................................................................... 39 2.5 The principle of conservation of linear momentum................. 40 2.6 Problem types............................................................................. 41 2.7 Elastic and inelastic collisions. ................................................. 46 Unit 3: Vertical projectile motion in one dimension. ............. 50 3.1 Revision: Graphs of velocity, acceleration and displacement.............................................................................. 51 © Department of Basic Education 2015 introduction v Mind the Gap Physics

© Department of Basic Education 2015 3.2 Free fall....................................................................................... 52 3.3 Graphs of Projectile Motion Type 1: Dropping a projectile. ..... 53 3.4 Graphs of Projectile Motion Type 2: Projectile shot up, then falls down........................................................................... 55 3.5 Type 2a: A projectile projected vertically upwards which falls back to the same level....................................................... 56 3.6 Type 2b: A projectile projected vertically upwards which falls below the original level. ..................................................... 58 3.7 Graphs of Projectile Motion Type 3: A bouncing ball. .............. 60 Unit 4: Work, energy and power. ..............................................68 4.1 Work............................................................................................ 68 4.2 Energy. ........................................................................................ 73 4.3 Power. ......................................................................................... 77 Unit 5: Doppler Effect. ................................................................ 81 5.1 Waves: Revision.......................................................................... 81 5.2 The Doppler Effect. .................................................................... 84 5.3 Applications of the Doppler Effect with ultrasound waves...... 88 5.4 Applications of the Doppler Effect with light. ........................... 88 Unit 6: Electrostatics...................................................................90 6.1 Definition: Electrical change & electric force. .......................... 90 6.2 The Law of Conservation of Charge.......................................... 91 6.3 Coulomb’s Law. .......................................................................... 93 6.4 Electric fields around charged objects. .................................... 96 6.5 Electric field strength................................................................. 98 Unit 7: Electric circuits..............................................................103 7.1 Factors influencing the resistance of a wire conductor.........104 7.2 Ohm’s Law. ...............................................................................104 7.3 Voltage (Potential Difference) and Electromotive Force (emf)..........................................................................................106 7.4 Internal Resistance..................................................................106 7.5. Electric energy..........................................................................108 7.6 Power. .......................................................................................108 Unit 8: Electrodynamics: Electrical machines (generators and motors). ....................................................116 8.1 Motors and generators. ........................................................... 117 8.2 Alternating current circuits. .....................................................126 Unit 9: Optical phenomena and properties of materials. ...134 9.1 Electromagnetic waves and visible light: Revision.................134 9.2 The photoelectric effect...........................................................135 Unit 10: Emission and absorption spectra............................144 10.1 Continuous emission spectra..................................................144 10.2 Atomic emission spectra. ........................................................145 10.3 Atomic absorption spectra.......................................................146 vi CONTENTS  © Department of Basic Education 2014 Mind the Gap Physics

© Department of Basic Education 2015 Dear Grade 12 learner We are confident that this This Mind the Gap study guide helps you to prepare for the end-of-year Mind the Gap study CAPS Grade 12 exam. guide can help you to prepare well so that you The study guide does NOT cover the entire curriculum, but it does focus on pass the end-of-year core content of each knowledge area and points out where you can earn exams. easy marks. You must work your way through this study guide to improve your understanding, identify your areas of weakness and correct your own mistakes. To ensure a good pass, you should also cover the remaining sections of the curriculum using other textbooks and your class notes. Overview of the Grade 12 exam The following topics make up each of the TWO exam papers that you write at the end of the year: Cognitive Description Paper 1 (Physics) level 1 Remembering/Recall 15% 2 Understanding/Comprehension 35% 3 Applying and analysing 40% 4 Evaluating and creating (synthesis) 10% Paper Type of questions Duration Total Date Marking Physics 10 multiple-choice 1 questions – 20 marks 3 hours 150 October/November External Structured questions – 130 marks Paper 1: Physics Focus Content Marks Total Duration Weighting of cognitive levels Mechanics 63 Waves, sound and light 17 150 marks 3 hours 15 35 40 10 Electricity and magnetism 55 Matter and materials 15 © Department of Basic Education 2015 introduction vii Mind the Gap Physics

© Department of Basic Education 2015 Look out for these icons in the study guide. How to use this study guide This study guide covers selected parts of the different topics of the CAPS Grade 12 curriculum in the order they are usually taught during the year. The selected parts of each topic are presented in the following way: PAY SPECIAL ATTENTION • An explanation PAY ofSPECIAL terms and concepts; NB ATTENTION NB • Worked NBexamples to explain and demonstrate; NB • Activities with questions for you to answer; and PAY SPECIAL ATTENTION • Answers NBfor you to use to check yourNB own work. HINT! HINT! hint hint Hints to help you remember a concept NB PAY SPECIAL ATTENTION Pay special attention NBhint HINT! orEG - guide you in e.g. EG - worked examples Worked examples e.g. worked examples solving problems exams HINT! Step-by-step e.g. exams EG - worked examples Refers you to the Activities with hint questions for you instructions exemplar paper to answer exams e.g. EG - worked examples activity • The activities are based on exam-type questions. Cover the answers activity ACTIVITIES boy and girl-left and provided and do each activity ACTIVITIESonboy your own. Then check your answers. right of page exams and girl-left and Reward yourself for thingsrightyou of page get right. If you activity get any incorrect activity answers, make sure you understand where you went wrong before activity ACTIVITIES boy moving on to the next section. and girl-left and right of page • In these introduction pages, we will go through the mathematics activity that you need to know, in particular, algebra and graphs. These are Step by step comment comment activity Use this Step by step comment comment study guide as a workbook.ACTIVITIES boy crucial skills that you will need for any subject that makes use of Make notes, draw picturesand right of page mathematics. Make sure you understand these pages before you go girl-left and and highlight important any further. comment activity concepts. Step by step comment • Go to www.education.gov.za to download past exam papers for you to practice. Step by step comment comment viii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 Top 10 study tips 1. Have all your materials ready before you begin studying – pencils, pens, highlighters, paper, etc. Try these study tips to 2. Be positive. Make sure your brain holds on to the information you make learning are learning by reminding yourself how important it is to remember easier. the work and get the marks. 3. Take a walk outside. A change of scenery will stimulate your learning. You’ll be surprised at how much more you take in after being outside in the fresh air. 4. Break up your learning sections into manageable parts. Trying to learn too much at one time will only result in a tired, unfocused and anxious brain. 5. Keep your study sessions short but effective and reward yourself with short, constructive breaks. 6. Teach your concepts to anyone who will listen. It might feel strange at first, but it is definitely worth reading your revision notes aloud. 7. Your brain learns well with colours and pictures. Try to use them whenever you can. 8. Be confident with the learning areas you know well and focus your brain energy on the sections that you find more difficult to take in. 9. Repetition is the key to retaining information you have to learn. Keep going – don’t give up! 10. Sleeping at least 8 hours every night, eating properly and drinking plenty of water are all important things you need to do for your brain. Studying for exams is like strenuous exercise, so you must be physically prepared. If you can’t explain it simply, you don’t understand it well enough. Albert Einstein © Department of Basic Education 2015 introduction ix Mind the Gap Physics

© Department of Basic Education 2015 Mnemonics A mnemonic code is a useful technique for learning information that is Mnemonics difficult to remember. encode information and make it easier Here’s the most useful mnemonic for Mathematics, Mathematical Literacy to remember and Physical Science: BODMAS: B – Brackets O – Of or Orders: powers, roots, etc. D – Division M– Multiplication A – Addition S – Subtraction Throughout the book you will be given other mnemonics to help you remember information. The more creative you are and the more you link your ‘codes’ to familiar things, the more helpful your mnemonics will be. Education helps one cease being intimidated by strange situations. Maya Angelou x introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 Mind maps There are several mind maps included in the Mind the Gaps guides, summarising some of the sections. Mind maps work because they show information that we have to learn in the same way that our brains ‘see’ information. As you study the mind maps in the guide, add pictures to each of the branches to help you remember the content. You can make your own mind maps as you finish each section. How to make your own mind maps: 1. Turn your paper sideways so your brain has space to spread out in all directions. Mind 2. Decide on a name for your mind map that summarises the mapping your information you are going to put on it. notes makes them more interesting 3. Write the name in the middle and draw a circle, bubble or picture and easier to around it. remember. 4. Write only key words on your branches, not whole sentences. Keep it short and simple. 5. Each branch should show a different idea. Use a different colour for each idea. Connect the information that belongs together. This will help build your understanding of the learning areas. 6. Have fun adding pictures wherever you can. It does not matter if you can’t draw well. © Department of Basic Education 2015 introduction xi Mind the Gap Physics

© Department of Basic Education 2015 On the day of the exam 1. Make sure you have all the necessary stationery for your exam, i.e. pens, pencils, eraser, protractor, compass, calculator (with new batteries). Make sure you bring your ID document and examination admission letter. 2. Arrive on time, at least one hour before the start of the exam. 3. Go to the toilet before entering the exam room. You don’t want to waste valuable time going to the toilet during the exam. 4. Use the 10 minutes reading time to read the instructions carefully. This helps to ‘open’ the information in your brain. Start with the question you think is the easiest to get the flow going. 5. Break the questions down to make sure you understand what is being asked. If you don’t answer the question properly you won’t get any marks for it. Look for the key words in the question to know how to answer it. Lists of difficult words (vocabulary) is given a bit later on in this introduction. 6. Try all the questions. Each question has some easy marks in it so make sure that you do all the questions in the exam. 7. Never panic, even if the question seems difficult at first. It will be linked with something you have covered. Find the connection. 8. Manage your time properly. Don’t waste time on questions you are unsure of. Move on and come back if time allows. Do the questions that you know the answers for, first. GOOD LUCK! 9. Write big and bold and clearly. You will get more marks if the marker can read your answer clearly. 10. Check weighting – how many marks have been allocated for your answer? Take note of the ticks in this study guide as examples of marks allocated. Do not give more or less information than is required. xii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 Question words to help you answer questions It is important to look for the question words (the words that tell you what to do) to correctly understand what the examiner is asking. Use the words in the table below as a guide when answering questions. Question word/phrase What is required of you Analyse Separate, examine and interpret Calculate This means a numerical answer is required – in general, you should show your working, especially where two or more steps are involved Classify Group things based on common characteristics Compare Point out or show both similarities and differences between things, concepts or phenomena Define Give a clear meaning Describe State in words (using diagrams where appropriate) the main points of a structure/process/ phenomenon/investigation Determine To calculate something, or to discover the answer by examining evidence Differentiate Use differences to qualify categories Discuss Consider all information and reach a conclusion Explain Make clear; interpret and spell out Identify Name the essential characteristics PAY SPECIAL ATTENTION Label Identify on a diagram or drawing In every exam List Write a list of items, with no additional detail question, put a CIRCLE Mention Refer to relevant points around the question word and underline any other important Name Give the name (proper noun) of something key words. These words tell State Write down information without discussion you exactly what is being asked. Suggest Offer an explanation or a solution Tabulate Draw a table and indicate the answers as direct pairs © Department of Basic Education 2015 introduction xiii Mind the Gap Physics

© Department of Basic Education 2015 Vocabulary apply (v). Make a formal application; be relevant to; work hard; place on. The following vocabulary consists of all the approximate (v. & adj.). Come close to (v); difficult words used in Mind the Gap Mathematics, roughly, almost, not perfectly Mathematical Literacy, and Physical Science. We accurate, close but not exact. The suggest that you read over the list below a few verb is pronounced “approxi-mayt” times and make sure that you understand each and the adjective is pronounced term. Tick next to each term once you understand “approxi-mitt”. it so you can see easily where the gaps are in your ascending (adj). Going up. knowledge. C KEY Abbreviation Meaning cause (v). Make something happen. (v) verb: doing-word or action word, such cause (n). The person or thing that makes as “walk” something happen; an aim or (n) noun: naming word, such as “person” movement to which a person is (adj) adjective: describing word such as committed. “big” causality (n). Someone or something (adv) adverb: describing word for verbs, responsible for a result. such as “fast” clockwise (adj). In the direction a clock’s (prep) preposition: a word describing a hands move. position, such as “on”, “at” collide (v). To crash into; to hit. (sing) singular: one of consecutive (adj). One after another without any (pl) plural: more than one of gaps or breaks. (abbr) abbreviation consider (v). Think about. contrast (v). Show the difference between; (n) something that is very different General terms from what it is being compared with. Term Meaning conversely (adv). The opposite of. A counteract (v). Act against something in order to stop it. counter- see anticlockwise. abbreviate (v). Make shorter. clockwise account for (v). Explain why. adjacent (adj). Next to something. D affect (v). Make a difference to; touch the feelings of. Do not confuse with effect. See effect. data (pl), (n). Information given or found. datum (sing) analyse (v). Examine something in detail. deduce (v). To work something out by ante- (prep). Before (e.g., ante-natal – reasoning. before birth) deduction (n). Conclusion or idea that anti- (prep). Against (e.g., anti-apartheid someone has worked out. – against apartheid). define (v). Give the meaning of a word or anti- (adv. and adj.). In the opposite words. clockwise direction to the way a clock’s hands move. definition (n). The meaning of a word or words. apparent (adj). Clearly visible; the way something seems to be or the way deliver (v). To bring and hand over. it appears. denote (v). To refer to or mean something. appear (v). Come into sight; seem to be. descending (adj). Going down. xiv introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 determine (v). Work out, usually by experiment H or calculation. direct (v). Instruct or tell someone what horizontal (adj). Across, from left to right or to do; to control a process or right to left. (From “horizon”, the movement; to go straight. Moving line dividing the earth and the sky). from one place to another in the straightest and quickest way (adj). hover (v). Float just above a surface. hypothesis (n). A theory or proposed explanation. E hypothetical (adj). Theoretical or tentative; waiting for further evidence. effect (n). Result. effect (v). Carry out, do, enact. I eject (v). Force or throw something or someone out violently or suddenly. identify (v). Recognise or point out. elapse (v). Pass by or finish, e.g., time. illustrate (v). Give an example to show what establish (v). Show or prove, set up or create. is meant; draw. excluding (prep). Not including. impair (v). Weaken or damage. exclusive (adj). Excluding or not admitting imply (v). Suggest without directly saying other things; reserved for one what is meant. particular group or person. indicate (v). Point out or show. exemplar (n). A good or typical example. inter- (adj). Can be swapped or exempt (v). To free from a duty. changeable exchanged for each other. exempt (adj). Be freed from a duty. investigate (v). Carry out research or a study. exemption (n). Being freed from an obligation. issues (v). Comes out of. expel (v). Force someone or something to issues (n). An important problem or a topic leave a place. Eject. for debate. extent (n). The area covered by something. M F macroscopic (adj). Visible without being made factor (n). A circumstance, fact or bigger. influence that contributes to a magnitude (adj). Size. result; a component or part. manipulate (v). Handle or control (a thing or a factory (n). A place where goods are made person). or put together from parts. microscopic (adj). Very small, not visible without find (v). Discover or locate. being made bigger. find (n). Results of a search or discovery. motivate (v). Give someone a reason for finding (n). Information discovered as the doing something. result of an inquiry. mount (v). Attach, place upon (“mount fixed (adj). Not able to move, attached; or the picture”); climb on (“mount the repaired, not broken. chair”); begin (“mount an attack”). format (n). Layout or pattern; the way mount (n). Mountain; frame or attachment something is laid out. that you can mount things on. multiple (adj). Many. G N global (adj). Found all over the world (globe). negligible (adj). Small and insignificant; can be ignored. From “neglect” (ignore). © Department of Basic Education 2015 introduction xv Mind the Gap Physics

© Department of Basic Education 2015 numerical (adj). Relating to or expressed as a relative (adj). Considered in relation to number or numbers. something else; compared to. numerous (adj). Many. relative (n). A family member. represent (v). Be appointed to act or speak for O someone; amount to. resolve (v). Finalise something or make it clear; bring something to a observe (v). Look at; watch carefully. conclusion. obtain (v). Get. respect (v). Admire something or someone; occur (v). Happen. consider the needs or feelings of operate (v). Work; drive; control. another person. optimal (adj). Best; most favourable. respectively (adj). In regards to each other, in relation to items listed in the same optimum (adj). Best; (n) the most favourable order. situation for growth or success. orientation (n). Position or layout relative to other things or to compass points; S getting used to the position or layout of things. simul- (adv). At the same time. taneously P site (n). Place. phenomenon (n). A fact or situation that is seen T to exist or happen. phenomena (n). Plural of phenomenon. tendency (n). An inclination to do something prefix (n). Part of a word that is attached in a particular way; a habit. to the beginning of many different transmission (n). The act of sending words, changing their meaning, (transmitting) something. e.g., prehistoric – before written records were kept. transmit (v). Send across. principal (n). Head of a school. transverse (adj). Extending across something. principal (adj). Main or most important. truncated (adj). Cut short. principle (n). A basic truth that guides the way a person behaves. U provide (v). Make available for use; supply. uniform (n). Standardised clothing. Q uniform (adj). Remaining all the same at all times; unchanging. quality (n). The standard of something unimpeded (adj). Free to move. compared to other similar things; universal (adj). Found everywhere; true a characteristic of someone or everywhere; applicable everywhere. something. universe (n). Everything that exists. R V reciprocal (adj). Given or done in return. verify (v). Show to be true; check for truth; record (v). Make a note of something in confirm. order to refer to it later (pronounced vice versa (adv). The other way round. ree-cord). versus (prep). Against. Abbreviated “vs” record (n). A note made in order to refer and sometimes “v”. to it later; evidence of something; a copy of something (pronounced vertical (adj). Upright; straight up; standing. rec-cord. via (prep). By way of; by means of; travelling through. xvi introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 Technical terms atm (n). Abbreviation: atmospheres of pressure (1 atm = 101,3 kPa). A The pressure of the air at sea level. Same as “bar” (barometric pressure). absorption (n). To take into; the process of ATM (n). Abbreviation: automatic teller taking something in. machine. accelerate (v). To increase speed (or velocity) atmosphere (n). The air or the gases per second, measured in metres surrounding a planet; the sky; as a per second per second (m·s–2 or unit of measurement, see atm. m/s2). See also velocity, speed, atmospheric (adj). To do with the atmosphere. decelerate. atom (n). The smallest unit of a chemical algebra (n). A mathematical system element, which, if broken down where unknown quantities are further, no longer behaves in the represented by letters, which same way chemically. Consists can be used to perform complex of a nucleus or centre part which calculations through certain rules. is positively charged, and an ammeter (n). A device to measure electron cloud (negatively charged) amperage. See amp, amperage. which surrounds the nucleus. See amperage (n). The number of amperes. nuclear. amp, ampere (n). One coulomb of charge attract (v). To bring something closer. passing one point in one second. average (n). Mathematics: The sum of parts See coulomb for more. Technically, divided by the quantity of parts. In 1 ampere is that constant current common use: neither very good, which, if maintained in two straight strong, etc., but also neither very parallel conductors of infinite weak, bad, etc; the middle. In length, of negligible circular cross Physical Science and Mathematics: section, and placed 1 metre if you are asked to find the apart in a vacuum, would produce average, you always have to between these conductors a force calculate it using the information of 2 × 10−7 N/m. you have. For example, the amplitude (n). The size of something average of (1;2;3) is 2, because (common usage). Physics: the (1+2+3)/3 = 2. See also mean, maximum extent of a vibration median and mode. in the direction perpendicular to axis (sing), (n). A line along which points the direction of propagation; or, in axes (pl, can be plotted (placed), showing simpler language: the furthest a pronounced how far they are from a central vibration moves (left/right) when a “akseez”) point, called the origin. See origin. wave is travelling forward. “Vertical axis” or “y-axis” refers angel (n). In Abrahamic religions, a to how high up a point is above messenger from God. Note the the origin (or how far below). spelling. “Horizontal axis” or “x-axis” refers to how far left or right a point is angle (n). The difference in position away from the origin. between two straight lines which meet at a point, measured in degrees. Note the spelling. B anions (n). A negative ion. See cation, ion. apparatus (n). Equipment; parts of a scientific experiment. battery (n). A collection of cells connected in series (end-to-end). See cell. In area (n). Length x breadth (width). In common use, “battery” is used to common usage: a place. mean the same as “cell” (e.g. a armature (n). Part of an electrical engine or penlight or AA cell), but this use is generator (dynamo); the part which incorrect except for a car battery, has coils of wire wrapped around which consists of a series of cells. it, which is attached to a central axle (rod). © Department of Basic Education 2015 introduction xvii Mind the Gap Physics

© Department of Basic Education 2015 bias (n). To be inclined against Cartesian (adj). Anything believed or something or usually unfairly proposed by Rene Descartes. opposed to something; to not In particular, the x-and-y axis accurately report on something; to coordinate system. favour something excessively. cation (n). A positively charged ion. See bi- (prefix). Two. anion, ion. BODMAS (abbr.). Brackets, of/orders cell (n). An apparatus that generates (powers, squares, etc), division, electricity using electrochemistry. multiplication, addition, An AA or Penlight battery, as it is subtraction. A mnemonic commonly called, is a cell. A car (reminder) of the correct order battery consists of a number of in which to do mathematical cells inside a single container. operations. CFL (n). Compact Fluorescent Light; a body (n). Physics: any object. small fluorescent tube curled up inside a standard lightbulb shape. boil (v). Physics: to cause a liquid’s vapour pressure to exceed charge (n). Chemistry: having too the pressure of the gas in the many or too few electrons container, usually by heating it, (most commonly), resulting in but it can be done by lowering a substance ionising. A positive the pressure of the gas in the charge results from too few container, too. See vapour electrons, and a negative charge pressure. In common usage, to from too many electrons. Physics: make a liquid hot until it bubbles. a basic feature of all physical electromagnetic particles, except, bond (n). A connection. In physics and e.g. neutrons and photons, which chemistry, between atoms and have zero charge. All protons have molecules. a positive charge, all electrons brake vs (n & v). To brake means to stop. have a negative charge. break To break means to destroy. Brakes chart (v). To draw a diagram comparing are devices on vehicles which values on Cartesian axes. cause the vehicle to stop. breadth (n). How wide something is. From circuit (n). A track or pathway which the word “broad”. meets end-to-end. Electricity: a closed series of wires or cabling brush (n. & v.). Anything that contacts which starts at a power source and anything else lightly in a sweeping ends at the same power source motion (n); to sweep over which enables the flow of current. something whilst touching lightly (v). Electrodynamics: the item circum- (n). The distance around the outer that contacts the armature (the ference rim of a circle. part of the motor that turns). See coefficient (n). A constant value placed armature. next to an algebraic symbol as a multiplier. Same as constant (see below). Or: a multiplier or factor C that measures a property, e.g. coefficient of friction. coil (n). Electricity: a loop of wire or calibrate (v). To adjust a measuring tool series of loops of wire used to or measurement against a create a magnetic field. In a car known accurate measurement to engine, it refers to the device ensure that the measuring tool or which sends a spark of electricity measurement is accurate; to check to the spark plugs. General use: a measurement or measuring any loop of any wire, rope or string. tool’s accuracy; to mark with commutator (n). The connectors or rings which accurate measurements using a contact the brushes in an electrical standard scale like cm, mm, mℓ, engine. etc. Common use: to assess or compressed (adj). Subjected to pressure, evaluate carefully. squashed. xviii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 condensation (n). When a vapour or gas cools coordinate (n). The x or y location of a point on down and starts to collect into a Cartesian graph, given as an x or larger droplets; changing phase y value. Coordinates (pl) are given from vapour or gas to liquid. as an ordered pair (x, y). Condensation reaction: to produce correlate (v). To see or observe a relationship a larger molecule from two smaller between two things, without ones. showing that one causes the other. conditions (n). Physics and Chemistry: how correlation (n). That there is a relationship (STP) the environment is: temperature between two things, without and pressure. STP (Standard showing that one causes the other. Temperature and Pressure is 25˚C and 1 atm). correspond (v). To pair things off in a conduct (v). Electricity: to allow correlational relationship. For two electricity through a substance. things to agree or match. E.g. A Thermodynamics: to allow heat corresponds to 1, B corresponds to through a substance. 2, C corresponds to 3, etc. conductivity (n). How well a substance allows coulomb (n). A measure of charge, the heat or electricity through it; the quantity of electrons carried by a opposite of resistance. 1-amp current past a point in one conductor (n). A substance that allows second. 6,2415 × 1018 electrons. electricity through. A poor conductor counteract (v). Oppose or resist. obstructs or resists electricity. crest (n). The top of a wave in a conservation (n). A law which describes transverse wave. See transverse, something that does not change. trough, wave. E.g. the conservation of matter- energy says that matter-energy cubed (adj). The power of three; cannot be created or destroyed, multiplied by itself three times. only transformed from one form cubic (adj). Shaped like a cube; having into another. There are a number been multiplied by itself three of other conservations, e.g. times. momentum and torque. current (n). Flowing electrons. constant (n). See coefficient. Means cylinder (n). A tall shape with parallel sides “unchanging”. and a circular cross-section – think continuous (adj). Mathematics: having no of a log of wood, for example, or a breaks between mathematical tube. See parallel. The formula for points; an unbroken graph or curve the volume of a cylinder is πr2h. represents a continuous function. See function. control (n. and v.). To ensure something D does not change without being allowed to do so (v); an experimental situation to decelerate (v). To slow down. Opposite of which nothing is done, in order accelerate. to compare to a separate experimental situation, called the denominator (n). See divisor. In popular speech: ‘experiment’, in which a change a common factor. is attempted. The control is then density (n). How much mass is contained compared to the experiment to see in a particular volume; how if a change happened. compact something is. Measured control (n). A variable that is held constant in kg/m3 variable in order to discover the relationship depend (v). To be controlled or determined between two other variables. by something; to require something “Control variable” must not be to happen or exist first. confused with “controlled variable” (see independent variable). conventional (adj). In a standard way. Electricity: conventional current flows from + to -. In reality, electrons flow from - to +. © Department of Basic Education 2015 introduction xix Mind the Gap Physics

© Department of Basic Education 2015 dependent (adj/n). A variable whose value Doppler (n). The compression of a wave (variable) depends on another; the thing (effect) (increase in its frequency) as an that comes out of an experiment, object approaches an observer the effect; the results. See also and the spreading out of a wave independent variable and control (or lowering of frequency) as variable. The dependent variable the object moves away from an has values that depend on the observer. independent variable, and we plot ductile (adj). Able to be stretched, usually it on the vertical axis. into a thin wire. Said of metals. determine(s) (v). To cause; to ensure that; to set (causation) up so that; to find out the cause of. durable (adj). Tough; something that can endure. di- (prefix). Two. dynamic (adj). Changing often. Relating diagonal (adj. & n.). A line joining two to forces that produce motion. opposite corners of an angular Opposite of static. See static and shape. electrostatic. diameter (n). The line passing through the dynamo (n). A mechanical device structured centre of a shape from one side the same way as an electric of the shape to the other, esp. a motor, however, instead of taking circle. Formula: d = 2r. See radius, in electrical energy so as to turn, radii, circumference. mechanical energy is used to turn difference (n). Mathematics: subtraction. it and it generates electricity. A Informally: a dissimilarity. How machine that converts mechanical things are not the same. energy into electrical by rotating metal coils (usually copper) diffraction (n). The process of spreading within a magnetic field. Same as out light or splitting light into its generator. See generator, motor. component wavelengths. See wavelength. dimension (n). A measurable extent, e.g. E length, breadth, height, depth, time. Physics, technical: the base units that make up a quantity, e.g. earth (n., v.). The planet upon which we mass (kg), distance (m), time (s). live (n); Electricity: to connect an diode (n). A semiconductor device with electrical circuit to the earth (v); two terminals (electrodes), usually the wire that connects an electrical allowing current to flow in one circuit to the earth, having zero direction only. potential difference (n). Used to prevent circuit overload in the case discharge (v., n.). Release (v); something that of excess current. has been released (n). Electricity: to release a charge (v). ecliptic (n). A circle in outer space, representing the sun’s apparent displace (v). To move or relocate something. path during the year. displacement (n). A distance moved; the process Ek (n, abbr.). Kinetic energy. The of being moved. energy of motion. distance (n). See displacement. elastic (adj). Able to stretch and return distribution (n). How something is spread to its original shape. Physics: a out. Mathematics: the range and collision which does not transfer variety of numbers as shown on a Ek from one object involved in the graph. collision, to the other object(s). divisor (n). The number below the line electric (adj). Containing electricity in a fraction; the number that is (electrons). dividing the other number above electrode (n). General use: the point where the fraction line. See numerator, electrons enter or exit a power denominator. source or a circuit. Specifically domain (n). The possible range of x-values (Electrochemistry): Part of a circuit for a graph of a function. See dipped into a solution to receive or range. release electrons. See anode and cathode. xx introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 electro- (n). The study of motion caused by empirical (adj). Relating to the senses or to dynamics electricity, or electricity caused by things that you can see, touch, motion. taste, etc. Chemistry: empirical electro- (adj). That electricity causes formula: a formula giving the magnetic magnetism and vice versa; the proportions of the elements relationship between electricity present in a compound but not the and magnetism. actual numbers or arrangement of atoms; the lowest ratio of electro- (n). A coil of wire that becomes elements without giving structure magnet magnetised when electricity or quantities. passes through it. energetic (adj). Having a lot of energy; electromotive (adj). Usually electromotive force performing a lot of work. or emf. The potential difference caused by electromagnetism, energy (n). Work or the ability to do which causes current to flow. work. There are various forms of Producing a current with energy: motion (Ek), light energy electromagnetism. See emf. (photons), electrical energy, heat, etc. Energy can neither be created electron (n). A fundamental physical nor destroyed, but only converted particle bearing a negative charge, from one form to another. See weighing approximately 9 × 10−28g, conservation. which is found around atomic nuclei in areas called ‘orbitals’. engine (n). A machine that transfers or Responsible for electricity and converts energy, or converts power chemical reactions. Symbol e–. (electrical, chemical) into motion. See proton, nucleus. Ep (n, abbr). Potential energy. See electroscope (n). A device to measure the potential and Ek. Energy that strength of a charge of static a system has due to torsion electricity or ionisation of the air. (twisting), extension (stretching), See static. or gravitation (being placed at a height above a large body). When electrostatic (adj). Relating to electrons or a system which is under these electric fields which are not flowing conditions is released, it releases as current. See static. the Ep in the form of Ek. Examples: element (n). Mathematics: part of a set a compressed spring, someone of numbers. Physics: a pure about to jump. substance made only of atoms of estimate (n., v.). To give an approximate one type, with the same number value close to an actual value; an of protons in each nucleus. An imprecise calculation. element cannot be broken down further without losing its chemical evaporate (v). To change phase of matter properties. Each element has a from liquid to gas. Compare unique atomic number which is the sublimate and boil. number of protons in the nucleus. excited (n). The state of being in a higher See nucleus, atom, isotope. energy level (higher than ground Popular use: part of. state). elevation (n). Science: height above the exert (v). To impose or place pressure or ground or sea level. Architecture: a force upon something; to make an face of a building as viewed from a effort. certain direction on an architect’s exponent (n). When a number is raised to plan of the building. See plan. a power, i.e. multiplied by itself emf (abbr). Same as electromotive as many times as shown in the force. Always written in lowercase power (the small number up above (small letters). the base number). So, 23 means emission (n). Something released, e.g. gas, 2 × 2 × 2. See also cubed. light, heat. exponential (adj). To multiply something many emit (v). To release. times; a curve representing an exponent. © Department of Basic Education 2015 introduction xxi Mind the Gap Physics

© Department of Basic Education 2015 extrapolation (n). To extend the line of a graph frequency (n). How often. Physics: the further, into values not empirically number of crests and troughs of a documented, to project a future wave passing a point per second. event or result. In plain language: See wavelength, crest, trough. to say what is going to happen Experienced by humans as high- or based on past results which were low-pitched sounds (high or low obtained (gotten) by experiment frequency), or different colours and measurement. If you have (visible light). See also Doppler a graph and have documented effect. certain results (e.g. change vs friction (n). The act of rubbing two things time), and you draw the line further together; heat generated or lost in the same curve, to say what due to objects rubbing together; a future results you will get, that is force; the resistance between two called ‘extrapolation’. See predict. surfaces moving whilst in contact. function (n). Mathematics: when two F attributes or quantities correlate. If y changes as x changes, then y = f(x). See correlate, graph, fahrenheit (n). A temperature scale based Cartesian, axis, coordinate. Also: on human body temperature. a relation with more than one Water freezes at 32°F and variable (mathematics). Chemistry: boils at 212°F under standard functional group: part of a conditions. The conversion molecule that gives the substance formula to centigrade/celcius is: its chemical properties in common 5 with other similar chemicals. (˚F – 32) × ​ __ 9 ​= ˚C fuse (n., v.). Electricity: a wire that melts field (n). Physics: an area in which a when an excessive current passes force is experienced (felt) and through it (n). The rating of the is caused by either electricity, fuse in amps tells you what the gravity, or magnetism. E.g. an maximum current can be before electric field is one in which a the fuse melts and breaks the test charge (a free particle) would circuit. Common use: to join or move spontaneously. A magnetic merge two things (v). field is one in which iron filings automatically orient themselves along regular lines. G fluid (n). Any substance that can flow and take the shape of a container; liquid, some gels, and gas. gas (n). The third phase of matter. flux (n). A flow of energy or ions; a When a solid is heated it turns into substance used to lower another liquid, and when a liquid is heated substance’s melting point so that it turns into gas. it can be shaped or used more gaseous (adj). In a gas form. easily, e.g. glass and lead. generator (n). See dynamo. force (n). The exertion of energy. An gradient (n). A slope. An increase or influence tending to change the decrease in a property or motion of a body or produce measurement. Also the rate of motion or stress in a stationary such a change. In the formula for body. The size of such an influence a line graph, y = mx + c, m is the is often calculated by multiplying gradient. the mass of the body by its acceleration (F = ma). See energy, gradually (adv). To change or move slowly. exert. graph (n). A diagram representing experimental or mathematical values or results. See Cartesian. graphic (n., adj.). A diagram or graph (n). Popular use: vivid or clear or remarkable (adj.). xxii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 graphically (adv). Using a diagram or graph. I Popular use: to explain very clearly. gravitation (n). A force of attraction exerted ideal (adj). Not as seen in real on all bodies by all bodies. The life; theoretical. Ideal gas: a strength of the force is proportional hypothetical gas whose molecules to the mass of the body; the more occupy negligible space and massive the body, the larger the have no interactions, and which G(m1m2) consequently obeys the gas laws gravity. F = ​ _______​. See body, r2 (PV = nRT) exactly. force. ignition (n). The start of a combustion gravity (n). Same as gravitation. Popular reaction. Common use: to start use: seriousness. a car (which has an internal ground (n). Same as earth. combustion engine). See engine, combustion. illuminate (v). To explain or light up. H impact (n., v.). To hit hard (v); a strong influence or blow (n). heat (n). Physics: a measure of impair (v). Prevent; hinder; slow down. the average kinetic energy of impulse (n). Physics: A shock wave or the molecules or atoms in a impact which acts on a body for a substance; enthalpy; the energy short period of time, resulting in a of an object as molecular motion. change in the body’s momentum. Alternatively, infra-red radiation Effectively the same as (heat radiation) coming off a body. momentum. Or: a jolt of electricity. See body. Popular use: to have a sudden heavy (adj). Massive; having a great careless urge to do something. mass; weighty. See weight, mass. incandescent (adj). Giving off light as a result of hertz (n). A measure of frequency. being heated. hf (equation). E = hf is a measure incidence, (n., adj.). An event, something that of the energy of a photoelectron incident happened (n). As of light, falling (see photoelectron). h is Planck’s onto a surface (adj). “Incident constant, 6,626 × 10−34 J·s. f is light”. the frequency of incident light (see incline (n. & v.). Slope. See gradient (n); to incident). hf standing alone can lean (v). also represent light, a photon or independent (n). The things that act as input photoelectron. See photon. (variable) to the experiment, the potential horsepower (n). An old unit of power, causes. Also called the controlled approximately 750 W. See watts, variable. The independent variable power. is not changed by other factors, Huygens’ (n). Every point on a wave-front and we plot it on the horizontal principle may be considered a source of axis. See control, dependent secondary spherical wavelets variable. which spread out in the forward indigo (n). The colour between violet and direction at the speed of light. The blue; purplish-blue. new wave-front is the tangential induce (v). To cause or start. Electricity: to surface to all of these secondary cause a current to flow by moving wavelets. Used to explain a conductor (metal) in a magnetic refraction. field. hyperbola (n). Mathematics: a graph of a induction (n). The process of inducing a section of a cone with ends going current. off the graph; a symmetrical (both sides the same) open curve. inelastic (adj). Opposite of elastic; a collision in which energy is lost or hypotenuse (n). The longest side of a right- transferred. angled triangle. © Department of Basic Education 2015 introduction xxiii Mind the Gap Physics

© Department of Basic Education 2015 inertia (n). Physics: The same as J momentum; the unwillingness of an object to move or change direction when moving. joule (n). Unit of energy. infrared (n., adj.). Electromagnetic radiation (light) with a frequency just lower than visible red light. Heat K radiation. 800 nm to about 1 mm. Not visible to the naked eye. kelvin (n). Unit of temperature, with insulate (v). To separate or isolate from. absolute zero being the point Electricity: to prevent the flow of where no molecular motion occurs, electricity. at –273,15˚C. Hence, the freezing insulation (n). The state of being separated point of water is 273,15 K. Note from. Electricity: a substance or that there is no degree sign surface that stops the flow of before K. electricity. kinetic (adj). Pertaining to motion. (About intensity (n). How strong a force or field is; movement). how bright a light is. kWh (abbr). Unit of power (kilowatt interference (n). When two forces or things hours) that electricity suppliers interact in a way that prevents charge for. See power, watt. the other from acting. Physics: 1000 watts used in 1 hour specifically when two waves = 1kWh = 1 unit. So e.g. a 2000 W interact and either reinforce or heater uses 2 units per hour. cancel each other out. interval (n). Gap. A difference between two measurements. L intra- (adj). Within or inside a molecule. molecular See molecule, intermolecular. law (n). In Physical Science, a inverse (n). The opposite of. Mathematics: formula or statement, deduced 1 one divided by. E.g. ​ __ 2 ​is the (discovered) from observation inverse of 2. (watching). The formula or inversion (n). Chemistry: turning something statement will then predict that upside down. under the same conditions the same thing will always happen. E.g. ion (n). An atom or molecule or part of the first law of thermodynamics a molecule which has an electrical says that matter and energy charge due to gaining or losing one cannot be destroyed, but only or more electrons. changed from one form to another. ionic (bond) (adj.). A bond in which electrons LED (abbr). Light emitting diode. A have been transferred from one type of light bulb, used to make side of the molecule to another computer screens and modern resulting in a cation and anion, torches. which then attract. E.g. NaCℓ. linear (adj). In a line. Mathematics: ionisation (n). The process of ionising. See in a direct relationship, which, ionise. when graphed with Cartesian ionise (v). To turn into an ion. See ion. coordinates, turns out to be a isotope (n). An element which has a straight line. different number of neutrons from logarithm (n). Mathematics: a quantity the usual number of neutrons in representing the power by which a the element. E.g. 12C has 6 protons fixed number (the base) must be and 6 neutrons, but 14C has raised to produce a given number. 8 neutrons and 6 protons, and The base of a common logarithm is is radioactive. 10, and that of a natural logarithm is the number e (2,7183…). A log graph can turn a geometric or exponential relationship, which is normally curved, into a straight line. xxiv introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 longitude (n). Lines running north to south medium (sing) (n). Common use: the average. on the earth, measuring how far Science: the substance that east or west one is, in degrees, transmits something else (e.g. from Greenwich in the UK. glass, air), or allows something “Longitudinal” (adj) means from to pass through it, e.g. light, north to south, or top to bottom. information, etc. Plural is media. Running lengthwise. Physics: metal (n). A substance which is malleable a wave whose vibrations move (can be hammered flat), is ductile in the direction of propagation (can be drawn into a wire), which (travel). Example: sound. Statistics: conducts electricity and heat well a study in which information is and which is reflective (most light gathered about the same people striking it is emitted again). Most or phenomena over a long period elements are metals except the of time. few on the right hand side of the periodic table starting at Boron (B) and running diagonally down to M Astatine (At). meter (n). A device used to measure mach (n). A measure of speed. Mach 1 is something. You might see this the speed of sound, approximately spelling used in American books 340 m/s. for metre. See metre. macroscopic (adj). Large enough to be visible metre (n). The SI unit of length, 100 cm. to the unaided human eye; big metric (adj). A measurement system, enough to be seen. using a base of 10 (i.e. all the magenta (n). A bright purple/pink colour. units are divisible by 10). The USA uses something known as the magnitude (n). Size. Imperial system, which is not used manipulate (v). To change, or rearrange in science. The Imperial system is something. Usually in Mathematics based on 12. Examples: 2,54 cm it means to rearrange a formula to (metric) = 1 inch (imperial). solve for (to get) an answer. 1 foot = 12 inches = approx. mass (n). The amount of substance in a 30 cm; 1 metre = 100 cm. 1 Fl.Oz body. Do not confuse with weight, (fluid ounce) = approx 30 mℓ. which is the amount of force that a microscopic (adj). Too small to be seen by the mass exerts on a surface. The mass unaided human eye. of an object is constant everywhere microwave (n). Electromagnetic radiation in the universe (as the amount (light) with a lower frequency than of substance does not change). red light, and lower than infrared. However, the weight of an object Wavelength between 1 mm and changes according to the strength 30 cm. Used to heat food and in of the gravitational field on it. E.g. cellphone towers. you would weigh about 38% of your 1 modal (adj). Pertaining to the mode, or weight on Mars, and about ​ __6 ​ on method. Can mean: about the the moon, because these bodies mathematical mode or about the are smaller than earth. (Advanced method used. See mode. point: mass increases as you approach light speed). mode (n). Mathematics: the most common number in a series of matter (n). Substance; stuff. Opposite of numbers. See also mean, median. vacuum (nothing). model (n). A general or simplified way mean (n). See average. to describe an ideal situation, mechanical (n). By means of physical force. in science, a mathematical media (pl) (n). More than one medium, or way description that covers all cases of of transmitting or sending. the type of thing being observed. A representation. median (n). Mathematics: the number in the middle of a range of numbers written out in a line or sequence. © Department of Basic Education 2015 introduction xxv Mind the Gap Physics

© Department of Basic Education 2015 momentum (n). Same as impulse; roughly how O (sing), much force something can exert momenta (pl) when it collides with something else, or, more technically, how ohm (n). Unit of resistance, symbol Ω, much inertia it has (how unwilling written as R in a formula. it is to move or change direction). ohmic (adj). Follows Ohm’s law (V = IR). Specifically, where p is momentum: p = mv (mass × velocity). See omega (n). Last letter of the Greek velocity, mass, inertia, impulse. alphabet, Ω. Symbol for ohms. Momentum is conserved; if one opaque (adj). Not transparent; not allowing object collides with another, the sharp vision through, but allowing momentum of both objects before light through. and after the collision is the same. optical (adj). Pertaining to light or vision or moon (n). Any small spherical body that the eye. orbits a planet. See planet. optimal (adj). Best, most. orbit (n). The (approximate) circle traced by planets travelling around a star N (sun). Biology: the eye socket. origin (n). Mathematics: the centre of neutron (n). A subatomic particle with a Cartesian coordinate system. no charge, mass approximately General use: the source of the same as a proton, found in anything, where it comes from. the nucleus of an atom. Symbol outlier (n). Statistics: a data point which n0. If there are too many neutrons lies well outside the range of in a nucleus, the substance related or nearby data points. will be radioactive as it releases alpha particles (helium nuclei, 2p+ + 2n0). P newton (n). Unit of force, symbol N, equal to the force that would give a mass of one kilogram an acceleration of packet (n). Physics: a quantum. one metre per second per second parallel (adj). Keeping an equal distance (1 kg·m/s2). 1 kg mass exerts a along a length to another item force of 9,8 N on earth. (line, object, figure). Mathematics: nichrome (n). An alloy of nickel and chrome, two lines running alongside each with some iron; a wire of high other which always keep an equal resistance. distance between them. nonconserva- (adj). Not conserving; see particle (n). Any small part, e.g. a proton, tive conservation. an atom, a molecule. normal (n., adj.). Mathematics and particular (adj). A specific thing being pointed Physics: a force, vector or line that out or discussed; to single out or acts at right angles to another point out a member of a group. force, vector or line or object. (n). pascal (n). The unit of pressure, Common use: Regular or standard abbreviated Pa, units: N/m2 (adj). perimeter (n). The length of the outer edge; nucleus (n.) The centre of something the outer edge of a shape. (sing), nuclei (generally), specifically the centre (pl), nuclear of an atom, consisting of at least period (n). The time between crests or (n.) one proton (hydrogen), or two troughs of waves; the time gap protons and two neutrons (helium). between events; a section of time. Plural nuclei is pronounced “noo- periodic (adj). Regular; happening regularly. klee-eye”. perpendicular (adj). Normal; at right angles to numerator (n). The opposite of a denominator; (90˚). the number on top in a fraction. xxvi introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 phase (n). Time, period; a state of matter power (n). Power is the rate at which (solid, liquid, gas); the relationship (how fast) work is done or energy in time between the cycles of a is transferred or transformed system (such as an alternating (work/time or J/s). It is a scalar electric current or a light or sound quantity, measured in watts (W). wave) and either a fixed reference Mathematics: an exponent. See point or the states or cycles of exponent. another system with which it predict (v). General use: to foresee. may or may not be synchronised Physical Science: to state what will (simultaneous). I.e. if two systems happen, based on a law. See law. vibrate at the same time at the same rate, they’re “in phase”. pressure (n). A continuous force exerted on an object over a certain area, in phi (n). The Greek letter Φ (f) used to pascals, Pa. N/m2. See pascal. represent a photon (particle of light). probability (n). How likely something is. See likely. Probability is generally a photocell (n). A device which captures mathematical measure given as a light energy and converts it to decimal, e.g. [0] means unlikely, electrical energy. Makes use but [1,0] means certain, and of the photoelectric effect. See [0,5] means just as likely versus photoelectric. unlikely. [0,3] is unlikely, and [0,7] photoelectric (adj). The effect in which high- is quite likely. The most common energy light, such as UV, can cause way to express probability is as a electricity to flow in a metal, by frequency, or how often something 1 ejecting electrons from the metal comes up. E.g. an Ace is ​ ___ 13 ​ or as photoelectrons. 0,077 likely, in a deck of cards, photoelectron (n). An electron ejected from a because there are 4 of them in a metal by light. set of 52 cards. photon (n). A light particle. project (n. & v.). A project (n., pronounced PRODJ-ekt) is a plan of action pi (n). π, the Greek letter p, the ratio of or long-term activity intended to the circumference of a circle to its produce something or reach a goal. diameter. A constant without units, To project (v., pronounced prodj- value approximately 3,14159. EKT), is to throw something, or to plane (n). A flat surface. guess or predict (a projection). To planet (n). A large spherical astronomical project a result means to predict a body which orbits a star (sun). See result. See extrapolate. moon. projectile (n). An object that has been plot (v). To place points on a Cartesian projected (thrown or shot). coordinate system; to draw a graph. Projectiles will usually start at velocity (speed) 0 m/s, and polarity (n). To have poles or distinct accelerate, then fall to earth at the ends, e.g. positive and negative same rate as they accelerated. (electricity), north and south propagate (v). Send or deliver, spread. Said (magnetism); having two opposite of roots, information, and light or and contradictory properties. fields. positive (adj). Having many protons not propagation (n). The state of travelling paired with electrons; a lack of or spreading. “Direction of electrons. propagation” means the direction potential (n). Having the ability to do work, in in which something is travelling or particular, Ep (potential energy, the spreading. tendency to fall or start moving, proportion (n). To relate to something else as in a spring), or emf (voltage). in a regular way, to be a part of General use: potential exists when something in relation to its volume, there is an energy difference size, etc; to change as something between two points, e.g. due to else changes. See correlate and gravity or electrical charge. In the respectively. context of electricity, read it as “voltage”. © Department of Basic Education 2015 introduction xxvii Mind the Gap Physics

© Department of Basic Education 2015 proton (n). The positively-charged particle radiate (v). Physics: To send out radiation that forms the centre of an atomic (electromagnetic emissions, nucleus, weighing 1 836 times as light, or radioactivity). See emit. much as an electron, but having Mathematics: To spread out from a the same and opposite charge. central point. See radius. Symbol p+. See also nucleus, radiation (n). Radioactivity or neutron, electron. electromagnetic emissions. See pull (n). A force towards the source of radioactive. the force. Measured in newtons, N. radio (n). Electromagnetic radiation pulse (n). A single vibration or wave with a wavelength larger than crest/trough. microwaves. pump (n). A machine that uses energy radioactive (adj). Giving out harmful radiation, to transfer a fluid from one especially alpha (helium nuclei), place to another. In Biology one beta (high-energy electrons) finds cellular pumps, which are or gamma (high-frequency biological machines for transferring electromagnetic radiation). ions and nutrients. radius (sing), (n). The distance between the push (n). The opposite of a pull. radii (plur) centre of an object, usually a circle, and its circumference or Pythagoras’s (n). The square on the hypotenuse outer edge. Plural is pronounced Theorem is equal to the sum of the squares “ray-dee-eye.” on the other two sides of a right- angled triangle. Where h is the rainbow (n). A visible spectrum produced hypotenuse, a is the side adjacent by sunlight diffracting through to the right angle, and b is the raindrops. See diffraction. other side: h2 = a2 + b2. random (n). Unpredictable, having no cause or no known cause. Done without planning. Q range (n). Mathematics: the set of values that can be supplied to a function. The set of possible y-values in a qualitative (adj). Relating to the quality graph. See domain. or properties of something. A rate (n). How often per second (or per qualitative analysis looks at any other time period). Physics: changes in properties like colour, number of events per second; see that can’t be put into numbers. frequency. Often contrasted with quantitative. quantitative (adj). Relating to, or by comparison ratio (n). A fraction; how one number to, quantities. Often contrasted relates to another number; exact with qualitative. A quantitative proportion. If there are five women analysis is one in which you for every four men, the ratio of compare numbers, values and women to men is 5:4, written measurements. with a colon (:). This ratio can be 5 1 represented as the fraction __ ​ 4 ​ or 1​ __ 4​ quantity (n). Amount; how much. or 1,25; or we can say that there quantum (n). The smallest amount of energy are 25% more women than men. possible in a wave of a particular ray (n). Physics: A single straight line frequency, a discrete quantity of of electromagnetic radiation. energy E proportional in magnitude Mathematics: a line from amongst to its frequency. In a formula, a set of lines passing through the E = hf. See hf. same central point. See radius. reflective (adj). Giving off all light shone R upon it. Popular use: someone who thinks a lot. radar (n). A device which uses radio to detect moving vehicles, especially aircraft. xxviii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 refraction (n). Bending light when it travels S from one medium (e.g. air) into another medium (e.g. water or glass). Changing the direction of satellite (n). Any body that orbits any other propagation of any wave as a result body. Astronomy: a moon is a of its travelling at different speeds satellite of a planet. A planet is at different points along the wave a satellite of a star. Engineering: front. See Huygens’ principle, a man-made machine that diffraction. orbits the earth to provide repel (v). To push away from. Physics: telecommunications or military like charges repel, unlike charges services. attract. scalar (n). A quantity that does not have a repulsion (n). The process of repelling direction, only a magnitude (size). something. Compare vector. Example: mass. resistance (n). How much a conductor (usually scale (n). A system of measurement, with metal) slows down or impedes regular intervals or gaps between the flow of current. See current, units (subdivisions) of the scale. impede, ohm. series (n). Physics: components of resistor (n). A device which resists current. a circuit arranged end-to-end without any branches, so that rest (n). Physics: A state of lowest current passes through all items energy. in the circuit one after the other. resultant (n). Physics: A force, velocity, or Chemistry: a range of elements on other vector quantity which is the periodic table with common equivalent to the combined effect properties. of two or more component vectors SI (abbr). Système International. The acting at the same point. The international system of metric result of mechanical forces pulling units used by scientists. See in different directions. See vector. metric, IUPAC. It is calculated using trigonometry, with two of the vectors being sides simplify (v). To make simpler. Mathematics: of a right-angled triangle, and the to divide throughout by a common resultant being the third side. factor (number or algebraic letter) that will make the equation easier revolution (n). Physics: A turn; to turn over. to read and calculate. Hence common usage in politics: to overturn the government. solve (v). To come up with a solution (answer). Show your working. revolve (v). To turn around a point; usually said of something moving at a sound (n). Vibrations that are transmitted distance from the central point or through air or another medium, another object. Compare to rotate. which one can hear (detect with rhombus (n). A quadrilateral (four sided) the ear). It travels at approximately figure (diagram or shape) which mach 1. See mach. Old use: to has equal sides, but no right- be safe or intact, as in “safe and angles (90˚ angles). sound” (compare “asunder”, and “sundry”, meaning in parts). rms (abbr). Root mean square. Always lowercase. The square root of the space (n). Astrophysics: the vacuum arithmetic mean (average) of the between planets, stars, etc. squares of a set of values. See spectroscope (n). A device used to split light up average. Used to get the average into a spectrum, to analyse the voltage of an AC current. components (elements) that make rotate (v). Same as revolve, but usually up the light source. Used to tell said of a solid object turning what elements make up stars. around its own centre point. © Department of Basic Education 2015 introduction xxix Mind the Gap Physics

© Department of Basic Education 2015 spectrum (n). A series of continuous substitution (n). The process of substituting. (sing), spectra wavelengths which make Mathematics: to replace an (pl) up a portion of the range of algebraic symbol in a formula with electromagnetic radiation. a known value or another formula, More simply, a section of so as to simplify the calculation. electromagnetic radiation, or light. See simplify. Chemistry: to cause a E.g. the microwave spectrum, substituent to bond to a substance. the radio spectrum, the visible sum (n., v.). To add things up. light spectrum, etc. A rainbow is Represented by Greek Sigma (s): a spectrum of visible light. See ∑ or the plus sign (+). rainbow. sun (n). Our nearest star. See star. speed (n). How much distance is covered in a certain amount of time; the system (n). Any closely associated and scalar of velocity. Measured in inter-related or inter-dependent metres per second (m/s). See group of things; a set of things velocity. working together. Chemistry: a vessel (container) which contains a sphere (n). A perfectly round three- chemical reaction. dimensional shape. A ball. square (n). Mathematics: a shape or figure with four equal sides and only right T angles; the exponent 2 (e.g. the square of 4 is 42 = 16). squared (adj). Having been multiplied by tangent (n). Mathematics: a straight line itself, put to the exponent 2. See touching a curve only at one point, square. indicating the slope of the curve at that point; the trigonometric function stable (adj). Chemistry and Nuclear of the ratio of the opposite side of Physics: not likely to break down or a triangle to the adjacent side of a react further. triangle in a right-angled triangle; a star (n). A large spherical body in outer curve that goes off the chart. space undergoing continuous temperature (n). A measurement of heat. See nuclear fusion. E.g. the sun. See heat, enthalpy. SI unit is kelvin. sun. The second-nearest star to us is Alpha Centauri, at 4,7 light- tensile (adj). How strong a material is years. (strength) when stretched; how much force it static (adj). Not in motion, opposite can withstand before breaking. of dynamic. Also short for static tension (n). A force of stretching in a electricity. material. stationary (n). Not moving. terminal (n). Final; end point. stationery (n). Pens, pencils, paper, ink, files, theory (n). A usually mathematical folders, etc. representation of an explanation statistics (n). The mathematics of chance for something in the sciences, and probability. which does not depend on the thing being explained. A theory steam (n). Water vapour, microscopic differs from a law in that theories droplets of water. Not a gas, a are prone to empirical (visible or suspension of water droplets in air. measurable) refutation (rejection); See suspension, gas, liquid, phase, meaning that they can be aerosol. discarded if evidence comes in STP (abbr). Standard temperature and that they are wrong. See law. pressure; 101,3 kPa and 25˚C. threshold (n). Physical Science: the sublimate (v). To change phase of matter magnitude or limit of something, from solid straight to gas without which, if exceeded, will cause the intermediate phase of liquid. something else, e.g. release of See the case of dry ice (CO2). radiation, a chemical reaction, etc; the minimum amount of energy substance (n). Matter. Physical things. required to cause something. substitute (v). To replace. Medicine: the maximum safety level of a dose. xxx introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 thrust (n). See push. unit (n). A subdivision of a scale. See tides (n). Motion of the earth’s oceans scale. due to the gravitational pull of the unstable (n). Chemistry or Nuclear Physics: moon. prone to disintegrating or reacting. torque (n). Angular momentum; a force UV (abbr). See ultraviolet. that causes rotation; the amount of force delivered through a rotation. See momentum. V trajectory (n). The path that a projectile will take. See projectile, project (v). vacuum (n). The absence of matter. See transformer (n). Physics: a device used to matter. Vacuums do not have increase (step-up) or decrease suction powers. Suction is really a (step-down) the voltage of an AC result of a higher pressure pushing current. objects into the area of low transistor (n). Physics: a device which can pressure (vacuum). hold current. vapour (n). The pressure above a liquid transparent (adj). Allowing most light through. pressure caused by molecules evaporating Compare opaque. from the surface of its liquid form, transverse (adj). A wave whose vibrations move when in phase equilibrium (i.e. as (wave) perpendicular to the direction of many molecules leaving the liquid propagation (left/right). Example: surface are condensing back into light. Compare longitudinal. the liquid). travel (n). Movement. variable (n., adj.). A letter used to represent an unknown quantity in algebra trends (n). Mathematics: regular patterns (n); a quantity that changes (n); within data. subject to change (adj). trigonometry (n). Mathematics: the relationship vector (n). A mathematical quantity with and ratios between sides and direction and magnitude (size). angles within a right-angled Example: velocity. Compare speed, triangle. scalar. trough (n). The low point in a transverse velocity (n). The amount of distance wave. See crest, wave, transverse. covered per unit time, in a “Trough” is pronounced “troff”. specified direction. Compare tungsten (n). An element (metal) used in speed. Units: m/s. incandescent (hot) light bulbs, visible (adj). Able to be seen by the human symbol W. eye, opposite of invisible. Compare microscopic, macroscopic. U viscosity (n). The thickness of a fluid. A viscous fluid flows slowly, e.g. syrup. Pronounced “viss-KOSS- ultrasound (n). Sound with too high a itee” and “viss-k’s”. frequency for the human ear to volatility (n). How easily something detect, used in scanning the body, evaporates. E.g. Ether (C2H5OC) is e.g. the heart for obstructions for more volatile than water. blood flow, or foetuses to image volt (n). Unit of potential difference (draw, depict) them on a scanning in electricity. The difference of machine. potential (Ep) that would carry one ultraviolet (n). Light with a higher frequency ampere of current against one than violet light, not visible to ohm resistance. Same as emf. See the human eye. Present in emf, resistance, ohm, ampere. sunlight, causes sunburn. Can voltage (n). The measurement of volts. be stopped by Ozone (O3) in the atmosphere. In darkness it volume (n). A measure of the space reflects off white clothing as violet occupied by an object, equal to light. Also called “black light”. length x breadth x height. Wavelength 400 nm - 10 nm. © Department of Basic Education 2015 introduction xxxi Mind the Gap Physics

© Department of Basic Education 2015 W weber (n). The SI unit of magnetic flux, causing the emf of 1 volt in a circuit of one coil when generated watt (n). Unit of power. See power. or removed in one second. W P = ​ __ t ​(joules per second). weight (n). The force exerted on the wattage (n). The amount of power being mass of a body by a gravitational used, usually rated in kWh. See field. See mass, gravity, force, kWh. acceleration. W = mg, where g = 9,8 m/s2, m is the mass of wave (n). A series of increasing and the body. decreasing energy concentrations, e.g. sound, light, heat radiation, work (n). Energy, measured in joules. movements on liquid surfaces, Usually in the sense of the amount vibrations. A periodic vibration of energy required to move an either in a substance (e.g. air, object or change its course water, solids) which disturbs (direction of motion). or moves the particles in that substance, or a vibration or fluctuation (regular change) in X electromagnetism. Waves have compressions and rarefactions (dense and looser areas) in air x-rays (n). Electromagnetic radiation of or water or solids when acting higher energy than UV but lower as sound. When acting on the energy than Gamma radiation. surface of a liquid, or in light or Wavelength 0,01 to 10 nm. Used electromagnetic waves, waves to see into the body, but emitted by have crests and troughs (high stars as well. points and low points). See trough, crest, propagate. Waves have a period (time range between crests/ Y troughs), a frequency (how many crest/trough combinations pass a point per second), as well as a yard (n). Old Imperial measurement of wavelength. See wavelength below. length, approximately equal to a metre (1,09 m). wavelength (n). The distance between crests/ troughs, written as Greek L (lambda): λ. xxxii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 The maths you need This section gives you the basic mathematical skills that you need to pass any subject that makes use of mathematics. Whether you’re studying for Physical Science, Mathematics, or Mathematical Literacy, these basic skills are crucial. Do not go any further in this book until you have mastered this section. 1. Basic Pointers • If a formula does not have a multiplication (x) sign or a dot-product (⋅), and yet two symbols are next to each other, it means “times”. So, m1m2 means mass 1 times mass 2. You can also write it as m1 × m2, or m1⋅m2 • Comma means the same as decimal point on your calculator (i.e. 4,5 = 4.5). Do not confuse the decimal point with dot product 1 (multiply): 4.5 = 4​ __ 2 ​but 4⋅5 = 20. Rather avoid using the dot product for this reason. • In science it is common to write divisors with an exponent. This means, for example, that 0.5 metres per second is usually written 0.5 m⋅s–1 rather than 0.5 m/s. Both notations are perfectly correct, however, and you may use either. It is important, however, that you either use –1 or / . If you just put 0.5 ms, that means 0.5 milliseconds, which is not a velocity (speed in a direction); it is a time. • A variable is something that varies (means: changes). So, for example, the weather is a variable in deciding whether to go to the shops. Variables in science and mathematics are represented with letters, sometimes called algebraic variables. The most common you see in maths is x, probably followed by y, z. In science, variables are given their letter symbols specifically depending on what they stand for; so, for example, M or m are used for mass (amount of substance in kilograms); v is used for velocity (speed in a certain direction); a is used for acceleration (change in velocity), etc. You can guess for the most part what a variable is for by what its letter is; so V is voltage, R is resistance, P is pressure, and so on. © Department of Basic Education 2015 introduction xxxiii Mind the Gap Physics

© Department of Basic Education 2015 PAY SPECIAL ATTENTION NB NB 2. Subject of Formula or Solving For Very often in science and mathematics you have to “make something the subject of a formula” HINT! or “solve for something”. This refers to finding the hint value of an unknown quantity if you have been given other quantities and a formula that shows the relationship between them. e.g. Worked example 1 EG - worked examples If John has 5 apples, and he gives some to Joanna, and he has two apples left, how many did he give to Joanna? Well, the formula would be something like this: exams 5–x=2 To solve for x, we simply have to swap the x and the 2. What we’re actually doing is adding “x” to both sides: 5–x+x=2+x this becomes: activity 5 =PAY2 SPECIAL + x ATTENTION then NB we subtract 2 from both sides to move the 2 ACTIVITIES boyover: NB and girl-left and 5–2=2-2+x right of page 5–2=x activity 3=x ... so John gave Joanna three apples. HINT! The same procedures apply no matter how complex the formula looks. Just either add, hint subtract, square, square root, multiply, or divide throughout to move the items around. Step by step comment comment e.g. Worked example 2 EG - worked examples Let’s take an actual example from Electricity: V = IR. This means, the voltage in a circuit is equal exams to the current in the circuit times the resistance. Suppose we know the voltage is 12 V, and the resistance is 3 Ω. What is the current? V = IR 12 = 3 × I divide throughout by 3 so as to isolate the I () ​  12 ​= ​_​  3 ​  ​ I __ 3 3 rememberactivity that anything divided by itself is 1, so: ​  ​= (1) × I ... and __ __ 12 ​ 12 ​= 4 ... so 3 3 ACTIVITIES boy 4 = I or and girl-left and I = 4 A ... The circuit has a current of 4 amperes. right of page It is possibleactivity to remember how to solve for these equations using a triangle mnemonic as shown: V If you’re solving for V, cover V with your hand. Then, I next to R means times R, or IR. So, V = IR. If you’re solving for R, cover R with your hand. V is over I. So R = ​ V_​. While I I R this is an easier wayStep to do it, remember thatcomment by step many formulas do not consist of only three comment parts, so it is better to know how to make something the subject of a formula, or solve for something. xxxiv introduction  © Department of Basic Education 2015  Mind the Gap Physics

HINT! © Department of Basic Education 2015 hint e.g. Worked example 3 EG - worked examples Here’s a more tricky example. Given exams Kc = 4,5 [SO3] = 1,5 mol/dm3 [SO2] = 0,5 mol/dm3 (x – 48) [O2] = ​ ______​ mol/dm3 64 solve for x. [SO ]2 (1,5)2 Kc = _______ ​  23 ​ ∴ 4,5 = _________ ​  ​ [SO ] [O ] (x – 48) activity 2 2 (0,5)2 ______ ​  ​ 64 ∴ x = 176 g ACTIVITIES boy and girl-left and How did we get that answer? right of page activity Step by Step Let’s see how it works. First, solve for the Step exponents by step(powers): comment comment 2,25 4,5 = _________ ​  (x – 48) ​ (0,25) ______ ​  ​ 64 Now, we can see that 2,25 and 0,25 are similar numbers (multiples of five), so let’s divide them as shown. 4,5 = (2,25 / 0,25) × ( (x – 48) / 64 ) That leaves us with 4,5 = 9 × ( (x – 48) / 64 ) But if we’re dividing a divisor, that second divisor can come up to the top row. Here’s a simple example: 1 ÷ (2 ÷ 3) = ​ ___12 ​ ​  ​ 3 = ​  1____ ×3 ​ 2 = ​ _3 ​= 1,5 2 If you doubt this, try it quickly on your calculator: 1 ÷ (2 ÷ 3) ... this means, one, divided by two-thirds. Well, two-thirds is 0,6667, which is almost one. So how many “two- thirds” do you need to really make up one? The answer is one and a half “two-thirds”... i.e. 0,6667 + (0,6667 ÷ 2) = 1. Hence the answer is 1,5. So, back to the original problem, we can bring the 64 up to the top line and multiply it by nine: 4,5 = 9 × ( (x – 48) / 64 ) 4,5 = ​  9 × _____64 ​ x – 48 4,5 = ​  576 ​ _____ x – 48 Now we can inverse the entire equation to get the x onto the top: ___​ = x – ​ 48 ​  1 ___​ 4,5 576 © Department of Basic Education 2015 introduction xxxv Mind the Gap Physics

© Department of Basic Education 2015 Now we multiply both sides by 576 to remove the 576 from the bottom row (x – 48) 576 ​  576 ___​ = _________ ​  ​ 4,5 576 and we cancel the 576’s on the right hand side as shown above. Now, if 576 ÷ 4,5 = 128, then 128 = x – 48 Now we add 48 to both sides to move the 48 across 128 + 48 = x – 48 + 48 ... hence, 128 + 48 = x = 176. 3. Statistics Many experiments in science and many reports in economics make use statistics. You should therefore at least know the following: Dependent variable: The thing that comes out of an experiment, the effect; the results. Independent variable(s): The things that act as input to the experiment, the potential causes. Also called the controlled variable. Control variable: A variable that is held constant in order to discover the relationship between two other variables. “Control variable” must not be confused with “controlled variable”. It is important to understand that correlation does not mean causation. That is, if two variables seem to relate to each other (they seem to co- relate), it doesn’t mean that one causes the other. A variable only causes another variable if one of the variables is a function f(x) of the other. We will see more about this when we look at graphs, below. Mean: The average. In the series 1, 3, 5, 7, 9, the mean is 1 + 3 + 5 + 7 + 9 divided by 5, since there are 5 bits of data. The mean in this case is 5. Median: The datum (single bit of data) in the precise middle of a range of data. In the series 1, 3, 5, 7, 9, the median value is 5. Mode: The most common piece of data. In the series 1, 1, 2, 2, 3, 3, 3, 4, 5, the mode is 3. Often in scientific formulas it is said that things are proportional to each other. However, we cannot calculate the value of a force or energy output or mass etc., if we only know what things are in proportion (i.e. which things correlate). Let’s take momentum for example. Momentum (how forcefully something moves, more or less), is proportional to velocity (speed in a direction). So the faster something’s moving, the more momentum it has. But p (momentum) can’t be calculated if we only know velocity; we need to know mass as well. Why? Because momentum is also proportional to mass; the more massive something is, the more momentum it has. Thus, to get rid of the proportionality sign (∝), we have to come up with a formula. Many experiments in science serve to find out what the relationship is between two variables, i.e. if they’re merely correlated — proportional — or if they’re causally related. In the case of momentum, it’s easy, because there are no further variables: p = mv. However, in the case of gravity or xxxvi introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 electric or magnetic field strengths, it’s not that easy. In those cases, we have to introduce something called a “constant”. A constant is a fixed value that is always multiplied into an equation. Constants are often written k. However, some specific constants, such as in the Law of Gravity, have their own symbol, in this case, G. These constants are given in the tables later in this book. 4. Triangles Science makes a lot of use of triangles, e.g. vector force diagrams, areas under graphs, and so on. b The area of a triangle is half the base times the height: a = ​ __ 2 ​(h). A triangle with a base of 5 cm and a height of 3 cm will have an area of 2,5 × 3 = 7,5 cm2. This is useful when considering graphs of motion/ acceleration. A = 7.5 b Base 5 a hb c hb Height 3 PAY SPECIAL ATTENTION γ NB NB b Lengths of Triangle Sides You can calculate the lengths of sides of right-angled triangles using Pythagoras’ Theorem. HINT! The square of the hypotenuse is equal to the sum of the squares of the other two sides: In this diagram, b = base, hb = height, andhint c = the hypotenuse: c2 = hb2 + b2. e.g. Worked example 4 EG - worked examples In the triangle shown, the hypotenuse, marked “?”, can be obtained by squaring both sides, exams ? adding them, and then square-rooting them for 3cm the length of the hypotenuse. That is: 32 + 52 = 9 + 25 = 34. Since in this case 34 = hyp2 it follows that the square root of 34 gives 5cm the value of “?”, the hypotenuse. That is, 5,83 cm. activity This is useful when you get to vector addition, e.g. if the known sides were vectors and we wanted to know the value ACTIVITIES boy and strength of the force, and the direction in which it would go.and girl-left and right of page activity 5. Trigonometry You can use trigonometry to calculate triangle sides’ sizes if you do not have enough information, e.g. you do not have the sizes of at least two sides (but do have the angle). Step by step comment comment sin = opposite / hypotenuse sin = O/H cos = adjacent / hypotenuse cos = A/H tan = opposite / adjacent tan = O/A © Department of Basic Education 2015 introduction xxxvii Mind the Gap Physics

© Department of Basic Education 2015 PAY SPECIAL ATTENTION NB NB It’s probably easiest to remember this as SOHCAHTOA (soak a toe or soccer toe). Hypotenuse is the longest side next to to the angle, usually represented with a theta (θ). “Opposite” HINT! means the side of the triangle directly opposite the angle. “Adjacent” means the side adjacent to (next to) the angle, that hint is not the hypotenuse. e.g. Worked example 5 EG - worked examples hyp oten opposite us 3cm e In this triangle, the side opposite the angle θ is 3cm long. The side adjacent to the angle θ, and the hypotenuse, are unknown. Theta, the angle, is 30 degrees. exams θ = 30° How do we calculate the hypotenuse? Well, adjacent sin θ = ​ O __​= 3 cm ÷ H. H sin 30˚ = 0,5 (you can get this from your calculator, or memorise it). thus 0,5 = 3 / H solving for H, we multiply throughout by H to make H the subject of the formula: activity H × 0,5 = 3 × H ÷ H ACTIVITIES boy H × 0,5 = 3 and girl-left and now divide throughout by 0,5 toright of page isolate the H: activity H × 0,5 ÷ 0,5 = 3 ÷ 0,5 H = 3 ÷ 0,5 ∴ H = 6 cm Let’s try work out what the adjacent side is equal to, assuming we don’t know the hypotenuse. tan θ = O/A Step by step comment comment tan 30˚ = 3 cm ÷ A 0,57735 = 3 ÷ A A × 0,57735 = 3 × A ÷ A A × 0,57735 = 3 A = 3 ÷ 0,57735 A = 5,196 cm ≈ 5,2 cm. Let’s check this with Pythagoras. Suppose we want to prove that the opposite side is 3 cm. We have H = 6 and A = 5,2. So, Pythagoras tells us that A2 + O2 = H2. So, 5,22 + O2 = 62 O2 = 62 – 5,22 O2 = 36 – 27 O2 = 9 So the square root of O2 will give us O .... namely, O = 3 cm. The trigonometric calculation is correct. Lastly, there are three other operations you can do in trigonometry, but they’re just inverses of the first three: cosecant, secant, and cotangent. Cosec, sometimes abbreviated csc, is the reciprocal (inverse) of sine. Sec is the inverse of cosine. And cot is the inverse of tangent. So this means if sin = O/H, then cosec = H/O, and so on. xxxviii introduction  © Department of Basic Education 2015  Mind the Gap Physics

HINT! © Department of Basic Education 2015 hint e.g. Worked example 6 EG - worked examples unknown Earth orbits the sun at a distance of 149 597 870 700 metres or 149 597 870,7 km (one hundred and forty nine million km). This distance is called the AU, or astronomical exams unit. The flat disc that corresponds to earth’s orbit is called the ‘ecliptic’. Suppose that on θ 21 December, an unknown object is observed at an angle of 88˚ to the ecliptic, and that sun 2 AU on June 21 the same object is observed at 92˚. How far is the unknown object in AU? earth ecliptic Step by Step activity Step 1. Ignore extra information. Since the earth orbits the sun, the angle to the ACTIVITIES boy unknown object relative to the earth is the and same girl-left in andboth cases; it’s just that on one date, the earth is on one side of the unknownright of page object, and on the other date, it is on the other side. activity From the angles given, you can tell that the unknown object is at 90˚ to the sun relative to the earth. Step 2. We know the angle to the unknown object, and the distance to the sun. So, earth if we draw a triangle where the sun is the right-angle, the earth is at the tip of the θ Step by step comment comment hypotenuse, and the distant unknown object is opposite the sun, we get the following ? triangle. So, we want the hypotenuse. We know that triangles add up to 180˚, so the difference sun ? between θ and the given angle of 88˚ is 2˚. That means that the angle the unknown unknown object makes in reference to earth is 2˚. Thus: O sin = ​ __ ​ H sin 2˚ = 1 AU ÷ H = 149 597 870,7 km ÷ H 0,035 = 149 597 870,7 km ÷ H H = 149 597 870,7 km ÷ 0,035 H = 4 286 533 756,4964 km = 28,6 AU This means that the unknown object is 4,2 billion km away, or 28,6 AU away. 6. Graphs A lot of work in science involves interpreting graphs. You get graphs of motion, graphs of rates of chemical reactions, graphs of distance-relative strengths of force fields, and so on. Before you can understand these graphs, it’s probably best to start from scratch with Cartesian Coordinates. “Coordinates” are numbers that refer to the distance of a point along a line, or on a surface, or in space, from a central point called the “origin”. Graphs that you will use have only two dimensions (directions). The positions of points on these graphs are described using two coordinates: how far across (left-to-right) the point is, called the x-coordinate, and how far up-or-down on the page the point is, called the y-coordinate. © Department of Basic Education 2015 introduction xxxix Mind the Gap Physics

HINT! © Department of Basic Education 2015 hint e.g. Worked example 7EG - worked examples Consider the following graph. It shows six points in a straight line. exams y 3 2 1 activity -x x –2 –1 1 2 3 ACTIVITIES boy and girl-left and right of–1 page activity –2 –3 -y The coordinates shown can be described using what are called “ordered pairs”. For example, the furthestStep by step point in this graph is 3comment comment units across on the “x-axis” or horizontal line. Likewise, it is also 3 units up on the y-axis, or vertical (up and down) line. So, its coordinates are (3;3). The point just below the midpoint or “origin”, is one unit down of the x-axis, and one unit left of the y-axis. So its coordinates are (–1;–1). Note that anything to the left or below of the origin (the circle in the middle), takes a minus sign. In most cases in science, you’ll only have graphs showing positive axes (plural of axis, pronounced aks-eez), since most graphs are of time. This series of dots look like they’re related to each other, because they’re falling on a straight line. If you see a result like this in an experimental situation, it usually means that you can predict what the next dot will be, namely, (4;4). This kind of prediction is called “extrapolation”. If you carry out the experiment, and find that the result is (4;4), and then (5;5), you’ve established that there is a strong relation or correlation. You can therefore start thinking about a formula to describe your findings. For example, this might be a graph showing a measurement of voltage (x) against a measurement of resistance (y). Now, another way of saying that x relates to y, or x is proportional to y, is to say that y is a function of x. This is written y = f(x). So, in the example given above, voltage is a function of resistance. But how is y related to x in this graph? Well, it seems to be in a 1 to 1 ratio: y = x. So the formula for this graph is y = x. In this case, we’re only dealing with two factors; x and y. In other graphs you’ll find that sometimes more factors are involved, such as acceleration graphs, which have units of m/s2. Don’t worry about that; you treat them the same way (for example, m/s2 vs. time). xl introduction  © Department of Basic Education 2015  Mind the Gap Physics

HINT! © Department of Basic Education 2015 hint e.g. Worked example 8 EG - worked examples Now, let’s take a slightly more complex case, illustrated below. exams In this next graph, we see that wherever x is equal to something, y is one more. So, trace your finger from the bottom left dot upwards. It meets the x-axis at the point –3. Do the same for the same point towards the y-axis. You’ll see it meets the y-axis at –2. You’ll see the next coordinates are (–2;–1), then (–1;0), then (0;1), (1;2), and finally (2;3). From this we can see that whatever x is, y is one more. So, y = x + 1 is the formula for this line. y activity 3 ACTIVITIES boy 2 and and girl-left right of page activity 1 –x x -3 -2 -1 1 2 3 PAY SPECIAL ATTENTION NB Step by step comment -1 NB comment -2 HINT! –y -3 hint e.g. Worked example 9 EG - worked examples Let’s take another case. In this next case, we see the following values: where x has a certain exams value, y has double that value. Let’s tabulate it. y 3 2 x y 1,5 3 1 activity 1 2 ACTIVITIES boy –x and girl-left and x 0,5 1 -3 -2 -1 1right of page 2 3 0 0 activity -1 –0,5 –1 –1 –2 -2 –1,5 –3 -3 Step by step –y comment comment © Department of Basic Education 2015 introduction xli Mind the Gap Physics

© Department of Basic Education 2015 So, when x is 1,5, y is 3, when x is 1, y is 2. Thus, the formula for this line is: y = 2x. This value next to x is called the “gradient” or “slope” of the line. The larger the value next to x is, i.e. the larger the gradient, the steeper the slope. The gradient is usually abbreviated as “m” when it is unknown. Now, how this applies to science is simple: if we are looking, for example, at a case of a graph of a chemical reaction, we will usually have the x-axis as time. And the y-axis will usually be the quantity (amount) PAY SPECIAL of substances produced. So, if we have a graph ATTENTION NB of a chemical reaction with a large gradient, it means that NBthe reaction is fast; a lot of substance (y) i s produced in a short time (x). If, for example, we heated the reaction, and saw that the gradient increased even more, that would show that the chemical reaction was sped up by heat, or, that reaction rate is proportional to heat. Likewise, if the gradient sloped downwards, HINT! it would show that the reaction slowed down over time, because y, the amount of substance produced, was decreasing, as x (time) increased, hint e.g. because the reactants were being used up. e.g. Worked example 10 EG - worked examples Let’s do one more case. In this case, we can see that y is a function of x, since it’s a straight exams line graph. However, it’s not that easy to see the relationship between x and y. We can see that the slope is the same as the previous graph, so it must be something like y = 2x. However, it doesn’t quite make sense, since 2(–1,5) is not –2. We see that where x is zero (at the origin), y is at 1. But the slope is the same, so it must be y = 2(0) + 1. So the formula is: y = 2x + 1. y 3 activity 2 As you can see, this is ACTIVITIES boy where mathematicians get and girl-left and right of page the general equation for a 1 straight line of y = mx + c. (“c” stands for “constant”). activity –x x -3 -2 -1 1 2 3 -1 Step by step comment comment -2 –y -3 x y 2x + 1 –1,5 –2 2(–1,5)+1 = –3+1 = –2 –1 –1 2(–1)+1 = –2+1 = –1 –0,5 0 2(–0,5) +1 = –1+1 = 0 0 1 2(0)+1 = 0+1 = 1 0,5 2 2(0,5)+1 = 1+1 = 2 1 3 2(1)+1 = 2+1 = 3 xlii introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 7. Circles d • Diameter is the width of a circle (2r); radius is half the diameter (​ __ 2 ​). n ce The edge of a circle is called the “circumference”. “Diameter” means m fere to “measure across”. Compare “diagonal” which means an angle u Cir across a square or rectangle, so “dia-” means “across” (Greek). c “Circumference” means to “carry in a circle” (Latin); think of how the earth carries us in a circle or orbit around the sun. To remember Diameter the difference between these things, just remember that the sun’s rays radiate out from the sun in every direction, so the radius is the Ra distance from the centre of a circle, e.g. the sun, to the outer edge of dius a circle surrounding it, e.g. earth’s orbit (the circumference). • Area of a circle = π r2 • Circumference = 2 π r • You can use the above to solve for radius or diameter. Resource Sheets The following information sheets will be supplied to you in the exam. You do not need to memorise them. SI Units: Multipliers Prefix Symbol Value Value written in full tera T 1012 1 000 000 000 000 giga G 109 1 000 000 000 mega M 10 6 1 000 000 kilo k 10 3 1 000 hecto h 10 2 1 00 deka da 101 1 deci d 10–1 0,1 If you are studying centi c 10 –2 0,01 CAT/IT, do not be confused when you notice that these milli m 10 –3 0,00 1 same multipliers are given in multiples of 1024, not 1000. micro µ 10 –6 0,00 000 1 This is because computers use nano n 10 –9 0,00 000 000 1 binary (powers of 2). pico p 10 –12 0,00 000 000 000 1 femto f 10 –15 0,00 000 000 000 000 1 © Department of Basic Education 2015 introduction xliii Mind the Gap Physics

© Department of Basic Education 2015 Constants Name SI Unit Symbol Approximate Value Easier to Understand Acceleration due to g 9,8 m/s2 Your speed increases by about gravity 10 metres per second, every second you fall Gravitational constant G 6,67 × 1011 N⋅m2 / kg2 667 000 000 000 units STP (Standard not applicable, 1 atm (101,3 kPa), 25˚C You generally put two atm (bar) pressure Temperature and two conditions (298 Kelvin (K)) in car tyres, i.e. the pressure in a car tyre Pressure), (in Physics). is twice atmospheric pressure Speed of light c 3 × 108 m/s 300 000 000 metres per second, or about a billion kilometres per hour Speed of sound at sea v 340,29 m/s – 343 m/s One kilometre in three seconds. You level in air at STP can use this to work out how far away a lightning strike is. Planck’s constant h 6,626 × 10–34 J⋅s 0,0000000000000000000000000000 000006626 Js 1 Charge on electron e, __ ​  C ​ –1,6 × 10–19 C 0,00000000000000000016 coulombs (coulombs) is the charge on one electron Electron rest mass me 9,109 × 10–31 kg 0,0000000000000000000000000009 109 g is the weight of one electron Proton rest mass, mp u or µ 1,67 × 10–27 kg 0,00000000000000000000000167 g Atomic Mass Unit is the weight of one proton Gas constant R 8,3 J / mol⋅K The R in PV = nRT Molar gas volume at V0 22,4 dm3 / mol 22,4 Litres of gas is made by a mole of a STP substance Coulomb’s law constant k 9 × 109 N⋅m2 / C2 9 000 000 000 units xliv introduction  © Department of Basic Education 2015  Mind the Gap Physics

© Department of Basic Education 2015 Formulas Forces and Motion: Doppler Effect: V ± VL (fs) V (fs) u, v0 = velocity (initial) fL = ​ _______ V±V ​ OR fL = ​ _____ (V– V ) ​ s s v, v1 = velocity (final) s, x = displacement/distance travelled Electricity and Transformers: t = time a = acceleration E = energy (joules) Δ = change in q, Q = charge p = momentum v, V = voltage r = distance between bodies F = force I = current in amperes F = ma R = resistance in ohms (Ω) F = G(m1m2)/r2 (Newton’s Law of Gravitation) s/p = secondary/primary coils v a = Δv/t v = s/t E = ​ __ d ​ E = F/q kQ v = u + at or v1 = v0 + aΔt E = ___ ​ r2 ​ 1 1 s = ut + ​ __ __ 2 ​ at or Δx = v0Δt + ​ 2 ​aΔt 2 2 k(q1q2) F = ​ ______ r2 ​ (compare to Newton’s Law v2 = u2 + 2as or v12 = v02 + 2as of Gravitation) F = k(I1I2)L/r (where in this case, p = mv p 1v1 = p 2v2 k = 2 × 10–7) Replace “a” with g (9.8 m/s2) if falling to earth. V = IR I = Q/Δt QV Vt Energy, Work, Power, Waves: P = IV = ​ ___ t ​ E = VIt = __ ​ R ​ R = r1 + r2 + r3 ... (in series) E, W = energy or work 1 __ __ __ 1 1 Ek = kinetic energy ​  R ​ = ​  r1 ​ + ​  r2 ​+ ... (in parallel) W0 = work function Vs VpIp = VsIs ​ __ V ​= Ns/Np P = power p H = height off the ground h = Planck’s constant λ = wavelength (Greek letter L) f = frequency c = speed of light m = mass v = velocity T = time/period W Notes: E = W P = ​ __ t ​ Ep = mgH In most cases, a subscript of “1” means W = F∆xcosθ “before” and a subscript of “2” means “after”, Ek = ​ __ 1 so, p1v1=p2v2 means “momentum at time 1 2 ​ mv (at speeds much less than light) 2 times velocity at time 1 (before collision), is E = mc2 (light speed, nuclear reactions) equal to momentum at time 2, times velocity at E = hf time 2 (after collision)”. To convert from km/h to m/s, divide by 3,6. v = fλ (at light speed, use c = fλ) 1 f = ​ __T ​ © Department of Basic Education 2015 introduction xlv Mind the Gap Physics

© Department of Basic Education 2015 Notes

Unit 1 Mechanics: Force and Newton’s Laws Summary This section covers important aspects from the Grade 11 work. Make sure that you revise it well so that you can apply this knowledge to your Grade 12 work. In order to be successful in this section, you need to revise trigonometry and Pythagoras’ theorem. You must know how to: • Draw a sketch of parallel and perpendicular vectors. • Determine the resultant vector graphically using the head to tail method as well as by calculation. • Resolve a vector into its parallel and perpendicular components. You must remember: • Newton’s Laws and how to apply them. • Different kinds of forces. • Force diagrams and free body diagrams. Definitions and Laws you must remember: 1. A force is a push or pull upon an object resulting from the object’s interaction with another object. 2. Gravitational force is the force of attraction that objects exert on other objects in virtue of having mass. It is the force that makes all things fall and causes tides in the ocean. The greater the mass of an object, the greater its gravitational pull. 3. The normal force is a perpendicular force that a surface exerts on an object with which it is in contact. 4. The resultant (net) force acting on an object is the vector sum of all the forces acting on the object. The vector sum is the sum of all vectors (all the forces added up, taking their directions into account). 5. Newton’s First Law of Motion: An object will remain at rest or continue moving at a constant velocity (or at constant speed in a straight line) unless acted upon PAY SPECIAL ATTENTION by a non-zero external resultant force. NB 6. Newton’s Second Law of Motion: If a resultant (net) force acts on an object, the object will accelerate in the direction of the resultant force. The acceleration is directly proportional to the resultant force and inversely proportional to the mass of the object. HINT! 7. Newton’s Third Law of Motion: When object A exerts a force on object B, object hint B simultaneously exerts a force on object A, which is of equal magnitude but opposite in direction. “Universal” 8. Newton’s Law of Universal Gravitation: A force of gravitational attraction exists means the between any two particles or objects anywhere in the universe. The magnitude e.g. statement is valid for any object in EG - worked examples of this force is directly proportional to the product of the objects’ masses and is the universe. inversely proportional to the square of the distance between their centres. exams © Department of Basic Education 2015 UNIT 1 MECHANICS: FORCE AND NEWTON’S LAWS 1 Mind the Gap Physics

1 Unit 1.1NB Revision: VectorsNB PAY SPECIAL ATTENTION A vector is a quantity that has both magnitude (size) and direction. • We can      use bold type to represent a vector, R, or an arrow above the → letter ​R ​. HINT! • Vectors may be added or subtracted graphically by laying them head hint to tail / head to head on a set of axes. e.g. Worked example 1 EG - worked examples exams activity PAY SPECIAL ATTENTION NB ACTIVITIES boy NB →           →      → and girl-left and      →      →      → ​A ​+ ​C  ​= ​D ​ right of page ​A ​– ​C ​= ​E ​ • Just as one vector is the sum of two vectors, we can also find two vectors to make activity up one vector. HINT! • In mechanics, it is often useful to break up a vector into two component vectors, one hint horizontal and the other vertical. We use basic trigonometry to find the components. Step by step comment comment e.g. Worked example 2 EG - worked examples      →      → Example: Vector ​R ​makes an angle θ with the x-axis. ​R ​is broken into component      →      → exams vectors ​R ​x and ​R ​y. adj R cos θ = ___ ​  ​ = __ ​  x ​ hyp R ∴ Rcos θ = Rx activity opp Ry sin θ = ___ ​  ​ = __ ​  ​ ACTIVITIES boy hyp R and girl-left and ∴ Rsin θ = Ry right of page      →      → activity      →      →      →      → ​R ​x = ​a ​ cos θ and ​R ​y = sin ​R ​y, where ​R ​y is the angle between ​a ​and the x-axis      → (​R ​x ). Step by step comment comment 2 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 1 e.g. Worked example 3 EG - worked examples      → If vector ​R ​has a magnitude of 5 and is at an angle of θ = 36,86°, the components are      →      → exams ​F ​x = 5 cos 36,86° = +4 and ​R ​y = 5 sin 36,86° = +3. 1.2 What is force? When objects interact with each other, they exert forces on each other. activity If a force acts on an object, it can cause a change to the object. Some of the possible changes are: ACTIVITIES boy and girl-left and • the shape of an object right of page • the object’s state of rest • activity the velocity of the object • the direction in which the object moves PAY SPECIAL ATTENTION • the object’s acceleration. NB NB      → Definition Force (​F ​) is a vector quantity. Step by step This means comment comment it has magnitude and direction. Repulsion: a force between • It may be represented by an arrow in a vector diagram. The length of HINT! objects that tends to separate hintshows its direction. the arrow shows its magnitude and the angle them • It is measured in the SI unit newton (N). Attraction: a force between      → objects that brings them We show the force vector using ​F ​. together F without an arrow represents the size of the force vector only. e.g. EG - worked examples Example      → represents a force (​F ​) of 12 N to the right. 12 N exams • Objects exert push (repulsion) or pull (attraction) forces on each other. • A force can be classified as either a contact force or a non-contact force • Objects can exert a force on each other when they are in contact (touching each other) e.g. friction and normal forces OR activity ACTIVITIES boy • Objects can exert a force on each other when they are not in and girl-left and contact (i.e. are apart from each other) e.g. magnetic, electrostatic right of page and gravitational forces. activity 1.3 Different types of forces We study these different forces: Step by step comment comment      →      → 1. Gravitational force or weight (​F ​g or ​w ​)      →      → 2. Normal forces (​F ​N or ​N ​)      → 3. Frictional forces (​F ​f) 4. Applied forces (push or pull)      →      → 5. Tension (​F ​T or ​T ​) © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 3 Mind the Gap Physics

1 Unit      →      1. Gravitational force (​F ​g or → ​a ​): • Gravitational force is the force of attraction that the Earth exerts on an object above its surface. • Gravitational force acts downwards towards the centre of the Earth.      → • The     weight (​w ​) of an object      is the same as the gravitational force → →      → (​F ​g) on the object, so ​F ​g = ​w ​ • The weight of an object is the product of the mass and the normal: In Physics, normal gravitational acceleration of the Earth.      →      →      → means perpendicular so ​w ​= m​g ​where m is mass and ​g ​is the acceleration due to to. It does not mean gravity. ‘ordinary’.      →           → ∴ ​F ​g = ​→ w ​= m​g ​      → where ​F ​g is gravitational force      → w ​ is the weight of an object ​      → g ​ m​ is mass × gravitational acceleration      →      → 2. Normal force (​F ​N or ​N ​): When an object rests on a surface, the surface exerts a force on the object, called a normal force. It is a contact force that acts at a right angle (90°) upwards from the surface. In the diagrams below, you will see a free body diagram and a force PAY SPECIAL ATTENTION NB diagram. In aNB force diagram, you show the object that is experiencing forces. The forces act on the body at its “centre of gravity”. In a free Definition body diagram, you do not show the object that is experiencing forces; i.e. you treat the object as a single point. A “centre of gravity” is a point from which the weightHINT! hint of a body or system may be 2.1. When an object is resting or moving on a horizontal surface considered to act. In uniform the normal force will have the same magnitude, but an gravity it is the same as the opposite direction to the weight of the object or gravitational center of mass. force. e.g. EG - worked examples An object resting on a horizontal surface exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 4 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 2.2. When an object is resting or moving on an inclined plane (surface), the normal force will have the same magnitude, but an opposite direction to the perpendicular component of the weight of the object or gravitational force. An object resting on an inclined plane (surface)      →      → 3.1 Frictional force (​F ​f or ​f ​): • Frictional force opposes motion. So it works against the movement of an object. • Frictional force acts in the opposite direction to an object’s motion or intended motion. • The rougher the surface, the more friction there is between the object and the surface. The less rough the surface, the less friction there is between the object and the surface. This means that the greater the magnitude of the normal force acting on the object, the greater the magnitude of the frictional force. Think of grinding something here. The harder you press, the more “normal” (perpendicular) force there is. Hence, when you are grinding something, e.g. crushing maize for making pap, it experiences strong normal (perpendicular) forces and thus strong frictional forces; hence it is ground up. • If an object is at rest, then there is a static frictional force. µ is the Greek • If the object is moving, then there is kinetic frictional force. letter mu. 3.2 The coefficient of friction (µ) The coefficient of friction depends on the material of the two surfaces that are in contact. Examples • Steel on wet ice has a low coefficient of friction (slides easily). • Rubber on tar has a higher coefficient of friction (more grip, less sliding). • When an object is at rest on a horizontal surface and no force is applied to it, then there is no static friction. • When a small force is applied to an object at rest, then the force of static friction increases as the applied force increases. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 5 Mind the Gap Physics

1 Unit • As the force increases, the static friction continues to increase. • This continues until the static friction reaches a maximum value – it cannot increase further. Eventually maximum static friction force is exceeded and the object moves. • The friction then     → decreases to a smaller value called the kinetic friction (​f ​k). • The kinetic friction remains constant while the object moves at a constant speed. • The kinetic friction remains smaller than the maximum static friction. fs ≤ µsFN and fk = µkFN PAY SPECIAL ATTENTION NB NB When an object moves along a surface inclined at an angle θ, the normal force is multiplied by the kinetic coefficient of friction to find the frictional force. HINT! The kinetic coefficient is calculated using cos θ: hint      →      →      → ​F ​N = ​F ​g⊥ = m​g ​·cos θ      → fk = µk ​F ​N 4. Applied forces e.g. EG - worked examples An applied force is a force that a person or object applies to another object. If a person is pushing a cart along the ground, then there is an applied force acting upon the object. exams activity      →      → 5. Tension (​F ​T or ​T ​): ACTIVITIES boy When an object is pulled by a rope (or string or cable), or hanging from and girl-left and right of page a ceiling, the rope applies a force on the object. This force is called activity tension. It is a contact force and acts in the opposite direction to the ‘pull’. If an object hangs from a rope, the direction of the tension is always upwards in the rope. This force complies with Newton’s Third Law, i.e. it is the reaction to the action of the pulling. Step by step comment comment 1.4 Force diagrams and free body diagrams PAY SPECIAL ATTENTION NB NB 1. In force and free body diagrams we consider forces acting on ONE (the same) object 2. When you answer questions about force, you must: - name the forces HINT! hintwhich object exerts a force on which object - state - state the directions of the forces. e.g. EG - worked examples 6 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics exams

Unit 1 Example A man pushes a loaded trolley along a horizontal floor. We can identify the following forces that are acting on the trolley:      → • the weight or ​F ​g of the trolley and load, i.e. the force exerted downwards on the trolley by gravity/ the earth      → • the normal force, ​F ​N , exerted upwards on the trolley by the floor • the applied force that the man exerts on the trolley, which acts in a forwards direction      → • the frictional force, ​F ​f , in the opposite direction to the motion. Forces acting on an object can be represented by force diagrams or by free body diagrams. Force diagram Free body diagram • The object is represented as a block and the • The object is represented as a dot and the forces as vectors. forces as vectors. • The vectors start at the point of application. • The vectors start at the dot and they all point • Weight is drawn from the object’s centre away from the dot. of gravity, downwards. • If the object is on an inclined surface, the weight vector can be resolved into two component vectors.           → → ​F ​N ​F ​N →      →      → ​F ​g ​F ​g Force symbols in diagrams We use these symbols to help represent forces in force diagrams and free body diagrams:      →      → • ​F ​or ​F ​applied : applied force, in the direction applied      →      → • ​F ​f or ​F ​: friction force, surface on object, opposite to direction of motion      →      → • ​F ​g or ​w ​: gravitational force or weight, force exerted by the earth on object, downwards      → • ​F ​N or N: normal force, surface on object; perpendicularly upwards from the surface      →      → • ​F ​T or ​T ​: tension, cable or rope on object, in direction of motion. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 7 Mind the Gap Physics

1 Unit Steps for drawing force or free body diagrams Follow the steps in this example. Example: An object on a horizontal surface (plane): Sipho exerts a force to the right while pushing a car along a rough, flat road. Draw a force diagram and a free body diagram to represent the situation. Force diagram: Free body diagram: Step 1. If there is a surface, draw a line to represent it. Step 2. Force diagram: draw a block to represent the object. Step 3. Free body diagram: draw a dot to represent the object.      → Step 4. Draw a vector to represent the weight (​F ​g) of the object. Step 5. If the body rests on a surface draw an arrow to represent the normal      force, upwards from and perpendicular to the surface. → (​F ​N) Step 6. Draw an arrow to represent each applied force. Step 7. Draw an arrow to represent friction (if there is friction). Step 8. If you are drawing a free body diagram, erase the line representing the surface. 8 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 1 e.g. Worked example 4EG - worked examples Draw a force diagram and a free body diagram exams for an object hanging from a rope or a cable. Solution activity Force diagram: Free body diagram: ACTIVITIES boy and girl-left and right of page activity Step by step comment comment Component vectors in free body diagrams Free body diagrams are useful for showing all the forces involved in a situation. When an object rests on an inclined plane, the force due to gravity may be shown by two vectors: • one representing the component parallel to the surface • the other representing the component perpendicular to the surface. Force diagram: Free body diagram: © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 9 Mind the Gap Physics

HINT! 1 Unit hint e.g. Worked example 5 EG - worked examples If an object with a mass of 40 kg slides down a surface which has a coefficient of kinetic examsfriction μk = 0,14, and a slope of 15°, what is the net force on the object as it slides down the surface? Use the diagram to help you. activity ACTIVITIES boy and girl-left and right of page Solutionactivity Weight component down the slope      → PAY SPECIAL ATTENTION sin θ = mg sin 15° = 40 × 9,8 × 0,26 = 101,92 = ​F ​gNB NB N. Frictional force up the slope      → Step by step comment comment ​F ​g·μk cos θ = 40 × 9,8 × 0,14 × 0,96 = 52,68 N ∴ Net force down the slope is: HINT! hint 101,92 + (–52,68) = 49,24 N. e.g. Worked example 6 IAL ATTENTION EG - worked examples NB Direction of frictional force: Calculate the components of the weight of an object when it is resting on a surface The friction acts against which slopes at an angle of θ. exams the object to prevent it from sliding down the slope so it Solution acts upwards parallel to the θ + a = 90° and a + c = 90°. So c = θ. slope. The component of the weight perpendicular to the surface with a slope of θ is      →      →      →      → ked examples ​F ​g⊥ = ​F ​g cos θ and parallel to the surface is ​F ​g|| =​F ​g sin θ. activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment ACTIVITIES boy and girl-left and right of page 10 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015 ep  comment comment Mind the Gap Physics

Unit 1 1.5 Resultant (net) force When a number of forces act on an object, we need to determine the resultant or net force acting on the object. PAY PAY SPECIAL SPECIAL ATTENTION ATTENTION NBNB NB NB The resultant (net) force acting on an object is the vector sum of all the forces acting on the object.      →      →      →      → →      HINT! ​F ​net = ​F ​res = ∑​F ​= ​ F ​1 + ​a ​2 + … HINT! hint hint      → ∑​F ​is the sum of all the forces acting on the object e.g. Worked example 7 e.g. EG - worked examples EG - worked examples John exerts a force of 100 N to the right on a box resting on a rough, horizontal surface. Sarah exams exerts a force of 50 N to the left on the box. The friction between the box and Remember: exams the surface is 5 N. Draw a force diagram and calculate the resultant force acting on the • Force is a vector. box. • Indicate the direction of the force with a + or – sign. • Interpret the answer in activity words as the final step activity ACTIVITIES boy in your solution and girl-left and ACTIVITIES right of boy page and girl-left and activity right of page activity Solution Note: Since the force      →      →      →      →      → ​F ​net = Σ​F ​= (+​F ​John) + (−​F ​Sarah) + (−​F ​f) applied by Sarah and                comment the frictional force are → → →Step by step comment = ​F ​John – ​F ​Sarah – ​F ​f opposite in direction to the Step by step      → comment comment force     → applied by John, = 100 – 50 – 5 = 45 N ∴ ​F ​net = 45 N to the right ​F ​Sarah    → = –50 N and ​F ​f = –5 N Now consider a situation where a box slides down a slope. The force that makes the box slide down the slope is the component of the box’s weight that acts parallel to the slope.      → ∴ ​F ​gll = mg·sin α where α is the angle between the slope and the horizontal. Always calculate this force first. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 11 Mind the Gap Physics

1 Unit Activity 1 A box of mass 100 kg slides down a rough slope which forms an angle of 30° to the horizontal. The friction that acts on the box is 20 N. Draw a free body diagram representing all the forces acting on the object and calculate the resultant force acting on the box and causing it to slide. Perpendicular forces may be ignored.  (12) Solution Let the direction down the slope be positive.      → F ​f = –20N 3 Then ​ 3           →      →      → → 3 ​F ​net = Σ​F ​ = (+ ​F ​gll) + (​F ​f) 3      →      → = ​F ​gll + ​F ​f 3      → = mg·sin α + ​F ​f 3 3 3 3 = (100)(9,8)(sin 30°)+( – 20) 3 = 490 – 20 = 470 N ∴ 470 N, down the slope 3 3 [12] Activity 2 R and S are two positively charged spheres. P is a negatively charged sphere. Sphere R exerts an electrostatic force of 0,2 N on P and sphere S exerts a force of 0,6 N on sphere P. Draw a free body diagram for sphere P and then calculate the resultant force on sphere P. (5) Remember: Opposite charges attract. Therefore Solution R attracts P and S attracts P. 3 3      →      →      →      → ​F ​net = Σ​F ​= (​F ​S on P) + (​F ​R on P) 3 = 0,6 + (–0,2) 3      → Note: ​F ​R = –0,2 N = 0,4 N ∴ 0,4 N towards S 3 [5] 12 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 Activity 3 PAY SPECIAL ATTENTION NB Three identical spheres X, Y and Z are in the same horizontal plane. Spheres X and Z are both positive and sphere Y is negative. Sphere Y exerts an electrostatic force of 450 N on sphere X and sphere Z exerts an electrostatic force of 350 N on sphere X. HINT! 1. Draw a free body diagram for sphere X hint and indicate the electrostatic forces X and Z are positive acting on it.  (2) ∴ Z repels X and 2. Calculate the magnitude of the resultant electrostatic force on sphere X.  (8) e.g. EG - worked examples Y is negative and Z is positive ∴ Y attracts X [10] exams Solutions 3 1. Free body diagram: 3 (2)      →2      →2 3      →2 activity 2. ​F ​ net = ​F ​ Z on X + ​F ​ Y on X (pythagoras) ACTIVITIES boy 3 3 and girl-left and = 350 + 4502 2 = 325 000 ________ right of page      → = ∴ ​F ​net = ​√325 000 ​= 570,09 N 3 3 activity 3 Step by step comment 3 3 (8) [10] © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 13 Mind the Gap Physics

1 Unit 1.6 Newton’s First Law of Motion (Law of Inertia) PAY SPECIAL ATTENTION NB NB Inertia • Inertia is the property of an object that resists any change in the state of rest or PAY SPECIAL ATTENTION NB HINT! uniform motion. NB hint If the object is at rest, it resists any change to a state of motion. Definition If it is in motion, it resists any change to the speed and direction of its motion. Resists: • Inertia is determined by the object’s mass. The greater an object’s mass, the greater opposes, prevents, worksHINT! its inertia. e.g. hint against EG - worked examples Example exams e.g. A box lying in the boot of a car will move forwards when the car brakes. EG - worked examples The box’s inertia resists the change in movement and allows the box to continue moving in the direction in which the car was moving before it exams stopped. This is why you must wear seatbelts! activity PAY SPECIAL ATTENTION NB ACTIVITIES boy and girl-left and NB right of page activity activity Newton’s 1st Law of Motion An ACTIVITIES object willboy remain at rest or continue moving at a constant velocity HINT! (in and a straight line) unless acted upon by a non-zero external resultant force. girl-left and hint right of page      → Fnet = 0 N ∴ ​a ​= 0 m·s–2 activity Step by step comment comment e.g. EG - worked examples Step by step comment comment exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 14 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 1 e.g. Worked example 8EG - worked examples The diagram shows the forces on a trolley moving examswith constant velocity. 1. A man pushes a loaded trolley with constant velocity along a horizontal floor. The trolley and load have a mass of 56 kg and the friction of the moving trolley is 2,1 N. Calculate the force the man exerts to push the trolley along the floor.activity 2. If he then pushes the trolley with a force of 2,5 ACTIVITIES boy N to the right, calculate the and girl-left and acceleration of the trolley. right of page activity Solutions 1. The acceleration = 0, so the net force is equal to zero. The force the person pushes with is equal and opposite to the force of kinetic friction on the trolley. Step by step comment comment Fpush = Ftrolley = 2,1 N 2. The diagram shows the free body diagram. The net force is 0,4 N in the forward direction. The trolley accelerates forward:      →      → ​F ​net ___ 0,4 ​a ​ = ​ ___ m ​ = ​  56 ​= 7,14 × 10 m·s –3 –2 forward / to the right. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 15 Mind the Gap Physics

1 Unit 1.7 Velocity and acceleration: Revision These equations are listed on the data • Velocity (v) is the rate of change in position (displacement). It is a sheet in the exam paper. vector. Speed is a scalar. You don’t have to      →      → Δ​x ​ memorise them, but you ​V ​average = ​ ___ Δt ​ ... Δx is the displacement; rate is shown by change must know how to use in time Δt them.      → • Acceleration (​a ​) is the rate of change of velocity.      →      →      →      → ​v ​f – ​v ​i Δ​v ​ ______ ​a ​ = ​ ___ Δt ​ = ​  Δt ​ ... Δv is change in velocity: final velocity (vf) – initial velocity (vi) • Equations of motion: In Grade 10 you learnt these equations that describe the relationships between velocity, acceleration, displacement and time:      →      →      → • ​v ​f = ​v ​i + ​a ​Δt      →2      →      → • ​v ​f = ​v ​i2 + 2​a ​Δx      →      → • Δ​x ​= ​v ​i Δt + ½Δt2 1.8 Newton’s Second Law of Motion in terms of acceleration When the resultant force acting on an object is NOT zero, the object’s state of motion will change. It may:      →      → • start moving (then ​v ​i = 0 m·s−1 and ​v ​f ≠ 0 m·s−1);      → • stop moving (come to rest, then ​v ​f = 0 m·s−1); • move faster (accelerate); move slower (decelerate); or • the direction in which it moves will change. 16 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 PAY SPECIAL ATTENTION NB NB Newton’s Second Law of Motion If a resultant (net) force acts on an object, the object will accelerate in the direction of the resultant force. TheHINT! acceleration produced is directly proportional to the resultant hint force and inversely proportional to the mass of the object. In other words, acceleration is the amount of change in speed (or velocity), per second, hence, it is metres per second change per second, or m·s–2      →      →      →      →      For any objectEG​-a ​ ∝ ​F ​net 1 and ​a ​ ∝ ​ __ m​ ∴ ​F ​net = m​→ a ​ e.g.      → worked examples      → where ​a ​is acceleration (m·s ), ​F ​is force (N) and m is mass (kg) −2      → →      exams ​ a ​ ∝ ​F ​net ∴ straight line through origin activity      → When​a ​ ∝ m the graph is a hyperbola      → with​ a ​and ACTIVITIES boy m on the axes. and girl-left and right of page activity      → 1 When ​a ​ ∝ ​ __ ​the graph is a straight →m 1      line with ​a ​ and ​ __ m ​on the axes. PAYStep SPECIAL ATTENTION by step comment comment NB NB If different forces are applied to the same object and its mass stays constant, then​      → →      a ​ ∝ ∑​F ​. The bigger the net resultant force acting on the object, the more the HINT! object will accelerate. hint NOTE: e.g. Worked example 9 EG - worked examples If a constant non-zero resultant force is applied to →      →           → 1 A resultant force ​F ​is applied to an object of mass m and the object accelerates at ​a ​. two objects, then ​a ​ ∝ ​ __ m​ . exams What will the object’s acceleration be if the resultant force acting on the object is The object with the smaller tripled? mass will accelerate more than the object with the Solution                bigger mass. Think about it:      → → → → m is constant ∴ ​a ​ ∝ ​F ​and if the force is tripled (from ​F ​to 3​F ​), the it’s easier to make a lighter      → object move further and acceleration will also triple ∴ the object will accelerate at 3 ​a ​. faster. activity ACTIVITIES boy and girl-left and right of page activity © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 17 Mind the Gap Physics

HINT! 1 Unit hint e.g. Worked example 10 EG - worked examples →      A constant resultant force ​F ​is applied to objects of masses m and 2 m. If the object of →      mass m accelerates at ​a ​, what will the acceleration of the other object be? exams Solution      →      → 1 ​F ​is constant ∴ ​a ​ ∝ ​ __ m​ If the mass doubles (from m to 2 m), the acceleration will halve 1      ∴ the object of mass 2 m accelerates at __ → ​  ​ ​ a ​. 2 activity ACTIVITIES boy and girl-left and Steps to solve problems on right of page activity Newton’s Laws PAY SPECIAL ATTENTION NB Step 1: NB Read the problem as many times as you need. Step 2: Sketch the situation if it is necessarily. Step 3: Draw a force diagram for the situation. comment Step 4: Draw aStep by step free body diagram;comment you must resolve the forces into HINT! components on the Cartesian plane if necessary. hint Consider this example. You are told that the force F acts at an Sometimes the forces you are angle of 60˚ to the normal or 30˚ to the horizontal plane. What given in a diagram are not are its vertical and horizontal components? e.g. at right angles toEGeach other, - worked yet when you draw a force examples diagram on the Cartesian Plane, the forces must be exams drawn at right angles to each other. In order, then, to find out what the vertical and horizontal parts of a force are, when that force is at an angle, we have to use trigonometry. activity ACTIVITIES boy and girl-left and right of page Well, the y-component is opposite the angle, and the hypotenuse activity (10 N) is known, so since sine is O/H, sin 30˚ × 10 N = the y-component: 5 N. Likewise, the x-component is adjacent to the angle, so since cosine is A/H, cos 30˚ × 10 N = 8,67 N. So your x-component is 8,67 N and your y-component is 5 N. Step 5: List all the given information and convert the units if necessary. Step by step comment comment Step 6: Determine which physical principle (law) can be applied to solve the problem. Step 7: Use the principle (law) to answer the question, often by substituting numerical values into an appropriate equation. Step 8: Check that the question has been answered and that the answer makes sense. 18 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 Activity 4 Two boxes, A and B, are lying on a table and are connected by a piece of string. The mass of box A is 3 kg and the mass of box B is 2 kg. Assume that the mass of the string is very small, so we can ignore it. A 30 N pulling force, pointing to the right, is applied to box B, causing the two boxes to move. The surface acts with a frictional force of 5,9 N on box A and 4,1 N on box B. 1. Calculate the acceleration of boxes A and B. (14) 2. Calculate the magnitude of the tension on the string. (5)  [19] Solution 1. We are going to take the whole system as a unit. 3 Data: mA = 3 kg, mB = 2 kg 33 3 3 mB = 2 kg mT = mA + mB = 3 kg + 2 kg = 5 kg 3 FA = 30 N to the right FfA = 5,9 N to the left 3 FfB = 4,1 N to the left 3 3 FfT = FfA + FfB = 5,9 + 4,1 = 10 N to the left a = ? to the right T=? (9) FRTx = mTa (from F = ma) 3 3 FA + FfT = mTa 3 3 (30) + (–10) = 5a a = 4 m/s2 to the right 3 (5) 2. To calculate the tension you may use box A or box B Tension using box A FRAx = mAa 3 T + FfA = mAa 3 T – 5,9 = (3)(4) 33 T = 12 + 5,9 T = 17,9 N 3 (5) © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 19 Mind the Gap Physics

1 Unit Tension using box B FRBx = mBa 3 FA + T + FfB = mBa 3 30 – T – 4,1 = (2)(4) 33 –T + 25,9 = 8 –T = –17,9 T = 17,9 N 3 (5) [19] Activity 5 The sketch below shows a block of 8,5 kg at equilibrium on an inclined (sloping) plane (surface). Calculate: 1. The magnitude of the tension in the cord.  (12) 2. The magnitude of the normal force acting on the block. (6) 3. The magnitude of the block’s acceleration, If the cord is cut.  (4) [22] Solution 1. Data m = 8,5 kg ; α = 300 ; vi = 0 ; ai = 0 Let’s make the free body diagram of forces. (6) 3 3 The gravitational force is not in the 3 direction of any axis then we have to determine its components on the x-axis and y-axis. 3 3 3 20 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 1. Applying Newton’s First Law      → ∑ ​F ​= 0      →      → ​F ​g + ​T ​= 0 3 Working with the projections of the forces on the x-axis we get: Fgx – T = 0 Fg·sin α – T = 0 3 m·g·sin 30° – T = 0 3 3 3 8,5 × 9,8 × 0,5 – T = 0 41,65 – T = 0 T = 41,65 N 3 (6) 2. Working on the y-axis N – Fgy = 0 N – (Fg·cos α) = 0 3 3 N – (m·g·cos 300) = 0 3 3 3 N – (8,5 × 9,8 × 0,866 ) = 0 N – 73,1 = 0 N = 73,1 N 3 (6) 3. Applying Newton’s Second Law.      →      → ∑ ​F ​x= m​a ​x 3 If the cord is cut there is no tension force acting on the block and there is only one force acting on the direction of the x-axis, causing acceleration to the block. Working with the projections Fgx = m·a Fg·sin α = m·a mg·sin 30° = m·a 3 Simplifying: g·sin 30° = a a = g·sin 30° a = 9,8 × 0,5 3 a = 4,9 m·s–2 3 (4) [22] © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 21 Mind the Gap Physics

1 Unit Activity 6 Two blocks of 25 kg and 15 kg are connected by a light string on a horizontal surface. Assume that the string cannot stretch. A force of magnitude 240 N is applied to the block of 15 kg forming an angle of 60° with the horizontal as shown in the sketch below. The coefficient of kinetic friction is 0,20. 1. State Newton’s Second Law of Motion in words. (7) 2. Draw a free body diagram for each block. (8) 3. Calculate the magnitude of the acceleration of the blocks. (14)  [29] Solutions 1. If a resultant force 3 acts on a body, it will cause the body to accelerate 3 in the direction of the resultant 3 force. The acceleration of the body will be directly 3 proportional to the resultant 3 force and inversely 3 proportional to the mass 3 of the body. (7) 2. 3 3 3 3 3 3 3 3 (8) 3. Option 1 Taking the objects as a system FRx = ma 3 Fx + FfT = max 3 Note: In this series of solutions we have Fx – FfT = mTax omitted (taken out) the vector arrow above Fx – (Ff1 + Ff1) = (m1 + m2)ax 3 F and a; this is just to make it easier to read the solution. Fx – (µN1 + µN2) = (m1 + m2)ax 22 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 We have to calculate the normal force for both blocks N1 = Fg = m 1g N2 = m2g – F·sin 60˚ 3 3 3 F·cos 60˚ – [µm1g + µ(m2g – F·sin 60˚)] = (m1 + m2)ax 3 3 (240·cos 60˚) – [(0,2)(25)(9,8) + (0,2)[(15)(9,8)] – 240·sin 60˚]] = (25+15)ax 3 3 3 120 – [49 + (0,2)(147 – 207,85)] = 40ax 3 83,17 = 40ax 3 ax = 2,08 m/s2 3 Option 2 Applying Newton’s Second Law of motion to each object individually FR = m1a For object 1: T = Ff1 = m1ax 3 T – Ff1 = m1ax T – µm1g = m1ax 3 For object 2: FR2x = m2a 3 Fx + T + Ff2 = m2ax 3 Fx – T – Ff2 = m2ax 3 F·cos 60˚ – T – µN1 = m1ax F·cos 60˚ – T – µ(m2g – F·sin 60˚) = m2ax 3 Adding equation (1) and (2). T – µm1g + F·cos 60˚– T – µ(m2g –F·sin 60˚) = m1ax + m2ax Taking out T and ax: – µm1g + F·cos 60˚– µ(m2g –F·sin 60˚) = (m1 + m2)ax 3 3 [–(0,2)(25)(9,8)] + [240·cos 60˚– (0,2)[(15)(9,8) – (240·sin 60˚)] = (25 + 15)ax 3 3 3 3 (–49 + 120) – (0,2)(147 – 207,85) = 40ax 71 + 12,17 = 40ax 3 83,17 = 40ax ax = 2,08 m/s2 3 (14) [29] © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 23 Mind the Gap Physics

1 Unit Activity 7 The sketch below shows two blocks connected by a string of negligible mass that passes over a frictionless pulley also of negligible mass. The arrangement is known as Atwood’s machine. One block has mass m1= 2 kg and the other has mass m2= 4 kg. The blocks have just this instant been released from rest. m1 1. Draw a free body diagram of all the forces acting on each block. (6) 2. Calculate the magnitude of the acceleration of the system. (7) m2 3. Calculate the magnitude of the tension in the string.  (4) 4. Compare the magnitude of the net force on m1 with the net force on m2.(1) Write down only GREATER THAN, SMALLER THAN or EQUAL TO. 5. Will the pulley rotate clockwise or anticlockwise?  (1)  [19] Solutions 1. See diagram below: 3 3 Note: In this series 3 of solutions we have omitted (taken out) the vector arrow above F 3 and a; this is just to 3 make it easier to read the solution. 3      (6) →      → 2. Ʃ​F ​net = m​a ​ 3 For the 2 kg block (+ upwards) T – Fg1 = m1a T –m1g = m1a 3 T – 2 × 9,8 = 2a 3 For the 4 kg block (+ downwards) –T + m2g = m2a –T+ 4 × 9,8 = m2a 3 Solving the system of equations 3 3 T – 2 × 9,8 –T + 4 × 9,8 = (2 + 4)a 2 × 9,8 = 6 a a = + 3,27 m·s–2 (upwards) a = 3,27 m·s–2 3 (7) 24 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 3. Option 1 ( 2m1m2 T – ​ ______ ) ​  m1 + m2 ​  ​× g 33 (2 × 2 × 4) T = ​ _______ ​  2 + 4 ​  ​ × 9,8 T = 26,13 N 3 3 Option 2 3 T – Fg1 = m1a OR T= m1(a+g) 3 T = 2 (3,27 + 9,8) 3 T = 26,14 N 3 Option 3 3 3 3 –T + m2g = m2a OR – T = m2a – m2g OR T = – m2a + m2g T = 4(–3,27 + 9,8) 3 3 T = 26,12 N 3 (4) 4. Smaller than. 3 (1) 3 5. Anticlockwise. (1) [19] Activity 8 A 4 kg block on a horizontal, rough surface is connected to a 8 kg block by a light string that passes over a frictionless pulley as shown below. Assume that the string cannot stretch. The coefficient of kinetic (dynamic) friction between the block of 4 kg and the surface is 0,6. 1. Draw a free body diagram of all the forces acting on both blocks. (6) 2. Calculate the acceleration of the system. (10) 3. Calculate the magnitude of the tension in the string. (3) 4. Calculate the magnitude of the frictional force that acts on the 4 kg block. (4) 5. Calculate the apparent weight of the 8 kg block. (4) 6. How does the apparent weight of the 8 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN.  (1) 7. How does the apparent weight of the 4 kg block compare with its true weight? Write down only GREATER THAN, EQUAL TO or LESS THAN.  (1) [29] © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 25 Mind the Gap Physics

1 Unit Solutions 1. 3 3 3 3 3 Positive to the right 3 (6) Let’s take the direction of motion as positive 2. Let’s apply Newton’s Second Law of Motion to each block.      →      → ∑​F ​= m​a ​ 3 Note: In this series of Block of 4 kg (A) solutions we have omitted In the x direction (horizontal) (taken out) the vector arrow T – ff = mAa above F and a; this is just to T – µN = mAa 3 make it easier to read the solution. T – µmAg = mAa (call this Equation 1) Block of 4 kg (A) In y direction (up/down) N – Fg = 0 (not moving horizontally) N = Fg = mg 3 Block of 8 kg (B) In the x direction (horizontal) –T + Fg = mBa 3 –T + mBg = mBa (call this Equation 2) Solving the system of equations (adding Equation 1 and 2) T – µmAg – T + mBg = mAa + mBa 3 3 Removing T and isolating a: –µmAg + mBg = (mA + mB)a 3 3 3 –(0,6)(4)(9,8) + (8)(9,8) = (4 + 8)a 54,88 = 12a a = 4,57 m/s2 3 (10) 3. Using Equation 2 –T + mBg = mBa 3 3 –T = 8 × 4,57 – (8 × 9,8) T = 41,84 N 3 (3) 26 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 1 Using Equation 1 T – µmAg = mAa T – (0,6)(4)(9,8) = (4)(4,57) 3 3 T = (0,6)(4)(9,8) + (4)(4,57) T = 41,8 N 3 (3) 4. ff = µN N = mg ff = µmg 3 3 3 ff = 0,6 × 4 × 9,8 ff = 23,52 3N (4) 5. –T + mBg = mBa 3 3 3 –T = –8 × 4,57 + (8 × 9,8) Apparent weight = T = 41,84 3 N (4) 6. Less than 3 (1) 7. Equal to 3 (1) [29] 1.9 Newton’s Third Law of Motion PAY SPECIAL ATTENTION NB NB For a third law forces pair: • the forces are equal in magnitude When pairs of objects interact • the forces act in the same straight line but in opposite directions on they exert forces on each different objects HINT! other. If object A exerts a force • the forces do not cancel each other, as they act on different objects. hint on object B, object B will exert an equal force on object A but Example in the opposite direction. The force diagram shows the pair of forces when a brick rests on a table. For any two    objects A and B; ​ →      → (Note: these are the contact forces) e.g. EG - worked examples F ​A on B = −​F ​B on A exams Opposite direction is indicated by a negative (–) sign. Example activity The reaction force of the weight of an object is ACTIVITIES boy the force that the object exerts on the earth, and girl-left and right of page upwards. (These are not contact forces, they act at a distance.) activity Step by step comment comment © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 27 Mind the Gap Physics

1 Unit 1.10 Newton’s Law of Universal Gravitation PAYPAY SPECIAL ATTENTION SPECIAL ATTENTION NB NB NB NB Newton’s Law of Universal Gravitation states that: Each body in the universe attracts every other body with a force that is directly proportional to the product of their masses and is inversely proportional to the HINT! HINT! square of the distance between their centres. hint hint ​  12 ​ ∴ F = G ​  1 For any two objects: F ∝ m1·m2 and F ∝ __ r ____2 ​ 2 mm r DEFINITION F: magnitude of force (N) Universal means that the m: mass (kg) e.g. e.g. EG -EG - statement is valid worked examples worked anywhere examples r: distance between centres of the objects (m) in the universe. G: universal gravitation constant (6,67 × 10–11 N·m2·kg–2) exams exams activity activity 1 1) F ∝ m1.m2 ACTIVITIES ACTIVITIESboyboy 2) F ∝ __ ​  r2 ​ andand girl-left andand girl-left right ∴ straight line of page right of page ∴ hyperbola through the activity activity origin Step by by Step step step comment comment comment comment 28 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 1 e.g. Worked example 11 EG - worked examples exams A force of gravitational attraction exists between the earth with mass m1 and a      →      → person with mass m2. The force on m1 is ​F ​1 and the force on m2 is ​F ​2. Compare the magnitudes (sizes) of these forces and state the name of the law which explains youractivity answer. ACTIVITIES boy Solution and girl-left and right of page      →      → ​F ​1 = – ​F ​2 according to Newton’s Third Law of Motion: activity The force between the earth (m1) and a person (m2) standing on its surface: mm ∴ m2g = G ​ _____ 1 2 2 ​ r m ∴ g = G ​ ___ 1 2 ​ Step by step comment comment r The gravitational acceleration on earth (or on any planet) is: • dependent on the mass of the earth (planet) (m1) • dependent on the distance between the centre of the object and the centre of the earth (planet) • independent of the mass of the object on the planet on which the force acts. On a different planet, acceleration due to gravity is different. Application of the law of Universal Gravitation: This law enables us to calculate the size (mass) of astronomical bodies like planets, stars, etc. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 29 Mind the Gap Physics

1 Unit 1.11 The difference between mass and weight PAY SPECIAL ATTENTION NB NB Mass Weight It is important to understand • Mass is the amount of • Weight is determined by the force and be able to compare mass matter in an object. of attraction the earth exerts on and weight. HINT! • Mass determines the the object. hint object’s inertia. • Weight depends on the object’s • Mass remains constant. distance from the centre of the • Mass is measured in earth. kilograms (kg). • Weight depends on the masses of e.g. EG - worked examples • Mass is a scalar quantity PAY SPECIAL NB magnitude, but not • ATTENTION the earth (planet) and the object. NB (with Weight is measured in Newton (N). direction). • Weight is a vector quantity, so it has magnitude and direction.      exams →      →      → ​F ​g = m ·​g ​ or = m ·​g ​ where      → HINT! ​g ​= gravitational acceleration hint (9,8 m·s−2 on earth). activity e.g. Worked example 12EG - worked examples ACTIVITIES boy and girl-left and right of page The diagram shows a ball A of mass 0,01 kg which is 1 m (measured from centre to activity centre) exams from another ball B of mass 520 g. Calculate the magnitude of the force of ball A on ball B. Step by step comment comment activity Solution ACTIVITIES boy m1m2 6,67 × 10 –11 × 0,01 ×and 0,52 girl-left and F = G ​ _____ r ____________________ 2 ​ = ​    12 right of ​page = 3,57 ×activity 10–14 N Step by step comment comment 30 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 1 e.g. Worked example 13 EG - worked examples An object weighs 720 N on earth. It orbits the Earth in a satellite at a height equal to the earth’s diameter, above the surface of theexams Earth. What does the object weigh on the satellite? Hint: diameter = 2 × radius Step by step Step 1. Determine the number of radii On the Earth’s surface, the object is activity from the centre of the Earth. 1 radius from the Earth’s centre. ACTIVITIES boy and girl-left and Step 2. Determine how many times the right of page In orbit, the object is 1 diameter = 2 radii distance between the object above the surface. activity and the centre of the Earth has increased. Step 3. Square this number (multiply it by So the object is 3 radii from the centre. itself). (3)2 = 9 Step by step comment comment Step 4. The force has decreased this ∴ the gravitational force on the object number of times because has decreased 9 times because      → __ 1 ​ F ​ ∝ ​  r2 ​      → 1 ​F ​ ∝ ​ __ r2 ​ Step 5. Divide the value of the force (or ∴ its weight on the satellite is weight) by the value calculated in 720 N ÷ 9 = 80 N step 3. © Department of Basic Education 2015 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS 31 Mind the Gap Physics

1 Unit PAY SPECIAL ATTENTION NB NB Gravitational acceleration on planets other than earth Newton’s universal law of gravitation can be used to calculate the acceleration due to the force of gravity on any planet. HINT!      → If the mass and radius of a planet are known, we can calculate ​g ​for that hint planet. e.g. Worked example 14 EG - worked examples The Mars Rover is an automated vehicle that has been sent to explore the surface of theexams planet Mars.      → If the value of acceleration due to gravity on the planet Mars is ​g ​Mars = 3,7 m·s–2. Calculate the weight of the Mars rover on Mars if it has a mass of 174 kg. Solution      →      → ​w ​Mars = ​g ​Mars × mobject = 3,7 × 174 = 643,8 N towards the centre of the planet Mars. activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment Keep going! 32 unit 1 MECHANICS: FORCE AND NEWTON’S LAWS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Momentum and impulse Summary Problem types: Application 1. Two objects collide and continue to move as separate Definitions objects after the collision: Newton’s second law →      →      • momentum ∑​p ​i = ∑​p ​f • Newton’s laws →      →      →      →      m1​v ​1i + m2​v ​2i = m1​v ​1f + m2​v ​2f • Impulse Conservation • Elastic of linear 2. Two objects collide and unite: • Inelastic collisions →      →      A momentum formula: momentum ∑​p ​i = ∑​p ​f p = mv →      →      →      m1​v ​1i + m2​v ​2i = (m1 + m2)​v ​f Elastic and inelastic 3. Two moving objects that are collisions initially joined, then separate: Change in momentum →      →          momentum ∑​p ​i = ∑​p ​f → ​v ​f – m        ∆​p ​= m→ → ​v ​i →      →      →      (m1 + m2)​v ​1 = m1​v ​1f + m2​v ​2f Unit for momentum 4. Two stationary objects that are Newton’s second kg⋅m⋅s–1 law in terms of initially joined, separate (e.g. momentum during an explosion): →      →      Impulse-momentum Fnet = ∆p /∆t ∑​p ​i = ∑​p ​f Application theorem →      →      (m1 + m2)​v ​1 = m1f1f + m2​v ​2f ∆p = Fnet ∆t →      0 = m1v1f + m2​v ​2f →      →      m1​v ​1f = – m2​v ​2f 5. An object falls vertically onto another object that is moving horizontally below it: ∑pi = ∑pf m1v1i + m2v2i = (m1 + m2)vf m1v1i + m2(0) = (m1 + m2)vf m1v1i = (m1 + m2)vf © Department of Basic Education 2015 UNIT 2 MECHANICS: MOMENTUM AND IMPULSE 33 Mind the Gap Physics

2 Unit 2.1 Momentum Momentum is a vector quantity with the same direction as the object’s velocity. You need to remember the differences between speed and velocity. Speed Velocity • Speed is the distance covered per • Velocity is the rate at which an object unit time is displaced. • Speed is a scalar quantity (thus has • Velocity is a vector (thus has magnitude and no direction). magnitude and direction). →      • Symbol: v • Symbol: ​v ​ PAY SPECIAL ATTENTION NB NB The formula for momentum is: p = mv • where HINT! p = momentum hint m = mass v = velocity • mass is measured in kilograms (kg) • velocity is measured in m⋅s–1 e.g. EG - worked examples • the unit of momentum is: kg⋅m⋅s–1 exams Conservation means to keep things The momentum of an object is defined as the product of its mass and velocity →      →      (forces) the same (in a (​p ​= m​v ​) closed system). Linear momentum refers to Momentum is directly proportional Momentum is also directly the momentum of objects to the mass of the object: proportional to the velocity in a straight line. of the object: A closed system is a system activity that does not experience any external forces. ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 34 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Examples of momentum: PAY SPECIAL ATTENTION The motion that is a result of objects colliding with each other, an object NB exploding, or a bullet being fired is described by momentum. 2. Impulse: Impulse is the product of the net force acting on an object and the time HINT! that the force is applied to an object. (Impulse = F∆t). Think of the term “impulsive” or “having an impulse to do something”. This might help you to hint remember what it means. Definitions and principles or laws have certain key words 3. Newton’s second law of motion in terms of momentum: e.g. that should not be EG -left out.examples worked These are written in bold in The net (resultant) force acting on an object is equal to the object’s rate of ∆p each of the above definitions. change of momentum. In a formula: Fnet = ​ ___ ∆t ​ exams 4. The law of conservation of linear momentum: The total linear momentum of an isolated (closed) system remains constant (is conserved). 2.2 Change in momentum activity When an object’s velocity changes in magnitude (size) or direction, its ACTIVITIES boy momentum will also change. Since an object's mass remains constant and girl-left and right of page during a collision (assuming it does not break up or approach light speed), it follows that the change in its velocity is what causes a change in its activity momentum. We only study objects moving in straight lines, for example, backwards and forwards, left and right or up and down. Step by step comment Change in velocity Change in momentum →           →      → →           →      → ∆​v ​=​ v ​f –​v ​i ∆​p ​= ​p ​f – ​p ​i →      →      →      where ∆​p ​= m​v ​f – m​v ​i →      →      →           → ∆​v ​: change in velocity in m·s–1 ∆​p ​= m(​v ​f – ​v ​i) →      ​v ​f: final velocity in m·s–1 where →      →      ​v ​i: initial velocity in m·s–1 ∆​p ​: change in momentum in kg·m·s–1. →      ​p ​f: final momentum in kg·m·s–1. →      ​p ​i: initial momentum in kg·m·s–1 m: mass in kg © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 35 Mind the Gap Physics

2 Unit Momentum is a vector quantity, Steps to follow when solving so you must give the direction in all your problems answers. 1. Make a sketch (on your rough work page) of the situation. 2. Always choose and indicate direction and write it down clearly. It is recommended that you choose a positive direction (e.g. to the right PAY SPECIAL ATTENTION NB is positive). NB 3. Write down the information in symbols. Remember to include the correct signs for the directions of the initial and final velocity. 4. Choose the correct formula from the information sheet. 5. Substitute theHINT! values into the formula. 6.hint Solve for the unknown variable. e.g. Worked example 1 EG - worked examples 1. A car has a momentum of 20 000 kg·m·s–1. What will the car’s new momentum be if its mass is doubled (by adding more passengers and a greater load) and it travels exams at the same velocity? 2. What will the velocity be if the momentum is 60 000 kg·m·s–1 and the mass of the car is 2 000 kg? 3. A truck has a mass of 6 000 kg and travels at 80 km·h–1. How does the momentum change if the truck is loaded with 1 200 kg and then travels at 60 km·h–1? activity Solutions →      →      1. The formula for momentum is ​p ​=boy ACTIVITIES m​v ​, so the momentum will double and will be equal to and 40 girl-left and 000 right of page kg·m·s–1 in the same direction as before. →      2. ​p ​= m​activity →      v ​ →      60 000 = 2 000 × ​v ​ → ∴      ​v ​= 30 m·s–1 in the same direction as the momentum. 3. Convert both velocities to m·s–1: →      →      ​v ​i = 22,22 m·s–1 & ​v ​f = comment Step by step comment 16,67 m·s–1 →      →      ​p ​i = mi​v ​i = (6000 × 22,22) = 133 320 kg·m·s–1 →      →      ​p ​f = mf ​v ​f = (7200 × 16,67) = 120 024 kg·m·s–1 →      →      →      ∆​p ​= ​p ​f – ​p ​I = 133 320 – 120 024 = 13296 kg·m·s–1 36 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 2.3 Newton’s Second Law of PAY SPECIAL ATTENTION NB Motion in terms of momentum Newton’s Second Law of Newton’s Second Law of Motion PAY SPECIAL can be used to find the object’s acceleration ATTENTION Motion states that: The NB due to the net force, and the object’s change in NB momentum due to the net resultant/net force acting HINT! force. hint on object is equal to We know that the object’s change in momentum is always: the rate of change of      → momentum and the change • directly proportional HINT! to the net force acting on the object Δ​p ​ ∝ Fnet is in the direction of the hint • directly proportional to the time that the net force acts on the object      → Δ​p ​ ∝ Δt in the direction of the net force acting on the object. e.g. resultant/net force. EG - worked examples e.g. Worked example 2 EG - worked examples exams Why is it less painful for a high jumper to land on foam-rubber carpet than on the ground? exams Solution →      ∆p ​F ​net = ​ ___ ∆t ​ activity →      →      The ​F ​net needed to bring the jumper to rest (​v ​f = 0 m·s−1) depends on NOTE: ACTIVITIE →      and girl-l ∆​p ​and Δt. WhenPAYheSPECIAL lands ATTENTION on the foam-rubber, he comes to rest over a NB NB The same reasoning explains right of p longer period of time (Δt) than if he lands on the ground. activity activity why a cricket player would • So time taken Δt to change his momentum increases draw his hands back to catch →      →      1 ACTIVITIES boy a fast ball and why modern • ​F ​net decreases (​F ​net ∝ __ ​  Δt ​) and girl-left and →      right of page motor vehicles are designed • The magnitude HINT!of ​F ​ determines the amount of pain net with air bags and crumple hint activity experienced, so it is less painful to land on foam-rubber. zones. Step by step com e.g. Worked example 3 EG - worked examples Step by step comment comment NOTE: 1. A spaceship has a mass of 1 000 kg. The rocket engines discharge for 5 s and • As the question is asking increase the rocket’s velocity from 25 to 30 m·s–1. Calculate the force exerted by exams for a vector quantity the engines to cause this change in momentum. (force) the answer must 2. Assume the direction of the initial velocity is positive and the answer you obtain in have both magnitude and the above problem is negative, what would be the direction of the exerted force? direction. • Since the answer is Solutions positive and it was 1. Let the direction of the initial velocity be positive. decided the direction →      activity →      mv – mv ∆​p ​ (1000)(30) – (1000)(25) of the initial velocity is F ​net = ​ ___ _______ __________________ f i ​ ∆t ​ = ​  ∆t ​ = ​    ACTIVITIES5boy ​ positive the direction of and girl-left and = 1 000 N in the initialright of direction page of motion. the force exerted is the same as that of the initial activity 2. The same i.e. in the initial direction of motion. velocity. Step by step comment comment © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 37 Mind the Gap Physics

2 Unit Activity 1 Study the diagrams below showing the movement of a 150 g baseball thrown at a wall at right angles. Convert mass to kg! Diagram A: m = 150 ÷ 1000 = 0,15 kg →      ​v ​i = + 18 m·s−1 →      ​v ​f = −12 m·s−1 Diagram B: m = 0,15 kg →      ​v ​i = + 18 m·s−1 →      ​v ​f = 0 m·s−1 Let the direction towards the wall be positive. 1. Calculate the momentum of the baseball as it strikes the wall in Diagram A.  (3) 2. Calculate the momentum of the baseball as it leaves the wall in Diagram A. (3) 3. Calculate the change in momentum during the collision in Diagram A. (3) 4. Calculate the force exerted by the wall on the baseball in Diagram A and in Diagram B, if each collision lasts 0,1 s. (6) 5. Draw a vector diagram to illustrate the relationship between the →      →      initial momentum (​p ​i), the final momentum (​p ​f) and the change →           →      → in momentum (∆​p ​= ​p ​f –​p ​i) for the baseball in Diagram A. (6) [21] 38 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Solutions 3 →      → 3      →      →      1. ​p ​i = m​v ​i m = 150 g = 0,15 kg 2. ​p ​f = m​v ​f m = 150 g = 0,15 kg →      →      ​v ​i = 18 m·s−1 ​v ​f = −12 m·s−1 →      →      →      →      ​p ​i = m​v ​i = (0,15)(18) 3 ​p ​f = m​v ​f = (0,15)(−12) = −1,8 kg·m·s–1 →      →      3 ​p ​i = 2,7 kg·m·s−1 towards the wall (3) ​p ​f = 1,8 kg·m·s−1 away from the wall (3) 3 3 3. Here are two ways to answer: →      →      →      →      →      ∆​p ​= m·∆​v ​ m= 0,15 kg OR ∆​p ​= ​p ​f −​p ​i 3 →      3 →      ∆​p ​= (0,15)(−12 −18) 3 ∆​p ​= −1,8 − 2,7 3 →      →      ∆​p ​= −4,5 kg·m·s−1 ∆​p ​= −4,5 kg·m·s−1 ∴ ∆​→      3 ∴ ∆​→      3 p ​is 4,5 kg·m·s−1 away from the wall p ​is 4,5 kg·m·s−1 away from the wall (3) 3 3 →      →      →      ∆​p ​ ________ m​v ​f – m​v ​i (0,15)(–12) – (0,15)(18) 4. Diagram A: Fnet = ​ ___ ∆t ​ = ​  ∆t ​ = ​ __________________    0,1 ​= − 45 N 3 Force exerted by the wall is 45 N away from the wall 3 3 3 →      →      →      ∆​p ​ m​v ​f – m​v ​i (0,15)(0) – (0,15)(18) Diagram B: Fnet = ​ ___ ∆t ​ = ​  ________ ∆t ________________ ​ = ​    0,1 ​= − 27 N 3 Force exerted by the wall is 27 N away from the wall (6) →      3 3 5. ​p ​f = –1,8 kg·m·s–1 away from the wall 3 →      ​p ​i = 2,7 kg·m·s–1 towards the wall 3 →      ∆​p ​f = 4,5 kg·m·s–1 away from the wall 3 3 (6) [21] 2.4 Impulse Impulse is another way to define momentum. Impulse is a measure of the amount of force applied to an object, for a certain period of time. Think of it as a measure of the shock experienced by an object when another object collides with it. The formula for Impulse is: Impulse = F∆t where F is force in newtons N ∆t is change in time in seconds Impulse and momentum are in fact the same thing. We can show this by dimensional analysis, that is, by working out what the units of impulse are, and comparing the units to the units of momentum. F = ma F ∴ a = ​ __ m​ ∆s a = ___ ​ t2 ​ ∆s ∴ F = ​ ___ t2 ​×m ∆s Impulse = ___ ​ t2 ​ × m × ∆t ∆s Impulse = ___ ​  t ​ × m p = mv = vm ∆s p = ​ ___ t ​×m ∴ p = Impulse © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 39 Mind the Gap Physics

2 Unit Remember: 1N·s = 1 kg·m·s–1 Activity 2 1 newton second = 1 kilogram metre A cricket ball of mass 175 g is thrown horizontally towards a player at per second 12 m·s−1. It is hit back in the opposite direction with a velocity of 30 m·s−1. The ball is in contact with the bat for a period of 0,05 s. Calculate: 1. The impulse of the ball. (4) 2. The force exerted on the ball by the bat.  (3) 3. The force exerted on the bat by the ball. Motivate your answer by referring to a Law of Motion. (5) [12] Solutions →      3     → 3 3 1. impulse = ​F ​∆t = m∆​v ​= (0,175)[(–30)–(12)] = –7,35 N·s therefore 7,35 N·s away from the 3 bat (4) →      3 –7,35 3 2. ​F ​∆t = –7,35 = F(0,05) ∴ F = ​ _____ 0,05 ​= –147 N therefore 147 N away from the bat 3 (3) 3 3 3. 147 N towards the bat. According to Newton’s Third Law of Motion the force of the bat on the ball is equal to the force of 3 the ball on the bat, but in the opposite direction, 3 (5) 3 Fbat on ball = −Fball on bat. [12] 2.5 The principle of conservation PAY SPECIAL ATTENTION of linear momentum NB NB The principle of conservation Steps for solving problems on of linear momentum states conservation of linear that: The total linear HINT! momentum hintin a closed momentum system remains constant Step 1. Choose a direction as positive. (is conserved) Step 2. Sketch the situation – draw a block to represent each object. Step 3. Write down the equation for the Conservation of Momentum: e.g. EG - worked examples →      →      Ʃ​p ​i = Ʃ​p ​f Step 4. Expand this equation according to the type of collision. Step 5. Substitute the known values into the equation. Remember to exams check the direction of the objects’ velocities and to use the correct signs for the directions. Step 6. Calculate the answer. Step 7. Write the answer, include units and indicate the direction. Very Important activity • Always remember to include units in your answer • Remember that the +/- signs represent direction ACTIVITIES boy and girl-left and right of page activity 40 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 We can solve problems about the conservation of linear momentum according to the nature of the collision or separation (explosion) of the PAY SPECIAL ATTENTION objects involved. We usually solve problems in which two objects are NB involved. 2.6 Problem types HINT! 1. Two objects collide and continue to move as separate objects after hint the collision: →      →      This looks scary, but it’s not ∑​p ​i = ∑​p ​f →      →      →      →      really! It’s saying that the sum m1​v ​1i + m2​v ​2i = m1​v ​1f + m2​v ​2f e.g. of the momenta EG remains the - worked examples same, that is, before collision 2. Two objects collide and unite: →      →      and after collision, the total ∑​p ​i = ∑​p ​f momentum before and after →      →      →      m1​v ​1i + m2​v ​2i = (m1 + m2)​v ​f exams is the ame. So, ∑pi is the sum of all the initial momenta. 3. Two moving objects that are initially joined, then separate: ∑pf is the sum of all final →      →      ∑​p ​i = ∑​p ​f momenta. To calculate the →      →      →      initial momentum sum, just (m1 + m2)​v ​1 = m1​v ​1f + m2​v ​2f add up the momenta of all the 4. Two stationary objects that are initially joined, separate (e.g. during objects. an explosion): activity →      →      ACTIVITIES boy ∑​p ​i = ∑​p ​f and girl-left and →      →      right of page (m1 + m2)​v ​1 = m1f1f + m2​v ​2f →      0 = m1v1f + m2​v ​2f activity →      →      m1​v ​1f = – m2​v ​2f 5. An object falls vertically onto another object that is moving horizontally below it: ∑pi = ∑pf Step by step comment m1v1i + m2v2i = (m1 + m2)vf m1v1i + m2(0) = (m1 + m2)vf m1v1i = (m1 + m2)vf © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 41 Mind the Gap Physics

2 Unit Problem Type 1: Two objects collide and continue to move as separate objects after the collision Activity 3 In a railway shunting yard, a locomotive (train engine) of mass 4 000 kg, travels due east at a velocity of 1,5 m·s−1. The train driver tries to link it to a stationary wagon of mass 3 000 kg by letting them collide. Instead, the wagon moves due east with a velocity of 2,8 m·s−1. Calculate the magnitude and direction of the velocity of the locomotive immediately after the collision. Remember:  (11) If an object is stationary, its velocity is 0 m·s-1; "stationary" means "not moving". Solution Let the locomotive be object 1 and the wagon be object 2. Let motion to the east be positive Then: m1 = 4 000 kg and m2 = 3 000 kg v1i = +1,5 m·s−1 v2i = 0 m·s−1 v1f = ? v2f = + 2,8 m·s−1 →      →      ∑​p ​i = ∑​p ​f 3 3 →      →           → →      m1​v ​1i + m2​v ​2i = m1​v ​1f + m2​v ​2f 3 3 3 (4000)(1,5) + (3000)(0) = (4000)v1f + (3000)(2,8) 3 3 →      3 3 4000​v ​1f = 6000 – 8400 →      ​v ​1f = 0,6 m·s–1 3 ∴ 0,6 m·s–1 to the west 3 [11] 42 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Problem Type 2: Two objects collide and unite Activity 4 Remember: The question asks A boy of mass 40 kg runs at 5 m·s–1 east and jumps onto a skateboard of for the speed of the mass 2 kg moving at 3 m·s–1 east. skateboard. Speed is a scalar, so you don’t Calculate the speed at which the boy and skateboard move together. (7) need the direction of the motion in the answer. Solution Let the boy be object 1 and the skateboard be object 2. Let motion to the east be positive. Then: m1 = 40 kg and m2 = 2 kg v1i = 5 m·s–1 v2i = 3 m·s–1 v(1 + 2)f = ? ∑pi = ∑pf 3 3 (40)(5) + (2)(3) = (40 + 2)vf 3 3 33 42vf = 206 vf = 4,9 m·s–1 3 ∴ the boy and skateboard move together at 4,9 m·s–1 3 [7] Problem Type 3: Two moving objects that are initially joined, then separate Activity 5 Hendrik is an amateur rocket builder. He launches a two-stage rocket as shown in the diagram. Section A (stage 1) contains the rocket engine and fuel. Section B (stage 2) has a mass of 2 kg. 1. Hendrik says that Newton’s Third Law of Motion is used to explain why the rocket moves upwards during flight. Identify one action- reaction pair of forces involved with the rocket’s motion.  (1) At a certain height, when the rocket has a velocity of 5 m·s–1 upwards, the last fuel is used up, and section A has a mass of 3 kg. To get section B even higher, a small explosion separates section B from section A at this point and increases the upwards velocity of section B to 8 m·s–1. 2. State the Law of Conservation of Linear Momentum in words. (4) 3. Calculate the velocity of section A after the explosion. (11) [16] © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 43 Mind the Gap Physics

2 Unit Solutions 1. The force of the rocket on the expelled (pushed out) gases and the force of the expelled gases on the rocket. 3 (1) 2. The total linear momentum of a closed system remains constant in magnitude and direction. 3 (4) 3 3 3 3. Let upwards be positive. For section A: mA = 3 kg and vAf = ? For section B: mB = 3 kg and vBf = 8 m·s−1 Before the explosion: V(A+B)i = 5 m·s–1 ∑pi = ∑pf 3 3 3 (mA + mB)vi = mAvAf + mBvBf (3 + 2)(5) = (3)(vAf) + (2)(8) 3 33 3 vAf = 25 - 16 3 33 9 ∴ vAf = ​ __ 3 ​= –3 m·s –1 3 ∴ 3 m·s–1 upwards 3 (11) [16] Problem Type 4: Two stationary objects that are initially joined, and then separated (e.g. during an explosion) When two objects are forced apart by an explosion or as a result of a compressed spiral spring released between them, they move in opposite directions after the explosion e.g. when a gun fires a bullet, the bullet moves forwards and the gun moves backwards. Activity 6 A gun of mass 1 kg is attached to a trolley of mass 4 kg and is loaded with a bullet of mass 2 g. The system is at rest on a frictionless horizontal surface. The gun is fired by remote control and the bullet has a muzzle velocity of 350 m·s−1. Calculate the velocity of the trolley and gun after the bullet has been fired.(8) Solution Let the direction of the bullet’s motion be positive. Let the trolley and gun be object 1: m1 = 1 + 4 = 5 kg v1f = ? ∑pi = ∑pf 3 Let the bullet be object 2: (m1 + m2)vi = m1v1f3+ m2v2f3 m2 = 2 g = 0,002 kg 0 = m1v1f + b2v2f v1f = 350 m·s−1 m1v1f = – m2v2f 3 3 Before the explosion: (5)v1f = –(0,002)(350) 3 –(0,002)(350) v(1+2)i = 0 m·s−1 v1f = ___________ ​  5 ​= –0,14 m·s–1 3 Therefore the gun and trolley move at 0,14 m·s−1 in the direction opposite to that of the bullet, after the explosion. [8] 3 44 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 PAY SPECIAL ATTENTION Problem Type 5: An object falls vertically onto another NB object that is moving horizontally below it Activity 7 HINT! hint A trolley of mass 3 kg moves at 4 m·s , west along a frictionless horizontal −1 If an object falls vertically, its path. A brick of mass 1 kg drops vertically onto it. The brick lands on the horizontal velocity is zero. trolley at a vertical velocity of 0,5 m·s−1. Calculate the velocity of the brick and trolley system after the collision. e.g. EG - worked examples  [13] exams activity Solution ACTIVITIES boy and girl-left and right of page The brick strikes the trolley vertically at 0,5 m·s−1. So the brick’s horizontal velocity is zero (0 m·s–1). 3 Momentum is conserved in a activity straight line. 3 So the brick’s vertical velocity is ignored when applying the Law of Conservation of Linear Momentum. 3 (3) Let motion west be positive Step by step comment Let the trolley be object 1: Let the brick be object 2: m1 = 3 kg m2 = 1 kg v1i = +4 m·s v2i = 0 m·s−1 –1 After the collision: v1 + v2 = ? ∑pi = ∑pf 3 3 3 m1v1i + m2v2i = (m1 + m2)vf (3)(4) + (1)(0) = (3 + 1)vf Zero, because the brick is not 3 3 3 3 3 moving horizontally 12 = (4)vf vf = 3 m·s–1 3 3 ∴ brick and trolley system have a velocity of 3 m·s−1 west (horizontally) (10) [13] © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 45 Mind the Gap Physics

2 Unit 2.7 Elastic and inelastic collisions 2.7.1 Revision Linear momentum is always conserved in a closed system. Kinetic energy is, however, always conserved, and is often transformed into other forms PAY SPECIAL ATTENTION NB of energy, NB like heat and sound, or potential energy. 2.7.2 Differentiating between elastic and HINT! inelastic collisions hint Elastic Collisions: Inelastic Collisions: Collisions are classified as • linear momentum is • linear momentum is conserved either elastic collisions or as conserved • colliding objects are joined or • colliding objects remain change their shapes e.g. inelastic collisions. EG - worked examples separate and are not changed • total kinetic energy is not in any way conserved: • total kinetic energy is ∑ Eki > ∑ Ekf exams conserved: • Ek before collision > Ek after collision ∑ Eki = ∑ Ekf • some of the initial kinetic • Ek before collision = Ek after collision energy is transformed into other • the initial kinetic energy is not forms of energy e.g. heat, light, transformed into any other sound. forms of energy. activity Remember that for objects moving much below the speed of light (e.g. bullets, trains, ACTIVITIES boy and girl-left and people, bricks), right of page E = _​ 1 ​ mv2 activity 2 thus, if p = mv, then E = _​ 1 ​ pv 2 Step by step comment Steps for solving problems on comment elastic and inelastic collisions Step 1. Calculate the sum of the kinetic energies of all the objects before the collision 1 1 Ʃ Eki = ​ __ __ 2 ​m1v1i + ​ 2 ​m2v2i 2 2 Step 2. Calculate the sum of the kinetic energies of all the objects after the collision 1 1 Ʃ Ekf = ​ __ __ 2 ​m1v1f + ​ 2 ​m2v2f 2 2 Step 3. Compare the total kinetic energy of the system before the collision to the total kinetic energy of the system after the collision. Step 4. If Ʃ Eki = Ʃ Ekf ∴ Ʃ Ek before the colliison = Ʃ Ek after the collision therefore the collision was elastic If Ʃ Eki ≠ Ʃ Ekf ∴ Ʃ Ek before the collision ≠ Ʃ Ek after the collision therefore the collision was inelastic 46 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Activity 8 Collisions between vehicles take place on the roads in our country daily. In one of these collisions, a car of mass 1 600 kg, travelling at a speed of 30 m·s–1 to the left, collides head-on with a minibus of mass 3 000 kg, travelling at 20 m·s–1 to the right. The two vehicles move together as a unit in a straight line after the collision. 1. Calculate the velocity of the two vehicles after the collision.  (8) 2. Do the necessary calculations to show that the collision was inelastic.(13) 3. New cars have a crumple zone to help minimise injuries during accidents. Air bags and padded interiors can also help to reduce the chance of death or serious injury. Use principles in Physics to explain how crumple zones and air bags can reduce the chance of death or injury. (9) [30] Solutions 1. Let the motion to the left be positive, and treating the minibus as object 1 and the car as object 2: 3 3 p before collision = p after collision m1v1i + m2v2i = (m1 + 3m2)vf Remember: (1 600)(30) + (3 000)(–20) = (1 600 + 3 000)vf Velocity is a vector 3 3 3 3 quantity. ∴ include the 48 000 – 60 000 = (4 600)·vf direction in your answer. vf = –2,6 m·s–1 ∴ vf = 2,6 m·s–1 to the right 3 (8) 2. Before the collision: 1 1 ∑Ki = ​ __ __ 2 ​mv1v 1i + ​ 2 ​m2 v 2i 2 2 3 1 3 3 1 3 3 2 = ​ __ __ 2 ​(1 600)(30) + ​ 2 ​(3 000)(20) 2 = 720 000 + 600 000 = 1,32 × 106 J3 After the collision: 1 1 ∑Kf = ​ __ __ 2 ​mv1v 1f + ​ 2 ​ m2 v 2f 2 2 1 3 3 3 = ​ __ 2 ​(1 600 + 3 000)(2,6) 2 = 15 548 J 3 ∑Ki > ∑K 3f 3 ∴ Ek before collision > Ek after collision ∴ the collision is inelastic 3 (13) © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 47 Mind the Gap Physics

2 Unit 3. Crumple zones in a car ensure that the car comes to rest over a longer period of time (∆t) during an accident, while air bags ensure 3 that the driver / passenger comes to rest over a longer period of time inside the car. 3 3 ∴ ∆t to change the momentum of the car and of the driver OR passenger increases →      →      1 ∴ ​F ​net decreases (​F ​net ∝ __ ​  ∆t ​) and 3 3 →      the magnitude of ​F ​net determines the extent of the passengers’ injuries 3 3 ∴ crumple zones and air bags decrease the extent of injuries during accidents. 3 3 (9) [30] Activity 9 A bullet of mass 10 g, moving at a velocity of 300 m·s–1, strikes a wooden block of mass 1,99 kg resting on a flat horizontal surface as shown in the diagram below. The bullet becomes embedded in the block. Ignore the effects of air friction. 1. Write down in words the principle of conservation of linear momentum. (2) 2. Calculate the speed of the block-bullet system immediately after the collision. (4) 3. Is this collision elastic or inelastic? Give a reason for the answer. (2) The floor exerts a constant frictional force of 8 N on the block-bullet system as it comes to rest. 4. Calculate the distance that the block-bullet system moves after the collision. (5) [13] 48 unit 2 MECHANICS: MOMENTUM AND IMPULSE  © Department of Basic Education 2015  Mind the Gap Physics

Unit 2 Solutions 1. The total (linear) momentum remains constant/is conserved 3 in an isolated/a closed system/the absence of external forces. 3 (2) 2. To the right as positive ∑pbefore = ∑pafter 3 (0,01)(300) 3+ (1,99)(0) = (0,01 + 1,99)vf2 3 vf2 = 1,5 m·s–1 3 (4) 3. Inelastic 3 Kinetic energy is not conserved.3 (2) 4. Fnet = ma 3 ∴ (–8) = 2a 3 ∴ a = –4 m·s–2 Vf2 = Vi2 + 2aΔx 3 02 = (1,5)2 + 2(–4)Δx 3 Δx = 0,28 m 3 (5) [13] Keep going! © Department of Basic Education 2015 unit 2 MECHANICS: MOMENTUM AND IMPULSE 49 Mind the Gap Physics

3 Unit Vertical projectile motion in one dimension Summary You must remember that: acceleration, velocity and displacement (change in position or place) of a projectile occurs if it is: • dropped from a certain height i.e. an object falling freely from rest; • projected (thrown) upwards and then falls back to the same level as the original level; • projected upwards and then falls back to a level below the original level; • a falling object that bounces on a surface. You need to be able to: • Describe the motion for the different types of projectiles mentioned above; • Draw and interpret graphs of the rate of change of position, velocity and Learn the acceleration; summary very well. • Use the graphs to calculate displacement or acceleration (using a velocity- time graph). You must remember: • displacement = moving to another place • velocity = speed in a specific direction • acceleration = change in velocity within a certain time. 50 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 3.1 Revision: Graphs of velocity, acceleration and displacement 1. The slope of a displacement-time graph gives the velocity of an object: →      ∆​y ​      → slope = ​ ___ ∆t ​= ​v ​ (m/s) →      ​v ​ →      ​y ​ →      ∆​y ​= m∆t t where m is the velocity 2. The slope of a velocity-time graph gives the acceleration of the object: →      → ∆​v ​      slope = ​ ___ ∆t ​= ​a ​ (m/s ) 2 →      ​a ​ →      ​  v​ →      ∆​v ​= m∆t t where m is the acceleration 3. The area below a velocity-time graph gives the displacement of the object: 1 Area of Triangle = __ ​ 2 ​ base × ⊥(perpendicular)height →      ​a ​ →      ​  v​ 1 __ ​  2 ​ base × ⊥height = area 1 __ →      ​  2 ​ ∆t × ∆ ​v ​= s t where s is displacement © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 51 Mind the Gap Physics

3 Unit PAY SPECIAL ATTENTION NB NB 3.2 Free fall PAYHINT! SPECIAL ATTENTION NB NB hint projectile: an object (e.g. DefinitionS: stone, ball or bullet) that • Free fall is the unimpeded motion of an object in the absence of air friction e.g. travels throughHINT! the air while EG - worked examples (resistance) where only gravitational force influences the object. hint gravity is the only force acting • Gravitational acceleration: on it. →      • Gravitational acceleration (​g ​) is the constant acceleration of a free falling object unimpeded: without being due to gravity. exams opposed or obstructed or e.g. disturbed. EG - worked examples • All objects experience the same gravitational acceleration (if we ignore the effects of air resistance). Hence, if there were no air a feather would fall at the same speed as a stone. See http://www.youtube.com/watch?v=5C5_dOEyAfk (where O is "oh", not zero). exams • It is always directed downwards. →      →      activity • On earth it is ​g ​earth = 9,8 m·s–2 and on the moon: ​g ​moon = 1,6 m·s–2 ACTIVITIES boy and girl-left and Formulas right of page activity All the following formulas are useful when you calculate projectile activity motion: → ACTIVITIES boy     →      →      • ​v ​ = ​v ​i + ​a ​∆t and girl-left and f where: → right of page      →      →           → →      • ​v ​f2 = ​v ​i2 + 2​a ​·∆​y ​ ​v ​i is initial velocity (m·s−1) activity Step by step → comment      comment →      → 1      • ∆​y ​= ​ v ​i ∆t + __ ​ 2 ​ ​ a ​∆t2 →      ​v ​f is final velocity (m·s−1) →      →      →      → ​v ​1 + ​v2 ​      ∆​y ​is displacement (m) • ∆​y ​= ​ ______ 2 ​ × ∆t →      ​a ​is acceleration (m·s−2) Step by step comment comment Δt is time (s) Tips for calculations: • Ignore air resistance for all calculations in Grade 12 unless the question states that there is air resistance. • Free falling objects experience a constant downward acceleration →      equal to the gravitational acceleration, (​g ​= 9,8 m·s–2). • Choose a direction (downwards or upwards) as positive and keep this unchanged throughout the problem. • Indicate the direction you have chosen as positive clearly at the start of your answers. e.g. REMEMBER: →      Gravitational acceleration “​g ​” is constant and always directed downwards even when the object is moving upwards. Therefore: →      • If upwards direction is chosen as positive gravitational acceleration (​g ​) will be negative. →      • If downwards direction is chosen as positive gravitational acceleration (​g ​) will be positive. 52 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 3.3 Graphs of Projectile Motion Type 1: Dropping a projectile NOTE: If the object is thrown When a projectile is dropped (from rest) from a certain height, then: →      downwards (not dropped) • Initial velocity ​v ​i = 0 m·s–1 from a certain height, then • The velocity increases as the object falls downwards initial velocity is not zero. →      →      • The velocity is a maximum (​v ​f) as the object hits the ground. ​v ​i ≠ 0 →      ​v ​i = the velocity at which the If you choose downwards If you choose upwards object is thrown. as positive as positive →      →      ​v ​i = 0 ​v ​i = 0 →      →      ​a ​= +9,8 ​a ​= –9,8 →      positive →      positive ∆ ​y ​ +? ∆ ​y ​ –? ∆t = ? ∆t = ? →      →      ​v ​f = + max ​v ​f = – max Graphs and Projectile Motion: Type 1 – Dropping a projectile Downwards as positive Upwards as positive The distance (y) NOTE: starts increasing Note that the ground is taken slowly and then as the zero reference for more quickly. these graphs. The velocity (v) increases at the same rate from start to finish. The accelertion is constant at 9,8 m/s2 whether it is in a positive direction or not. Only the sign changes. © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 53 Mind the Gap Physics

PAY SPECIAL ATTENTION NB NB 3 Unit HINT! hint Activity 1 The mass of a falling object A ball of mass 200 g is dropped from the roof of a 100 m high building. is irrelevant during free fall. Ignore air resistance and calculate: e.g. Ignore the givenEG - value! The worked value of the mass is only examples 1. the velocity of the ball when it hits the ground.  (4) relevant if you’re asked to 2. how long the ball is in the air before it hits the ground. (5) calculate the momentum with [9] exams which it strikes the ground. Remember the feather and hammer. Solutions (Calculations for ‘down positive’ and for ‘up positive’ are provided. You only need to do one way!) Let direction of motion down be positive →      →      activity ​v ​i = 0 m·s−1 ​ →      → v ​f = ? (a)      ∆​y ​= + 100 m ​a ​= + 9,8 m·s−2 ACTIVITIES boy ∆t = ? (b) and girl-left and right of page →2      →2      →      →      activity 1. ​v ​f = ​v ​i + 2​a ​· ∆​y ​ 3 = 0 + (2)(9,8)(100) 2 = 1960 3 3 3 →      _____ ​v ​f = ​√1960 ​= +44,27 m·s–1 Step by step comment ∴ 44,27 m·s–1, comment downwards 3 (4) →      →      →      2. ​v ​f = ​v ​i + ​a ​ ∆t 3 3 3 44,27 = 0 + (9,8) ∆t 44,273 _____ ∆t = ​  9,8 ​= 4,52 s 3 ∴ the ball is in the air for 4,52 s (5) [9] If we let direction of motion up to be positive the solution is the same, only the sign changes. This example shows that projectiles can have their motion described by a single set of equations for both upward and downward motion. It is not necessary to set motion in two directions for the same question. However it is important that you are able to solve problems using both approaches i.e. downwards - positive OR upwards - positive. 54 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 3.4 Graphs of Projectile Motion Type 2: Projectile shot up, then falls down Thinking about Physics A gun fires a bullet up into the air and the bullet leaves the gun at mach 1 (the speed of sound). Suppose the bullet flies up to a certain height and then falls down to earth. Will it hit the ground at mach 1, or will it reach a certain limiting velocity and not achieve mach 1 again? Answer: It will hit the ground at approximately mach 1 (ignoring air resistance). The reason is that as it flies up, it decelerates (gets gradually slower), until it reaches 0 m/s at the peak of its travel. It then has the same Ep (potential energy) as it did when it left the gun with the original kinetic energy (Ek). Thus, as it falls back to earth, the Ep is converted to Ek again. Since energy is conserved, the amount of energy with which it strikes the ground, must be the same as that energy that it had when it left the gun, namely, enough energy to reach mach 1. The deceleration of the bullet from mach 1 to 0 m/s is entirely due to gravity and air resistance; hence, when it falls back, its acceleration will be entirely due to gravity. NB: These three graphs below are of the SAME EVENT: an object thrown upwards, then falling back down. The displacement (y) increases in the positive direction until it reaches a maximum, and then it decreases. It becomes negative if it moves Displacement - below the starting point. time graph NB: This graph tracks the position of an object thrown up, then falling, in time. It DOES NOT mean that the object is thrown in an arc. The velocity (v) decreases until it reaches zero at the maximum height. Then the magnitude of the velocity increases, but in a negative Velocity - direction, hence it goes below the t axis. time graph NB: This is NOT a rock rolling down a hill. This is an object being thrown or shot up, reaching a maximum height, then falling back down again. The acceleration is constant at 9,8 m/s2 throughout, since we chose upwards as positive. Acceleration - time graph NB: This is NOT an object travelling in a straight line; it is an object experiencing -9,8 m/s2 uniform negative acceleration (deceleration) due to gravity. © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 55 Mind the Gap Physics

3 Unit PAY SPECIAL ATTENTION NB NB For the three graphs illustrated above • Upwards is positive. If downwards is taken as positive the graphs will be inversed (upside-down). HINT! hint • The original position is taken as the reference point. e.g. EG - worked examples 3.5 Type 2a: A projectile projected vertically upwards which falls exams back to the same level • We choose downwards as • The time taken to rise from A to positive B = time taken to return from B →      • Initial velocity (​v i​) at A. The to the original position C object leaves the starting point in an upwards direction. activity t (up) AB = t (down) BC • The total time taken through AB →      • ​v ​l is negative, as it is moving ACTIVITIES boy to C = time to rise from A to B + and girl-left and upwards. right of page time to return from B to original • The magnitude of the velocity position C. decreases as the object rises activity • The object’s displacement is • The velocity is zero when the zero (as it returns to its original object reaches the highest point position). →      at B. • The velocity (​v f​) down is a →      • ​v ​f (up) at B = 0 maximum as the object hits the →      = ​v ​i (down) at B ground at C. Step by step comment comment • The velocity increases as the • The acceleration of the object object falls down towards the is constant at →      ground. ​g ​= 9,8 m·s−2 downwards • The initial velocity up is equal in throughout the motion. magnitude to the final velocity down.           • ​→ → v i​ (up) at A= ​v f​ (down) at C If downwards is chosen as positive Note: vi (up) at A = vf (down) at C If upwards is chosen as positive →      because level A is the same as level C →      ​g ​= +9,8 m·s ALWAYS. −2 ​g ​= –9,8 m·s−2 ALWAYS. 56 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 Activity 2 A ball is thrown vertically upwards at 4 m·s−1 and returns to the thrower’s hand. Let the direction of motion downwards be positive. Calculate: 1. The maximum height reached by the ball. (4) 2. The time taken for the ball to reach the highest point in its trajectory. (5) 3. The total time that the ball is in the air. (3) 4. The ball’s total displacement during the motion.  (1) [13] Solutions Let the direction of motion down be positive →      ​v ​i = −4 m·s−1 →      ​v ​f = 0 m·s−1 →      ∆​y ​= ? (a) m At the highest point →      of the trajectory the ​a ​ = + 9,8 m·s−2 velocity is 0 m·s–1 while ∆t = ? (b) the acceleration is still 9,8 m·s–2 downwards. →2      →2      →      →      1. ​v ​f = ​v ​i + 2​a ​· ∆​y ​ 3 3 3 →      0 = (–4)2 + (2)(9,8) ​y ​∆ →      19,6 ∆​y ​= –16 →      ∆​y ​= –0,82 m 3 ∴ the ball reaches a height of 0,82 m above the starting level (4) →      →      →      2. ​v ​f = ​v ​i + ​a ​∆t 3 3 3 0 = (–4) + (9,8)∆t 4 ∆t = ​ ___ 9,8 ​3 = 0,41 s 3 ∴ the ball takes 0,41 s to reach the highest point in its trajectory (5) 3 3 3. Time upwards = time downwards ∴ total time in the air is (2)(0,41) = 0,82 3 s (3) →      4. Total displacement = ∆​y ​= 0 m 3 (1) Displacement is measured in a straight line from the initial position (the thrower’s line from the original to the final position (the thrower’s hand is the initial and final position). [13] © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 57 Mind the Gap Physics

3 Unit 3.6 Type 2b: A projectile projected vertically upwards which falls below the original level • We choose upwards as • The time taken to rise from A to B positive. = time taken to return from B to →      • Initial velocity (​v ​i) at A the original position C. • is velocity as object leaves the • The total time taken to move starting point in an upwards though ABC to D = time to rise direction. It is positive. from A to B + time to return from • The magnitude of the velocity B to original position C + time decreases as the object rises. from C to the ground D. • The velocity is zero when the • The object’s displacement is object reaches the highest downwards (as it passes the point at B i.e. starting point and falls further →      downwards). ​v f​ (up) at B = 0 →      • The object’s displacement (A to = ​v ​i (down) at B. D) is equal in magnitude to the • The magnitude of the velocity height from which it was released increases as the object falls →      downwards. • The final velocity (​v ​f) is a maximum as the object hits the • The magnitude of the initial ground at D. velocity upwards at A is equal to magnitude of velocity down • The acceleration of the object →      at the starting level, C. is constant, ​g ​= 9,8 m·s−2 downwards throughout the ​→ →           • v i​ (up) at A = − ​v f​ (down at motion. starting level) at C. If downwards is chosen as positive If upwards is chosen as positive →      −2PAY SPECIAL ATTENTION →      ​g ​= +9,8 m·s ALWAYS. NB NB ​g ​= –9,8 m·s−2 ALWAYS. Activity 3 HINT! hint Lerato throws a stone vertically into the air from the top of a cliff. The stone • Use a ruler to draw the strikes the ground below after 3 s. The velocity vs. time graph below shows axes and any straight the motion of the stone. Ignore the effect of air resistance. e.g. lines! EG - worked examples Drawing a sketch graph • The graph does not have to be to scale, but it must exams have the correct shape • The graph must show the physical quantity in words followed by the abbreviated unit (in brackets) e.g. velocity activity (m·s−1) • The graph must include ACTIVITIES boy and girl-left and the values asked for in the right of page question. activity Step by step comment comment 58 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 1. How long does the stone take to fall from the height of the cliff to the ground below? (2) 2. What is the maximum height that the stone reaches above the groud? (Hint: calculate the height the stone reaches above the cliff, then calculate the height of the cliff, and add these two numbers). (4) 3. Draw a graph of position versus time. Use upwards as negative. (6) [12] Solutions 1. 3 - 2,043 = 0,96 s3 (2) 2. Option 1 1 ∆y = vi ∆t + __ ​ 2 ​a ∆t2 3 1 = (10)(3) + ​ __ 2 ​(–9,8)(3)2 3 = 14,1 m ∆y = 14,1 m below the starting point vf2 = vi2 + 2a∆y 0 = 100 + 2(–9,8) ∆y 3 ∆y = 5,1 m Maximum height above the ground = 5,1 + 14,1 = 19,2m 3 Option 2 1 ∆y = vi ∆t + __ ​ 2 ​a ∆t2 3 1 = 0 3 + __ ​ 2 ​(–9,8)(3 – 1,02)2 3 = –19,21 m ∆y = 19,21 m (maximum height above the ground) 3 Option 3 vf2 = vi2 + 2a∆y 3 (–19,4)2 3 = 0 + 2 (–9,8) ∆y 3 ∆y = 19,2 m (maximum height above the ground) 3 (4) © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 59 Mind the Gap Physics

3 Unit 3. The mass of the ball is irrelevant. The ball is in free fall – this means that all effects of air resistance may be ignored. Marks for: correct shape 33; graph starts at zero 3; maximum height shown as –5,1 m 3; times indicated correctly 3; graph ends at 3 s 3.  (6) [12] 3.7 Graphs of Projectile Motion Type 3: A bouncing ball When a ball falls freely through the air, gravity is the only force that acts on it. The resultant force on the ball is downwards and it accelerates in the direction of the resultant force (Newton’s Second Law of Motion). • Consider a ball that bounces • The resultant force on the ball up from the ground (at A). Let is therefore upwards and the direction of motion upwards ball accelerates upwards, in be positive. other words it slows down and • The ball rises to a maximum stops. height while slowing down • While still in contact with (gravity accelerates it the ground, it starts to move downwards at 9,8 m·s−2) upwards and (at D) leaves the • At the highest point in its path ground. (at B) its velocity is • The collisions with the ground 0 m·s−1 are inelastic and some of • It still accelerates downwards the ball’s kinetic energy is at 9,8 m·s−2 due to gravity. transformed into e.g. sound • It falls to the ground and its and heat every time it strikes velocity increases until it the ground. strikes the ground (at C). • So the velocity as the ball →      • From C to D the ball is in leaves the ground (​v ​i upwards) is contact with the ground. less than the velocity at which • The ground exerts an upward it hits the ground. force on it which is greater • After each bounce, the height than the force of gravity. reached by the ball is less than during the previous bounce. 60 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 Activity 4 A hot-air balloon is rising upwards at a constant velocity of 5 ms–1. When the balloon is 60 m above the ground, a boy drops a ball from it and the ball falls freely. Assume that the balloon continues to move upwards at the same constant velocity. When the ball hits the ground, it bounces vertically upwards to a height of 8 m above the ground. It falls back to the ground and bounces again to reach a height of 5 m. Take upwards as positive. 1. Describe the motion of the ball from the moment it is dropped until it hits the ground.  (8) 2. Why does the ball not reach the same height during the second bounce as during the first? (7) 3. What is the magnitude and direction of the ball’s velocity at the moment when it is dropped?  (1) 4. Calculate maximum height reached by the ball. (5) 5. Where is the ball after 3 seconds? (6) 6. How far apart will the ball and the balloon be after 3 seconds? (7) 7. Calculate the time taken for the ball to reach the ground.  (7) PAY SPECIAL ATTENTION 8. Calculate the time the ball takes to reach the height of 8 m above NB the ground after its first bounce. (10) 9. Calculate the velocity at which the ball hits the ground the after the first bounce. (4) 10. Draw a sketch graph of velocity vs. time for the ball from the HINT! moment it is dropped until it reaches the height of 5 m after its hint first bounce.  (5) 11. Draw a sketch graph of position vs. time for the ball for the same The point of reference for a time as in (10). Use the position of the ball when it is dropped as position-time graph is placed the point of reference. (5) e.g. on the time axis,EGwhere y = 0 m. - worked examples 12. Draw a sketch graph of acceleration vs time for ball for the same time as in (10).  (4) [69] exams Solutions 1. Initially the ball and hot-air balloon will both move upwards at a constant velocity. 3 3 When the ball is dropped it continues to move upwards but decelerates constantly (at 9,8 m·s−2) 3 due to the gravitational 3 attraction force of the earth and slows down until it reaches activity the highest point in its trajectory (path) 3. It stops momentarily ACTIVITIES boy →      and girl-left and (​v ​= 0) 3 and then starts to accelerate downwards constantly 3 right of page (at 9,8 m·s−2). Its speed increases until it hits the ground at a maximum velocity 3. (8) activity 2. The collision between the ball and the ground is inelastic. Some of the ball’s kinetic energy is converted into heat 3 and sound energy and the ball is deformed during the collision. 3 Step by step comment The upward force of the ground on 3 the ball causes it to bounce upwards but the kinetic energy is less than before the collision, so the velocity 3 at which the ball leaves the ground is less than 3 the velocity at which it hit the ground and the height reached 3 is lower 3 than the previous bounce. (7) © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 61 Mind the Gap Physics

3 Unit 3. 5 ms–1 upwards 3 (1) →      →      →      4. ​v ​i = +5 ms–1 ​v ​i = 0 ms–1 ​a ​= –9,8 ms–2 →2      →2      →           → ​v ​f = ​v ​i + 2​a ​·∆​y ​ 3 →      02 = 52 + 2(–9,8)∆​y ​ 3 3→      –25 = –19,6∆​y ​ →      –25 ∆​y ​ = ​ _____ –19,6 ​= 1,28 m 3 3 ∴ the ball will reach a maximum height of (60 + 1,28) = 61,28 m above the ground. (5) →      →      5. ​v ​i = +5 ms –1 ∆t = 3 s ​a ​= –9,8 ms –2 →      →      __      1 →2 ∆ ​y ​ = ​v ​i ∆t + ​ 2 ​ ​a ​t 3 → 3 3 1 3 ∴∆​    y ​ = (5)(3) + ​ __ ​ (–9,8)(3) 2 2 → ∴∆​     y ​ = – 29,1 m 3 ∴ the ball is 29,1 m below the point from where it was released, or (60 – 29,1) = 30,9 m above the ground. 3 (6) 6. The hot-air balloon moved upwards at a constant velocity. →      1      → ∆​y ​ = vi ∆t + __ ​ 2 ​ ​ a ​∆t2 3 →      3 3 ∆​y ​ = (5)(3) + 0 →      ∆​y ​ = 15 m 3 3 ∴ After 3 s the hot-air balloon will be 15 m above the starting point. We know from Question 4 that the ball will be 29,1 m below the starting point after 3 s. 3 ∴ after 3 s the hot-air balloon and the ball will be (15 + 29,1) = 44,1 m apart. 3 (7) 7. NOTE: Always calculate the velocity at which the ball hits the ground first. →      →      ​v ​i = + 5 ms–1 ∆y = – 60 m ​a ​= –9,8 ms–2 →2      →2      →           → ​v ​f = ​v ​i + 2​a ​·∆​y ​ 3 →2      3 ​v ​f = (5)2 + 2(–9,8)(–60) 3 → ∴ ​     v ​f = 34,66 ms–1 downwards 3 →      →      →      then ​v ​f = ​v ​i + ​a ​·∆t –34,66 = 5 + (–9,8)∆t 3 3 ∴ ∆t = 4,05 s (7) 3 62 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 8. NOTE: The up and down displacement of the ball 3 from the first to the second contact with the ground, is the same in magnitude (size) and tup = tdown. 3 Consider the downward motion as negative as in 3 the previous calculations. For the downward part of the bounce: →      ​v ​i = 0 ms–1 3 →      ∆​y ​ =–8m →      ​a ​ = – 9,8 ms–2 →      →      1     → ∆​y ​ = ​v ​i ∆t + __​ 2 ​ ​a ​∆t2 3 1 –8 = 0 + ​ __ ​(–9,8)(∆t2) 3 2 3 ∆t = 1,28 s 3 but tup = tdown 3 ∴ the ball takes 1,28 s to reach a height of 8 m. 3 (10) → → →                9. ​v ​f = ​v ​i + ​a ​∆t 3 →      3 3 ​v ​f = 0 + (–9,8)(1,28) →      ​v ​f = –12,54 ms–1 3 ball hits the ground at 12,54 ms–1 downwards after the first bounce (4) 10. 3 3 3 3 3 (5) © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 63 Mind the Gap Physics

3 Unit 11. distance (m) 3 3 3 3 3 (5) 12. 3 3 3 3 (4) [69] 64 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 Activity 5 1. A 30 kg iron sphere and a 10 kg aluminium sphere with the same diameter fall freely from the roof of a tall building. Ignore the effects of friction. When the spheres are 5 m above the ground, they have the same ... A momentum. B Acceleration C kinetic energy D potential energy (2) 2. An object is thrown vertically into the air at 12 m·s–1 in the absence of air friction. When the object is at the highest point, the velocity of the object in m·s–1 is? A 0 B 9,8 downwards C 9,8 upwards D 12 (2) 3. An object is projected vertically upwards and then falls back to the ground level. The acceleration of the object is …. A Directed upwards throughout its movement. B Zero at the greatest height. C Directed downwards throughout its movement. D Directed upwards and then downwards. (2) [6] Solutions 1. B 33 (2) 2. A 33 (2) 3. C 33 (2) [6] © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 65 Mind the Gap Physics

3 Unit Activity 6 A ball of mass 0,15 kg is thrown vertically downwards from the top of a building to a concrete floor below. The ball bounces off the floor. The velocity versus time graph below shows the motion of the ball. Ignore the effects of air friction. TAKE DOWNWARD MOTION AS POSITIVE. 1. From the graph, write down the magnitude of the velocity at which the ball bounces off the floor. (1) 2. Is the collision of the ball with the floor ELASTIC or INELASTIC? Refer to the data on the graph to explain the answer. (2) 3. Calculate the: a. Height from which the ball is thrown (3) b. Size of the displacement of the ball from the moment it is thrown until time t (5) (in an exam, you might see the word “magnitude” — this means “size”). 4. Sketch a position versus time graph for the motion of the ball from the moment it is thrown until it reaches its maximum height after the bounce. USE THE FLOOR AS THE ZERO POSITION. Indicate the following on the graph: • The height from which the ball is thrown • Time t (4) [15] 66 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION  © Department of Basic Education 2015  Mind the Gap Physics

Unit 3 Solutions 1. 15 m·s–1 3 (1) 2. Inelastic 3 The speed/velocity at which the ball leaves the floor is less / different than that at which it strikes the floor OR The speed/ velocity of the ball changes during the collision. 3 Therefore the kinetic energy changes/is not conserved. (2) 3a. vf2 = vi2 + 2a∆y 3 (20)2 = (10)2 + 2(9,8)∆y 3 ∴ ∆y = 15,31 m 3 (3) 3b Displacement from floor to maximum height vf2 = vi2 + 2a∆y3 (0)2 = (–15)2 + 2(9,8)∆y 3 ∆y = –11,48 m 3 Total displacement = –11,48 + 15,3 3 = 3,82 m 3 or 3,83 m (5) 4. Marking criteria for graph: Correct shape as shown for first part. 3 Correct shape as shown for the second part up to t / 2,55 s. 3 Graph starts at –15,3 m at t = 0 s. 3 Maximum height after bounce at time t / 2,55 s. 3 Maximum height after bounce less than 15,3 m. (4) [15] Keep going! © Department of Basic Education 2015 UNIT 3 VERTICAL PROJECTILE MOTION IN ONE DIRECTION 67 Mind the Gap Physics

4 Unit Work, energy and power 4.1 Work Summary • Work is a scalar quantity and therefore does not have a direction. • The measuring unit of work is Joule. The symbol of Joule is J. • Net Work is the sum of all work done on an object. • Net Work is done by a Net Force. • Positive work is the work done on an object to move it in the Learn the direction of the force (or component of the force). Positive work summary very well. increases the kinetic energy of an object. • Negative work is the work done by an opposing force. Negative work decreases the kinetic energy of an object. • Work done by the man is positive. • Work done by the friction is negative. You must remember: • Work is defined as the product of the force parallel to the movement of an object and the displacement of the object. • Work can be defined Mathematically as: W = F ∆x cos θ • W is the magnitude of work • F is the magnitude of the applied force • ∆x is the magnitude of the displacement • θ is the angle between the applied force and the displacement of the object. • Work-energy theorem: The net work done on an object is equal to the change in the object’s kinetic energy OR work done on an object by a net force is equal to the change in the object’s kinetic energy. • Work-energy theorem formula: Wnet = ∆K = Kf – Ki This is just saying that the total work is the the difference between the initial and the final kinetic energy state. 68 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 Activity 1 Multiple Choice Questions Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (1.1 – 1.2). 1. An object moves in a straight line on a ROUGH horizontal surface. If the net work done on the object is zero, then A the object has zero kinetic energy. B the object moves at constant speed. C the object moves at constant acceleration. D there is no frictional force acting on the object. (2) 2. An object is pulled along a straight horizontal road to the right without being lifted. The force diagram below shows all the forces acting on the object. Which ONE of the above forces does POSITIVE WORK on the object? A W B N C f D component F (2) [4] Solutions 1. B 33 (2) 2. D 33 (2) [4] © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 69 Mind the Gap Physics

4 Unit Activity 2: Work A 220 N force is applied horizontally to a box of mass 50 kg which rests on a rough horizontal surface and the box moves 10 m. The kinetic friction between the surface and the box is 40 N. Calculate: 1. The work done on the box by the applied force.  (3) 2. The work done on the box by the normal force.  (3) 3. The work done on the box by the friction.  (3) 4. The net work done on the box.  (3) 5. The net force acting on the box.  (5) 6. The work done on the box by the net force. (3) [20] Solutions 3 3 1. WFapplied = Fapplied·∆x·cos θ = (220)(10)(cos 0°) = (220)(10)(1) = 2 200 J 3 (3) 3 3 3 2. WFnormal = Fnormal· ∆x· cos θ = mg· ∆x · cos 90°= (50)(9,8)(10)(0) = 0 J (3) 3 3 3. WFfriction = Ffriction·∆x · cos θ = (40)(10)(cos 180°) = (40)(10)(−1) = −400 3 J (3) 4. Wnet = ƩW = WFapplied 3 + WFfriction = (2 200) + 3 (−400) = 1 800 J 3 (3) 5. Let Fapplied act in the positive direction (∴ Ffriction acts in the negative direction) 3 3 3) = (220) + (− 40) = + 1803 Fnet = ƩF = (Fapplied ) + (−Ffriction N ∴ 180 N in the direction of the applied force 3 (5) 6. WFnet = Fnet·∆x · cos θ = (180)(10)(cos 0°) = (180)(10)(1) = 1 800 J 3 3 (3) 3 [20] 70 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 Activity 3: Work A crateof mass 70 kg slides down a rough incline that makes an angle of 20° with the horizontal, as shown in the diagram below. The crate experiences a constant frictional force of magnitude 190 N during its motion down the incline. The forces acting on the crate are represented by R, S and T. 1. Label the forces R, S and T. (3) 2. The crate passes point A at a speed of 2 m·s–1 and moves a distance of 12 m before reaching point B lower down on the incline. Calculate the net work done on the crate during its PAY SPECIAL ATTENTION motion from point A to point B. (5) NB [8] Solutions HINT! 1. R is the Normal Force3, S is the force of gravity3 and T is the hint Frictional force3 (3) 1. Draw a free body diagram 2. R and label all the forces. e.g. EG - worked examples 2. Resolve the Force T = Fcos180° of gravity into its components to determine the applied force acting exams down the incline. F|| = Fαcos70° 70° 3. To determine the net work apply the formula (W = 20° F∆x⋅cos(180˚)) to both the applied force and the frictional force and add to S = Fg = mg findactivity the net work done. ACTIVITIES boy and girl-left and Wnet = Fg∆x · cosθ + Ft · ∆x · cosθ 3 right of page Wnet = (686)(12)(cos 70˚) 3 + (190)(12)(cos 180˚) 3 activity Wnet = 2815,51 – 2280 3 Wnet = 535,51 J 3 (5)[8] Step by step comment © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 71 Mind the Gap Physics

4 Unit Activity 4: Work-energy theorem A rescue helicopter is stationary (hovers) above a soldier. The soldier of mass 80 kg is lifted vertically through a height of 20 m by a cable at a CONSTANT SPEED of 4 m·s–1. The tension in the cable is 960 N. Assume that there is no sideways motion during the lift. Air friction is not to be ignored. 1. State the work-energy theorem in words. (2) 2. Draw a labelled free body diagram showing ALL the forces acting on the soldier while being lifted upwards. (3) 3. Write down the name of a non-constant force that acts on the soldier during the upward lift. (1) 4. Use the WORK-ENERGY THEOREM to calculate the work done on the soldier by friction after moving through the height of 20 m. (5) 5. Identify TWO forces which do negative work. (2) [13] Solutions 1. The net (total) work done on an object3 is equal to the change in kinetic energy of the object. 3OR The work done on an object by a net (resultant) force3 is equal to the change in kinetic energy of the object. 3 (2) 2. (3) 3. Gravitational force or weight of the soldier.3 (1) 4. Solution as follows: Wnet = ∆K 3 WFgravity + Wtension + Wfriction = ∆K Fg∆y · cosθ + FT∆y · cosθ + Ff∆y · cosθ = ∆K (960)(20) · cos 0˚ 3 + (80)(9,8) · cos 180˚ 3 + Wf = 0 3 19200 – 15680 + Wf = 0 ∴ Wf = 3520 J 3 (5) 5. Air friction3 and Force of gravity3 (2) [13] 72 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 4.2 Energy Summary • Energy is a scalar quantity and therefore it does not have a direction • The measuring unit of energy is called Joule. The symbol of Joule is J. • The principle of conservation of mechanical energy states that the total mechanical energy (sum of gravitational potential energy and kinetic energy) in an isolated system remains constant. (A system is isolated when the resultant/net external force acting on the system is zero.) • Solve conservation of energy problems using the equation: Wnc = ∆Ek + ∆Ep 1 • The formula Ek = ​ __ 2 ​ mv is used to calculate the kinetic energy. 2 • The formula Ep = mgH is used to calculate the potential energy. • The Law of Conservation of Energy states that energy cannot be created or destroyed. Energy can only be transferred from one object to another or transformed from one type of energy to another type. Learn the summary for You must remember: Energy very well. • Energy is the ability to do work • A conservative force is defined as a force for which the work done in moving an object between two points is independent of the path taken. Examples are gravitational force, the elastic force in a spring and electrostatic forces (coulomb forces). • A non-conservative force is defined as a force for which the work done in moving an object between two points depends on the path taken. Examples are frictional force, air resistance, tension in a cable, etc. © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 73 Mind the Gap Physics

4 Unit Activity 5 Multiple Choice Questions: Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (5.1 – 5.2). 1. The kinetic energy of a car moving at a constant velocity v is K. The velocity of the car changes to 2v. What is the new kinetic energy of the car? A 0,25 K B 0,5 K C 2K D 4K (2) 2. A stone is dropped from the edge of a cliff. Which ONE of the following graphs best represents the change in kinetic energy of the stone during its fall? (2) [4] Solutions 1. D 33 (2) 2. A 33 (2) [4] 74 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 Activity 6: Work done by Non- Conservative Forces A box of mass 100 kg slides down a slope. Its velocity increases from 0 m·s−1 at point A to 4 m·s−1 at point B as in the diagram. Calculate the work done by the non-conservative force while it slides from A to B. PAY SPECIAL ATTENTION NB [4] HINT! Solution hint Remember that a frictional Wnc = ∆Ep + ∆Ek 3 force is an example of a non- 1 1 = [mghB – mghA] + [​ __ __ 2 ​mvB² − ​ 2 ​mvA²] 1 1 e.g. conservative EG - force,therefore worked examples = [(100)(9,8)(0) – (100)(9,8)(1)] 3 + [(​ __ __ Wnc = Wf 2 ​)(100)(4)² − (​ 2 ​)(100)(0)²] 3 = –980 + 800 = 180 J ∴Wfriction is 180 J 3 exams [4] activity ACTIVITIES boy and girl-left and right of page activity Step by step comment © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 75 Mind the Gap Physics

4 Unit Activity 7 A toy truck, mass 1,4 kg, moving down an inclined track, has a speed of 0,6 m·s–1 at point P, which is at a height of 1,5 m above the ground level QR. The curved section of the track, PQ, is 1,8 m long. When the truck reaches point Q it has a speed of 3 m·s–1. There is friction between the track and the truck. 1. State the principle of conservation of mechanical energy (2) 2. Is mechanical energy conserved? Explain. (2) 3. Assume that the average frictional force between the track PAY SPECIAL ATTENTION and the truck is constant along PQ and calculate the average NB NB frictional force experienced by the truck as it moves along PQ. (6) [10] HINT! Solutions hint 1. The total mechanical energy in an isolated system 3 remains • Use the formula constant or is conserved. 3 (2) Wnet = ∆EK + ∆Ep to 2. The mechanical energy is not conserved 3 due to the presence e.g. calculate theEGwork - worked examples by the non-conservative done of non-conservative force (frictional force). 3 (2) 3. Wnc = ΔEk + ΔEp 3 force (friction). 1 1 Wnc = (​ __ __ 2 ​ mvf – ​ 2 ​ mvi ) + (mghf – mghi) 3 2 2 • Use the work done by the exams non-conservative force to Wnc = (0,5)(1,4)(32) – (0,5)(1,4)(0,62) + 0 – (1,4)(9,8)(1,5) 3 calculate friction. Wnc = 6,3 – 0,252 + 0 – 20,58 • Remember, Work is a scalar, it does not have a Wnc = –14,532 = 15,532 J uphill 3 direction and therefore do Wf = f∆x ·cos θ 3 not leave your answer in a activity negative form. 14,532 = f(1,8) ·cos 0˚ ACTIVITIES boyf = 8,07 N 3 (6) and girl-left and right of page [10] activity Step by step comment comment 76 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 4.3 Power Learn the Summary summary for Power very well. • Power is the rate at which work is done or energy is transferred (or converted). • Power is an indication of the rate at which (how fast) work is done or energy is transferred or transformed and is a scalar quantity. You must remember: • The unit of Power is the watt • The symbol of watt is W. W • The formula P = ​ __ ∆t ​is used to calculate power, where P is power, W is work and ∆t is the change in time. • Average power can be calculated by applying the formula Pav = FVav, where Pav is the average power, F is the force or net force and Vav is the average velocity or average speed. Activity 8 Multiple Choice Questions: Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A–D) next to the question number (8.1 – 8.2). 1. Power is defined as the rate….. A of change in velocity. B at which work is done. C of change of momentum. D of change of displacement. (2) 2. Which ONE of the following physical quantities is equal to the product of force and average velocity? A Work B Average power C Energy D Average acceleration (2) [4] Solutions 1. B 33 (2) 2. B 33 (2) [4] © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 77 Mind the Gap Physics

4 Unit Activity 9: Power A car of mass 500 kg accelerates from 10 m·s–1 to 30 m·s–1 in 20 s. Calculate the power of the car. [5] NB PAY SPECIAL ATTENTION SolutionNB 1 1 Wnet = ∆ Ek = (​ __ __ 2 ​ mvf – ​ 2 ​ mvi ) 3 2 2 Wnet = (0,5)(500)(302) – (0,5)(500)(102) 3 3 HINT! Wnet = 200 000 J hint P W = ​ __ ∆t ​= 200 000 ÷ 20 3 1. First calculate the work P = 10 000 W 3 [5] done by applying the e.g.work-energyEGtheorem. - worked examples 2. Calculate the power (P). Activity 10: Power exams A pump is needed to lift water through a distance of 25 m from a borehole at a steady rate of 180 kg/min. What is the minimum power motor that could operate the pump if the velocity at the intake is 4 m·s–1 but at the outlet the water is moving with a speed of 9 m·s–1. activity ACTIVITIES boy and girl-left and right of page  [8] activity Solution PAY SPECIAL ATTENTION NB Step by step NB∆Ep 3 W = ∆Ekcomment nc comment + 1 1 Wnc = (​ __ __ 2 ​ mvf – ​ 2 ​ mvi ) + (mghf – mghi) 3 2 2 Wnc = (0,5)(3)(92) – (0,5)(3)(42) + (3)(9,8)(25) – 0 HINT! 3 3 3 Wnc = 121,5 – 24 + 735 – 0 hint Wnc = 832,5 J 3 In 1 second, the mass of water lifted up is But W3 e.g. 180 kg/60 s = 3EG -kgworked examples P = ​ __ ∆t ​= 832,5/1 = 832,5 W 3 [8] exams 78 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 4 Activity 11: Power A 0,5 horsepower electric pump is used to bring water out of a borehole that is 80 m deep. 1 horsepower = 745,7 W. Calculate the mass of water that is let out of the borehole in one minute. [5] Solution P = 0,5 × 745,7 = 372,85 W W P = ​ __ ∆t ​ 3 372,85 = W / 60 W = 22 371 J W = F∆y·cosθ 3 F = mg W = mg ∆y·cos 0° 3 22 371 = m × (9,8)(80)cos 0° 22 371 = 784 × m 3 ∴m = 28,53 3 kg [5] Activity 12: Power A motor car of mass 1 400 kg moves with a constant speed up a slope that makes an angle of 10° with the horizontal. The motor car experiences a frictional force of 700 N as it moves up the slope. 1. Draw a free body diagram to indicate the forces acting on the car. (7) 2. Calculate the applied force necessary to move the motor car up the slope at a constant speed. (5) 3. If the motor car moves at 80 km·h–1, calculate the power delivered by the motor car’s engine. (3) [15] © Department of Basic Education 2015 UNIT 4 WORK, ENERGY AND POWER 79 Mind the Gap Physics

4 Unit Solutions 1. 3 3 3 3 3 3 3 3 3 3 3 3 3 3 Fapplied: pulling force exerted by the engine 3 FN: normal force upwards on the car by the surface 3 Fg: gravitational attraction of the earth on the car vertically downwards 3 Fg||: component parallel to incline 3 Fg⊥: component perpendicular to incline 3 Ff: frictional force between the car and the surface, in the direction opposite to the motion 3 (7) 2. If v is constant, a = 0 m·s and Fnet = 0 N –2 Let direction up the incline be positive. Fnet = ma = ∑F 3 0 = Fapplied – Ff – Fg|| 3 3 0 = Fapplied – 700 – (1400)(9,8)(sin10°) 3 Fapplied = 700 + (1400)(9,8)(sin 10°) the applied force is 3 082,45 N up the incline. 3 (5) Remember First convert km/h to m·s–1. An easy way to do this is to divide by 3,6 3. v = 80km/h = (80 000 m) / (3600 s) = 22,22 ms–1 3 P = Fv = (3 082,45)(22,22) = 68 498,89 W (3) Keep going! 3 3 [15] 80 UNIT 4 WORK, ENERGY AND POWER  © Department of Basic Education 2015  Mind the Gap Physics

Unit 5 Doppler Effect 5.1 Waves: Revision • Vibrations cause waves and waves cause vibrations. There are two kinds of waves: transverse waves and longitudinal waves. Transverse waves Longitudinal waves • The disturbance of the medium is • The disturbance of the medium is parallel perpendicular to the direction in which the to the direction of propagation of the pulse. wave is propagated (transmitted). • Example: sound waves, slinky spring • Examples: water waves, electromagnetic waves (light, radio waves, X-rays etc.) © Department of Basic Education 2015 UNIT 5 DOPPLER EFFECT 81 Mind the Gap Physics

5 Unit 5.1.1 Wave properties • The amplitude (height) of a wave motion is the maximum displacement of the particles from their equilibrium (rest) position. The amplitude determines the volume of a sound wave. • The wavelength (λ) of a wave is the distance between two consecutive points in the wave which are in phase and is measured in metres (m). It is therefore also the distance between two successive crests or the distance between two successive troughs. • The frequency (f) of a wave motion is the number of complete waves passing a specific point per second and is measured in hertz (Hz). The frequency of a sound wave determines its pitch. The frequency of a light wave determines its colour. • The frequency of a wave determines the energy of the wave. • The higher the frequency, the higher the energy. So E ∝ f PAY SPECIAL ATTENTION NB NB The period (T) of a wave motion is the time taken for one complete wave to pass a fixed point. HINT! f = number of waves passing a point T = _​ 1 ​ and hint _______________________ ​      ​ time f f = ​ _1 ​ and T T is the period of the graph T= time _____________________ ​       ​ (s) f is the frequency (Hz) number of waves passing a point e.g. EG - worked examples The speed (v) of a wave is the rate at which the energy is propagated by the wave and is measured in m⋅s−1. exams v=fλ where v is speed (m·s–1) f is frequency (Hz) λ is wavelength (m) activity ACTIVITIES boy and girl-left and right of page activity 82 UNIT 5 DOPPLER EFFECT  © Department of Basic Education 2015  Mind the Gap Physics

Unit 5 5.1.2 Light PAY SPECIAL ATTENTION NB • The visible spectrum of light is just a small section of a much greater For light: c = f λ series of wavelengths called the electromagnetic spectrum. c is the speed of an • The speed of light (and all other electromagnetic radiation) is ·s–1) electromagnetic wave (mHINT! constant (3 × 108 m·s−1). hint c = 3× 108 m·s−1 • The colour of light depends on its frequency. • In the colour spectrum, red has the longest wavelength and lowest f is frequency (Hz) frequency and violet has the shortest wavelength and the highest λ is wavelength (m) frequency. e.g. EG - worked examples The Visible Light Spectrum exams activity ACTIVIT and girl right of activity Step by step co © Department of Basic Education 2015 UNIT 5 DOPPLER EFFECT 83 Mind the Gap Physics

5 Unit 5.2 The Doppler Effect PAY SPECIAL ATTENTION NB NB The Doppler Effect is the change in frequency or pitch of the sound or the colour of light that is detected when the wave source and the observer move relative to each other. HINT! hint Vocabulary: Frequency means “how often”. “Observer” means person who sees, hears, or otherwise comes to know through the senses. Examplee.g. EG - worked examples When a car approaches a listener: • the sound waves emitted by the car’s hooter are compressed in exams front of the car; • more sound waves reach the listener per second and • the pitch appears to be higher than the sound emitted by the source (the car’s hooter). The opposite is true when the car moves away from the listener. activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 84 UNIT 5 DOPPLER EFFECT  © Department of Basic Education 2015  Mind the Gap Physics

Unit 5 Activity 1 1. A sound source approaches a stationary (not moving) observer at constant velocity. Which ONE of the following describes how the observed frequency and wavelength differ from that of the sound source? Observed Wavelength Observed Frequency A. Greater than Greater than B. Less than Less than C. Greater than Less than D. Less than Greater than (2) 2. Which one of the following is the main principle applied when using the rate of blood flow or the heartbeat of a foetus in the womb? A. Doppler Effect. B. Photoelectric effect C. Huygens principle D. Diffraction (2) 3. An ambulance approaches an accident scene at constant velocity. The siren of the ambulance emits sound waves at a frequency of 980 Hz. A detector at the scene measures the frequency of the emitted sound waves as 1 050 Hz. a. Calculate the speed at which the ambulance approaches the accident scene. Use the speed of sound in air as 340 m·s–1 (6) b. Explain why the measured frequency is higher than the frequency of the source. (2) c. The principle of the Doppler Effect is applied in the Doppler flow meter. State ONE positive impact of the use of the Doppler flow meter on humans. (2) 4. The siren of a stationary (not moving) ambulance emits sound waves at a frequency of 850 Hz. An observer (person witnessing this) who is travelling in a car at a constant speed in a straight line, begins measuring the frequency of the sound waves emitted by the siren when he is at a distance x from the ambulance. The observer continues measuring the frequency as he approaches, passes, and moves away from the ambulance. The results obtained are shown in the graph below. © Department of Basic Education 2015 UNIT 5 DOPPLER EFFECT 85 Mind the Gap Physics

5 Unit a. The observed frequency suddenly changes at t = 6s. Give a reason for this sudden change in frequency. (1) b. Calculate b(1) The speed of the car (Take the speed of sound in air as 340 m·s–1)(5) b(2) Distance x between the car and the ambulance when the observer BEGINS measuring the frequency. (3) [23] Solutions 1. D 3 3 (2) 2. A 3 3 (2) 3. V ± VL (fs) 3 V (fs) 3 fL = ​ _______ V±V ​ OR fL = ______ ​  (V – V ) ​ 3 s s ∴ 1050 3 – (340 – 0) / (340 – Vs) × 980 3 ∴ Vs = 22,67 m/s 3 (6) a. Waves in front of the moving source are compressed. The observed wavelength decreases (3). For the same speed of sound (3), a higher frequency will be observed. (2) b. Determine whether arteries are clogged or narrowed (3), so that precautions can be taken to prevent heart attack or stroke (3), OR Determine the heartbeat (3) of a foetus to assure that the child is alive or does not have a heart defect (3). (2) 4.a The approaching observer (higher f) passes the source at t = 6 s and moves away (lower f) from the source. (3) (1) 4.b(1) Option 1. Approaching observer: V ± VL (fs) fL = V + VL (fs) fL = ​ _______ V±V ​ OR __________ ​  V ​ 3 s ∴ 900 3 = (340 + VL) (850) 3 / (340) ∴ VL = 20 m/s 3 (5) 86 UNIT 5 DOPPLER EFFECT  © Department of Basic Education 2015  Mind the Gap Physics

Unit 5 Option 2. Observer moving away: V ± VL (fs) fL = V - VL (fs) fL = ​ _______ V±V OR _________ ​ ​  V ​ 3 s 3 ∴ 800 3 = (340 – VL) (850) 3 / (340) ∴ VL = 20 m/s 3 (5) Notes: any other correct Doppler Effect formula gets maximum ​ 3_​ marks. 4 4b(2) Option 1. 1 ​ 2 ​a∆t2 3 ∆x = v1∆t + __ 1 = (20)(6) + __ ​ 2 ​(0)∆t2 3 = 120 m 3 Option 2. 3 ∆x = v∆t = (20)(6) 3 3 = 120 m (3) Note: accept s = ut or s = vt, as well as s = ut + _​ 1 ​ at2, as well as Δy = v1Δt + _​ 1 ​ aΔt2 2 2 [23] Activity 2 An ambulance moving at 40 m·s–1 approaches a traffic light where a blind man and his dog wait to cross a road. The siren of the ambulance (source) emits sound waves at a frequency of 350 Hz (fs). The pitch of the sound that the man hears increases as the ambulance moves towards him and decreases as the ambulance passes him and moves away. 1. If the speed of sound in air is 340 m·s–1, determine the frequency (fL) of the sound waves that the man hears while the ambulance approaches him. (3) 2. Explain how this effect can help a blind person waiting to cross the road. (2) [5] Solutions (V±VL) ______ 1. fL = ​ (V±VS)fS ​ 3 (340+0) fL = ​ _______________    (340–40) × 350 Hz ​ 3 fL = 396,67 Hz 3 (3) 2. When crossing a street, a blind person can determine whether a car is moving towards or away from him. If the pitch of a v ehicle decreases, 3 the person knows that the vehicle is moving away from him, and vice versa. 3 (2) [5] © Department of Basic Education 2015 UNIT 5 DOPPLER EFFECT 87 Mind the Gap Physics

5 Unit 5.3 Applications of the Doppler Effect with ultrasound waves The Doppler flow meter is used to measure the rate of blood flow in a patient’s blood vessels. • Ultrasound is a longitudinal wave with very high frequency of above 20 kHz that we cannot hear. • A catheter connected to a Doppler flow meter is inserted in a blood vessel. It gives out a sound wave at ultrasound frequency. The blood velocity through the heart causes a ‘Doppler shift’ in the frequency of the returning waves. The meter measures this and compares the frequencies. • The receiver detects the reflected sound and an electronic counter measures the reflected frequency. • From the change in frequency, the speed of the blood flow can be determined and narrowing of blood vessels identified. 5.4 Applications of the Doppler Effect with light The electromagnetic Spectrum The electromagnetic spectrum is the full range of types of electromagnetic radiation. Electromagnetic radiation consists of waves which have both an electric and magnetic component. They are transverse waves. They are emitted by many objects eg. the sun, lights, fires, stoves, persons. All forms of light, radio, and heat at a distance, are electromagnetic radiation. Note that electromagnetic radiation is NOT the same as radioactivity except for gamma waves, which come from nuclear reactions. AL ATTENTION NB The electromagnetic spectrum is shown on the next page. Visible light is part of the electromagnetic spectrum. Remember: • Stars, like the sun, emit light. The wavelength of light is • When a star moves away from the Earth, its spectrum shifts to longer usually measured in nm wavelengths (lower frequencies) – in other words, the red side of the (nanometers) and must be spectrum. The star appears red. converted to m (metres) before doing any calculations. • When a star moves towards the Earth, its spectrum shifts to shorter wavelengths (higher frequencies) – in other words, the blue side of 1 nm = 1 × 10–9 m ed examples the spectrum. The star appears blue. ACTIVITIES boy and girl-left and right of 88 page UNIT 5 DOPPLER EFFECT  © Department of Basic Education 2015  Mind the Gap Physics

Unit 5 Ultrasound machines use the Doppler Effect to get readings of unborn babies in the womb so that doctors can see if the baby is healthy. Keep going! © Department of Basic Education 2015 UNIT 5 DOPPLER EFFECT 89 Mind the Gap Physics

6 Unit Electrostatics Summary • Electrostatics is the study of static (rest or stationary) positive or negative charges. Think for example of how you can get little shocks from scuffing your feet on a carpet when wearing rubber-soled shoes. This kind of electricity is not flowing, unlike, for example, the electricity in a plug point. • Define electrostatics, electric field and electric field strength. • Give evidence for the existence of two kinds of electric charge (like charges repel, unlike charges attract). • Describe and demonstrate a method for determining whether an unknown charge is positive or negative. • Name the unit of charge, and discuss its size with respect to common electrostatic situations and in terms of the number of unit charges it represents. • Describe what it means to say that charge is conserved. • Coulomb’s Law • Drawing of electric field lines. • Application of Coulomb’s Law and electric field strength (by calculations). 6.1 Definitions: Electrical charge and electric force Electrical Charge • At the atomic level, charge is associated with protons and electrons. They have the same magnitude of charge, but their charge is opposite in sign. Protons have Positive charge and Electrons have Negative charge. The symbol for a proton is p+ and the symbol for an electron is e–. • Charge is measured in Coulombs, abbreviated C. It takes 6,25 × 1018 Static, rest or stationary have charges to make 1C of charge, i.e. 6 250 000 000 000 000 000 charged similar meanings particles. 6,25 × 1018 electrons will make –1C of charge. A coulomb is in science defined technically as one ampere-second (1 As), in other words, the amount of charge carried by one ampere in one second. • Coulomb’s Law is a measure of how strong the force is between two charged orbjects. Its formula is: kQ1Q2 F = _____ ​  r2 ​ • Electrical charges will exert forces on each other. Two positive charges will repel each other and two negative charges will repel each other. (Like charges, or charges of the same sign, repel each other.) A positive and a negative charge will attract each other. (Unlike charges attract each other.) • Electrostatic charge is a strong force. 90 UNIT 6 ELECTROSTATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 • As a result most objects usually have about the same amount of positive and negative charge. If they have exactly the same amount of positive and negative charge the net charge is zero and we say they are neutral. • For convenience, we can abbreviate “positive” and “negative” as +ve and -ve respectively. Electric Force • The force is proportional to the product of the charges and inversely proportional to the square of the distance between them. (If we double one charge the force doubles. If both charges are doubled then the force increases by four. If the distance between the charges is increased, the force decreases, and vice versa.) This is similar to Newton's Law of Gravitation, which has the same formula structure. • Electrical forces can do work and there is a potential energy associated with this force. 6.2 The Law of Conservation of Charge PAY SPECIAL ATTENTION NB • When two charged spheres are brought into contact with each other, electrons flow from the sphere with more electrons to the Charge cannot be destroyed sphere with fewer electrons. or created, but can only be • The symbol for charge is Q. Do not confuse this with current, Ι. transferred from one object HINT!to Q is measured in coulombs and Ι is measured in amperes. • If sphere B has more electrons than sphere A: hint another. e.g. REMEMBER EG - worked examples when charges are brought into contact they must be added and then exams divided by TWO to get an average. activity ACTIVITI and girl- right of p activity Step by step com © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 91 Mind the Gap Physics

HINT! 6 Unit hint e.g. Worked example 1EG - worked examples +5C –7C exams A B Two spheres A and B carry charges of +5 C and −7 C respectively. They are brought into contact and are then separated. 1. What is the nature of the force between the charges before they are allowed to touch? Explain. activity 2. In which direction are electrons transferred during the contact? Explain. ACTIVITIES boy 3. Calculate the total charge in the system. and girl-left and right of page 4. Calculate the charge on each sphere when they are separated. 5. Calculateactivity the change in the charge on A and on B. 6. Calculate the number of electrons transferred from one sphere to the other. Solutions 1. Attraction. Opposite charges (+ and −) attract. comment Step by step comment 2. From sphere B to sphere A. Electrons are transferred from the sphere with the most electrons (B in this case) to the sphere with the least electrons (A). 3. Qtotal = Qi (A) + Qi (B) = +5 + (−7) = −2 C total –2 4. Qnew on each = Q​ ____ ___ 2 ​ = ​  2 ​= −1 C = Qf 5. ∆QA = Qf – Qi = −1 – (+5) = −6 C ∴ 6 C charge was transferred from B to A ∆QB = Qf – Qi = −1 – (−7) = +6 C ∴ 6 C charge was transferred from B to A ∆Q(A) ∆Q(B) 6. No. of e– transferred _____ ​  Qe– ​= No. of e– transferred = _____ ​  Qe– ​ 6 6 = ​ _________ 1,6 × 10–10 ​ or = _________ ​ 1,6 × 10–19 ​ = 3,75 × 1019 electrons 92 UNIT 6 ELECTROSTATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 6.3 Coulomb’s Law Coloumb’s Law is the electrostatic force of attraction or repulsion between two charged objects is directly proportional to the product of the charges and inversely proportional to the square of the distance between their centres. CIAL ATTENTION NB For any two charges, where F is electrostatic force (N), Q is charge (C), r is the distance between the centres of the objects (m), and k is Coulomb’s constant (9 x 109 Nm2C–2) ​  12 ​ F ∝ Q1 · Q2 and F ∝ __ r kQ1Q2 ∴ F = ​ ____​ r2 ked examples 6.3.1 Using Coulomb’s Law PAY SPECIAL ATTENTION We apply Coulomb’s Law to determine how the electrostatic force between NB NB two charged objects (or point charges) changes when the charge on one or both of the objects changes and when the distance between their centres changes. ACTIVITIES boy and girl-left and PAY SPECIAL ATTENTION right of page We can also use Coulomb’s HINT! Law to calculate the electrostatic force between NB two charges, the distance between them, or the magnitudes (sizes) of the hint charges. HINT! ep commente.g.comment Worked example 2 EG - worked examples hint Remember to square the The original force between two charges is F. If both charges are doubled and the distance. distance exams is a third of the original distance, what is the magnitude of the new force e.g. EG - worked examples relative to the original force? Solution kQ1Q2 exams Foriginal = ​ _____ r2 ​ k(2Q1)(2Q2) Fnew = ​ _________ ​ activity 1 __ 2 (​  3 ​r) 4kQ1Q2 36kQ1Q2 ACTIVITIES boy Fnew = ​ ______ 1 __ 2 ​ = ​ _______ r2 ​= 36 Fand girl-left and original ​  9 ​r activity right of page activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 93 Mind the Gap Physics Step by step comment

HINT! 6 Unit hint ‘Charges are unknown’ this means I e.g. Worked example 3 EG - worked examples need to calculate them. I must remember just not forget to convert mm to m. exams Two small identical metal spheres carry equal but opposite charges. If their centres are 30mm apart,activity and the electrostatic force between them is 2.56 × 10–3N. Calculate: 1. The magnitude (size) of the charge on each sphere ACTIVITIES boy and girl-left and 2. The number of electrons that wouldright flowoffrom page the negatively charged sphere to the positively charged sphere if they were brought into contact. activity Solutions kQ1Q2 1. F = ​ _____ r2 ​ × 0,0009 2,56 × 10–3 Step by step (9 × 10–4comment ) = 9 × 10 9 ×Q×Q comment × 10 / ( 9 × 10 ) = Q 2,304___________ –6 9 2 Q = ​√2,56 × 2    10–16 ​; Q = 1,6 × 10–8C total charge 2. number of electrons = _________________ ​      charge on one electron ​ = (1,6 × 10–8) / (1,6 × 10–19) = 1 × 1011 electrons 94 UNIT 6 ELECTROSTATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 Activity 1 Two small identical metal spheres, A and B carrying charge of –4 × 10–12C and –3 × 10–12C respectively, are mounted on insulated stands as shown. The distance between the centres of the spheres is 5 cm. 1. Calculate the magnitude and direction of the force that A exerts on B.  (6) Sphere A is moved and makes contact with sphere B. It is then moved back to its original position. 2. Calculate the new charge on each of the spheres.  (3) 3. How does the magnitude of the force that the sphere A exerts on sphere B change? Answer by writing ONLY increases, decreases or remains the same.  (2) [11] PAY SPECIAL ATTENTION NB Solutions kQ1Q2 1. F = ​ _____ r2 ​ 3 3 (9 × 109) × (–4 × 10–12) × (3 × 10–12) = ​ __________________________       ​ (0,05)2 3 HINT! –1,08 × 10 = ​ ___________ –13 hint 0,0025 ​ Start by converting the = –4,32 × 10–11 N 3 distance 5cm to m. Look at = 4,32 × 10–11 N 3 Force of attraction 3 (6) e.g. the conversion EG table. - worked examples (Q1 + Q2) 2. Q = ​ _______ 2 ​ 3 (–4 × 10 ) + (3 × 10–12) –12 = ​ __________________    2 ​ 3 exams = –5 × 10–13C 3 This is a new charge (3) 3. Increases 3 3 (2) [11] activity ACTIVITIES boy and girl-left and right of page activity Step by step comment © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 95 Mind the Gap Physics

6 Unit 6.4 Electric fields around charged PAY SPECIAL ATTENTION objects NB • Electric fieldsNB are represented by field lines as illustrated in the diagrams below. Definition • Electric field is a vector quantity. An electric field is a region • An electric field line indicates the direction in which a positive test or space in which an electric HINT! charge would move if placed at a point in the electric field. chargehint experiences an electric force. 6.4.1 Properties of electric field lines e.g. EG - worked examples Electric field lines: • start and end perpendicular to the surface of a charged object • never cross each other • are closer where the electric field is stronger exams • are directed from positive to negative 6.4.2 Representing electric fields You must be able to draw simple diagrams to show the electric fields around charged objects. activity Remember: The arrows representing Around a) ACTIVITIES boy a positive point b) Around a negative point and girl-left and electric field lines charge right of page charge ALWAYS point away from activity a positive charge and towards a negative charge. Step by step comment comment c) Between two like point d) Between two like point charges of equal magnitude charges that are not equal in (both positive or both magnitude negative) If the point charges are The field lines are closer negative, the arrows point together when the electric inwards, towards the point field is stronger (around the charges. greater charge). 96 UNIT 6 ELECTROSTATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 e) Between two unlike (opposite) f) Between two oppositely point charges (one positive charged(one positive and one and the other negative) negative) parallel plates This electric field is uniform – it is equally strong everywhere between the two plates, so the electric field lines are equally spaced and parallel. © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 97 Mind the Gap Physics

6 Unit 6.5 Electric field strength PAY SPECIAL ATTENTION NB NB Step by step PAY SPECIAL ATTENTION Step 1. ALWAYS calculate NB Formula NB the electric field The electric field strength at a point is the electric force per unit positive charge strength at the HINT! experienced at a point in an electric field. hint given point (P in For any charge: this case) due to kQ HINT! E = ​ __F ​ and ∴ E = ​ __ ​ each of the point hint Q r2 charges first. The e.g. EG - worked examples negative sign for a negative charge is NOT used in this equation. e.g. Worked example 4 EG - worked examples exams Step 2. Then choose an electric field Two point charges, Q1 and Q2, at a distance of 3 m apart, are shown below. The charge direction as on Q1 exams is –14 µC and the charge on Q2 is +20 µC. positive and state Remember: this clearly. First convert µC to C: –14 µC = –14 × 10–6 C and 20 µC = 20 × 10–6 C Step 3. Then find the activity resultant (or net) electric ACTIVITIES boy field strength by and girl-left and adding the two right of page activity activity field strength ACTIVITIES boy values. Lastly, and girl-left and right of page remember the signs for the a) Define the activity electric field strength at a point. directions! b) Calculate the net (resultant) electric field at point P situated 2 m from Q2. Step by step comment comment Solutions a) Electric field strength at a point is the electric force per unit Step by step comment comment positive charge experienced at the point. The values substituted b) Electric field at P due to Q1: are positive (+), because kQ (9 × 109)(14 × 10–6) left is chosen as positive E = ​ ___ r2 ​ = ​ _______________    12 ​ and both the electric fields = 1,26 × 105 N·C–1 to the left are directed to the left. Electric field at P due to Q2: kQ (9 × 109)(20 × 10–6) E = ___ ​ r2 ​ = ​ _______________    12 ​ = 4,5 × 104 N·C–1 to the left Let ← E+ Enet = EQ1 + EQ2 = (+1,26 × 105) + (+4,5 × 104 N·C–1) = +1,71 × 105 N·C–1 ∴ 1,71 × 105 N·C–1 to the left 98 UNIT 6 ELECTROSTATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 Activity 2 Multiple Choice Questions: 1. The sketch below shows two small metal spheres, A and B, on insulated stands carrying charges of magnitude q and 2q respectively. The distance between the centres of the two spheres is r. Sphere A exerts a force of magnitude F on sphere B. What is the magnitude of the force that sphere B exerts on sphere A? 1 A ​ __ 2 ​F B F C 2F D 4 (2) 2. Two identical small metal spheres on insulated stands carry equal charges and are a distance d apart. Each sphere experiences an electrostatic force of magnitude F. The spheres are now placed a distance ½ d apart. The magnitude of the electrostatic force each spheres now experience is.. 1 A ​ __ 2 ​F 1 B F​ __ 2​ 1 C 2​ __ 2 ​F D 4F (2) 3. Three identical point charges, q1, q2 and q3, are placed in a straight line, as shown below. Point charge q2 is placed midway between point charges q1 and q3. X and Y are two points on the straight line as shown. X Y q2 q3 q1 © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 99 Mind the Gap Physics

6 Unit Which ONE of the following best describes how the electric field E at a point X compares to that at point Y? DIRECTION OF E MAGNITUDE OF E A Same EX > EY B Same EX < EY C Opposite EX > EY D Opposite EX < EY (2) [6] Solutions 1. B 33 (2) 2. D 33 (2) 3. D 33 (2) [6] Activity 3 A negative charge of 2 µC is positioned 10 cm from point P, as shown below. 1. Define the electric field at point P in words. (2) 2. Draw the electric field lines associated with this charge.  (2) 3. A positive charge of 5 µC is now positioned 15 cm from point P, as showed in the diagram below. Calculate the magnitude of the electric field at point P due to both charges.  (12) [16] 100 UNIT 6 ELE CTRO STATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 6 Solutions 3 3 1. The force per unit charge. (2) 2. 3 3  Marking criteria Shape of field lines 3 Direction of field lines (towards charge) 3 (2) 3. 3 3 3 kQ E 2 µC = ​ ___ r2 ​ 3 (9 × 109)(2 × 10–6) ______________ 3 = ​    (0,1)2 ​ = 1,8 × 106 N·C–1 3 towards the 2 µC 3 kQ E 5 µC = ​ ___ r2 ​ (9 × 109)(5 × 10–6) = ​ ______________    (0,15)2 ​ 3 = 2 × 106 N·C–1 3 away from the 5 µC charge _____________________ Eresultant = ​√(1,8 × 106)    2 + (2 × 106)2 ​ 3 pythagoras (12) = 2,69 × N·C–1 3  3 [16] © Department of Basic Education 2015 UNIT 6 ELECTROSTATICS 101 Mind the Gap Physics

6 Unit Activity 4 The centres of two small, charged conducting spheres, X and Y, on insulated stands, are separated by a distance of 60 mm. Sphere X initially carries a charge of +12 × 10–9 C. X and Y are brought into contact with each other and are separated again. After separation, each sphere carries EQUAL charges of +5 × 10–9 C. 1. Draw a neat diagram of the resultant electric field pattern that surrounds X and Y.  (4) 2. Calculate the number of electrons that must be added to Y to make it neutral.  (3) 3. Calculate the magnitude of the force which X exerts on Y once they are back in their original positions (after separation). (4) 4. Calculate the original charge on sphere Y.  (4) [15] Solutions 1. The field is curved; lines on the outside are important. Marks for: field lines between charges (3); field lines outside the charges (3); direction: away from X and Y (3); field lines not going into spheres but touching surface and not touching each other. (3)(4) Q 2. Number of electrons on Y = ​ __ qe ​ –5 × 10–9 (3) = ​ __________ ​ –1,6 × 10–19 (3) = 3,125 × 1010 electrons (3) (3) kq1q2 3. Force F = ​ _____ r2 ​ (3) (9 × 109)(5 × 10–9)(5 × 10–9) (3) = ​ _____________________       (0,06)2 (3) ​ = 6,25 × 10–5 N (3) If you forget to convert mm to metres, or forget to square the distance between the two (r2) you will lose a mark. (4) (Q1 + Q2) 4. Total charge Q = ​ _______ 2 ​ 3 (12 × 10–9) QY Q = ​ ___________ 2 ​ (where QY is the 3 charge on sphere Y) 2(5 × 109 C) = (12 × 10–9) QY 2(5 × 109 C) _________ ​  12 × 10–9 ​= QY = 8,3 × 1017 C (4) 3 3 [15] Keep going! 102 UNIT 6 ELE CTRO STATICS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 Electric circuits Summary You must remember: • For a current to flow, we need a source of electrical energy (cell or battery) and a closed circuit (or at least a magnetic field moving near a conductor). • The direction of the conventional current is from the positive pole or terminal of the cell through the circuit to the negative pole or terminal of the cell. This flow in one direction is called Direct Current. • The potential difference between two points in a conductor is the work done per unit charge to move a positive charge from one point to another. Potential difference is measured in volts (V) with a voltmeter which is connected in parallel in a circuit. PAY SPECIAL ATTENTION NB NB Potential difference = ​ work done ________ ​ charge ​ W ​ V = __ HINT! Q hint ​ W ​ E = __ Q Electric current is the amount of charge per second which flows past a point. It is measured in amperes (A) with an ammeter which is connected in series in a circuit. e.g. Q examples EG - worked __ I = ​ t ​ Resistance is a measure of how much a conductor opposes the flow of charge through it. It is measured in Ohms (Ω). A resistor is a component in a circuit that resists (opposes) the flow of current.exams activity ACTIVITIES boy and girl-left and right of page activity Step by step comment comment © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 103 Mind the Gap Physics

7 Unit Learn the factors 7.1 Factors influencing the influencing the resistance of a wire conductor. It will resistance of a wire conductor help when explaining your answer in an exam. • The length of the resistor (the longer the wire, the greater the resistance) • The diameter (thickness) of the resistor (the thinner the wire, the greater the resistance) • The temperature of the resistor (the higher the temperature, the greater the resistance) • The type of material that the resistor is made from. Different substances have different resistances, e.g. tungsten (W) has a very high resistance but copper (Cu) has a very low resistance. 7.2 Ohm’s Law • Potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. • The mathematical formula of Ohm’s Law V R = ​ __I ​ PAY SPECIAL ATTENTION NB NB Definition of the ohm: A conductor has a resistance of 1 ohm (1 Ω) if the potential difference of 1 volt HINT! (1 V) applied across its ends, causes a current of 1 ampere (1 A) to flow through it. hint 7.2.1 Circuit connection e.g. EG - worked examples Circuits can be connected into two ways: Series and Parallel. A series circuit requires the electricity to travel one path that does not split. A parallel circuit allows the electricity to go down different paths (there’s a split in the circuit). See the diagrams below showing resistors in series and exams parallel for an illustration. Resistors in a circuit activity ACTIVITIES boy and girl-left and right of page activity Step by step • comment comment Resistors are connected in series or in parallel. • Resistors, rheostats and light bulbs resist the flow of current. 104 UNIT 7 ELE CTRIC CIRCUITS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 • Resistors are connected in series or in parallel. • Resistors, rheostats and light bulbs resist the flow of current. • Work is done in a resistor when the electric energy is transformed to heat energy or to light energy. • Resistors in series are • The total current in a circuit potential (voltage) dividers. increases when more • The total resistance in a resistors are added in circuit increases when more parallel. resistors are added in series. • If one resistor burns out, • The current through all current still flows through the resistors in series is the other resistor (in a parallel same. circuit). • The total current in a circuit • Resistors in parallel are decreases when more current (amperage) dividers. resistors are added in series. • The total resistance in a • If one resistor burns out, circuit decreases when the circuit is broken and no more resistors are added in current flows (in a series parallel. circuit). 7.2.2 Comparison between series and parallel circuits Series connection Parallel connection I1 = I2 = I3… V1 = V2 = V3… The total current across the series The total voltage across the is the same. parallel component is the same. VT = V1 + V2 + V3… IT = I1 + I2 + I3… This is a potential divider. This is a current divider. 1 1 1 RT = r1 + r2 + r3… ​ __ ​ __​ __​ R = ​ r1 + ​ r2 + ... T Addition of resistors. Addition of the ratio of resistors. © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 105 Mind the Gap Physics

7 Unit 7.3 Voltage (Potential Difference) and Electromotive Force (emf) • Voltage in which the charge is losing energy is a potential difference, V. • Voltage in which the charge is gaining energy is an electromotive force (emf), ε • Therefore the voltage across the battery is an electromotive force (emf), while the voltage across each resistor is potential difference (p.d). 7.4 Internal Resistance • When the switch is open, the voltmeter reads the emf of the battery which is 12,5 V. This means the battery can transfer 12,5 J of energy for every 1 C of charge. • When the switch is closed, the voltmeter reads the p.d of the external circuit. • Internal resistance is found inside the cell or battery, which is the small amount of energy that is used up inside the cell or battery. • An ideal cell would have zero internal resistance, r = 0 Ω. • The volts used inside the cell are referred to as the lost volts, V’. 106 UNIT 7 ELE CTRIC CIRCUITS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 7.4.1 emf The emf (ε) of a cell is: emf = 1 (R + r) is the final formula to use • the electrical potential difference across the terminals (poles) of a when calculating emf cell while no current flows • the total amount of electric energy supplied by the cell per coulomb of charge • measured with a voltmeter connected in parallel over a cell or battery when no current flows (the switch is open)(see diagram below) • measured in volts (V). PAY SPECIAL ATTENTION NB NB Formula to calculate the emf Derivation: since R = V/I, V = IR, so if emf = V + V' (the sum of the two voltage measurements), then emf = IR + IR'. For convenience, we write IR' as Ir. Now, since I is common to both factors HINT!IR and Ir, we can take it out as a multiplier, like so: I(R+r). Hence,hint emf = I(R+r). given: emf = V + V’ ∴ emf = IR + Ir ∴ emf = I (R + r) e.g. EG - worked examples 7.4.2 Ohmic and non-ohmic conductors: differences and examples exams Ohmic conductor Non-ohmic conductor Obeys Ohm’s Law or Obeys Ohm’s Law when voltage Does not obeys Ohm’s Law when not? or current is varied voltage or current is varied Graph of the voltage activity vs. the current across conductors ACTIVITIES boy and girl-left and right of page activity Shape: Straight line from the origin V/I=R Shape: Curve Step by step comment comment V/I≠R Examples circuit resistors light bulb nichrome wire. diodes transistors © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 107 Mind the Gap Physics

7 Unit 7.5 Electric energy When current flows or charges move through a resistor: • electric energy is transferred from the moving charges to the particles in the resistor • the particles in the resistor gain kinetic energy • the temperature of the resistor increases as the kinetic energy of the particles increases. Formulas The work done (W) is equal to the energy (E) transferred. W = E (in joule J) V2t W = VQ = VIt = I2Rt = ​ ___ R​ W: work done (J) joule V: potential difference (V) volt I: current (A) ampere R: resistance (Ω) ohm t: time (s) seconds If you are asked to calculate the amount of energy transferred, use the correct formula to calculate the work done 7.6 Power PAY SPECIAL ATTENTION NB NB Definition Power is the rate at which work is done or energy is transferred. HINT! P = __W ​= VI = I2R = ​ V ​ ∆t 2 __​ hint R P: power (W) watt W: work done (J) joule V: potential difference (V) volt e.g. EG - worked examples I: current (A) ampere R: resistance (Ω) ohm t: time (s) seconds exams 7.6.1 The brightness of light bulbs The brightness of a light bulb is determined by the rate at which energy is transformed in the bulb, that is, by the power (P). • As the power increases, the brightness increases. • For bulbs in series: activity P ∝ R (power is directly proportional to the resistance of the bulb) ACTIVITIES boy and girl-left and ∴ if the resistance increases, the power increases and right of page ∴ the brightness increases. activity 108 UNIT 7 ELE CTRIC Step by step CIRCUITS  comment comment © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 • For bulbs in parallel: 1 __​ (power is inversely proportional to the resistance of the bulb) P ∝​  R ∴ if the resistance decreases, the power increases and ∴ thePAYbrightness increases. SPECIAL ATTENTION NB NB Note: the above applies to incandescent bulbs only (the type with a filament of wire). Compact Fluorescent Lights (CFL) come in a range of power/wattages but they will simply blow HINT! if overpowered by a significant increase in voltage/ hint amperage. e.g. Worked example 1EG - worked examples In this circuit the battery has an emf of 24 V and an internal resistance of 2 Ω. Voltmeter V1exams is connected as shown and voltmeter V2 is connected over the three resistors in parallel. The resistance of the connectors and of the ammeter may be ignored. activity ACTIVITIES boy and girl-left and right of page activity 1. Switch S is open. a) What is the reading on V1? b) What is the reading on V2? 2. Switch S is now Step closed. Calculate: by step comment comment a) the total resistance in the circuit. b) the current that flows through the 8 Ω resistor. c) the charge that flows past a cross section of the 8 Ω resistor in one minute. Solutions 1a) V1 = 24 V (when S is open, no current flows and V1 is connected across the battery) 1b) V2 = 0 V (S is open, no current flows, V is connected on one side of the switch, ∴ only to one pole of the battery 1 1 1 1 1 1 1 6+2+1 9 ​ ___ ​ __​ __​ __​ __ __ ___ _______ ___ R = ​ R + ​ R + ​ R = ​ 3 ​ + ​  9 ​ + ​  18 ​ = ​  18 ​ = ​ 18 ​ || 1 2 3 ∴ R|| = 18/9 = 2 Ω © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 109 Mind the Gap Physics

7 Unit Solutions 2a) Rexternal = Rseries + R|| = 8 + 2 = 10 Ω and Rexternal = Rext + rint = 10 + 2 = 12 Ω 2b) The 8 Ω resistor is connected in series ∴ the total current flows through it ∴ calculate the total current (I) that flows through the circuit. Vtot Rtot = ​ ___ I ​ tot ∴ 12 = ​ 24 ___​ Itot This is a 24 ___ ∴ Itot = ​  ​= 2 A 12 current Q that flows 2c) I = __ ​ ∆t ​ through all Q the circuits. ∴ 2 = ​ ______​ (1)(60) ∴ Q = (2)(1)(60) = 120 C Remember: Convert minutes to seconds: x 60 or hours to seconds: x 3600 Remember: Current is the rate at which electric charge passes a fixed point in a conductor. I ∝ Q Q I = ___ ​ ∆t ​ 110 UNIT 7 ELE CTRIC CIRCUITS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 Activity 1 In the circuit below the battery has an emf of 24 V and an unknown internal resistance. Voltmeter V1 is connected across the battery. The resistance of the connectors and of the ammeter may be ignored. When switch S is closed, voltmeter V2 reads 4 V and voltmeter V1 reads 20 V. 1. Calculate: a) The reading on ammeter A2.(2) b) The reading on ammeter A1.(5) c) The resistance of resistor R. (4) d) The internal resistance of the battery. (3) e) The energy converted in resistor R in 10 minutes. (3) 2. Switch S is now opened. Will the reading on voltmeter V1 increase, decrease or remain constant? Explain. (6) [23] Solutions 1a. Ammeter A2 reads the current that flows through the 4 Ω resistor. V4Ω 3 __ R4Ω = ​ ___​ ∴ ​  4 ​∴ Itot = ​ __ 4 I 4Ω I tot 4 ​= 1 A 3 (2) 3 3 1b. The resistance of the 16 Ω resistor is DOUBLE the resistance of the (4 + 4) = 8 Ω resistors ∴ the current that flows through the 16 Ω resistor is HALF the current that flows though the 3 (4 + 4) = 8 Ω resistor. 1 A flows through the (4 + 4) = 8 Ω resistor ∴ 0,5 A flows through the 16 Ω resistor 3 ∴ Itotal = 1 + 0,5 = 1,5 A 3 (5) 1 1 1 1 1 2 3 ​  R ​= ​ __ 1c. __ ​ __​ ___ ______ ___ ___ 3 R + ​ R = ​ 16 ​ + ​  (4 + 4) ​= 1 + ​ 16 ​ = ​  16 ​ || 1 2 16 R|| = ​ ___ 3 ​= 5,33 Ω 3 Vext 3 20 and Rext = ​ ___ ​ ___ I ∴ Rext = ​ 1,5 ​= 13,33 Ω 3 (4) ext 3 ∴ 24 = 20 + V 3 1d. emf = Vtot = Vexternal + Vinternal internal ∴ Vinternal = 24 – 20 = 4 V 3 (3) 1e. Energy transferred (or transformed) = E = work done = W W = I2RΔt 3 = (1,5)2(8)(10)(60) 3 = 10 800 J 3 (3) 2. When current flows, V1 reads the external potential difference: when S is opened there are fewer resistors in parallel 3 ∴ Rexternal increases 3 ∴ Itotal decreases 3 ∴ Vint decreases 3 and emf is constant ∴ Vexternal increases (6) 3 3 [23] © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 111 Mind the Gap Physics

7 Unit Activity 2 Multiple Choice questions: 1. Which ONE of the following is the unit of measurement for the rate of flow of charge? A watt B coulomb C volt D ampere (2) 2. The two resistors in circuit 1 below are identical. They are connected in series to a cell of emf V and negligible internal resistance. The power dissipated by each resistor is P. The two resistors are now connected in parallel, as shown in circuit 2 below. The power dissipated by each resistor in the circuit 2 is... A 2P B 4P C 8P D 16P (2) [4] Solutions 1. D 3 3 (2) 2. B 3 3 (2) [4] 112 UNIT 7 ELE CTRIC CIRCUITS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 Activity 3 In the circuit represented below, the battery has an emf of 10 V and an unknown internal resistance. Voltmeter V1 is connected across the battery and voltmeter V2 is connected across the open switch S. The resistance of the connecting wires and ammeter can be ignored. Switch S is open 1. What is the reading on V1? (2) 2. What is the reading on V2? (2) When Switch S is closed, the reading on V1 drops to 7,5 V. 3. What is the reading on V2? (2) 4. Calculate the reading on the ammeter. (8) 5. Calculate the internal resistance of the battery. (5) [19] Solutions 1. 10 V 33 (2) 2. 10 V 33 (2) 3. Zero or 0 V 3 3 (2) 1 1 1 ​  ​= ​ __​+ ​ __​ 4. __ 3 product Rp R1 R2 OR Rp = ​ ______ sum ​ (3) 1​ 1 __ 1 __ 1 __ ​ __ Rp = ​ 6 ​ (3) + ​ 3 ​ (3) = ​ 2 ​ (6 × 3) Rp = ​ ______ (6 + 3) ​ Rp = 2 Ω (3) Rp = 2 Ω (3) Now use 2 Ω in the next calculation... Rext = 2 Ω + 1 Ω = 3 Ω V I = ​ __ R ​ (3) = 7,5 (3) / 3 (3) = 2,5 A (3) (8) 5. emf = Vcirc + Vlost (3) 10 (3) = (7,5 + 2,5)r 3 3 r = 1 Ω (3) (5) [19] © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 113 Mind the Gap Physics

7 Unit Activity 4 The headlights of a car are connected in parallel to a 12 V battery, as shown in the simplified circuit diagram below. The internal resistance of the battery is 0,1 Ω and each headlight has a resistance of 1,4 Ω. The starter motor is connected in parallel with the headlights and controlled by the ignition switch, S2. The resistance of the connecting wires may be ignored. 1. State Ohm’s Law in words. (2) 2. With only switch S1 closed, calculate the following: a) Effective resistance of the two headlights (3) b) Potential difference across the two headlights (4) c) Power dissipated by one of the headlights (3) 3. Ignition switch S2 is now closed (whilst S1 is also closed) for a short time and the starter motor, with VERY LOW RESISTANCE, rotates. How will the brightness of the headlights be affected while switch S2 is closed? Write down only INCREASES, DECREASES or REMAINS THE SAME. Fully explain how you arrived at the answer. (9) [21] 114 UNIT 7 ELE CTRIC CIRCUITS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 7 Solutions 1. The current in a conductor is directly proportional to the potential difference across its ends at constant temperature. 33 OR The ratio of potential difference to current is constant at constant temperature  (2) 1 1 __ __ 1 2.a) ​ __ R ​ = ​  R1 ​ + ​ R2 ​ 3 1 ___ 1 ___ = ​ 1,4 ​ + ​  1,4 ​ 3 = 0,7 Ω 3 (3) 2.b) emf = I(R + r) 3 12 = I(0,7 + 0,1) 3 I = 15 A V R = ​ __I ​ V = 0,7 × 15 3 = 10,5 V 3 (4) 2.c) I(light) = 7,5 A (from 2b: 2 headlights = 15 A) 3 P = VI = (10,5)(7,5) 3 = 78,75 W 3 (3) 3. Decreases 3 (Effective/total ) resistance decreases. 3 (Total) current increases. 3 “Lost volts” / Vinternal / Ir increases, thus potential difference / V (across headlights) decreases. 3 P = V / R decreases. 3 P = W / Δt 3 60 = W / 2(60) W = 7200 J 3 Starter motor W = Vq 3 q = 7200 / 12 q = 600 C 3 (9) [21] Keep going! © Department of Basic Education 2015 UNIT 7 ELECTRIC CIRCUITS 115 Mind the Gap Physics

8 Unit Electrodynamics: electrical machines (generators and motors) Summary • Definitions • Faraday’s law of electromagnetic induction • Differences between motors and generators. Electrodynamics • Operations of motors and generators. is the study of the • Differences between direct current (DC) and alternating current (AC) in the relationship between cases of both motors and generators. electricity, magnetism • The graphs of AC and DC. and mechanical phenomena. • Right hand rule to determine the direction of the force on the conductor. • The use of motors in everyday life. • Calculations of the Root Mean Square. PAY SPECIAL ATTENTION NB NB DefinitionS AND LAWS YOU MUST REMEMBER Magnetic flux (Φ) is the product of the strength of a magnetic field and the surface area the field cuts perpendicularly.HINT! It is measured in Wb (weber) units. hint induction is when a magnet moves relative to a conductor, and the magnet’s Electromagnetic magnetic field is at right angles to the conductor, the maximum electric current is induced in the conductor. e.g. EG - worked examples • Faraday’s Law of Electromagnetic Induction –– The induced emf in a conductor is directly proportional to the rate of change of the magnetic flux in the conductor. exams ∆Φ –– So ε ∝ ___ ​  ∆t ​ ∆Φ –– ε = – N ​ ___ ∆t ​ where N is the number of turns in the coil ε is emf in (V) volts activity ∆Φ is change in magnetic flux in (Wb) weber ∆t is change in boy ACTIVITIES time in (s) seconds –– The negative sign shows that the emf creates a current and a magnetic and girl-left and right of page field B that opposes the change in the magnetic flux Φ. activity 116 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015 Step by step comment comment  Mind the Gap Physics

Unit 8 8.1 Motors and generators 8.1.1. Alternating current generators The principle of the AC generator We know that, according to the phenomenon of electromagnetic induction: • when an electric conductor moves in a magnetic field, there is a change in the magnetic flux which induces an emf that causes a current flow in the conductor; • the magnetic field strength (B) that passes perpendicularly through a surface area A (in m2) is called the magnetic flux (Φ) and is measured in weber (Wb). © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 117 Mind the Gap Physics

8 Unit NB: Use the Right Hand Rule to determine the direction of the force on the charges (F) in the conductor of the generator – the conventional current direction (I) of the induced current. The magnetic field (B) is in the North to South direction. Remember it like so: First Finger is Field; SeCond finger is Current; ThuMb is Movement or Thrust. “Fleming’s left-hand rule is used for electric motors, while Fleming’s right-hand rule is used for electric generators. Different hands need to be used for motors and generators because of the differences between cause and effect.” (Wikipedia). So, if you’re trying to work out the direction of current in a generator, you need to use the Right Hand Rule, and, vice versa, if you’re trying to work out which way an electric motor will turn, you need to use the Left Hand Rule. The fingers are the same; just the hand changes. Note also that an alternative hand positioning is to place all four fingers forwards (Field), and then the thumb indicates the thrust or motion, and a line perpendicular to the palm indicates the current. Step-by-step: The AC Generator Suggestion: watch the video http://www.youtube.com/watch?v=wpCYiSFBQ0U (where 0 is zero) Step I: Coil vertical: Step II: Coil horizontal: • ab and cd are parallel to the normal to the • Coil is rotated in a clockwise direction magnetic field • ab and cd cut through the magnetic field which is • which is directed from N to S directed from N to S. • and is parallel to the motion of sides ab and cd • ∴ there is a change in the magnetic flux and ab • there is no change in the magnetic flux moves down and an emf is induced across the ends of the coil which induces current in the coil. • ∴ no emf is induced in the coil and • ∴ conventional current direction is from b to a, so a and A are – (negative) • ∴ no current flows in AND the coil • cd moves up and an emf is induced across the ends of the coil which V = 0 V and I = 0 A induces current in the coil. So conventional current direction is from d to c, so d and D are + (positive) • V and I increase to Vmax and Imax when the coil is horizontal and it cuts the magnetic field perpendicularly • V and I decrease from Vmax and Imax as the coil turns further. 118 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 Step III: Coil vertical: Step IV: Coil horizontal: • ab and cd are again parallel to the normal to the • ab and cd again start to cut through the magnetic magnetic field field which is directed from N to S. • which is directed from N to S • ∴ there is again a change in the magnetic flux • and is parallel to the as ab moves up (thumb points up) and an emf is motion of sides ab induced across the ends of the coil which induces and cd current in the coil. • ∴  there is no change • ∴ conventional current direction is from a to b, so in the magnetic flux a and A are + (positive) • ∴  no emf is induced AND in the coil and • cd moves down • ∴  no current flows in (thumb points down) the coil and an emf is induced across the ends V=0 V and I = 0 PAY A SPECIAL ATTENTION of the coil which NB NB induces current in the coil. Palm faces forwards from c to d ∴ conventional current HINT! direction is from c to Because each side of the coil or d, so d and D are – hint armature is always connected to (negative) the same slip ring and brush, the • V and I increase to current changes direction. Vmax and Imax when the coil is horizontal and it cuts the magnetic field e.g. EG - worked examples perpendicularly. V and I decrease from Vmax and Imax as the coil turns further. exams The alternating current (AC) cycle In AC (alternating current), the current changes voltage (and direction) every cycle; that is, every time the generator or dynamo turns over through one revolution (full cycle). activity ACTIVITIES boy and girl-left and right of page activity When the coil is vertical When the coil is horizontal (in coil Step positions by step 1, 3 and 5) comment comment (in coil positions 2 and 4) • ab and cd are parallel to the normal to the • ab and cd are perpendicular to the normal to magnetic field, and do not cut through the magnetic the magnetic field, and therefore cut through the field magnetic field • There is a changing magnetic flux • There is no changing magnetic flux • ∴ emf and current are induced in the coil • ∴ no emf or current is induced in the coil • ∴ V = Vmax and I = Imax • ∴ V = 0 V and I = 0 A but the emf and current are reversed. © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 119 Mind the Gap Physics

8 Unit AL ATTENTION NB Increasing the induced emf and current The induced emf (and therefore the amount of induced current) increases if: • The conductor (wire) is rotated faster so that the rate at which the magnetic flux changes, increases; • the magnetic field is stronger (use stronger magnets); • there are more turns (loops) on the coil, so that the length of the conductor (wire) moving through the field is increased. ed examples The direct current (DC) cycle In DC (direct current), the current keeps the same voltage (and direction) in every cycle; that is, every time the generator or dynamo turns over through one revolution (full cycle). The direct current (DC) cycle ACTIVITIES boy and girl-left and right of page When the coil is vertical When the coil is horizontal p comment comment • ab and cd are parallel to the • ab and cd are perpendicular to the normal to the magnetic field, and normal to the magnetic field, and do not cut through the magnetic therefore cut through the magnetic field field • there is no changing magnetic flux • there is a changing magnetic flux • ∴ no emf or current is induced in • ∴ emf and current are induced in the coil the coil • ∴ V = 0 V and I = 0 A • ∴ V = Vmax and I = Imax • The emf and current are always positive. 120 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 8.1.2 The difference between AC and DC generators A direct current generator (dynamo) generates direct current instead of alternating current. Alternating current (AC) generator Direct current (DC) generator Similarities between AC and DC generators • Both convert mechanical energy to electrical energy. • The coils are turned mechanically (e.g. by steam, flowing water or wind). • The induced emf increases and decreases during each cycle. • When the coil cuts through the magnetic field, the changing magnetic flux induces an emf and electric current in the coil. • The induced V and I have maximum values twice during every cycle. • Carbon brushes collect the current. Differences between AC and DC generators AC generator DC generator • The coil is connected to slip rings. • The coil is connected to a split ring commutator. • The same part of the coil is always connected to the • A brush makes contact with a different half of same slip ring. the split ring commutator during each half of the • The current in the slip rings changes direction when rotation (cycle). the current in the coil reverses. • One brush always makes contact with the positive • The brushes collect the alternating current (AC) half of the split ring commutator and the other from the slip rings. brush always makes contact with the negative half of the split ring commutator. • The brushes collect DC from the split ring commutator. © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 121 Mind the Gap Physics

8 Unit 8.1.3 Electric motors Parts of the direct current (DC) motor Use Fleming’s Right Hand Rule to determine the direction of the force on the conductor –the direction in which the coil turns. Remember: the current is in the direction of the middle finger or palm, whereas the magnetic field is in the direction of the fingers (or index finger), and the thrust (motion) is in the direction of the thumb. 122 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 8.1.4 The working of a simple DC-motor Step 1: Coil horizontal at 0° Step 2: Coil vertical at 90° • the split ring commutator makes contact with the • the openings in the split ring commutator are brushes ∴ current flows through the coil opposite the brushes • ab and cd are at 90° to the magnetic field • ∴ no current flows through the coil • the magnetic field is from N to S (fingers left to right) • ∴ no resultant force on the coil • ab is connected to the + terminal • BUT coil’s momentum carries it past the vertical ∴ conventional current position from b to a (thumb from b to a) • ∴ palm faces up ∴ DOWNWARD force on ab • cd is connected to the − terminal ∴ conventional current from d to c (thumb from d to c) ∴ palm faces down ∴ UPWARD force on cd • the 2 forces cause a resultant torque (turning force) on the coil • ∴ it rotates anticlockwise Step 3: Coil horizontal at 180° Step 4: Coil vertical at 270° • The split ring commutator makes contact with the • The openings in the split ring commutator are brushes again ∴ current flows through the coil opposite the brushes again ∴ no current flows • ab and cd are at 90° to the magnetic field through the coil • the magnetic field is from N to S (fingers left to • ∴ no resultant force on the coil right) • BUT coil’s momentum carries it past the vertical • ab is now connected to the − terminal position back to the original position in Step 1 ∴ conventional current from a to b (thumb from a to • the first cycle is complete and the process is b) repeated. • ∴ palm faces down ∴ UPWARD force on ab • cd is connected to the + terminal ∴ conventional current from c to d (thumb from c to d) • ∴ palm faces up ∴ DOWNWARD force on cd • the 2 forces cause a resultant torque (turning force) on the coil • ∴ it continues to rotate anticlockwise Increasing the speed at which the DC motor rotates (turns) The coil will turn faster if: • the current in the coil increases; • the number of turns on the coil increases; • the strength of the magnetic field increases. © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 123 Mind the Gap Physics

8 Unit 8.1.5 The differences between AC and DC motors Alternating current (AC) motor Direct current (DC) motor Coil Split ring Carbon brush AC Battery • AC power supply • DC power supply (battery) • Fixed magnets supply a fixed magnetic field from N • Brushes contact with split ring commutator to to S and the brushes make contact with slip rings supply the DC to the coil to supply the AC to the coil OR • AC electromagnets supply a magnetic field that changes direction during each AC cycle and the brushes make contact with a split ring commutator to supply the AC to the coil • Used for heavy loads e.g. washing machines, • Used for small loads e.g. hair dryers, toy cars electric drills 124 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 8.1.6 Differences between a motor and a generator An electric motor and an electric generator are basically the same device. The primary difference is in the case of a motor, electricity is used to turn it, whereas in the case of a generator, turning it mechanically generates electricity. Direct Current (DC) Motors Direct Current (DC) Generators Coil Split ring Carbon brush Battery • Connected to a battery. The battery is a source of • NOT connected to a battery. electric energy. • Converts electrical energy to mechanical energy. • Converts mechanical energy to electrical energy. • The magnetic fields of the magnets and those • The coil is turned mechanically (e.g. by steam, around the current-carrying conductor, interact. flowing water or wind). • The resultant magnetic field exerts a magnetic force • When the coil cuts through the magnetic field of the on the coil. magnets, there is a change in the magnetic flux. • The coil turns due to the resultant magnetic force • According to Faraday’s Law of Electromagnetic acting on it. Induction, a change in magnetic flux induces an emf across the ends of the coil and current is induced in the coil. 8.1.7 Electrical motors in everyday life In practice, motors turn evenly at a high speed. The coil in a motor consists of a soft iron core, surrounded by coils. This coil forms the armature. Most armatures have many coils, which are placed at different angles. Each coil in the armature has its own commutator. This results in a bigger turning effect which makes the motor turn evenly. A very important example of an electric motor is the starter motor of a car, which turns the car engine over in order to start it. The purpose of the car battery is to power the starter motor (and other things like lights). When the car is running, the petrol motor turns a generator over which then recharges the battery. Some motors, e.g. an electric drill, can also use alternating current because they contain electromagnets and not permanent magnets. As the alternating current flows in the coil, the magnetic field changes direction. Thus the motor continues to turn in the same direction. © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 125 Mind the Gap Physics

8 Unit 8.2 Alternating current circuits • Frequency: The frequency (f) of an alternating current supply is the number of complete cycles per second and is measured in hertz (Hz). In South Africa electricity is supplied by Eskom power stations and has a frequency of 50 Hz. • Period: The period (T) of an alternating supply is the time taken to complete one cycle. If the frequency of the AC current is 50 Hz, 1 1 T = ​ __f ​ = ​ ___ 50 ​= 0,02 s 8.2.1 Voltage and current in an AC circuit emf induced by an AC generator Current induced by an AC generator V varies in cycles I varies in cycles – between zero and Vmax – between zero and I max – between + and – values – between + and – values as a function of time (i.e. over time) as a function of time (i.e. over time) Voltage changes polarity twice in one AC cycle Current changes direction twice in one AC cycle Vmax is read at the crest (top) of the voltage curve and Imax is read at the crest (top) of the current curve and is is the amplitude of the voltage curve the amplitude of the current curve Vmax is reached twice in 1 AC cycle Imax is reached twice in 1 AC cycle Vrms is root mean squared voltage, measured in volts (V) Irms is root mean squared current Vmax Vrms = ​ ___ __ ​ measured in amperes (A) ​√2 ​ Imax ___ Irms = ​  __ ​ ​√2 ​ PAY SPECIAL ATTENTION NB NB Definition Root mean squared voltage HINT! The root mean squared voltage (Vrms) is the equivalent DC voltage value that produces hint the same heating effect or power as the changing AC. V I Vrms = ___ ​  max __ ​ Irms = ___ ​ max__ ​ ​√2 ​ ​√2 ​ The root mean squared current (I examples ) is the effective current value of alternating current. e.g. EG - workedrms • Root mean square (rms) values are the AC equivalent of DC emf. • If a DC circuit has an emf of 100 V and an AC circuit has a Vrms of exams 100 V, the circuits would use the same amount of power. 126 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 8.2.2 Electric power in an AC circuit AL ATTENTION NB Definition Electrical power (P) is the rate at which energy is transferred or transformed from one type to another. Summary ed examples Remember: 1 megavolt = 1 × 106 volt 1 MV = 1 × 106 V ACTIVITIES boy and girl-left and right of page vmax vrms = ​ ___ __ ​= 100 V V = 100 V ​√2 ​ Vrms ___ V R = ​ I ​∴ Vrms = IrmsR R = ​ __I ​ ∴ V = IR rms Rrms2 V2 p P avg comment = VrmsIrms comment = I2rmsR = ​ ____ R ​ P = VI - I2R = ​ __ R​ Pavg : average power watt (W) Vrms : rms potential difference volt (V) Irms : rms current ampere (A) R : resistance ohm (Ω) © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 127 Mind the Gap Physics

8 Unit 8.2.3 Advantages of alternating current • The most important advantage of AC is the fact that the potential difference can be changed by using transformers. • Transformers can only function with alternating current. 1 • The power in a transformer remains constant and P = VI ∴ V ∝ __ ​  Ι ​ • At power stations step-up transformers are used to increase (step-up) the voltage which decreases the current. • The voltage is increased to between 130 and 750 MV • This allows electrical energy to be transmitted in electric cables over long distances while the current is low. • The loss in energy due to the heating effect of the cables is low when the current (Ι) is small: W = I2Rt ∴ Wtransformed to heat ∝ I2 • Conducting cables are thick to help decrease the energy lost as heat during transmission. • Factories need high voltage (± 10 kV). • In towns step-down transformers are used to decrease (step-down) the voltage to ± 220 V. You can see these at the side of the road in most suburbs; they are painted dark green. 128 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 Activity 1 A simplified sketch of a generator is shown below. 1. Is the output voltage AC or DC? Give a reason for your answer.  (2) 2. What type of energy conversion takes place in the above generator?(2) 3. State TWO effects on the output voltage if the coil is made to turn faster. (2) 4. What is the position of the coil relative to the magnetic field when the output voltage is a maximum? (1) [7] Solutions 1. AC — The generator has slip rings. 3 3 (2) 2. Mechanical 3 energy is converted to electrical energy. 3 (2) 3 3 3. Output voltage increases and the number of cycles per second increases. (2) 3 4. The coil position is parallel to the magnetic field. (1) [7] © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 129 Mind the Gap Physics

8 Unit Activity 2 Lights in most households are connected in parallel, as shown in the simplified circuit below. Two light bulbs rated at 100 W; 220 V and 60 W; 220 V respectively are connected to an AC source of rms value 220 V. The fuse in the circuit can allow a maximum current of 10 A. 1. Calculate the peak voltage of the source.  (3) 2. Calculate the resistance of the 100 W light bulb when operating at optimal conditions. (3) 3. An electric iron, with a power rating of 2 200 W, is now connected across points a and b. Explain, with the aid of a calculation, why this is not advisable.  (5)  [11] Solutions Vmax 1. Vrms = ​ ___ __ ​ 3 ​√2 ​ Vmax3 220 = ​ ___ __ ​ ∴ V max = 311,1 V 3  (3) ​√2 ​ V2rms 2. P = ​ ____ R ​3 2202 100 = ​ ____ R3 ​ ∴ R = 484 Ω3  (3) 3. Pave = VrmsIrms 3 2 200 = (220)Irms 3 2200 Irms = ​ _____ 220 ​ Irms = 10 A 3  The iron draws 10 A of current. Together with the lights the total current will exceed 10 A causing the fuse to blow. 3 3(5) [11] 130 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 Activity 3 The essential components of a simplified DC motor are shown in the diagram below. When the motor is functioning, the coil rotates in a clockwise direction as shown. 1. Write down the function of each of the following components: a) Split-ring commutator (1) b) Brushes  (1) 2. What is the direction of the conventional current in the part of the coil labeled AB? Write down only FROM A TO B or FROM B TO A.  (1) 3. Will the coil experience a maximum or minimum turning effect (torque) if the coil is in the position as shown in the diagram above?  (1) 4. State ONE way in which this turning effect (torque) can be increased.  (1) 5. Alternating current (AC) is used for the long-distance transmission of electricity. Give a reason why AC is preferred over DC for long-distance transmission of electricity. (2) 6. An electrical appliance with a power rating of 2 000 W is connected to a 230 V rms household mains supply. Calculate the: a) Peak (maximum) voltage (3) b) rms current passing through the appliance (3) [13] © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 131 Mind the Gap Physics

8 Unit Solutions 3 1. a) Reverses current direction in the coil every half cycle.  (1) 3 b) Connects external circuit to split ring commutator.  (1) 2. B to A 3(1) 3. Maximum 3(1) 4. Increase current strength 3/ Increase number of coils 3/ Use stronger magnets. 3 (any one) (1) 5. AC can be stepped up to high voltages and low current. 3 Less energy loss with low current (W = I2R∆t). 3(2) Vmax Vmax 6. a) Vrms = ​ ___ __ ​ 3 ∴ 230 3 = ​ ___ __ ​ ∴ V max = 325,27 V 3  (3) ​√2 ​ ​√2 ​ b) Pave = VrmsIrms 3 2 000 = (230)Irms 3 Irms = 2000 / 230 Irms = 8,695 A 3  (3) [13] Activity 4 Electric motors are important components of many modern electrical appliances. AC motors are used in washing machines and vacuum cleaners, and DC motors are used in toys and some tools. 1. What energy conversion takes place in electric motors? (2) 2. What is the essential difference in the design between DC motors and AC motors? (4) 3. List THREE ways in which the efficiency of the motor can be improved.(3) 4. Consider the diagram. The conventional current direction is indicated by the arrows. a) In which direction (clockwise or anti-clockwise) will the coiled armature rotate if the switch is closed? (1) b) Why does the armature continue moving in the same direction once it has reached the vertical position? (2) [12] 132 UNIT 8 ELE CTRO DYNAMICS: ELECTRICAL MACHINES  © Department of Basic Education 2015  Mind the Gap Physics

Unit 8 Solutions 3 3 1. Electric energy converted to mechanical energy. (2) 2. A DC motor reverses current direction with the aid of the 3 commutator whenever the coil is in the vertical position to ensure continuous rotation. 3 An AC motor, with alternating current as input, works without commutators3since the current alternates/slip 3 rings can be used. (4) 3 3. Increase the number of turns on each coil; increased current;3stronger magnets. 3 (3) 4. a) Anticlockwise  (1) b) The armature’s own momentum 3/ the split ring commutator changes direction of current, every time the coil reaches the vertical position. 3(2) [12] Activity 5 In the circuit the AC source delivers alternating voltages at audio frequency to the speaker. 1. What is the peak voltage that the source can deliver? (3) 2. Calculate the average power delivered to the speaker. (7) [10] Solutions __ 1. Vrms = Vmax/​√2 ​ 3 15 ∴ Vmax = ​ ___ __ ​ 3 = 21,21 V 3(3) ​√2 ​ 2. Rtotal = 8,2 + 10,4 3 = 18,6 Ω 3 V I = ​ __ R ​ 3 = 15 / 18,6 = 0,81 A 3 P = I2R 3 = (0,81)2(10,4) 3 = 6,76 W 3  (7) [10] Keep going! © Department of Basic Education 2015 UNIT 8 ELECTRODYNAMICS: ELECTRICAL MACHINE S 133 Mind the Gap Physics

9 Unit Optical phenomena and properties of materials 9.1 Electromagnetic waves and visible light: Revision 9.1.1 Electromagnetic waves Electromagnetic waves consist of electric and magnetic fields. These fields are perpendicular to each other and to the direction in which the wave is propagated (the direction in which it travels). You must remember: • Electromagnetic waves travel through a vacuum at 3 × 108 m·s−1 • They increase in frequency and energy from radio waves (lower frequency and less energy) to gamma rays (higher frequency and more energy) • They increase in wavelength from gamma rays (shorter wavelength) to radio waves (longer wavelength) • The visible light spectrum is part of the electromagnetic spectrum (shown below). The electromagnetic spectrum (Source:http://en.wikipedia.org/wiki/Electromagnetic_spectrum) 134 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 9 9.1.2 Visible light PAY SPECIAL ATTENTION NB You must remember: Visible light is part of the electromagnetic spectrum. NB • Visible light increases in frequency and energy from red (lower Dual nature of light: frequency, less energy) to violet (higher frequency, more energy) • Visible light increases in wavelength from violet (shorter wavelength) • Wave nature during to red (longer wavelength). HINT! propagation proved • hint Visible light has a dual nature because it has wave properties while by diffraction and interference it is propagated (transmitted) and it has particle properties when it strikes and interacts with other matter (the photoelectric effect). • Particle nature during interaction with e.g. EG - worked examples 9.2 The photoelectric effect matter proved by the photoelectric effect The photoelectric effect is used in solar panels to generate electricity. The photoelectric effect refers to the ability of light to cause metals to release exams electrons. You must remember: • Light energy is transmitted in ‘packages’ which are called photons. • Each photon consists of a certain amount of energy which is called a quantum. • The amount of energy (E) in a quantum is directly proportional to the frequency (f) of the light. activity • An electron needs a minimum amount of energy to be released from ACTIVITIES boy an atom. So the photon providing this energy must have a minimum and girl-left and right of page frequency before it will allow an electron to be released from the metal surface. activity • When light shines on a surface (like a metal), the photons collide with the atoms in the surface. PAY SPECIAL ATTENTION • All the energy of the photon (E = hf) is transferred to the atom with NB which the photon collides. • • If an electron in an atom on the surface of the metal gains sufficient Step by step comment • The formula comment c = fλ is energy during the collision, it is ejected from the metal surface and is used to calculate the called a photoelectron. speed of light, where c HINT! (3 hint × 10 m∙s ) is 8 –1 the speed of light in IAL ATTENTION metres per second, λ NB is the wavelength in DefinitionS e.g. meters (m) and f is the frequency in hertz (Hz). EG - worked examples • The photoelectric effect is the process whereby electrons are ejected from a metal • The formula E = hf is surface when light of suitable frequency is incident on (falls on) that surface. used to calculate the • The work function (W0) of a metal is the minimum energy that is required to emit a exams energy of radiation, photoelectron from the surface of the metal. where • The threshold frequency or cut-off frequency (f0) is the minimum frequency of E is the energy of ed examples the incident photons (light) that is required to emit a photoelectron from the surface radiation in Joule (J), h of the metal. is Planck's constant energy ∝ frequency and work function ∝ threshold (cut-off) frequency (6,63 × 10–34 J∙s–1) and ∴ E∝ f and W0 ∝ fo activity f is the frequency in ∴ E = hf and W0 = hfo hertz (Hz). ACTIV and g right activity ACTIVITIES boy Step by step and girl-left and right of© Department of Basic Education 2015 page UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS 135 Mind the Gap Physics

9 Unit 9.2.1 Changing the frequency and intensity of the incident light • The intensity of a light wave is measured by the power (wattage) of the light source. An 8 W (watt) CFL lamp is dim but a 16 W CFL lamp is bright. • It also depends on the type of light. So, for example, Light Emitting Diode (LED) lights use very little power but are very bright, Compact Fluourescent Lights (CFLs) use a mid-range amount of power and are bright for the amount of power they use, and Tungsten Filament or Incandescent Lamps use a lot of power for the amount of light they provide E.G. 60 W, 100 W. • This means that the same type of light with a higher wattage will be brighter than the same type of light with a lower wattage. When the frequency of the incident light (the light falling on the metal) is greater than the threshold frequency, changes to the intensity (brightness) and the frequency cause these changes: Increasing the frequency of Increasing the intensity of incident light incident light The frequency of the incident The intensity (brightness) of light is increased while its the incident light is increased intensity (brightness) remains while its frequency remains constant: constant: • The same number of • More photoelectrons are photoelectrons are emitted from the surface of emitted from the surface the metal of the metal • the energy of each photon • the kinetic energy of the remains constant as the photoelectrons increases frequency is constant and • the speed at which the • the kinetic energy and photoelectrons move speed of the emitted away from the metal photoelectrons remains increases constant so that BUT • the speed at which the • the same number of photoelectrons move away photons strike the metal from the metal remains surface per second and constant • the same number of BUT photoelectrons are • more photons strike the emitted from the metal metal surface per second surface per second; and i.e. the rate at which • more photoelectrons are photoelectrons are emitted from the metal emitted, remains surface per second; i.e. the constant. rate at which photoelectrons are emitted, increases. 136 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 9 9.2.2 Calculating energy of photoelectrons PAY SPECIAL ATTENTION PAY SPECIAL ATTENTION PAY SPECIAL ATTENTION NB NB NB NB NB NB HINT! HINT! HINT! 9.2.3 hint Conditions for emission hint of photoelectrons hint e.g. 1 f light < fo EG - worked examples e.g. 2 f light = fo EG - worked examples e.g. 3 f light > fo EG - worked examples exams Elight < W0 exams Elight = W0 exams Elight > Wo hflight < hfo hflight = hfo hflight > hfo Photoelectrons: activity Photoelectrons: activity Photoelectrons: activity • are not emitted. ACTIVITIES boy and girl-left and • are emitted onto the surface of theACTIVITIES boy and girl-left and • are emitted from the surface of theACTIVITIES boy and girl-left and metal metal right of page right of page right of page activity activity activity • don’t have kinetic energy • have kinetic energy • don’t move away. • move away. Step by step comment comment Step by step comment comment Step by step comment comment PAY SPECIAL ATTENTION NB NB E: energy of light (J) h: Planck’s constant (6,63 × 10–34 J·s–1) HINT! f: frequency (Hz) hint W : work 0 function (J) Ek: kinetic energy (J) v: velocity of electrons (m·s–1) e.g. m: mass of electron EG - worked examples (9,11 × 10−31 kg) exams © Department of Basic Education 2015 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS 137 Mind the Gap Physics

9 Unit 9.2.4 Another example of the photoelectric effect A photoelectric diode in an electric circuit is another example of the application of the photoelectric effect. When photons (light) with a frequency higher than the cut-off frequency of the metal cathode shines on the cathode, photoelectrons are emitted. Photoelectric diodes are used in: • smoke detectors • light meters in cameras • remote controls and • CD players. PAY SPECIAL ATTENTION NB NB HINT! hint A photoelectric diode in an electric circuit e.g. Worked example 1 EG - worked examples Calculate the energy of a light wave with a wavelength of 660 nm. When the wavelength exams is given, always calculate Solutions the frequency first. c = λf Convert nm to m 660 nm = 660 × 10–9 m ∴ 3 × 108 = 660 × 10–9 f Use c = λf f = 3 × 108 – 4,55 × 1014 Hz Then calculate the energy using E = hf. E = hf E = (6,63 activity × 10–34)(4,55 × 1014) E = 3,02 × 10–19 J ACTIVITIES boy and girl-left and right of page activity Step by step comment comment 138 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS  © Department of Basic Education 2015  Mind the Gap Physics

HINT! hint Unit 9 e.g. Worked example 2EG - worked examples A learner wants to demonstrate the photoelectric effect. exams He uses a disk of zinc placed on an electroscope. The work function (W0) of zinc is 6,9 × 10–19 J. 1. Define the concept work function. 2. Calculate the maximum wavelength of light that will eject electrons from the zinc. 3. The electroscope is negatively charged and then exposed to ultraviolet light from a mercury discharge lamp. One of the wavelengths of the light is 260 nm. Calculate activity the kinetic energy of an electron emitted from the zinc disk by a photon of this light. ACTIVITIES boy Solutions and girl-left and right of page 1. The work function (W0) of a metal is the minimum energy that is activity required to emit a photoelectron form the surface of the metal. 2. E = hf and c = λf 6,9 × 10 = (6,63 × 10 ) f –19 –34 3 × 108 = λ (1,05 × 1015) f = 1,04 × 1015 Hz λ = 2,88 × 10–7 m Step by step comment comment 3. E = W0 + Ek E = hf hf = W0 + Ek (6,63 × 10–19)(1,15 × 1015) = 6,9 × 10–19 + Ek Ek = (7,63 × 10–19) – (6,9 × 10–19) Ek = 7,3 × 10–20 J And c = λf 3 × 108 = (260 × 10–9) f f = 3 × 108 / (260 × 10–9) f = 1,15 × 1015 Hz Remember: Convert nm to m 260 nm = 260 × 10–9 m © Department of Basic Education 2015 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS 139 Mind the Gap Physics

9 Unit Activity 1 A metal surface is illuminated with ultraviolet light of wavelength 330 nm. Electrons are emitted from the metal surface. The minimum amount of energy required to emit an electron from the surface of this metal is 3,5 × 10–19 J. 1. Name the phenomenon illustrated. (1) 2. Give ONE word or term for the underlined sentence in the above paragraph.(1) 3. Calculate the frequency of the ultraviolet light. (3) 4. Calculate the kinetic energy of a photoelectron emitted from the surface of the metal when the ultraviolet light shines on it. (3) 5. The intensity of the ultraviolet light illuminating the metal is now increased. What effect will this change have on the following? a) Kinetic energy of the emitted photoelectrons. (Write down only INCREASES, DECREASES or REMAINS THE SAME.) (1) b) Number of photoelectrons emitted per second. (Write down only INCREASES, DECREASES or REMAINS THE SAME.) (1) [10] Solutions 1. Photoelectric effect. 3 (1) 2. Work 3 function. (1) 3. c = λf 3 3 × 108 = (330 × 10–9) f 3 3 f = 9,09 × 1014 Hz (3) 4. Ek = hf – WO 3 330 nm = 330 × 10–9 m = (6,6 × 10–34)(9,09 × 1014) – 3,5 × 10–19 = 6,0 × 10–19 – 3,5 × 10–19 3 = 2,5 × 0–19 3 J (3) 5. a) Remains the same. 3 (1) b) Increases. 3 (1) [10] 140 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 9 Activity 2 In the simplified diagram below, light is incident on the emitter of a photocell. The emitted photoelectrons move towards the collector and the ammeter registers a reading. 1. Name the phenomenon illustrated above. (1) 2. The work function of the metal used as emitter is 8,0 × 10–19 J. The incident light has a wavelength of 200 nm. Calculate the maximum speed at which an electron can be emitted. (6) 3. Incident light of a higher frequency is now used. How will this change affect the maximum kinetic energy of the electron emitted in the question above? Write down only INCREASES, DECREASES or REMAINS THE SAME.  (1) 4. The intensity of the incident light is now increased. How will this change affect the speed of the electron calculated in QUESTION 11.1.2? Write down INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2) 5. A metal worker places two iron rods, A and B, in a furnace. After a while he observes that A glows deep red while B glows orange. Which rod A or B has higher energy of radiation? Give a reason for your answer. (2) 6. Neon signs illuminate many buildings. What type of spectrum is produced by neon signs? (1) [13] © Department of Basic Education 2015 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS 141 Mind the Gap Physics

9 Unit Solutions 1. Photo-electric effect 3  (1) 2. E = WO + Ek 3 hf = hfO + Ek 3 hc 1 ​  λ ​= WO + ​ __ __ 2 ​mv 3 2 (6,63 × 1034)(3 × 108) 1 ________________ ​      200 × 10–9 ​ 3 = 8 × 10–19 3 + ​ __ 2 ​(9,11 × 10 )v 3(6) –31 2 3. Increases 3(1) 4. Remains the same 3 Intensity only affects number of photoelectrons emitted per second. 3  (2) 5. B3 Orange light has a higher frequency than red light. 3  (2) 6. Line emission (spectra) 3  (1) [13] Activity 3 During an investigation, light of different frequencies is shone onto the metal cathode of a photocell. The kinetic energy of the emitted photoelectrons is measured. The graph below shows the results obtained. 1. For this investigation, write down the following: a) Dependent variable (1) b) Independent variable (1) c) Controlled variable (1) 2. Define the term threshold frequency. (2) 142 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS  © Department of Basic Education 2015  Mind the Gap Physics

Unit 9 3. Use the graph to obtain the threshold frequency of the metal used as cathode in the photocell. (1) 4. Calculate the kinetic energy at E1 shown on the graph. (4) 5. How would the kinetic energy calculated in QUESTION 11.4 be affected if light of higher intensity is used? Write down only INCREASES, DECREASES or REMAINS THE SAME.  (1) [11] Solutions 1.a) Kinetic energy 3 (1) b) Frequency 3 (1) c) (Type of) metal 3 (1) 2. The minimum frequency needed to emit electrons 3 from the surface of a metal. 3 (2) 3. 9 x 1014 Hz 3 (1) 4. E = W0 + Ek 3 hf = hf0 + Ek (6,63 × 10–34)(14 × 1014) 3 = (6,63 × 10–34)(9 × 1014) + Ek 3 therefore Ek = 3,32 × 10–19 J 3 (4) 5. Remains the same 3 (1) [11] Keep going! © Department of Basic Education 2015 UNIT 9 OPTICAL PHENOMENA AND PROPERTIES OF MATERIALS 143 Mind the Gap Physics

10 Unit Emission and absorption spectra Summary All images on • When a light ray passes from one optical medium to another, the ray is pages 144, 145 refracted and its speed and direction change. and 146 are in • The pattern which forms when a ray of light is broken up into its component colour on the frequencies is called a spectrum. The light is broken up by the individual rays of different frequencies being refracted or having their path “bent” inside front cover. as they go through optical (transparent) media of different optical density (different degrees of transparency/light conductivity). • Spectra can be observed with a diffraction grating, a spectroscope or a prism, or in a rainbow after a storm. (Source: http://en.wikipedia.org/wiki/Prism) 10.1 Continuous emission spectra • The spectrum produced when white light passes through a prism is called a continuous spectrum. • The spectrum emitted by the sun is a continuous emission spectrum. • The colours in the spectrum follow on each other without any gaps between them. A familiar example of a spectrum is a rainbow that one sees after a thunderstorm. 144 UNIT 10 EMISSION AND ABSORPTION SPECTRA  © Department of Basic Education 2015  Mind the Gap Physics

Unit 10 10.2 Atomic emission spectra • Atomic emission spectra are produced when a gas is heated or by passing an electric current through it in a gas discharge tube. • The electrons in the atoms of the gas absorb the energy and become excited and move to a higher (excited) energy level. This high energy state is unstable. • When excited electrons return to the ground state or a lower energy level, the energy is released in specific energy packets called “photons”, or light particles. • The gas becomes incandescent (glowing). • The energy of the emitted photon equals the energy difference between the two energy levels. The energy of light is directly proportional to its frequency and the frequency of light determines its colour. • Only the frequencies (colours) of light that are in the visible range, that are emitted by the atoms, are seen by the eye. Colours out of the visible range, such as ultraviolet and infrared are not seen. The range of frequencies emitted by a particular substance are called a line emission spectrum as most substances do not emit the full spectrum; instead, they emit a particular pattern of frequencies. • The atoms of each element have a unique set of energy levels, so the line emission spectrum is a set of discrete coloured lines with dark spaces in between where those frequencies are not being emitted. • The line emission spectrum for each element is unique to that element, and can be used to identify that element. For example, amber street lamps have a sodium lamp in them, and thus produce an amber light, as sodium emits primarily in the yellow band. Likewise, fireworks’ colours are determined by the chemicals used in them. So, for example, bright red is produced by strontium (Sr), blue- green by copper (Cu), and so on. • Scientists are able to tell what elements are present on distant planets and stars by projecting their light through a prism and capturing the line emission spectrum. © Department of Basic Education 2015 UNIT 10 EMISSION AND ABSORPTION SPECTRA 145 Mind the Gap Physics

10 Unit 10.3 Atomic absorption spectra PAY SPECIAL ATTENTION NB NB spectrum is a continuous spectrum where certain An atomic absorption colours or frequencies are missing. These frequencies appear as dark DEFINITIONS lines in the spectrum. The region A-B in the diagram below is Infrared. The discrete: clear and individual, region B-C and part way to D is red. The region C-D is orange. The region separate HINT! D-E is green. The region E-F is cyan/light blue. The region around F is blue. hint glowing incandescent: The region F-G is indigo, and G-H violet. The region KH is ultraviolet. ground state: lowest stable energy state excited state: high and EG - worked examples e.g. unstable energy state exams (Source: Wikimedia Commons) You must remember: • Atomic absorption spectra are produced when light passes through a activity cold gas. • The electrons in the atoms of the gas absorb energy from the light ACTIVITIES boy andgirl-left and and become excited and move to a higher (excited) energy level. • The energy of the absorbed light energy equals the energy difference right of page activity between the two energy levels. • The energy of light is directly proportional to its frequency and the frequency of light determines its colour. • The light that has not been absorbed by the gas, reaches the eye and therefore shows the range of frequencies in the atomic absorption Step by step spectrum. comment comment • The atoms of each element have a unique set of energy levels, so the atomic absorption spectrum is a continuous spectrum with a few black lines. These lines represent the colours (and hence frequencies) of the light that were absorbed by the gas atoms’ electrons. • The atomic absorption spectrum for each element is unique to that element, and can be used to identify that element. • The dark lines represent the same frequencies of light that are emitted in the same element’s atomic emission spectrum. If an atomic emission spectrum and an atomic absorption spectrum are combined for a specific element, we see a continuous spectrum. 146 UNIT 10 EMISSION AND ABSORPTION SPECTRA  © Department of Basic Education 2015  Mind the Gap Physics

Unit 10 Activity 1 1. What is the approximate wavelength range of visible light? (2) 2. Name five wavelength ranges and their uses.  (5) 3. How can a scientist tell what elements are present on a star? (3) 4. What is the approximate wavelength of red light? And violet? (2) 5. What does the wavelength of UV tell you about its energy levels? (2) 6. Does microwave radiation or gamma radiation have more energy per photon?  (1) 7. Give one example of a colour in fireworks achieved through emission spectra.  (1)  [16] Solutions 1. 400 nm (3) to 700 nm (3) (One mark per correct value) (2) 2. Visbile light: to see (3); Xrays: to inspect bones without surgery (3); Gamma rays: to kill bacteria (3); UV: suntanning (3), helps bees navigate (3), powers photosynthesis (3); Infrared: night vision (3), heat radiation (3), some lasers (3); Microwaves: telecommunications (3), radar (3), ovens (3); Radio waves: telecommunications (3). (any 5) (5) 3. She can project the light from the star through a spectroscope (3) which splits it into its components (3). She can then compare the spectrum to known emission spectra of known elements (3). (3) 4. Any value 700-600 nm (3) (it's continuous); Any value near 400-450 nm (3). (2) 5. UV has a short wavelength (3) which means that it has high (3) energy levels. (2) 6. Gamma. (3) (1) 7. Cu / Copper: blue / green / cyan / blue-green / turquoise (3); Strontium / Lithium: Red (3); Iron / Sodium / Calcium / Na / Fe / Ca: orange / yellow (3); Magnesium / Mg / Aluminium / Aℓ: White (3); Potassium/K: lilac / violet (3); Green: Barium / Ba (3) (light green), possibly Copper / Cu (darker green) (3). (any one) (1) [16] Well done! © Department of Basic Education 2015 UNIT 10 EMISSION AND ABSORPTION SPECTRA 147 Mind the Gap Physics

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