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Grade 12 Physical Science Revision Booklet - FET Study Guide 2017
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Further Education and Training Grade 12 (FET) Bright Ideas Physical Science Revision Booklet Physical Science Grade 12 REVISION BOOKLET 1 2 1 Foreword 5 2 How to use this Revision Booklet 6 2.1 For every topic you need to: 6 2.2 General strategy for problem-solving in Physics 6 3 Key Physics Concepts 7 3.1 Newton’s Laws 7 3.2 Momentum and Impulse 8 3.3 Vertical Projectile Motion 9 3.4 Work Energy and Power. Work-Energy Theorem. Law of Conservation of Mechanical Energy. 10 3.5 Doppler Effect 11 3.6 Electrostatics 12 3.7 Electric Circuits 15 3.8 Electrodynamics 16 3.9 Photoelectric Effect 18 4 Revision Questions- Set 1 19 4.1 Newton’s Laws 19 4.2 Momentum and Impulse 22 4.3 Vertical Projectile Motion 27 4.4 Work Energy Power 29 4.5 Doppler Effect 34 4.6 Electrostatics 34 4.7 Electric Circuits 38 4.8 Electrodynamics 42 5 Revision Questions - Set 2 (master an additional 20%) 46 5.1 Newton’s Laws 46 5.2 Momentum and Impulse 47 5.3 Vertical Projectile Motion 48 5.4 Work Energy Power 50 5.5 Doppler Effect 51 5.6 Electrostatics 51 5.7 Electric Circuits 52 5.8 Electrodynamics 54 5.9 Photoelectric Effect 55 6 Check your answers - Set 1 6.1 Newton’s Laws 56 6.2 Solutions: Momentum and Impulse 58 6.3 Vertical Projectile Motion 58 6.4 Work Energy Power 62 6.5 Electrostatics 65 6.6 Electric Circuits 67 6.7 Electrodynamics 69 7 Check your answers - Set 2 7.1 Newton’s Laws 71 7.2 Momentum and Impulse 76 7.3 Vertical Projectile Motion 79 7.1 Work Energy Power 82 3 7.2 Doppler Effect 82 7.3 Electrostatics 83 7.4 Electric Circuits 85 7.5 Electrodynamics 87 7.6 Photoelectric Effect 88 Chemistry 4 1 How to use this book 89 2 Key subject Concepts 89 3 The Representation Model 89 4 Tips on specific topics 90 5 Study and examination tips 90 6 Message to Grade 12 learners from the writers 90 7 The Mole Concept 91 8 Stoichiometry 91 8.1 Excess and Limiting Reagents 92 9 Practice activities from Grade 10 93 10 Dynamic Chemical Equilibrium Graphs 93 11 Concentration-time graphs 94 12 fertilizers 96 13 ORGANIC CHEMISTRY 101 13.1 BASIC ORGANIC CHEMISTRY 101 13.2 NOMENCLATURE OF ORGANIC COMPOUNDS 102 14 ELECTROCHEMISTRY 124 1 Foreword Message from the Minister of Basic Education Message to Grade 12 Learners from the Minister of Basic Education Matric (Grade12) is perhaps the most important examination you will prepare for. It is the gateway to your future; it is the means to enter tertiary institutions; it is your opportunity to pursue the career of your dreams. It is not easy to accomplish but it can be done with hard work and dedication, prioritising your time and effort to ensure that you cover as much of the curriculum content as possible in order to be well-prepared for the examinations. I cannot stress the importance and value of revision in preparing for the examinations. Once you have covered all the content and topics, you should start working through the past examination papers, thereafter check your answers against the memoranda. If your answers are not correct, go back to the Mind the Gap (MTG) series and work through the content again. Then retest yourself, and continue with this process until you get all the answers right. The Bright Idea… getting exam ready booklet will allow you to do this in a systemic way. It has been developed to assist you to achieve a minimum of 40% in the examinations – if you work hard and follow the advice and guidance provided. I also urge you to continue with the next section in this booklet that deals with an additional 20%, which will ensure you have covered the basics required to achieve at least a 60% pass. Use this valuable resource which has been developed especially for YOU. Work hard, persevere, work every day, read and write every day to ensure that you are successful. I have faith that you can do this. Remember ‘SUCCESS’ depends on the second letter, ‘U’. Best Wishes MRS AM MOTSHEKGA, MP MINISTER OF BASIC EDUCATION DATE: 24/02/2017 5 2 How to use this Revision Booklet 2.1 For every topic you need to: • Learn the content. • Study the strategy you need to use to answer questions on that topic. • Study the worked examples. First try to answer the worked examples yourself by writing down your own solution. Then compare your solution with the solution given in the book. • Do the exercises without looking at the solutions. • Compare your solutions with the given solutions and identify what your mistakes were (if any). • Go back to your answer and correct any errors. • Do another exercise. You can find more exercises in this book, previous NSC question papers, Mind the Gap (MTG) study guides, the Grade 12 Siyavula textbook and many other sources. 2.2 General strategy for problem-solving in Physics (a) Read the whole question as many times as you need to, in order to understand what is given and what you need to find out or calculate. • If a diagram is given, look at the diagram while reading the question and try to understand the diagram. • (e) If a diagram is not given, represent the problem in a rough sketch and put all the information given into the diagram, e.g. if the velocity of a car is given, write it down, e.g. car = 2 m∙s-1 to the left Write the given information in symbolic form (data) using the correct scientific symbols, e.g. i for initial velocity, net for net force, etc. Select /choose the appropriate equation /formula /principal /law /theorem /concept. 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direction onequal the directly proportional toforce the product velocity unless arest ofaobject the force acceleration of their and inversely its state of rest or acts onat anan object, theinWhen object body, the second body exerts aother force ofEach equal every other body with a a force force that that is itsstate stateof ofrest rest orbody acts on an body, the second body exerts aforce force body with aforce that is motion at constant will accelerate in the direction magnitude inbody. the opposite direction on first directly proportional to the product A will remain inobject, When resultant/net force one body exerts force onthe aevery second body in masses the universe attracts motion at constant will accelerate in the direction magnitude in the opposite direction on theaafirst directly proportional to the product its state of rest or acts on an object, the object body, the second body exerts force ofaof equal every other body with force that or acts on an object, the object body, the second body exerts of equal every other body with aaof that isis less aits non-zero resulan acceleration directly propormasses and inversely proportional to non-zero directly proportional to the square the motion at constant will accelerate in the direction magnitude in the opposite direction on the first directly proportional to the product velocity unless a of the force at an acceleration body. of their masses and inversely non-zero directly proportional to the proportional to the square of the ⃗ ⃗ motion at constant will accelerate in the direction magnitude in the opposite direction on the first directly proportional to the product motion at constant will accelerate in the direction magnitude in the opposite direction on the first directly proportional to the product 𝑭𝑭the ⃗⃗an velocity aand ofunless the force at acceleration body. of their masses and inversely itsunless state of rest orthe acts on an object, body, the second body exerts a force of equal every body with a force that isbetween their 𝑭𝑭 𝑨𝑨𝑨𝑨 object ⃗𝑭𝑭 ⃗direction velocity unless a at of the force at an acceleration body. their masses and inversely motion atacts constant will accelerate the direction magnitude inthe the opposite onofthe the first directly proportional toother the product motion constant will accelerate inin the direction magnitude in opposite direction on first directly proportional to the product resultant/net force force inversely distance between their centres. 𝑩𝑩𝑩𝑩 ⃗ ⃗ tant/net force on it. tional to force and inversely the square of the distance velocity a of the force at acceleration body. of their masses and inversely 𝑭𝑭 𝑨𝑨𝑨𝑨 non-zero directly proportional to the proportional to the square of the resultant/net force force and inversely distance between their centres. 𝑩𝑩𝑩𝑩 velocity unless aacceleration of the force at an body. ofsquare their masses and inversely velocityunless unless theforce force an body. oftheir their masses and inversely ⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨acceleration non-zero directly proportional proportional toproportional the of at constant will accelerate inthe the direction in the opposite direction on the first 𝑮𝑮𝒎𝒎 to the thethe product 𝒎𝒎inversely ⃗⃗⃗magnitude non-zeroactsvelocity directly proportional to the proportional to the square ofdirectly the velocity aa ofofthe the force atat an acceleration body. of their masses and inversely amotion of at an acceleration body. of masses and onresultant/net it. unless mass of 𝟏𝟏 𝟐𝟐centres. 𝑭𝑭 non-zero proportional to the to square of proportional to the mass of tothe ⃗to force force and inversely distance between theirproportional centres. 𝑩𝑩𝑩𝑩 𝑮𝑮𝒎𝒎 ⃗directly ⃗𝑨𝑨𝑨𝑨 acts on it.aproportional proportional to the mass of 𝑭𝑭 ⃗⃗𝑩𝑩𝑩𝑩 𝟏𝟏 𝒎𝒎square 𝟐𝟐 non-zero directly proportional the 𝑭𝑭 proportional to the of the the non-zero directly to the proportional tothe the square ofthe the 𝑭𝑭𝑭𝑭 = 𝑭𝑭 ⃗ ⃗ ⃗ ⃗ 𝑨𝑨𝑨𝑨 𝑭𝑭 𝑭𝑭 ⃗ ⃗ resultant/net force force and inversely distance between their centres. velocity unless of the force at an acceleration body. of their masses and inversely 𝟐𝟐 resultant/net force force and inversely distance between their centres. non-zero directly proportional to the proportional to the square of the 𝑭𝑭 𝑭𝑭𝑭𝑭 = 𝑩𝑩𝑩𝑩 𝑨𝑨𝑨𝑨 ⃗ ⃗ non-zero directly proportional to the proportional to square of ⃗ ⃗ ⃗⃗ it. ⃗⃗acts theobject. object. resultant/net force force and inversely distance between their centres. 𝑮𝑮𝒎𝒎 𝒎𝒎 𝑩𝑩𝑩𝑩 𝑭𝑭 ⃗ ⃗ 𝒓𝒓 𝑭𝑭 ⃗ ⃗ 𝑭𝑭 = 𝟎𝟎 B A on proportional to the mass of 𝟐𝟐 𝟏𝟏 𝟐𝟐 𝑭𝑭 𝑨𝑨𝑨𝑨 𝑭𝑭 𝑨𝑨𝑨𝑨 𝒏𝒏𝒏𝒏𝒏𝒏 ⃗ ⃗ ⃗ ⃗ the object. ⃗ ⃗ ⃗ ⃗ resultant/net force force and inversely distance between their centres. 𝑩𝑩𝑩𝑩 𝒓𝒓 resultant/net force force and inversely distance between their centres. 𝑩𝑩𝑩𝑩 𝑭𝑭 = 𝟎𝟎 B 𝑭𝑭 ⃗ ⃗ 𝑭𝑭 ⃗ ⃗ A 𝑮𝑮𝒎𝒎 𝒎𝒎 𝒏𝒏𝒏𝒏𝒏𝒏 acts on it. proportional to the mass of 𝑭𝑭 non-zero proportional to 𝑨𝑨𝑨𝑨 the proportional to𝟏𝟏the𝟐𝟐 square of𝑮𝑮𝒎𝒎 the 𝟐𝟐 𝑨𝑨𝑨𝑨the = centres. 𝑭𝑭𝑩𝑩𝑩𝑩 𝑮𝑮𝒎𝒎between 𝒎𝒎𝟐𝟐 𝑭𝑭𝑭𝑭 acts on it. resultant/net proportional toacts the of ⃗directly resultant/net force force and inversely distance between their centres. force force and inversely distance their 𝑩𝑩𝑩𝑩 𝟏𝟏 ⃗𝑭𝑭 ⃗mass 𝟐𝟐 on it. proportional to of ⃗the mass = 𝒎𝒎𝒂𝒂 ⃗⃗ ⃗⃗⃗𝑨𝑨𝑨𝑨 ⃗⃗𝒏𝒏𝒏𝒏𝒏𝒏 the object. 𝑭𝑭𝑭𝑭 𝑮𝑮𝒎𝒎𝟏𝟏𝟏𝟏𝒎𝒎 𝒎𝒎𝟐𝟐 𝒏𝒏𝒏𝒏𝒏𝒏 𝒓𝒓=between 𝑮𝑮𝒎𝒎 𝒎𝒎 𝑭𝑭 =⃗resultant/net 𝟎𝟎 𝑭𝑭 𝑭𝑭𝑭𝑭 = acts on it. proportional toA mass of−𝑭𝑭 ⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨 B ⃗𝑩𝑩𝑩𝑩 ⃗𝑭𝑭⃗𝑩𝑩𝑩𝑩 acts onit.it.it. proportional tothe of ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = 𝒎𝒎𝒂𝒂 ⃗⃗the mass 𝟏𝟏 𝟐𝟐 = = ⃗proportional the object. force force andof inversely centres. 𝟐𝟐 𝑭𝑭𝑭𝑭distance 𝑮𝑮𝒎𝒎 𝒎𝒎 𝑮𝑮𝒎𝒎 ⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨 ⃗⃗𝑩𝑩𝑩𝑩 𝒓𝒓𝟐𝟐 their𝑭𝑭𝑭𝑭 ⃗𝟎𝟎⃗ on ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = the object. acts on to the mass of 𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗𝟎𝟎proportional B acts to mass A 𝟏𝟏𝟏𝟏𝒎𝒎 𝟐𝟐𝟐𝟐 𝑭𝑭𝑭𝑭 = 𝒓𝒓 = −𝑭𝑭 = ⃗ ⃗ B ⃗ ⃗ the object. A ⃗ ⃗ 𝒓𝒓𝒓𝒓𝟐𝟐𝟐𝟐 𝑭𝑭 = 𝟎𝟎 B A ⃗ ⃗ 𝟐𝟐 𝑭𝑭 = 𝒎𝒎𝒂𝒂 𝒏𝒏𝒏𝒏𝒏𝒏 ⃗ ⃗ 𝑭𝑭𝑭𝑭 = ⃗ ⃗ the object. 𝑭𝑭𝑭𝑭 = ⃗ ⃗ ⃗ ⃗ the object. 𝒏𝒏𝒏𝒏𝒏𝒏 𝑮𝑮𝒎𝒎 𝒎𝒎 𝒓𝒓 𝑭𝑭 = 𝟎𝟎 B A 𝑭𝑭 = 𝟎𝟎 dIFFERENT TyPES OF FORcES B ⃗⃗𝑨𝑨𝑨𝑨 = −𝑭𝑭 ⃗⃗ A of ⃗⃗OF it. = proportional 𝟏𝟏 𝟐𝟐 𝒏𝒏𝒏𝒏𝒏𝒏=⃗𝟎𝟎 𝑭𝑭 ⃗to ⃗ the mass ⃗dIFFERENT ⃗𝒏𝒏𝒏𝒏𝒏𝒏 𝑭𝑭 = 𝒎𝒎𝒂𝒂 ⃗⃗ acts on ⃗⃗𝒏𝒏𝒏𝒏𝒏𝒏 the object. ⃗𝟎𝟎 ⃗𝑭𝑭⃗𝑭𝑭 the object. ⃗⃗ 𝒏𝒏𝒏𝒏𝒏𝒏 𝒓𝒓𝒓𝒓𝟐𝟐𝟐𝟐 𝑭𝑭 𝒎𝒎𝒂𝒂 ⃗𝑭𝑭 𝒏𝒏𝒏𝒏𝒏𝒏 FORcES ⃗⃗⃗⃗B ⃗⃗𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 ⃗⃗𝑩𝑩𝑩𝑩 = 𝒎𝒎𝒂𝒂 𝒏𝒏𝒏𝒏𝒏𝒏= 𝑭𝑭𝑭𝑭 = ⃗𝑭𝑭⃗force ⃗B ⃗𝑩𝑩𝑩𝑩 𝑭𝑭 = −𝑭𝑭 𝒏𝒏𝒏𝒏𝒏𝒏 ⃗𝑭𝑭⃗A ⃗A ⃗TyPES ⃗𝒏𝒏𝒏𝒏𝒏𝒏 the −𝑭𝑭 ⃗⃗⃗⃗𝑨𝑨𝑨𝑨 = −𝑭𝑭 ⃗⃗⃗⃗𝑩𝑩𝑩𝑩 𝟐𝟐 == 𝒎𝒎𝒂𝒂 ⃗⃗ 𝑭𝑭 =𝒎𝒎𝒂𝒂 𝒎𝒎𝒂𝒂 NormaldIFFERENT Force weight Frictional ⃗𝟎𝟎⃗Force 𝑨𝑨𝑨𝑨 𝑭𝑭 ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏OF object. 𝒏𝒏𝒏𝒏𝒏𝒏 𝒓𝒓 ⃗ ⃗ TyPES FORcES ⃗ ⃗ = B A ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ 𝑭𝑭 = 𝒎𝒎𝒂𝒂 𝑭𝑭 = Normal weight Frictional force 𝑭𝑭 = −𝑭𝑭 𝑭𝑭 = −𝑭𝑭 𝒏𝒏𝒏𝒏𝒏𝒏 DIFFERENT TYPES OF FORCES 𝒏𝒏𝒏𝒏𝒏𝒏 𝑨𝑨𝑨𝑨 𝑩𝑩𝑩𝑩 𝑨𝑨𝑨𝑨=−𝑭𝑭 𝑩𝑩𝑩𝑩 ⃗⃗𝑨𝑨𝑨𝑨 ⃗𝑩𝑩𝑩𝑩 ⃗⃗𝑭𝑭 ⃗⃗⃗motion TyPES OF FORcES −𝑭𝑭 dIFFERENT TyPES OFForce FORcES 𝑭𝑭 Commented [U17]: P|se check/complete/ re-write to en ⃗ ⃗ 𝑨𝑨𝑨𝑨= 𝑩𝑩𝑩𝑩 dIFFERENT TyPES OF FORcES Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational ForcedIFFERENT orNormal the dIFFERENT This is force that opposes the of an object and which acts parallel to the ⃗ ⃗ 𝑭𝑭 = 𝒎𝒎𝒂𝒂 weight Frictional force Commented [U17]: dIFFERENT TyPES OF𝒏𝒏𝒏𝒏𝒏𝒏 FORcES Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational Force or the This is force𝑭𝑭 opposes to–the TyPES OFFORcES FORcES ⃗⃗that ⃗⃗𝑩𝑩𝑩𝑩 the motion of an object and which acts parallel legibility see example below in this block and elsewhere. (T = −𝑭𝑭 Normal Force weight Frictional force 𝑨𝑨𝑨𝑨 dIFFERENT TyPES OF FORcES dIFFERENT TyPES OF Normal Force weight Frictional force Normal Force Weight Frictional force legibility – see example component of a surface. force that the Earth exerts on Normal Force weight Frictional force Commented [U17]: P|se check/complete/ reWeight (𝑤𝑤 ⃗⃗⃗) is the gravitational Force or the This is force that opposes the motion of an object and which acts parallel to the author cannot always determine what is being said and re-wri component of a surface. Normal Force weight Frictional force force that the Earth exerts on NormalForce Forceor weight Frictional force Commented check/compl dIFFERENT TyPES OF FORcES Commented [U17]:legibility P|se check/complete/ re-write to ensure author cannot always d[ Weight (𝑤𝑤 ⃗⃗⃗) is theis gravitational Force the This isthe force that opposes theand motion of the an the object andto which acts parallel to the – see example [U17]: below in P|se this block and els Normal Force weight Frictional force weight Frictional force Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational Commented Force orforce theNormal This force that opposes motion of an object which acts parallel the Static frictional force ( f ) is the force that opposes tendency of motion of a which a the object of mass m. s Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational Force or the This is force that opposes motion of an object and which acts parallel to the not done as part of editing.) component of a force surface. force that the Earth exerts on Commented [ Commented [U17]: P|se check/complete/ re-writ legibility – see example below in this block Static frictional force ( f ) is the force that opposes the tendency of motion of a which a the object of mass m. s Force or the compoThis is force that opposes the motion of an object and which acts parallel to the surface. Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational Force or the This is force that opposes the motion of an object and which acts parallel to the legibility – see example below in this block and elsewhere. (The Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational not done as part of editi Force or the This is force that opposes the motion of an object and which acts parallel to the author cannot always determine what is being legibility – seesaid ext Normal Force weight Frictional force Weight ( ) is the gravitational Commented [U17]: P|se check/complete/ re-write component of a surface. Commented [U17]: P|se check/complete/ re-write force that the Earth exerts on component of a surface. Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational Force or the This is force that opposes the motion of an object and which acts parallel to the Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational force that the Earth exerts on Force or the This is force that opposes the motion of an object and which acts parallel to the surfaceforce exerts on ana ⃗⃗⃗ stationary object relative to a surface. ⃗⃗⃗⃗ 𝒘𝒘 = 𝒎𝒎𝒈𝒈 legibility –iselsewh seeisex legibility – see example below inand this block and component of a author always determine what b that the Earth exerts on surface. Static frictional force ( fs) isobject the force thatto opposes the tendency of motion of aauthor cannot always which theexerts object of mass m. force determine what isofbelow being said re-writing not–done as partcannot editing.) surface on an ⃗⃗⃗ stationary relative a surface. ⃗⃗⃗⃗ 𝒘𝒘 = 𝒎𝒎𝒈𝒈 author cannot al component of a surface. force that the Earth exerts on legibility – see example below in this block and elsewhe component of a legibility see example in this block and elsewher force that the Earth exerts on nent of a force which Commented [U17]: P|se ch 𝒎𝒎𝒎𝒎𝒎𝒎 Static frictional force ( f ) is the force that opposes the tendency of motion of a Weight (𝑤𝑤 ⃗⃗⃗) is the gravitational force which Force or the force thatopposes opposes motion ofofthe an object and which acts parallel toCommented theofmotion the object of mass m.object cannot al force that the Earth exerts onsurface. thestationary author cannot always determine whatauthor is being said an not done as partthis of editing.) 𝒎𝒎𝒈𝒈 Static frictional force ( m. fsobject ) This is theis force that motion a opposes force which a with component ofitaa surface. the object of mass m. component of force that the Earth exerts on force that the Earth exerts on object which 𝒇𝒇s𝒔𝒔 = the 𝝁𝝁the 𝑵𝑵tendency ⃗⃗⃗ a not done as part editing.) 𝒈𝒈 -an acceleration due to 𝒎𝒎𝒎𝒎𝒎𝒎 of ffs)) is the tendency of a force which a the of mass [G18]: like 𝒔𝒔force done as and part surface exerts on ⃗⃗⃗ relative to the a frictional surface. ⃗⃗⃗⃗ 𝒘𝒘 author cannot always determine what isbeing being said and author cannot always determine isnot said legibility –what see example below inr object with which it= 𝒇𝒇force =that 𝝁𝝁of 𝑵𝑵opposes ⃗⃗⃗ 𝒈𝒈 acceleration toon (frictional Static frictional thetendency force that the tendency ofmotion motion ofpart aa Better force which aforce the object ofdue mass m. object Static frictional force fStatic )isis force that the motionthe force which s is Commented [G18]: the object ofmass mass m. Static force (force fopposes ) is((the force opposes tendency ofof station𝒔𝒔 that not done as part not done as of editing.) a surface exerts an surface exerts on an ⃗⃗⃗ stationary object relative to a surface. component of surface. ⃗⃗⃗⃗ 𝒘𝒘 =- 𝒎𝒎𝒈𝒈 that the Earth exerts s Static frictional force )sis the force that opposes the tendency of𝒔𝒔 motion motion ofofaaa 𝒎𝒎𝒎𝒎𝒎𝒎 surface exerts on an ⃗⃗⃗a stationary relative toforce aforce surface. Static frictional the force that opposes the tendency of of force which aa which ⃗⃗⃗⃗ 𝒘𝒘 =surface 𝒎𝒎𝒈𝒈 the object of mass m. which aon the object of m. object of mass m. sfs)stationary Kinetic frictional (f the force that opposes the motion of a moving object is inforce contact, and notdone done as part of editing.) gravity. not as part of editing.) k()(fis exerts on an ⃗⃗⃗ object relative to a surface. ⃗⃗⃗⃗ 𝒘𝒘 = 𝒎𝒎𝒈𝒈 author cannot always determine object with it 𝒇𝒇 = 𝝁𝝁 𝑵𝑵 ⃗⃗⃗ 𝒈𝒈 acceleration due to Commented [G18]: Better like this Commented [G15]: Better like this Kinetic frictional force (f ) is the force that opposes the motion of a moving object is in contact, and 𝒔𝒔surface. 𝒔𝒔 to gravity. k𝒎𝒎𝒎𝒎𝒎𝒎 surface exerts on an ⃗⃗⃗ stationary object relative a surface. ⃗⃗⃗⃗ 𝒘𝒘 = 𝒎𝒎𝒈𝒈 surface exerts on an ⃗⃗⃗ stationary object relative to a surface. ⃗⃗⃗⃗ 𝒘𝒘 = 𝒎𝒎𝒈𝒈 𝒎𝒎𝒎𝒎𝒎𝒎 ary object relative to a toStatic frictional ( fs) is the opposes motion of[G18]: a Commented object with itadue 𝒇𝒇 force = 𝝁𝝁that 2 force which the object of -mass m.to a surface. ⃗⃗⃗ 𝒎𝒎𝒎𝒎𝒎𝒎 the tendency of 𝒈𝒈 -𝒘𝒘 due to stationary with which is as part of editing.) [G18]: Better like[G15]: this objectobject with which it exerts 𝒇𝒇 =isforce 𝝁𝝁the surface exerts on an stationary object relative to𝒔𝒔aa(fsurface. surface. ⃗⃗⃗ it- on 𝒔𝒔 𝑵𝑵the an object relative to ⃗⃗⃗⃗acceleration 𝒎𝒎𝒈𝒈 𝒈𝒈 acceleration to ⃗⃗⃗⃗ 𝒘𝒘 == 𝒎𝒎𝒈𝒈 thatsurface is perpendicular relative Commented Better Commented like this not done ⃗⃗⃗which 𝒈𝒈 ≈ 𝟗𝟗,is 𝟖𝟖 𝑚𝑚/𝑠𝑠 close to⃗⃗⃗⃗⃗⃗the 𝒔𝒔 𝑵𝑵 force with which itit𝒈𝒈 𝒇𝒇 = 𝑵𝑵 ⃗⃗⃗ 𝒈𝒈 acceleration 2 Kinetic 𝒎𝒎𝒎𝒎𝒎𝒎 that𝒔𝒔opposes frictional force motion of 𝝁𝝁 object is in contact, and Commented [ k)a that perpendicular relative to surface. ⃗⃗⃗to≈ 𝟗𝟗, 𝑚𝑚/𝑠𝑠 todue the object with which 𝒇𝒇𝒔𝒔𝒎𝒎𝒎𝒎𝒎𝒎 = 𝝁𝝁a𝒔𝒔𝒔𝒔moving 𝑵𝑵of a moving ⃗⃗⃗𝒘𝒘 𝒈𝒈 acceleration due to isfrictional object with which 𝒇𝒇𝒎𝒎𝒎𝒎𝒎𝒎 =𝝁𝝁surface. 𝝁𝝁𝑵𝑵𝑵𝑵 ⃗⃗⃗-object 𝒈𝒈 -gravity. acceleration dueto Commented [G15]: Better this Commented [U16]:[G18]: P|se check/complete/ re-write. Commented [ 𝒎𝒎𝒎𝒎𝒎𝒎 Commented Better likelike this 𝒔𝒔 object 𝒔𝒔the 𝒔𝒔 𝑵𝑵 force Kinetic force (ffrictional is that opposes the motion object isthat in and surface exerts on an ⃗⃗⃗ close object relative to= aforce ⃗⃗⃗⃗𝟖𝟖-= 𝒎𝒎𝒈𝒈 gravity. k )𝒇𝒇𝒇𝒇 Kinetic frictional (f the force that opposes the motion of a moving is in contact, and object with which = 𝝁𝝁 in contact, and is object with which itititsurface ⃗⃗⃗ ⃗⃗⃗𝒈𝒈 acceleration due to gravity. k )stationary 𝒈𝒈 --is acceleration due 2 Commented [U16]: ⃗⃗⃗ contact, 𝒇𝒇 = 𝝁𝝁 𝑵𝑵 Commented [G18]: Better like this acceleration due to gravity. Commented [G18]: Better like this Kinetic force (f ) is the force that opposes the motion of a moving object in contact, and of the Earth. 𝒔𝒔 𝒔𝒔 𝒔𝒔 𝒔𝒔 Commented [G15]: Better like this gravity. k to the surface. 𝑵𝑵 𝒌𝒌 𝒌𝒌 that is perpendicular relative to a surface. 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Kinetic frictional force (f ) is force opposes the motion object relathat is perpendicular object which it = 𝑵𝑵 relative to a surface. ⃗⃗⃗ acceleration due to ⃗⃗⃗ Commented [U16]: P|se check/complete/ re𝒈𝒈 ≈ 𝟗𝟗, 𝟖𝟖 𝑚𝑚/𝑠𝑠 close to the Commented [lii [U19]: Formatting: pse check ad hoc use of 2 Commented [G15]: Better like this Commented [G18]: Better Kinetic frictional force (f ) is the force that opposes the motion of a moving object that isperpendicular perpendicular contact, and Kinetic frictional force (f ) is the force that the motion of a moving relative to surface. isisisininincontact, and 𝒔𝒔 𝒔𝒔 ⃗⃗⃗ gravity. k 𝒈𝒈 ≈ 𝟗𝟗, 𝟖𝟖 𝑚𝑚/𝑠𝑠 close to the gravity. k You need to know that a frictional force: to the 𝑵𝑵 that is perpendicular relative a ⃗⃗⃗ 𝟖𝟖 𝑚𝑚/𝑠𝑠 close to the ⃗⃗⃗ Commented [U19]: 𝒇𝒇 a=kfrictional 𝝁𝝁𝒌𝒌 𝑵𝑵 2the Earth. 𝒈𝒈 Commented [G15]:Better Better likethis this Commented [G15]: like surface of Commented [U16]: P|se check/compl to is the surface. You need to to know that force: that is𝟖𝟖surface perpendicular relative to a surface. surface. [U16]: P|se check/complete/ re-write. that perpendicular ⃗⃗⃗ ≈ relative toaaa surface. 𝒈𝒈 ≈ 𝟗𝟗, 𝟗𝟗,relative 𝟖𝟖 𝑚𝑚/𝑠𝑠 2 to close toKinetic the ⃗⃗⃗ ≈ 𝟗𝟗, 𝑚𝑚/𝑠𝑠 close tothe the in text and standardise usage throughout. Commented [ 22 close ⃗⃗⃗ frictional force (fk )𝒌𝒌is the that opposes the𝑵𝑵motion of Commented a moving object is inof contact, and 𝒇𝒇𝒌𝒌 force = 𝝁𝝁𝒌𝒌 𝑵𝑵 gravity. of the Earth. tive to a surface. ⃗⃗⃗ to surface the surface. 𝑵𝑵 that perpendicular 𝒇𝒇 = 𝝁𝝁 𝑵𝑵 that isisperpendicular relative to surface. surface. ⃗⃗⃗𝒈𝒈 in text and standardise ⃗⃗⃗𝒈𝒈 𝟗𝟗, 𝟖𝟖 𝑚𝑚/𝑠𝑠 to the the Earth. 𝒈𝒈 ≈≈ 𝟗𝟗, 𝟖𝟖 𝑚𝑚/𝑠𝑠 close to Commented [li Commented [U19]: Formatting: pse check ad to thesurface. surface. 𝑵𝑵 ⃗⃗⃗ 𝒌𝒌 𝒌𝒌 Commented [U16]: P|se check/complete/ re-writ 𝒇𝒇 = 𝝁𝝁 Commented [G15]: Better surface of the Earth. close to thesurface surface You need to know that a frictional force: to surface. 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Commented [U19]: Formatt 6 You need to know that a frictional force: 6 6 in text UseUse thisthis equation 66 and standardise usage thr equationtotocalculate calculate the is •proportional to the normal 6 is proportional to the force normal force 66 acceleration of anyofother planet: is independent of the area of contact the acceleration any other 6 Commented [U20]: Formatting: pse insert bull planet: is •independent of the velocity of motion is independent of the area of contact Use this equation to calculate is proportional to the normal force lists. 𝑮𝑮𝑮𝑮 NOTE: 𝒈𝒈 = 𝟐𝟐 acceleration of any other is independent of thedoes areanot of contact • is the parallel velocity Commented [U21]: Formatting: pse standardis o the If a force ( Findependent ) applied to aof body to of themotion surface cause the object to 𝒓𝒓 Com planet: is independent of the velocity of motion with all equivalent items, including in tables and ot move, F is equal in magnitude to the static frictional force. lists. NOTE: 𝑮𝑮𝑮𝑮 is proportional NOTE: Use this equation to calculate to the normal force 𝑚𝑚𝑚𝑚𝑚𝑚 o The static𝒈𝒈frictional ) just before the object starts to move = 𝟐𝟐 force is a maximum (𝑓𝑓𝑠𝑠 Com o If a force ( F area ) applied to a body parallel to the surface does not cause the object to the acceleration of anythe other of the of contact 𝒓𝒓 is independent Commented [U22]: Pse check/complete. across with o surface. If a force ( F ) applied to aequal bodyinparallel to the surface does not cause the object to move, F is magnitude to the static frictional force. Commented [U20]: planet: is independent of the velocity of motion 𝑚𝑚𝑚𝑚𝑚𝑚 o If the applied force exceeds (𝑓𝑓𝑠𝑠 ), a resultant force accelerates the movement of 𝑚𝑚𝑚𝑚𝑚𝑚 lists. 𝑮𝑮𝑮𝑮 move, F is equal in static magnitude toforce the static frictional(𝑓𝑓𝑠𝑠force. NOTE: o The frictional is a maximum ) just before the object starts to move 𝒈𝒈 =the𝟐𝟐object. Com thetosurface. Commented [U21]: o If a force ( F across ) applied a body parallel to the surface does not cause the object to 𝒓𝒓 𝑚𝑚𝑚𝑚𝑚𝑚 o equal If thein applied force exceeds (𝑓𝑓𝑠𝑠frictional ),) ajust resultant force accelerates thetomovement ofwith all equivalent items move, F is magnitude to the static force. before the object starts move o The static frictional force is a maximum ( 3.2 Momentum and Impulse o The the static frictional force is a maximum (𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ) just before the object starts to move the object. across surface. Physical quantities Newton's Commented [U22]: across the surface. Impulsesecond law in momentum and o the If the applied force exceeds (𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ),),aaof resultant force accelerates the movement of 3.2 Momentum and Impulse momentum Law of conservation lineal momentum resultant force accelerates the movement of the o If applied force exceeds ( terms of change in Impulse theNewton's object. Physical quantities theorem object. momentum momentum Impulsesecond law in momentum and momentum Law of conservation Momentum is the The impulse of a The resultant/ An isolated (closed) system in physics is a system of lineal momentum ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 3.2 Momentum and Impulse ⃗⃗of 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = ∆𝒑𝒑 terms change in Impulse theorem product of an object force is the net force acting on which the resultant/net external force is zero. Physical quantities ⃗ Newton's momentum 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 momentum = 𝑚𝑚∆𝑣𝑣⃗ ⃗⃗ Impulse∆𝒑𝒑 mass and its product of the on an object is second ⃗⃗ ⃗ ⃗ ⃗ law in 𝐹𝐹 = 0 OR = 𝟎𝟎 momentum and ⃗ 𝑛𝑛𝑛𝑛𝑛𝑛 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛of ∆𝑡𝑡a The resultant/ Momentum is the The impulse An isolated (closed) system in physics is a system ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ 7 8 ∑ after is the total momentum after collision ∑ before is the total momentum before collision Commented [U20]: Formatting: pse inse planet: is of independent of the velocity of motion the acceleration of any other is independent the area contact Commented [U20]: Formattin planet: isof independent the velocity of motion the acceleration of any𝑮𝑮𝑮𝑮 other is independent theofarea of of contact lists. NOTE: Commented [U20]: Formatting: pse insert bullet points with planet:planet: is independent of the velocity of motion lists.Formatting: 𝑮𝑮𝑮𝑮 NOTE: Commented [U20]: pse insert bullet p is independent of the velocity of motion 𝒈𝒈 = 𝟐𝟐 Commented [U21]: Formatting: pse stan o If a force ( F ) applied to a body parallel to the surface does not cause the object lists. to = 𝟐𝟐 𝑮𝑮𝑮𝑮 𝑮𝑮𝑮𝑮 𝒓𝒓 𝒈𝒈NOTE: lists. NOTE: Commented [U21]: o If a force ( F ) applied to a body parallel to the surface does not cause the object to 𝒓𝒓 𝒈𝒈 = 𝟐𝟐𝒈𝒈 = with all equivalent items, includingFormattin in tables F to is equal inparallel magnitude to surface the static frictional force.the object to Commented [U21]: Formatting: standardise lineincluding spacing o If aoforce F )move, applied a body to the does not cause withFormatting: allpse equivalent 𝒓𝒓 F to is aequal magnitude the static frictional force. 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Thefrictional static frictional is a maximum ) just before the object starts to move If the applied force exceeds (𝑓𝑓𝑠𝑠𝑚𝑚𝑚𝑚𝑚𝑚 ),𝑠𝑠 a𝑚𝑚𝑚𝑚𝑚𝑚 resultant force accelerates the movement of Commented [U22]: Pse check/complete. acrossacross theosurface. 3.2 Momentum and Impulse If the applied force exceeds (𝑓𝑓 ), a resultant force accelerates the movement of Commented [U22]: Pse check/complete. theosurface. 𝑠𝑠 𝑚𝑚𝑚𝑚𝑚𝑚 the object. o If the force exceeds (𝑓𝑓𝑠𝑠 ),(𝑓𝑓a𝑚𝑚𝑚𝑚𝑚𝑚 resultant force accelerates the movement of o applied If the applied force exceeds ), a resultant force accelerates the movement of the object. 𝑠𝑠 the object. the object. Physical quantities 3.2 Momentum and Impulse 3.2 Momentum and Impulse Physical quantities Newton’s 3.2 Momentum and Impulse Newton'ssecond Physical quantities 3.2 Momentum and Impulse Impulse-momentum Newton's Impulselaw in terms of Law of conservation of lineal momentum Physical quantities Impulsesecond law in momentum and Newton's Momentum and change Physical quantities theorem second law in Newton's momentum Law of conservation of lineal momentum momentum and ImpulseImpulse momentum momentum Law of conservation of lineal momentum Impulseterms of change in Impulse second law in law interms of momentum and in momentum second theoremLaw of conservation of lineal momentum change in Impulse momentum and momentum momentum theorem Law of conservation of lineal momentum momentum momentumImpulse terms of change in momentum terms of change in momentum Impulse theorem momentum momentum Momentum is the The impulse of a The resultant/theorem 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 An isolated (closed) system in physics is a system ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ = ∆𝒑𝒑 momentum ⃗⃗ momentum is the of aresultant/ The resultant/ AnAn isolated (closed) system in physics is a system ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ Momentum is theMomentum The impulseThe of impulse The net isolated (closed) system in physics is a system on ⃗⃗ which 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = ∆𝒑𝒑 product ofThe an object force isThe theresultant/ net force 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 acting = ∆𝒑𝒑 on the resultant/net external force is zero. Momentum is the impulse of a An isolated (closed) in physics is a system ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 ⃗ isolated ofaThe an object force theresultant/ net force acting⃗𝐹𝐹 onsystem which the resultant/net external force is zero. Momentum isproduct the impulse of a isThe An (closed) system in physics isexternal a system ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ product of an object force is the force acting on an which resultant/net force is zero. ⃗ ⃗ ⃗ 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = ∆𝒑𝒑 ⃗ ⃗ ∆𝒑𝒑 𝐹𝐹 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 ⃗ and its product offorce the acting on an object is product of anmass object force is net on𝑛𝑛𝑛𝑛𝑛𝑛 which the resultant/net zero.is zero. ⃗⃗ OR external ⃗⃗ ∆𝒑𝒑⃗⃗force isforce 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = resultant/net 0 𝟎𝟎 ⃗𝑛𝑛𝑛𝑛𝑛𝑛object ⃗𝑛𝑛𝑛𝑛𝑛𝑛 andproduct itsthe is the product offorce the acting on 𝐹𝐹 an is product of anmass object force net the ∆𝑡𝑡to 𝑚𝑚∆𝑣𝑣 ⃗=∆𝑡𝑡𝑚𝑚∆𝑣𝑣 𝐹𝐹 ⃗⃗ = ⃗𝑛𝑛𝑛𝑛𝑛𝑛 mass andits its velocity. of the object is 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =∆𝒕𝒕0 OR external = ⃗𝟎𝟎⃗ 𝐹𝐹= ∆𝑡𝑡 ⃗ ∆𝑡𝑡 on which ⃗⃗ ∆𝒑𝒑 velocity. force acting on equal to equal the rate 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 mass and product of the on anacting object is ∆𝒕𝒕 OR ⃗ ⃗ ⃗⃗ ⃗ ⃗ ∆𝒑𝒑 ⃗ force on equal to the rate mass and itsvelocity. product of the on an object is 𝐹𝐹 = 0 OR = 𝟎𝟎 ⃗ = 𝑚𝑚𝑣𝑣⃗𝑓𝑓 − 𝑚𝑚𝑣𝑣 ⃗𝑖𝑖 𝐹𝐹⃗ = ∆𝒕𝒕0 ⃗⃗ OR =∑ ⃗⃗ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏 𝑛𝑛𝑛𝑛𝑛𝑛 𝐹𝐹of ∆𝑡𝑡 𝟎𝟎 force acting onand anto thethe 𝑛𝑛𝑛𝑛𝑛𝑛change an object of rate change of 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 𝑛𝑛𝑛𝑛𝑛𝑛⃗𝑖𝑖 Momentum is a vector = 𝑚𝑚𝑣𝑣 ⃗𝑓𝑓 − 𝑚𝑚𝑣𝑣 velocity. force acting on equal rate ∆𝒕𝒕 Momentum is a vector an object andto of change of ∑ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏 Momentum is a vector velocity. force acting on equal the rate = 𝑚𝑚𝑣𝑣 ⃗𝑓𝑓of −of 𝑚𝑚𝑣𝑣 ⃗𝑖𝑖 − 𝑚𝑚𝑣𝑣⃗ object and the time of momentum the time the momentum quantity. = 𝑚𝑚𝑣𝑣 ⃗ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 an object and andthe of change of ∑ when Momentum is a vector 𝑖𝑖 ⃗𝟎𝟎⃗ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = time 𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏 an object of the change of momentum 𝑓𝑓of ∑ 𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎𝒎 Momentum is quantity. a vector ∑𝒔𝒔𝒔𝒔𝒔𝒔/𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓𝒓/𝒏𝒏𝒏𝒏𝒏𝒏 quantity. ⃗ ⃗ force acts on the the theobject objectin in the the ⃗⃗ = the ⃗⃗ time when ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 = ⃗𝟎𝟎⃗ 𝒑𝒑 𝒎𝒎𝒗𝒗 acts on the momentum quantity. force acts onofthe the object in the ⃗⃗ =the ⃗⃗force the time the momentum of 𝒑𝒑 𝒎𝒎𝒗𝒗 quantity. ⃗𝟎𝟎 ⃗ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = Then When ∑ when −1 ⃗⃗ ⃗ ⃗ ∑ object. direction of the when 𝑭𝑭 𝟎𝟎 Then SI units⃗⃗[𝑘𝑘𝑘𝑘force ∙ 𝑚𝑚 ∙ 𝑠𝑠the ] object. acts on the object in the in direction ⃗⃗ = 𝒎𝒎𝒗𝒗 ⃗𝒑𝒑 𝒑𝒑 direction of the of the −1theonobject. 𝒆𝒆𝒆𝒆𝒆𝒆 = force acts the the object the ⃗⃗⃗ = 𝒎𝒎𝒗𝒗 SI units [𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚 ∙ 𝑠𝑠 ] ⃗⃗ Then ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 −1 in momentum resultant/net Impulse is a ⃗⃗ Then Change object. direction of the SI unitsSI[𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚 ∙ 𝑠𝑠 ] ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 Then force. resultant/net object. is a vector Impulsedirection is aresultant/net of the Change momentum units [𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚 ∙ 𝑠𝑠 −1 ] in Impulse ⃗⃗ ⃗⃗ ∆𝑝𝑝 ⃗ = 0 𝑝𝑝 ⃗ − 𝑝𝑝 ⃗ = 0 force. vector quantity. ⃗⃗ 𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑝𝑝⃗ 𝑇𝑇𝑇𝑇= 0 ∆𝑝𝑝⃗ = 𝑝𝑝⃗𝑓𝑓 − 𝑝𝑝⃗Impulse 𝑇𝑇𝑇𝑇 resultant/net is a ⃗⃗ Change in momentum 𝑓𝑓 𝑝𝑝⃗𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 0 𝑛𝑛𝑛𝑛𝑛𝑛 resultant/net force. is a vector quantity. ∆𝑝𝑝⃗ = 𝑝𝑝⃗𝑓𝑓 − 𝑝𝑝Impulse ⃗𝑓𝑓 Change in momentum ⃗⃗ ⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ∆𝒑𝒑 ⃗∆𝒕𝒕 𝑝𝑝⃗𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇 =−0 force. quantity. ⃗= ⃗ −vector 𝑚𝑚𝑣𝑣⃗ quantity. 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = ⃗𝑭𝑭force. ⃗⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇 ∆𝑝𝑝⃗ = 𝑝𝑝⃗∆𝑝𝑝 𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 0 𝑓𝑓 − ⃗⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ∆𝒑𝒑 ⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇 ⃗ =𝑝𝑝⃗𝑓𝑓𝑝𝑝⃗𝑓𝑓∆𝑝𝑝− 𝑝𝑝⃗𝑓𝑓𝑚𝑚𝑣𝑣 𝑇𝑇𝑇𝑇 ∆𝑝𝑝𝑓𝑓⃗ = 𝑚𝑚𝑣𝑣⃗𝑓𝑓𝑓𝑓 −vector 𝑚𝑚𝑣𝑣⃗𝑓𝑓 quantity.𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 = ⃗𝑭𝑭⃗∆𝒕𝒕 ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗⃗ ⃗⃗ 𝑭𝑭∆𝒕𝒕 ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ∆𝒑𝒑 𝑝𝑝⃗𝑖𝑖𝑖𝑖=+𝑝𝑝⃗𝑝𝑝⃗𝑖𝑖𝑖𝑖 = 𝑝𝑝⃗𝑓𝑓𝑓𝑓 + 𝑝𝑝⃗𝑓𝑓𝑓𝑓 𝑝𝑝⃗𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑝𝑝⃗𝑇𝑇𝑇𝑇 ⃗∆𝒕𝒕units⃗⃗ [𝑁𝑁 ∙ 𝑠𝑠] SI 𝒏𝒏𝒏𝒏𝒏𝒏 = ∆𝑝𝑝 𝑚𝑚(𝑣𝑣 ⃗𝑓𝑓 −𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 𝑣𝑣⃗𝑓𝑓 ) 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 ∆𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣 ⃗⃗𝑓𝑓 = − 𝑚𝑚𝑣𝑣 𝑚𝑚𝑣𝑣 = ⃗𝑭𝑭 ⃗ ⃗ ⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗⃗ ∆𝒑𝒑 𝑝𝑝 ⃗ + 𝑝𝑝 ⃗𝑖𝑖𝑖𝑖 = 𝑝𝑝⃗𝑓𝑓𝑓𝑓 + 𝑝𝑝⃗𝑓𝑓𝑓𝑓 ⃗ ⃗ ∆𝑝𝑝 ⃗⃗𝑓𝑓⃗𝑓𝑓 = − 𝑚𝑚𝑣𝑣 ⃗ SI units [𝑁𝑁 ∙ 𝑠𝑠] = 𝑭𝑭 ∆𝒕𝒕 𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇 ∆𝒕𝒕 ∆𝑝𝑝𝑓𝑓⃗ = 𝑚𝑚(𝑣𝑣⃗𝑓𝑓 − 𝑣𝑣⃗𝑓𝑓 ) 𝑖𝑖𝑖𝑖 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = ⃗𝑭𝑭⃗ = 𝑚𝑚𝑝𝑝⃗𝑝𝑝𝐴𝐴⃗𝑓𝑓𝑓𝑓 𝑣𝑣⃗𝑖𝑖𝑖𝑖++𝑝𝑝⃗𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑓𝑓𝑓𝑓 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑓𝑓𝑓𝑓 𝑝𝑝⃗𝑖𝑖𝑖𝑖 + 𝑝𝑝⃗𝑝𝑝𝑖𝑖𝑖𝑖 =+𝑝𝑝⃗𝑝𝑝⃗𝑓𝑓𝑓𝑓 + 𝒏𝒏𝒏𝒏𝒏𝒏 units ∙ 𝑠𝑠] [𝑁𝑁 ∙ 𝑠𝑠] ∆𝒕𝒕 ⃗⃗)= SI ⃗⃗ SI[𝑁𝑁units 𝒎𝒎∆𝒗𝒗 ∆𝑝𝑝⃗ = 𝑚𝑚(𝑣𝑣 𝑣𝑣⃗𝑓𝑓⃗) −∆𝒑𝒑 ⃗ = ∆𝒕𝒕 𝑚𝑚 ⃗𝑖𝑖𝑖𝑖 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑓𝑓𝑓𝑓 + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑓𝑓𝑓𝑓 ∆𝑝𝑝⃗⃗𝑓𝑓=−𝑚𝑚(𝑣𝑣 𝑣𝑣 ⃗ 𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓 ⃗⃗ = 𝒎𝒎∆𝒗𝒗 ⃗⃗ ∆𝒑𝒑 𝐴𝐴 𝑣𝑣 𝑓𝑓 𝑓𝑓 of states: The 𝑚𝑚𝐴𝐴 𝑣𝑣The ⃗𝑖𝑖𝑖𝑖 + 𝑚𝑚⃗The ⃗𝑖𝑖𝑖𝑖conservation = 𝑚𝑚 + 𝑚𝑚 ⃗momentum ⃗⃗ SI⃗⃗units of ∆𝒑𝒑 ⃗⃗ = 𝒎𝒎∆𝒗𝒗 ∆𝒑𝒑 𝐵𝐵 𝑣𝑣 𝑓𝑓𝑓𝑓 𝐵𝐵 𝑣𝑣 𝑓𝑓𝑓𝑓 𝑚𝑚law 𝑚𝑚 ⃗𝐴𝐴𝑖𝑖𝑖𝑖𝑣𝑣⃗conservation = 𝑚𝑚 ⃗of ⃗momentum of states: The ⃗⃗ ⃗⃗ = 𝒎𝒎∆𝒗𝒗 SI⃗⃗−1 units of ∆𝒑𝒑 ∆𝒑𝒑 𝐴𝐴 𝑣𝑣 𝑖𝑖𝑖𝑖 +law 𝐵𝐵 𝑣𝑣 𝐴𝐴 𝑣𝑣 𝑓𝑓𝑓𝑓 + 𝑚𝑚of 𝐵𝐵 𝑣𝑣 𝑓𝑓𝑓𝑓 total linearofmomentum of an isolated (closed) ∙ 𝑚𝑚 ∙ 𝑠𝑠 ] The law of conservation momentum states: The ⃗⃗[𝑘𝑘𝑘𝑘 ∆𝒑𝒑 SI unitsSIofunits −1 total linearofmomentum an isolated (closed) The law of conservation momentum of states: The ⃗ ⃗ of ∆𝒑𝒑 [𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚 ∙ 𝑠𝑠 ] system remains constant(closed) (is conserved). total linear momentum of an isolated [𝑘𝑘𝑘𝑘 ∙ 𝑚𝑚[𝑘𝑘𝑘𝑘 ∙ 𝑠𝑠 −1 ] ∙ 𝑠𝑠 −1 ] system remains constant(closed) (is conserved). total linear momentum of an isolated ∙ 𝑚𝑚 ∑ ∑ ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 ⃗(is ⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 𝒑𝒑 = 𝒑𝒑 system remains constant ∑ 𝒑𝒑 ∑ 𝒑𝒑 ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 ⃗(is ⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 system remains constant conserved). =conserved). ∑ 𝒑𝒑 ∑∑ ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 ⃗⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 𝑝𝑝⃗=before is the total momentum before collision 𝒑𝒑 ∑ 𝒑𝒑 ∑ ⃗⃗= ⃗ ⃗ 𝒑𝒑 ∑𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 𝑝𝑝 ⃗ beforelaw is the total momentum collisionstates: The total 𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 The of conservation ofbefore momentum ∑ 𝑝𝑝⃗ before the total before before collision ∑ is 𝑝𝑝⃗ before is themomentum total momentum collision linear momentum of an isolated (closed) system 7 7 remains constant (is conserved). 7 7 VERTICAL PROJECTILE MOTION Free Projectile Projectile fall gravitational force. acceleration due to constant 0 free fall. is (constant acceleration) (speed decreases) Vertically upwards 00 (speed decreases) (speed decreases) increases) (speed decreases) (speed decreases) (speed decreases) experiences (constant (constant (constant (constant object (constant acceleration) acceleration) acceleration) acceleration) negligible air acceleration) (constant acceleration) (constant (constant (constant (constant acceleration) acceleration) acceleration) acceleration) resistance 0 0 and 𝑡𝑡 0 𝑡𝑡 00 0 𝑥𝑥⃗ 0 𝑡𝑡 𝑡𝑡0 𝑡𝑡 0 00 𝑡𝑡 𝑡𝑡𝑡𝑡 0 𝑡𝑡 0 𝑡𝑡 𝑡𝑡 0 0 0 𝑡𝑡 𝑣𝑣⃗ 0 𝑡𝑡 𝑡𝑡0 𝑡𝑡 0 00𝑡𝑡 𝑡𝑡 𝑡𝑡 0 𝑡𝑡 0 𝑡𝑡 0𝑡𝑡 𝑡𝑡 0 𝑔𝑔⃗ 9 𝑡𝑡 𝑡𝑡0 𝑡𝑡 0 9 𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑡𝑡 99 9 𝑡𝑡 9 9 9 9 9 Vertical Motion Vertical 3.3 Projectile Projectile Vertical 3.3VERTIcaL Motion Projectile VerticalPROjEcTILE Motion Projectile Motion mOTION VERTIcaL VERTIcaL PROjEcTILE PROjEcTILE mOTION mOTION VERTIcaL PROjEcTILE mOTION Equations of motion mOTION VERTIcaL PROjEcTILE Projectile Equations of motion Projectile Projectile Equations Equations of motion of motion Projectile Equations of motion Equations of motion VERTIcaL PROjEcTILE mOTION VERTIcaL VERTIcaL PROjEcTILE VERTIcaL PROjEcTILE mOTION PROjEcTILE mOTION mOTION Projectile is an object upon Position Projectile Displacement Velocity Acceleration Position Velocity acceleration [U23]: Formatting: pseComment ensure all Projectile Position Position displacement displacement Velocity Velocity acceleration acceleration acceleration Commented Commented [U23] Position displacement Velocity displacement acceleration Commented [U23]: Formatting: pse ens Projectile Projectile is anisProjectile andisplacement Projectileis isan an Position Velocity is an Projectile Equations of motion Projectile Projectile Projectile Equations Equations of motion Equations of motion of motion appearappear equivalent, including capital letters or not. appearcapital equivalent, appear equi incl equivalent, including letters o 1 1 1 1 which the only force acting is the 𝑦𝑦⃗𝑓𝑓 =𝑦𝑦⃗𝑦𝑦𝑓𝑓⃗𝑖𝑖 =+ 𝑦𝑦∆𝑦𝑦 Magnitude constant: ⃗ 𝑣𝑣 = 𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude constant: 𝑦𝑦 ⃗ = 𝑦𝑦 ⃗ 𝑦𝑦 ⃗ + = ∆𝑦𝑦 ⃗ 𝑦𝑦 ⃗ + ∆𝑦𝑦 ⃗ 𝑣𝑣 = 𝑣𝑣 𝑣𝑣 + = 𝑔𝑔∆𝑡𝑡 𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude Magnitude constant: constant: ⃗ + ∆𝑦𝑦 ⃗ 𝑣𝑣 = 𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude constant: 𝑓𝑓 ∆𝑦𝑦 2 𝑦𝑦 𝑓𝑓 ∆𝑦𝑦 Position 𝑓𝑓 𝑖𝑖 Position 𝑖𝑖 1 𝑓𝑓 𝑖𝑖 𝑖𝑖 𝑓𝑓𝑖𝑖 2 𝑓𝑓 2 𝑦𝑦 ⃗𝑓𝑓 = + 𝑣𝑣𝑓𝑓acceleration = 𝑣𝑣𝑖𝑖 + 𝑔𝑔∆𝑡𝑡acceleration Magnitude constant: object uponupon which object upon which which object which object Position displacement Commented Comm Projectile is an Position displacement Velocity Velocity acceleration acceleration Commented [U23]: [U23] = 𝑣𝑣which Projectile Projectile is upon an Projectile is an is𝑖𝑖∆𝑦𝑦 ∆𝑦𝑦 = ⃗𝑣𝑣𝑖𝑖 𝑖𝑖∆𝑦𝑦 ∆𝑡𝑡 = +⃗ 𝑣𝑣 ∆𝑡𝑡 displacement +𝑖𝑖2 𝑔𝑔∆𝑡𝑡 =an𝑣𝑣+ +displacement 𝑔𝑔∆𝑡𝑡 𝑖𝑖 ∆𝑡𝑡 𝑖𝑖𝑔𝑔∆𝑡𝑡 𝑖𝑖 ∆𝑡𝑡𝑔𝑔∆𝑡𝑡 object upon ∆𝑦𝑦Velocity = 𝑣𝑣𝑖𝑖 ∆𝑡𝑡 + Velocity 𝑔𝑔∆𝑡𝑡 2 2 2 2 2 appear equivalent, appear incl appear equivalent, incl force of gravity. 2 𝑔𝑔∆𝑡𝑡 𝑦𝑦𝑦𝑦 ⃗⃗𝑓𝑓𝑓𝑓 = ⃗⃗𝑖𝑖𝑖𝑖 + Magnitude constant: = 𝑦𝑦𝑦𝑦 + ∆𝑦𝑦 ∆𝑦𝑦 𝑦𝑦⃗⃗𝑓𝑓⃗ = 𝑦𝑦⃗𝑖𝑖 + ∆𝑦𝑦 𝑦𝑦⃗𝑓𝑓⃗ = 𝑦𝑦⃗𝑖𝑖 + ∆𝑦𝑦⃗2 11 2 2 22 2 1 𝑣𝑣𝑣𝑣𝑓𝑓𝑓𝑓2= = 𝑣𝑣𝑣𝑣𝑖𝑖1 + 𝑔𝑔∆𝑡𝑡 𝑣𝑣 2 = 𝑣𝑣2𝑖𝑖 + 𝑔𝑔∆𝑡𝑡 𝑣𝑣𝑓𝑓 =−2 Magnitude 𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude constant: Magnitude constant: constant: 𝑖𝑖 + the only forceforce acting 2 2𝑓𝑓 the ∆𝑦𝑦 only force acting force acting the only acting object upon which object upon object which upon object = 𝑔𝑔⃗2𝑔𝑔∆𝑦𝑦 =𝑣𝑣𝑖𝑖𝑔𝑔 9,8 ∙ 𝑠𝑠𝑚𝑚2∙ 𝑖𝑖𝑠𝑠 −22𝑔𝑔⃗ = 9,8𝑔𝑔⃗𝑚𝑚=∙ 9,8 ∆𝑦𝑦 = 𝑣𝑣𝑣𝑣𝑖𝑖𝑖𝑖∆𝑡𝑡 ∆𝑡𝑡𝑣𝑣+ ∆𝑦𝑦 + = 𝑔𝑔∆𝑡𝑡 𝑣𝑣𝑖𝑖+ ∆𝑡𝑡𝑣𝑣2𝑔𝑔∆𝑦𝑦 ∆𝑦𝑦 + =2𝑔𝑔∆𝑦𝑦 𝑔𝑔∆𝑡𝑡 𝑣𝑣𝑖𝑖 ∆𝑡𝑡 𝑠𝑠 −2𝑚𝑚 ∙ 𝑠𝑠 −2𝑔𝑔⃗ = 9,8 𝑚𝑚 ∙ 𝑠𝑠 −2 ⃗+ =𝑚𝑚 9,8 ⃗ the − Displacement, the only force acting 2 ∆𝑦𝑦⃗ −∆𝑦𝑦 ∆𝑦𝑦⃗which −upon Displacement, ∆𝑦𝑦 ⃗ − which Displacement, ∆𝑦𝑦 ⃗ only − Displacement, Displacement, = 𝑣𝑣 𝑣𝑣𝑓𝑓2+=2𝑣𝑣𝑔𝑔∆𝑡𝑡 2𝑔𝑔∆𝑦𝑦 𝑣𝑣𝑔𝑔∆𝑡𝑡 which 𝑓𝑓 2 𝑖𝑖𝑣𝑣𝑓𝑓+= 𝑓𝑓 𝑖𝑖2= 𝑖𝑖 + Displacement, 𝑣𝑣 = 𝑣𝑣 + 2𝑔𝑔∆𝑦𝑦 2is 2 2 2 2 2 2 2 𝑓𝑓 𝑖𝑖 𝑣𝑣𝑓𝑓 −𝑣𝑣𝑖𝑖𝑣𝑣𝑓𝑓 −𝑣𝑣𝑖𝑖 2 𝑣𝑣𝑓𝑓 −𝑣𝑣𝑖𝑖 𝑣𝑣𝑓𝑓 −𝑣𝑣𝑖𝑖 which isofchange the 𝑣𝑣 2 −𝑣𝑣 which which is the is change change which is change the change −2 −2 −2 is theisforce is the isthe the force gravity. of gravity. the of gravity. 22 𝑖𝑖 2 2 2 the only force acting the only force the only acting force the only force which change in position. ∆𝑦𝑦 = ∆𝑦𝑦is= the ∆𝑦𝑦 = 2𝑔𝑔⃗⃗ ∆𝑦𝑦 = the 𝑔𝑔𝑔𝑔 ⃗⃗ = 9,8 = 9,8 𝑚𝑚 𝑚𝑚 ∙𝑔𝑔 ∙⃗𝑠𝑠𝑠𝑠= 9,8 𝑚𝑚 𝑔𝑔 ∙⃗𝑠𝑠= 9,8 𝑚𝑚 ∙ 𝑠𝑠 −2 is∆𝑦𝑦 the of gravity. The motion of force aof gravity. projectile isforce ⃗⃗acting − Displacement, ∆𝑦𝑦 −force Displacement, ∆𝑦𝑦 ⃗acting − ⃗ − Displacement, = + 2𝑔𝑔∆𝑦𝑦 ∆𝑦𝑦 = 2 𝑓𝑓𝑣𝑣𝑣𝑣 = 𝑣𝑣𝑣𝑣𝑖𝑖2𝑖𝑖2Direction + 𝑣𝑣 2𝑔𝑔∆𝑦𝑦 2𝑔𝑔∆𝑦𝑦 = 𝑣𝑣𝑖𝑖2always + 2𝑔𝑔∆𝑦𝑦 2𝑔𝑔⃗⃗Displacement, ⃗⃗22 2𝑔𝑔⃗⃗∆𝑦𝑦 𝑓𝑓𝑓𝑓 𝑓𝑓 = 𝑣𝑣𝑖𝑖is+ 𝑣𝑣 𝑓𝑓 Commented [U24]: Formatting: pseComment standardis in position. always 222𝑔𝑔 2 2⃗⃗ Direction Commented [U24] 2 Commented [U24]: Formatting: pse stan in position. in position. Direction is always isDirection always is in position. isDirection 2𝑔𝑔 𝑣𝑣 −𝑣𝑣 𝑣𝑣 −𝑣𝑣 𝑣𝑣 −𝑣𝑣 𝑣𝑣 −𝑣𝑣 in position. always 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝑖𝑖 which is change which isVertical the which change isa the which change is= change motion of is a The motion Theisof motion of of amotion motion ofthe including in tables, etc.(togravity. including tables, etc in including in tables, etc.including in is theaforce force ofthe gravity. forceaisof the gravity. force ofthe gravity. ∆𝑦𝑦 ∆𝑦𝑦 =the2𝑔𝑔 ∆𝑦𝑦 = ∆𝑦𝑦 = 3.3 Motion downwards (towards the The of Projectile Direction is always downwards calledThe freeThe fall. downwards downwards (towards (towards the the downwards (towards the downwards (towards the ⃗⃗ ⃗⃗ 2𝑔𝑔 2𝑔𝑔⃗⃗ 2𝑔𝑔⃗⃗ Commented [U24]: in Direction is always Commented Comm [U24] in position. position.in position. Direction Direction is always Direction is always is always 𝑣𝑣𝑖𝑖 + 𝑣𝑣𝑣𝑣in 𝑣𝑣𝑖𝑖 + 𝑣𝑣𝑓𝑓 𝑣𝑣𝑖𝑖 + 𝑣𝑣𝑓𝑓 𝑣𝑣𝑓𝑓 𝑖𝑖𝑓𝑓 +position. 𝑣𝑣 + 𝑣𝑣 centre of the Earth). centre centre of the Earth). of the Earth). centre of the Earth). 𝑖𝑖 𝑓𝑓 wards the centrethe of the Earth). projectile is called projectile projectile is of called isaaThe called projectile is The called motion The motion The ofmotion ofmotion a ∆𝑦𝑦is=∆𝑦𝑦 ofcalled centre Earth). ( =a( ) ∆𝑡𝑡 ) ∆𝑦𝑦 including etc. including in intables, tables, includin etc. ∆𝑡𝑡 ∆𝑡𝑡 = ( ∆𝑦𝑦 = () ∆𝑡𝑡 )∆𝑦𝑦 projectile downwards (towards the downwards downwards (towards downwards (towards the of the(towards the ) ∆𝑡𝑡 mOTION 2 2 2𝑣𝑣 + 𝑣𝑣 = 𝑣𝑣(PROjEcTILE 𝑣𝑣𝑣𝑣𝑖𝑖𝑖𝑖 + +2𝑣𝑣𝑣𝑣𝑓𝑓𝑓𝑓 VERTIcaL + 2𝑣𝑣𝑓𝑓 𝑖𝑖 𝑓𝑓 𝑖𝑖 Freefree fall free is the motion in which centre of the Earth). centre of the centre Earth). of the centre Earth). of the Earth). fall. fall. free fall. free projectile is called projectile projectile isfall. calledProjectile projectile is Graphs ∆𝑦𝑦 ∆𝑡𝑡 ∆𝑦𝑦 = = (( ∆𝑦𝑦 )= ∆𝑡𝑡 ∆𝑦𝑦 = ∆𝑡𝑡 Graphs )Graphs ( Graphs ) ∆𝑡𝑡 ( )Equations freecalled fall. is called of motion Graphs Graphs 22 vs 2timeVelocity 2Velocity Direction of motion Position vs time Velocity vs time Acceleration vs time Direction Direction of motion of motion Position Position vs time time vs time vs time Acceleration vs timeacceleration vsAcceleration time Direction of motion Position vs time Velocity vs Acceleration vsAcceleration time an object experiences negligiPosition displacement Velocity Projectile is an Direction of motion Position vs time Velocity vs time vs time FreeFree fall fall is is the Free Free fallfreeisfall isFree the the free fall. free fall. fall.thefree fall. fall is the Graphs Graphs Graphs Graphs Vertically downwards Vertically Vertically downwards downwards Vertically downwards Direction of motion Position vs time Velocity vs time Acceleration vs time 1 𝑦𝑦 ⃗ = 𝑦𝑦 ⃗ + ∆𝑦𝑦 ⃗ 𝑣𝑣 = 𝑣𝑣 + 𝑔𝑔∆𝑡𝑡 Magnitude constant: Vertically downwards 𝑓𝑓 𝑖𝑖 𝑓𝑓 𝑖𝑖 2 ble airmotion resistance and constant Direction Position time Velocity time Acceleration Direction of Direction motion Direction motion Position vs time vsPosition time Velocity vs time time vs𝑔𝑔⃗time Acceleration vs time timeAcceleration vs𝑔𝑔⃗time vs time 𝑔𝑔 ⃗ 𝑔𝑔⃗ vsVelocity 𝑥𝑥⃗ of upon which 𝑔𝑔⃗ vs 𝑥𝑥⃗ of vs 𝑥𝑥motion ⃗𝑣𝑣⃗Position ⃗𝑖𝑖 ∆𝑡𝑡time 𝑣𝑣⃗ Velocity 𝑥𝑥 ⃗ motion 𝑣𝑣⃗ ∆𝑦𝑦 𝑣𝑣vs +vs 𝑔𝑔∆𝑡𝑡 in which an motion inFree which which anobject motion in which an motion Free fall is fall is anfall the Free fall is in the the Free is thean of 𝑥𝑥⃗ = 𝑣𝑣⃗ Acceleration motion in which 2 Vertically downwards Vertically Vertically downwards downwards Vertically downwards Vertically downwards acceleration due to gravitational (speed increases) (speed (speed increases) increases) (speed increases) onlyin force acting ⃗⃗2 + 2𝑔𝑔∆𝑦𝑦 𝑔𝑔⃗ 𝑣𝑣𝑣𝑣 ⃗⃗ 𝑔𝑔⃗ 𝑔𝑔⃗ = 9,8 𝑚𝑚 ∙ 𝑠𝑠 −2 𝑥𝑥⃗ 𝑥𝑥⃗ 𝑣𝑣⃗ 𝑣𝑣⃗ 𝑣𝑣 2 = 𝑣𝑣𝑔𝑔𝑔𝑔 object experiences object object experiences experiences object experiences motion in which an increases) motion in motion which in anthe motion which an which an ∆𝑦𝑦(speed ⃗ −𝑥𝑥𝑥𝑥⃗⃗Displacement, object experiences 𝑓𝑓 𝑖𝑖 2 2 𝑣𝑣 −𝑣𝑣 force.negligible which the change (speed (speed increases) (speed increases) (speedisincreases) isnegligible the force of(speed gravity. increases) ∆𝑦𝑦 = 𝑓𝑓2𝑔𝑔⃗⃗ 𝑖𝑖 air negligible negligible air air negligible air object experiences object experiences object increases) experiences object experiences (constant (constant (constant air in(constant (constant position. Direction is always The motion of air a acceleration) acceleration) acceleration) resistance resistance resistance and and resistance and and negligible air negligible negligible air negligible airacceleration) downwards (towards the acceleration) resistance and (constant acceleration) (constant (constant (constant (constant 𝑣𝑣𝑖𝑖 + 𝑣𝑣𝑓𝑓 centre of the Earth). called acceleration) acceleration) constant = (𝑡𝑡𝑡𝑡 constant constant constant resistance and resistance resistance andprojectile resistance and0is acceleration) constant 𝑡𝑡 0 𝑡𝑡 00 0 𝑡𝑡∆𝑦𝑦 0 2𝑡𝑡 0) ∆𝑡𝑡0 𝑡𝑡 𝑡𝑡 0 𝑡𝑡 𝑡𝑡 0 𝑡𝑡 0 𝑡𝑡 0 𝑡𝑡 0 and acceleration) 0 0 𝑡𝑡 𝑡𝑡 fall. Graphs acceleration due due to acceleration due tofree due to acceleration to acceleration constant constant constant constant acceleration due to 00 𝑡𝑡 00 𝑡𝑡 𝑡𝑡 00𝑡𝑡 𝑡𝑡 0 𝑡𝑡0 0 𝑡𝑡 𝑡𝑡0 0𝑡𝑡 𝑡𝑡0 𝑡𝑡 𝑡𝑡 Vertically upwards Vertically Vertically upwards upwards Vertically upwards Vertically Direction Position 𝑣𝑣vs vs𝑔𝑔⃗time 𝑔𝑔⃗ Acceleration vs time 𝑥𝑥⃗ 𝑥𝑥 𝑥𝑥⃗ ofupwards 𝑥𝑥motion ⃗𝑣𝑣⃗ 𝑣𝑣⃗ ⃗ time𝑣𝑣⃗ 𝑔𝑔⃗ 𝑔𝑔⃗ Velocity ⃗due the Free fall is gravitational force. 𝑔𝑔⃗ gravitational gravitational force. force. gravitational force. 𝑥𝑥⃗ 𝑣𝑣⃗ acceleration due to acceleration acceleration due to acceleration due to to gravitational force. Vertically upwards Vertically downwards Vertically Vertically upwards upwards Vertically upwards upwards (speed decreases) (speed decreases) (speed decreases) 𝑔𝑔⃗ 𝑥𝑥⃗ 𝑣𝑣⃗ 𝑔𝑔𝑔𝑔 ⃗⃗ 𝑥𝑥𝑥𝑥⃗⃗ decreases) 𝑣𝑣𝑣𝑣 ⃗⃗ 𝑔𝑔⃗ 𝑥𝑥⃗ 𝑣𝑣⃗ motion in (speed which an (speed gravitational force. gravitational gravitational force. gravitational force. force.decreases) Projectile 3.3 3.3 3.3 3.3Vertical Projectile Motion 3.3 3.3 Vertical Vertical Projectile Projectile Motion Motion Vertical Projectile Motion 3.3 Vertical Projectile Motion 3.3 Vertical Projectile Motion 3.3 Vertical Projectile Motion VERT an Position 𝑦𝑦⃗𝑓𝑓 = 𝑦𝑦⃗𝑖𝑖 + ∆𝑦𝑦⃗ object upon which is the only force acting ∆𝑦𝑦⃗ − Displacement, is the force of gravity. which is the change in position. The motion of a projectile is called 9 the Direction of motion disp ∆ ∆𝑦𝑦 Posi ∆ 3.4 10 10 Commented [U25]: Formatting: pse remove blank spaces/ Work Energy and Power. Work-Energy Theorem. Law of Conservation of Mechanical Energy. pages in text. 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Commente has due to its 𝑾𝑾 gravitational expended. scalar served). aposition gravitational field in the=on kinetic product of the force tial energy plusenergy 𝟐𝟐a has ∆𝒕𝒕 In an energy isolated system, equal to the kinetic of position inhas a potential 𝑾𝑾 potential energy 𝑬𝑬𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) =Commented 𝑬𝑬𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) 𝟐𝟐 to equal toproduct the is equal force acting an system remains 𝟐𝟐 energy is the kinetic energy of an𝑬𝑬 object work is done or change equal thethe in 𝟐𝟐quantity 𝑬𝑬 𝑬𝑬𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) force multiplied by [G29]: It must be (liki the change in the conserved). Commented [U28 defined as due to itsenergy 𝟐𝟐quantity gravitational expended. 𝑬𝑬 𝒎𝒎𝒗𝒗 [G30]: BetterCommented like this product 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝒎𝒎𝒗𝒗 Commented [G29]: It must be like this = 𝒎𝒎𝒗𝒗 of the of𝑬𝑬the 𝑲𝑲 = 𝟏𝟏 𝟐𝟐 equal the 𝑲𝑲 = to 𝑷𝑷 = Physical work energy and Law of conservation Comme 𝟏𝟏 𝑾𝑾 𝑲𝑲 𝑷𝑷 = 𝑾𝑾 the change in the conserved). Comment kinetic energy of has due to its gravitational expended. position in a potential energy 𝑬𝑬 = 𝑬𝑬 𝑾𝑾 kinetic energy of position in a If the body 𝑬𝑬 = 𝒎𝒎𝒗𝒗 potential energy 𝑬𝑬 = 𝑬𝑬 relative to some energy. 𝑾𝑾 𝟐𝟐 kinetic energy of position in a potential energy 𝑬𝑬 = 𝑬𝑬 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) SI unit: joule [J] 𝟐𝟐 𝟐𝟐 Commented [G29 equal to the only conservative the object. 𝑲𝑲 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) gravitational field plus the kinetic 𝟐𝟐 Commented [G29]: It must be like this ( into bracke product of the 𝒏𝒏𝒏𝒏𝒏𝒏 to some energy of the object. multiplied byforce the kinetic energy. SI unit: joule [J] relative 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) object is equal to constant (is the object. Commente energy an object product of themultiplied due its motion. to the sum of the energy is gravitational field ∆𝒕𝒕 In an isolated system, plus the kinetic 𝑾𝑾 product of the kinetic energy of ∆𝒕𝒕 In an isolated system, position in a 𝑬𝑬 = 𝒎𝒎𝒗𝒗 𝑷𝑷 = potential energy 𝑬𝑬 = 𝑬𝑬 the displacement scalar quantity 𝑷𝑷 = in inserts force multiplied by 𝟐𝟐 𝑷𝑷 = 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) Commented [G30]: Better like tht the principle of of mechanical energy product of the Commented [G30]:Commented Better like this [G29 𝑾𝑾are SI unit: [J] kinetic energy of an thean object. position afield potential energy 𝑬𝑬an = 𝑬𝑬 plusIfthe kinetic SI by unit: [J] of the object. gravitational field possesses =𝑾𝑾 plus the kinetic SIjoule unit: joule [J] reference point. 𝑬𝑬𝑲𝑲 + 𝑬𝑬𝒑𝒑energy. 𝟐𝟐𝑬𝑬 ∆𝒕𝒕𝒏𝒏𝒏𝒏𝒏𝒏 In isolated system, the object. = 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 gravitational plus the𝑲𝑲kinetic 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) ∆𝒕𝒕 In isolated system, Comment Ifbody body forces acting. 𝟏𝟏 relative to some ∆𝒕𝒕 In isolated system, energy. 𝑴𝑴 𝑲𝑲 gravitational product the Potential mechanical the change in the conserved). the Commente forcejoule multiplied by has duefield toinits gravitational expended. relative to = some only conservative 𝑾𝑾𝑷𝑷𝒏𝒏𝒏𝒏𝒏𝒏 forceofmultiplied reference point. the [G30 SIby unit: joule [J] the object. force multiplied gravitational field 𝑷𝑷 = only conservative Commented [G30]: Better likeCommented this plus the kinetic 𝟐𝟐 Commente ofdisplacement the point equal to the ∆𝒕𝒕 In an isolated system, thebydisplacement theofdisplacement conservation of work Kinetic energy Power force multiplied by𝒎𝒎𝒎𝒎𝒎𝒎 to some SI unit: [J] 𝑬𝑬 = joule 𝒎𝒎𝒗𝒗relative If the body the object. gravitational field [G30 plus the energy kinetic to energy. 𝑾𝑾𝒏𝒏𝒏𝒏𝒏𝒏 conservative rectilinear 𝑾𝑾 𝑲𝑲 If the body 𝑬𝑬[J] ∆𝒕𝒕 In anconservative isolated system, Commented only conservative − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 only SI unit: joule [J]𝑬𝑬relative possesses OR reference point. 𝒏𝒏𝒏𝒏𝒏𝒏 𝑬𝑬𝑬𝑬𝑴𝑴𝑲𝑲to = 𝑬𝑬the +body 𝑬𝑬possesses only 𝒑𝒑 =relative = 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 𝒏𝒏𝒏𝒏𝒏𝒏 forces energy energy force multiplied kinetic energy of position in a SI unit: joule potential 𝑬𝑬𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) = 𝑬𝑬𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) reference point. are acting. =some +Ifsome = body 𝜟𝜟𝑬𝑬 the Comment 𝑲𝑲 𝒑𝒑 energy. 𝑲𝑲𝑾𝑾 If the theapplication displacement 𝟐𝟐energy. forces are acting. relative some 𝑾𝑾𝒏𝒏𝒏𝒏𝒏𝒏 𝑾𝑾 point of 𝑴𝑴to 𝒑𝒑 energy. the displacement 𝑲𝑲 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 Commente only conservative application of the product ofreference the by the of point of displacement energy of theof point 𝑷𝑷 = total If − the body possesses the displacement relative topossesses some reference point. energy. 𝑬𝑬𝑴𝑴[J] =plus 𝑬𝑬𝑴𝑴 + 𝑬𝑬 𝑾𝑾 = 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 point. uniform motion 𝑬𝑬 = 𝑬𝑬 + 𝑬𝑬 = 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 possesses only conservative forces are acting. reference point. = 𝑬𝑬 + 𝑬𝑬 𝑾𝑾 = 𝑬𝑬 − 𝑬𝑬 = 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 𝑲𝑲 𝒑𝒑 𝒏𝒏𝒏𝒏𝒏𝒏 forces are acting. 𝑲𝑲 SI unit: joule [J] If the body posThe mechanical rectilinear 𝑴𝑴 𝑲𝑲 𝒑𝒑 𝑬𝑬 = 𝒎𝒎𝒎𝒎𝒎𝒎 forces are acting. 𝑲𝑲 SI unit: joule [J] 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝑲𝑲 𝒑𝒑 𝑲𝑲 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 𝑲𝑲𝑲𝑲 the object. rectilinear gravitational field the kinetic SI unit: joule In an isolated sys𝑬𝑬 = 𝒎𝒎𝒎𝒎𝒎𝒎 OR the displacement − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝒑𝒑 of point of ∆𝒕𝒕 an acting. isolated system, SI unit: joule [J] 𝑬𝑬𝑴𝑴 = 𝑬𝑬Total reference point. = 𝜟𝜟𝑬𝑬𝑲𝑲OR (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 ofapplication 𝒑𝒑Kinetic of themultiplied point forces are of the force by the 𝑲𝑲 +=𝑬𝑬𝑬𝑬 𝒑𝒑 +possesses force byof thethe point by ofthe theforce Commente energy Power is the work done by In Total mechanical Gravitational Work doneof𝑬𝑬by a 𝒎𝒎𝒎𝒎𝒎𝒎 possesses application of theof rectilinear reference point. 𝑬𝑬 𝑬𝑬+𝒑𝒑 𝑊𝑊rectilinear 𝑬𝑬 = 𝒎𝒎𝒎𝒎𝒎𝒎 =The 𝜟𝜟𝑬𝑬 the point of rectilinear − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 (v constant), SI unit: [J] rectilinear forces are acting. − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 OR 𝑬𝑬[J] 𝑴𝑴 motion 𝑲𝑲motion 𝒑𝒑relative joule 𝑲𝑲 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 of an object uniform 𝒑𝒑 = SIjoule unit: [J] OR 𝑊𝑊 = 𝑊𝑊 𝒑𝒑= = − 𝑬𝑬− sesses If𝒏𝒏𝒏𝒏𝒏𝒏 the𝑾𝑾 body uniform to𝒎𝒎𝒎𝒎𝒎𝒎 some SIjoule unit:SI joule [J] energy. The total mechanical 𝑾𝑾 𝑾𝑾 =energy 𝑬𝑬𝑲𝑲𝑲𝑲=−𝑬𝑬OR 𝑬𝑬 of the of 𝑛𝑛𝑛𝑛𝑛𝑛joule 𝑐𝑐rectilinear 𝑛𝑛𝑛𝑛 SI unit: [J]unit:𝑬𝑬 tem, only conseronly conservative 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 𝑲𝑲𝑲𝑲𝒏𝒏𝒏𝒏𝒏𝒏𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 The total mechanical =potential 𝒎𝒎𝒎𝒎𝒎𝒎 application of point the of − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 ofby the 𝑲𝑲𝑲𝑲 SI unit: joule [J] OR application the 𝒑𝒑 SI unit: joule [J] cosine ofapplication the cosine of theforce angle the displacement mechanical the resultant (net) energy in an isolated force by the is the energy rate at which rectilinear uniform motion constant force is 𝑬𝑬 = 𝒎𝒎𝒎𝒎𝒎𝒎 the motion − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝑾𝑾 = 𝑬𝑬 − 𝑬𝑬 then: application of the SI unit: joule [J] SI unit: joule [J]uniform uniform motion The total mechanical OR 𝑾𝑾 = 𝑬𝑬 − 𝑬𝑬 𝒑𝒑 joule SI unit: joule [J] The total mechanical 𝑾𝑾 = 𝑬𝑬 − 𝑬𝑬 (v = constant), remains constant as 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 𝑲𝑲𝑲𝑲 SI unit: [J] The total mechanical 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 𝑲𝑲𝑲𝑲 Using the workpossesses (v = constant), reference point. an object 𝑬𝑬𝑴𝑴 = 𝑬𝑬𝑲𝑲uniform + 𝑬𝑬 𝒏𝒏𝒏𝒏𝒏𝒏𝑲𝑲energy 𝑲𝑲𝑲𝑲 𝜟𝜟𝑬𝑬 (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 motion (vmotion =𝑊𝑊𝑛𝑛𝑛𝑛 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = + 𝑊𝑊= uniform motion forces areforces acting.are energy of anof object application of the =force 𝑬𝑬 𝑬𝑬𝑲𝑲𝑲𝑲 𝑊𝑊 = 𝑊𝑊 + force by the vative SI unit: joule [J]=(v The mechanical 𝑐𝑐𝑾𝑾 𝑛𝑛𝑛𝑛 force the force the 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 − 𝑲𝑲𝑲𝑲 𝑛𝑛𝑛𝑛𝑛𝑛 angle formed bycosine energy is𝑊𝑊equal acting on antotal system remains of theby point is𝑭𝑭𝒗𝒗 the an joule object [J] has work is𝑐𝑐object done or formed by by the linecosine uniform (v𝒑𝒑 ==constant), = constant), defined asofthe SI unit: 𝑾𝑾𝑊𝑊 𝑬𝑬 𝑬𝑬𝑲𝑲𝑲𝑲 𝑷𝑷𝒂𝒂𝒂𝒂𝒂𝒂 ofofthe of the SIenergy joule [J] SIenergy (v =Using constant), The total mechanical energy of object force by the energy of an object 𝑊𝑊 = + 𝑊𝑊 the moves, then:joule 𝒏𝒏𝒏𝒏𝒏𝒏 energy of an object 𝑊𝑊 +the 𝑊𝑊 Commented [G27]: Space corrected theorem: rectilinear 𝑊𝑊 == 𝑊𝑊 +− 𝑊𝑊 𝑬𝑬unit: 𝑛𝑛𝑛𝑛𝑛𝑛−remains 𝑐𝑐 remains 𝑛𝑛𝑛𝑛 constant asan 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐 then: 𝑛𝑛𝑛𝑛 (v constant), the workunit: [J]= OR constant as 𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐𝑲𝑲𝑲𝑲 𝑛𝑛𝑛𝑛 𝒑𝒑 = 𝒎𝒎𝒎𝒎𝒎𝒎𝒂𝒂𝒂𝒂𝒂𝒂then: constant), force by the Using workenergy of an object cosine of the 𝑊𝑊 = 𝑊𝑊 + 𝑊𝑊 acting. cosine of the cosine of the object is equal to constant (is energy an object due its motion. to the sum of the energy is the line of action 𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐 𝑛𝑛𝑛𝑛 (v = provided constant), of the then: formed by application then: action of theangle forceangle SI unit: watt [W]𝑷𝑷 𝑊𝑊 = energy of an then: remains constant asobject formed by cosine remains constant as 𝑊𝑊 =𝑬𝑬work𝑊𝑊 + 𝑬𝑬 𝑊𝑊 ofscalar the quantity Using the workthatUsing the net 𝑷𝑷 ==𝑭𝑭𝒗𝒗 remains constant as[G27]:[G27]: Using the workuniform motion + 𝑊𝑊 ∆𝐸𝐸 𝑭𝑭𝒗𝒗 the 𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐moves, 𝑛𝑛𝑛𝑛 the object moves, 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 𝑾𝑾 = − Commented Space correc then: energy theorem: SI unit: joule [J] The total mechanical the object 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 Commented Space corrected energy theorem: 𝒏𝒏𝒏𝒏𝒏𝒏 𝑲𝑲𝑲𝑲 𝑲𝑲𝑲𝑲 remains constant as cosine of the angle formed by Using the work𝟏𝟏 angle formed byofline the change inthe theobject conserved). Comme has due to𝑷𝑷𝒂𝒂𝒂𝒂𝒂𝒂 its = 𝑭𝑭𝒗𝒗gravitational angle formed expended. then: 𝑷𝑷 = 𝑭𝑭𝒗𝒗 ofand the the force and force by thetheby the of action remains constant as corrected moves, 𝑷𝑷 = 𝑭𝑭𝒗𝒗 𝟐𝟐 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 equal to direction of the object moves, Commented [G27 the line action Using the workenergy theorem: 𝒂𝒂𝒂𝒂𝒂𝒂 SI unit: watt [W] work done by the object moves, Commented [G27]: Space angle formed by energy theorem: 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 (v = constant), Commente SI unit: watt [W] But: 𝑊𝑊 = −∆𝐸𝐸 energy theorem: provided that the net 𝑬𝑬 = 𝒎𝒎𝒗𝒗 𝑷𝑷 = 𝑭𝑭𝒗𝒗 energy of an object + 𝑊𝑊 = ∆𝐸𝐸 provided that the net 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝+ 𝑊𝑊 = 𝑊𝑊 + 𝑊𝑊 𝑲𝑲 𝑊𝑊 𝑊𝑊 = ∆𝐸𝐸 Using the workenerthe object moves, 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 Commented [G27 angle formed by energy theorem: 𝑾𝑾 𝑛𝑛𝑛𝑛𝑛𝑛 𝑐𝑐 𝑛𝑛𝑛𝑛 the line of action 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 OR the line of action kinetic energy of position in a potential energy 𝑬𝑬 = 𝑬𝑬 SI unit: watt [W] the line of action 𝑷𝑷 = 𝑭𝑭𝒗𝒗 SI unit: watt [W] 𝟐𝟐 the direction of cosine of the 𝑴𝑴(𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊𝒊) 𝑴𝑴(𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇𝒇) SI unit: watt [W] the object moves, provided that the net Comme of the force and 𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 Comment provided that the net product of the energy theorem: external non𝑊𝑊 + 𝑊𝑊 = ∆𝐸𝐸 of the force and provided that the net the displacement. 𝑊𝑊 +SI 𝑊𝑊 =then: ∆𝐸𝐸 the line of action 𝑊𝑊 + 𝑊𝑊 = ∆𝐸𝐸 work done by 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 𝑷𝑷 = −∆𝐸𝐸 + 𝑊𝑊 = ∆𝐸𝐸 unit: watt [W] 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 remains constant as But: 𝑊𝑊 = −∆𝐸𝐸 work done by 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 Using the workBut: 𝑊𝑊 = −∆𝐸𝐸 provided that the net 𝑝𝑝 𝑛𝑛𝑛𝑛 𝐾𝐾 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑊𝑊 𝑝𝑝the gy theorem: + 𝑊𝑊 = ∆𝐸𝐸 the line of action 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝 SI unit: joule [J] of the force and object. gravitational field plus the kinetic of the force and SI unit: watt [W] 𝑐𝑐 𝑛𝑛𝑛𝑛 𝐾𝐾 of the force and ∆𝒕𝒕 In an isolated system, the displacement. provided work work done by that formed by by work done the direction 𝑊𝑊𝑐𝑐 + 𝑊𝑊 conservative forces But: 𝑊𝑊 =by −∆𝐸𝐸 done by the net force multiplied But: 𝑊𝑊𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =𝑷𝑷−∆𝐸𝐸 −∆𝐸𝐸 the direction of of of 𝑭𝑭𝒗𝒗 Comme external 𝑊𝑊 =𝑝𝑝∆𝐸𝐸 −∆𝐸𝐸 the angle force and 𝑛𝑛𝑛𝑛 = 𝐾𝐾non𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝 + the object moves, external non𝒂𝒂𝒂𝒂𝒂𝒂 𝒂𝒂𝒂𝒂𝒂𝒂 𝑝𝑝 Commente energy theorem: work done by −∆𝐸𝐸 𝑊𝑊= = 𝑊𝑊 ∆𝐸𝐸 But: 𝑊𝑊But: =𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 −∆𝐸𝐸 watt [W] 𝑝𝑝body 𝑛𝑛𝑛𝑛 𝐾𝐾 of the force and the relative to some Hence: energy.SI unit: 𝑾𝑾 the direction of The total mechan𝑝𝑝If+ 𝑛𝑛𝑛𝑛 𝐾𝐾= ∆𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝= −∆𝐸𝐸 the 𝒄𝒄𝒄𝒄𝒄𝒄 direction of only conservative direction of 𝒏𝒏𝒏𝒏𝒏𝒏 work done by external non𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝜽𝜽 displacement. the line external nonis zero. But: 𝑊𝑊 the displacement. external nonthe displacement SI unit: watt [W] −∆𝐸𝐸 + 𝑊𝑊 = ∆𝐸𝐸 the conservative forces 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝 −∆𝐸𝐸 + 𝑊𝑊 = ∆𝐸𝐸 the direction ofof action provided that the net conservative forces −∆𝐸𝐸 + 𝑊𝑊 = ∆𝐸𝐸 𝑝𝑝 𝑛𝑛𝑛𝑛 𝐾𝐾 𝑾𝑾 = ∆𝑬𝑬 + ∆𝑬𝑬 𝑊𝑊 +𝑝𝑝𝜟𝜟𝑬𝑬 𝑊𝑊𝑛𝑛𝑛𝑛∆𝐸𝐸 =𝑛𝑛𝑛𝑛 ∆𝐸𝐸𝐾𝐾 𝐾𝐾externalforces 𝑛𝑛𝑛𝑛 Hence: 𝐾𝐾 non𝒏𝒏𝒏𝒏 = 𝑬𝑬𝑲𝑲 + 𝑲𝑲𝑝𝑝 𝑬𝑬Hence: reference point. 𝑬𝑬𝑴𝑴 −∆𝐸𝐸 +𝑐𝑐∆𝑬𝑬 𝑊𝑊 the direction of = (𝒘𝒘𝒘𝒘𝒘𝒘𝒘𝒘 the are acting. ical energy of an displacement. 𝒑𝒑 𝒑𝒑 possesses 𝑛𝑛𝑛𝑛 = external nonforces the displacement. SI unit: the joule [J]𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 the force and 0 =conservative ∆𝑬𝑬𝑲𝑲𝑝𝑝+ conservative 𝑾𝑾 =𝒄𝒄𝒄𝒄𝒄𝒄 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄displacement. 𝜽𝜽of −∆𝐸𝐸 +𝑲𝑲forces 𝑊𝑊𝐾𝐾= ∆𝐸𝐸conservative is zero. of the point of 𝒑𝒑 𝜽𝜽the Hence: work done byforces is zero. displacement. 𝑝𝑝𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑛𝑛𝑛𝑛 = 𝐾𝐾 Hence: can But: 𝑊𝑊 −∆𝐸𝐸 Hence: conservative forces This equation 𝑾𝑾∆𝑬𝑬 =+∆𝑬𝑬 + ∆𝑬𝑬 𝑝𝑝 rectilinear 𝑾𝑾 = ∆𝑬𝑬 𝑬𝑬 = 𝒎𝒎𝒎𝒎𝒎𝒎 𝒏𝒏𝒏𝒏 𝑲𝑲 𝒑𝒑 − 𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆𝒆 𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕𝒕 the displacement. SI unit: joule [J] Hence: OR 𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 𝒏𝒏𝒏𝒏 𝑲𝑲 𝒑𝒑 𝒑𝒑 conservative forces is zero. 𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 object remains is zero. 𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 is zero. the direction of SIjoule unit: joule 0 = ∆𝑬𝑬 + ∆𝑬𝑬 unit: [J] 𝑾𝑾 [J] application of the SI unit: jouleSI[J] Hence: 𝑾𝑾 = ∆𝑬𝑬 + ∆𝑬𝑬 external non0 = + ∆𝑬𝑬 𝑾𝑾 = ∆𝑬𝑬 + ∆𝑬𝑬 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 𝑲𝑲 𝒑𝒑 =+𝑊𝑊 ∆𝑬𝑬 +∆𝐸𝐸 ∆𝑬𝑬 𝒑𝒑 𝒑𝒑 motion −∆𝐸𝐸 + 𝒏𝒏𝒏𝒏 𝒑𝒑 zero. This equation 𝑝𝑝𝑲𝑲 𝑛𝑛𝑛𝑛𝑲𝑲 𝑾𝑾 𝑬𝑬=𝒑𝒑𝒑𝒑𝑲𝑲𝑲𝑲𝑲𝑲 −𝐾𝐾is𝑬𝑬 This𝑲𝑲uniform equation can SI unit: joule [J] be used to𝒏𝒏𝒏𝒏solve 𝑾𝑾𝒏𝒏𝒏𝒏 =𝑾𝑾∆𝑬𝑬 ∆𝑬𝑬 ∆𝑬𝑬 SI unit: joule [J] 𝒄𝒄𝒄𝒄𝒄𝒄 𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝜽𝜽 zero. SI unit: joule [J] 𝒏𝒏𝒏𝒏𝒏𝒏 =isThe ∆𝑬𝑬 +∆𝑬𝑬 ∆𝑬𝑬𝑲𝑲mechanical SI unit: joule 𝒏𝒏𝒏𝒏 𝑲𝑲 0can = += ∆𝑬𝑬 constant as the 0𝑲𝑲=total ∆𝑬𝑬 𝒑𝒑+ the displacement. 𝒑𝒑 𝑾𝑾𝒏𝒏𝒏𝒏𝑲𝑲 = ∆𝑬𝑬 ∆𝑬𝑬𝒑𝒑𝑲𝑲𝑲𝑲0 𝒑𝒑 forces force by SI unit: joule [J]the [J] equation can 𝑲𝑲 +can This equation 0 conservative = energy ∆𝑬𝑬 + ∆𝑬𝑬 conservation of used This equation (v constant), 𝒑𝒑an be=can used toThis solve Hence: be to solve of SI unit: joule [J] 𝑊𝑊 = 𝑊𝑊 + 𝑊𝑊 This equation can𝑐𝑐 0𝑲𝑲 = ∆𝑬𝑬 + object ∆𝑬𝑬𝒑𝒑 𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 𝑲𝑲 object moves, 𝑾𝑾cosine = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 is zero. of the This equation can be used to solve used to solve energybe problems be used to solve then: conservation of 𝑾𝑾 = ∆𝑬𝑬 + ∆𝑬𝑬 conservation of remains constant as 𝒏𝒏𝒏𝒏 solve the𝑲𝑲work- 𝒑𝒑 be usedUsing to SIangle unit: joule 0 = ∆𝑬𝑬𝑲𝑲 + ∆𝑬𝑬𝒑𝒑the provided that be used conservation ofto solve formed[J]by conservation and is called the conservation of 𝑷𝑷of 𝑭𝑭𝒗𝒗 energy equation can energy problems the object moves, 𝒂𝒂𝒂𝒂𝒂𝒂 = problems 𝒂𝒂𝒂𝒂𝒂𝒂 This Comme energy theorem: conservation of conservation of energy problems net work done by energy problems the line of action principle of and energy problems SI unit: [W] and is watt called the beproblems used to solve is called the energy 𝑊𝑊 This equation can 𝑐𝑐 + 𝑊𝑊 𝑛𝑛𝑛𝑛 = ∆𝐸𝐸 𝐾𝐾 be provided that the net energy problems and is called the and is principle called theof of conservation of principle and is 𝑊𝑊 called the of the force and external of work donenon-conby andused isconservation called the But: −∆𝐸𝐸 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 =conto and called principle ofissolve of conservation energy.principle of the 𝑝𝑝 the direction of ofprinciple servative forces is energy problems conservation of external nonprinciple of −∆𝐸𝐸 + 𝑊𝑊 = ∆𝐸𝐸 𝑝𝑝 𝑛𝑛𝑛𝑛 𝐾𝐾 servation of energy principle conservation of of the conservation of conservation of energy. the displacement. and is called energy. conservative forces zero. conservation of Hence: conservation energy. problems and isof energy. energy. 𝑾𝑾 = 𝑭𝑭∆𝒙𝒙 𝒄𝒄𝒄𝒄𝒄𝒄 𝜽𝜽 principle of is zero. energy. 𝑾𝑾 = ∆𝑬𝑬 + ∆𝑬𝑬 𝒏𝒏𝒏𝒏 principle 𝑲𝑲 energy. called the SI unit: joule [J] conservation of 𝒑𝒑 0 = ∆𝑬𝑬𝑲𝑲 + ∆𝑬𝑬𝒑𝒑 10 This equation can of energy. conservation of be used to solve 10 10 energy. 10 conservation 10 of 10 10 10 energy problems and is called the 10 principle of conservation of energy. 11 Commented [U31]: Formatting: pse remove bla Commented pages in text. [U31]: Formatting: pse remove bla pages in text. Commented [U31]: Formatting: pse remove blank sp Commented [U31]: Formatting: pse remove bla pages in text.in text. [U31]: Formatting: pse remove blan pages Commented pages in text. 12 12 Commented [U33]: Pse check. 3.5 3.5Doppler Effect Doppler Effect Sound Light dOPPLER EFFEcT 3.5 Doppler Effect dOPPLER EFFEcT Sound Light Theory Equations Applications Red shift Blue shift dOPPLER EFFEcT dOPPLER EFFEcT Sound Light Equations applications Red shift Blue shift Commented [U31]: Formatting: pse remove blan 3.5Theory Doppler Effect dOPPLER EFFEcT Sound Equations Light LightBlue shift applications Red shift pages in text. [U31]: Formatting: pse remove blank Commented 3.5Theory Doppler Effect Sound (dEFINITION) (DEFINITION) Sound Light Theory Equations applications Red shift Blue shift (dEFINITION) Theory Equations applications Red shift Blue shift pages in text. dOPPLER EFFEcT Theory applications Red shift shift ± 𝑣𝑣𝐿𝐿 (dEFINITION) The Doppler effect is a Equations 𝑣𝑣𝑣𝑣 ± If the light source is moving If the is source moving The (dEFINITION) Doppler effecteffect is Sound ais If the light source is movingBlue away fromsourceIf the • find To is moving towards the ob To thefind rate the 𝑣𝑣𝐿𝐿 ) 𝑓𝑓𝑠𝑠 The Doppler a 𝑓𝑓𝐿𝐿 = ( dOPPLER To find the EFFEcT rate If the light source is movingLight If the source is moving (dEFINITION) 𝑓𝑓 = 𝑓𝑓 ( ) 𝑣𝑣 ± 𝑣𝑣 𝐿𝐿 𝑣𝑣 ± 𝑣𝑣 𝑠𝑠 To find of blood flow Light 𝑠𝑠 𝑣𝑣 ± 𝑣𝑣 The Doppler effect is(or aSound If the light source is moving If the source is moving change in frequency The Doppler effect is a Equations 𝐿𝐿 the rate change in frequency the observer (positive velocity), then the rate of blood server (negative velocity), the observed away from the observer towards the observer If the light source is moving If the source is moving Theory applications Red shift Blue shift 𝐿𝐿 To find the rate 𝑣𝑣 ± 𝑣𝑣 of blood flow away from the observer change in effect frequency 𝑓𝑓𝐿𝐿 =𝑓𝑓(𝐿𝐿 = 𝑣𝑣 the is observer ( ±)𝑣𝑣𝑓𝑓𝐿𝐿𝑠𝑠𝑠𝑠 ) 𝑓𝑓𝑠𝑠 The Doppler is a Equations If theshift light source is moving Iftowards the moving Theory applications Red Blue shiftsource (Doppler To find the rate (dEFINITION) 𝑣𝑣 ± 𝑣𝑣 𝑣𝑣 ± 𝑣𝑣 of blood flow of blood flow 𝑠𝑠 (Dopplerflow (Doppler 𝑓𝑓𝐿𝐿 = ( )𝑠𝑠 𝑓𝑓 change in frequency away(positive fromfrom the observer towards observer pitch) of sound observed frequency and the the the frequency isthe higher and the wavelength (or pitch) of sound change in detectfrequency velocity), then theis lower (negative velocity), away the observer towards observer (dEFINITION) where (or pitch) of sound 𝑣𝑣 ± 𝑣𝑣𝑠𝑠 𝑠𝑠 (positive velocity), then the (negative the velocity), the ofscanning) blood flow away where (Doppler change in effect frequency fromsource the observer towards observer (Doppler where 𝑣𝑣 ± 𝑣𝑣 scanning) The Doppler is a If the light is moving If the source is moving 𝐿𝐿 To find the rate (oradetected pitch) ofby sound ed by listener because observed wavelength greater (red-velocity), scanning) is shorter (blue-shift). (positive velocity), then the a listener (or pitch) of sound observed frequency is lower observed frequency isthe higher (positive velocity), then the is(negative (negative velocity), the 𝑓𝑓𝐿𝐿 where = (𝑣𝑣where (Doppler To see the ± 𝑣𝑣𝐿𝐿 ) 𝑓𝑓𝑠𝑠 The Doppler is a the lightvelocity), source isthen moving If(negative the source is moving detected byeffect observed frequency is the lower observed frequency is higher scanning) find the rate scanning) To see the Ifaway 𝑣𝑣 = ± 𝑣𝑣 (or pitch) ofa listener sound ofTo blood flow (positive velocity), the 𝑠𝑠) 𝑓𝑓𝑠𝑠 𝑓𝑓 = ( 𝑣𝑣 𝑓𝑓𝑓𝑓 change in frequency from the observer towards the observer 𝐿𝐿 where scanning) the sound source and the shift). detected by aby listener observed frequency is wavelength lower frequency is higher detected a listener because the sound observed frequency is lower observed observed frequency higher and the observed and the wavelength isisshorter 𝑣𝑣 ± = 𝑣𝑣𝑓𝑓𝑓𝑓 blood flow To the unborn child of To see the 𝑠𝑠 change frequency away the observer towards the observer •see To see the becauseinbythe sound For a stationary (Doppler and from the observed wavelength and the wavelength is higher shorter Commented [U34]: Pse check – perhaps: For a s listener unborn child detected a listener observed frequency is lower observed frequency is 𝑣𝑣 stationary = 𝑓𝑓𝑓𝑓 Commented [U34]: Pse check – perhaps: For a s For a listener To see the 𝑣𝑣 = 𝑓𝑓𝑓𝑓 (or pitch) of sound (positive velocity), then the (negative velocity), the (Doppler listener: listener have different because the sound and the observed wavelength and the wavelength is shorter source and the listener is greater (red-shift). 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Red shifts are evidence that the than observer is higher 𝐿𝐿 the frequency of the the observer is higher For aFor source the shifts Universe is expanding. source away (v𝑣𝑣+moving: 𝑓𝑓 (𝑣𝑣moving: )𝑣𝑣𝑠𝑠𝑓𝑓)𝑠𝑠 than the frequency of the ± scanning) 𝐿𝐿a= Red evidence that the is higher 𝑓𝑓away +𝐿𝐿 )𝑣𝑣𝑠𝑠𝑓𝑓 ) towardsThe theobserver observer and isisare expanding. 𝑣𝑣moving: heard by For a𝐿𝐿 source = ( (v the Universe expanding. the Universe is expanding. 𝑠𝑠 thansource thefrequency frequency of the when the source than the frequency of the away (v + 𝑣𝑣 ) • away away (v + 𝑣𝑣 ) The frequency heard by source when the source towards (v 𝑠𝑠 − v 𝑠𝑠s ) the Universe expanding. than the frequency of the away lower the when the is source Red shifts areisevidence that towards − observer higher (v(v + 𝑣𝑣𝑠𝑠v) s ) source when the source For a source moving: moves towards the source when thehigher source Red shifts are evidence that towards (v − v ) the observer is moves towards the towards (v − v ) Listener moving: s s For a source moving: moves than away from the obsource when the source the Universe is expanding. thetowards frequency oflower the • Listener towards towardsmoving: moves observer and the moves towards the away (v(v+−𝑣𝑣𝑠𝑠v)s ) the Universe is expanding. the frequency the Listener moving: observer and oflower moving: Listener towards (v𝑣𝑣+) vL ) server.than moves towards the away (v + 𝑠𝑠 vL ) towards (v + source when thelower source moving: observer and observer and lower Listener when the source moves towards (v − source the source Listener towards (v +−− vvLvv+ )s)))vL ) when when the source moves towards towards (v away (v moving: Ls) observer and lower (v moves towards the away (v − v L towards (v + when the from source Commented [G32]: This is the correct word away themoves observer. when the source moves Listener moving: vL ) moves towards the away Commented [G32]: This is the correct word awaythe from the observer. (v − vL− ) vL ) away (v when source moves Listener moving: observer and lower Commented [U33]: Pse check. Commented [G32]: This isPse the check. correct word word away from the observer. towards away (v − Commented [G32]: This is the correct away fromand the observer. L )vL ) (vv+ Commented [U33]: observer lower • towards Commented [G32]: This is the correct word away the fromsource the observer. when moves towards (v + vL ) Commented [U33]: Pse check. Commented [U33]: Pse check. 12 when the source moves away (v − vL ) Commented [U33]: Pse check. 12 away (v − vL ) Commented [G32]: This is the correct word away from the observer. • away 12 12 Commented [G32]: This is the correct word away from the observer. Commented [U33]: Pse check. 12 3.5 Doppler Effect DOPPLER EFFECT 3.5 Doppler Effect 3.5 Doppler Effect 12 r2 F F r2 r2 3.6 Electrostatics Electrostatics3.6Force Electric field Electrostatics r r2 charge. 𝐸𝐸⃗⃗ 𝐸𝐸⃗⃗ 13 13 Principle of superposiPrinciple of superposiElectric field at a Lines point tion of forces tion of fields Electric field is an The electric field Electric field lines The force that a sysThe electric field Coulomb`s law ELECTROSTATICS ELECTROSTATICS area of space in at a point is the are IMAGINARY tem of point charges strength at a point Electric field atfield a at a Principle of Principle of The magnitude of the elec-Force which Electric Principle of Principle of due to a system of Electrostatics Electric field Lines an electric electrostatic LINES along which a exerts on another Electrostatics Force Electric field point Lines superposition of forces superposition of fields point superposition of forces superposition of fields trostatic force exerted bylaw one charge experiforce experienced small POSITIVE test point charge is equalfield point charges is equal coulomb`s Electric field is The electric field The force that a The electric Electric field lines are coulomb`s law Electric field is The electric field The force that a The electric field Electric field lines are point chargeThe (Q magnitude ) on anotherof the ences an area of space at a point is the system of point strength at a point IMAGINARY LINES a an force. per unit positive charge would move. toofthe vector addition the vector addition area of space at a point is the system point strength at ato point IMAGINARY LINES 1 The magnitude of the in which an ofan electrostatic charges exerts on the due todue aeach system ofof all force force along along which which a small point chargeelectrostatic (Q ) electrostatic is directly in which electrostatic charges exerts onforces to a system of the electric field a small The direction charge placed at of all 2 electric charge force force POSITIVE another point point charge point point charges is by onebyof point electric charge POSITIVE another charge charges is The test forcetest experienced proportionalexerted to theexerted product one point the electric field that point. one exerts on it. strengths of each one experiences a per per charge wouldwould move.move. is equal to theto vector equalequal to theto vector charge (Q1) on experiences aexperienced experienced charge is equal the vector the vector (Qanother ) on another by the positive test the magnitudes ofcharge the charges 1 a force. point is the atthe a specific point. The unit positive addition of all of theall the addition of all of theall force. The unit positive The force addition addition point charge (Q2) is(Q at The force point charge ) is direction charge inforces and inversely proportional to of theof the charge placed at forces each one electric field field experienced by is thealways 2direction in which direction charge placed at each one electric experienced by the directly proportional F2 + ... +F net = F1 + strengths directly proportional to electric field at a at that positive test charge exerts on it. F ofn each the direction of theexerts the square of the distance (r) toa positive electric a point. that point. positive test charge on it. strengths of each test field the product of the of the the product point point is theis the E =F F one at a specific is always in the in the F net F F ... F E net = E1 + E2 + ... + En one at a specific is always tangent to the field between them: 1 2F F n... F F magnitudes of the net F charge would 1 2 n magnitudes of the direction in direction of the of the point. E = Eq= direction in direction point. charges and inversely q line charges and inversely which a positive tangent to the to field q move if placed at E net EEnet EE ...E E which a positive tangent the field 1 2 n proportional to the to the test charge Fora apoint point 1 2 ... En For proportional line For a point test charge line that point. squaresquare of the of distance charge: the distance wouldwould movemove if They start They start on a onon charge: ifcharge: They start a a posi(r) between them: them: placed (r) between at thatat that positive charge and and tive charge placed positive charge and end point.point. KQ KQ end onend a negative KqQ on a negative E = on a negative charge. KqQ KqQ E2 = F= charge. ELECTROSTATICS 3.6 Electrostatics 13 Electric field pattern Positive like charges Unlike charges 14 I= I Rr R + r ε 15 15 15 15 15 15 15 3.7 3.7 Electric Circuits CURRENT EMF & Electric PARALLEL Ohm’s Law Work Power 3.7Electric Circuits Circuits SERIES 3.7 Electric Circuits 3.7 Electric Circuits EMF & cuRRENT SERIES PaRaLLEL Ohm’s Law work Power Commented [U39]: Formatting: pse ensure n cuRRENT SERIES SERIES PaRaLLEL PaRaLLEL Ohm’s Ohm’s Law Law work work Power Power Commented Commented [U39]: Formatting: pse EMFSERIES EMF & & cuRRENT PaRaLLEL Ohm’s Law work Power Commented [U39]: Formatting: pse [U39]: ensure no text isFormatti obscure EMF cuRRENT &POTENTIAL – pse check entire text – repetitive problem seen. POTENTIaL –repetitive pse check – pse entire check text entire – repetitive textis– obscure repetitiv probl – pse check entire text –Formatting: problem seen. cuRRENT SERIES PaRaLLEL Ohm’s Law work Power Commented [U39]: pse ensure no text & POTENTIaL POTENTIaL 3.7 ElectricEMF Circuits POTENTIaL DIFFERENCE dIFFERENcE Commented [G40]: corrected – pse check entire text – repetitive problem seen.[G40]: POTENTIaL dIFFERENcE Commented Commented [G40]: corrected correcte dIFFERENcE dIFFERENcE Commented [G40]: corrected cuRRENT SERIES PaRaLLEL Ohm’s Law work Power The rate Commented [U39]: Formatting: pse ensure no text is obscur EMF & atrate R1 The total Work done General: dIFFERENcE Commented [G40]: corrected The The at rate at R R The The total Work Work done done R1 General: General: Work done (energy General: The rate The total The rate at R R The total Worktotal done General: R 1 1 1 R – pse check entire text – repetitive problem seen. 3 2 R R R R R R R R POTENTIaL thatcharge W VItWThe 1 2 3 3 2R 3 1 The totalthatcharge done General: charge that(energy that(energy (energy which VIt Wrate VItatwhichwhichwhich at which charge (energy W VIt 1 transferred) per unit charge that Work R11 R22 R3 dIFFERENcE Commented [G40]: corrected electrical passes transferred) per Series: which charge that (energy electrical electricalelectrical passes passes transferred) transferred) per per W VIt electrical passes transferred) per Series: Series: Series: charge to movetothe passes through R2 2 work is The rate at R The total Work done General: through a unit charge W = VIt R R 1 2 is 2 R2 W I Rt 2 electrical work is is passes transferred) per 2 2 through through a to a unit charge unit R charge to to work R Series: through a unit charge R W which IWRt I Rtdone or work 1 WWIVItRt 3 I 2 charge from work is a conductor per charge (energy conductor per move the the charge I I R2 I 2 Parallel: work is through athat charge to per donedone or or conductor conductor per move move the charge the charge done or conductor per unit move the charge W I Rt Parallel: Parallel: Parallel: electrical electrical passes transferred) per unit of time. from the negative negative electrode to unit of time. Series: I 2 electrical electricaldone or conductor per move the charge unit ofunit time. of time.from the fromnegative the negative electrical unit of time. from negative 2 or Series: 2 V done 2 R3 R R The potential Parallel: R2 2 𝑄𝑄 the V V work is energy is V through a unit charge to electrode to the 3.7 Electric Circuits R t W 3 3 the positive electrode W I Rt 𝑄𝑄 𝑄𝑄 3 𝑄𝑄 electrical energy energy is iselectrical unit of time. 𝐼𝐼 = from the negative electrode to theto the TheW potential energy electrode to the electrode 2 t The potential The potential done t W t oris transferre V I RWenergy 𝐼𝐼to= R3 𝐼𝐼 = 𝑄𝑄 per electrode positive electrode conductor move charge difference across aR across Δ𝑡𝑡𝐼𝐼 =inthe R R Parallel: is the transferre transferre The potential positive positive electrode electrode the battery. energy is Commented [U39]: difference difference across a a transferre positive electrode difference across a t W Δ𝑡𝑡 cuRRENT Δ𝑡𝑡 SERIES PaRaLLEL Ohm’s Law work Power EMF & whole circuit: The potential difference 𝐼𝐼 = electrical d.= I 2 d. unit of Δ𝑡𝑡 time. from the negative in the in battery. conductor is directly 2circuit: Wcircuit: R t d. R whole whole circuit: whole transferre positive electrode theinbattery. thePOTENTIaL battery. difference across a conductor conductor is directly is directly d. in the battery. conductor is directly Δ𝑡𝑡𝑄𝑄 V transferred.– pse check entire text – R RRT R R RR2 𝑾𝑾 𝑊𝑊 energy to the The potential proportional to the across a circuit: conductor is to t to W W It whole R R312 RR23 3R3 𝑾𝑾 𝑾𝑾 R1 𝑊𝑊 RT13 R 𝑾𝑾isIt 𝑷𝑷 = d. 𝑊𝑊 inelectrode the battery. conductor is directly the the T1 proportional to proportional the proportional 𝐼𝐼 = 2 𝜀𝜀 = R𝜀𝜀T =𝑊𝑊 dIFFERENcE Commented [G40]: W W It W It R 𝜀𝜀 = I R I transferre 𝑷𝑷 = 𝑾𝑾Parallel: ∆𝒕𝒕𝑷𝑷 = 𝑷𝑷 = 𝜀𝜀 =electrode positive ina the differencecurrent across Δ𝑡𝑡 𝑞𝑞IT IR directly 𝑊𝑊 proportional current current in the inproportional the 2 I213 II321 I 2 current in the ∆𝒕𝒕 The ∆𝒕𝒕 rate at 𝑞𝑞21IIR 𝑞𝑞31 W It ∆𝒕𝒕General: 𝑞𝑞 total R R1to the The Work done 3 RI 3 1R 1 1 1 1 whole circuit: 𝑷𝑷 = = Potential 𝑷𝑷 = 𝑽𝑽𝑽𝑽 d. in the𝜀𝜀battery. conductor is directly conductor at theatcurrent 1 1in the R +2R21V + R 11 3 1 2 11 R3 1 𝑷𝑷 = 𝑽𝑽𝑽𝑽 𝑷𝑷 =which 𝑽𝑽𝑽𝑽 to conductor at in the 𝑷𝑷 = ∆𝒕𝒕 𝑽𝑽𝑽𝑽W VIt V V I2(energy conductor at1 conductor Potential T = 1 V Potential 𝑞𝑞charge thatIV 1R 32 1R 3VTR proportional VV23 R1EV3 1R1 R2 current V 𝑾𝑾 𝑊𝑊 Potential VR11I VTR 𝑽𝑽𝟐𝟐 toconstant the constant 𝟐𝟐 𝟐𝟐 T1 1 V1 is RRR3 R constant 𝟐𝟐 TTthe 22the 33VV 1 W It 𝑷𝑷 = 𝑽𝑽𝑽𝑽 conductor at 𝑽𝑽 𝑽𝑽 𝑽𝑽 Potential constant defference defference is the is R R R R 𝑷𝑷 = conductor at constant 𝜀𝜀 =defference defference is the R R R R passes VT V transferred) (Resistors act asact 𝑷𝑷 = 𝑷𝑷 = 𝑷𝑷 =electrical Series: E1 21 32 3 E as 1 as2 3E V (Resistors Vper current in the temperature. (Resistors act 𝟐𝟐 1 2 3 Potential (Resistors act as I I I ∆𝒕𝒕 𝑷𝑷 = 𝑞𝑞 work done per unit 𝑹𝑹 1 done 2 unit 3 unit 𝑽𝑽 R2 temperature. temperature. defference is the RE1 VR11 VV R1 VR2 VV 2 2 temperature. work work done per per work done through per unit a (Resistors unit charge 𝑹𝑹 work 𝑹𝑹 is potential 321 1constant temperature. V V W IV Rt 𝑽𝑽𝑽𝑽 act conductor 𝑷𝑷𝑷𝑷== 𝑹𝑹 dividers.) 1 V 32 1V3 1at potential dividers.) 1 2 321 V31 Potential =charge ItoV2dividers.) =potential I 3dividers.) charge between defference isper the V V VI temperature. VTbetween move Vbetween I1the Vas potential work done per unit charge charge charge between done or = + + 𝑹𝑹 conductor I I I I 𝟐𝟐 1 2 3 V V V V W = t T 1 2 3 𝑽𝑽Parallel: potential dividers.) constant II21R I 32 RRI 3 R R I1TRE I21R1 I32R2 IIR defference is points theper two in a points 1RR 3RT3E IIT work unit charge between points two in a the in two points indone a oftwo electrical unit time. from V (Resistors actanegative as 𝑷𝑷 = 2R 1 temperature. 2 3 (Resistors act as I I I I I circuit. work donein per unit R3 (Resistors T 3 𝑹𝑹 V I The I potential (Resistors act Iact as Ract as V1 V12 current V23 as(Resistors two points a between 𝑄𝑄 circuit. circuit. circuit. energy is charge twoelectrode to the + V potential dividers.) t W dividers.) WholeWhole circuit: 𝑊𝑊 V = V + V charge between 𝐼𝐼 = (Resistors act asV current current dividers.) dividers.) I current dividers.) V Whole circuit: circuit: T 1 2 3 Whole circuit: 𝑊𝑊 𝑊𝑊 Whole 𝑊𝑊 circuit. V R circuit: transferre = V = V positive electrode I I I I difference across a 𝑉𝑉circuit. = 𝑉𝑉 = 𝑉𝑉 = (Resistors act as Δ𝑡𝑡 points in a 1 2 3 T 1 2 3 A real battery has R potential 𝑉𝑉 = 𝑊𝑊in a two points current dividers.) 𝑄𝑄 Whole circuit:has A realAR battery real battery has hasis directly =conductor A real battery whole circuit: d. in 𝑄𝑄 the𝑄𝑄battery. (Resistors act as I resistance. internal resistance. 𝑉𝑉 = 𝑄𝑄 circuit. dividers.) A real battery has internal internal resistance. resistance. internal I R R R R 𝑾𝑾 𝑊𝑊 𝑄𝑄𝑊𝑊 proportional to the T 1 2 3 W It current dividers.) Whole circuit: 𝑷𝑷 = 𝜀𝜀 = internal resistance. I = I + I + I current in the 𝑉𝑉 = T 1 2 3 I I I ∆𝒕𝒕 𝑞𝑞 Whole circuit: I 1 has real battery 1 2 3 𝑄𝑄 IA 1 1 current 𝑷𝑷 = 𝑽𝑽𝑽𝑽 at Potential RI1r RI rRconductor act resistance. as VT V1 V2(Resistors V3 Iinternal r R r 𝑽𝑽𝟐𝟐 defference is the Rr1 R2 R3 A realconstant dividers.) RER battery has internal (Resistors act as 𝑷𝑷 = temperature. work done per unit 𝑹𝑹 IV1 V2 V3 potential dividers.) charge between RI r I I resistance.V I T 1 2 3 R two points in a (Resistors act as I circuit. current dividers.) Whole circuit: 𝑊𝑊 𝑉𝑉 = A real battery has 𝑄𝑄 internal resistance. 3.7 Electric Circuits • Strategy to Solve Problems on Electric Circuits • Read the problem carefully as many times as needed. • If not given, draw a circuit diagram. • Write down the data in symbolic form. • Indicate the conventional direction of the current from high-potential to low-potential (+ to -). • Re-draw the circuit diagram to simplify it if necessary. • Identify the type of connection (series/parallel). Analysing circuits: 1 – The algebraic sum of the changes in potential in a complete transversal of any loop of a circuit must be zero. (ε= Ir+ Ir) 2. The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction. • i = i1+i2+…+ in • Write down the formula/ equation that solves the question. • Find the unknowns if needed (multi-concept problems). • Do the calculations and write down the final answer. - Check your answer. Check if it makes sense, e.g.: - Is the unit correct? Is the numerical value reasonable? 15 16 Elements of the AC generator: Split-ring (commutator). Two poles of magnet (permanent magnets) 3. The thumb shows the direction of the force, the first finger shows the direction of current and the second finger shows the direction of the magnetic field. The thumb, the first finger and the second finger must be at a right angle to (900) each other. Fleming’s left hand (motor) rule Carbon brushes. 2. The DC generator or dynamo is similar to the AC generator, but the slip rings are replaced by split ring (commutator) to get direct current (current that flows in one direction). The DC generator (Dynamo) The thumb represents the direction of motion of the conductor. The first finger represents the direction of the magnetic field (north to south). The second finger represents the direction of the induced or generated current (the direction of the induced current will be the direction of conventional current - from positive to negative). Fleming’s right hand (dynamo) rule (for generators) 4. Two poles of magnet (permanent magnets). 3. Carbon brushes. 2. Slip-rings. 1. Armature. The generator is based on the principle of “electromagnetic induction” Elements of the DC-motor: Armature. Generators convert mechanical energy into electrical energy. Motors convert electrical energy into (rotational) mechanical energy. 1. AC and DC Generators Direct current motor (DC–motor) Electrical machines 3.8 Electrodynamics 17 18 18 18 18 18 Alternating current (AC) is an oscillating electric current thatthat varies sinusoidally with time, reversing its its direction of flow Alternating current (AC) is an oscillating electric current varies sinusoidally with time, reversing direction of flow aLTERNaTING cuRRENT aLTERNaTING cuRRENT periodically. periodically. Alternating current (AC) is an oscillating electric current that varies sinusoidally with time, reversing its direction of flow Alternating currentGraphs (AC) is an oscillating electric current that varies sinusoidally with time, reversing its direction Average of flow power rms values aLTERNaTING cuRRENT a periodically. periodically. Graphs a rms values Average Graphs aelectric current that varies sinusoidallyrms power Alternating current (AC) is an oscillating withvalues time, reversing its direction ofAverage flow power Advantages of AC over Advantages of AC overGraphs a Advantages of AC over DC periodically. rms values Average power Graphs values Average power DCDC Currenta versus time Voltage versus time rms Current Voltage Voltage Voltage Advantages Current versusCurrent time versus time Voltage versus timeversus time CurrentCurrent Voltage AdvantagesofofAC ACover over Graphs aversus time rms values Average power DC Current Voltage versus time Voltage Current versus time cycle Voltage versus time Current Current Voltage The root-meanThe rootForFor one complete For oneone complete 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 DC Easy to be transformed For one complete cycle • Easy to be transformed up ortransformed For one complete cycle The root-meanThe root-meanThe root-meanThe rootone complete cycle For complete = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 (step Easy to Advantages ofbe AC over square current mean-square cycle (step up or step down square current mean-square cycle step down usinga DC transformer). square current square voltage (step up or step down Current versus time Voltage versus time Current Voltage The root-meanThe rootFor one complete cycle For one complete 𝑃𝑃 = 𝐼𝐼 𝑉𝑉 totobe transformed 𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 The root-meanThe rootFor one complete cycle For one complete 𝑃𝑃 = 𝑉𝑉=𝑟𝑟𝑟𝑟𝑟𝑟 𝐼𝐼𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 Easy Easy be is the value voltage is the 𝑟𝑟𝑟𝑟𝑟𝑟𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 using a transformer). isvalue the value voltage is the using atransformed transformer). issquare the ofof of is the value of 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟𝐼𝐼𝑎𝑎𝑎𝑎𝑎𝑎 square current mean-square cycle (step up oror step down 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝑅𝑅 𝑅𝑅 current mean-square cycle (step up step down the current in value of the using Easier convert from ACFor the current in value of the 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 AC Easier to convert from AC one complete cycle • Easier to convert from tototransformed DC The root-meanThethe rootFor one complete 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 𝐼𝐼 𝑉𝑉 is value of isisthe Easy to be the current in voltage in𝐼𝐼 OR aatransformer). 𝑉𝑉= OR 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 isthe the value of anvoltage voltage using transformer). an AC circuit voltage inthe an 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟== 𝑅𝑅 to DC than from DCDC to to an AC circuit voltage in an𝑟𝑟𝑟𝑟𝑟𝑟 toto DC than from square current mean-square cycle the current in value of the (step up or step down than from DC to AC. 𝑅𝑅 𝑉𝑉 = 𝐼𝐼 𝑅𝑅 𝑅𝑅 Easier convert from AC AC circuit that an AC circuit 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 = 𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 the current in value ofcircuit the thatthat OR Easier to convert from AC that willwill have ACAC circuit AC.a that AC. OR= 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 is the value ofhavevoltage is the an AC circuit voltage in an using transformer). totoDC than from DC to 𝐼𝐼 will have the that will have anthe AC circuit voltage in an ==𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 than from DC to same will have thethe𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝑅𝑅𝐼𝐼 𝑅𝑅𝑅𝑅 𝑟𝑟𝑟𝑟𝑟𝑟 the same will have AC.Easier to generate. 𝑉𝑉 For DCEasier to generate. the current in value of the that will have AC circuit that Easier to convert from AC series 𝑟𝑟𝑟𝑟𝑟𝑟 • Easier to generate. For series that will have AC circuit that same heating the same heatOR𝑟𝑟𝑟𝑟𝑟𝑟 AC. heating effect same heating heating effect same heating OR It can be transmitted at an AC circuit voltage inthe an Itthan can be transmitted at the same will have DC from DC to connection: to Easier to generate. connection: the same will have the 𝑉𝑉 = 𝐼𝐼 𝑅𝑅 For series effect as a DC ing effect as a as a DC circuit. effect as a DC Easier to generate. 𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 as have a DC circuit. effect as a DC high voltage andand low For series that will AC circuit that high voltage heating same heating • It can be transmitted at high voltage AC. ItItcan be atatlow heatingeffect effect same heating connection: circuit. circuit. DC circuit. cancurrent betransmitted transmitted circuit. current over long connection: the samecircuit. will have a the For series overlow long as effect )2 𝑅𝑅 2 Easier todistances generate. 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎𝑃𝑃 = (𝐼𝐼𝑟𝑟𝑟𝑟𝑟𝑟 (𝐼𝐼connection: high voltage and and low current over long asaaDC DC circuit. effectas as aDC DC For series 𝑎𝑎𝑎𝑎𝑎𝑎 = 𝑟𝑟𝑟𝑟𝑟𝑟 ) 𝑅𝑅 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 high voltage and lowless distances with less energy heating effect same heating distances with energy circuit. 𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓𝑰𝑰𝒓𝒓𝒓𝒓𝒓𝒓 = = Itcurrent can beover transmitted at 2 circuit. with less energylost. connection: 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 𝑃𝑃𝑃𝑃 ==(𝐼𝐼(𝐼𝐼 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 2 current overlong long 𝑟𝑟𝑟𝑟𝑟𝑟 ) )𝑅𝑅 lost. OROR √𝟐𝟐 √𝟐𝟐effect 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 as a DC circuit. as a=DC 𝑅𝑅 lost. 𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟 𝑽𝑽𝒓𝒓𝒓𝒓𝒓𝒓 For𝑎𝑎𝑎𝑎𝑎𝑎 parallel connection: high voltage and lowenergy distances with less 𝑽𝑽 = 𝑰𝑰 For parallel connection: 𝒓𝒓𝒓𝒓𝒓𝒓 𝒎𝒎𝒎𝒎𝒎𝒎 𝑰𝑰 = distances with less energy OR where 𝒓𝒓𝒓𝒓𝒓𝒓 2 High frequency used in √𝟐𝟐 OR where 𝑽𝑽 circuit. 𝑰𝑰 = 2 High frequency used in 𝒎𝒎𝒎𝒎𝒎𝒎 √𝟐𝟐 𝒓𝒓𝒓𝒓𝒓𝒓 𝑉𝑉 2 current over long lost. 𝑽𝑽 𝑉𝑉 OR √𝟐𝟐 𝑟𝑟𝑟𝑟𝑟𝑟 (𝐼𝐼connection: 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = • High frequency usedlost. in AC makes it = 𝒎𝒎𝒎𝒎𝒎𝒎 For 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 OR √𝟐𝟐 𝒓𝒓𝒓𝒓𝒓𝒓Where 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 𝑽𝑽Where 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = )=𝑅𝑅 𝑟𝑟𝑟𝑟𝑟𝑟 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 AC makes suitable for for 𝑽𝑽 Forparallel parallel connection: 𝑃𝑃𝑟𝑟𝑟𝑟𝑟𝑟 AC makes it suitable 𝒓𝒓𝒓𝒓𝒓𝒓 = √𝟐𝟐 𝑎𝑎𝑎𝑎𝑎𝑎 OR where distances withitless energy 2 𝑰𝑰 = High frequency used in 𝑅𝑅 𝑰𝑰 = For parallel connection: 𝒎𝒎𝒎𝒎𝒎𝒎 𝑰𝑰 = OR where 𝑅𝑅 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹 2 suitable for motors. 𝒓𝒓𝒓𝒓𝒓𝒓 High frequency used in √𝟐𝟐 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 =𝒎𝒎𝒎𝒎𝒎𝒎 𝑰𝑰= 𝑹𝑹 𝑹𝑹 motors. 𝑽𝑽 𝑽𝑽 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 𝒎𝒎𝒎𝒎𝒎𝒎 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝑹𝑹 Where motors. 𝑽𝑽√𝟐𝟐 lost. 𝒎𝒎𝒎𝒎𝒎𝒎 OR 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 AC 𝑽𝑽 𝑽𝑽𝒓𝒓𝒓𝒓𝒓𝒓 =𝒎𝒎𝒎𝒎𝒎𝒎 For parallel Where 𝒎𝒎𝒎𝒎𝒎𝒎 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎=connection: = 𝑅𝑅 ACmakes makesititsuitable suitablefor for OR 𝑰𝑰 = 𝒎𝒎𝒎𝒎𝒎𝒎 where 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑹𝑹 𝑽𝑽𝑽𝑽 𝑹𝑹 2 𝑅𝑅 High frequency used in OR motors. 𝒎𝒎𝒎𝒎𝒎𝒎 ==𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 OR 𝑉𝑉𝑟𝑟𝑟𝑟𝑟𝑟 𝑰𝑰√𝟐𝟐 𝑹𝑹 motors. 𝒎𝒎𝒎𝒎𝒎𝒎 𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 Where 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 = AC makes it suitable for 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑅𝑅 𝑽𝑽𝒎𝒎𝒎𝒎𝒎𝒎 = 𝑰𝑰𝒎𝒎𝒎𝒎𝒎𝒎 𝑹𝑹 𝑹𝑹 motors. cuRRENT Alternating current (AC) isaLTERNaTING anaLTERNaTING oscillatingcuRRENT electric current that varies sinusoidally with time, reversing its direction of flow periodically. ALTERNATING CURRENT 18 Photoelectric Effect 19 19 Photoelectric EffectOPTIcaL PHENOmENa 3.9 3.9Photoelectric Effect OPTIcaL PHENOmENa ENERGy RadIaTION EFFEcT SPEcTRa 3.9 Photoelectric EffectPHOTOELEcTRIc OPTICAL PHENOMENA OPTIcaL PHENOmENa ENERGyENERGY RadIaTION PHOTOELEcTRIc EFFEcT SPEcTRa Theory Einstein’s equation Emission RADIATION PHOTOELECTRIC EFFECT absorption SPECTRA 3.9 Photoelectric Effect ENERGy RadIaTION PHOTOELEcTRIc EFFEcT SPEcTRa Theory Einstein’s absorption Emission effect: This Theory is the E = WoOPTIcaL +equation Ek(max) PHENOmENa The energy radiated by hot Photoelectric An atomic An atomic Einstein’s equation Absorption Emission process whereby electrons are ejected Theory Einstein’s equation absorption Emission Photoelectric effect: This is the E = W The energy radiated by hotin ENERGy An atomic o + Ek(max) objects is liberated the emission spectrum RadIaTION PHOTOELEcTRIc EFFEcT SPEcTRa The energy radiated by hot obEAn = Wabsorption + Eatomic An atomic absorption An atomic emission specPhotoelectric effect: process whereby from a metal surface are when light This of ais the where 𝑬𝑬 = 𝒉𝒉𝒉𝒉 o k(max) process whereby electrons ejected Photoelectric effect: This is the E = W + E The energy radiated by hot An atomic An atomic o k(max) objects is liberated in the absorption emission spectrum [U43]: Pse check. form of separate packets of suitable frequency is incident on that spectrum is is formed when electrons are ejected from a metal surface when light of Theory Einstein’s equation absorption Emission jects is liberated infrom the a form spectrum is formed when Commented trum is formed when certain metal surface whenprocess light ofwhereby a where 𝑬𝑬 = 𝒉𝒉𝒉𝒉 electrons are ejected Where surface. a suitable frequency is incident on that surface. Commented [G44]: It is exactly like in exam guide lines objects is liberated in the absorption emission spectrum hc Commented [U43]: Pse check. form of separate packets of suitable frequency is incident on that spectrum is is formed when energy called quanta. formed when certain Photoelectric effect: This is the E = W + E Theof energy radiated by hot An atomic An atomic o k(max) from a metal surface when light of a of separate packets energy certain frequencies of frequencies of electromagwhere 𝑬𝑬 = 𝒉𝒉𝒉𝒉 Threshold frequency ( fo ) is the OR E surface. Commented [G44]: It is exactly like in exam guide lines electrons ejected Commented [U43]: Pse ch formisof separate packets of process suitable whereby frequency is incident on that formed spectrum of is emission is formedspectrum when hc are energy quanta. A called quantum of energy certainwhen certain frequencies objects is liberated in the absorption called quanta. electromagnetic radiation netic radiation are emitted minimum frequency light to minimum freThreshold frequency (surface f )Eis the Threshold frequency ( of fsurface. ) is the OR from a needed metal when light of a where 𝑬𝑬 = 𝒉𝒉𝒉𝒉 o Commented [G44]: It is ex o hc energy called quanta. formed when emit frequency electrons aneeded certain A quantum of energy isform certain frequencies of that pass called photon. of electromagnetic quencyoffrom ofoflight light needed to emit electrons certain Commented [U43]: Pse ch of separate packets suitable frequency is incident spectrum isthrough iscertain formed whendue to an atom’s electrons Efrequencies OR a mediThreshold frequency (from fo )aon is that the minimum tometal maximum kinetic OR surface. surface. quantum energy calledof metal electrons fromsurface. aiscertain metal A emit quantum energy certain are frequencies Commented [G44]: It is ex energy ) to frequencies calledA photon. of electromagnetic The energy ofofa photon is:is called electromagnetic radiation hc frequency of light( E needed K (max) energy formed certain um when (e.g. a cold gas) are ofmaking a transition from a maximum kinetic Workquanta. function ( W o Threshold )minimum is the minimum E OR frequency ( f ) is the surface. Maximum kinetic energy ( o emit electrons from a certain metal photon. called photon. frequencies of electromagnetic ( EK (max) ) The energy electromagnetic radiation are due 𝑬𝑬 = 𝒉𝒉𝒉𝒉of a photon is:A Work radiation that emitted to an kinetic energy an )function electron in( the absorbed. quantum of that energy is the certain frequencies ofhigh-energy state to a lower function (W minimum Work W frequency )metal isenergy the minimum energy that minimum of light needed to maximum surface. o 2 o metal needs to be emitted from the E energy ( ) m v The energy of a photon is: electromagnetic radiation are energy state. 𝑬𝑬 = 𝒉𝒉𝒉𝒉 that emitted due to an max the e metal ) K (max) Eradiation electrons electron in the needsfrom be from pass through a atom's electrons Work function (toWE )aemitted iscertain the minimum energy that ananelectron inemit themetal metal called photon. of electromagnetic K (max) o K (max) 2 maximum kinetic frequencies The energy of a photon is: surface. surface. needs to be emitted fromenergy the metal mein v maxthe2 metal energy metal surface. 𝑬𝑬 = 𝒉𝒉𝒉𝒉 Effect radiation that radiation emitted due that an pass medium through a atom's electrons a making a transition The energy of The the ( EK (max)2 )(e.g. of intensity on the energy of a photon is: Work electromagnetic are to an E K( electron function W o )is the the minimum (max) surface. needs to be emitted from metal m v 2 photoelectric effect. The number of Commented [G45]: Better to include it e max a pass through atom's electrons medium a transition The The energy the cold (e.g. gas)2 making are from a high-energy energyof is Effect radiation isof the 𝑬𝑬 = 𝑵𝑵𝑵𝑵𝑵𝑵 of intensity on that the 𝑬𝑬radiation =Effect 𝒉𝒉𝒉𝒉 radiation thata emitted due to an of intensity onan the photoelectric ef- E work in function energy electron the metal K (max) surface. electrons emitted per unit time is m2 v max Commented photoelectric effect. The number of Commented [G45]: [U46]: Better toFormatting: include it pse standardise font type, medium (e.g. aa atom's makingelectrons a transition energy of The the needs to be emitted from the Effect of work intensity on the E m v 2eare fect. number ofincident electrons emitted permetal unit cold absorbed. gas) from astate high-energy radiation is 𝑬𝑬 =number 𝑵𝑵𝑵𝑵𝑵𝑵The where Nisisthe the to athrough lower where N ofnumber proportional the intensity of bold/not bold, etc. with all equivalent items, including paragrap pass function (max) = e max K(max) E electrons emittedto per unit time is K Commented [U46]: Formatting: standardise font[G45]: type, size, surface. photoelectric The of Commented Better text, text in tables, headings,pse sub-headings, etc. is𝑵𝑵𝑵𝑵𝑵𝑵 proportional to theeffect. intensity W0 h of fnumber light. is time 2 2 state energy 0 incident cold gas) from a bold/not high-energy radiation 𝑬𝑬 =intensity emitted. absorbed. tomedium a lower whereof Nphotons is theemitted. number ofThe state. work function photons proportional to the of incident bold, etc. with all equivalent items, including paragraph (e.g. are a making a transition energy of the Effect of intensity on the electrons emitted per unit time is Commented [U46]: Forma Effect light. of frequency on the text, text in tables, headings, sub-headings, etc. h fThe light. N is the number of photoelectric 0 intensity 0 absorbed. stateatohigh-energy a lower number of Work function Commented [G45]: Better where toWeffect. the of incident bold/not energy cold state. photons emitted. photoelectric effect.proportional The Commented [G47]: Better include it bold, etc. with all equ gas) are from radiation is frequency 𝑬𝑬 = 𝑵𝑵𝑵𝑵𝑵𝑵 work function Effect of on maximum theemitted OR text, text in tables, headings, s W h f electrons per unit time is The power of the radiating light. Commented [U46]: Format kinetic energy of of the frequency photoelectrons Effect onis the photoelectric W00 = h 0f 0 energy state. photons emitted. The power of where the photoelectric effect. maximum absorbed. state toCommented a lower [G47]: Better include itbold/not bold, etc. with all equi Nlinearly is the proportional number ofThe proportional the intensity ofon incident Effect of toOR frequency the to the frequency of sources: effect. The maximum kinetic energy of the Commented [G50]: Better in thistext wayin tables, headings, su kinetic energy of the photoelectrons is text, hc W photoelectric effect. Commented [G47]: Better 0 h f0 The radiating power sources: of thephotons the incident radiation.light. OR energy state. W0 The maximum photoelectrons is linearly proportional to the linearlyemitted. proportional to the frequency of Effect of frequency on the OR kinetic energy of the photoelectrons is Commented [G50]: Better in this way hc Significance photoelectric 0 The power of of thethe ∆𝑵𝑵 radiating the incident radiation. frequency of the incident radiation. Weffect. photoelectric Thefrequency maximum Commented [G47]: Better i 𝑷𝑷 = sources: 𝒉𝒉𝒉𝒉 linearly proportional of OR 0 to the effect. The photo-electric effect Commented [G48]: Better include it ∆𝒕𝒕 Commented [G50]: Better hc Significance of the photoelectric 0 ∆𝑵𝑵 radiating sources: kinetic energy of the photoelectrons is the incident radiation. W establishes the quantum theory and Theeffect. power of the 0 𝑷𝑷 = ∆𝒕𝒕 𝒉𝒉𝒉𝒉 Commented [U49]: Pse standardise appearance of all The Significance photo-electric Commented [G48]: Better include it ofeffect the photoelectric effect. 0 to thephotoelectric frequency of Significance ∆𝑵𝑵 illustrates the particlelinearly nature proportional of light. of the equivalent items, including hyphens or not – variations Commented [G50]:seen. Better i hc 𝑷𝑷 = 𝒉𝒉𝒉𝒉 establishes the quantum theory and radiating sources: Commented [U49]: Pse standardise appearance of all The photo-electric effect establishes effect the W0 the incident radiation. effect. The photo-electric Commented [G48]: Better ∆𝒕𝒕 illustrates the particle nature of light. equivalent items, including hyphens or not – variations seen. 0 Significance of quantum thethe photoelectric establishes the theory naand quantum theory and illustrates particle ∆𝑵𝑵 Commented [U49]: Pse st 𝑷𝑷 = ∆𝒕𝒕 𝒉𝒉𝒉𝒉 effect. The photo-electric effect Commented [G48]: Better illustrates the particle nature of light. equivalent items, including hy i ture of light. establishes the quantum theory and Commented [U49]: Pse sta 19 illustrates the particle nature of light. equivalent items, including hyp 19 3.9 4 Revision Questions- Set 1 4.1 Newton’s Laws Strategies for problem-solving questions on Newton’s Laws Step 1: Read the problem as many times as needed. Step 2: Sketch the problem, if necessary. Step 3: Draw a force diagram for the situation. Step 4: Draw a free-body diagram and resolve the forces into component on the Cartesian plane. Step 5: List all the given information and convert these to SI units, if necessary. Step 6: Determine which physical principle can be used to solve the problem. Step 7: Use the principle to solve 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. 19 Worked Example 1 A block of mass M is held stationary by a rope of negligible mass. The block rests on a frictionless plane that is inclined at 30o to the horizontal. Worked Example 1 A block of mass M is held stationary by a rope of negligible mass. The block rests on a frictionless plane that is inclined at 30o to the horizontal. 1.1 State Newton’s first law of motion in words. 1.2 Draw a free-body diagram that shows all the forces acting on the block. 1.3 Calculate the magnitude of the gravitational force of the block when the force in the rope is 8 N. 1.4 1.2 Draw a free-body shows all theon forces on the block. reaching the horizontal rough Now the rope is cut and thediagram block that slides down the acting inclined plane, 1.3 Calculate the magnitude of the gravitational force of the block when the force in the rope-2is 8 N. surface and then continuing to move with constant acceleration of magnitude 1,5 m·s . 1.1 State Newton’s first law of motion in words. 1.4 Now the rope is cut and the block slides down on the inclined plane, reaching the horizontal rough 1.4.1 -2 and then continuing to move withinconstant Statesurface Newton’s second law of motion words.acceleration of magnitude 1,5 m·s . Commented [U53]: Pse ch as the editor is an English langu There are many problems with specialist must check the revis ensure it is also technically cor Comment applies to all sugges 1.4.1 State Newton’s second law of motion in words. 1.4.2 Calculate the acceleration of block the block whileititis is sliding sliding down thethe inclined plane.plane. 1.4.2 Calculate the acceleration of the while down inclined 1.4.4. Calculate the coefficient of kinetic friction between the block and the surface. 1.4.4 Calculate the coefficient of kinetic friction between the block and the surface. Solutions for Worked Example 1 Example 1 Solutions for Worked 1.1 An object continues a state inofa rest uniform motion (motion constant velocity), it is 1.1 An objectincontinues state or of rest or uniform motion (motion with with aaconstant velocity), unless unless it acted on by an(net unbalanced (net or resultant) acted on by anisunbalanced or resultant) force. force. To calculate the gravitational we can use 1.3 1.3 To calculate the gravitational force, we canforce, use the the parallel component: parallel component: 𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 We do not have the component, but we can calculate this We do not have the component, but we can calculate this by applying Newton’s second law in are going to calculate only the magnitude, because the the x direction. We are going to calculate only the direction is known. magnitude, because the direction is known. by applying Newton's second law in the x direction. We 21 On the x-axis: On the x-axis: 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝑇𝑇 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0 𝑇𝑇 − 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 = 0 8 − 𝐹𝐹⃗ 𝑔𝑔 sin 30𝑜𝑜 = 0 𝐹𝐹𝐹𝐹 = 8 sin 30𝑜𝑜 𝐹𝐹𝐹𝐹 = 16 𝑁𝑁 1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant force. The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. 201.4.2 Free body diagram, coordinate system and components. Applying Newton’s second law of motion Commented [G54]: Better like this 𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝑜𝑜 8 −𝑅𝑅𝑅𝑅𝐹𝐹⃗ 𝑔𝑔 sin 30 =0 𝑇𝑇 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0 8 𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin𝑜𝑜 𝜃𝜃 = 0 𝑇𝑇 − sin 30 𝐹𝐹𝐹𝐹 =𝐹𝐹⃗16 𝑁𝑁 30𝑜𝑜 = 0 8− 𝑔𝑔 sin 8 𝐹𝐹𝐹𝐹 = If a𝑜𝑜 resultant force acts on a body, it will cause the body to accelerate in the direction of the resultant 1.4.1 sin 30 1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the The acceleration of the body will be directly proportional to the resultant force and inversely 𝐹𝐹𝐹𝐹 = 16force. 𝑁𝑁 resultant force. The acceleration of the body will be directly proportional to the resultant force and proportional to the mass of the body. inversely proportional to the mass of the body. 1.4.1 If a resultant force acts on a body, it will cause the body to accelerate in the direction of the 1.4.2 Free body diagram, coordinate system and components. 1.4.2 diagram, coordinate system and components. resultantFree force.body The acceleration of the body will be directly proportional to the resultant force and inversely proportional to the mass of the body. Applying Newton’s second law of motion 1.4.2 Free body diagram, coordinate system and components. 1.4.3 1.4.3 Applying Newton’s second law of motion Free body diagram 1.4.3 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ (1) There is only one force acting on the block, Applying Newton’s which is the frictionalsecond force: law of motion 𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗ (2) 𝐹𝐹⃗ = 𝑚𝑚𝑎𝑎⃗ (1) 𝜇𝜇𝐾𝐾𝑅𝑅𝑁𝑁 = 𝑚𝑚𝑚𝑚 (3) There is only one force acting on the block, The normal force is unknown and we need to which is the frictional force: calculate it. 𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗ (2) (3) 𝜇𝜇𝐾𝐾 𝑁𝑁 = 𝑚𝑚𝑚𝑚 The normal force is unknown and we need to calculate it. ⃗⃗ 𝑁𝑁 Free body diagram ⃗⃗ 𝑁𝑁 𝑓𝑓⃗𝑓𝑓 𝐹𝐹⃗𝑔𝑔 Commented [G54]: Better like this 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ Applying in the x direction ApplyingitNewton’s second law of motion 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗ 𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚30 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ = 𝑚𝑚𝑎𝑎⃗ 𝑜𝑜 𝑎𝑎⃗Applying = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔30 it in the x direction 𝑜𝑜 𝑎𝑎⃗𝐹𝐹⃗= = 9,8𝑚𝑚𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠30 ⃗ 𝑔𝑔𝑔𝑔 𝑎𝑎⃗𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚30 = 4.9 m𝑜𝑜∙ 𝑠𝑠=−2𝑚𝑚𝑎𝑎⃗ 𝑎𝑎⃗ = 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔30𝑜𝑜 𝑎𝑎⃗ = 9,8 𝑠𝑠𝑠𝑠𝑠𝑠30𝑜𝑜 𝑎𝑎⃗ = 4.9 m ∙ 𝑠𝑠 −2 Free body diagram 𝐹𝐹⃗𝑔𝑔 Commented [G54]: Better like this 𝑓𝑓⃗𝑓𝑓 working working in in the the y-direction: y-direction: Commented Commented [U55]: [U55]: Pse Psecheck check––perhaps perhaps“Work “Workininthe thedire dire of ofy.” y.” Pse Psecheck checkall allsuch suchinstances. instances. working in thevertical y-direction: [U55]: Pse[U55]: check – perhaps “Work in the“Work direction working in thedirection, y-direction: Commented Pse check – perhaps in the dire In the the of In the vertical direction, the acceleration acceleration is is zero, zero, therefore therefore we we can can apply apply Newton’s Newton’s first first law law22 ofCommented of y.” of y.” Pse check all such instances. motion. Pse check all such instances. motion. Working in the y-direction: In thevertical vertical direction, the is acceleration iswe zero, therefore canlaw apply Newton’s first law of motion. In the vertical direction, the acceleration zero, is therefore can we apply first of law In the direction, the acceleration zero, therefore canNewton’s applywe Newton’s first of motion.motion. ⃗𝟎𝟎⃗⃗⃗ ⃗𝑭𝑭⃗𝑭𝑭 ⃗⃗𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) = 𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) = 𝟎𝟎 22 ⃗⃗⃗⃗ ⃗⃗⃗ ⃗⃗𝒈𝒈 = 𝑵𝑵 𝟎𝟎 ⃗⃗⃗ 𝑵𝑵 + + ⃗𝑭𝑭⃗𝑭𝑭 𝒈𝒈 = 𝟎𝟎 ⃗ ⃗+ ⃗y-positive ⃗𝑭𝑭⃗𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚) Taking ⃗𝟎𝟎⃗ ⃗⃗⃗ ⃗⃗ direction ⃗⃗⃗ ⃗𝑭𝑭⃗𝒈𝒈 = 𝟎𝟎 =⃗𝑭𝑭⃗⃗𝟎𝟎 𝑵𝑵 + ⃗𝑭𝑭⃗𝒈𝒈 the = = positive 𝑵𝑵 𝟎𝟎 𝒏𝒏𝒏𝒏𝒏𝒏(𝒚𝒚)as Taking as the y-positive direction Taking aspositive positive the y-positive direction Commented Commented [U56]: [U56]: Pse Psecheck. check. Commented Commented [G57]: [G57]: Beter Beterlioke liokethis this Taking 𝑵𝑵 as the𝑵𝑵 y-positive Taking as𝟎𝟎𝟎𝟎positive the = y-positive direction 𝑭𝑭𝑭𝑭𝒈𝒈 = − 𝑵𝑵 − −positive = 𝑵𝑵 − 𝒎𝒎𝒎𝒎 𝒎𝒎𝒎𝒎 = 𝟎𝟎𝟎𝟎direction 𝒈𝒈 𝑵𝑵 − 𝑭𝑭𝒈𝒈 = 𝑵𝑵 − =− 𝟎𝟎 𝒎𝒎𝒎𝒎 = 𝟎𝟎 𝑵𝑵 = 𝟎𝟎 𝒎𝒎𝒎𝒎 𝑵𝑵 𝒈𝒈 𝒎𝒎𝒎𝒎 ∴∴𝟎𝟎− 𝑵𝑵 𝑵𝑵𝑭𝑭= = 𝒎𝒎𝒎𝒎 (4) (4) Substituting 3: 4 in 3:Hence Substituting 𝜇𝜇𝜇𝜇 𝐾𝐾𝐾𝐾𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚4= =in𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚 Hence 𝜇𝜇𝜇𝜇𝐾𝐾𝐾𝐾𝑔𝑔𝑔𝑔 = = 𝑎𝑎𝑎𝑎 4 3: in 3:(4) ∴ 𝑵𝑵 = Substituting 𝒎𝒎𝒎𝒎 ∴ 𝑵𝑵 = 𝒎𝒎𝒎𝒎 44 in Substituting Substituting in 3: Commented [U56]: Pse[U56]: check. Pse check. Commented Commented [G57]: Beter lioke this Commented [G57]: Beter lioke this (4) 𝜇𝜇𝐾𝐾 𝑚𝑚𝑚𝑚 =Substituting 𝜇𝜇𝐾𝐾 𝑔𝑔 = 𝑎𝑎𝜇𝜇𝐾𝐾 𝑔𝑔 = 𝑎𝑎 𝜇𝜇𝑚𝑚𝑚𝑚 HenceHence 𝐾𝐾 𝑚𝑚𝑚𝑚 = 𝑚𝑚𝑚𝑚Hence Substituting Substituting 𝜇𝜇Substituting (9,8) = = 1,5 1,5 𝜇𝜇𝐾𝐾𝐾𝐾(9,8) Substituting 𝜇𝜇𝜇𝜇𝐾𝐾𝐾𝐾 = = 0,153 0,153 = 𝜇𝜇𝐾𝐾 (9,8) Exercise (9,8) =11𝜇𝜇 1,5 𝜇𝜇𝐾𝐾 = 0,153 𝜇𝜇𝐾𝐾1,5 𝐾𝐾 = 0,153 Exercise Exercise AExercise A 1block block of of1mass mass 10 10 kg kg is is pulled pulled to to the the right right at at constant constant velocity velocity with with aa force force of of magnitude magnitude 30 30 N N that that oo acts at an angle of 37 with the horizontal. acts at an angle of 37 with the horizontal. A block A of block mass of 10mass kg is 10 pulled the right at constant velocity velocity with a force magnitude 30 N that kg istopulled to the right at constant with of a force of magnitude 30 N that acts at an angle of angle 37o with theo with horizontal. acts at an of 37 the horizontal. ⃗⃗𝑭𝑭 ⃗⃗ = 𝑭𝑭 = 𝟑𝟑𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝟑𝟑 10 10 kg kg 10 kg 1.1 1.1 10 kg 37o o 37 ⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑𝑭𝑭 ⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑 37o𝑭𝑭 37o Draw Draw aa free-body free-body diagram diagram with with all all the the force force action action on on the the block. block. Use Use aa Cartesian Cartesian plane plane and and 21 uting ) = 1,5 se 1 𝜇𝜇𝐾𝐾 = 0,153 Exercise 1 k of mass 10 kg is pulled to the right at constant velocity with a force of magnitude 30 N that block of mass 10 kg is pulled to the right at constant velocity with a force of magnitude 30 N that acts at an an angle of 37o A with the horizontal. angle of 37o with the horizontal. 10 kg ⃗⃗ = 𝟑𝟑𝟑𝟑𝟑𝟑 𝑭𝑭 37o Draw a free-body withaall the force action on the block. Cartesian plane 1.1diagram Draw free-body diagram with all Use the aforce action onand the block. Use a Cartesian plane and represent represent the component of component the force. the of the force. Calculate the magnitude and direction of the normal force. Calculate the coefficient of friction. the magnitude and direction of the normal force. 1.2 Calculate The force is removed and the block comes to rest after a time. Calculate the acceleration of 1.3forceCalculate the coefficient of friction. the block after the has been removed if the frictional force remains constant. 1.4 The force is removed and the block comes to rest after a time. Calculate the acceleration of the block after the force has been removed if the frictional force remains constant. 23 Exercise 2 Exercise 2 o o to the30 horizontal. A block, massis20pulled kg, is pulled upwards force 𝐹𝐹⃗ along with an incline alonga plane a plane with an 30 incline to the horizontal. It A block, of mass 20of kg, upwards bybyforce -22 m·s-2. The surface of the incline exerts a friction force on the It accelerates along the incline plane at accelerates along the incline plane at 2 m·s . The surface of the incline exerts a friction force on the block. block. 2.1 Draw a diagram showing all the forces acting on the block (force diagram) as it moves up the incline. 2.1 aDraw a diagram showingfor all the acting on the block diagram) as Cartesian it moves up the Also draw free-body diagram theforces block. Represent the(force forces on the plane and resolve incline. Also draw a free-body diagram for the block. Represent the forces on the Cartesian the gravitational force into components. plane and resolve the gravitational force into components. 2.2 Calculate component of the gravitational force exerted theblock block parallel toincline. the incline. 2.2 the Calculate the component of the gravitational force exertedon on the parallel to the 2.3 2.3 Calculate the normal force. Calculate the normal force. 2.4 Calculate the magnitude of the friction force if the magnitude of is equal to 200 N. 2.5 Calculate the coefficient of kinetic friction for the block. 2.5 Calculate the coefficient of kinetic friction for the block. Calculate the magnitude of the friction force if the magnitude of 𝐹𝐹⃗ is equal to 200 N. 2.4 4.2 Momentum and Impulse 4.2 Momentum and Impulse Momentum and impulse problem solving strategy Momentum and impulse problem solving strategy 1 Read 1the Read problem as many you to; to; then draw ordiagram diagram situation described the problem as time many as time as need you need then drawaasketch sketch or of of thethe situation Commented [U58]: Formatting: Pse all equivalent items, including line spaci described if not provided. if not provided. etc. 2 2 From the reading, collect the data and write it in symbol form (not only numbers are part of the data) e.g. the object starts moving from rest means the initial speed is zero. Commented [U59]: Not understood 3 Draw a freecollect body diagram for each define the (not system younumbers are going toare work with. Commented From the reading, the data andobject writeand it inclearly symbol form only part of the data) e.g.[G60]: It could be inclu If possible, choose a system that is isolated ( F net 0 ) and closed (m = constant). If the the object starts moving from rest means the initial speed is zero. interactions are sufficiently short and intense you can ignore external forces. If it is not possible to choose an isolated system, try to divide the problem into parts (scenarios). Commented [U61]: Pse introduce b 3 to ensure legibility. Draw a free body diagram for each object and clearly define the system you are going to work statement with. applies Comment to all such instances. • Commented [U62]: Formatting: pse numbers, including font type, font size, alignment, etc. Repetitive problem seen If choose the mathematical is based law closed of conservation of momentum (m = constant). If the interactions If possible, a systemrepresentation that is isolated ( F net in = 0the) and , write it in component form. p p initial and are sufficiently short finalintense you can ignore external forces. • the text, as the editor will not know all o If it is not possible to choose an isolated system, try to divide the problem into parts (scenarios). 4 Select the law (principle), theorem, equation or formula that will answer your question. 5 Substitute the values into the equation or formula (the system of units must be homogeneous). Commented [U63]: Pse check all sc 6 Check your answer Is the value of the answer reasonable? Is the unit correct? Commented [G64]: I did the correc worked example 1 22 24 4 Select the law (principle), theorem, equation or formula that will answer your question. If the mathematical representation is based in the law of conservation of momentum • (∑ p initial = ∑ p final ), write it in component form. 5 Substitute the values into the equation or formula (the system of units must be homogeneous). 6 Check your answer • Is the value of the answer reasonable? • Is the unit correct? Worked example 1 Trolley A, with a mass of 120 kg, is moving at 20 m·s-1 ; it collides head-on with trolley B, with a mass of 150 kg, moving at 30 m·s-1 in the opposite direction. If A rebounds at 25 m·s-1: State the law of conservation of momentum in words. 1.1 Trolley A, with a mass of 120 kg, is moving at 20 m·s-1 ; it collides head-on with trolley B, with a mass -1 in the opposite direction. If A rebounds at 25 m·s -1: of kg, moving at 30 m·s 1.2150 the rebound velocity of-1 B. Trolley A, withCalculate a mass of 120 kg, is moving at 20 m·s ; it collides head-on with trolley B, with a mass of 150 kg, moving at 30 m·s-1 in the opposite direction. If A rebounds at 25 m·s -1: 1.1 State the law of conservation of momentum in words. 1.3 Calculate the velocity impulse 1.2 Calculate the rebound of of B. the force exerted by trolley B on trolley A. 1.1 1.3 State the law the of conservation of force momentum in by words. Calculate impulse of the exerted trolley B on trolley A. 1.2 Calculate the rebound velocity of B. 1.3Solution: Calculate the impulse of the force exerted by trolley B on trolley A. Solution: Solution: data Data data ma = 120 kg ma =𝒗𝒗 kg m · 𝑠𝑠 −1 to the right ⃗⃗ 120 = 20 𝒊𝒊𝒊𝒊 −1 ⃗⃗𝒊𝒊𝒊𝒊 m =B20 m · 𝑠𝑠kg to the right 𝒗𝒗 = 150 150 kg = 30 m · 𝑠𝑠 −1 to the left mB =𝒗𝒗 ⃗⃗(𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃)𝑩𝑩 ⃗⃗(𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃)𝑩𝑩 = 30 m · 𝑠𝑠 −1 −1 to the left 𝒗𝒗 ⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 to the left 𝒗𝒗 ⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 −1 to the left 𝒗𝒗 ⃗⃗(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂)𝑩𝑩 =? 𝒗𝒗 ⃗⃗(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂)𝑩𝑩 =? 𝒗𝒗 1.1 The total linear momentum of an isolated system is constant. 1.11.1 TheThe total linear momentum of an isolated system is constant. total linear momentum an isolated system is constant. 1.2 When we work in Physics it is veryof important to represent a problem using a diagram; here we are working with vector quantities, so it is important to indicate direction with all elements (positive or 1.2 When we work in Physics it is very important to represent a problem using a diagram; here we are negative) 1.2 When work in Physics it is very important represent problem(positive using aordiagram; here we are working working withwe vector quantities, so it is important to indicate to direction with alla elements negative) with vector quantities,Positive so it direction is important to indicate direction with all elements (positive or negative) y y direction -1 VA = 20 m.sPositive VB = 30 m.s-1 VA = 20 m.s-1 VB = 30 m.s-1 Before B VA = 25 m.s-1 B A A VA = 25 m.s-1 B A A After B x After x Before In order to find the velocity of trolley B after the collision, the law of conservation of momentum must be applied. Learners' attention must be addressed to the general characteristics of this kind of problem, In order to find velocity of trolleyof B trolley after theBcollision, thecollision, law of conservation momentum mustofwe i.e.order when anthe interaction between two bodies takes Firstly, wethe have to identify which bodies In to find the velocity afterplace. the law ofof conservation momentum must be applied. be applied. Learners' attention be addressed to theimportant general characteristics of this kind of problem, are going to consider as ourmust “system”. This is a very step, since momentum is only conserved attention musttwo beIn addressed to the characteristics of this kind of problem, i.e. when an i.e.Learners’ when an interaction between bodies takes place. Firstly, weBhave to identify which Once bodies in closed and isolated systems. this problem, trolleys Ageneral and integrate the system. wewe have are identified going to consider as of ourthe “system”. This is a very important step, since between momentum is only conserved the bodies system, we have to identify the interaction the bodies of the system interaction between two bodies takes trolleys place. AFirstly, we have to identify which bodies we are going to consider in closed and isolated with systems. In this problem, and integrate the system. Once we have a and the interaction external bodies. Learners know thatBwhen we have any kind of interaction, identified the bodies of the system, we have to identify the interaction between the bodies of the system force will appear (as a measurement of the intensity of any particular interaction). During the collision, andthe thetwo interaction with external bodies. Learners knowmove that when wethe have anyand, kindofofcourse, interaction, a trolleys interact with the surface where they on, with Earth with each Commented [G65]: corrected force will appear (as aacting measurement of the intensity of any particular interaction). other. The forces on each trolley are represented in the next diagram: During the collision, Commented [U66]: Pse clarify (surface of what?) the two trolleys interact with the surface where they move on, with the Earth and, of course, with each Commented [G65]: corrected 23 other. The forces acting on each trolley are represented in the next diagram: Commented [U66]: Pse clarify (surface of what?) 25 as our “system”. This is a very important step, since momentum is only conserved in closed and isolated systems. In this problem, trolleys A and B integrate the system. Once we have identified the bodies of the system, we have to identify the interaction between the bodies of the system and the interaction with external bodies. Learners know that when we have any kind of interaction, a force will appear (as a measurement of the intensity of any particular interaction). During the collision, the twoytrolleys interact with theFsurface where FBA AB they move on, with the Earth and, of course, with each other. The forces acting on eachNtrolley are represented A y FBA y in the next diagram: FAB FBAFAB yy y NNA FAB FAB y A yy y NA AA y FBA FBA FAB A BB FFABABFABNA FFBABAFBA NA NB NNAANA NNB BNB F F F BA BAAB NNB B A B A NA NB BB BNB AA A A B FAB FgA NB A NB B B FgA B FgB FBA x FgB x FgA F NgB xx x FFN A BFF FgA FgB gAgA gBgB xx FgA FgB FgA FgB wA wB x FgA FgB x FgA FgB B A w wB A wA wB wwAw wwBw A A B B wwA A wwB B wA wB The forces FAB and FBA are the forces exerted by the trolleys on each o wA wB x F F and gA FBA are the forces exerted by the gB trolleys on each other during the interaction, so The are forces FABthe FBA The forces FAB and forces exerted by the trolleys on each other during the interaction, so trolleys on FFABAB FFBABA The forces are the forces the each other during the interaction, so exerted by Fand FBA The The forces forces and and are are the the forces forces exerted exerted byby the the trolleys trolleys ontrolleys on each each other other during during the the interaction, interaction, so so forces N A and Nthey they are internal forces. The so-called “norma AB The forces and are the forces exerted by the on each other during the interaction, so are B are N N and they are internal forces. The forces are so-called “normal forces”, that is, the reaction and during the internal forces. A interaction, B “normal the are FF exerted by the on each so Nother Nduring they The forces so-called andFF are theforces forces exerted by thetrolleys trolleys on each interaction, so forces”, that is, the reaction AB BA Aother B arethe ABand BA are N N and they are internal forces. The forces are so-called “normal forces”, that is, the reaction Fare FBAforces. forces exerted by the surface where the trolleys move[G67]: on. These NAN NBeach NBtrolleys and and they internal internal forces. The forces forces are are so-called so-called “normal “normal forces”, forces”, that that is, is,the the reaction reaction The and are theThe forces exerted by the on other during the interaction, Fthey The forces FAB andforces are forces exerted by trolleys on other during the interaction, so forces exerted by the surface where the trolleys move on. These two forces are insoequilibrium with Commented Better like two this AN AB B ABthe BA forces. w weach internal The forces and so-called forces”, that the reaction forces exerted N exerted the surface where trolleys move on. These two forces in with Commented [G67]: Better likeby this B“normal inequilibrium is, the are surface Awhere Nexerted Nby and forces are so-called “normal forces”, that is, the reaction A Nexerted NBby andby einternal internalforces. forces.The Theforces forces are so-called “normal forces”, that is, the reaction two are A Aexerted Bby forces the the trolleys move on. These forces are equilibrium with Commented [G67]: Better like this forces forces the the surface surface where where the the trolleys trolleys move move on. on. These These two two forces forces are are inis, inequilibrium equilibrium with withforces Commented Commented [G67]: [G67]: Better Better likelike thisthis Pse check. Commented [U68]: N N and they are internal forces. The forces are so-called “normal forces”, that the reaction F F and . Finally, the gravitational force applied by the Earth on trolleys A and B, w N N and they are internal forces. The forces are so-called “normal forces”, that is, the reaction F the gravitational force applied by the Earth on trolleys A and B, A Commentedforce [U68]: Pse check. A forces Bare gA gB Better Aon. gA Fwith F and . Finally, forces gravitational force applied byBmove thetwo Earth on trolleys A and B, with w A xerted by where the trolleys move Commented [G67]: like this exerted bythe thesurface surface where the trolleys move on.These These two forces areininequilibrium equilibrium the the surface where the trolleys on. These two forces are in equilibrium with the gravitational applied Commented [G67]: Better like this gA gB Commented [U68]: Pse check. Commented Commented [U68]: [U68]: PsePse check. check. FFgAequilibrium Fare Finally, forces the gravitational applied by the Earth trolleys AAThese and B, wwAw the on the onon Fand FgBgB F.gBin and and .with Finally, .equilibrium Finally, forces forces the the gravitational gravitational force force applied applied byby the Earth Earth trolleys trolleys Aand and B, B, A A [G67]: exerted the surface where trolleys move two forces with Commented [G67]: Better like this force surfacebywhere forces exertedforces by the the trolleys move on. These twoon. forces are in Commented Better like this gA gA Commented [U68]: Pse check. Commented [U68]: Pse check. and are the forces applied by the trolleys on the horizontal surface (sometimes called weights). w Ftrolleys FFgBthe B forces tational force by trolleys AA and B, wwother FgAgAand and .Finally, Finally, forces vitational forceapplied applied byEarth the Earth on trolleys and B,exerted AA FBA FEarth The forces andon are the forces by on each during the interaction, so and the forces applied by the trolleys the horizontal surfac w B .are by the trolleys A and and .trolleys Finally, forces are the forces applied by theonPse gB and the forces applied by B, the on the horizontal surface (sometimes weights). wthe B are and .B, AB on w A wcalled Commented check. Commented Pse check. [U68]: Fand forces the gravitational force by the Earth on trolleys A B,surface w A trolleys Fhorizontal FgB and .F Finally, forces the gravitational force by theapplied A and w force trolleys Earth on and are the forces applied by the trolleys the (sometimes called weights). AB wwBw [U68]: the on and and are are the the forces forces applied applied by by the trolleys trolleys on on the the horizontal horizontal surface surface (sometimes (sometimes called called weights). weights). gA and gB Finally, gA Bapplied B F F N N All the external forces ( , , and ) are in equilibrium, so there is no net external gA gB A B F F N N All the external forces ( , , and ) are in equilibrium, so there is no net external force NB are so-called they aretrolleys internalon forces. The forces “normal forces”, that is,forces the reaction are by the (sometimes called arethe theforces forcesapplied applied bythe the trolleys on thehorizontal horizontal surface (sometimes calledweights). weights). N All the external forces the horizontal surface weights). Allso the external (force inequilibrium, so gB gA N A and B surface A (sometimes A , NB , FgA and FgB )) are in F,gAWe Fcalled the external forces ( (N , ,AN ,the and )the ininequilibrium, there is net external w acting Bthe forces on All the Fon FgB Fhorizontal N N All All the external external forces forces ( AN ,BBN ,the and )are are ) horizontal are inequilibrium, equilibrium, so(sometimes so there there isno isnono net net external external force force gB and are the applied by trolleys surface weights). and w B are forces applied by the trolleys on surface (sometimes gA gAand gB A BF on the trolleys. can then say that: exerted by the surface where the trolleys move on. These twocalled forcesweights). are in called equilibrium with Commented [G67]: Better like this on the trolleys. We can then say that: F F Nforces xternal forces , and ) are in equilibrium, so there is no net external force F F NAA, ,acting N external forces( (Nequilibrium, , and ) are in equilibrium, so there is no net external force gB gAgAon so BB acting onWe thecan trolleys. then say that: gB there isWe no net external force acting on theso trolleys. then We saycan that: acting the trolleys. can then say that: acting acting on the the trolleys. trolleys. We We can then say say that: that: FgB N Ncan All the external forces , then insoequilibrium, is noforce net external force All the external forces (onN , FgA( applied and )Fare in equilibrium, there netthere external gA and A, F Bby Commented [U68]: Pse check. gB Bforce A, N the gravitational the Earth on) are trolleys A andisB,noF n the then say gA and FgB . Finally, forces w A on thetrolleys. trolleys.We Wecan can then saythat: that: ⃗ ⃗ onWe thecan trolleys. We can acting on the acting trolleys. then say that:then say that: 0 = 𝑭𝑭 ∑ 𝒆𝒆𝒆𝒆𝒆𝒆 ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0 ∑on and w B are the forces applied by the trolleys the horizontal surface (sometimes called weights). ⃗𝑭𝑭⃗⃗𝑭𝑭 ⃗𝒆𝒆𝒆𝒆𝒆𝒆 ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 00 0 === ∑ ∑ 𝒆𝒆𝒆𝒆𝒆𝒆 ∑ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0 ∑ ⃗ ⃗ ⃗ ⃗ 0 = 𝑭𝑭 ∑ 0 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 𝒆𝒆𝒆𝒆𝒆𝒆 (=N A∆𝑝𝑝 All the external∑ forces , ⃗NB ,⃗⃗FgA and FgB ⃗⃗) are in0equilibrium, so there is no net external force ∆𝑝𝑝⃗ ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 = 0 ∑ 𝑭𝑭𝒆𝒆𝒆𝒆𝒆𝒆 = Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 Since: = ⃗ ⃗∆𝑝𝑝⃗ ∆𝑝𝑝 ∆𝑡𝑡 then say that: ∆𝑝𝑝 acting on We can ⃗∑ ∆𝑡𝑡trolleys. ⃗𝑒𝑒𝑒𝑒𝑒𝑒 ⃗the Since: ∑ Since: 𝐹𝐹 = ∆𝑝𝑝⃗ ∑ Since: Since: 𝐹𝐹 𝐹𝐹 = = 𝑒𝑒𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒 ∆𝑡𝑡∆𝑡𝑡∆𝑡𝑡 ⃗⃗ ∆𝑝𝑝 ∆𝑝𝑝 Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = = = 𝑒𝑒 ∆𝑡𝑡 𝑒𝑒𝑒𝑒𝑒𝑒 ∆𝑡𝑡∆𝑡𝑡 ∆𝑝𝑝⃗ ∆𝑝𝑝⃗ Then: Commented [U69]: Pse check all for ⃗ Since: ∑ 𝐹𝐹𝑒𝑒𝑒𝑒𝑒𝑒 = Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒Then: = Commented [U69]: Pse check all formulas and equations i ∆𝑡𝑡 ∆𝑡𝑡 text, Pse including brackets (1 bracket and 2 Then: Commented [U69]: check allall and equation ∑ ⃗𝑭𝑭⃗𝒆𝒆𝒆𝒆𝒆𝒆 = 0 Then: Then: Commented Commented [U69]: [U69]: PsePse check check all formulas formulas and equatio equa text, including brackets (1 bracket and 2formulas brackets), as and the edito Then: Then: not know what isand correct and what not. including (1(1bracket and 22brackets), asasthe ede text, text, including including brackets brackets (1 bracket bracket and brackets), 2 brackets), as thethe Commented and equations inin Commented[U69]: [U69]:Pse Psecheck checkallallformulas formulas and equations nottext, know what isbrackets correct and what not. Content editing isnot. required. not know what isrequired. and what not. ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 Then: Commented [U69]: Pseand check all formulas not not know know what what iscorrect correct is correct and and what what not. text, (1(1bracket 2 2brackets), as the editor will Commented Pse check all formulas equations in and equations text,including includingbrackets brackets bracketand and[U69]: brackets), as the editor will Content editing is ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 Content editing is2isrequired. ∆𝑝𝑝⃗ ∆𝑝𝑝 text, brackets bracket andeditor 2 brackets), Content Content editing editing required. is (1 required. not and what text, including brackets (1including bracket and brackets), as the will as the edito ⃗𝑛𝑛𝑛𝑛𝑛𝑛 00 0 === notknow knowwhat whatisiscorrect correct and whatnot. not. ⃗𝑛𝑛𝑛𝑛𝑛𝑛 ⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑝𝑝 Since: ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = ∆𝑝𝑝 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 not know what not. is correct and what not. Content not know what is correct and what Contentediting editingisisrequired. required. ∆𝑡𝑡 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 0− 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 0 ⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑝𝑝 ⃗𝑛𝑛𝑛𝑛𝑛𝑛==00 ∆𝑝𝑝 ⃗ 0 ∆𝑝𝑝 = ⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 00 0 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 −− == Content editing is required. 𝑛𝑛𝑛𝑛𝑛𝑛 Content editing is required. ⃗0𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑝𝑝𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑝𝑝 =𝑝𝑝− ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ==00 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇−−𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ==𝑝𝑝= ⃗ ⃗ ⃗ ⃗ 𝑝𝑝 𝑝𝑝 Then: [U69]: Pse check all formulas and equations ⃗ ⃗ ⃗ ⃗ + 𝑝𝑝 = 𝑝𝑝 + 𝑝𝑝 𝑝𝑝 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗ ⃗ 0Commented − 𝑝𝑝 = 𝑝𝑝 ⃗⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 = 0 = 𝑝𝑝⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝𝑝𝑝 𝑝𝑝⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 0 𝑝𝑝⃗+𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 − 𝑝𝑝⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 =− 𝑝𝑝⃗=𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝⃗𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇= ⃗𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 text, including brackets (1 bracket and 2 brackets), as the edit ⃗𝑝𝑝𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗⃗𝑝𝑝𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑝𝑝𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝+ == 𝑝𝑝 ++ 𝑝𝑝⃗𝑝𝑝𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑚𝑚 ⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝑝𝑝⃗𝑝𝑝⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝑝𝑝⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝⃗𝐴𝐴𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑣𝑣+ +=𝑚𝑚 𝑣𝑣= =𝑝𝑝+ 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚𝐴𝐴+ 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗ 𝑝𝑝 = ⃗ ⃗ ⃗ ⃗ ⃗ ⃗ 𝑚𝑚 𝑣𝑣 + 𝑚𝑚 𝑣𝑣 𝑚𝑚 𝑣𝑣 + 𝑚𝑚 𝑣𝑣 𝑝𝑝 = 𝑝𝑝 ⃗ ⃗ 𝑝𝑝 = 𝑝𝑝 not know what is correct and what not. 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗𝑚𝑚 ⃗+ + 𝑝𝑝⃗𝑝𝑝 ⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)= ⃗𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗+𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)++𝑝𝑝⃗𝑝𝑝𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) =𝑝𝑝 𝑝𝑝 +𝑝𝑝𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑝𝑝𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑣𝑣𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑣𝑣𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚 𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ==𝑚𝑚 ++𝑚𝑚 𝑚𝑚 ⃗𝑝𝑝 ⃗𝐵𝐵𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐴𝐴𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐵𝐵𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) =𝑚𝑚𝐴𝐴𝑚𝑚 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵𝑚𝑚 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚 𝐵𝐵𝑚𝑚 Content editing is required. 𝐴𝐴𝑣𝑣 𝐴𝐴⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵 𝐴𝐴𝑣𝑣 𝐵𝐵𝑣𝑣 0𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = ⃗ ⃗ ⃗ + 𝑝𝑝 = 𝑝𝑝 + 𝑝𝑝 ⃗ ⃗ ⃗ ⃗ + 𝑝𝑝 = 𝑝𝑝 + 𝑝𝑝 𝑝𝑝 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗ ⃗ ⃗ + 𝑝𝑝 = 𝑝𝑝 𝑝𝑝 𝑣𝑣⃗𝑣𝑣 𝑣𝑣⃗𝑣𝑣 ++𝑚𝑚𝑚𝑚𝐵𝐵𝐵𝐵𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)++𝑚𝑚𝑚𝑚𝐵𝐵𝐵𝐵𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏)==𝑚𝑚𝑚𝑚 ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐴𝐴 When the data inwe the thesubstituting in the problem, have: ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)we will have: 𝑚𝑚 𝑣𝑣⃗ +data 𝑚𝑚𝑝𝑝⃗𝐵𝐵𝑚𝑚 𝑣𝑣⃗given = 𝑣𝑣⃗𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) +will 𝑚𝑚problem, +𝑚𝑚 𝑚𝑚𝐴𝐴𝐵𝐵𝑣𝑣⃗𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) +𝑚𝑚 𝑚𝑚𝐴𝐴𝐵𝐵given 𝑚𝑚𝐴𝐴 𝑣𝑣⃗When 𝐵𝐵 𝑣𝑣 ⃗ 0 − = 𝑝𝑝 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎)substituting 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 ⃗ ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚 𝑣𝑣 = 𝑚𝑚 When substituting the data given ininthe problem, we will have: 𝐴𝐴 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴 𝑣𝑣 When When substituting substituting the the data data given given inthe the problem, problem, we we will will have: have: ⃗ ⃗ 𝑝𝑝 = 𝑝𝑝 tuting the we will have: 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 tituting thedata datagiven givenininthe theproblem, problem, we will have: (120)(20) = (120)(−25) + (150)(−30) = 150𝑣𝑣 + 150𝑣𝑣⃗ (120)(20) (120)(−25) ⃗𝐵𝐵(𝑎𝑎+ + (150)(−30) + When substituting the given in we the problem, we𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) will have:𝐵𝐵(𝑎𝑎 When substituting the data in data the have: 𝑝𝑝⃗(120)(−25) =will 𝑝𝑝⃗in 𝑝𝑝⃗(150)(−30) 𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) (120)(20) (150)(−30) When substituting the+problem, data the we willWhen have:substituting the data given in the problem, we will have: ⃗+ ++given = + 150𝑣𝑣 (120)(20) (120)(20) (150)(−30) (120)(−25) ⃗problem, ⃗150𝑣𝑣 + = =(120)(−25) + + 150𝑣𝑣 150𝑣𝑣 𝐵𝐵(𝑎𝑎 2400 −given 4500 = −300 𝐵𝐵(𝑎𝑎 𝐵𝐵(𝑎𝑎 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗ 2400 − 4500 = −300 + 150𝑣𝑣 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵= 𝑣𝑣⃗=𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 𝑚𝑚 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵−1𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚+𝐴𝐴+𝑣𝑣 ⃗𝐵𝐵(𝑎𝑎 150𝑣𝑣 (150)(−30)==(120)(−25) (120)(−25) ⃗− +(150)(−30) 150𝑣𝑣 𝐴𝐴 𝐵𝐵(𝑎𝑎 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 2400 4500 −300 ++ 150𝑣𝑣 ⃗ ⃗ 2400 2400 − − 4500 4500 = −300 −300 + 150𝑣𝑣 150𝑣𝑣 ⃗ 𝑣𝑣 = +6 𝑚𝑚 ∙ 𝑠𝑠 −1 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) (120)(20) + (150)(−30) ⃗ =+(120)(−25) (120)(20) (150)(−30) (120)(−25) ⃗ 𝑚𝑚 ∙ 𝑠𝑠+ 150𝑣𝑣 150𝑣𝑣 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = +6 −1 −1𝐵𝐵(𝑎𝑎 ⃗= 2400 −−4500 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 2400 4500=+ =−300 −300++150𝑣𝑣 150𝑣𝑣 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ==𝐵𝐵(𝑎𝑎 +6 𝑚𝑚𝑚𝑚∙𝑚𝑚 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑣𝑣 =+6 +6 ∙𝑠𝑠𝑠𝑠∙−1 𝑠𝑠 (120)(20) + (150)(−30) = (120)(−25) + 150𝑣𝑣⃗𝐵𝐵(𝑎𝑎 2400 −the 4500 = −300 ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 2400 −+6 4500 −300 + 150𝑣𝑣 −1 −1 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) we will have: When substituting data given + in 150𝑣𝑣 the ⃗problem, 𝑣𝑣⃗𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ==+6 𝑚𝑚𝑚𝑚∙ ∙𝑠𝑠𝑠𝑠= ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) Since the is positive, wetrolley can say thatmove trolley will move direction in the original direction of trolley ⃗a Since the answer positive, can B will in B the original trolley 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = +6 𝑚𝑚say ∙ 𝑠𝑠 −1that 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) =is+6 𝑚𝑚 answer ∙ 𝑠𝑠 −1we 2400 −of4500 =a −300 + 150𝑣𝑣𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) Since the answer isispositive, we can say that trolley BBwill move ininthe original direction ofofof trolley aaa before the collision. Since Since the the answer answer ispositive, positive, we we can can say say that that trolley trolley Bwill will move move inthe the original original direction direction trolley trolley before the collision. (120)(20) (150)(−30) (120)(−25) ⃗ + = + 150𝑣𝑣 ⃗ 𝑣𝑣 = +6 𝑚𝑚 ∙ 𝑠𝑠 −1 𝐵𝐵(𝑎𝑎 nswer is positive, we can say that trolley B will move in the original direction of trolley a answer is positive, we before can say that trolley B will move in the original direction of trolley a the collision. 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) before before the the collision. collision. Since the answer is positive, we can say that trolley B will move in the original direction of trolley a Since the answer is positive, we can say that trolley B will move in the original direction of trolley a 2400 − −300 + original 150𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) he hecollision. collision. =4500 6original 𝑚𝑚 ∙ = 𝑠𝑠 −1 in the direction of trolley a 𝑣𝑣⃗in 𝑓𝑓𝑓𝑓the = 6 𝑚𝑚 ∙the 𝑠𝑠 −1−1 direction of trolley a before collision. −1 before the𝑣𝑣⃗𝑓𝑓𝑓𝑓 collision. 𝑣𝑣⃗original =direction +6 𝑚𝑚 ∙ of 𝑠𝑠of ==6=6𝑚𝑚6 ininthe direction trolley aaa 𝑣𝑣⃗𝑣𝑣𝑓𝑓𝑓𝑓 𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝑓𝑓𝑓𝑓 𝑚𝑚∙𝑚𝑚 ∙𝑠𝑠𝑠𝑠∙−1 𝑠𝑠 −1 inthe the original original direction oftrolley trolley 𝑣𝑣⃗𝑓𝑓𝑓𝑓 Since the answer is positive, we can say that trolley B will move in the −1 −1in the originalSince the direction ofofanswer trolley aais positive, we can say that trolley B will move in the original direction of trolley A before the 𝑠𝑠 in the original direction trolley 1.3 −1 −1 before the collision. 6 𝑚𝑚 ∙ 𝑠𝑠 indirection the original direction 𝑣𝑣⃗𝑓𝑓𝑓𝑓 in = the original of trolley a of trolley a 𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 1.3 Since 1.3 1.3 1.3the answer is positive, we can say that trolley B will move in the original direction of trolley a collision. before the collision. 𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 −1 in the original direction of trolley a 1.3 1.3 26 26 −1 in the theoriginal original direction of trolley A direction of trolley a 𝑣𝑣⃗𝑓𝑓𝑓𝑓 = 6 𝑚𝑚 ∙ 𝑠𝑠 in 26 2626 26 26 1.3 26 26 1.3 Then: 1.3 ma = 120 kg mA = 120 kg ⃗⃗𝒊𝒊𝒊𝒊 = 20 m · 𝑠𝑠 𝒗𝒗 −1 26 to the right ⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 −1 to the left 𝒗𝒗 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 =? Let’s take positive to the right: 24𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = ∆𝑝𝑝 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝑚𝑚(𝑣𝑣𝑓𝑓 − 𝑣𝑣𝑖𝑖 ) Commented [U70]: Pse check. ⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) = 25 m · 𝑠𝑠 𝒗𝒗 to the left ⃗⃗𝒊𝒊𝒊𝒊 = 𝒗𝒗 20 m · 𝑠𝑠 −1 to the right 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 =? ⃗⃗𝑨𝑨(𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂) 𝒗𝒗 = 25tomthe · 𝑠𝑠 −1 to the left Let’s take positive right: Commented [U70]: Pse check. 𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰𝑰 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = take ∆𝑝𝑝 =?positive to the right: Let’s Let’s − 𝑣𝑣𝑖𝑖 ) to the right: 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = take 𝑚𝑚(𝑣𝑣𝑓𝑓positive Commented [U70]: Pse check. 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 ∆𝑝𝑝 − 20) 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 120=(−25 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 − ∙𝑣𝑣𝑠𝑠𝑖𝑖 ) 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼= =−5400 𝑚𝑚(𝑣𝑣𝑓𝑓 𝑁𝑁 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 ∙ 𝑠𝑠 𝑡𝑡𝑡𝑡 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼= =5400 120𝑁𝑁(−25 − 𝑡𝑡ℎ𝑒𝑒 20)𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = −5400 𝑁𝑁 ∙ 𝑠𝑠 Worked Example 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 =2 5400 𝑁𝑁 ∙ 𝑠𝑠 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 -1 Two Worked cars (S andExample T) are travelling 2 on a straight road. They approach a robot at velocities of 10 m·s East and 18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction. worked Example 2 Commented [U71]: Formatting: pse standardise appearance of all equivalent items, including font colour. Commented [G72]: Better like this Two cars (S and T) are travelling on a straight road. They approach a robot at velocities of 10 m·s-1 East and 18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction. Commented [U71]: Formatting: pse standardise appeara all equivalent items, including font colour. Commented [G72]: Better like this Car T suddenly stops and car S collides with car T. After the collision, the two cars move off together -1 cars (S andmass T) are travelling straight road. as a Two unit. The combined of each car withon theadriver is 1500 kg. They approach a robot at velocities of 10 m·s East and 18 m·s-1 East, respectively, as shown in the sketch below. Ignore the effect of friction. 2.1 State the law of conservation of linear momentum in words. (2) 2.2 Calculate the speed of the immediately the collision. (4) off together Car T suddenly stops andtwo car cars S collides with carafter T. After the collision, the two cars move Research has shown that a force greater than 000the N during collision fatal the injuries. as aT unit. The combined mass of each car85with driver 1500 kg. may Car suddenly stops and car S collides with carisa T. After the cause collision, two cars move off together as a The collision described above lasts for 0,08 s. unit. The the combined mass of each car with theindriver 1500 kg.in a fatal injury.(4) 2.3 Determine, by means of calculations, the collision aboveis could result 2.1 State law of conservation of whether linear momentum words. (2) 2.2 Calculate the speed of the two cars immediately after the collision. (4) The cars have crumple zones, belts, air bags interiors that canmay reduce thefatal chance Research has shown thatseat a force greater thanand 85 padded 000 N during a collision cause injuries. 2.1 State theduring law of conservation (2) of death serious injury accidents. Commented [U73]: Part of 2.3 or 2.4? Theorcollision described above lasts for 0,08 s.of linear momentum in words. 2.4 Use of by Physics toof explain how airwhether bags can reduce theabove risk ofcould injuryresult or death. 2.3 principles Determine, means calculations, the collision in a fatal injury.(4) 2.2 Calculate the speed of the two cars immediately after the collision. (4) Solution The cars have crumple zones, seat belts, air bags and padded interiors that can reduce the chance of death or serious injury during accidents. Commented [U73]: Part of 2.3 or 2.4? 2.1 The momentum of an and closed system remains magnitude and cause fatal injuries. The 2.4 total Uselinear principles of Physics toaisolated explain air bags can theconstant risk of injury or death.may Research has shown that force how greater than 85reduce 000 N during aincollision direction. OR In an isolated system, the total linear momentum of a system before a collision/interaction is equal to the total linear collision above lasts formomentum 0,08 s. of the system after the collision. Solution described Commented [U74]: Formatting/re-typing: pse standardise line 2.2 What to look for: breaks, indentation, etc. of all equivalent items. 2.1 The total linear momentum of an isolated and closed system remains constant in magnitude and 27 result in a fatal injury.(4) 2.3 Determine, by means of calculations, whether collision could direction. OR In an isolated system, the total linear momentum of athe system beforeabove a collision/interaction is equal to the total linear momentum of the system after the collision. Commented [U74]: Formatting/re-typing: pse standardis 2.2 What to look for: breaks, indentation, etc. of all equivalent items. The cars have crumple zones, seat belts, air bags and padded interiors that can reduce the chance of death 27 or serious injury during accidents. 2.4 Use principles of Physics to explain how air bags can reduce the risk of injury or death. Solution 2.1 The total linear momentum of an isolated and closed system remains constant in magnitude and direction. OR In an isolated system, the total linear momentum of a system before a collision/interaction is equal to the total linear momentum of the system after the collision. 2.2 What to look for: (i) In which direction are the cars travelling? (ii) What is the mass of each car? (iii) Is there a possibility of the cars moving together or moving separately after the collision? (iv) What will happen to the speed of the cars after collision? (v) What is the most appropriate equation for the scenario? 25 h direction are the cars travelling? the mass of each car? (i) In which direction are the cars travelling? a possibility of the cars moving together or moving separately after the (ii) What is the mass of each car? ? Is there possibility of the cars moving together or moving separately after the ill happen to the speed of(iii) the cars afteracollision? (i) In which direction are the cars travelling? collision? the most appropriate equation for the scenario? (ii) What will is the mass to of the each car? of the cars after collision? (iv) What happen speed (iii) Is there possibility of the cars moving together or is: moving separately after the general equation for conservation of momentum (v) is: What is athe most appropriate equation for the scenario? or conservation The of momentum collision? (i) In which direction are the cars travelling? (iv) equation happen speed of the cars (ii) What will isfor the mass to of the each car? The general conservation of momentum is: after collision? ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 ⃗𝒑𝒑⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 (iii) Is there possibility of the cars moving for together or moving separately after the (v) = ∑ What is athe most appropriate equation the scenario? ∑ 𝒑𝒑 collision? ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 𝒑𝒑 = cars ∑momentum ∑ 𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 collision? Thetwo general equation forhappen conservation of is: ⃗𝒑𝒑⃗after (iv) were What will to the speed of the blem statement the cars moving separately before collision but However, from the problem statement the two cars were moving separately before collision but moved (v) What is the most appropriate equation for the scenario? nit after the collision so the most appropriate equation to use is: Commented [U75]: Pse check/re-write – not understood. However, from problem the separately before collision butis: ⃗⃗𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 𝒑𝒑 ∑ two ∑ ⃗𝒑𝒑⃗moving together as athe unit afterstatement the collision socars the=were most appropriate equation use 𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 Commented [G76]:to I think it is better in this way general equation for conservation of momentum is:appropriate equation moved together as a unit after the collision so the most to use is: Commented [U75]: Pse check/re-write – not understood. 𝑚𝑚𝑠𝑠 + 𝑚𝑚 𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) The Commented [G76]: I think it is better in this way = 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) (1500+1500) However, from the problem statement the two cars were moving separately before collision but ⃗𝒑𝒑⃗𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂𝒂 −1 𝑚𝑚𝑠𝑠 𝑣𝑣𝑣𝑣𝑠𝑠 + 𝑚𝑚 𝑇𝑇 𝑣𝑣𝑖𝑖 𝑖𝑖 𝑇𝑇 = ( 𝑚𝑚𝑠𝑠 + 𝑚𝑚 𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) ∑ 𝒑𝒑 = ∑appropriate 𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃𝒃 Commented [G77]: moved together as a unit after the collision⃗⃗so the most equation to use is: Symbols corrected Commented [U75]: Pse check/re-write – not understood. (𝑆𝑆𝑆𝑆) = 5 𝑚𝑚 ∙𝑠𝑠 (1500)(10)+(1500)(0) = 𝑣𝑣 (1500+1500) hat the accident was fatal, the net force has𝑓𝑓(𝑆𝑆𝑆𝑆) to be calculated using the Commented [G76]: I think it is better in this way −1 =statement 5 𝑚𝑚 ∙𝑠𝑠 Commented [G77]: Symbols corrected 𝑓𝑓(𝑆𝑆𝑆𝑆) However, 𝑚𝑚𝑠𝑠 𝑣𝑣𝑣𝑣𝑠𝑠 + 𝑚𝑚from = problem (𝑣𝑣𝑚𝑚 𝑇𝑇 𝑣𝑣𝑖𝑖 𝑖𝑖 𝑇𝑇the 𝑠𝑠 + 𝑚𝑚 𝑇𝑇 )𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) the two cars were moving separately before collision but 2.3 In order to prove that the accident was fatal, the net force has to be calculated using the moved together as a unit after the collision so the most appropriate equation to use is: Commented [U75]: Pse check/re-write – not understood. (1500)(10)+(1500)(0) = 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) (1500+1500) equation: −1 N, then the accident em statement, if the answer obtained 𝑣𝑣exceeds 85 000 Commented [G76]: think it is better in this way = 5 𝑚𝑚 ∙𝑠𝑠 [G77]: ISymbols corrected 𝑓𝑓(𝑆𝑆𝑆𝑆) ⃗𝑛𝑛𝑛𝑛𝑛𝑛 ⃗𝑖𝑖 𝑖𝑖mass ∆𝑡𝑡 = 𝑚𝑚 𝑚𝑚∆𝑣𝑣 𝐹𝐹 𝑚𝑚 𝑚𝑚𝑠𝑠prove +the 𝑚𝑚accident In to that the was fatal, net force has tothe be calculated using the equation: 𝑠𝑠 𝑣𝑣𝑣𝑣 𝑠𝑠 + 𝑇𝑇 𝑣𝑣 𝑇𝑇 = (and 𝑇𝑇 )𝑣𝑣 𝑓𝑓(𝑆𝑆𝑆𝑆) he accident was 2.3 not fatal. The velocity of either offatal, the cars before 2.3 In order toorder prove that wasaccident the net force hasthe to be calculated using According to the problem statement, if the answer obtained exceeds 85 000 N, then the accident (1500+1500) (1500)(10)+(1500)(0) = 𝑣𝑣𝑓𝑓(𝑆𝑆𝑆𝑆) be used. equation: was fatal. If not, then𝑣𝑣the accident was not fatal. The mass and velocity of either of the cars before −1 = 5 𝑚𝑚 ∙𝑠𝑠 Commented [G77]: Symbols corrected ⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣⃗ 𝑓𝑓(𝑆𝑆𝑆𝑆) 𝐹𝐹 and after collision can be used. 2.3 In ordertotothe prove that the accidentif was fatal, theobtained net force has to be using the According problem statement, the answer exceeds 85 calculated 000 N, then the accident equation: ) was fatal. If not, then the accident was not fatal. The mass and velocity of either of the cars before 𝐹𝐹⃗ ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣⃗ ⃗to the and after collision canproblem be used. statement, if the answer obtained exceeds 85 000 N, then the accident was fatal. If ∆𝑡𝑡 = 𝑚𝑚∆𝑣𝑣 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 According 𝑛𝑛𝑛𝑛𝑛𝑛 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) = (5-0) statement, if the answer obtained exceeds 85 000 N, then the accident According to 1500 the problem ⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) not, then accident was not TheThe mass either of cars the before cars before and after collision can 𝐹𝐹 was fatal. If=the not, then the accident wasfatal. not fatal. massand and velocity velocity ofof either of the ⃗7500 𝑚𝑚∆𝑣𝑣 𝐹𝐹 𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 7500 ⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) and after collision can be used. 𝐹𝐹 = 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛used. be 0,08= 1500 (5-0) greater than 85 000 N, so the accident will be fatal. 7500 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) = 93 = 750 𝑁𝑁 𝑚𝑚∆𝑣𝑣⃗ ⃗ 𝐹𝐹⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 = 7500 𝐹𝐹 = than 85 000 N, so the accident will be fatal. In both cases, 𝐹𝐹depends 𝑛𝑛𝑛𝑛𝑛𝑛the collision 𝑛𝑛𝑛𝑛𝑛𝑛 is greater of the impact during Commented [U78]: Pse check. 0,08 ⃗ 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 (0,08) = 1500 (5-0) on the relationship between the Fnet ollision when the𝐹𝐹⃗⃗𝑛𝑛𝑛𝑛𝑛𝑛 momentum is constant. When the air bag inflates during the = 93 750 𝑁𝑁 [G79]: I think it is better in this way (0,08) = 7500 of the impact during the collision depends on theCommented 𝐹𝐹 𝑛𝑛𝑛𝑛𝑛𝑛 The magnitude 2.4 relationship between the Fnet Commented [U78]: Pse check. tact time of the passenger or driver with the airthan bag 85 is longer without air will be fatal. 7500 iscollision greater 000 N,than so the accident In both cases, 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛of ⃗𝑛𝑛𝑛𝑛𝑛𝑛 𝐹𝐹 = and the time when the momentum is constant. When the air bag inflates during the Commented [G79]: I think it is better in this way force on the passenger0,08 or driver is reduced according to the equation (F net collision, the contact time of the passenger or driver with the air Commented bag is longer than air [U80]: Psewithout check. ⃗𝑛𝑛𝑛𝑛𝑛𝑛 collision. =The 93magnitude 750 𝑁𝑁 p is constant for 𝐹𝐹 the 2.4 of the impact during the collision depends on the relationship between the F Commented [U78]: Pse check. net bag and thus the force on the passenger or driver is reduced according to the equation (F net ⃗𝑛𝑛𝑛𝑛𝑛𝑛ofiscollision and the time when the momentum is constant. When the air bag inflates during the greater than 85 000 N, so the accident will be fatal. In both cases, 𝐹𝐹 ∆𝑝𝑝 Commented [G79]: I think it is better in this way [U80]: Pse check. In both greater than N, so the accident will be fatal. = ∆𝑡𝑡 cases, ), becauseisΔp is constant for85 the000 collision. collision, the contact time of the passenger or driver with the air bag is longer than without air and thus below theofforce on theinitially passenger or driver is reduced according to thebetween equationthe (F net 2.4 bag The magnitude the during the collision depends on the relationship Fnet Commented [U78]: Pse check. s time graph is shown in the graph forimpact a car moving horizontally ∆𝑝𝑝 the time of collision when the momentum is constant. When the air bag inflates during the Commented [U80]: check. shown below. 1.4 and The magnitude of the impact during the collision depends on the relationship between the[G79]: Fnet Pse = ∆𝑡𝑡 ), because Δp is constant for the collision. Commented Iand think it the is better in this way The momentum versus time graph is passenger shown in the forair a bag car initially moving horizontally collision, the contact time of the or graph driver below with the is longer than without air towards the EAST. Is below. time of collision the momentum constant. When theequation air bag(Finflates during the collision, the bag and thus theshown forcewhen on the passenger or driver is is reduced according to the net ∆𝑝𝑝 Commented [U80]: Pse check. = ), because Δp is constant for the collision. contactversus time of thegraph passenger driver with theforair bag is longer without air bag and thus the force The momentum time is shown or in the graph below a car initially movingthan horizontally ∆𝑡𝑡 towards the EAST. Is shown below. on the passenger or driver is reduced according to the equation (Fnet = ), because Δp is constant for the The momentum versus time graph is shown in the graph below for a car initially moving horizontally collision. towards the EAST. Is shown below. The momentum versus time graph is shown in the graph below for a car initially moving horizontally towards the EAST. Is shown below. 28 28 28 28 26 1.1 Define momentum. (2) 1.2 What is the value of the initial momentum? (1) 1.3 In which time interval is the momentum of the car constant. (1) 1.4 At what time (s) did the car stop? (3) 1.5 Compare the direction in which the car is moving from: 0 to10 s; 10 to 15 s; and 15 to 25 s. (2) 1.6 There are many reasons for the change in momentum. Give any two possible reasons that could have caused the momentum of the car to change from 10 s to 20 s. (2) 1.7 Define impulse of a force. (2) 1.8 Calculate the change in momentum of the car from the starting point to 15s. (3) [16] 4.3 Vertical Projectile Motion PROJECTILE MOTION PROBLEM SOLVING STRATEGY Step 1: Read the problem as many times as you need to, then draw a sketch of the situation (translate the words into a sketch). Step 2: List all the given information and convert it to SI units, if necessary. Step 3: Draw a free-body diagram. • If the only force acting on the object is the gravitational force, then it is a projectile and the equations of uniformly accelerated motion can be used. Step 4: Select the formula or equation that will be used to answer the question. In some cases, more than one equation is needed to calculate the answer. Step 5: Substitute numerical values into an appropriate equation. Step 6: Check your answer once you have and answer, check if it makes sense: • Does it has the correct unit? • Is the numerical value reasonable? 27 Exercise 1 A boy is standing on top of a building. He throws a ball vertically upwards from a position 3.5 m above the ground, with an initial velocity of 10 m∙s-1. Ignore the effects of air resistance and answer the following questions: 1.1 What is the magnitude and direction of the acceleration of the ball? 1.2 Calculate the maximum height above the ground that the ball reaches. 1.3 What was the velocity of the ball at its maximum height? 1.4 Calculate the time taken by the ball to reach its maximum height. 1.5 How much time did the ball take to reach its maximum position and return to the position from which it was thrown? 1.6 Calculate the total time taken by the ball to reach the ground. 1.7 Calculate the velocity with which the ball hits the ground. 1.8 Draw a rough sketch of the velocity-time graph. Show relevant points on the velocity and time axis. 1.9 Draw a position-time graph. Exercise 2 A boy is standing on the top of a building and throws a tennis ball vertically upwards. The graph below shows the velocity of the ball from the moment it is thrown until it hits the ground for the second time. Ignore air resistance. Graph of velocity versus time 2.1 Describe the motion of the tennis ball in words. 2.2 What is the initial velocity of the tennis ball? 2.3 Calculate the height of the building. 2.4 Is the collision of the tennis ball with the ground elastic or inelastic? Explain. 28 2.5 Draw the position-time graph. 4.4 Work Energy Power PROBLEM-SOLVING STRATEGY FOR WORK ENERGY POWER STEP 1: Read and model the situation • Read the problem as many time as you need to, in order to understand it. • Draw a sketch of the situation (if not given) and identify which objects are parts of the system. Some problems may need to be sub-divided into two or more parts (scenarios). STEP 2: Visualize STRATEGY FOR WORK ENERGY POWER Draw aPROBLEM-SOLVING free-body diagram that show all the forces acting on the object (objects). Identify the type of force acting on the object (conservative or non-conservative) PROBLEM-SOLVING STRATEGY FOR WORK ENERGY POWER STEP 1: Read and model the situation Read the problem STRATEGY as many time youENERGY need to,POWER in order to understand it. PROBLEM-SOLVING FOR as WORK STEP Solve STEP Readofand the(ifsituation 3: Draw a1: sketch themodel situation not given) and identify which objects are parts of the system. Some needastomany be sub-divided into twoto, or in more parts (scenarios).it. problems Read the may problem time as you need order to understand STEP 2:sketch Visualize STEP Draw a1: of themodel situation (if not given) and identify which objects are parts of the system. Some Read and the situation • Collect the data andbewrite it in symbol form. Draw athe free-body diagram that allinto thetwo forces acting onunderstand the object (objects). Identify the type of problems may need sub-divided (scenarios). PROBLEM-SOLVINGSTRATEGY FOR WORK ENERGY POWER Read problem asto many timeshow as you need to,orinmore orderparts to it. force acting on the object (conservative or non-conservative) STEP 2: Visualize Draw a sketch of the situation (if not given) and identify which objects are parts of the system. Some • Select equation, law (principle) orforces theorem that you can(objects). apply to solve the problem. STEP Solve Draw a3:the free-body diagram that show all thetwo acting on the object Identify the type of problems may need to be sub-divided into or more parts (scenarios). model Collect the data and write it in symbol form. force acting on the object (conservative or non-conservative) STEP 1: Read and the situation STEP 2: Visualize • asIfSelect the asystem isdiagram isolated non-dissipative, theobject total mechanical energy STEP 3: Solve the equation, law to, (principle) or theorem thatit. youthen can to solve the problem. Read the problem many time as you need inand order toall understand Draw free-body that show the forces acting onapply the (objects). Identify the typeisofconserved: situation If the system is given) isolated and non-dissipative, thenare theparts totalof mechanical energy Collect the data and write it in symbol form. Draw a sketch of the (if not and identify which objects the system. Someis conserved: force acting on the object (conservative or non-conservative) 𝐸𝐸3: =equation, 𝐸𝐸𝐾𝐾𝐾𝐾 +into 𝐸𝐸𝑃𝑃𝑃𝑃law = 𝐸𝐸(principle) = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 to𝐸𝐸 Select the or theorem that you can apply to solve the problem. problems may need be+sub-divided two or parts (scenarios). 𝐾𝐾𝐾𝐾 𝑃𝑃𝑃𝑃 𝑀𝑀 more STEP Solve STEP 2: Visualize the system is isolated then theortotal mechanicalthen energy conserved: IfIf there are data non-conservative forces acting dissipative), use is equation: Collect the and writeand it innon-dissipative, symbol form.(external Draw a free-body show all the forces acting ontheorem the object typethe of problem. 𝐸𝐸𝐾𝐾𝐾𝐾 +that 𝐸𝐸the =equation, 𝐸𝐸𝐾𝐾𝐾𝐾 +=𝐸𝐸 =+ 𝐸𝐸𝑀𝑀 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑊𝑊 ∆𝐸𝐸 ∆𝐸𝐸= 𝑃𝑃𝑃𝑃are 𝑃𝑃𝑃𝑃 𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐾𝐾 𝑝𝑝 • diagram IfSelect there non-conservative forces acting (external dissipative), then use equation: law (principle) or that(objects). you can Identify applyor tothe solve force acting on the object (conservative or non-conservative) If the there are non-conservative forces acting (external or dissipative), then For both conservative and forces acting, you can use theuse work energy theorem: If system is isolated andnon-conservative non-dissipative, then the total mechanical energy isequation: conserved: STEP 3: Solve 𝑊𝑊 ∆𝐸𝐸= +𝑀𝑀∆𝐸𝐸 ∆𝐸𝐸=𝐾𝐾 𝐸𝐸𝐾𝐾𝐾𝐾 + = 𝑊𝑊 𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐾𝐾 𝐸𝐸 𝑛𝑛𝑛𝑛𝑛𝑛 𝐸𝐸 +=𝐸𝐸𝑃𝑃𝑃𝑃 𝐸𝐸𝑃𝑃𝑃𝑃 =𝑝𝑝𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐾𝐾𝐾𝐾 Collect the data and it inare symbol form. and Kinematics equations other laws(external must be acting, used with problems. For both conservative andsome non-conservative forces yousome can use theuse work energy theorem: write If there non-conservative forces acting or dissipative), then equation: Select the equation, (principle) or theorem that you canequation. apply to solve the problem. law Substitute the values in the selected = ∆𝐸𝐸 𝑊𝑊 𝑛𝑛𝑛𝑛𝑛𝑛 𝐾𝐾 𝑊𝑊 = ∆𝐸𝐸 + ∆𝐸𝐸 𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐾𝐾 𝑝𝑝 non-conservative forces acting, you can use the work energy theorem: For both conservative and If the system•is isolated and non-dissipative, thensome the total energy conserved:problems. Kinematics equations and and othermechanical laws must beacting, usediswith For both conservative non-conservative forces you some can use the work energy theorem: 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 Substitute = 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾the values in the selected equation. If there are non-conservative forces acting (external dissipative), use equation: Kinematics equations and someorother laws mustthen be used with some problems. wORKEd ExamPLE 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐾𝐾 + ∆𝐸𝐸𝑝𝑝 the ∆𝐸𝐸 Substitute values in the selected equation. • Kinematics equations and some other laws must be used with some problems. For both conservative and non-conservative forces acting, you can use the work energy theorem: 1. The sketch ExamPLE below shows a 2,0 kg block sliding from A to B along a frictionless surface. When the block wORKEd 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 reaches B, other it the continues to slide along a horizontal surface (BC), where a kinetic frictional force acts on • Substitute values inbethe selected equation. Kinematics equations and some laws must used with some problems. the block. As a result, theablock slows at C.aThe kinetic energy the block at A 1. wORKEd The sketch below shows 2,0 kg blockdown, slidingcoming from Ato torest B along frictionless surface.ofWhen the block Substitute the values in the selected equation. ExamPLE is 40,0 J;B, theit height of Atoand B is 14,0am and 8,0 m above(BC), the ground, reaches continues slide along horizontal surface where arespectively. kinetic frictional force acts on WORKED EXAMPLE the block. As a result, thea block to B rest at C. The kinetic surface. energy of the block at A 1. The sketch below shows 2,0 kgslows block down, slidingcoming from A to along a frictionless When the block is 40,0 J;B,the height of A is 14,0 m and 8,0 surface m above(BC), the ground, reaches it continues toand slideBalong a horizontal where arespectively. kinetic frictional force acts on WORKED EXAMPLEthe block. As a result, the block slows down, coming to rest at C. The kinetic energy of the block at A 1. The sketch below shows a 2,0 kg block sliding from A to B along a frictionless surface. When the block is 40,0 J; the height of A and BAis 14,0 m and 8,0 m above the ground, respectively. 1. The sketch below shows a 2,0 kg sliding fromtoA slide to B along a frictionless surface. When the blockwhere a kinetic frictional force acts on reaches B,block it continues along a horizontal surface (BC), reaches B, it continues to slide along a horizontal force acts on A m(BC), where a kinetic frictional hAsurface = 14,0 C kinetic energy of the block at A is B As adown, result, the block slows down, rest at atC.A The the block. As a result,the theblock. block slows coming to rest at C. The kineticcoming energy oftothe block hB= hc= 8,0 m is 40,0 J; the height of A and B is 14,0 m and 8,0h m above ground, respectively. 14,0 40,0 J; the height of A and B ismthe 14,0 m and 8,0 m above A= A C respectively. B the ground, hB= hc= 8,0 m 1.1. Is the total mechanical hA=energy 14,0 mof the block conserved as Bthe block goes from C A to B? Why or why not? h B= hc= 8,0 m A the block reaches point B, has its kinetic energy INCREASED, DECREASED or 1.2. 1.1. When Is the total mechanical energy of the block conserved as the block goes from A to B? Why or REMAINED why not? THE SAME? Provide a reason for your answer. = 14,0 the mtheblock A 1.3. Calculate speed of the block when it reaches B. Cenergy INCREASED, DECREASED or 1.2. hIs When reaches point B,Bblock has its kinetic 1.1. the total mechanical energy of the conserved as the block goes from A to B? Why or hB=conserved hfor manswer. 1.4. Is the total mechanical energy of the block as the block goes from B to C? Justify c= 8,0 REMAINED THE SAME? Provide a reason your why not? your answer. 1.3. When Calculate speed of the block it reaches B. energy INCREASED, DECREASED or 1.2. thethe block reaches point when B, has its kinetic 1.5. How does the kinetic do during the block BC segment of the 1.4. REMAINED Is themuch total work mechanical energy of frictional the blockforce conserved as the goes from B totrip? C? Justify THE SAME? Provide a reason for your answer. your answer. 1.1. Is the total mechanical energy of block as the block goes 1.3. Calculate thethe speed of conserved the block when it reaches B. from A to B? Why or SOLuTION: 1.5. Is How work does the kinetic frictional force do during theblock BC segment of B the why not? 1.4. themuch total mechanical energy of conserved as the from totrip? C? Justify 1.1. Is the total mechanical energy ofthe theblock block conserved as thegoes block goes from A to B? Why or why not? STEP 1: Read and 1.2. When the block reaches point B,model has its kinetic energy INCREASED, DECREASED or your answer. There is a Provide sketch ofa the situation, so answer. we do not need to draw the sketch. SOLuTION: REMAINED THE SAME? reason for your 1.5. How much work does the kinetic frictional force do during the BC segment of the trip? The block isblock interacting the surface and the Earth. STEP and model 1.3. Calculate of1: theRead when itwith reaches B. B, has 1.2.the speed When the block reaches point its kinetic energy INCREASED, DECREASED or REMAINED THE There is a sketch of the situation, so we do need to draw 1.4. Is the total mechanical energy of the block conserved as thenot block goes from the B tosketch. C? Justify SOLuTION: 31 The block is interacting with the surface and the Earth. SAME? Provide a reason for your answer. your answer. STEP 1: Read and model 1.5. How much workThere does is thea kinetic force dosoduring segment ofthe thesketch. trip? sketch frictional of the situation, we dothe notBC need to draw 31 The block is interacting with the surface and the Earth. SOLUTION: STEP 1: Read and Model 31 There is a sketch of the situation, so we do not need to draw the sketch. 29 The block is interacting with the surface and the Earth. 31 1.3. Calculate the speed of the block when it reaches B. 1.4. Is the total mechanical energy of the block conserved as the block goes from B to C? Justify your answer. 1.5. How much work does the kinetic frictional force do during the BC segment of the trip? SOLUTION: STEP 1: Read and Model There is a sketch of the situation, so we do not need to draw the sketch. The block is interacting with the surface and the Earth. System: block-surface-Earth STEP 2: Visualize System: block-surface-Earth STEP 2: Visualize Free-body Free-body diagram diagram Part 1 (from A to B) the only two forces acting are gravitational force and normal force. Commented [U88]: Pse check/clarify. System: block-surface-Earth Sliding down Sliding up Commented [GI89]: Better like this Part 1 (from A to B) the only two forces acting are gravitational force and normal force. STEP 2: Visualize ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 Free-body diagram 𝑁𝑁 𝑁𝑁 Part 1 (from A to B) the only two forces acting are gravitational force and normal force. Commented [U88]: Pse check/clarify. Sliding down Sliding (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 up 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Sliding(𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 down Sliding up 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 Commented [GI89]: Better like this ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑁𝑁 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐹𝐹⃗𝑔𝑔 - Conservative ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑁𝑁 𝐹𝐹⃗𝑔𝑔 - Conservative (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐹𝐹⃗𝑔𝑔 - Conservative From A to B there are \only conservative forces acting therefore the system is isolated. Part 2 Sliding𝐹𝐹⃗on a horizontal surface (from B to C) - Conservative 𝑔𝑔 From A to B there are \only conservative forces acting therefore the system is isolated. ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑁𝑁 Part 2 Sliding on a horizontal surface (from B to C) ⃗⃗⃗⃗ 𝑓𝑓𝑓𝑓 Non- conservative ⃗⃗⃗⃗ 𝑓𝑓𝑓𝑓 Non- conservative (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑁𝑁 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐹𝐹⃗𝑔𝑔 - Conservative Commented [U90]: Formatting: pse ensure that no text is obscured on a page or disappears off a page. Commented [U91]: Pse check/clarify. Commented [GI92]: like thispse ensure that no text is Commented [U90]:Better Formatting: obscured on a page or disappears off a page. Commented [U91]: Pse check/clarify. Commented [GI92]: Better like this From BFrom to C the frictional force is acting the block which is not acting conservative therefore system is isolated. A to B there are \only on conservative forces therefore thethe system is not isolated. Commented [U93]: Pse check/clarify. 𝐹𝐹⃗𝑔𝑔 - Conservative STEP 3: Solve Commented [GI94]: Better like this Data Part 2 Sliding on a horizontal surface (from B to C) From B to C the frictional force is acting on the block which is not conservative therefore the system m= 2,0 kg is not isolated. Commented [U93]: Pse check/clarify. EKA= 40 J STEP 3: Solve Commented [GI94]: Better this m B to C the frictional force is acting on the block which is not conservative therefore hA= 14,0 From the system is like not Data hB= hc= 8,0 m m= 2,0 kg vB-? isolated. EKA= 40 J Wff-? h = 14,0 m 1.1.A The total mechanical energy is conserved if the net work done by the non- conservative forces = 8,0 m3: Solve hB= hSTEP isc zero, or only conservative forces act on the object, or W nc = 0J. Only two forces act on the vB-? block during its trip from A to B: the gravitational force, which is conservative; and the normal Wff-? force, which is conservative in this case. Thus, we conclude that W nc = 0 J, with the result being 1.1. Data The total mechanical energy is conserved if the net work done by the non- conservative forces that the total mechanical energy is conserved during the AB part of the trip. is zero, or only conservative forces act on the object, or W = 0J. Only two forces act on the 1.2. As we have seen, the total mechanical energy is the sum of thenckinetic and gravitational potential block during its trip from A to B: the gravitational force, which is conservative; and the normal energy, and remains constant from A to B. Therefore, as one type of energy decreases, the m= 2,0which kgit is force, conservative in this case. Thus, we conclude that W nc = 0 J, with the result being other must increase, in order for the sum to remain constant. Since B is lower than A, the that the total mechanical energy is conserved during the AB part of the trip. gravitational potential energy at B is less than that at A. As a result, the kinetic energy at B must 1.2. EAs = we40 have the total mechanical energy is the sum of the kinetic and gravitational potential J seen, be KA greater than that at A. energy, and it remains constant from A to B. Therefore, as one type of energy decreases, the 1.3. From A to B, total mechanical energy is conserved. other must increase, in order for the sum to remain constant. Since B is lower than A, the 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 that at A. As a result, the kinetic energy at B must = 14,0 m potential energy at B is less than hgravitational 1 A be greater than that at A. 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵 2 1.3. From A to B, total mechanical energy is conserved. 1 𝐸𝐸 = 𝐸𝐸 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐸𝐸𝐾𝐾𝐾𝐾𝐸𝐸+ + 𝑚𝑚𝑚𝑚ℎ = 𝑚𝑚𝑣𝑣=2 𝐸𝐸 +𝑀𝑀𝑚𝑚𝑚𝑚ℎ 𝑃𝑃𝑃𝑃 𝐾𝐾𝐾𝐾 𝐾𝐾𝐾𝐾 𝐴𝐴 𝐵𝐵 2 1 𝐵𝐵 2 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵 2 32 30 1 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵 2 A to B there are \only conservative forces acting therefore the system is isolated. Sliding on a horizontal surface (from B to C) Commented [U90]: Formatting: pse ensure that no text is obscured on a page or disappears off a page. Commented [U91]: Pse check/clarify. Commented [GI92]: Better like this ⃗⃗ − 𝐷𝐷𝐷𝐷𝐷𝐷𝐷𝐷 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑁𝑁 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 hB= hc= 8,0 m ⃗⃗⃗⃗ 𝑓𝑓𝑓𝑓 Non- conservative vB-? Wff-? 𝐹𝐹⃗𝑔𝑔 - Conservative B to C the frictional force is acting on the block which is not conservative therefore the system isolated. Commented [U93]: Pse check/clarify. 1.1. The total mechanical energy is conserved if the net work done by the non- conservative forces is zero, 3: Solve Commented [GI94]: Better like this 0 kg 40 J 4,0 m = 8,0 m or only conservative forces act on the object, or Wnc = 0J. Only two forces act on the block during its trip from A to B: the gravitational force, which is conservative; and the normal force, which is conservative in this case. Thus, we conclude that Wnc = 0 J, with the result being that the total mechanical energy is conserved during the AB part of the trip. The total mechanical energy is conserved if the net work done by the non- conservative forces 1.2. As we have seen, the total mechanical energy is the sum of the kinetic and gravitational potential is zero, or only conservative forces act on the object, or W nc = 0J. Only two forces act on the block during its trip from A to B: the gravitational force, which is conservative; and the normal energy, and it remains constant from A to B. Therefore, as one type of energy decreases, the other must force, which is conservative in this case. Thus, we conclude that W nc = 0 J, with the result being that the total mechanicalincrease, energy is conserved the sum AB part the trip. constant. Since B is lower than A, the gravitational potential in orderduring for the to ofremain As we have seen, the total mechanical energy is the sum of the kinetic and gravitational potential energy from at BAistoless than thatasatone A.type As aof result, the kineticthe energy at B must be greater than that at A. energy, and it remains constant B. Therefore, energy decreases, other must increase, in order for the sum to remain constant. Since B is lower than A, the gravitational potential energy at B is less than that at A. As a result, the kinetic energy at B must 1.3. From A to B, total mechanical energy is conserved. be greater than that at A. From A to B, total mechanical energy is conserved. 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝑃𝑃𝑃𝑃 = 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 1 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵 2 1 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚ℎ𝐴𝐴 = 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚ℎ𝐵𝐵 2 32 1 × 2 × 𝑣𝑣𝐵𝐵2 + (2 × 9,8 × 8) 2 40 + 274,4 = 𝑣𝑣𝐵𝐵2 + 156,8 314,4 − 156,8 = 𝑣𝑣𝐵𝐵2 1 𝑣𝑣𝐵𝐵2 = 157,6 40 + (2 × 9,8 × 14) = × 2 × 𝑣𝑣𝐵𝐵2 + (2 × 9,8 × 8) 2 𝑣𝑣𝐵𝐵40 =+ √157,6 274,4 = 𝑣𝑣 2 + 156,8 40 + (2 × 9,8 × 14) = 𝑣𝑣𝐵𝐵 = 12,55 m·s-1 𝐵𝐵 314,4 − 156,8 = 𝑣𝑣𝐵𝐵2 𝑣𝑣𝐵𝐵2 = 157,6 1.4. During the trip from B to C, the frictional force acts on the block. This force is non-conservative 1.4. During the trip from B to C, the frictional force acts on the block. This force is non-conservative and does 𝑣𝑣 𝐵𝐵 = and does work on the block, consequently. The net√157,6 work done by the non–conservative force -1 𝑣𝑣 = 12,55 m·s 𝐵𝐵 is not zero (W ≠ 0), so the total mechanical energy is notby conserved during the BC part force of the is not zero (W ), so work on the block,ncconsequently. The net work done the non–conservative nc trip. the mechanical isBmechanical not during the part of the trip. 1.4. During theenergy trip to C,conserved the frictional on BC the block. This is non-conservative 1.5. total During the BC part, thefrom total energyforce is notacts conserved, because aforce kinetic frictional and does work on the block, consequently. The net work done by the non–conservative force force is present. is not zero (W nc≠ 0), so the total energy is not conserved during the BC part of the 𝑊𝑊𝑛𝑛𝑛𝑛mechanical = ∆𝐸𝐸𝑀𝑀 1.5. During the BC energy is not conserved, because a kinetic frictional force is trip.part, the total mechanical 𝑊𝑊𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝 1.5. During the BC part, the=total energy is not conserved, because a kinetic frictional present. (𝐸𝐸𝐾𝐾𝐾𝐾mechanical − 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ 𝑊𝑊𝑛𝑛𝑛𝑛 𝐶𝐶 − ℎ𝐵𝐵 ) force is present. 1 1 𝑊𝑊𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2𝑊𝑊+ = ∆𝐸𝐸𝐶𝐶𝑀𝑀− ℎ𝐵𝐵 ) 𝑛𝑛𝑛𝑛𝑚𝑚𝑚𝑚(ℎ 2 2 𝑊𝑊𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝 1 1 2 × (12,55)2 + 2 × 9,8 × (8 − 8) 𝑊𝑊𝑛𝑛𝑛𝑛 = ( 2 × 02 −𝑊𝑊𝑛𝑛𝑛𝑛 2 2 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 ) 12 + 02 𝑊𝑊 = 01− (12,55) 2 𝑊𝑊𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶 − 𝑚𝑚𝑣𝑣𝐵𝐵 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 ) 2 𝐽𝐽 𝑊𝑊𝑛𝑛𝑛𝑛 =2 −157,50 1 1 2 points in the opposite direction of The work done by frictional force𝑊𝑊 is –157,50 J, 2because this force 𝑛𝑛𝑛𝑛 = ( 2 × 0 − 2 × (12,55) + 2 × 9,8 × (8 − 8) 2 2 the displacement. 𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 + 0 Alternative solution: 𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽 Applying the work energy theorem: The work done by frictional force is 𝑊𝑊 –157,50 J, because this force points in the opposite direction of 𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 the displacement. + 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊J, (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾this ) The work done by frictional force 𝑊𝑊 is𝑓𝑓𝑓𝑓 –157,50 force points in the opposite direction of the 𝑁𝑁 =because Alternative solution: 1 1 2 0 0 displacement.Applying the work 𝑊𝑊 90 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁90 = ( 𝑚𝑚𝑣𝑣𝐶𝐶 − 𝑚𝑚𝑣𝑣𝐵𝐵2 ) energy theorem: 𝑓𝑓𝑓𝑓 + 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 2 2 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛1 = ∆𝐸𝐸𝐾𝐾 1 2 (12,55) ( − 𝑊𝑊𝑓𝑓𝑓𝑓 + 0 + 0 = 2 × 0 2 × 𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊𝑁𝑁 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 2𝐸𝐸𝐾𝐾𝐾𝐾 ) 2 2 2 1 1 −0(12,55) 𝑛𝑛𝑛𝑛 = 090 𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁900 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 ) 2 2 𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽 1 1 2 opposite direction of The work done by frictional force is –157,50 in the 0 = ( 2this − 2points 𝑊𝑊𝑓𝑓𝑓𝑓 + 0J,+because × 02force × (12,55) 2 2 the displacement. 𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 STEP 4: Assess −157,50 𝐽𝐽 they are reasonable and answer 𝑛𝑛𝑛𝑛 =for work (J), The results have the correct units for speed (m·s-1)𝑊𝑊and The work done by frictional force is –157,50 J, because this force points in the opposite direction of the questions. the displacement. STEP 4: Assess The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer the questions. 31 𝑊𝑊𝑛𝑛𝑛𝑛 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 ) 1 1 𝑊𝑊𝑛𝑛𝑛𝑛 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 + 𝑚𝑚𝑚𝑚(ℎ𝐶𝐶 − ℎ𝐵𝐵 ) 2 2 1 1 𝑊𝑊𝑛𝑛𝑛𝑛 = ( 2 × 02 − 2 × (12,55)2 + 2 × 9,8 × (8 − 8) 2 2 𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 + 0 𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽 Alternative solution: The work done by frictional force is –157,50 J, because this force points in the opposite direction of the displacement. Alternative solution: Applying the work energy theorem: Applying the work energy theorem: 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝐸𝐸𝐾𝐾 𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑊𝑊𝐹𝐹𝐹𝐹 + 𝑊𝑊𝑁𝑁 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 ) 1 1 𝑊𝑊𝑓𝑓𝑓𝑓 + 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 900 + 𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁900 = ( 𝑚𝑚𝑣𝑣𝐶𝐶2 − 𝑚𝑚𝑣𝑣𝐵𝐵2 ) 2 2 1 1 𝑊𝑊𝑓𝑓𝑓𝑓 + 0 + 0 = ( 2 × 02 − 2 × (12,55)2 2 2 𝑊𝑊𝑛𝑛𝑛𝑛 = 0 − (12,55)2 𝑊𝑊𝑛𝑛𝑛𝑛 = −157,50 𝐽𝐽 The work done by frictional force is –157,50 J, because this force points in the opposite direction of the displacement. STEP Assess force is –157,50 J, because this force points in the opposite direction of the The work done by4:frictional The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer displacement.the questions. STEP 4: Assess The results have the correct units for speed (m·s-1) and for work (J), they are reasonable and answer the questions. 33 32 Exercise 1 A donkey pulls a cart of mass 600 kg from rest along a horizontal road. The donkey applies a constant force of magnitude 191,7 N at an angle of 30° to the horizontal. The cart accelerates and reaches a speed of 3 m.s-1 in 5 minutes. The average frictional force that acts on the cart is 160,02 N. Ignore the effect of the rotation of the wheels of the cart. Exercise 1 A donkey pulls a cart of mass 600 kg from rest along a horizontal road. The donkey applies a constant force of magnitude 191,7 N at an angle of 30° to the horizontal. The cart accelerates and reaches a speed of 3 m.s-1 in 5 minutes. The average frictional force that acts on the cart is 160,02 N. Ignore the effect of the rotation of the wheels of the cart. 1.1 State the work-energy theorem in words. (2) 1.2 Use the WORK-ENERGY THEOREM to calculate the distance covered by the cart in 5 minutes. (5) 1.3 Calculate the power developed by the donkey in 5 minutes. (3) The donkey now applies a force of the same magnitude on the cart, during the same time period and over the same distance, but at a SMALLER ANGLE to the horizontal. 1.1 1.2 State the work-energy theorem in words. (2) Commented [U95]: Plse check. 1.4 Use the WORK-ENERGY How does the THEOREM power developed donkey nowbycompare developed to calculateby thethe distance covered the cart into the power Commented [U96]: Formatting/re-typing: pse standardise 5 minutes. (5) SMALLER appearance of terminology in text. Repetitive problems seen. by the donkey in QUESTION 1.3? Write down only GREATER THAN, 1.3 Calculate the power developed by the donkey in 5 minutes. (3) THAN or EQUAL TO. (2) The donkey now applies a force of the same magnitude on the cart, during the same time period and over the same distance, but at a SMALLER ANGLE to the Give a reason for the answer. (No calculations are required.) horizontal. 1.4 How does the power developed by the donkey now compare to the power developed by the donkey in QUESTION 1.3? Write down only GREATER THAN, SMALLER THAN or Exercise 2 EQUAL TO. Give a reason for the answer. (No calculations are required.) [12] (2) [12] A 2 kg block, initially at 4 m height, is released and slides downhill from rest on a frictionless ramp, and then Exercise 2 moves along a horizontal surface. It then moves on a 10 m length rough horizontal surface with coefficient of A 2 kg block, initially at 4 m height, is released and slides downhill from rest on a frictionless ramp, =0,2, until reaches roughonramp therough same coefficient of friction and slides on it for 5 m kinetic friction and then moves along aµkhorizontal surface. It thenamoves a 10 mwith length horizontal surface with coefficient of kinetic friction µk =0,2, until reaches a rough ramp with the same coefficient of friction until it stops. and slides on it for 5 m until it stops. 5m θ A 2.1. 2.2. 2.3. 2.4. 10 m h B State the law (principle) of conservation of mechanical energy in words. State the lawCONSIDERATIONS (principle) of conservation ofspeed mechanical energy in words. Use ENERGY to calculate the of the block at the bottom of the ramp. State the work energy theorem in words. Use the WORK ENERGY THEOREM to calculate the speed of the block when passing position 2.1. B. Use ENERGY CONSIDERATIONS to calculate the speed of the block at the bottom of the ramp. 2.3. Use the WORK ENERGY THEOREM to calculate the speed of the block when passing position B. 2.4. 34 Use the LAW OF CONSERVATION OF ENERGY to calculate the height reached by the block until it comes to rest. 2.5. Calculate the angle at θ. 2.5. Use the LAW OF CONSERVATION OF ENERGY to calculate the height reached by the block 2.2. Statetothe until it comes rest.work energy theorem in words. 2.6. Calculate the angle at θ. 33 4.5 Doppler Effect 4.5 Doppler Effect Worked Example worked Example Doppler Effect AAman is standing on a pavement when he hears an ambulance approaching. The siren of the ambulance is man is standing on a pavement when he hears an ambulance approaching. The siren of the ked Example emitting ambulance is emitting a wave with a frequency of 500 Hz.ambulance The ambulance is moving a wave sound withsound a frequency of 500 Hz. The is moving at ata aconstant speed of 30 m·s-1. constant speed of 30 m·s-1. Calculate the frequency of the sound the man hears. Take the speed of the frequency of the sound the man hears. Take the speed of the sound in the air as 340 m·s-1. an is standing Calculate on pavement theasound in thewhen air ashe 340hears m·s-1.an ambulance approaching. The siren of the ulance is emitting a wave sound with a frequency of 500 Hz. The ambulance is moving at a stant speed of 30Solution m·s-1. Calculate the frequency of the sound the man hears. Take the speed of Solution sound in the air as 340 m·s-1. 𝑣𝑣 ± 𝑣𝑣𝐿𝐿 𝑓𝑓𝐿𝐿 = ( ) 𝑓𝑓 ution 𝑣𝑣 ± 𝑣𝑣𝑠𝑠 𝑠𝑠 ( 𝑣𝑣 ± 𝑣𝑣𝐿𝐿 ) 𝑓𝑓 𝑣𝑣 ± 𝑣𝑣𝑠𝑠 𝑠𝑠 ( fL ( 340 0 ) 500 340 30 340 0 f L 1,10 500 ) 500 340 30 f L 550Hz 1,10 500 550Hz 4.6 Electrostatics 4.6 Electrostatics worked example Worked example Electrostatics QuESTION 1 (Question 3 - Spring School material, Northern cape) ked example Two identically1 charged spheres (a and B) are 30 material, cm apart, with a chargeCape) of +3 x 10-4 C and QUESTION (Question 3 - Spring School Northern ESTION 1 (Question 3 Spring School material, Northern cape) -4 -8 –2 x 10 C, respectively. A small positive charge (c) of +10 C is placed 10 cm from sphere a, as Two identically charged (A with andaB) are 30 cmx apart, with a charge of +3 x 10 shown in the diagram below. identically charged spheres (a and B) are spheres 30 cm apart, charge of +3 10-4 C and C and –2 x 10-4 C, Bis placed respectively. A small of +10 10ascm from sphere A, as shown in the diagram 10-4 C, respectively. A small positive charge (c) ofCcharge +10-8 C is(C) placed 10 cmCfrom sphere a, A positive wn in the diagram below. below. -4 -8 A 1.1 1.2 C 10 cm B 30 cm 10 cm State Coulomb’s law in words. 30 cm Calculate the magnitude of the force exerted on the small positive charge c by charge a. 1.3law in Calculate State Coulomb’s words. the magnitude of the force exerted on the small positive charge c by charge B. 1.1 State Coulomb’s law in words. 1.4 Calculate the magnitude ofthe thesmall resultant forcecharge exerted small Calculate the magnitude of the force exerted on positive c on by the charge a.positive charge c. 1.5 Define electric at a on point words. Calculate the magnitude of the forcefield exerted theinsmall positive charge c by charge B. 1.2 Calculate the magnitude of the force exerted on the small positive charge C by charge A. 1.6 Calculate the magnitude of the electric at positive point c,charge 10 cm from Calculate the magnitude of the resultant force exerted on thefield small c. a, as a result of charge a. thein small charge c is removed. Calculate the resultant electric field at a point situated at Define electric1.7 field atNow a point words. 1.3 Calculate the magnitude of the force exerted on the small positive charge C by charge B. the of centre point between a andc,B.10 cm from a, as a result of charge a. Calculate the magnitude the electric field at point Now the small charge c is removed. Calculate the resultant electric field at a point situated at 1.4 Calculate the magnitude of the resultant force exerted on the small positive35charge C. 1.5 Define electric field at a point in words. 1.6 Calculate the magnitude of the electric field at point C, 10 cm from A, as a result of charge A. 1.7 Now the small charge C is removed. Calculate the resultant electric field at a point situated at the centre point between A and B. 1.8 Charge C makes contact with charge B until the total charge between B and C is distributed. Calculate the new charge on C. the centre point between a and B. 34 35 SOLUTION 1.8 Charge c makes contact with charge B until the total charge between B and c is distributed. Charge c makesof contact with charge B force until the total charge between B and c is distributed. 1.11.8 Calculate The magnitude theonelectrostatic exerted by one point charge (Q1) on another point charge (Q2) the new charge c. Calculate the new charge on c. is directly proportional to the product of the magnitudes of the charges and inversely proportional to the SOLuTION square of the distance (r) between them. SOLuTION 1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point 1.1 The magnitude of the electrostatic force exerted by one point charge (Q 1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and 1.2 Data charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. inversely proportional to the square of the distance (r) between them. 1.2 Data 1.2 QAData = +3x10-4 C -8 A = +3x10 qc =Q+10 C -4 C q = +10 C x 10-2 m rAC =c 10 cm-8=10 9 cm =10 2 r x2 10-2 m AC = 10 k = 9 x10 N9 m C 2 2 k ==9?x10 N m C FAonC FAonC = ? KQAqC FAonC KQ 2 q FAonC (rAC ) A 2C (rAC ) 9 x10 9 3 10 4 10 8 FAonC 9 x10 9 3 21024 10 8 FAonC (10 10 )2 2 (10 10 ) 27 10 3 3 FAonC 27 10 FAonC100 10 4 100 10 4 FAonC = 0, 27x101N = 2,70 N FAonC = 0, 27x101N = 2,70 N 1.3 Data Data 1.31.3Data QB = – 2 x 10-4 C QB = – 2 x 10-4 C qc = +10-8 C = +10 rBCq=c 20 cm-8=C20 x 10-2 m 9 cm = 220 x210-2 m r = 20 BC m C k = 9 10 N 9 2 2 k = =9 FAonC ? 10 N m C FAonC = ? KQB qC FBonC KQ (rBC ) 2 B q2C FBonC (rBC ) 9 x10 9 2 10 4 10 8 FBonC 9 x10 9 2 21024 10 8 FBonC (20 10 )2 2 (20 10 ) 18 10 3 3 FBonC 18 10 4 FBonC 400 10 4 400 10 FBonC = 0,045x101 N = 0,45 N FBonC = 0,045x101 N = 0,45 N 1.4 There are two electrostatic forces acting on point charge C; the gravitational force is 1.4 There are two electrostatic forcesit acting on point charge to C;the theelectrostatics gravitational force is ignored because it is ignored because is too small in comparison forces, 36 therefore we too small in comparison toapply the electrostatics forces, therefore we must apply the principle of superposition of 36 must the principle of superposition of forces. forces. F F AC F BC (it is a vector equation) F R = F AC + F BC (it is Ra vector equation) We must draw a free body diagram showing the electrostatic forces acting on the point chargediagram and choose the positive We must draw a free body showing the direction. electrostatic forces acting on the point charge and choose the positive direction. 𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 Xx Working with the magnitudes Working with the magnitudes of the forces:of the forces: FR = FAC + FBC Substitution: FR = 2,70 N + 0,45 N 𝐹𝐹 R = 3,15 N 35 𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹⃗𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝐹𝐹⃗𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 Xx Xx Working with the magnitudes of the forces: Working with the magnitudes of the forces: FR = FAC + FBC FR = FAC + FBC FR = FAC + FBC Substitution: Substitution: Substitution: = 2,70 0,45 N N NN++0,45 F =FR2,70 FR = 2,70 RN + 0,45 N 𝐹𝐹 R = 3,15 N = 3,15 N 𝐹𝐹 R = 3,15 RN 2.3 2.4 2.3 The electric field at a point is the electrostatic force experienced per unit of positive The field at a point is athe electrostatic force experienced per unit of positive 2.3electricThe electric fieldatat charge placed thatpoint point.is the electrostatic force experienced per unit of positive charge placed at that point. charge placed at that point. 2,7 F -1 -1 = 2,7 108 N∙C x 10x8 N∙C 2.3F2.4 E = 2,7 E = 8 E = E82,7 E = q8 E 10 = 2,7 x 10 N∙C-1 E= 10 q Each charge has an electrostatic field around it, therefore there are two electric fields at 1.71.7 Each charge has an electrostatic field around it, therefore there are two electric fields at the point, and we 1.7 Each charge has an electrostatic field around it, therefore there are two electric fields at the point, and we must apply the principle of superposition of fields. must apply principle of superposition the point, and we mustthe apply the principle of superpositionofoffields. fields. EA A+ EBB (it is a vector equation) E R E=R E is a vector equation) equation) E R E A E B (it is a vector (it We must draw the vector electric field of each charge at a point. must draw electric the vector electric field at ofaeach We mustWe draw the vector field of each charge point.charge at a point. 𝐸𝐸⃗⃗𝐴𝐴 𝐸𝐸⃗⃗𝐴𝐴 𝐸𝐸⃗⃗𝐵𝐵 X x 𝐸𝐸⃗⃗𝐵𝐵 X x Both electric point in the same direction, therefore we add them together: Commented [U97]: Pse check – not understood. Both electricfiled filed vectors vectors point in the same direction, therefore we add them together: Both electric filed vectors point in the same therefore we add them together: Commented [U97]: Pse check[G98]: – not understood. 9 2direction, 2 4 5 2 Commented Corrected KQ EA E =R = E2AA+ EB EA = ER = EA + EB KQ r 9 x10 Nm C x3x10 C (15 x10 2 m) 2 9 x10 9 Nm 2 C 2 x3x10 4 C EA = 27 x10 Nm C 225 x10 4 m 2 Commented [G98]: Corrected 27 x10 5 Nm 2 C A EA = KQ KQAEA = 9 x1099xNm 10 9 2Nm x3x410 275x4Nm 10 52 2Nm C 22xC3x210 C 4ECA = 27 x10 C 2C EA =E225 EA =ErA2 =2 A 2 E9 A =EA = (15 x810 2 m2-1) 2 2 2 A =x10 m 24 2 4 EA = 0,12 (10 15 x10 ) right r rx 10 = 1,2 x 10 (15 xN∙C mto ) mthe 225 x225 10 x10 m m right EA = 0,12 x 109 = 1,29 x 108 N∙C9-18 to the 2 -1 to2-1 = 0,12 = x1,2 10 N∙C to the x 10x9 10 = 1,2 N∙CNm the right EA =E0,12 AKQ 910xx810 C x2 xright 10 4 C B 18 x105 Nm2C 37 37 EB = EB = EB = 2 4 2 2 (15 2 x10 2m 4 )2 225 m2 KQBr 9 x10 9 Nm C C 18 x105 Nm C x10 92 x 22 x10 2 4 9 2 4 EB = KQ KQBEB = 9 x109 xNm B= 10 Nm x2 x10 185x4Nm 10522Nm C xC22 x10 C EC 18 x10 C 2C B 2 2 EB =ErB =2 2 EB =EB = (15 x10 m2 ) 2 2 EB =E 225 B =x10 m 4 15tox10 ) r rx 1099 = 0,8 x (10 225 x225 10 x410 m2 m2 158x8(10 m ) mright EEBB== 0,08 the 0,08 x 10 = 0,8 x 10 to the right EB = 0,08 x 109 = 0,89 x 108 to the right = x0,8 to right the right EB =E0,08 x 10x98810 = 0,8 10x8 10 to888the B = 0,08 10 EERR== 1,2 x 10 0,8xx10 10 ++ 0,8 ER = 1,2 x 108 + 0,8 x 108 1,2 + x0,8 ER =E1,2 10x8 10 +80,8 10x8 108 R= x N∙C-1-1 to tothe theright right EER== 2 x 108 N∙C R 2 x 108 N∙C-1 to the right ER = 8 N∙C -1 to-1 2 x8 10 to right the right ER =E2R x= 10 N∙C the 1.8 Since Since system issystem isolated, then weapply must apply the law of conservation of charge. Commented [U99]: Alignment 1.8 Since isolated, then we apply theof law of conservation of charge. 1.8 thethe system isthe isolated, theniswe must the lawmust of conservation charge. Commented [U99]: Alignment problem / missing item problem / missing problem? Since the system is isolated, we must apply the law of conservation of charge. problem? Commented [U99]: Alignment problem / missing 1.8 1.8 Since the system is isolated, thenthen we must apply the law of conservation of charge. Commented [U99]: Alignment problem / missing item item problem? problem? Commented [U100]: Pse check/complete. Commented [U100]: Pse check/complete. Commented [U100]: Pse check/complete. Commented [U100]: Pse check/complete. Commented [G101]: Corrected [G101]: Corrected Commented Commented [G101]: Corrected Commented [G101]: Corrected Commented [U102]: Pse check. Commented [U102]: Pse check. Commented [U102]: Pse check. Commented [U102]: Pse check. = 𝑄𝑄 +𝑐𝑐𝑄𝑄+ 𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛 𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛 =𝑐𝑐 𝑄𝑄 𝑄𝑄𝐵𝐵𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝐵𝐵 = += 𝑄𝑄𝐵𝐵𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛𝑄𝑄= 𝑛𝑛𝑛𝑛𝑛𝑛𝑄𝑄𝑐𝑐=+𝑄𝑄𝑄𝑄 𝑐𝑐 𝐵𝐵 After being in contact (𝑄𝑄𝑐𝑐 =(𝑄𝑄𝑄𝑄𝐵𝐵 = = 𝑄𝑄 𝑄𝑄 = 𝑄𝑄 After being contact being in contact (𝑄𝑄𝑐𝑐𝑐𝑐𝑄𝑄𝐵𝐵= = 𝑄𝑄𝐵𝐵 AfterAfter being inin (𝑄𝑄𝑐𝑐 (= 𝐵𝐵𝑄𝑄= 𝑄𝑄 ) After being incontact contact 𝑄𝑄𝑐𝑐 + 𝑄𝑄𝐵𝐵 = 2𝑄𝑄 + 𝑄𝑄 = 2𝑄𝑄 𝑄𝑄 +𝑄𝑄𝑄𝑄 =𝐵𝐵2𝑄𝑄 2𝑄𝑄 𝑄𝑄𝑐𝑐𝑐𝑐 + 𝑐𝑐𝑄𝑄𝐵𝐵 𝐵𝐵= 𝑄𝑄𝑐𝑐 + 𝑄𝑄𝐵𝐵 𝑄𝑄 = 𝑄𝑄 + 𝑄𝑄𝐵𝐵 𝑄𝑄𝑄𝑄 𝑐𝑐𝑄𝑄 𝐵𝐵 𝑄𝑄2𝑐𝑐𝑐𝑐++ 𝐵𝐵 𝑄𝑄 = 𝑄𝑄 =𝑄𝑄 =2 2 2 (110 8 )8 ( 2 10 4 )4 04,000110 4 4 2 410 4 4 1,9999 10 4 4 4 4 - 1,00 x10-4 C -4 Qc = (110 (110 (10 2 10 0,0001 2 10 1 ,9999 =10 )8 8() 2 )=4 )0,0001 10 1 , 9999 1010 2 42 10 = -41,00 = - 10 1,00 x10-4x10 C C Qc =Q(c1=10 2) (2 10 =) = 0,0001 2 10 4 2 10 1 , 9999 2 2 2 2 = = - 1,00 x10-4 C Qc = 2 2 2 2 2 Exercise 1 Exercise Exercise 1 1 Two small, metal spheres (a and B), carrying charges of +30nC and -14nC, Exercise 1identical small, identical metal spheres (a and carrying charges of +30nC -14nC, TwoTwo small, identical metal spheres (a and B), B), carrying charges of +30nC and and -14nC, respectively, are mounted on insulated stands; these are placed at a certain distance r from respectively, mounted on insulated stands; these are placed a certain rand from-14nC, Two small, identical metal spheres (a and B), of distance +30nC respectively, are are mounted on insulated stands; these arecarrying placed at charges aat certain distance r from each other, as shown below. each other, as shown below. each other, asare shown below. on insulated stands; these are placed at a certain distance r from respectively, mounted each other, as shown below.r r 30 nC 30 nC30AnC A A 36 30 nC A r r B B -B14 nC - 14 nC - 14 nC B - 14 nC Qc = Exercise 1 2 (110 8 ) (2 10 4 ) 0,000110 4 2 10 4 1,9999 10 4 = = - 1,00 x10-4 C 2 2 2 Exercise 1 Two small, identical metal spheres (a and B), carrying charges of +30nC and -14nC, respectively, insulated stands; these are placed at a certain distance r from Two small, identical metal spheresare (Amounted and B),on carrying charges of +30nC and -14nC, respectively, are mounted each other, as shown below. on insulated stands; these are placed at a certain distance r from each other, as shown below. r 30 nC A B - 14 nC 1.1. Define electric field in words. 1.1. Define electric field in words. 1.2. The contact magnitude of the electrostatic force between them is now 6,4original x 10 N.position. The Sphere B is moved position. and makes with sphere A. It is then moved back to its magnitude of the1.2.1 electrostatic force between them is 6,4 x 10-4 N. Sketch the field pattern surrounding thenow charges. 1.2. Sphere B is moved and makes contact with sphere a. It is then moved back to its original -4 1.2.2 Calculate the magnitude of the charge on each sphere after contact. 1.2.1 Sketch the field pattern surrounding the charges. 1.3. Determine the original distance between the spheres. 38 1.2.2 Calculate the magnitude of the charge on each sphere after contact. Exercise 2 1.3. Determine the original spheres. Two smalldistance objects (abetween and B) arethe equally positively charged; these are placed in a vacuum, as Exercise 2 shown in the figure below. B A 200 cm Two small objects (A and B) are equally positively charged; these are placed in a vacuum, as shown in the Charge B repels charge a with an electrostatic force of 3,0 x 10-6 N. figure below. 2.1 State Coulomb’s law in words. 2.2 What is the magnitude of the electrostatic force that A exerts on B? Charge B repels charge A with an electrostatic force of 3,0 x 10-6 N. 2.3 Draw the resultant electric field lines of the electric field around charges a and B. 2.1 2.4law Calculate the magnitude of charge a and B. State Coulomb’s in words. 2.2 What is the magnitude of the electrostatic force that A exerts on B? 2.3 STRATEGY SOLVE PROBLEMS ON ELECTRIC CURRENT Draw the resultant electric fieldTO lines of the electric field around charges A and B. 2.4 Calculate the magnitude of charge A and B. 2.5 Calculate the electric field strength atinasymbolic point inform. the centre of charge A and B. Write down the data 2.5 Calculate the electric field strength at a point in the centre of charge a and B. 4.7 Electric Circuits Read the problem carefully, as many times as you need to. If not given, draw a circuit diagram. Commented [U103]: Pse c Indicate the conventional direction of the current, from high-potential to low-potential Commented [GI104]: corr (+ to -). Re-draw the circuit diagram to simplify it, if necessary. Identify the type of connection (series/parallel). Analysing circuits: 1 – The algebraic sum of the changes in potential in a complete transversal of any loop of a circuit must be zero (ε= Ir+ Ir). 2. The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction: 39 37 2.6 4.7 Electric Circuits STRATEGY TO SOLVE PROBLEMS ON ELECTRIC CURRENT Read the problem carefully, as many times as you need to. If not given, draw a circuit diagram. Write down the data in symbolic form. Indicate the conventional direction of the current, from high-potential to low-potential (+ to -). Re-draw the circuit diagram to simplify it, if necessary. Identify the type of connection (series/parallel). Analysing circuits: 1 – The algebraic sum of the changes in potential in a complete transversal of any loop of a circuit must be zero (ε= Ir+ Ir). 2. The sum of the currents entering any junction must be equal to the sum of the currents leaving that junction: i = i1+i2+…+ in Write down the formula/ equation that solves the question. Find the unknowns, if needed (multi-concept problems). Do the calculations and write down the final answer. Check your answer. Check if it makes sense, i.e.: - Is the unit correct? - Is the numerical value reasonable? 38 Do the calculations and write down the final answer. Check your answer. Check if it makes sense, i.e.: WORKED EXAMPLE - Is the unit correct? - Is the numerical value reasonable? wORKEd ExamPLE In the circuit diagram below, the voltmeter V reads 12 V when switch S is open. When switch S is closed, the In the circuit diagram below, the1 voltmeter V1 reads 12 V when switch S is open. When switch S V. is closed, the readingofdrops to 10 V. The Ammeter and the wires is reading drops to 10 The resistance the Ammeter andresistance the wiresofisthe negligible. negligible. 𝑉𝑉1 1 𝑉𝑉2 S R A 3A 6Ω 𝑉𝑉3 1.1 4Ω What is the reading of the ammeter when the switch is open? 1.2 What is the emf of the battery? Commented [GI105]: corrected 1.1 What is the reading of the ammeter when the switch is open? 1.2 1.3 ofWhat is the reading of voltmeter 2? What is the emf the battery? 1.3 What is the reading of voltmeter 2? 1.4 Calculateresistance the internal of resistance of the battery? Calculate the1.6 equivalent the resistors connected in parallel. 1.5 Determine the reading of voltmeter 3? 1.8 What is the resistance of the resistor R? 1.6 How would the reading of voltmeter V2 change if the 6 Ω resistor is removed from the Calculate the1.9 internal resistance of the battery? 1.7 answer. Determine the total resistance of the circuit? 1.8 What is the resistance of the resistor R? 1.9 1.4 Calculate the equivalent resistance of the resistors connected in parallel. 1.5 Determine the reading of voltmeter 3? 1.7 Commented [U106]: Formatting/ points from blank lines in text – repet Pse use either bullets or numbers – no items/sub-items. Repetitive problems Determine the total resistance of the circuit? circuit? Write down INCREASE, DECREASE or REMAIN THE SAME. Explain your 40 How would the reading of voltmeter V2 change if the 6 Ω resistor is removed from the circuit? Write down INCREASE, DECREASE or REMAIN THE SAME. Explain your answer. SOLUTION Commented [U107]: Formatting: pse use formatting to ma 1.1. Zero (when the circuit is open there is no flow of charges therefore no current in the SOLUTION items legible – repetitive problems seen, including use of bullet point/not using bullet points, line spacing, alignment, indentatio standardisation of formatting, etc. (It is difficult to understand w belongs with what in the text because of the lack of attention to formatting.) 12 V (when (when there is no electric current flowing the circuit, the reading of the voltmeter Commented [U108]: Pse check ‘bullet’ points at items 1.1.1.2.Zero the circuit is open there is noin flow of charges therefore no current in the throughout – pse standardise appearance of all with equivalent 1.2. connected across the terminals of the battery is equal to the emf of the battery, because items and remove bullet points from items that are not bullet points. There appear to be repetitive problems in the text. 12 V (when there is no electric current flowing in the circuit, the reading of the voltmeter connected across Commented [GI109]: corrected there is no energy dissipated. the terminals of the battery is equal to the emf of the battery, because there is noCommented energy dissipated. [U110]: Pse check/ complete. Pse re-check the text in its entirety: the editor is an English specialist, not a Physics specialist, but there appear to be multip problems with the drafting of the text. Zero (there in the but thebut resistance is negligible: V=IR=(I)(0)=0 V) 1.3.1.3.Zero (thereisiscurrent current in circuit, the circuit, the resistance is negligible: V=IR=(I)(0)=0 V) Commented [U111]: Pse check. of of 6 Ω6and 4 Ω are in parallelin parallel 1.4.1.4.The The resistors resistors Ω and 4 Ωconnected are connected Commented [U112]: Formatting required: text should not b obscured – repetitive problem seen in text. 1 solve 1 1 the following way Where there are two resistors we can in = + 𝑅𝑅𝑃𝑃 𝑅𝑅1 𝑅𝑅2 𝑅𝑅1 𝑅𝑅2 (6)(4) 𝑅𝑅𝑃𝑃 = = = 2,4 Ω 𝑅𝑅1 + 𝑅𝑅2 6+4 1 1 5 1 = + = 𝑅𝑅𝑃𝑃 6 4 12 𝑅𝑅𝑃𝑃 = 2,4 Ω 1.5 𝑉𝑉𝑝𝑝 = 𝐼𝐼𝐼𝐼 𝑉𝑉𝑝𝑝 = 3 × 2,4 𝑉𝑉𝑝𝑝 = 7,2 𝑉𝑉 39 1.5 𝑃𝑃 𝑅𝑅 + 𝑅𝑅 ++4 4 1𝑅𝑅1 + 2𝑅𝑅2 6 6 1 1 5 1 = + = 𝑅𝑅𝑃𝑃 6 4 12 11 11 11 55 == ++ == 4 4 1212 𝑅𝑅 6Ω 𝑅𝑅𝑃𝑃𝑅𝑅𝑃𝑃= 6 2,4 𝑃𝑃 = 𝐼𝐼𝐼𝐼 2,4 2,4ΩΩ 1.5 𝑅𝑅 𝑉𝑉𝑝𝑝𝑃𝑃𝑅𝑅= 𝑃𝑃 = 1.5 = 𝐼𝐼𝐼𝐼 1.5𝑉𝑉 𝑉𝑉𝑝𝑝𝑝𝑝𝑉𝑉𝑝𝑝==3 𝐼𝐼𝐼𝐼 × 2,4 𝑉𝑉𝑝𝑝𝑉𝑉𝑝𝑝==3 3××2,4 𝑉𝑉𝑝𝑝 = 7,2 𝑉𝑉 2,4 𝑉𝑉𝑝𝑝𝑉𝑉𝑝𝑝==7,2 7,2𝑉𝑉 𝑉𝑉 of the voltmeter 3 is 7,2 V. The reading The reading of the voltmeter 3 is 7,2 V. ofofthe The reading thevoltmeter voltmeter3 3isis7,2 7,2V.V. 1.6 The 𝜀𝜀 =reading 𝑉𝑉 𝑒𝑒𝑒𝑒𝑒𝑒 + 𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖 1.6 1.6 ++𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖 𝜀𝜀 𝜀𝜀===𝑉𝑉𝜀𝜀𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉− 𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖 1.6𝑉𝑉𝑖𝑖𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒𝑒𝑒𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉 =− 𝜀𝜀 𝜀𝜀−𝑉𝑉−𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉= 𝐼𝐼𝐼𝐼𝑖𝑖𝑖𝑖𝑖𝑖 𝑖𝑖𝑖𝑖𝑖𝑖𝜀𝜀= 𝑒𝑒𝑒𝑒𝑒𝑒𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝐼𝐼𝐼𝐼 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 𝜀𝜀−− 𝑉𝑉 3𝑟𝑟𝐼𝐼𝐼𝐼= ==𝜀𝜀12 − 10 𝑒𝑒𝑒𝑒𝑒𝑒 3𝑟𝑟 =1212− −1010 𝑟𝑟 3𝑟𝑟 ==0,67 Ω 𝑟𝑟 𝑟𝑟==0,67 0,67ΩΩ 1.7 𝜀𝜀 = 𝐼𝐼𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1.7 1.7 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1.7𝜀𝜀 𝜀𝜀==𝐼𝐼𝑅𝑅𝐼𝐼𝑅𝑅 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 12 = 3𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 1212==3𝑅𝑅 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 3𝑅𝑅 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 41 Rtotal = 4 Ω Rtotal = 4 Ω 4141 1.8 1.8 Option 1 Option 2 𝑅𝑅𝑇𝑇 = 𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒 + 𝑟𝑟 𝑉𝑉𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑉𝑉2 = 𝑉𝑉1 𝑅𝑅𝑒𝑒𝑒𝑒𝑒𝑒 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑅𝑅𝑃𝑃 𝑅𝑅𝑇𝑇 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 𝑅𝑅𝑃𝑃 + 𝑟𝑟 4 = 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 2,4 + 0,67 𝐼𝐼𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 7,2 = 10 3𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 + 7,2 = 10 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = 0,93 Ω 𝑅𝑅𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = 0,93 Ω INCREASE 1.8 1.8 INCREASE If the bulb of 6 Ω is removed, the total resistance of the circuit increases, then the If the bulb of 6 Ω is removed, the total resistance of the circuit increases, then the reading of the ammeter reading of the ammeter decreases, because the current decreases due to the decreases, because the current decreases due to the increasing external resistance; but the reading of the increasing external resistance; but the reading of the voltmeter increases, because the voltmeter increases, because the drop of potential in the internal resistance also decreases; but the emf of the drop of potential in the internal resistance also decreases; but the emf of the battery is battery isconstant, constant, according ε -load Vint==εε -- V Ir.int = ε - Ir. according to Vloadto= V Exercise 11 Exercise Three resistors ( 2 Ω, 4 Ω and Rx ) are connected to a battery, as shown in the circuit diagram below. With switch S1 open and S2 closed, the reading on the voltmeter is 10 V. With both switches closed, the reading on the voltmeter is 8 V and the ammeter is 1 A. V A S1 4Ω S2 40 Rx 2Ω 1.1 Write down the value of the emf of the battery. 1.2 Calculate the: (1) 1.2.1 resistance of the unknown resistor Rx. (7) 1.2.2 internal resistance of the battery. (3) drop of potential in the internal resistance also decreases; but the emf of the battery is constant, according to Vload = ε - Vint = ε - Ir. Exercise 1 2 xΩ,) are 4 Ω connected and Rx ) are to connected to aas battery, as in shown in the circuit Three resistors (Three 2 Ω, resistors 4 Ω and( R a battery, shown the circuit diagram below. With diagram below. With switch Son S2 closed, theV.reading on the voltmeterclosed, is 10 V.the reading on the 1 open switch S1 open and S2 closed, the reading the and voltmeter is 10 With both switches With both switchesisclosed, voltmeter is 8 V and the ammeter 1 A. the reading on the voltmeter is 8 V and the ammeter is 1 A. V A S1 4Ω S2 Rx 2Ω 1.1 Write down the value of the emf of the battery. 1.2 Calculate the: 1.1 Write down the value of the emf of the battery. 1.2 Calculate the: (1) 1.2.1 resistance of the unknown resistor Rx. (7) 1.2.1 resistance of the unknown resistor Rx. 1.2.2 internal resistance of the battery. (7) (3) 1.2.2 internal resistance of the battery. 1.3 (1) (3) How will the reading on the voltmeter be affected if the switch 42 S1 is closed while S2 is opened. 1.3 will the reading on the voltmeter be affected if the switch S 1 is closed while Write How down only INCREASES, DECREASES or REMAINS THE SAME. (4) Briefly explain answer. Write down onlythe INCREASES, DECREASES or REMAINS THE SAME. [15] S2 is opened. Exercise 2 Briefly explain the answer. (4) [15] Exercise 2 In the circuit diagram below, the emf of the battery is 6 V and its internal resistance is 0,10 Ω. The resistance In the circuit diagram below, the emf of the battery is 6 V and its internal resistance is 0,10 Ω. (R) is UNKNOWN. The resistance (R) is UNKNOWN. V 6V 0,10 Ω A 4Ω R 2.1 Explain the term internal resistance. 2.2 Write down an equation for the terminal potential difference using the values given. 2.3 2.2 a sketch Write down an equation for the terminal potential difference the values given.the follow(2) Draw graph of terminal potential difference versususing current. Indicate (3) ing2.3in theDraw graph: a sketch graph of terminal potential difference versus current. Indicate the (3) Commented [U113]: Pse check. 2.1 Explain the term internal resistance. (2) (2) (2) following in the graph: 2.4 • Theofvalue of the emf. The value the emf. • The current at which the terminal potential difference is zero. 2.4 The current at which the terminal potential difference is zero. The energy dissipated in 4 Ω resistance is 40 J; the energy dissipated in resistance R The energy dissipated in 4 Ω resistance is 40 J; the energy dissipated in resistance R is is 60 J. 60 J. Calculate: Calculate: 2.5 2.4.1 Resistance R (4) 2.4.2 Total current in the circuit (3) 2.4.3 Reading of the voltmeter (3) A 7 Ω resistor is now connected in parallel to the 4 Ω resistor. How will this action affect the reading of the voltmeter? Write down only INCREASES, DECREASES or 41 2.5 2.4.1 Resistance R (4) 2.4.2 Total current in the circuit (3) 2.4.3 Reading of the voltmeter (3) A 7 Ω resistor is now connected in parallel to the 4 Ω resistor. How will this action affect the reading of the voltmeter? Write down only INCREASES, DECREASES or REMAINS THE SAME. Briefly explain your answer. (4) [21] (4) [21] 4.8 Electrodynamics 4.8 Electrodynamics Problem Solving Strategy for Electrodynamics Problem Solving Strategy for Electrodynamics With machines, look at the energy to identifytothe type ofthe electrical • electrical With electrical machines, look at theconversions energy conversions identify type of electrical machine. machine. Re-draw the circuit diagram to simplify if necessary. • Re-draw the circuit diagram toit,simplify it, if necessary. Identify the type of connection (series/parallel). • Identify the type of connection (series/parallel). Worked example Worked example The simplified diagrams below represent an electric motor and a generator. The simplified diagrams below represent an electric motor and a generator. Y X 1.1 1.1 Which ONE the diagramsabove aboverepresents representsa asimplified simplifieddiagram diagramofofan anelectric electricmotor? Give Which ONE of of the diagrams motor? for Give a reason for your answer. a reason your answer. 1.2 1.2 What type of of generator (AC oror DC) is is represented What type generator (AC DC) representedininthe thesimplified simplifieddiagrams diagramsabove? Give a reason for Give your aanswer. above? reason for your answer. 1.3 1.3 1.4 1.4 1.5 1.5 State ONE method of of increasing the induced State ONE method increasing the inducedemf emfofofthis thisgenerator. generator. Write down ONE use of electric motors. Write down ONE use of electric motors. The maximum potential difference produced by this generator is 12 V The maximum potential difference produced by this generator is 12 V and the and the frequency is 50 Hz. frequency is 50 Hz. 1.1. Sketch a graph of the induced potential difference versus time. 1.1. Calculate Sketch a graph of the induced rms potential differencedifference. versus time. 1.5.2. the induced potential 1.5.2. Calculate thethe induced rms potential difference. 1.5.3. Calculate average power dissipated if a 5 Ω resistor is 1.5.3.connected Calculate the average dissipated if a 5 Ω resistor is connected to this to thispower generator. SOLUTION generator. SOLUTION 42 44 1.1 X. It converts electrical energy into mechanical energy. 1.2 AC generator. It has slip rings. W electrical X.ItItconverts converts electricalenergy energyinto intomechanical mechanicalenergy. energy. X. 1.1 1.1 1.31.2 ANY ONE of the following: ACgenerator. generator. has sliprings. rings. AC ItIthas slip W W Increasing the speed/frequency of rotation ANYONE ONEofofthe thefollowing: following: ANY 1.2 1.3 1.3 Increasingthe the speed/frequency rotation Increasing the number of coils Increasing speed/frequency ofofrotation Increasingthe thenumber numberofofcoils coils Increasing Increasing the strength of the magnetic field Increasingthe thestrength strengthofofthe themagnetic magneticfield field Increasing Insert a soft iron core Insertaasoft softiron ironcore core Insert 1.4 ANY ONE of the following: 1.4 1.4 ANYONE ONEofofthe thefollowing: following: ANY Pumps Pumps Pumps Fans Fans Fans Compressors Compressors Compressors Hairdryers dryers Hair dryers Hair 1.5.1 1.5.1 1.5.1 1.5.2 1.5.2 1.5.2 VVmax/ max/maks maks VVrms rms/ wgk / wgk 22 12 Vrms / wgk 12 Vrms / wgk 2 2 Vrms / wgk 8,49 V Vrms / wgk 8,49 V 1.5.3 1.5.3 1.5.3 45 45 V 2 / wgk 2 Pave / gem V rms Pave / gem rmsR/ wgk R (8,49) 2 Pave / gem (8,49) 2 5 Pave / gem 5 Pave / gem 14,42 W Pave / gem 14,42 W Exercise 1 Exercise 1 A team of grade 12 learners conducted a practical investigation to investigate the A team of grade 12 learners conducted a practical investigation to investigate the relationship between the root mean square value of voltage generated and the frequency relationship between the root mean square value of voltage generated and the frequency of rotation of the armature of the generator using the device shown in the figure below. of rotation of the armature of the generator using the device shown in the figure below. 43 ave / gem 5 Pave / gem 14,42 W Exercise 1 Exercise 1 A team of grade 12 learners conducted a practical investigation to investigate the A team of grade 12 learners conducted a practical investigation to investigate the relationship between the root relationship between the root mean square value of voltage generated and the frequency mean square value of voltage generated and the frequency of rotation of the armature of the generator using of rotation of the armature of the generator using the device shown in the figure below. the device shown in the figure below. The following table shows thetable results obtained. The following shows the results obtained. Frequency Hz) Frequency Hz) Vrms(V) Vrms(V) 1.1. 0 00 0 10 1 10 1 20 2 30 20 3 2 40 4 50 305 3 40 50 4 5 1.1. What was type used of generator used during by thethe learners during the practical Give a reason for What type of generator by the was learners practical investigation? investigation? Give a reason for your answer. your answer. 1.2. Write an investigative question for this investigation. 1.2. Write an investigative this investigation. 1.3. Whatquestion could the for learner’s hypothesis have been for this experiment? 1.3. What could1.5. the learner’s thispaper experiment? Use the hypothesis data to draw ahave graphbeen on thefor graph provided. 1.4. What is the dependent variable in this experiment? 1.5. Use the data to draw a graph on the graph paper provided. 1.6. Write the conclusion for the results obtained. 1.7. Calculate the root mean square value of the voltage when the amplitude of the voltage is 5 V. 1.8. If a resistor of 5 Ω is connected to the generator, determine the root mean square value of the current through the resistor when the frequency of rotation of the armature is 50 Hz. 1.9. Calculate the average power dissipated through the resistor when the frequency of rotation of the armature is 50 Hz. 1.4. 1.6. What is the dependent variable in this experiment? Commented [GI114]: corrected Commented [U115]: Pse check. Write the conclusion for the results obtained. 46 1.10. Draw the graph of instantaneous voltage vs time when the armature rotates with constant frequency of 50 Hz. 1.11. List the advantages of alternating current (AC) over direct current (DC). 44 Exercise 2 The diagram below shows one of the electrical machines studied in class. 2.1 Is this electrical machine a DC motor or a DC generator? Give a reason for your answer. 2.2 What is the direction of rotation of the loop: clockwise or anti-clockwise? 2.3 Label the parts of this electrical machine indicated by A, B and C. 2.4 Write three applications of this electrical machine. 45 5 Revision Questions - Set 2 (master an additional 20%) 5. 5. Revision Questions - Set 2 (master an additional 20%) Revision Questions - SetLaws 2 (master an additional 20%) 5.1 Newton’s 5.1 Newton’s5.1 Laws Newton’s Laws Exercise 1 Exercise 1 Exercise 1 Two blocks of the are same materials are connected by a light, inelastic rope. BlockofA A has of of 5 kg; block Two blocks blocks ofofthe same materials connected by a light, inelastic rope. Block Arope. has a mass Two the same materials are connected by a light, inelastic Block hasaamass mass 5ofkg; block mass of 3fixed kg. rope isB fixed toforce block a100N forceis (is 𝐹𝐹⃗ applied ) of 100Nhorizontally. is The B5 has a mass kg. Another to additional block B and a aforce (( B𝐹𝐹⃗ and kg; block a3mass ofB3has kg. aAnother rope is Another fixed to block and ))ofof100N 5.B has Revision Questions -rope Set 2is (master an 20%) 5.1 Newton’s Laws applied horizontally. The blocks are moving along a horizontal frictionless surface. appliedare horizontally. The blocks are moving along a horizontal frictionless surface. blocks moving along a horizontal frictionless surface. Exercise 1 𝐹𝐹⃗ 𝐹𝐹⃗ aB a Two blocks of the same materials are connected byBa light, inelastic rope. Block A has a mass of 1.1 1.1 1.2 1.3 1.2 5 kg; block B has a mass of 3 kg. Another rope is fixed to block B and a force ( 𝐹𝐹⃗ ) of 100N is 1.1 Draw separate labelled free-body diagrams all the action surface. on the blocks. Draw separate free-body diagrams all the force’saof action onforce’s the blocks. appliedlabelled horizontally. The blocks areofmoving along horizontal frictionless Draw separate labelledthe of all the force’s action on the blocks. 1.2 Calculate acceleration of the blocks. Calculate the acceleration offree-body the blocks.diagrams Calculateofthe the tension in the rope. Calculate1.3 the magnitude themagnitude tension inof the rope. a Calculate the acceleration of the blocks. 𝐹𝐹⃗ B Thenow twopulled blocksover are now pulled over another surfaceexperience and the blocks experience friction. The blocks The two blocks are another surface and the blocks friction. The blocks 1.3 -2. The in Calculate the magnitude them·s tension the rope. exerted by the onN. block A is 62,5 N. accelerate at of 8,972 force exerted byforce the rope on block A rope is 62,5 accelerate at 8,972 m·s-2. The 1.1 Draw separate labelled free-body diagrams of all the force’s action on the blocks. 1.2 Calculate the acceleration of the blocks. The blocks1.4 arekinetic now pulled another blocks experience friction. The blocks accelerate Calculate theover kinetic coefficient of friction forthe block A. 1.4 twoCalculate the coefficient of friction for surface block A. and 1.3 Calculate the magnitude of the tension in the rope. -2 .1.5 Theblocks force exerted by the rope onsame block surface A is block 62,5 N. at1.5 8,972If m·s If thehave two the blocks have the area, will block B have a different the two same surface area, will B have a different 1.4 1.6 1.7 coefficient friction? Explain your answer. coefficient of blocks friction? your over answer. The two areExplain nowofpulled another surface and the blocks experience friction. The blocks Calculate thefrictional kinetic coefficient offorce friction foron block Calculate theexacted frictional exacted byblock the A. surface on block B. Calculate1.6 the force by the surface B. -2 accelerate at 8,972 m·s . The force exerted by the rope on block A is 62,5 N. 1.7action-reaction Name two pairs action-reaction pairs in the system. Name two in the system. If the two blocks have the same surface area, will block B have a different coefficient of 1.4 Calculate the kinetic coefficient of friction for block A. Explain your answer. Exercise 2 Exercise 2 1.5 If the two blocks have the same surface area, will block B have a different 1.5 friction? coefficient of friction? Explain your answer. kg frictional block on a force horizontal, surface is tobyan block by a light inextensible A 4 kgCalculate block on Aa 4horizontal, rough surface isrough connected anconnected 8 kg block a8 light 1.6 the exacted by thetosurface on block B.kg inextensible 1.6 Calculate the frictional force exacted by the surface on block B. string passes over a frictionless shown below.of The coefficient of kinetic (dynamic) string that passes overthat a frictionless pulley, as shownpulley, below.asThe coefficient kinetic (dynamic) 1.7 Name two action-reaction pairs in the system. 1.7 Name two action-reaction pairs in the system. between of 4 kg and the surface is 0,6. friction between friction the block of 4 kgthe andblock the surface is 0,6. Exercise 2 Exercise 2 4 kg 4 kg A 4 kg block on a horizontal, rough surface is connected to an 8 kg block by a light inextensible A 4 kg block on a horizontal, rough surface is connected to an 8 kg block by a light inextensible string that string that passes over a frictionless pulley, as shown below. The coefficient of kinetic (dynamic) passes over a frictionless pulley, as shown below. The coefficient of kinetic (dynamic) friction between the friction between the block of 4 kg and the surface is 0,6. block of 4 kg and the surface is 0,6. 8 kg 8 kg 4 kg 49 49 8 kg 2.1 Draw a free-body diagram of all the forces acting on both blocks. 2.2 Write down Newton’s second law of motion in words. 2.3 Calculate the acceleration of the system. 46 49 2.4 Calculate the magnitude of the tension in the string. 2.5 Calculate the magnitude of the frictional force that acts on the 4 kg block. 2.6 Calculate the apparent weight of the 8 kg block. 2.7 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. 2.8 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. Exercise 3 ZACUBE-1 is the first South African satellite. It was designed and built by postgraduate students, following the CubeSat Programme at the French South African Institute of Technology (F’SATI), in collaboration with the South African National Space Agency (SANSA). ZACUBE-1 was lunched on 21 November 2013 and is orbiting at a height of 600 km from the surface of Earth. Weighing 1.2 kg, this CubeSat is about 100 times smaller than Sputnik - the first satellite launched into space in 1957. 3.1 State Newton’s Law of Universal Gravitation in words. 3.2 (2) How would the magnitude of the gravitational force exerted by the Earth on ZACUBE-1 compare with the magnitude of the gravitational force exerted by the Earth on Sputnik, if both satellites were orbiting at the same height? 3.3 3.4 Write down only LARGER THAN, SMALLER THAN or THE SAME. Explain the answer. (4) Calculate the gravitational force exerted by the earth on the satellite ZACUBE-1. (5) Calculate the height the satellite must be orbiting at for the value of the acceleration due to gravity to be 25% of the gravitational acceleration on the surface of the Earth. (6) [17] 5.2 Momentum and Impulse Exercise 1 (Integration of: Law of Conservation of Momentum, Newton’s Laws and Equations of Motion) The sketch below shows a missile launcher of mass 4800 kg at rest on a frictionless horizontal surface to thebelow base of a smooth inclined launcher plane. It fires rocket4800 of mass horizontally, Thenext sketch shows a missile of amass kg100 at kg rest on a frictionless horizontal surface next and recoils up the inclined plane, rising to a height of 6 m. to the base of a smooth inclined plane. It fires a rocket of mass 100 kg horizontally, and recoils up the inclined plane, rising to a height of 6 m. 6m 𝜃𝜃 1.1. State the law of conservation of linear momentum in words. 1.1. State the law of conservation of linear momentum in words. the system formed by the missile the rocket is said to be isolated? 1.2.1.2. Why Why is the is system formed by the missile launcher andlauncher the rocket and is said to be isolated? 1.3. Calculate the initial speed of the rocket. Answer: 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1 Exercise 2 47 1.1. 1.2. 1.3. 1.3. State the law of conservation of linear momentum in words. Why is the system formed by the missile launcher and the rocket is said to be isolated? Calculate thethe initial speed of theof rocket. Calculate initial speed the rocket. Answer: 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1 Answer: Exercise 2 Exercise 2 P= kg·m·s-1 A Aconstant tothe theright rightis is applied to Trolley a mass 10 kg3during 3 s and starts moving from rest constant force force to applied to Trolley A; it A; hasitahas mass 10 kg during s and starts onmoving a horizontal surface. Aftersurface. the 3 s,After thethe trolley moves constant velocity and collides with Trolley from restfrictional on a horizontal frictional 3 s, the trolleywith moves with constant B,velocity whichand hascollides a mass 8 kg B, and is athas rest. After the After two the trolleys stick together and move as a withofTrolley which a mass of 8the kg collision, and is at rest. collision, system. The graph below shows how the momentum of Trolley A changes with time just before and after the the two trolleys stick together and move as a system. The graph below shows how the momentum collision. of Trolley a changes with time just before and after the collision. Momentum versus time graph for trolley A P= kg·m·s-1 80 70 4,2 3,0 4,0 80 70 0,0 Momentum versus time graph for trolley A 7 51 t (s) 2.1 the term Define the of term impulse of a force in words. 2.1 Define impulse a force in words. 2.2 Use the information in the graph to calculate: 2.2 Use the information in the graph calculate: 0,0 3,0 4,0to4,2 7 t (s) 2.2.1. The impulse of the force exerted on the trolley during the three first seconds. Commented [U119]: Is this one item? Is this complet? Pse check all – repetitive problems are suspected, with items that belong together being split into different items. 2.2.2 Velocity of Trolley A before the collision. 2.2.1 The impulse of the force exerted on the trolley during the three first seconds. Commented [GI120]: corrected 2.2.3 Velocity of the2.1 system formed by Trolley and B after collision. Define the term impulseAof a force in the words. 2.2.4 The average force exerted by trolley B on trolley A during the 2.2.2 2.2 Velocity of Trolley A before the collision. Use the information in the graph to calculate:collision. Commented [U121]: Pse check/correct. Commented [GI122]: corrected [U119]: Is this one ite 2.2.1. The impulse of the force exerted on the trolley during the three first seconds. Commented [U123]: Are theseCommented sub-items of something? 2.2.3 Velocity of the system formed by Trolley A and B after the collision. Pse check all – repetitive problems are Vertical Projectile2.2.2 Motion belong together being split into differe Velocity of Trolley A before the collision. 5.3 Exercise 1 2.2.4 Commented [GI120]: corrected 2.2.3 Velocity of the system formed by Trolley A and B after the collision. The average force exerted by trolley B on trolley A during the collision. 2.2.4 The average force exerted by trolley B on trolley A during the collision. A falling stone takes 0,28 s to travel past a window that is 2,2 m in height. Ignore air friction. Commented [U121]: Pse check/co 5.3 Vertical Projectile Motion Exercise 1 5.3 Commented [GI122]: corrected Commented [U123]: Are these su Vertical Projectile Motion Exercise 1 A falling stoneAtakes to travel a window that isthat 2,2ism2,2inmheight. Ignore falling0,28 stonestakes 0,28 spast to travel past a window in height. Ignoreair air friction. friction. 1.1. From what height above the top of the window was the stone dropped? 52 1.1. From what height above the top of the window was the stone dropped? 1.1. From what height above the top of the window was the stone dropped? 52 48 1.2. Draw the position vs time graph for the motion of the stone for the period of time from when it was dropped until it reached the base of the window. 1.3. Draw the velocity vs time graph for the motion of the stone for the period of time from when it was dropped until it reached the base of the window. Exercise 2 A very small rocket (A) is launched vertically upwards with an initial velocity of 100 m.s-1. At the same time, a stone (B), which is initially at a height of 150 m, is dropped from the top of the very high building. Ignore air resistance. 2.1 Calculate the velocity of stone B when it hits the ground. 2.2 Calculate the time taken for A and B to pass each other. 2.3 Calculate the fly time of the small rocket (A). 2.4 Draw the velocity versus time graph for the motion of the small rocket (A) from the moment it is launched until it strikes the ground. Indicate the respective values of the intercepts on your velocity-time graph. 49 5.4 Work Energy Power QUESTION 1 The diagram below illustrates illustrates a roller-coaster track. Sections A, B and C are frictionless. The diagram below a roller-coaster track. Sections A, B and C are There is friction on sections D and E. The sled has a total mass of 24,7 kg and is at rest at frictionless. There is friction on sections D and E. The sled has a total mass of the top left position. 24,7 kg and is at rest at the top left position. 1.1 1.1 How does the total mechanical energy of the sled at the top left position compare to the total mechanical energy of the sled at the end of section D. Write down GREATER THAN, LESS THAN or EQUAL TO. Then explain your answer. (3) 1.2 How State thethe work-energy theoremenergy in words. does total mechanical of the sled at the top left position compare (3) (2) the speed total mechanical of the sled at theBend of section 1.3 to The of the sled atenergy the beginning of section is 24,25 m·s-1 ; D. theWrite speeddown of the -1 sled at the end of section D is 18 m·s . GREATER THAN, LESS THAN or EQUAL TO. Then explain your answer. 1.3.1 Use the work-energy theorem to calculate the frictional force that acted on the sled in section D. 1.3.2 Use the LAW OF CONSERVATION OF ENERGY (ENERGY PRINCIPLE) to calculate the minimum length of section E required for the sled to come to rest, if it has the same frictional force as calculated in 1.3.1. (4) (5) (6) [16] 5.5 Doppler Effect Exercise 1 1.2 1.3 State the work-energy theorem in words. (2) The siren of a stationary police car emits sound waves at a frequency of 620 Hz. A stationary listener watches car approaching at constant onof a straight the speed the (4) The speed of the sled atthe thepolice beginning of section B him is 24,25 m·s-1 ; velocity -1 -1 sled at the end 18 m·s . is 340 m·s . road. Assume that of thesection speedDofissound in air 1.1 1.2 50 How does the wavelength of the sound waves heard by the listener compare to the wavelength of the sound produced by the siren of the police car when it approaches the listener? Write down LONGER THAN, SHORTER THAN or EQUAL TO. Then explain your answer. 1.3.1 Use the work-energy theorem to calculate the frictional force that Name the phenomenon observed in QUESTION 1.1. acted on the sled in section D. 1.3.2 Use the LAW OF CONSERVATION OF ENERGY (ENERGY PRINCIPLE) to calculate the minimum length of section E required for the sled to come to rest, if it has the same frictional force as calculated in 1.3.1. (4) (5) (6) 54 [16] (1) 5.5 Doppler Effect Exercise 1 The siren of a stationary police car emits sound waves at a frequency of 620 Hz. A stationary listener watches the police car approaching him at constant velocity on a straight road. Assume that the speed of sound in air is 340 m·s-1. 1.1 How does the wavelength of the sound waves heard by the listener compare to the (4) wavelength of the sound produced by the siren of the police car when it approaches the listener? Write down LONGER THAN, SHORTER THAN or EQUAL TO. Then explain your answer. 1.2 Name the phenomenon observed in QUESTION 1.1. (1) 1.3 Calculate the wavelength of the sound waves detected by the stationary listener if the police car moves toward him at a speed of 110 km·h-1. (6) 1.4 How will the answer to QUESTION 7.3 change if the police car moves away from the listener at 120 km·h-1? Write down INCREASES, DECREASES or REMAINS THE SAME. (1) 1.5 Write down ONE application of the Doppler effect in medicine. (1) [13] 5.6 Electrostatics Exercise 1 A group of grade 12 learners obtained the following set of values when investigating Coulomb’s law. Distance between charges (m) 1 2 4 8 Electrostatic force (N) 1600 400 100 25 1.1 Write down an investigative question for this investigation. (2) 1.2 Write down the dependent variable. (1) 1.3 Write down one controlled variable in this investigation. (1) 1.4 Using the results in the table above, plot a graph of electrostatic force versus distance between the charges. (2) 51 1.5 1.6 Write down a conclusion for this investigation. (2) Two small identical positively charged spheres are used in this investigation. 1.6.1 Draw the electric field pattern for the system of the two (2) charged spheres. 1.6.2 Calculate the magnitude of the charges used in this (4) investigation. 1.6.3 One of the charges is removed and the other one is mounted on an insulated stand, as shown in the sketch below. Calculate the electric field at a point 25 cm from the charge. + (3) 25 cm [19] (3) 5.7 Electric Circuits Exercise 1 + (3) 25 cm In the circuit diagram, below the emf of the battery is 12 V and the internal resistance is 0,4 Ω. The ammeter reading is 3 A. The resistance of the three resistors is the same. Ignore the [19] 5.7 Electric Circuits resistance of the wires. Exercise 1 5.7 Electric Circuits Exercise 1 [19] V V In the circuit diagram, emf of the battery is 12 and the the internal resistance is 0,4 Ω. In the circuit diagram, below thebelow emf the of 12 the battery is 12 V Vand internal resistance is 0,4 Ω. The Theisammeter reading is 3 A. of Thethe resistance of the three is the same. the Ignore the ammeter reading 3 A. The resistance three resistors isresistors the same. Ignore resistance of the S 2 resistance of the wires. wires. A V R1 R2 R3 12 V S2 S1 1.1 A Write down TWO differences between electromotive force (emf) and terminal potential difference. 1.1 R1 R2 R3 1.2 State ohm's law in words. (2) 1.3 Calculate the current passing through resistor R 2. (6) 1.4 S1 Determine the reading of the voltmeter. (3) 1.5 How will the reading on the voltmeter be affected if switches S 1 and S2 are 1.1 Write down TWO differences between electromotive force (emf) and terminal (2) Write closed? down TWO differences between electromotive force (emf) and potential difference. Write down INCREASES, DECREASES or REMAINS THE SAME. 1.2 State ohm's law in words. terminal potential difference. Then explain the answer. 1.3 Calculate the current passing through resistor R 2. 1.2 State Ohm’s law in words. 1.3 Calculate the current passing through resistor R2. 1.4 (2) 1.4 Exercise 1.5 2 Determine the reading of the voltmeter. (2) (4) [17] (6) (2) (3) (6) How will the reading on the voltmeter be affected if switches S 1 and S2 are Determine the reading of the voltmeter. (3) closed? Write down INCREASES, DECREASES or REMAINS THE SAME. Then explain the answer. 56 (4) [17] Exercise 2 52 (2) 56 1.5 How will the reading on the voltmeter be affected if switches S1 and S2 are closed? Write down INCREASES, DECREASES or REMAINS THE SAME. Then explain the answer. (4) [17] Exercise 2 Thabo is a grade 12 learner. He is investigating the relationship between potential difference and Thabo is a grade 12 learner. is investigating relationship between potential and current using current using a He resistor with unknownthe resistance. He sets up the circuit showndifference in the diagram. a resistor with unknown resistance. setscircuit up the circuit shown He in the diagram. Thabo adjusts Thabo adjusts the currentHe in the using the rheostat. takes the ammeter reading while the current in the circuit usingthe the rheostat. He takes the reading while the voltmeter is the disconnected. He then voltmeter is disconnected. He ammeter then measures the potential difference across unknown measures the potential difference across the unknown resistor for each current value I. From the readings resistor for each current value I. From the readings taken from the ammeter and voltmeter, he taken from the ammeter and the voltmeter, could complete the following table. could complete following he table. Experiment Current Potential difference number (A) 1 0,02 6 2 0,04 12 3 0,06 18 4 0,08 24 (V) 2.1. Writeinvestigative a possible investigative this practical investigation. 2.1. Write a possible questionquestion for thisforpractical investigation. 2.2. What could Thabo’s hypothesis have been for this investigation? Commented [GI125]: Better lik 2.2. What could Thabo’s hypothesis have been for this investigation? 2.3. What is: 2.3. What is: 2.3.1. the dependent variable Commented [GI126]: corrected 2.3.1. the dependent variable Commented [U127]: Pse check. 2.3.2. the2.4. independent Plot a graph variable? of current vs. potential difference with the experimental data collected. Commented [GI128]: Better in 2.3.2. the independent variable? Commented [U129]: Pse check 2.5. Calculate thevs. gradient of thedifference graph. 2.4. Plot a graph of current potential with the experimental data collected. 2.6. What quantity does the gradient of the graph represent? 2.5. Calculate the gradient of the graph. 2.7. Write a conclusion for this investigation. Commented [GI130]: Better fo 2.6. What quantity does the gradient of the graph represent? 5.8 Electrodynamics Exercise 1 for this investigation. 2.7. Write a conclusion 57 53 5.8 Electrodynamics Exercise 1 The diagram below represents an electrical machine; P is a split ring (commutator). The diagram below represents an electrical machine; P is a split ring (commutator). P 1.1 1.1 Identify theoftype of electrical write conversion down thethat energy Identify the type electrical machine andmachine write downand the energy takes conversion that takes place in this electrical machine. place in this electrical machine. (2) 1.2 1.2 Explain the function of the component P. Explain the function of the component P. 1.3 1.3 The ring (commutator) by slip rings. of the following voltThe splitsplit ring (commutator) is replacedis byreplaced slip rings. Which ONE of theWhich followingONE voltageage-time graphs (Graph A or Graph B)with corresponds with the above change? time graphs (Graph A or Graph B) corresponds the above change? (2) (2) (2) Voltage time Voltage 12 V 12 V time Graph B Graph A Explain your answer. Explain your answer. 1.4 1.4 (3) The light bulb shown in the circuit dissipates energy of 6 J per second. An identical light (5) The light bulb showntoin the circuit 6 J the pernew second. An identical bulb is connected in parallel it. Calculate the dissipates rms current inenergy the circuitofunder light bulb is connected in parallel to it. Calculate the rms current in [12] the circuit under conditions. Assume the emf remains unchanged. the new conditions. Assume the emf remains unchanged. 58 54 (3) (5) [12] 5.9 Photoelectric Effect Exercise 1 The work function of potassium is 3,52 x 10-19 J. An electromagnetic radiation strike on the surface of this metal produces the photoelectric effect. The maximum kinetic energy of the electrons ejected is 4,20 x 10-19 J. 1.1 Define work function. 1.2 Define photoelectric effect. 1.3 Calculate: 1.4 1.3.1 The frequency of the incident radiation. 1.3.2 The wavelength of the incident radiation. 1.3.3 The threshold/cut-off frequency. The intensity of the incident radiation on the metal plate is increased, whilst maintaining constant frequency. State and explain what effect this change has on the following: 1.4.1 kinetic energy of the emitted photo-electrons 1.4.2 number of photo-electrons emitted. a 55 mber of photo-electrons emitted. Commented [GI131]: Better in this way 6 Check your answers - Set 1 r answers - Set 6.1 1 Laws Newton’s Laws Exercise 1 Commented [U132]: Pse check right-hand item below. 1.1 1.2 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ Applying Newton's second law on the y-direction. y-axis. (upwards as positive y-axis ⃗⃗ 𝒇𝒇𝒌𝒌 ⃗⃗⃗ 𝑵𝑵 ⃗⃗𝑨𝑨𝑨𝑨 𝑭𝑭 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦 ⃗⃗ ⃗ ⃗⃗ +𝑭𝑭𝐹𝐹𝑨𝑨⃗y-axis 𝑁𝑁 𝐴𝐴𝐴𝐴 + 𝐹𝐹 𝑔𝑔 = 0 59 y-axis 𝟑𝟑𝟑𝟑𝒐𝒐 𝑁𝑁⃗⃗+ 𝐹𝐹 sin 37𝑜𝑜 − 𝑚𝑚𝑚𝑚 = 0 ⃗⃗⃗x-axis 𝑭𝑭𝑨𝑨𝑨𝑨 𝑵𝑵 ⃗𝑭𝑭⃗𝑨𝑨 𝑜𝑜 ⃗𝑁𝑁 ⃗𝑨𝑨𝑨𝑨= 𝑚𝑚𝑚𝑚⃗⃗⃗ 𝑵𝑵− 𝐹𝐹 sin 𝑭𝑭 ⃗𝑭𝑭⃗37 𝑨𝑨 ⃗⃗𝑨𝑨𝑨𝑨 𝑭𝑭 ⃗⃗ 𝒇𝒇𝒌𝒌 𝒐𝒐 𝟑𝟑𝟑𝟑 ⃗⃗ 𝒇𝒇𝒌𝒌 𝟑𝟑𝟑𝟑𝒐𝒐 y-axis ⃗𝑭𝑭⃗𝑨𝑨𝑨𝑨 x-axis ⃗⃗𝒈𝒈 𝑭𝑭 ⃗⃗ ⃗⃗⃗ 𝑵𝑵 x-axis 𝑭𝑭 ⃗⃗ 𝒇𝒇𝒌𝒌 1.3 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 𝑁𝑁 𝑨𝑨𝑨𝑨 x-axis ⃗⃗𝒈𝒈 𝑭𝑭 ⃗⃗𝒈𝒈 𝑭𝑭 ⃗⃗𝒈𝒈 𝑭𝑭 To calculate the frictional force, we apply Newton's second law in the x-direction. 𝑓𝑓 1.3 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 𝑁𝑁 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑘𝑘 1.3 as 𝑓𝑓𝑘𝑘 positive) 1.3 x-axis. (Right 𝑁𝑁 Commented [U133]: Pse check (including punctuation). (all) To calculate the frictional force, we apply Newton's second law in the x-direction. Exercise 2 ⃗𝑅𝑅𝑅𝑅 To thefrictional frictional force, we apply Newton’s second law in the x-direction. To𝐹𝐹calculate calculate force, we apply Newton's second law in the x-direction. = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 the 𝑓𝑓⃗𝑘𝑘 x-axis. + 𝐹𝐹⃗𝐴𝐴𝐴𝐴 =(Right 0 2.1 as positive) x-axis. (Right positive) x-axis. (Rightasas positive) 𝑜𝑜 −𝑓𝑓𝑘𝑘 + 𝐹𝐹 cos 37 = 0 ⃗ 𝑜𝑜= 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗⃗⃗ 𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos𝐹𝐹𝑅𝑅𝑅𝑅 37 𝑵𝑵 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗ ⃗ + cos 𝐹𝐹𝐴𝐴𝐴𝐴37 = 𝑜𝑜0 𝑓𝑓𝑘𝑘 =𝑓𝑓𝑘𝑘30 𝑓𝑓⃗𝑘𝑘 + 𝐹𝐹⃗𝑜𝑜𝐴𝐴𝐴𝐴 = 0 −𝑓𝑓𝑘𝑘 + 37 𝑁𝑁= 0 23,96 𝑓𝑓𝑘𝑘𝐹𝐹=cos −𝑓𝑓𝑘𝑘 + 𝐹𝐹 cos 37𝑜𝑜 = 0 ⃗⃗ 𝒇𝒇𝒔𝒔 𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos 37𝑜𝑜 𝑓𝑓𝑘𝑘 = 𝐹𝐹 cos 37𝑜𝑜𝑜𝑜 𝑓𝑓𝑘𝑘 23,96 = 0,31 𝜇𝜇𝑘𝑘 = = 𝑓𝑓𝑘𝑘 = 30 cos 37 𝑁𝑁 79,95 𝑓𝑓𝑘𝑘 𝒐𝒐= 30 cos 37𝑜𝑜 𝑓𝑓𝟑𝟑𝟑𝟑 𝑘𝑘 = 23,96 𝑁𝑁 𝑓𝑓𝑘𝑘 = 23,96 𝑁𝑁 ⃗𝑭𝑭⃗𝒈𝒈 𝑓𝑓 1.4.1 Commented [U133]: Pse check (including punctuation). (a Commented [U133]: Pse check (including punctuation). ( ⃗⃗ 𝑭𝑭 23,96 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 = 23,96= 0,31 𝑁𝑁 79,95 𝜇𝜇𝑘𝑘 = 𝑘𝑘 =2.2 = 0,31 𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 𝑁𝑁 79,95 We do not know the magnitude of the gravitational force, butx we can calculate it. 1.4.1 We apply Newton's second law in the direction. 1.4.1 1.4.1 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (20)(9,8) = 196 𝑁𝑁 (Right as positive) Now we can calculatex-axis the component: 𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30𝑜𝑜 = 196 sin 30𝑜𝑜 = 98apply 𝑁𝑁 Newton's second law in the x direction. 60 We 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 We apply Newton's second law in the x direction. 2.3 56 𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 x-axis (Right as positive) x-axis (Right as positive) −23,96 = 10𝑎𝑎⃗𝑥𝑥 2.4 We apply Newton's second 60 law in the We apply Newton's second 60 ⃗𝑥𝑥 y𝐹𝐹⃗𝑅𝑅𝑅𝑅law = in 𝑚𝑚𝑎𝑎the ⃗ x-direction. 𝐹𝐹𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 direction. 𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝑓𝑓⃗𝑘𝑘 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦 −23,96 = 10𝑎𝑎⃗𝑥𝑥 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 −23,96 = 10𝑎𝑎⃗𝑥𝑥 ⃗⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 0 𝑁𝑁 𝐹𝐹⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 61 𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0 200 + (−98) + 𝐹𝐹⃗𝑓𝑓 = (20)(2) 𝑁𝑁 − 𝐹𝐹𝐹𝐹 cos 30𝑜𝑜 = 0 Commented [U134]: Below and all instance Commented [GI135]: Corrected (in in both Commented [U136]: Formatting: pse align correctly (with the number). Repetitive problem correct all. y-axis y-axis y-axis y-axis ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ ⃗⃗⃗ 𝑵𝑵 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵y-axis ⃗⃗ 𝒇𝒇y-axis ⃗⃗⃗ 𝒌𝒌 𝑵𝑵 ⃗⃗ 𝒇𝒇𝒌𝒌 𝒇𝒇 ⃗⃗ ⃗⃗ 𝒇𝒇𝒌𝒌𝒌𝒌 ⃗⃗ 𝒇𝒇𝒌𝒌 ⃗⃗⃗ 𝑵𝑵 ⃗⃗ 𝒇𝒇𝒌𝒌 x-axis x-axis x-axis x-axis x-axis x-axis 𝑭𝑭 ⃗⃗𝒈𝒈 ⃗𝑭𝑭⃗𝒈𝒈 ⃗𝑭𝑭⃗𝒈𝒈 ⃗𝑭𝑭 ⃗𝒈𝒈 ⃗𝑭𝑭⃗𝒈𝒈 ⃗⃗𝒈𝒈 𝑭𝑭 Exercise 2 Exercise Exercise 2 2 Exercise 2 2 Exercise Exercise 22.1 2.1 2.1 2 Exercise 2.1 2.1 2.1 2.1 ⃗⃗ 𝒇𝒇 ⃗⃗ 𝒇𝒇𝒔𝒔𝒔𝒔 ⃗⃗ 𝒇𝒇𝒔𝒔 ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵 ⃗⃗⃗ 𝑵𝑵 ⃗⃗ 𝒇𝒇𝒔𝒔 ⃗⃗ 𝒇𝒇𝒔𝒔 ⃗⃗ ⃗𝑭𝑭⃗ 𝑭𝑭 ⃗⃗ 𝑭𝑭 ⃗⃗ 𝑭𝑭 ⃗⃗ 𝑭𝑭 ⃗𝑭𝑭⃗ 𝒐𝒐 ⃗⃗ 𝟑𝟑𝟑𝟑 𝟑𝟑𝟑𝟑𝒐𝒐𝒐𝒐 𝟑𝟑𝟑𝟑𝒇𝒇𝒐𝒐𝒔𝒔 ⃗𝑭𝑭⃗𝒈𝒈 𝟑𝟑𝟑𝟑 ⃗⃗ ⃗𝑭𝑭⃗𝒈𝒈 𝑭𝑭 ⃗𝑭𝑭 ⃗𝒈𝒈 𝒈𝒈 𝟑𝟑𝟑𝟑𝒐𝒐 ⃗⃗𝒈𝒈 𝑭𝑭 𝟑𝟑𝟑𝟑𝒐𝒐 2.2 𝐹𝐹𝑔𝑔𝑔𝑔 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 ⃗ ⃗sin 2.2 𝐹𝐹 = 2.2 𝐹𝐹 = 𝐹𝐹𝐹𝐹 𝒈𝒈𝐹𝐹𝐹𝐹𝜃𝜃𝜃𝜃sin 𝜃𝜃 2.2 𝑔𝑔𝑔𝑔 = 𝑔𝑔𝑔𝑔 2.2 𝐹𝐹𝑔𝑔𝑔𝑔 𝐹𝐹𝐹𝐹𝑭𝑭sin 2.2 𝐹𝐹We 𝐹𝐹𝐹𝐹not sinknow 𝜃𝜃 𝑔𝑔𝑔𝑔 =do the magnitude of the gravitational force, but we can calculate it. We do not know the magnitude ofof the gravitational force, butcan we calculate can we calculate We do not know the magnitude of the gravitational force, but we it. it. We do not know the the gravitational can 2.2 = 𝐹𝐹𝐹𝐹 sin 𝜃𝜃 magnitude 𝑔𝑔𝑔𝑔 We do not𝐹𝐹𝐹𝐹𝐹𝐹 know the magnitude force, but force, we can but calculate it. calculate it. = 𝑚𝑚𝑚𝑚 = (20)(9,8)of = the 196gravitational 𝑁𝑁 (20)(9,8) 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = = 196 𝑁𝑁 (20)(9,8) 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 =know = 196 𝑁𝑁 of the gravitational force, but we can calculate it. We not= the magnitude (20)(9,8) 𝐹𝐹𝐹𝐹 =do𝑚𝑚𝑚𝑚 196 𝑁𝑁 the component: Now we can=calculate Now we can calculate the We do not know the magnitude of the gravitational force, but we can calculate it. Now we can calculate the component: (20)(9,8) 𝐹𝐹𝐹𝐹 =we 𝑚𝑚𝑚𝑚can = calculate=the 196 𝑁𝑁 component: 𝑜𝑜 Now 𝐹𝐹𝐹𝐹𝐹𝐹 =𝑜𝑜 𝐹𝐹𝐹𝐹 𝑜𝑜sin 30component: = 𝑜𝑜196 𝑜𝑜sin 30𝑜𝑜 = 98 𝑁𝑁 Now we calculate the 𝐹𝐹𝐹𝐹 sin 30196 196 sincomponent: 3098=𝑁𝑁98 𝑁𝑁 (20)(9,8) 𝐹𝐹𝐹𝐹 𝑚𝑚𝑚𝑚 == 196 𝑁𝑁 𝐹𝐹𝐹𝐹𝐹𝐹=𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹=can sin 30 = sin 30 = Now we the 𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹can sincalculate 30𝑜𝑜 = 196 sincomponent: 30𝑜𝑜 = 98 𝑁𝑁 𝑜𝑜 𝑜𝑜 Now we can calculate the component: 𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30 = 196 sin 30 = 98 𝑁𝑁 𝐹𝐹𝐹𝐹𝐹𝐹 = 𝐹𝐹𝐹𝐹 sin 30𝑜𝑜 = 196 sin 30𝑜𝑜 = 98 𝑁𝑁 Commented [U134]: Below and all instance 2.3 2.4 We apply Newton's second law in the Commented [U134]: all instances : in or Commented [U134]: Below Below and all and instances : in or on? 2.3 2.3 We apply Newton's second law in the 2.4 apply We apply Newton's second lawthe in the 2.3 2.4y-We Newton's second law in Commented [U134]: Below[GI135]: and all instances : in (in or on? 2.3 We apply Commented Corrected in both We apply Newton's second law in the yNewton's second law in the y2.4 We apply Newton's second law in the x-direction. Commented [GI135]: Corrected (in incases) both cases) Commented [GI135]: Corrected (in in both We apply Newton's second law in the ydirection. Commented [U134]: Below and all : in or on? 2.3 [GI135]: Corrected (ininstances inFormatting: both cases) Commented [U136]: pse align x-direction. x-direction. 2.4 We apply Newton's second law in the direction. x-direction. direction. Commented [U136]: Formatting: pseitems alignvertically items ve We apply Newton's second law in the yCommented [U136]: Formatting: pse align correctly (with the number). Repetitive problem Commented [GI135]: Corrected (in inalign both cases) [U136]: Formatting: pse items vertically Commented [U134]: Below and all instances : in or on? 2.3 direction. 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑦𝑦 correctly (with the number). Repetitive problems seen inP correctly (with the number). Repetitive problems seen in text. correct all. 2.4 We apply Newton's second law in the x-direction. correctly (with all. the number). Repetitive problems seen in text. P We second law in the y𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝑚𝑚𝑎𝑎⃗ 𝐹𝐹⃗ 𝑚𝑚𝑎𝑎= direction. ⃗ Newton's 𝐹𝐹⃗𝑅𝑅𝑅𝑅 apply = correctcorrect all. Commented [U136]: Formatting: pse align items vertically Commented [GI135]: Corrected (in in both cases) ⃗ 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑅𝑅𝑅𝑅 𝑚𝑚𝑎𝑎⃗𝑦𝑦𝑦𝑦 ⃗⃗ 𝑦𝑦 ⃗ ⃗ correct all. 𝐹𝐹 𝑚𝑚𝑎𝑎= ⃗ 𝑚𝑚𝑎𝑎⃗ 𝐹𝐹𝑅𝑅𝑅𝑅 = x-direction. correctly (with the number). Repetitive problems seen in text. P 𝑁𝑁 + 𝐹𝐹𝑔𝑔𝑔𝑔 = 0 𝐹𝐹⃗𝑅𝑅𝑅𝑅 = 𝑅𝑅𝑅𝑅 𝑚𝑚𝑎𝑎⃗𝑥𝑥𝑥𝑥 ⃗ 𝑥𝑥 ⃗ direction. ⃗⃗ += Commented [U136]: Formatting: pse align items vertically ⃗⃗𝑔𝑔𝑔𝑔 ⃗𝑚𝑚𝑎𝑎 𝐹𝐹 + 𝐹𝐹𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗𝑓𝑓 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗𝐹𝐹⃗𝑦𝑦𝑔𝑔𝑔𝑔 𝐹𝐹 correct all. += 𝑁𝑁 𝑁𝑁 𝐹𝐹 0 =0 61 (with the number). Repetitive problems seen in text. P ⃗ ⃗ ⃗ ⃗𝑁𝑁⃗𝑅𝑅𝑅𝑅+ ⃗ ⃗ ⃗ ⃗ ++⃗𝐹𝐹𝑥𝑥𝑔𝑔𝑔𝑔 𝐹𝐹 = 0 𝐹𝐹⃗𝑅𝑅𝑅𝑅+ = 𝑚𝑚𝑎𝑎 ⃗𝑥𝑥 𝑚𝑚𝑎𝑎⃗𝑥𝑥 𝐹𝐹𝐹𝐹 𝐹𝐹⃗𝑓𝑓 + =𝐹𝐹𝑚𝑚𝑎𝑎 𝑓𝑓 = 𝑔𝑔𝑔𝑔 ⃗𝑔𝑔𝑔𝑔 61 correctly 𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔 = 0 61 ⃗ 𝐹𝐹 + 𝐹𝐹 + 𝐹𝐹 = 𝑚𝑚𝑎𝑎 ⃗⃗⃗𝑅𝑅𝑅𝑅 =⃗𝑔𝑔𝑔𝑔 𝑓𝑓 𝑥𝑥 ⃗ 𝐹𝐹 𝑚𝑚𝑎𝑎 correct all. 61 𝑁𝑁 −=𝐹𝐹𝑦𝑦𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗𝑓𝑓 = (20)(2) 𝑁𝑁 + 0 =0 − 𝐹𝐹 ⃗⃗𝑅𝑅𝑅𝑅 =⃗ 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗200 + (−98) 𝐹𝐹 ⃗(20)(2) 𝑁𝑁 − 𝐹𝐹𝑔𝑔𝑔𝑔 𝐹𝐹 + 𝐹𝐹 ++𝐹𝐹𝑓𝑓(−98) = 200 𝑔𝑔𝑔𝑔 = 0 200 + + 𝑚𝑚𝑎𝑎 𝐹𝐹⃗⃗𝑓𝑓⃗+ = 𝑔𝑔𝑔𝑔(−98) 𝑥𝑥 𝐹𝐹 𝑓𝑓 = (20)(2) 𝑁𝑁 −𝑜𝑜 𝐹𝐹𝐹𝐹 cos 30𝑜𝑜 = 0 61 (−98) (20)(2) 200 + + 𝐹𝐹 = ⃗𝑁𝑁 ⃗ ⃗ 𝑜𝑜 = 0030 𝑔𝑔𝑔𝑔 𝑁𝑁 cos = 300 = 0 𝑁𝑁 + − 𝐹𝐹 𝐹𝐹 =𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹 cos +𝑓𝑓𝐹𝐹 ⃗ + 𝐹𝐹⃗𝑔𝑔𝑔𝑔 + 𝐹𝐹⃗102 𝑔𝑔𝑔𝑔− ⃗⃗𝑥𝑥𝑓𝑓 = 40 𝐹𝐹 = 𝑚𝑚𝑎𝑎 𝑁𝑁 − 𝐹𝐹𝐹𝐹 cos𝑁𝑁30=𝑜𝑜 𝐹𝐹𝐹𝐹 = 0cos 30𝑜𝑜 (−98) 200 61 102 =𝐹𝐹⃗40 102 + 𝐹𝐹⃗⃗𝑓𝑓 + =𝑓𝑓𝐹𝐹⃗40 𝑓𝑓 = (20)(2) 𝑓𝑓+ 𝑜𝑜 𝑜𝑜 102 + 𝐹𝐹 = 40 𝑜𝑜 30 𝑁𝑁 − 030 𝐹𝐹𝐹𝐹 cos 𝑓𝑓 =𝐹𝐹𝑁𝑁 𝐹𝐹𝐹𝐹 cos 𝑔𝑔𝑔𝑔== 𝑁𝑁 ⃗𝑓𝑓 + 𝐹𝐹 = 40 − 102 ⃗ 𝑁𝑁 − = 𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹 cos cos 30 30𝑜𝑜 = 0 (−98) (20)(2) 200 + 𝐹𝐹 = 𝑓𝑓 ⃗𝑓𝑓−= 102 40 𝐹𝐹⃗40 40 − 102 𝐹𝐹⃗ =+ 102 𝑜𝑜 𝑓𝑓𝐹𝐹= 𝑁𝑁 − 𝐹𝐹⃗𝑓𝑓𝑓𝑓 = 40 − 102 = 𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹cos cos30 30𝑜𝑜 = 0 ⃗ 102 𝐹𝐹𝑓𝑓−=102 40 𝐹𝐹⃗𝑓𝑓 =+40 𝑁𝑁 = 𝐹𝐹𝐹𝐹 cos 30𝑜𝑜 ⃗ 𝐹𝐹𝑓𝑓 = 40 − 102 𝑓𝑓 2.5 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑓𝑓𝑘𝑘𝑘𝑘 = 𝑁𝑁𝑘𝑘 2.5 𝑓𝑓 𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 OR 𝜇𝜇𝑘𝑘 = 𝑁𝑁 2.5 62 𝜇𝜇𝑘𝑘62= = 0,365 169,7 𝜇𝜇𝑘𝑘 = = 0,365 169,7 6.2 Solutions: Momentum and Impulse Solutions: Momentum and Impulse Exercise 1 Solutions Exercise 1 Solutions 1.1 Momentum is the product of an object’s mass and its velocity. 1.1 1.2 Momentum is-1the product of an object’s mass and its velocity. 25 kgm.s [Read the graph; initial momentum is where the graph starts]. -1 1.2 1.3 25 kgm.s [Read graph; initial is where theisgraph starts].or decrease; the slope From 1s to 10the s the graph is amomentum horizontal line – there no increase 1.3 Fromof1s tograph 10 s the graph is a horizontal line – there is no increase or decrease; the slope the is zero. of the 1.4 Atgraph 15 s, is 25zero. s and 30 s [These points correspond to the 0 value for both the x and y axis.] 1.4 1.5 At 15From s, 250-15 s and 30 sobject [These points correspond the 0 value for both x andaty15 axis.] s the is moving in the sametodirection (East); thenthe it stops s; BUT it 1.5 Frommoves 0-15 sinthe is moving in the same direction theobject opposite direction from 15s to 25 s. (East); then it stops at 15 s; BUT it moves the opposite direction from 15sstopped to 25 s. and changed direction. / The object could have 1.6 Theinobject could have decelerated, 1.6 The collided object could decelerated, stopped and changed / The objectmoved could in have with have another object at 10s, and stopped for 5 sdirection. from 10-15 s, then the collided with another 10s,s.and stopped for 5 s from 10-15 s, then moved in the opposite directionobject from at 15-25 opposite directionoffrom 15-25acting s. 1.7 The product the force on an object and the time the force acts on the object. 1.7 1.8 The F product force acting on anfrom object the time the force acts on the object. = the ∆p [using the values theand graph] net ∆t of 1.8 FnetF∆t = ∆p [using the values from the graph] net ∆t = 0-25 6.2 Commented [U137]: Pse check item, includ Commented [U137]: Pse check item, including bra Commented [GI138]: corrected Commented [GI138]: corrected 57 6.2 Solutions: Momentum and Impulse Exercise 1 Solutions 1.1 Momentum is the product of an object’s mass and its velocity. 1.2 25 kgm.s-1 [Read the graph; initial momentum is where the graph starts]. 1.3 From 1s to 10 s the graph is a horizontal line – there is no increase or decrease; the slope of the graph is zero. 1.4 At 15 s, 25 s and 30 s [These points correspond to the 0 value for both the x and y axis.] 1.5 From 0-15 s the object is moving in the same direction (East); then it stops at 15 s; BUT it moves in the opposite direction from 15s to 25 s. 1.6 The object could have decelerated, stopped and changed direction. / The object could have collided with another object at 10s, and stopped for 5 s from 10-15 s, then moved in the opposite direction from 15-25 s. 1.7 The product of the force acting on an object and the time the force acts on the object. 1.8 Fnet ∆t = ∆p [using the values from the graph] Fnet ∆t = 0-25 = -25 kgm.s-1 Impulse = 25 kgm.s-1 to the West 6.3 Vertical Projectile Motion Exercise 1 Solutions It is important to make a sketch of the situation and to select positive direction. In this case, let’s use positive direction upwards. If a drawing is not provided, drawing the situation will help in solving the problem. 1.1 a =g= 9,8 (m.s-2 ) downwards. 1.2 H= hi + ∆y v2= vi2+2g∆y At the maximum height, the velocity of the projectile is zero: 58 02 = (10)2+2 (-9,8) ∆y -100 = - (19,6) ∆y ∆y = 5,10 m H = 3,5 + 5,10 = 8,60 m 1.3 1.1 Zero. a turn, here so it stops momentarily.] -2 ) downwards. a =g=[The 9,8 ball (m.smakes 1.2 H= hi + ∆y 2 1.4 The timevrequired is from point 0 of projection to the maximum height where the stone turns back = vi2+2g∆y At the maximum height, thev velocity of the projectile is zero: = 0 m.s-1 towards the ground and the f 2 2 0 = (10) +2 (-9,8) ∆y -100 = - (19,6) ∆y vf =(m.s vi +-2 )gdownwards. Δt 1.1 a =g= 9,8 ∆y = 5,10 m 1.2 H= hi + ∆y 2 H = 3,5 + 5,10 = 8,60 m = vi2+2g∆y 1.1 a =g= 9,8 (m.s-2 0 )vdownwards. = 10+ (-9,8) At the maximum height,Δt the velocity of the projectile is zero: 1.2 1.3 1.4 1.5 1.6 H= hi + ∆y 2 2 = (10) +2 (-9,8) 0Zero. v2=1.3 vi2+2g∆y [The ball ∆y makes a turn, here so it stops momentarily.] Commen -100 = (19,6) ∆y of the square brac At the maximum the Δt velocity projectile is zero: -10height, = (-9,8) 1.4 The time required is from point 0 of projection to the maximum height where the stone turns back ∆y (-9,8) = 5,10∆y m items. 02 = (10)2+2 -1 towards the ground and the vf = 0 m.s H = 3,5 -100 = - (19,6) ∆y + 5,10 = 8,60 m Commen Δtm= 1,02 s ∆y = 5,10 1.3 HZero. ball= makes Commented [U139]: Formatting/re-typing: pse check u vf = mvia+turn, g Δthere so it stops momentarily.] Commen = 3,5[The + 5,10 8,60 square brackets – standardise use throughoutseen text with 1.4 The time required0is=from 0 ofΔt projection to the maximum height where the stone turns back withequ c 10+point (-9,8) items. -1 1.5 Time taken from the point of projection the maximum height and the time from the maximum height towards the ground and the v = 0 m.s f stops momentarily.] Zero. [The ball makes a -10 turn,=here so it Commented [U139]: Formatting/re-typing: pse check use of (-9,8) Δt Commented corrected square brackets – standardise[GI140]: use throughout text with equivalent The time required is from point 0 of projection the maximum height where the stone turns back the point ofsprojection aretoequal items. Δt = 1,02 -1 v = v + g Δt Commented [U141]: Re-typing required – repetitive pr f i towards the ground and the vf = 0 m.s with character spacing – pse standardise throughout. Commentedseen [GI140]: corrected 0 = 10+ (-9,8) Δt -10 (-9,8) Δt +tdown tΔt ==tup vf = vi + g Commented [U141]: Re-typing required – repetitive problems = 1,02 s Δt taken from the point of projection the maximum height and the seen with character spacing – pse standardise throughout. 0 =Δt 10+ (-9,8) 1.5 Time time from the maximum Commen -10 = (-9,8)t height Δt= t when the ball reaches the projection point use of italic up downthe point of projection are equal Δt = 1,02 s t = t +t Commen up down 1.5 Time taken from the point of projection the maximum height and the time from the maximum Commented [U142]: Formatting: Pse check need for a t = 2tup of projection are equal use of italics (pattern not seen by editor). height the point tup= tdown when the ball reaches the projection point Commen = tuppoint +tdownof [U143]:Pse Pse check. Time taken fromt t= the projection the maximum height and the time from the maximum CommentedCommented [U142]: Formatting: check need for ad hoc t == 2t2,04 up x 1,02 use of italics (pattern not seen[GI144]: by editor).corrected =2tdown when the ballsreaches the projection point height the point oftupprojection are equal Commented t = 2 x 1,02 = 2,04 s t = tup+tdownt = 2tup Commented [U143]: Pse check. 1.6 when The total time for the motion of the stone is the time taken by the stone from the point of projection t = 2total x 1,02 =reaches 2,04 t the ball themotion projection up= tdown The 1.6 time for sthe of point the stone is the time taken by the stone from the point of corrected projection to Commented [GI144]: to the maximum height, plus the time for the stone from the maximum height to the ground. Commen 1.6 t The total time for the motion of the stone is the time taken by the stone from the point of projection = 2tup the maximum height, plus the time for the stone from the maximum height to the ground. to the maximum height, plus the time for the stone from the maximum height to the ground. Commented [U145]: Pse check. t = 2 x 1,02 = 2,04 s 1 Commen 2 Δyof viΔt + is g 1=the The total time for the motion stone theΔt time taken by the stone from the point of projection Commented [GI146]: corrected Δy = viΔt + g Δt2 2 to the maximum height, plus 2the time for the stone from the maximum height to the ground. Commented [U145]: Pse check. theheight maximum height to the ground: 1 the2 From Commented [GI146]: corrected From maximum to the ground: Δy = viΔt + g Δt 1 1 2 2 2 -8,60 = (0-8,60 xheight Δt) += (0 (-9,8) Δt xthe Δt) + (-9,8) Δt From themaximum maximum to ground: From the height to the 2 ground: 2 2 1 Δt -8,60 =- 4,9 -8,60 = (0 x Δt) + -8,60 (-9,8) Δt =-2 4,9 Δt2 8,60 2 1,76 8,60 4,9t t Δt2 -8,60 =-4,9 t 1.7 8,60Δt = 1,33 s 4,9 1,76 4,9 tt = t1 (upwards) + t 2 (downwards) Δt = 1,33 s 1,76 = 2,35 s Δt = 1,33 stt = 1,02+ t1,33 t = t1 (upwards) + t2 (downwards) tt = t1 (upwards) + t2 (downwards) tt =s 1,02+ 1,33 = 2,35 s 1.7 tvt f==1,02+ vi + g 1,33 Δt = 2,35 vf = 0 + (-9,8)1,33 v f = vi + g 1.7 Δt vf = vi + g Δt 1.7 vf = 0 + (-9,8)1,33 vf = 0 + (-9,8)1,33 vf = 0 – 13,03 vf = - 13,03 m.s-1 vf = 13,03 (m.s-1 ) downwards downwards 63 63 63 Thetime timetaken takenby by the the object object to to go go up up is is less less than than the the time time taken taken for The for itit to to come come down down from from the the maximum maximum height. Therefore, the area of the triangle representing motion of the stone up will be height. Therefore, area of the representing triangle representing of from the stone up will be less than the area of less than the areathe of the triangle the objectmotion moving maximum height to the ground. the triangle representing the object moving from maximum height to the ground. 1.8 59 ground. maximum height. Therefore, the area of the triangle representing motion of the stone up will be less than the area of the triangle representing the object moving from maximum height to the ground. 1.8 The position vs time graph is actually a mind-map of the actual path travelled by the object. 10 1.9 1.9 The Theposition position graph is actually a mind-map of path the actual travelled vs vs timetime graph is actually a mind-map of the actual travelledpath by the object. by the object. 9 8 10 7 9 6 8 5 7 4 6 3 y (m) 1.9 1.8 y (m) 1.8 5 2 4 1 3 t (s) 2 0 0 0.5 1 Exercise 2 1 1.5 2 2.5 t (s) 0 0 0.5 1 1.5 2 2.5 Solution:Exercise 2 Exercise 2 2.1 Solution: Solution: 2.1 64 64 2.1 From 0 to 1 s, the velocity of the ball decreases from 10 m∙s-1 to zero, and the acceleration of the ball is constant. From 1 to 2 s, the velocity of the ball increases from zero to (– 10 m∙s-1) with constant acceleration. At 2 s, the ball is at the point of projection. From 2 to 3 s, the velocity of the ball increases from (– 10 m∙s-1) to (– 20 m∙s-1). At 3 s, the ball hits the ground and bounces back. From 3 to 4 s, the velocity of the ball decreases to zero at the maximum height. From 4 to 5 s, the velocity of the ball increases with constant acceleration, until it hits the ground for the second time. 60 ball ballhits hitsthe theground groundand andbounces bouncesback. back. -1 From 0 to1From s,From the velocity ballvelocity decreases 10 m∙s to zero, and to the acceleration 3 3toto4of4s,the s,the the velocityoffrom ofthe the ball balldecreases decreases tozero zeroatatthe themaximum maximumheight. height. of the ball is constant. From From4 4toto5 5s,s,the thevelocity velocityofofthe theball ballincreases increaseswith withconstant constantacceleration, acceleration,until untilit ithits hitsthe the -1 From 1 to 2 s, the velocity of the ball increases from zero to (– 10 m∙s ) with constant ground ground for for the the second second time. time. acceleration. At 2 s, the ball is at the point of projection. From 2 to 3 s, the velocity of the ball increases from (– 10 m∙s-1) to (– 20 m∙s-1). At 3 s, the -1 -1 -1 ball hits the ground andupwards bounces back. 2.2 1010 m∙s upwards 2.2 2.2 10 m∙s m∙s upwards From 3 to 4 s, the velocity of the ball decreases to zero at the maximum height. From 4 to 5 s, the velocity of the ball increases with constant acceleration, until it hits the 2.3 Option 11 1 time. Option 2 ground for the second 2.3 2.3 Option Option Option Option 22 |∆𝑦𝑦| |∆𝑦𝑦| 𝐻𝐻𝐻𝐻= -1 = 10 m∙s upwards 2.2 𝑣𝑣𝑓𝑓𝑣𝑣𝑓𝑓++𝑣𝑣𝑖𝑖𝑣𝑣𝑖𝑖 2.3 Option 1 𝐻𝐻𝐻𝐻==( ( ) ∆𝑡𝑡 ) ∆𝑡𝑡 𝐻𝐻 = |∆𝑦𝑦| 22 𝐻𝐻 = ( Option 2 𝑣𝑣𝑓𝑓 + 𝑣𝑣𝑖𝑖 ) ∆𝑡𝑡 −20 (−10) −20++(−10) 2 (3(3 𝐻𝐻𝐻𝐻==( ( 22 −20 + (−10) 𝐻𝐻 = ( ) (3 − 2) 2 𝐻𝐻𝐻𝐻==1515𝑚𝑚𝑚𝑚 ) ) −−2)2) 𝐻𝐻 = 15 𝑚𝑚 11 22 𝐻𝐻𝐻𝐻==𝐴𝐴 𝐴𝐴==𝐴𝐴1𝐴𝐴1++𝐴𝐴2𝐴𝐴2==𝑏𝑏ℎ𝑏𝑏ℎ++ 𝑏𝑏ℎ𝑏𝑏ℎ 1 𝐻𝐻 = 𝐴𝐴 = 𝐴𝐴1 + 𝐴𝐴2 = 𝑏𝑏ℎ + 𝑏𝑏ℎ 2 1 𝐻𝐻 = (1)(−10) + (1)(−10) 2 11 (1)(−10)++ (1)(−10) (1)(−10) 𝐻𝐻𝐻𝐻==(1)(−10) 22 (−5)==1515𝑚𝑚𝑚𝑚 𝐻𝐻 = −10 + (−5) = 15 𝑚𝑚 𝐻𝐻𝐻𝐻== −10 −10++(−5) 2.4 Inelastic, because after bouncing, the speed of the ball is less than the speed it hits the 2.4 2.4 Inelastic, Inelastic, because because after after bouncing, bouncing, thespeed speed ofof the the ball ball isisthan less lessthe than than the thespeed itPseithits hits the thethen 2.4 Inelastic, after bouncing, the the speed of the ball is less speed itspeed hits the ground Commented [U147]: check – not understood. ground then kinetic energy isbecause not conserved. 2.5 ground ground then thenkinetic kineticenergy energyisisnot notconserved. conserved. kinetic energy is not conserved. 2.5 2.5 2.5 Position vs. time (taking ground as level zero) Position Positionvs. vs.time time(taking (takingground groundasaslevel level zero) zero) 25 20 y (m) 15 2525 10 2020 5 1515 y (m) y (m) 0 -5 0 1 2 1010 3 4 5 55 Or 00 -5-50 0 Or 6 t (s) 11 Position vs. time (taking top of building as level zero) 22 33 44 65 5 5 66 t (s) t (s) 10 5 Y (m) 0 -5 0 1 2 3 4 5 6 6565 -10 -15 -20 t (s) Work Energy Power Exercise 1 1.1 The net (total) work (done on an object/particle) is equal to the change in kinetic energy (of the object/particle). 1.2 Data Free-body diagram m = 600 kg FA = 191,7 N Θ = 300 ff = 160,02 N 61 Commented Commente -20 6.4 Work Energy Power Exercise 1 6.4 t (s) Work Energy Power Exercise 1.1 The net (total) work (done on1an object/particle) is equal to the change in kinetic energy 6.4 Work Energy 1.1 Power The net (total) work (done on an object/particle) is equal to the change in kinetic energy (of the object/particle). Exercise 1 1.1 (of the object/particle). The net (total) work (done on an object/particle) is equal to the change in kinetic energy (of the object/particle). 1.2 Free-body diagram 1.2 Data Data Free-body diagram 1.2 Data m = 600 kg mkg = 600 kg m = 600 FA = 191,7 N FA = 191,7 N Θ = 300 FA = 191,7 N ff = 160,02 N Θ = 300 Δx = ? Θ =N 30 ff = 160,02 0 Δx = ? ff = 160,02 N Δx = ? The work-energy theorem can be applied for both conservative and non-conservative forces: W net = ΔEK Wfr +be W Aapplied +W g+ Wfor N = ΔE K The work-energy theorem can both conservative and non-conservative forces: mv2f mvi2 2 2 ΔE W The=work-energy theorem can be applied for both conservative and non-conservative forces: ff Δx cos 1800 + FΔx cos 300 + FgΔx cos 900 + NΔx cos 900 = net K W net = ΔEK Wfr fr++WW +W = ΔE A K +W +g+ WW =NΔE W A g N K ff Δx cos 1800 + FΔx cos 300 + FgΔx cos 900 + NΔx cos 900 = mv2f 2 66 2 i mv 2 600(0) 6003 160,02 Δx (-1) +(191,7) Δx (cos 30 ) + 0 + 0 = 2 2 2 2 0 (-160,02+ 166.017...) Δx= 2700 (5,997...) Δx =2700 Δx = 450,22 m (5,997...) Δx =2700 66 (-160,02+ 166.017...) Δx= 2700 6003 2 160,02 Δx (-1) +(191,7) Δx (cos 300is ) +done: 0+0P =W 1.3= 450,22 Power is the rate at which work Δx m 2 t 600(0) 2 2 2 (-160,02+ 166.017...) Δx= 2700 600(0) 2 W0 6003 Fxcos 160,02 Δx (-1) +(191,7) Δx (cos 30 ) + 0 + 0 = answer P = Alternative 1.3 Power is the rate at which work is done: P (5,997...) Δx =2700 2 2 ∆t t Δx = 450,22 m (-160,02+ 166.017...) Δx= 2700 We must convert seconds: P=F v θ minutes F∆xcos (5,997...) into Δx =2700 W = Power=is300 1.3 the srate at which work is done: P 5P minutes Δx = 450,22 m ∆t t cos 191 , 7 450 , 22 30 o is the rate at which work is done: P W 30 1.3 Power F xcos P=191,7answer cos 300 P ) t Alternative P 300 We must convert minutesFinto 2 xcosseconds: t Alternative answer P P = must 249,03 W minutes t seconds: We convert into P=F v Greater than P=F v W P = 249,03 5 minutes == 300 ss We must convert minutes into seconds: 51.4 minutes 300 minutes = 300 s The work done will 5be greater. o 30 191,7 450,22 cos 30,7 450,22 cos 30 o 0 0 3 As increases; therefore, will greater. P=191,7 cosbe 30 P θ decreases, cosP θ 191 30 0) P=191,7 cos the horizontal component ) 300 300 2 2 P = 249,03 W P = 249,032W Exercise 1.4 Greater than P = 249,03 W P = Greater 249,03 W 1.4 than P = 249,03 2.1 The total mechanical energy sum of gravitational potential energyW and kinetic energy) in The work done will(the be greater. The work done will be greater. As θ decreases, cos θ increases; therefore, the horizontal component will be greater. an isolated system remains constant. 1.4 thancos θ increases; therefore, the horizontal component will be greater. As θGreater decreases, Exercise 2 2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in Exercise 2 The work done will beangreater. isolated system remains constant. 2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in an θ isolated system constant. As decreases, cosremains θ increases; therefore, the horizontal component will be greater. 2.2 From top to bottom of the inclined plane: Data 2.2 From top to bottom of the inclined plane: m= 2 kg Free body diagram Data hi=62 4m m= 2 kg Free body diagram hi= 4 m hf = 0 m 2.2 From top to bottom of the inclined plane: hf = 0 m vi= 0 m·s-1 Data vi= 0 m·s-1 300 5 minutes = 300 s We must convert minutes seconds: P = 249,03 W into cos 191,7 = 450 ,Greater 22 30 o 5Pminutes 300 s 1.4 than 2 P=F v 30 PP=191,7 = 249,03 W cos 300 ) The work done willo be greater. 300 cos 30cos θ increases; therefore, the horizontal component will be greater. 0 32 0 191,7 450 ,θ22 As decreases, P=191,7 cos 30 ) PP =249,03 W 2 300 1.4 Greater than P = 249,03 W 2.1will Thebe totalgreater. mechanical energy (the sum of gravitational potential energy and kinetic energy) in The work done 1.4 than an isolated system remains constant. P = 249,03 W As θGreater decreases, cos θ increases; therefore, the horizontal component will be greater. 2.1 Thedone total will mechanical energy (the sum of gravitational potential energy and kinetic energy) in an The work be greater. As θ decreases, cos θremains increases; therefore, the horizontal component will be greater. isolated system constant. Exercise 2 Exercise 2 Exercise 2 W P = 249,03 2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in Exercise 2 system 2.2 From top to bottom of the inclined plane: an isolated remains constant. 2.2 From top to bottom of the inclined plane: 2.1 The total mechanical energy (the sum of gravitational potential energy and kinetic energy) in Data an isolated system remains constant. Free body diagram m= 2 kg Data hi= 4 m hf = 0 m m= 2 kg vi= 0 m·s-1 g= 9,8 m·s-2 v 2.2 From topi=? to bottom of the inclined plane: hi= 4 m Data 2.2 From top to bottom of the inclined m= 2 kg Freeplane: body diagram h= 0 m Data hi= 4f m Only conservative energy acting the diagram object, the system is isolated therefore mechanical m= kg Free on body hf= 02 m -1 energy is conserved. = 0 m·s v hi= m -1 vi= 04i m·s 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 h = 0 m f g= 9,8 m·s-2 We can apply the principle of conservation of mechanical energy to solve the problem: 9,8-1m·s-2 𝐸𝐸𝑀𝑀𝑀𝑀 = 𝐸𝐸𝑀𝑀𝑀𝑀 vvii= =?0g=m·s 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝐾𝐾𝐾𝐾 = 𝐸𝐸𝐾𝐾𝐾𝐾 + 𝐸𝐸𝐾𝐾𝐾𝐾 g= 9,8 m·s-2 𝑚𝑚𝑣𝑣𝑓𝑓2 𝑚𝑚𝑣𝑣𝑖𝑖2 vi=?vi=? + 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 = + 𝑚𝑚𝑚𝑚ℎ𝑓𝑓 2 Commented [U148]: Pse check – used so that all readers can easily un – repetitive problems seen. 2 Only conservative energy acting on the object, the system is isolated therefore mechanical energy is Only conservative energy acting on the object, the system is isolated therefore mechanical conserved. 67 energy is conserved. Only 𝐸𝐸𝑀𝑀 =conservative 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 energy acting on the object, the system is isolated therefore mechanical energy conserved. We canisapply the principle of conservation of mechanical energy to solve the problem: 𝐸𝐸𝑀𝑀 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = 𝐸𝐸𝑀𝑀𝑀𝑀 𝑀𝑀𝑀𝑀can We apply the principle of conservation of mechanical energy to solve the problem: We can the of conservation of mechanical energy to solve the problem: 𝐸𝐸𝐾𝐾𝐾𝐾apply = 𝐸𝐸𝐾𝐾𝐾𝐾 + principle 𝐸𝐸𝐾𝐾𝐾𝐾 𝐸𝐸𝐾𝐾𝐾𝐾 + 2 2 𝐸𝐸 𝑀𝑀𝑀𝑀 = 𝐸𝐸𝑀𝑀𝑀𝑀 𝑚𝑚𝑣𝑣𝑓𝑓 𝑚𝑚𝑣𝑣 𝑖𝑖 +𝐸𝐸𝑚𝑚𝑚𝑚ℎ + 𝑚𝑚𝑚𝑚ℎ𝑓𝑓 = 𝑖𝑖 + = 𝐸𝐸 𝐸𝐸𝐾𝐾𝐾𝐾 𝐾𝐾𝐾𝐾 𝐾𝐾𝐾𝐾 +2 𝐸𝐸𝐾𝐾𝐾𝐾 22 2 𝑚𝑚𝑣𝑣 𝑚𝑚𝑣𝑣𝑖𝑖 𝑓𝑓 + 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 = + 𝑚𝑚𝑚𝑚ℎ𝑓𝑓 2 2 2𝑣𝑣𝑓𝑓2 2 × 02 + 2 × 9,8 × 4 = 2 2𝑣𝑣𝑓𝑓 2 + 2 × 9,8 × 4 2 2 2×0 + 2 × 9,8 × 4 = + 2 × 9,8 × 4 2 0 + 78,4 = 𝑣𝑣𝑓𝑓2 + 0 2 67 67 v2 = 78,4 v= Commented used so that al – repetitive pro Commente used so that – repetitive p 0 + 78,4 = 𝑣𝑣𝑓𝑓2 + 0 v = 78,4 v = 8,85 m∙s-1 v2 = 78,4 = The 8,85net/total m∙s-1work done on an object is equal to the change in the object’s kinetic energy. 78,4 v2.3 2.4 From A to B. Free 2.3 The net/total work done an object is equalistoequal the change the object’s energy. 2.3 The net/total work on done onbody andiagram object to the in change in thekinetic object’s kinetic energy. Data: 2.4 = 0,2 From Aµkto B. 2.4 Data: Δx = 10 m From A m∙s to B. -1 vi = 8,85 Free body diagram vi = Data: k = 0,2 0,2 µk = µ Δx = 10 m m∙s-1 vi = 8,85 Δx = 10 m vi = The work energy theorem can be applied to both conservative and non-conservative forces. vi = 8,85 m∙s-1 vi = Wnet Ek Wfr WFg WN Ek f Eki f K x cos180 0 Fg cos 90 0 N cos 90 0 Ek f Eki K Nx(1) mv 2f 2 mvi2 2 2 mv 2f mvcan The work energy theorem be applied to both conservative and non-conservative forces. i K mg.x 2 Wnet E k g.x v v 2 f K 2 i 2 63 2.3 The net/total work done on an object is equal to the change in the object’s kinetic energy. 2.4 From A to B. Free body diagram The work energy theorem canapplied be applied both conservative and non-conservative forces. Data: The work energy theorem can be to bothtoconservative and non-conservative forces. µk = 0,2 Δx = 10 m m∙s-1E k viW = 8,85 net vi = Wfr WFg WN Ek f Eki mv 2f mvi2 f K x cos180 Fg cos 90 N cos 90 Ek f Eki K Nx(1) 2 2 2 2 The work energy theorem be applied to both conservative and non-conservative forces. mv f canmv K mg.x i 2 2 Wnet Ek 2 2 Wfr WFg WN vEkf f Evki K g.x i mv 2f mvi2 0 0 2 f K x cos180 Fg cos 902 N cos 90 0 Ek Eki K Nx(1) 0 0 0 f v 2f (vi2 22g.x2 ) 2 2 mv f mv 2K mg.x 2 i v f (8,85) 2 2 2 0,2 9,8 10) v 2f vi2 2 vfK g.78 x ,32 39 ,2 2 2 v 2v vi2 39 2g,12 .x) f ( f v2f (8,85)2 2 0,21 9,8 10) v f 6,25m.s v 2f 78,32 39,2 2.5. From bottom to top. 2.5. From bottom to top. Data: 1 v f 39,12 v f 6,25m.s ⃗⃗ (𝐼𝐼𝐼𝐼 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 𝑁𝑁 2.5. From bottom to top. Data: i Data: v 6,25m.s 1 ⃗⃗ (𝐼𝐼𝐼𝐼 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑛𝑛𝑛𝑛𝑛𝑛 𝑑𝑑𝑑𝑑 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑖𝑖𝑖𝑖 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 𝑁𝑁 vi 6,25m.s 1 𝑓𝑓⃗𝑓𝑓 𝑓𝑓⃗𝑓𝑓 (𝑛𝑛𝑛𝑛 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) (𝑛𝑛𝑛𝑛 − 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) m = 2 kg 𝐹𝐹⃗𝑔𝑔 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) 68 68 𝐹𝐹⃗𝑔𝑔 (𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐) hi = 0 m vi 0 m. s 1 ==?2 kg hm f 𝜇𝜇𝑘𝑘 = 0,2 hi = 0 m When hf = ? there are non-conservative forces acting on a system, we can apply the principle of conservation of When there acting on a system, we can apply the principle of energy (laware of non-conservative conservation offorces energy). conservation of energy (law of conservation of energy). 𝑊𝑊𝑛𝑛𝑛𝑛𝑛𝑛−𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 = ∆𝐸𝐸𝐾𝐾 + ∆𝐸𝐸𝑝𝑝 𝑓𝑓𝑓𝑓 ∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = (𝐸𝐸𝐾𝐾𝐾𝐾 − 𝐸𝐸𝐾𝐾𝐾𝐾 ) + (𝐸𝐸𝑃𝑃𝑃𝑃 − 𝐸𝐸𝑃𝑃𝑃𝑃 ) 𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = ( 𝑚𝑚𝑚𝑚𝑓𝑓2 2 𝜇𝜇𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = ( 0,2 × 2 × 9,8 × 5(−1) = ( 2.6 6.6 𝑚𝑚𝑚𝑚𝑓𝑓2 2 − − 𝑚𝑚𝑚𝑚𝑖𝑖2 ) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 ) 2 𝑚𝑚𝑚𝑚𝑖𝑖2 ) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 ) 2 2×02 2×(6.25)2 − ) + (2 × 9,8 × ℎ𝑓𝑓 − 2 × 9,8 × 0) 2 2 −19.6 = 0 − 39,0625 + 19,6ℎ h = 0,99 m θElectrostatics Exercise 1 Opposite side 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = ℎ𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ℎ = ∆𝑥𝑥 sin 𝜃𝜃 0,99 = 5 sin 𝜃𝜃 sin 𝜃𝜃 = 0,198 θ = 11,450 1.1 electric field is a region of space in which an electric charge experiences a force. The 64 direction of the electric field at a point is the direction that a positive test charge would move if placed at that point. 𝜇𝜇𝜇𝜇𝜇𝜇∆𝑥𝑥𝑥𝑥𝑥𝑥𝑥𝑥 180 = ( 0,2 × 2 × 9,8 × 5(−1) = ( 𝑓𝑓 − 𝑚𝑚𝑚𝑚𝑖𝑖 ) + (𝑚𝑚𝑚𝑚ℎ𝑓𝑓 − 𝑚𝑚𝑚𝑚ℎ𝑖𝑖 ) 2 2×02 2×(6.25)2 − ) + (2 × 9,8 × ℎ𝑓𝑓 − 2 × 9,8 × 0) 2 2 −19.6 = 0 − 39,0625 + 19,6ℎ h = 0,99 m 2.6 2.6 6.6 2 θElectrostatics Exercise 1 Opposite side 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = ℎ = ∆𝑥𝑥 sin 𝜃𝜃 0,99 = 5 sin 𝜃𝜃 sin 𝜃𝜃 = 0,198 θ = 11,450 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ℎ𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 6.5 1.1Electrostatics Electric field is a region of space in which an electric charge experiences a force. The direction of the electric field at a point is the direction that a positive test charge would move Exercise 1 if placed at that point. 1.2.1Electric field is a region of space in which an electric charge experiences a force. The direction of the 1.1 electric field at a point is the direction that a positive test charge would move if placed at that point. 1.2.1 1.2.2 The system is isolated, so we must apply the law of conservation of charge: 69 1.2.2 The system is isolated, so we must apply the law of conservation of charge: 𝑄𝑄𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑄𝑄𝑄𝑄 𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑄𝑄1 + 𝑄𝑄2 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 After being in contact (𝑄𝑄1 = 𝑄𝑄2 = 𝑄𝑄 Afterbeing beinginincontact contact(𝑄𝑄 (𝑄𝑄1==𝑄𝑄𝑄𝑄2==𝑄𝑄𝑄𝑄 After 1 2 𝑄𝑄1 + 𝑄𝑄2 = 2𝑄𝑄 1 + 𝑄𝑄2 = 2𝑄𝑄 𝑄𝑄𝑄𝑄 1 + 𝑄𝑄2 = 2𝑄𝑄 𝑄𝑄1 + 𝑄𝑄2 𝑄𝑄 =𝑄𝑄𝑄𝑄1++𝑄𝑄𝑄𝑄2 1 22 𝑄𝑄 = 𝑄𝑄 = 22 Commented [U149]: Pse check. (brackets) Pse check all ite Commented [U149]: Pse check. (brackets) Pse check all item in text. Commented [U149]: Pse check. (brackets) Pse check all items in text. in text. 30nC 14nC 16nC 2 2 30nC 14 14nC nC 16 16nC 8nC 8 109 C (Charge on each) Q =30 nC (Chargeon oneach) each)QQ== nC 2 2 8nC 8 109 C (Charge 2 2 8nC 8 10 C 9 1.3. To calculate the electrostatic force exerted by one charge on the other we must apply the 1.3. To calculate theelectrostatic electrostatic forceexerted exerted byone onecharge charge onthe the otherwe wemust must applythe the 1.3. To calculate the electrostatic forceby exerted by one charge on the apply other we must apply the mathematical 1.3. To calculate the force on other mathematical expression of Coulomb’s law: mathematical expression of Coulomb’s law: mathematical expression of Coulomb’s law: expression of Coulomb’s law: kQ QB kQAA Q Q FE kQ 2 B FFE E A r2r2 B r 9 1099 8 1099 8 x1099 6,4 1044 9910 109 8810 1029 88x10 x109 10 4 r2 66,4,410 2 r r r = 0,03 m 0,03mm r r ==0,03 Because the charges are equal, we can also solve this in the following way: Becausethe thecharges chargesare are equal, wecan can alsosolve solve this the following way: Because the charges are equal, wealso can alsothis solve this in theway: following way: Because equal, we ininthe following kQ QB kQ2 kQAA Q QB FE kQ QA=QB FE kQ kQ222 2 A=QB E r2 FFE E A r2r2 B QQ FF A=QB E r r r2 9 2 9 9 109 (8 109 )2 6,4 1044 9910 109 (8(82 10 109 ) 2) 4 6 , 4 10 r2 6,4 10 2 r r r = 0,03 m 0,03mm r r==0,03 Exercise 2 Exercise22 Exercise 2.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another onanother another 2.1 The Themagnitude magnitudeofofthe theelectrostatic electrostaticforce forceexerted exertedby byone onepoint pointcharge charge(Q (Q)1)on 2.1 1 of the charges point charge (Q2) is directly proportional to the product of the magnitudes directlyproportional proportionaltotothe theproduct productofofthe themagnitudes magnitudesofofthe thecharges charges pointcharge charge(Q (Q)2)isisdirectly point 2 and inversely proportional to the square of the distance (r) between them. andinversely inverselyproportional proportionaltotothe thesquare squareofofthe thedistance distance(r)(r)between betweenthem. them. and 2.2 According to Newton’s third law, when one body exerts a force on a second body, the 2.2 According AccordingtotoNewton’s Newton’sthird thirdlaw, law,when whenone onebody bodyexerts exertsaaforce forceon onaasecond secondbody, body,the the 2.2 second body exerts a force of equal magnitude in the opposite direction on the first second body exerts a force of equal magnitude in the opposite direction on the first second body exerts a force of equal magnitude in the opposite direction on the first body, therefore: body, therefore: 65 Exercise 2 2.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. 2.2 According to Newton’s third law, when one body exerts a force on a second body, the second body exerts a force of equal magnitude in the opposite direction on the first body, therefore: 2.3 2.4 2.4 2.4 F= kQ1Q2 2 kQ1Q 2 Fr = r2 QA = QB = Q QA = QB = Q (9,0 109 )Q 2 3,0 10 6 (29,0 109 )Q 2 3,0 10 6 2 22 (9,0 109 )Q 2 3,0 10 6 (29,0 109 )Q 2 3,0 10 6 2 22 6 (3,0 10 ) 4 Q2 (3,09 10 6 ) 4 Q92,0 10 9,0 10 9 QA = QB = 3,65 x 10-8 C QA = QB = 3,65 x 10-8 C 2.5 2.52.5 We must draw the electric field vectors at the middle point: We must draw the electric field vectors at the middle point: We must draw the electric field vectors at the middle point: 𝐸𝐸⃗⃗𝐵𝐵 𝐸𝐸⃗⃗𝐵𝐵 𝐸𝐸⃗⃗𝐴𝐴 𝐸𝐸⃗⃗𝐴𝐴 X x X x E R E A E B (It is a vector equation.) E R E A E B (It is a vector equation.) ER= EA + EB is a vector equation.) The vectors are pointing in(Itdifferent directions and positive is taken in the positive direction of The vectors are pointing in different directions and positive is taken in the positive direction of the X axis; therefore: The vectors are pointing in different directions and positive is taken in the positive direction of the X axis; the X axis; therefore: E R Etherefore: A BB E R EA BB R = E A −ofBthe B charges are equal and the distance from the charge to the middle TheE magnitude The magnitude of the charges are equal and the distance from the charge to the middle point are the same; hence: point are the same; hence: 71 66 71 The magnitude of the charges are equal and the distance from the charge to the middle point are the same; EA BB hence: EA BB KQ E 2 r KQ E 2 r 9 109 3,65 108 E 12 9 109 3,65 108 E 12 E R 3 28,5 3 28,5 E R 3 28,5 -13 28,5 E R 0 N C E R 0 N C-1 6.6 Electric Circuits 6.7 Electric Circuits Exercise 1 Exercise 1 Circuits 6.7 Electric Exercise 1.1 When1switch S1 is open, there is no flow of charges, and the current in the circuit is zero; 1.1 When switch S is open, there is no flow of charges, and the current in the circuit is zero; therefore, 1 reads the emf of the battery, which is 10 V. therefore, theSvoltmeter 1.1 When switch 1 is open, there is no flow of charges, and the current in the circuit is zero; the voltmeter reads emf of theinbattery, which is 10 V. 1.2.1 Resistors 2 Ω and 4 Ωthe are parallel and both therefore, the voltmeter reads the connected emf of the battery, which is 10 are V. connected in series with the resistor Rx. 1.2.11.1.1 Resistors 2 Ω and 42ΩΩ areand connected in parallel and both connected seriesare withconnected in series with the Resistors 4 Ω are connected inare parallel andinboth resistor Rext = Rx+ Rthe (1) Rx. R . parallel resistor x Rext = Rx+ Rparallel (1) R1 R2 (1) =R+R RRext (2) ext R xx parallel R R R2 R 1 Rext R x 1 2 R1 R 2(2) (2) R2 R1 Rext = R x + We have to calculate the equivalent external resistance and the equivalent resistance for the R2 + R1 parallel connection. We have calculate equivalent external resistance and the equivalent resistance for the parallel connection. We have totocalculate thethe equivalent external resistance and the equivalent resistance for the parallelVconnection. 8 Rext Rext = 8 Ω Rext I 1 V 8 Rext Rext = 8 Ω Rext 1 1 I 1 1 R parallel R1 R2 1 1 1 R parallel R1 R2 1 1 1 Rp 2 4 1 1 1 Rp 2 4 Rparallel = 1,33 Ω Rparallel = 1,33 Ω We have only two resistors in parallel, so we can use: We have only two resistors in parallel, so we can use: RR Rparallel = 1 2 R R R21 R2 1 R2 R1 2 4 Rparallel = = 1,33 Ω 4 2 2 4 Rparallel = = 1,33 Ω 4 2 Rparallel = 72 Substituting the values of the external resistance and equivalent resistance in parallel in equation 1. 72 8 = Rx+ 1,33 Rx = 8 - 1,33 = 6,67 Ω 1.2.2 ε = Vext + I r 10 = 8 + 1r r = 2 Ω 67 equation 1. Commented [U150]: Pse check/complete. 8 = Rx+ 1,33 Rx = 8 - 1,33 = 6,67 Ω 1.3 INCREASES 1.2.2 ε = Vext + I r Total10 resistance increases. = 8 + 1r r = 2 Ω Current in the circuit decreases. 1.3 INCREASES Total resistance increases. The drop of potential within the battery (Vi= Ir) decreases, but the emf remains constant; therefore, according Current in the circuit decreases. to the equation (Vext = ε- Ir), the voltage increases. The drop of potential within the battery (Vi= Ir) decreases, but the emf remains constant; therefore, Exercise 2 according to the equation (Vext = ε- Ir), the voltage increases. Exercise 2 2.1 When current flows through a voltage source (battery/generator), a resistance to current flow arises due to the resistance of through the materials which tothe source 2.1 When current flows a voltage(chemicals/conductors) source (battery/generator), from a resistance current flowis made. arises due to the resistance of the materials (chemicals/conductors) from which the source 2.2 εis=made. Vext +Ir +I(0,10) 6 V=extV+Ir 2.2 ε = ext 6 = Vext +I(0,10) Vext = 6- (0,10)I Vext = 6- (0,10)I 2.3 2.3 V (V) To calculate the current at which terminal potential difference is zero, we need to substitute Vext with zero in the equation. 6 Vext = 6 - (0,10)I 0 = 6- (0,10)I I (A) 60 I= 6 =60 A 0,1 2.4.1 Resistors 4 Ω and Rx are connected in series, therefore the current is the same for both. 2.4.1 Resistors 4 Ω and Rx are connected in series, therefore the current is the same for both. W = I2RΔt W = I 2RΔt W = I RΔt (1) (1) For the resistor (1) of 4 Ω 2 For the resistor of 4WΩ 4 = I RΔt = 40 For the resistor of 4 Ω 2 2 W W44 == II2RΔt RΔt==40 40 (4)Δt == 40 40 I22(4)Δt I22 == I2(4)Δt = 40 I 2 10 = ∆𝑡𝑡 (2) (3) (2)(2) (4) (3) (3) W R = I2(4) RΔt(4) (5) Substituting 4 in 5 The current currentand andthe thetime timeare arethe thesame, same, therefore resistor The therefore forfor resistor Rx :Rx : 10 (10)Δt = 60 ∆𝑡𝑡 I RΔt R == I2RΔt WW R (5)(5) 2 same for both resistors, we can solve the problem As the current and as thefollows: time are the same both resistors, we can W for = I2RΔt solve the problem as follows: 2 W4 I R4 t ε = I(Rext+r) (1) 2 2 2 (4)t W4 40 I 2 RI 42 t 60 I RR2t 2 WR I RR t RR = 6 Ω I 2 (4)t 40 2 60 I RR t R = 6 Ω To calculate the emf, we need to know the value of theRexternal resistance, as the resistors are connected in series: RR = 6 Ω RR = 6 Ω 68 RR = 6 Ω Substituting 4 in 5 Substituting 4 in 5 2.4.2 2.4.2 2.4.2 As the current and the time are the alternative solution I RR t The current and the time are the same, therefore for resistor RW x : =W I2RRΔt 10 ∆𝑡𝑡 10 (10)Δt = 60 ∆𝑡𝑡 10)Δt = 60 73 alternative solution 𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅 (3) = 4 + 6 = 10 𝛺𝛺 ε = I(Rext+r) 𝑅𝑅𝑠𝑠 (1) ε = I(Rext+r) (1) Substituting the value of the total resistance I series in (1) To calculate the emf, we need to know the value of the external resistance, as the 6 = I(10+ 0,10) resistors are connected in series: 𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅 2.4.3 I = 0,594 A (3) We need to calculate the value of the terminal potential, as we have all the data. Commented [U151]: Pse RR = 6 Ω 2.4.2 ε = I(Rext+r) (1) Tocalculate calculatethe the emf, need to know value of the external resistance, asresistors the To emf, wewe need to know thethe value of the external resistance, as the are connected in resistors are connected in series: series: (3) 𝑅𝑅𝑠𝑠 = 𝑅𝑅4 + 𝑅𝑅𝑅𝑅 𝑅𝑅𝑠𝑠 = 4 + 6 = 10 𝛺𝛺 Substitutingthe thevalue value total resistance I series in (1) Substituting of of thethe total resistance I series in (1) Commented [U151]: Ps I(10+0,10) 0,10) 66==I(10+ I = 0,594 A I = 0,594 A 2.4.3 We need to calculate the value of the terminal potential, as we have all the data. 2.4.3 We need to calculate the value of the terminal potential, as we have all the data. Vext = IRext Vext = IRext Vext = (0,594)(10) Vext = (0,594)(10) Vext = 5,94 V = 5,94 V V ext 6.8 Electrodynamics Exercise 1 6.7 Electrodynamics 1.2. What is the relationship between the root mean square voltage and the frequency of rotation 1.1. ACExercise generator/alternator. It has slip-rings. 1 of the armature /coil of the generator? 1.1. ACroot generator/alternator. It has slip-rings. 1.3. The mean voltage generated when the frequency of rotation 1.2. What is thesquare relationship between the rootincreases/decreases mean square voltage and the frequency of rotation 1.2. What is the relationship between thethe root mean square voltage and thethe frequency of of rotation 1.2. What is the relationship between root mean square voltage and frequency rotation of the armature /coilgenerator of the generator? of the armature /coil of/coil the increases/decreases. of of theis armature of of thethe generator? the armature /coil generator? 1.2. What the relationship between the root mean square voltage and the frequency of rotation of the armature /coil the generator? 1.3. The rootsquare meanofsquare voltage generated increases/decreases when the frequency of rotation 1.4. Root mean voltage. 1.3. The root mean square voltage generated increases/decreases when thethe frequency of rotation 1.3. The root mean square voltage generated increases/decreases when frequency of rotation 1.3. 1.5. 74 of the armature /coil of the generator increases/decreases. of of thethe armature /coil of of thethe generator increases/decreases. armature /coil generator increases/decreases. The root mean square voltage generated increases/decreases when the frequency of rotation of the 1.4. Root mean square voltage. armature /coil of the generator increases/decreases. 1.4. Root mean square voltage. 1.4. Root mean square voltage. 6 1.5. 5 4 6 6 6 5 5 5 4 4 4 3 3 3 2 2 2 1 1 1 0 0 10 20 30 40 0 0 0 10 20 30 (Hz) 40 Frequency 0 0 10 10 20 20 30 30 40 40 Frequency (Hz) Frequency (Hz)(Hz) Frequency Root mean square voltage Rootmean meansquare squarevoltage voltage Root (V) (V) (V) Root mean square voltage (V) 1.5. 1.4. 1.5. Root 1.5. mean square voltage. 3 2 1 0 50 60 50 50 50 60 60 60 1.6. The root mean square voltage generated is directly proportional to the frequency of rotation 1.6. The root mean square voltage generated is directly proportional to the frequency of rotation The root mean square voltage generated is directly to to thethe frequency rotation 1.6. The root mean square voltage generated is directly proportional frequency rotation 1.6. 1.6. The root mean square voltage generated is proportional directly proportional to of theof frequency of rotation of the of the of armature /coil/coil of the generator. the armature of the generator. /coil ofofthe generator. ofarmature thethe armature /coil thethe generator. of armature /coil of generator. V Vmax 5 5 Vmax 5 53,54 3,V 54V =V 1.7=Vrmsmax 1.7 V1.7 rms 3,54 = =max Vrms 3,V 54V Vrms rms1.7 2 2 22 2 2 22 V 5 V Vrms 5 5 1A VIrmsrms 5 rms I rms rms 1.8 1.8 I1.8 RR 5155A1A1A 1.8 rms rms R 1.8 I R 5 1.9. Pave = IrmsVrms = 1 x 5 = 5 W 1.9. 1.9. Pave = I= = 1=x15x=5 5=W 1.9. Pave Vrms 5W rmsIV rms rms 1.9. Pave = IrmsVrms = 1 x 5 = 5 W 1.10. Vmax = 2Vrms 2 5 7,07V 1.10. Vmax = = 2V2 2 2557,07 VV 1.10. Vmax Vrms 7,07 rms 1.10. Vmax = 2Vrms 2 5 7,07V 69 1.8 I rms Vrms 5 1A R 5 1.9. Pave = IrmsVrms = 1 x 5 = 5 W 1.10. Vmax = 2Vrms 2 5 7,07V 1.10. 75 1.11ADVANTAGES ADVANTAGES OF OVER DC DC 1.11 OFACAC OVER Easy to be transformed (step up or step down using a transformer). Easier to convert from AC to DEC than from DC to AC. Easy to betotransformed (step up or step down using a transformer). Easier generate. It can be transmitted at high voltage and low current over long distances with less Easier to convert from AC to DEC than from DC to AC. energy lost. High used in AC makes it suitable for motors. Easier tofrequency generate. • • • • • Exercise 2 It can be transmitted at high voltage and low current over long distances with less energy lost. 2.1 2.2 DC motor. Electrical energy is converted into mechanical energy. High frequency used in AC makes it suitable for motors. Clockwise. Exercise 2 2.3 A - Armature 2.1 DC motor.brushes Electrical energy is converted into mechanical energy. B - Carbon 2.2 Clockwise. 2.4 2.3 C - Split ring commutator (commutator) Electric motors are used in pumps, fans, compressors, etc. A - Armature B - Carbon brushes C - Split ring commutator (commutator) 2.4 Electric motors are used in pumps, fans, compressors, etc. 76 70 7.1 Newton’s Laws SOLUTIONS 1.1 7 Check your answers - Set 2 1.1 Free-body diagram for block A: 7. Free-body diagram for block B: 7.1 Newton’s Laws 7. Check your answers - Set 2 Check your answers 7. -Check Set 2 your answers - Set 2 7.1 x-axis Newton’s Laws7.1 SOLUTIONS SOLUTIONS ⃗⃗⃗ 𝑵𝑵𝑨𝑨 1.1 1.1 7.1 Newton’s Laws SOLUTIONS Newton’s Laws SOLUTIONS 1.1 1.1 x-axis ⃗⃗⃗𝑩𝑩 𝑵𝑵 ⃗⃗ 𝑭𝑭𝑨𝑨 1.1 Free-body diagram for block A: Free-body diagram for block B: 1.1 Free-body diagram 1.1 for Free-body block A: diagram Free-body for diagram block A: for Free-body block B: diagram for B: B: 1.1 Free-body diagram for block A: Free-body diagram block for block x-axis x-axis x-axis x-axis ⃗⃗ x-axis x-axis x-axis𝑻𝑻 x-axis ⃗⃗⃗𝑨𝑨 ⃗⃗⃗𝑩𝑩 𝑵𝑵 𝑵𝑵 ⃗⃗⃗ ⃗⃗⃗ ⃗⃗⃗ ⃗⃗⃗ 𝑵𝑵𝑨𝑨 𝑵𝑵𝑨𝑨 𝑵𝑵𝑩𝑩 𝑵𝑵 ⃗𝑭𝑭⃗𝒈𝒈𝑨𝑨 ⃗𝑭𝑭⃗𝒈𝒈𝑩𝑩 𝑩𝑩 ⃗𝑭𝑭⃗𝑨𝑨 ⃗⃗ 𝑭𝑭 𝑻𝑻 ⃗⃗𝑨𝑨 ⃗⃗𝑨𝑨 𝑭𝑭 ⃗𝑻𝑻⃗ ⃗𝑻𝑻⃗ x-axis x-axis ⃗⃗ 𝑻𝑻 x-axis x-axis ⃗x-axis ⃗ ⃗⃗ x-axis 𝑻𝑻 𝑻𝑻 1.2 ⃗⃗ 𝑻𝑻 ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 ⃗⃗𝒈𝒈𝑩𝑩 𝑭𝑭 ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 We apply Newton's second law to each 1.2 1.2 1.2 object: 1.2 ⃗⃗𝒈𝒈𝑩𝑩 𝑭𝑭 ⃗⃗𝒈𝒈𝑩𝑩 𝑭𝑭 Solving the system We apply Newton's second We apply Newton’s second law to each object: T +law FAto- each T = m1a + m2a We apply Newton's second We apply lawNewton's to each second law to each object: Solving the system Solving the system Solving the system object: object: FA = (m1 + m 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ T 2+ )a FA - T = m1a + m2a T + FA - T = m1a + m2aT + FA - T = m1a + m2a FA = (m1 + m2 )a 100F ==(5 + 3) a 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ ⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ (right positive) FA = (m1 + m2 )a (m1 + m2 )a A 𝐹𝐹x-axis 𝐹𝐹⃗𝑅𝑅 = 𝑚𝑚𝑎𝑎⃗ 100 = (5 + 3) a x-axis (right positive) 100 = (5 + 3) a (5 + 3) a· 𝑠𝑠 −2 to the right 𝑎𝑎⃗100 𝑥𝑥 == 12,5𝑚𝑚 x-axis (right positive) 𝑎𝑎⃗𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 −2 to the right For block −2 x-axis (rightApositive) x-axis (right positive) 𝑎𝑎⃗𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 −2 to the𝑎𝑎⃗right 𝑥𝑥 = 12,5𝑚𝑚 · 𝑠𝑠 to the right For block A T =block mAa A For TT== mAm a Aa For block B For block B For block B a FFAA --TT==mm 2a2 FA - T = m2a 1.3 1.3 For block A T = mAa For block B FA - T = m2a For block A T = mAa For block B FA - T = m2a 1.3 To calculate the magnitude of the tension, we can use block A or block B. To calculate the 1.3magnitude To calculate of the tension, the magnitude we canofuse theblock tension, A orwe block canB. use block A or block B. To calculate the magnitude of the tension, we can use block A or77block B. 77 77 77 71 1.3 To calculate the magnitude of the tension, we can use block A or block B. (UsingA)A) (Using (Using B) (Using A) (Using A) (Using A) x-axis (right positive) x-axis x-axis x-axis x-axis ⃗⃗⃗ 𝑵𝑵𝑨𝑨 ⃗⃗⃗ 𝑵𝑵𝑨𝑨 ⃗⃗⃗ ⃗⃗⃗ 𝑵𝑵𝑨𝑨 𝑵𝑵𝑨𝑨 ⃗⃗ 𝑻𝑻 ⃗⃗ 𝑻𝑻 ⃗⃗ ⃗⃗ 𝑻𝑻 𝑻𝑻 x-axis 𝑇𝑇 = (5)(12,5) 𝐴𝐴 𝑥𝑥 ⃗⃗⃗𝑨𝑨𝑻𝑻 ⃗⃗ 𝑵𝑵 x-axis x-axis ⃗⃗ 𝑻𝑻 ⃗⃗𝒈𝒈𝑩𝑩 ⃗⃗ 𝑭𝑭 𝑻𝑻 x-axis ⃗⃗⃗ 𝑵𝑵𝑩𝑩 ⃗⃗⃗ 𝑵𝑵𝑩𝑩 ⃗⃗𝑨𝑨 𝑭𝑭 ⃗⃗𝑨𝑨 𝑭𝑭 ⃗⃗ 𝑻𝑻 ⃗⃗ 𝑻𝑻 x-axis x-axis ⃗⃗𝒈𝒈𝑩𝑩 𝑭𝑭 ⃗⃗𝒈𝒈𝑩𝑩 𝑭𝑭 ⃗⃗𝒈𝒈𝑩𝑩x-axis 𝑭𝑭 (Using B) x-axis ⃗𝑭𝑭⃗𝑨𝑨 ⃗⃗𝑨𝑨 𝑭𝑭 𝑅𝑅𝑅𝑅𝑅𝑅 ⃗⃗ 𝑻𝑻 𝐵𝐵 𝑥𝑥 1.4 1.4 𝑇𝑇 = 62,5𝑁𝑁 −𝑇𝑇 = 37,5 − 100 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥 𝐹𝐹 𝑚𝑚𝐴𝐴𝐴𝐴𝑎𝑎𝑎𝑎⃗𝑥𝑥 𝑇𝑇 −⃗𝑅𝑅𝑅𝑅𝑅𝑅 𝑓𝑓𝑘𝑘 ==𝑚𝑚 (5)(8,972) 62,5 − 𝑓𝑓 = 𝑇𝑇 − 𝑓𝑓𝑘𝑘 = 𝑇𝑇 − 𝑘𝑘 𝑚𝑚𝐴𝐴 𝑎𝑎 62,5 𝐴𝐴 𝑎𝑎 (5)(8,972) − 𝑓𝑓𝑓𝑓𝑘𝑘𝑘𝑘 = = 𝑚𝑚 𝑅𝑅𝑅𝑅𝑅𝑅 Data: 𝑎𝑎 = 8,972𝑚𝑚 · 𝑠𝑠 −2 Data: Data: 𝑎𝑎 = 8,972𝑚𝑚 · 𝑠𝑠 −2 −2 62,5𝑁𝑁 · 𝑠𝑠 −2 𝑎𝑎 𝑎𝑎 𝑇𝑇==8,972𝑚𝑚 𝑇𝑇 = = 8,972𝑚𝑚 62,5𝑁𝑁 · 𝑠𝑠 1.4 Data: x-axis ⃗𝑭𝑭⃗𝒈𝒈𝑩𝑩 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 𝑇𝑇 𝐹𝐹𝐴𝐴 = − 62,5𝑁𝑁 𝑇𝑇 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 𝑇𝑇 = 62,5𝑁𝑁 100 − 𝑇𝑇 = (3)(12,5) 1.4 1.4 𝐹𝐹⃗ = 𝑚𝑚 𝑎𝑎⃗ 1.4 Data: Data: ⃗⃗𝑨𝑨 𝑭𝑭 x-axis (right positive) 𝐹𝐹⃗𝐴𝐴𝑅𝑅𝑅𝑅𝑅𝑅 − 𝑇𝑇==𝑚𝑚𝑚𝑚 ⃗𝑥𝑥𝑎𝑎⃗𝑥𝑥 𝐹𝐹 𝐹𝐹⃗− 𝐵𝐵 𝐵𝐵 𝑎𝑎 𝑇𝑇 = = 𝑚𝑚 𝑚𝑚𝐵𝐵𝐵𝐵 𝑎𝑎 𝑎𝑎⃗⃗𝑥𝑥𝑥𝑥 x-axis 𝐹𝐹(right positive) 𝐴𝐴 𝑅𝑅𝑅𝑅𝑅𝑅 (3)(12,5) 100 − 𝑇𝑇 = 𝐹𝐹𝐴𝐴 − 𝑇𝑇 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 100 𝐹𝐹𝐴𝐴 − − 𝑇𝑇𝑇𝑇 = = (3)(12,5) 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 𝑇𝑇 = 62,5𝑁𝑁 −𝑇𝑇 = 37,5 − 100 𝑇𝑇 − = 100 62,5𝑁𝑁 ⃗ (3)(12,5) (3)(12,5) 100 − 𝑇𝑇 = 100 − 𝑇𝑇 = 𝐹𝐹𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥 −𝑇𝑇 = 37,5 𝑇𝑇 = 62,5𝑁𝑁 𝑇𝑇 =−62,5𝑁𝑁 100 ⃗𝑥𝑥= 37,5 − 100 −𝑇𝑇 = 37,5 𝑇𝑇 = 𝑚𝑚−𝑇𝑇 𝐴𝐴 𝑎𝑎 𝑇𝑇 = (5)(12,5) ⃗⃗⃗ 𝑵𝑵𝑩𝑩 x-axis x-axis x-axis positive) ⃗𝑭𝑭⃗𝒈𝒈(right x-axis (right positive) 𝑨𝑨 x-axis (right positive) x-axis (right positive) 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑥𝑥 𝐹𝐹⃗ = 𝑚𝑚 𝑎𝑎⃗ x-axis (right positive) x-axis (right positive) 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥 ⃗𝑅𝑅𝑅𝑅𝑅𝑅 ⃗ 𝑎𝑎 ⃗ 𝑎𝑎 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅𝑇𝑇==𝑚𝑚𝑚𝑚 𝐹𝐹 = 𝑚𝑚 𝑚𝑚𝐴𝐴𝑎𝑎⃗𝑎𝑎⃗𝑥𝑥 𝐴𝐴 𝑥𝑥 𝐴𝐴 𝑥𝑥 𝑇𝑇 = 62,5𝑁𝑁 ⃗𝑥𝑥 𝑇𝑇 𝐴𝐴 𝑎𝑎 (5)(12,5) 𝑇𝑇𝑇𝑇= ==𝑚𝑚 (5)(12,5) 𝑇𝑇𝑇𝑇==62,5𝑁𝑁 ⃗⃗⃗ 𝑵𝑵𝑩𝑩 x-axis x-axis (right positive) x-axis (right positive) ⃗𝑥𝑥 𝑇𝑇 𝑇𝑇==𝑚𝑚(5)(12,5) 𝐴𝐴 𝑎𝑎 x-axis ⃗⃗⃗ 𝑵𝑵𝑩𝑩 (Using A) ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 (Using (right B) x-axis positive) x-axis x-axis (Using B) x-axis ⃗⃗𝒈𝒈𝑨𝑨 𝑭𝑭 (Using (Using B) B) 𝐴𝐴 𝑥𝑥 𝑇𝑇𝐹𝐹⃗− 𝑓𝑓𝑘𝑘==𝑚𝑚𝑚𝑚𝐴𝐴𝐴𝐴𝑎𝑎⃗𝑎𝑎𝑥𝑥 𝑅𝑅𝑅𝑅𝑅𝑅 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑥𝑥 44,8662,5 − 62,5 (5)(8,972) 62,5 − −𝑓𝑓 𝑓𝑓𝑘𝑘 𝑘𝑘==(5)(8,972) − 𝑓𝑓𝑘𝑘𝑘𝑘 = = 44,86 − 62,5 −𝑓𝑓 𝑇𝑇 − 𝑓𝑓𝑘𝑘 = 𝑚𝑚𝐴𝐴 𝑎𝑎 78 78 17,64𝑁𝑁 44,86 − 62,5−𝑓𝑓𝑓𝑓𝑘𝑘 = − 62,5 −𝑓𝑓𝑘𝑘𝑓𝑓= 𝑘𝑘 = = 44,86 17,64𝑁𝑁 𝑇𝑇 = 62,5𝑁𝑁 62,5 − 𝑓𝑓𝑘𝑘 = (5)(8,972) 𝑓𝑓𝑘𝑘 = 17,64𝑁𝑁 = 17,64𝑁𝑁 𝑎𝑎 = 8,972𝑚𝑚 ·𝑓𝑓𝑠𝑠𝑘𝑘−2 −𝑓𝑓𝑘𝑘 = 44,86 − 62,5 𝜇𝜇𝑘𝑘 =? Calculating the normal force:the normal force: Calculating 𝑇𝑇 = 62,5𝑁𝑁 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 𝑓𝑓𝑘𝑘 = 17,64𝑁𝑁 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 Calculating the normal force: the normal force: Calculating 𝜇𝜇𝑘𝑘 =? 𝑓𝑓𝑘𝑘 𝑓𝑓 𝜇𝜇 𝑁𝑁 = 𝜇𝜇 𝑁𝑁 𝑓𝑓𝑘𝑘𝜇𝜇= 𝑓𝑓 y-axis 𝑘𝑘 = 𝑘𝑘 𝑘𝑘𝑁𝑁 y-axis 𝜇𝜇𝑘𝑘𝑘𝑘 = 𝑘𝑘 𝑁𝑁 𝑓𝑓 𝑓𝑓 Calculating the normal force: y-axis y-axis 𝜇𝜇𝑘𝑘 = 𝑘𝑘 𝜇𝜇𝑘𝑘 = 𝑘𝑘 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 𝑁𝑁 𝑁𝑁 𝐹𝐹⃗ = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦 We must calculate frictional force and the 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦 Wethe must calculate the frictional force and𝑅𝑅𝑅𝑅𝑅𝑅 the force. 𝑓𝑓𝑘𝑘 normal We must calculate the frictional force and y-axis 𝜇𝜇𝑘𝑘 =the ⃗ 𝑁𝑁 𝑁𝑁 ⃗⃗and 𝐹𝐹𝑅𝑅𝑅𝑅𝑅𝑅 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚 𝑎𝑎⃗ = 𝑚𝑚 𝑎𝑎⃗ normal + 𝐹𝐹⃗the We must force. calculateWe themust frictional force the andfrictional the calculate force 𝑔𝑔 = 0𝐴𝐴 𝑦𝑦 ⃗⃗ + normal force. 𝐹𝐹⃗𝑔𝑔 = 0 𝐴𝐴 𝑦𝑦 𝑁𝑁 ⃗⃗𝑁𝑁+−𝐹𝐹⃗𝑔𝑔𝐹𝐹𝐹𝐹==0 0 ⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0 normal force. the normal force. 𝑁𝑁 Calculating frictional force: 𝑁𝑁𝑁𝑁− 𝐹𝐹𝐹𝐹 force = 0 and the 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐴𝐴 𝑎𝑎⃗𝑦𝑦 We must calculate the frictional Calculating the frictional force: Calculating the frictional force: (5)(9,8) = 49𝑁𝑁 𝑁𝑁𝑁𝑁−= 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹 == 0 𝑚𝑚𝑚𝑚 = 𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0 𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁 ⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0 normal force. 𝑁𝑁 Calculating the frictional force:the frictional force: Calculating x-axis 𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8) 𝑁𝑁 = 𝐹𝐹𝐹𝐹==49𝑁𝑁 𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁 x-axis x-axis 𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0 𝑓𝑓𝑘𝑘 17,64 𝑓𝑓𝑘𝑘 17,64 𝜇𝜇𝑘𝑘 =the = 49 =𝜇𝜇0,355 Calculating force: 𝑁𝑁 frictional 𝑘𝑘 = 𝑁𝑁 = 49 = 0,355 x-axis x-axis 𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (5)(9,8) = 49𝑁𝑁 𝑓𝑓 17,64 𝑓𝑓 17,64 Calculating the normal force: 𝜇𝜇𝑘𝑘 = 𝑁𝑁𝑘𝑘 = 49 = 0,355 𝜇𝜇𝑘𝑘 = 𝑁𝑁𝑘𝑘 = 49 = 0,355 x-axis 78 𝜇𝜇𝑘𝑘 = 𝑓𝑓𝑘𝑘 = 17,6478 = 0,355 y-axis 49 𝑁𝑁 𝑇𝑇𝜇𝜇𝑘𝑘==? 62,5𝑁𝑁 = 62,5𝑁𝑁 𝜇𝜇𝑇𝑇 𝑘𝑘 =? 𝜇𝜇𝑘𝑘 =? 𝑘𝑘 1.5 1.5 78 No, the boxes are made of the same material. 1.5 No, the boxes are made of the same material. 1.6 1.6 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 No, the boxes are made of the same material. 1.6 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 We must calculate the normal force: We must calculate the normal force: 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎⃗𝑦𝑦𝐹𝐹⃗ 1.7 1.7 ⃗𝑦𝑦 𝑅𝑅𝑅𝑅𝑅𝑅 = 𝑚𝑚𝐵𝐵 𝑎𝑎 ⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0 𝑁𝑁 𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0 ⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0 𝑁𝑁 (3)(9,8) = 29,4𝑁𝑁 𝑁𝑁 =𝐹𝐹𝐹𝐹 𝐹𝐹𝐹𝐹= = 𝑁𝑁 − 0 𝑚𝑚𝑚𝑚 = 𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁 Now we can calculate the frictional force: Now we can calculate the frictional force: (0,355)(29,4) = 10,44 to the left. 𝑓𝑓𝑓𝑓𝑘𝑘 ==(0,355)(29,4) = 10,44 to the left. 𝑘𝑘 The force from box A downward on the floor (weight); and the force upwards from the floor on box 72The force from box A downward on the floor (weight); and the force upwards from the floor on box A (normal force). A (normal force). 1.6 e material. ⃗⃗ + 𝐹𝐹⃗𝑔𝑔 = 0 𝑁𝑁 𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 0 𝑓𝑓𝑘𝑘 = 𝜇𝜇𝑘𝑘 𝑁𝑁 We must calculate the normal force: We ⃗⃗ +must 𝑁𝑁 𝐹𝐹⃗𝑔𝑔 = 0calculate the normal force: 𝑁𝑁 − 𝐹𝐹𝐹𝐹 = 𝐵𝐵0𝑎𝑎⃗𝑦𝑦 = 𝑚𝑚 𝐹𝐹⃗𝑅𝑅𝑅𝑅𝑅𝑅 1.7𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁 𝑁𝑁 = 𝐹𝐹𝐹𝐹 = 𝑚𝑚𝑚𝑚 = (3)(9,8) = 29,4𝑁𝑁 Now we can calculate the frictional force: 𝑓𝑓𝑘𝑘 = (0,355)(29,4) = 10,44 to the left. Now we the the frictional force: force: Now wecan cancalculate calculate frictional The force from box A downward on the floor (weight); and the force upwards from the floor on box : A (normal force). 𝑓𝑓𝑘𝑘 = (0,355)(29,4) = 10,44 totothe theleft. left. The force from block A to the left by the rope on block B; and the force to the right by the rope on 1.7 block A. the floor (weight); and the force upwards from the floor on box The force from box A downward on the floor (weight); and the force upwards from the floor on box A (normal force). QuESTION 2 the rope on block B; and the force to the right by the rope on 2.1 The force from block A to the left by the rope on block B; and the force to the right by the rope on block A. QUESTION 2 2.1 Let’s take positive in the direction of motion. Commented [U152]: Pse check. 79 f motion. Commented [U152]: Pse check. 79 Let’s take positive in the direction of motion. 2.2 2.3 If a resultant force acts on a body, theforce bodyacts to accelerate direction ofto accelerate t h in the e direction of 2.2it willIf cause a resultant on a body, it in willthe cause the body resultant force. The acceleration of the the body will beforce. directly to the the body will resultant force and resultant Theproportional acceleration of be directly proportional to the inversely proportional to the mass of the body. force and inversely proportional to the mass of the body. resultant Let’s apply Newton’s second law to each block. 2.3of motion Let’s apply Newton’s second law of ∑ F = ma Block of 4 kg (A) In the x direction: T − f f = mAa T − µN = mAa motion to each block. F ma Block of 8 kg (B) In the y direction: - T Fg mB a Block of 4 kg (A) T mB g mB a (Equation 2) In the x direction: Solving the system of equations (adding 1 T f f mAa plus 2) T N mAa In the Y direction N Fg 0 N Fg mg T mA g T mB g mAa mBa mA g mB g (mA mB )a (0,6)(4)(9,8) (8)(9,8) (4 8)a 54.88 12a 73 a 4,57 m∙s-2 Block of 4 kg (A) Block of 4 kg (A) Solving the system of equations (adding 1 In the x direction: In the x direction: plus 2)In the x direction: T f m a T f f f mAAa T f f mAa T N m a T T N mAA amA g T mB g mAa mBa T N mAa mA g mB g (mA mB )a T mBBg mBBa (Equation 2) T mB g mB a (Equation 2) Solving the system of equations (adding 1 Solving the system of equations (adding 1 plus 2) Solving the system of equations (adding 1 plus 2) plus 2) T m g T m g m a m a T mAAg T mBBg mAAa mBBa T m g T m g m a mBa m g mA g (m Bm )a A mAAg mBBg (mAA mBB)a mA g mB g (mA mB )a In the Y direction In the Y direction (0,6In )(4the )(9,Y 8)direction (8)(9,8) (4 8)a (0,6)(4)(9,8) (8)(9,8) (4 8)a (0,6)(4)(9,8) (8)(9,8) (4 8)a (0,6)(4)(9,8) (8)(9,8) (4 8)a N F 0 N Fgg 0 N Fg 0 54.88 12a N Fg mg -2 N Fg mg N Fg mga 4,57 m∙s 54.88 12a 54.88 12a 54m∙s .88 -2 12a a 4,57 a 4,57 m∙s-2 a 4,57 m∙s-2 T m g m a (Equation 1) T mAAg mAAa (Equation 1) T mA g mAa (Equation 1) 2.4 2.4 2.4 2.4 using equation 1 using equation 2 Usingequation equation 2 using 2 T mA g mAa using equation 2 T m g m a T m Bg m Ba using equation 1 Using using equation 1 equation 1 using equation 1 T m g m a T m Ag m Aa A T) B mB g mB a T Am g m a T (0,6)( 4)(9,8) (4)(B-4T.57 8 4,57 8 9,8 T (0,6)( 4)(9,8) (4)( 4.A57) A T (0,6)( 4)(9,8) (4)( 4.57) - T 8 4,57 8 9,8 4),57 8 9,8 T (0,6)( 4)(9,8) (4)( 4.57) T (0,6)(T4)( ,8) -(TN 4)(48.57 = 941.84 T (0,6)( 4)(9,8) (4)( 4.57) T = 41.84 N T (0,6)( 4)(9,8) (4)( 4.57) T = 41.84 N T (0,6)( 4)(9,8) (4)( 4.57) T 41.8 0 N T 41.8 0 N T 41.8 0 N T 41.8 0 N 2.5. f N 2.5. f 2.5. 2.5. ,mg ff 0 6 4 9,8 ff ff 23 0,6.52 4N 9,8 f f 23.52 N 2.6 80 f N f f f N f f N f f mg f f mg f f mg 80 80 80 2.6 2.6 T mB g mB a T mB g mB a T 8 4,57 8 9,8 T 8 4,57 8 9,8 Apparent weight = T = 41.84 N Apparent weight = T = 41.84 N Apparent weight = T = 41.84 N 2.7 Less than. 2.8 Equal to. Exercise 3 3.1 Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. (2) 74 81 2.8 Equal to. 3.1 Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to Exercisethe 3 square of the distance between their centres. 3.1 Each particle in the universe attracts every other particle with a gravitational force that is directly proportional to the product of their masses and inversely proportional to Smaller than. 3.2 3.2 (2) (2) Smaller than. the square of the distance between their centres. The distance is the same as well as the mass of the Earth in both cases. The Commented [U153]: Pse check The distance is the same as well as the mass of the Earth in both cases. The magmagnitude of the gravitational force exerted on the satellites is directly proportional to Smaller nitude of than. the gravitational force exerted on the satellites is directly proportional to the the mass of the satellites. The mass of ZACUBE-1is issmaller smallerthan thanthe themass mass of of Sputmass the satellites. The mass of mass ZACUBE-1 The of distance is the same as well as the of the Earth in both cases. The Commented [U153]: Pse check. Sputnik. (4) nik. (4) magnitude of the gravitational force exerted on the satellites is directly proportional to 3.2 the mass of the satellites. The mass of ZACUBE-1 is smaller than the mass of GMm FSputnik. r2 3.3 3.3 3.3 F GMm (4) 24 11 6,67 r 2 10 6 10 1,2 F (6,38 10116 600 2410 3 ) 2 6,67 10 6 10 1,2 6 3 2 Earth. F =F9,85 Towards the (6,N 38 10 600 10 ) F = 9,85 N Towards the Earth. (5) F = 9,85 N Towards the Earth. 3.4 (5) (5) On the surface of the Earth. 3.4 On the surface of the Earth. 3.4 On the surface of the Earth. GmM F GmM 2 F R 2 R GmM 2 (1) OR/OF R g GM mg GmM mg R 2 2 (1) OR/OF R 2 g GM R AA height heighth.h. GmM GmM F1F1 2 ( R( Rhh))2 GmM mg1 GmM 2 (2) OR/oF ( R h) 2 g21 GM mg1 ( R h)2 (2) OR/oF ( R h) g1 GM ( R h) From (1) and (2) From (1) and (2) g1 R2 (3) g ( R h) 2 82 82 g1 = 0,25 g 0,25 R2 1 R2 OR/OF 2 ( R h) 4 ( R h) 2 From (3) we get ℎ2 + 2𝑅𝑅ℎ + (𝑅𝑅 2 − ℎ = −𝑅𝑅 ± ℎ = −𝑅𝑅 + 𝑅𝑅 2 )=0 0,25 𝑅𝑅 √0,25 𝑅𝑅 0,5 ℎ = −3,38 × 106 + (𝑅𝑅 + ℎ)2 = 4𝑅𝑅 2 𝑅𝑅 2 + 2𝑅𝑅ℎ + ℎ2 = 4𝑅𝑅 2 ℎ2 + 2𝑅𝑅ℎ − 3𝑅𝑅 2 = 0 (ℎ – 𝑅𝑅)(ℎ + 3𝑅𝑅) = 0 3,38 ×106 0,5 ℎ = 𝑅𝑅 = 3,38 x 106 m 7.2 (ℎ – 𝑅𝑅)(ℎ + 3𝑅𝑅) = 0 ∴ ℎ = 𝑅𝑅 𝑜𝑜𝑜𝑜 − 3𝑅𝑅 (6) (6) [17] [17] Momentum and Impulse Exercise 1 1.1. The total linear momentum of an isolated/a closed system remains constant (is conserved). 1.2. The resultant/net external force is zero. 1.3 F 75 ext = 0 ℎ = 𝑅𝑅 = 3,38 x 106 m ∴ ℎ = 𝑅𝑅 𝑜𝑜𝑜𝑜 − 3𝑅𝑅 (6) [17] 7.2 and Impulse 7.2 Momentum Momentum and Impulse Exercise Exercise1 1 1.1.The Thetotal totallinear linear momentum of an isolated/a closed system remains constant (is conserved). 1.1. momentum of an isolated/a closed system remains constant (is conserved). 1.2.The Theresultant/net resultant/net external force is zero. 1.2. external force is zero. 1.3 1.3 F ext = 0 Therefore: Δ p = 0 Hence: p p after before p = p - = 0 before after ( p r + pb ) before = ( pr + pb )after 𝑚𝑚𝐿𝐿 𝑣𝑣⃗𝐿𝐿(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) + 𝑚𝑚𝑅𝑅 𝑣𝑣⃗𝑅𝑅(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 𝑚𝑚𝐿𝐿 𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝑅𝑅 𝑣𝑣⃗𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚𝐿𝐿 0 + 𝑚𝑚𝑅𝑅 0 = 4800𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) ⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚 0 +4800𝑣𝑣 𝑚𝑚𝑅𝑅 0 =⃗4800𝑣𝑣 + 100𝑣𝑣𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 100 0 𝐿𝐿= 𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 0 = 4800𝑣𝑣⃗𝐿𝐿(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 83 Wehave have two unknowns, soneed we need to speed calculate the ofspeed ofafter thelauncher missile launcher afterwas thefired. rocket We two unknowns, soneed we to calculate the speed the missile the rocket We have two unknowns, so we to calculate the of the missile launcher the rocket after was wasfired. fired. Using equations of motion Using equations of motion Using equations of motion 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑥𝑥 𝑣𝑣𝑓𝑓2 do = not 𝑣𝑣𝑖𝑖2 know + 2𝑎𝑎∆𝑥𝑥 We the acceleration of the launcher on the inclined plane, therefore we must calculate it. Commented [U154]: Is this a heading? (Pse re-check all formatting and standardise all equivalent items to improve th Com legibility of the text. form The for the missile launcher on the inclined planeinclined is: Wefree-body do not diagram know the acceleration ofmoving the launcher on the plane, therefore we must calculate it. The free-body diagram for the missile launcher moving on the inclined plane is: 𝐹𝐹⃗𝑔𝑔 ⃗⃗ 𝑁𝑁 𝐹𝐹⃗𝑔𝑔 𝐹𝐹⃗𝑔𝑔 𝐹𝐹⃗𝑔𝑔𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ In the x direction 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ⃗𝑔𝑔 −𝐹𝐹𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 𝐹𝐹 ⃗⃗ 𝑁𝑁 𝐹𝐹⃗𝑔𝑔 −𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 ⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ 𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃𝐹𝐹 In the x direction 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 ℎ = ∆𝑥𝑥 sin 𝜃𝜃 −𝐹𝐹 𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ −𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 02 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ 𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃 𝑣𝑣𝑖𝑖2 = 2𝑔𝑔ℎ 𝑣𝑣 𝑣𝑣𝑖𝑖2𝑓𝑓2==2(9,8)6 𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥 𝑣𝑣𝑖𝑖 = 10.84 𝑚𝑚 ∙ 𝑠𝑠 −1 ℎ = ∆𝑥𝑥 sin 𝜃𝜃 76Now we can calculate the initial speed of the rocket: 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ 84 legib We have two unknowns, so we need to calculate the speed of the missile launcher after the rocket was fired. Using equations of motion Commented [U154]: Is this a heading? (Pse re-check all formatting and standardise all equivalent items to improve t legibility of the text. 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2𝑎𝑎∆𝑥𝑥 We do not know the acceleration of the launcher on the inclined plane, therefore we must calculate it. We do not know the acceleration of the launcher on the inclined plane, therefore we must calculate it. Thefree-body free-body diagram for the launcher moving onmoving the inclined planeinclined is: The diagram formissile the missile launcher on the plane is: ⃗⃗ 𝑁𝑁 𝐹𝐹⃗𝑔𝑔 𝐹𝐹⃗𝑔𝑔 𝐹𝐹⃗𝑔𝑔 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑚𝑚𝑎𝑎⃗ In the x direction 𝐹𝐹⃗𝑔𝑔𝑔𝑔 = 𝑚𝑚𝑎𝑎⃗𝑥𝑥 −𝐹𝐹𝑔𝑔 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 −𝑚𝑚𝑚𝑚 sin 𝜃𝜃 = 𝑚𝑚𝑚𝑚 𝑎𝑎 = −𝑔𝑔 sin 𝜃𝜃 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 + 2(−𝑔𝑔 sin 𝜃𝜃)∆𝑥𝑥 ℎ = ∆𝑥𝑥 sin 𝜃𝜃 𝑣𝑣𝑓𝑓2 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ 02 = 𝑣𝑣𝑖𝑖2 − 2𝑔𝑔ℎ 𝑣𝑣𝑖𝑖2 = 2𝑔𝑔ℎ 𝑣𝑣𝑖𝑖2 = 2(9,8)6 𝑣𝑣𝑖𝑖 = 10.84 𝑚𝑚 ∙ 𝑠𝑠 −1 can calculate the initial speed of the rocket: InNow thewe x direction 0 = 4800 × 10.84 + 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) Now we can calculate the initial speed of the rocket: 100 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = −52032 84 Exercise 2 Solution 𝑣𝑣𝑅𝑅(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 520,32 𝑚𝑚 ∙ 𝑠𝑠 −1 Exercise 2 2.1 Impulse is the product of the resultant/net force (force) acting on an object and the time the resultant/net force (force) acts Solution on the object. 2.2 resultant/net force (force) acts on the object. Data 2.2 Data 2.1 Impulse is the product of the resultant/net force (force) acting on an object and the time the Commented [G155]: corrected Commented [U156]: Pse check. 𝐹𝐹⃗ = 10 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 Δt= 3 s Impulse -? Let’s take positive to the right 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +10 × 3 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +30 𝑁𝑁 ∙ 𝑠𝑠 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 30 𝑁𝑁 ∙ 𝑠𝑠 to the right 2.3 Data 𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1 𝑚𝑚 = 10 𝑘𝑘𝑘𝑘 𝑣𝑣⃗ =? Let’s take positive to the right: 𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣⃗ + 80 = 10𝑣𝑣⃗ 𝑣𝑣⃗ = + 8 𝑚𝑚 ∙ 𝑚𝑚−1 2.4 Data 𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘 𝑚𝑚𝐵𝐵 = 8 𝑘𝑘𝑘𝑘 = 8 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right 𝑣𝑣⃗ 77 Commented [U157]: Formatting: pse ch and standardise appearance of all eqivalent i 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡 𝐹𝐹⃗ = 10 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑟𝑟𝑟𝑟𝑟𝑟ℎ𝑡𝑡 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +10 × 3 Δt= 3 s Impulse -? 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +30 𝑁𝑁 ∙ 𝑠𝑠 Let’s = take to the right 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 30positive 𝑁𝑁 ∙ 𝑠𝑠 to the right 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝐹𝐹∆𝑡𝑡 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = +10 × 3 2.3 2.3 Data Data 𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 𝑘𝑘𝑘𝑘 +30 𝑁𝑁 ∙ 𝑠𝑠 𝑚𝑚 = 10 𝑣𝑣⃗ =? 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 = 30 𝑁𝑁 ∙ 𝑠𝑠 to the right Let’s take positive to the right: 2.3 Data 𝑝𝑝⃗ = 𝑚𝑚𝑣𝑣⃗ 𝑝𝑝⃗ = 80 𝑘𝑘 · 𝑚𝑚 · 𝑠𝑠 −1 𝑚𝑚 = 10 𝑘𝑘𝑘𝑘⃗ + 80 = 10𝑣𝑣 𝑣𝑣⃗ =? Let’s𝑣𝑣⃗ take = + 8positive 𝑚𝑚 ∙ 𝑚𝑚−1to the right: 𝑝𝑝⃗ =Data 𝑚𝑚𝑣𝑣⃗ 2.4 𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘 + = 10𝑣𝑣⃗ to the 𝑚𝑚𝐵𝐵right = 80 8 𝑘𝑘𝑘𝑘 𝑠𝑠 −1 to the right 𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 8 𝑚𝑚 ∙ −1 𝑣𝑣⃗ = + 8 𝑚𝑚 ∙ 𝑚𝑚 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 0 2.4 Data =? 𝑣𝑣⃗𝐴𝐴𝐴𝐴 Data 2.4 𝑚𝑚𝐴𝐴 = 10 𝑘𝑘𝑘𝑘 For this problem, we are going to select right as positive. 𝑚𝑚𝐵𝐵 = 8 𝑘𝑘𝑘𝑘 𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 8 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = 0= 0 𝑣𝑣⃗𝐴𝐴𝐴𝐴 = ? Commented [U157]: Formatting: pse check use of and standardise appearance of all eqivalent items. Commented [U157]: Formatting: pse check use of and standardise appearance of all eqivalent items. 85 For this problem, we are going to select right as positive. ∑ 𝐹𝐹⃗𝑒𝑒𝑒𝑒𝑒𝑒 = 0 ∆𝑝𝑝⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0 ⃗𝑛𝑛𝑛𝑛𝑛𝑛 = − 0 𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 0 𝑝𝑝⃗∆𝑝𝑝 𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 85 = 𝑝𝑝⃗𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 − 𝑝𝑝⃗𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 0 𝑝𝑝⃗𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝⃗𝑝𝑝⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 𝑝𝑝𝑝𝑝⃗⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝑇𝑇(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 + 𝑇𝑇(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) = 𝑝𝑝⃗𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑣𝑣⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) = 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚 ⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) + 𝑝𝑝 = 𝑝𝑝⃗𝐴𝐴 (𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑝𝑝⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑝𝑝⃗𝐴𝐴𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) After𝑚𝑚the collision,+the trolleys=move together as𝐵𝐵 𝑣𝑣one object; therefore, the velocity is the same: ⃗𝐴𝐴(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) ⃗𝐵𝐵(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) 𝑚𝑚𝐵𝐵two 𝑣𝑣⃗𝐵𝐵(𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏) 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝐴𝐴(𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎) + 𝑚𝑚 𝐴𝐴 𝑣𝑣 𝑣𝑣⃗𝑖𝑖𝑖𝑖 collision, + 𝑚𝑚𝐵𝐵 𝑣𝑣⃗𝑖𝑖𝑖𝑖 the = 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 ) 𝑚𝑚𝐴𝐴the 𝐴𝐴 + 𝑚𝑚𝐵𝐵move After two(𝑚𝑚 trolleys together as one object; therefore, the velocity is the same: (10)(+8) +𝐵𝐵(8)(0) 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 + 𝐵𝐵8) ⃗𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣⃗𝑖𝑖𝑖𝑖 = 𝑣𝑣= (𝑚𝑚(10 ) 𝑚𝑚𝐴𝐴 𝑣𝑣⃗𝑖𝑖𝑖𝑖 + 𝑚𝑚 𝐴𝐴 + 𝑚𝑚 +80 = 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓+(18) (10)(+8) (8)(0) = 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 (10 + 8) = 80 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 (18) 𝑣𝑣⃗+80 𝑓𝑓𝑓𝑓𝑓𝑓 = 80 18 −1 𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣⃗= += 4,44 18 𝑚𝑚 ∙ 𝑠𝑠 . −1 −1 𝑣𝑣⃗𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 = 4,44 𝑚𝑚 ∙ 𝑠𝑠 to .the right. = + 4,44 𝑚𝑚 ∙ 𝑠𝑠 𝑓𝑓𝑓𝑓𝑓𝑓 𝑣𝑣⃗𝑓𝑓𝑓𝑓𝑓𝑓 = 4,44 𝑚𝑚 ∙ 𝑠𝑠 −1 to the right. 2.5 We can apply Newton’s second law in terms of momentum: ∆𝑝𝑝⃗ 2.5 apply Newton’s second in terms of momentum: cancan apply Newton’s second lawlaw in terms of momentum: 𝐹𝐹⃗2.5 = WeWe 𝑛𝑛𝑛𝑛𝑛𝑛 ∆𝑡𝑡 ∆𝑝𝑝⃗ 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = ∆𝑡𝑡 5.5 Vertical Projectile Motion Exercise 5.5 1Vertical Projectile Motion Exercise 1 1.1 1.1 𝑝𝑝𝑓𝑓 − 𝑝𝑝𝑖𝑖 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 𝑝𝑝𝑓𝑓∆𝑡𝑡 − 𝑝𝑝𝑖𝑖 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 =70 − 80 ∆𝑡𝑡 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 700,2 − 80 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −10 0,2 ⃗ 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛 = −10 0,2 𝐹𝐹⃗ = ⃗ 𝐹𝐹𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛= −50 0,2𝑁𝑁 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −50 𝑁𝑁 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 1 2,2 vi (0,28) (9,8)(0,28) 2 21 2,2 vi (0,28) (9,8)(0,28) 2 2 2,2 – 0,38 = vi (0,28) 2,2 – 0,38 = vi (0,28) vi = 6,5 m·s-1 vi = 6,5 m·s-1 1 y viAt gt 2 21 y viAt gt 2 2 781.2. 1.2. v 2f vi2 2 gy v 2f vi2 2 gy (6,5) 2 02 2(9,8)y (6,5) 2 02 2(9,8)y (6,5) 2 02 2(9,8)y 2 (6,=5)2,16 0m2 2(9,8)y Δy Δy = 2,16 m y = 2,16 + 2,2 y = 2,16 + 2,2 86 86 𝑝𝑝𝑓𝑓 − 𝑝𝑝𝑖𝑖 ∆𝑡𝑡 70 − 80 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0,2 −10 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 0,2 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = −50 𝑁𝑁 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 7.3 Vertical Projectile Motion Exercise 1 5.5 Vertical Projectile Motion Exercise 1 1.1 𝐹𝐹⃗𝑛𝑛𝑛𝑛𝑛𝑛 = 50 𝑁𝑁 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 1.1 1 2,2 vi (0,28) (9,8)(0,28) 2 2 2,2 – 0,38 = vi (0,28) vi = 6,5 m·s-1 v 2f vi2 2 gy (6,5) 2 02 2(9,8)y y viAt 1 gt 2 2 (6,5) 2 02 2(9,8)y Δy = 2,16 m 1.2. y = 2,16 + 2,2 1 ∆y = viA ∆t + g∆t 2 2 86 1.2. y = 2,16 + 2,2 y = 4,3 m downwards y = 4,3 m downwards 1 2,16 0t (9,8)t 2 2 t 2 Commented [G158]: Corrected Afrikaans word remove 2 2,16 9,8 Δt = 0,66 s Δttotal = Δt1+ Δt2 Δttotal = 0,66 + 0,28 Δttotal = 0,94 s 5 y (m) 4 3 2 1 0 0 0.2 0.4 0.6 0.8 1 t(s) 79 1.4 v f vi gt 1.4 1.4 v vf 1f v0i g9t,8 0,66 1 ,8 0s,66 v vf 1f 1 06,59m v f 1 6,5m s 1 1.4 v vf f22 0v,944 f i gt 00 9,89,80v, 944 v1f 0 9,8 0,66 1.4 m s 1s 1 v vf f22 9v9,f21 ,21 vi m gt v f 1 6,5m s 1 v f 1 0 9,8 0,66 v f vi gt 1 v (m·s-1) 6,5m v 0 v9f,18 0,66v f2s 0 9,8 0,944 1.4 f1 v (m·s-1) 1 v f 2 9,21 m s 1 5)m s f 1 6,-1 vv (m·s v f 22 0 9,8 0,944 Exercise m s 1 v fPositive 0 v9fdownwards ,28 90,,21 944 2 v (m·s-1) Exercise Exercise2 2 v f 2.1 9,21 m s 2 Commented [U159]: Formatting: pse check use of italics; standardise with all equivalent items. 1 v (m·s ) Exercise 2 Positive Positivedownwards downwards -1 Option 1 -1 v (m·s 𝑣𝑣⃗𝑓𝑓2 =) 𝑣𝑣⃗𝑖𝑖2 + 2𝑔𝑔 ⃗Δ𝑦𝑦⃗ downwards Positive 2.1 2 Exercise 2 2.1v f 02 (2)(9,8)(150) 2.1 Exercise 2 2 Positive downwards Option v f 1 2940 Option 1 Option 1 Comme standard Option 2 𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 + 𝑔𝑔⃗𝛥𝛥𝛥𝛥 t Commented [U159]: Formatting: pse ch standardise with all equivalent items. 2y g Commented [U159]: Formatting: pse check use of ita standardise with all equivalent items. Option 2 2 x150Option Option 22 5,53s 𝑣𝑣 ⃗ = 𝑣𝑣 ⃗ 9,8 𝑣𝑣⃗𝑓𝑓 𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 𝑖𝑖 ++𝑔𝑔⃗𝛥𝛥𝛥𝛥𝑔𝑔⃗𝛥𝛥𝛥𝛥 t Commented [U159]: Formatting: pse check use of italics; 2.1 Positive 2 2 ⃗ 𝑓𝑓 downwards =𝑣𝑣 𝑚𝑚 𝑠𝑠 −1 𝑣𝑣⃗ 2∙ = 𝑣𝑣⃗downwards + 2𝑔𝑔⃗Δ𝑦𝑦⃗ standardise with all equivalent items. 𝑖𝑖 ⃗ 𝑣𝑣⃗𝑓𝑓2 𝑣𝑣= ⃗54,22 ⃗Δ𝑦𝑦 𝑖𝑖 + 𝑓𝑓 2𝑔𝑔 9,8 x5,53 2y Option 2 2.1 2 Option 1 v 2 02 (2)(9,8)(150)vf = 0+ f -t y 2 2 2 vf = 54,22 downwards 𝑣𝑣⃗𝑓𝑓 = m.s 𝑣𝑣⃗ 𝑔𝑔⃗𝛥𝛥𝛥𝛥 2 g 𝑖𝑖 + f 𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 +2 2𝑔𝑔⃗Δ𝑦𝑦⃗ t1 Option 1 2 Option 2 g 2 v f 2940 ⃗0Δ𝑦𝑦 ⃗ (2)(9,8)(150) −1𝑣𝑣⃗𝑓𝑓 = 𝑣𝑣⃗𝑖𝑖 +t 𝑔𝑔⃗𝛥𝛥𝛥𝛥2yt 2 x150 𝑣𝑣⃗𝑓𝑓2 =2 𝑣𝑣⃗𝑖𝑖2 +v f 2𝑔𝑔 5,53s g 𝑣𝑣 ⃗ = 54,22 𝑚𝑚 ∙ 𝑠𝑠 downwards 𝑓𝑓 2.22 f 9,8 2y 2 2 2 x 150 v f 0 v f(2)(9,82940 )(150) t = 0+ 9,8 x5,53 5,53s vf 2 xt150 gt 5,53m.s −1 downwards ⃗ 𝑓𝑓 =1𝑣𝑣⃗ 𝑓𝑓54,22 𝑚𝑚𝑚𝑚∙ ∙𝑠𝑠𝑠𝑠−1 2 𝑣𝑣 54,22 Option = 54,22 downwards 9,s28 -1 downwards 9v,8f =Option v When f 2940 2 x 150 the two objects meet their position is the same: 150 = |ΔyA| +|ΔyB| 150 = the vf = 0+ 9,8 t 5v ,53=x5,53 s 0+ 9,8 x5,53 𝑣𝑣⃗ 𝑓𝑓 =yA54,22 = yB 𝑚𝑚 ∙ 𝑠𝑠 −1 downwards addition of the absolute values of 9v,8f = 54,22f m.s 1 downwards 2.2 1 v = 54,22 m.s 1 downwards the displacements of A and B. Commented [U160]: Pse check. 1 f 2 2 v = 0+ 9,8 x5,53 y Ai viAt gt y Bi viBft gt 1 vf = 54,22 2.2 2 2 m.s 1 downwards 150 = | viAt gt 2 | + | Option 1 2Option 2 Level2.2 zero at the projecting point of A (reference When the two objects meet their position is the same: 150 = the 1 point). 2 150 = |ΔyA| +|ΔyB| viB t Option gt | 2 Option 1 y A = yB 2.2 addition of the absolute values of 2.2 2 1 Option 11 Option 2 0 100t x9,8 xt 2 1501 0t 2 x9,8 xt 2 the displacements of A and Commented [U160]: Pse check. 1 1 2 2 When the two objects meet their position is the When the same: 150 = the B. A|t+|Δy 2 two 150 g- v150 = B| viAtmeet their 2 y Bi isvthe y objects gt position tiB t= |Δy g iB t 1 2 Option 1 Option 2addition yA = yB Ai 2of 2 2 150 = | | + | same: the absolute values of v t g t iA 100t 150 Option 1 objects Option 2 2 When the two their position is the same: 150 = |Δy +|Δy 150of= Athe 1A| displacements Level meet zero the projecting point of1A (reference 2 B| the and B. Commented [U160]: Pse check. 2 𝑠𝑠 2v t 𝛥𝛥𝛥𝛥1 at = 1,5 g t iA y v t g t y v t g t 1 yA = ythe point). 2of|ΔyA| +|ΔyB| B Ai two iA objects meetBitheir iB position When is the 150 = 150 = the addition of2same: the absolute values 1 t ggt2t| +| | 2 2 iBt 150 = | viAv 2 B. the displacements of A and Commented [U160]: Pse check. 1 1 1 1 2 2 2 2 2 100 t 150 y = y at the projecting point Av tzero B 0 x9v,iB 8 xt tofA150 addition of2 the absolute values of t y Bi t 0t x9,8 xt 1 y Ai Level g100 t g(reference iA 2 = 1,5 𝑠𝑠 1 point). 2 2 2 = | viAv -|2 + 2 150 v| iB| t gt = t 1g150 t𝛥𝛥𝛥𝛥 g t iB t the displacements of A and B. Comme 2 2 22 Level zero at the projecting point 2of2 A100 (reference 1 t 150 1 2 0 100 t x 9 , 8 x t 150 0 t x 9 , 8 x t 1 1 Ai iA Bi iB point). 1 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠2 viAt1gtg2== t 2| viB t 150 gt- 2v| 2 viAt gt 2 | + | iB t 150 2.3 2 1 1 2 2 2 2 100t 0150 0 100zero t at x9,8the xtprojecting 150 tpoint x9of ,8 xA t (reference 1 Level 100t 1502 2 150 - viB vt g1g t 2t=2 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 2 iA t 2 1𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 point). 2 100t 150 viB t 88gt 2 | 1 2 100t 1502 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 viAt gt 2 2 2 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 1 2.3 2 100t 150 v 0 (2)(9,8)(150) v 2940 y v t 1 1 gt y v t gt 2 2 1 1 x9,8 xt 150 0t x9,8 xt 2 2 150 - v t gt = 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 iB 2 100t 150 2.3 0 100t 2.3 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 viAt 1 gt 2 2 88 88 88 100t 150 𝛥𝛥𝛥𝛥 = 1,5 𝑠𝑠 2.3 88 80 2.3 Option 1 𝑣𝑣⃗𝑓𝑓 = 11𝑣𝑣⃗𝑖𝑖 + 𝑔𝑔⃗𝛥𝛥𝛥𝛥 Option Option Option 1 100 =𝑣𝑣⃗𝑣𝑣⃗𝑖𝑖−100 𝑣𝑣⃗𝑣𝑣𝑓𝑓⃗𝑓𝑓 == + 𝑔𝑔⃗𝑔𝑔𝛥𝛥𝛥𝛥 ⃗𝛥𝛥𝛥𝛥+ 9,8 𝛥𝛥𝛥𝛥 𝑖𝑖 + 𝛥𝛥𝛥𝛥 = 20,41 100 100 == −100 −100++𝑠𝑠9,8 9,8𝛥𝛥𝛥𝛥 𝛥𝛥𝛥𝛥 𝛥𝛥𝛥𝛥 𝛥𝛥𝛥𝛥 == 20,41 20,41𝑠𝑠𝑠𝑠 Option 2 Option2 2 Option 1 gt 2 2 11 2 2 yy viAv t ggtt iAt 1 22 011 100 t 9,8t 2 2 2 2 0 0 100 tt 9,89 100 ,8tt Option 2 y viAt 22 100 0t 100 t 4,4 9,9 t2 t 2 t 4,9t 0 0 100 2 2 2 100 t t100 4,49, t 100 9 tt 4,9t 2 ΔtΔt == 20,41 s sΔt = 20,41 s 20,41 2.4 2.4 2.4 2.4 Positive Positiveupwards upwards Positive upwards Positive downwards Positive downwards Positive downwards 5.6 Work Energy Power 5.6 Work Energy Power Exercise 1 Exercise 1 1.1 Greater than. 1.1 Greater than.(non-conservative) force acting; mechanical energy is not conserved. There is friction There is friction (non-conservative) force acting; mechanical energy is not conserved. (3) (3) net work done on an object is equal to the change in the kinetic energy of the object. 1.2 The 1.2 The (2) net work done on an object is equal to the change in the kinetic energy of the object. 5.61.3 Work Energy Power (2) 1.3 1.3.1 Wnet E f E i Exercise 1.3.1 Wnet1 E f E i 1.1 1.2 Commented [U161]: Pse check – only 2 Commented [U161]: Pse check – o Commented [U162]: Pse complete. Suggest cross-referencing of entire text is re Commented [U162]: Pse complete Suggest cross-referencing of entire text 89 Greater than. There is friction (non-conservative) force acting; mechanical energy is not 89 conserved. (3) The net work done on an object is equal to the change in the kinetic energy of the object. (2) Comm 1.3 1.3.1 Wnet E f E i Comm Sugges 81 89 7.1 Work Energy Power Exercise 1 1.1 5.6 Greater than. Work Energy Power There Exercise 1 is friction (non-conservative) force acting; mechanical energy is not conserved. (3) Greater than. 1.2 1.1 The net work done on an object is equal to the change in the kinetic energy of the object. (2) 1.3 1.2 1.3 There is friction (non-conservative) force acting; mechanical energy is not conserved. (3) The net work done on an object is equal to the change in the kinetic energy of the object. (2) Commented [U161]: Pse check – only 2 ticks. Commented [U162]: Pse complete. Suggest cross-referencing of entire text is required. 1.3.1 1.3.1 Wnet E f E i (5) 1 Fnet x cos 1m(v 2f 2 vi2 2) Fnet x cos 2 m(v f vi ) 0 2 1 2 89 2 Fnet (10) cos 180 0 1(24,7)(18 2 24,25 2) Fnet (10) cos 180 2 (24,7)(18 24,25 ) 2 Fnet 326,12 N frictional force / wrywingskrag Fnet 326,12 N frictional force / wrywingskrag Commented [U163]: Pse check. Commented [U163]: Pse check. (5) (5) 1.3.2 1.3.2 Wnc E K E P 1.3.2 Wnc E K E P 1 f x cos 1m(v 2f 2 vi2 2) mgh f x cos 2 m(v f vi ) mgh 2 1 326,12 x cos1800 0 1(24,7)(0 18 2 2) 24,7(9,8)(x sin 50 0 0) 326,12 x cos180 2 (24,7)(0 18 ) 24,7(9,8)(x sin 50 ) 326,12 x 4001,4 2185,4287 x 326,12 x 4001,4 185,4287 x x 7,82 m x 7,82 m (6) (6) (6) [16][16] [16] 7.5 Doppler Doppler EffectEffect 7.2 7.5 Doppler Effect Exercise 1 Exercise 1 1.1. Exercise 1 Lower than 1.1. 1.1. 1.2 1.2 Lower As the than speed of sound is constant wavelength is inversely proportional to As the speed of sound is is higher. constant wavelength is inversely proportional to frequency and frequency frequency and frequency is higher. Lower than OR OR than Lower Lower than 1 is constant wavelength is inversely proportional to frequency As the speed of sound When v is constant λ 1f and frequency is higher. When v is constant λ f and frequency is higher. and frequency is higher. OR OR than Lower OR Lower thanheard is higher, the wavelength is smaller, because when the speed As frequency frequency heard is higher, theand wavelength is smaller, because when the speed ofAs sound is constant, wavelength frequency are inversely proportional. of sound is constant, wavelength and frequency are inversely proportional. Lower than Doppler When veffect is constant λ Doppler effect OR 1.3 1.3 1.2 1 α (4) (4) (1) (1) f and frequency is higher. Option 1 Optionthan 1 Lower (6) (6) v vlistener f l vsound sound vlistener f s f l vsound vsource f s is higher, the wavelength is smaller, because when the speed of As frequency heard vsound vsource 0 wavelength and frequency are inversely proportional. sound is 340 constant, f l 340 0 x 620 f 340 30 .56 x 620 l Doppler effect 30.56 340 f l 681,23 Hz f l 681,23 Hz 90 90 82 (4) (1) (4) 1.2 Doppler effect (1) Option 1 1.3 (6) Option 1 1.3 (6) v v f l sound listener f s vsound vsource 340 0 fl x 620 340 30.56 f l 681,23 Hz v f 90 v f 340 681,23 340 681,23 0,50 m 0,50 m Option 2 v Option f 2 v Option f s f 2 sound 340 l 681 , 23 v v sound vsound source f s f l vsource 340 sound vm 0f,l50 x 620 30.56 340340 fl x 620 340 30.56 Option 2 f l 681 23 Hz v,sound f s f l f l v681 vHz sound,23 source v f 340 vf l f x 620 340 340 681,2330 . 56 340 681,23 0,50 m f l 681 ,23m Hz 0,50 681.23 v 340 f 340 .23 0,681 50 m 340 681,23 0,50 m 7.6 Electrostatics 0,50 m Exercise 1Solutions 7.6 Electrostatics 7.3Exercise Electrostatics 1 Solutions 1.1 What is the relationship between the electrostatic force between two charges and the 340 681.23 distance between those charges? 1.1 the relationship Exercise 1isSolutions 1.2 What Electrostatic between the electrostatic force between two charges and the m 0,50force distance between those charges? 1.3 Magnitude of the charges 1.2 Electrostatic force 1.1 What is the relationship between the electrostatic force between two charges and the distance between those charges? 1.4 1.3 Magnitude of the charges 7.6 Electrostatics Exercise 1 Solutions 1.4 2000 1.2 Electrostatic force 1000 500 1.4 5000 2000 0 0 0 1500 F(N) 1.4 F(N) F(N) 1.1 What is the relationship between the electrostatic force between two charges and the 1500 distance2000 between those charges? 1.2 Magnitude Electrostatic force 1.3 of thecharges 1500 1000 1.3 Magnitude of the charges 2 4 2 4 d(m) 6 8 10 6 8 10 d(m) 1000 500 0 0 2 4 6 d(m) 8 10 91 91 91 83 CRITERIA CRITERIA MARKS/PUNTE MARKS/PUNTE CRITERIA CRITERIA Correct shape Correct shape MARKS/PUNTE Correctshape shape Correct Three or or more coordinates plotted correctly Three more coordinates plotted correctly Three or more coordinates plotted correctly Three or more coordinates plotted correctly Commented [U164]: Pse Pse check. Commented [U164]: check. Pse Pse check all instances where another language is included andand it it check all instances where another language is included Commented [U164]: Pse check. appears as though something has has been copies andand pasted but but not not appears as though something been copies pasted Pse check all instances where language is document. included and it over-written/corrected so that it fits this this document. over-written/corrected so another that it with fits with appears as though something has been copies and pasted but not over-written/corrected so that it fits with this document. MARKS/PUNTE (2) (2)(2) (2) 1.5The Themagnitude magnitude ofelectrostatic the electrostatic force two between two chargesproportional is inversely proportional to the square of 1.51.5 of the force between charges is inversely to to thethe square The magnitude of the electrostatic force between two charges is inversely proportional square the distance between the charges. of the distance between charges. 1.5of The magnitude of thethe electrostatic force the distance between the charges. between two charges is inversely proportional to the square of the distance between the charges. 1.6.1 1.6.1 1.6.1 1.6.1 CrITerIA CrITerIA Shape Shape CrITerIA Direction of of thethe arrows Direction arrows Shape Direction of the arrows Marks Marks Marks (2)(2) 1.6.2 1.6.2 1.6.2 1.6.2 2 kQ 2 F F 2 kQ r r2 2 kQ F 2 r 9 9 2 2 (9 (910 10)Q )Q 1600 = = 1600 92 2 2 ( 1 ) (9 10(1))Q 1600 = (1) 2 -4 Q Q= 4.22 x 10 = 4.22 x 10-4C C Q = 4.22 x 10 -4 C (4)(4) 1.6.3 1.6.3 kQkQ E E 2 r r2 kQ E 2 r 4 9 10 9 910 109 4,22 10 4 4,22 E =E = 9 (0,25) 2 2 4 ,25) 10 9 10 4(0,22 E= (0,25) 2 [19] (3)(3) (3) E =E 6,07 x 10 = 6,07 x 10 N·C-1 -1 [19] [19] E = 6,07 x 10 N·C [19] -11 -11 -11 N·C -1 92 92 92 84 Commented [U165]: Formatting: pse pse alignalign items on right-hand Commented [U165]: Formatting: items on right-hand margin. Repetitive problems seen. margin. Repetitive problems seen. Commented [U165]: Formatting: pse align items on right-hand margin. Repetitive problems seen. (4) 1.6.3 1.6.3 (2) (4) 7.4 Electric Circuits Exercise 1 1.1 Differences between emf and potential difference: Electromotive force 1. Electromotive force transmits current both inside and outside the cell. 2. Electromotive force emf is the cause. 3. Electromotive force is always greater than potential difference. 4. Electromotive force creates potential difference entire the circuit. 5. Electromotive force does not depend on the resistance of the circuit. 6. Electromotive force remains constant. 7. The part of the circuit where electrical energy is created from any other energy then that part contains the source of Electromotive force. Potential difference: 1. Potential difference current transfers between any two points in the circuit. 2. Potential difference is the result. 3. Potential difference is always less than electromotive force. 4. Potential difference takes place between any two points in the circuit. 5. Potential difference of two points depends on the resistance of those points. 6. It does not remain constant. 7. Potential difference exists in the part of the circuit where electrical potential energy is transformed into another form of energy. 1.2 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. 85 1.2 The potential difference across a conductor is directly proportional to the current in the conductor at constant temperature. 1.3 1.3 First we must identify the type of connection; for this purpose we must indicate the direction of the First we must identify the type of connection; for this purpose we must indicate the direction of the current. current. 12 V S2 A + R1 - R2 + R3 93 S1 To current, we apply mustOhm’s applylawOhm’s law To calculate calculate current, we must Commented [U168]: Pse complete (theory?) Commented [U168]: Pse complete (theory?) To calculate current, we must apply Ohm’s law 𝑉𝑉2 𝐼𝐼To 2 = Commented [U168]: Pse complete (theory?) calculate current, we must apply Ohm’s law 𝑅𝑅22 𝑉𝑉 𝐼𝐼2 = 𝑅𝑅 𝑉𝑉22 𝐼𝐼The 2 = resistors are connected parallel: 𝑅𝑅2 Commented [U168]: Pse complete (theory?) To calculate current, we mustinapply Ohm’s law The resistors connected in parallel: The resistors are are connected in parallel: 𝑉𝑉 1 = 21 1 1 𝐼𝐼The resistors are connected in parallel: 2 = + + 2 𝑅𝑅1𝑒𝑒𝑒𝑒 𝑅𝑅𝑅𝑅 𝑅𝑅 𝑅𝑅 11 12 13 = + + 𝑅𝑅𝑒𝑒𝑒𝑒 𝑅𝑅11 𝑅𝑅12 𝑅𝑅13 1 = + + The resistors are 𝑅𝑅 𝑅𝑅 𝑅𝑅3 connected in parallel: 𝑒𝑒𝑒𝑒 =𝑅𝑅1 𝑅𝑅2 𝑅𝑅 𝑒𝑒𝑒𝑒 𝑅𝑅3 𝑅𝑅1𝑒𝑒𝑒𝑒 = 31 1 1 == 𝑅𝑅 + + 𝑅𝑅 𝑅𝑅 𝑅𝑅1three 𝑅𝑅2 resistors 𝑅𝑅3 𝑒𝑒𝑒𝑒the 𝑒𝑒𝑒𝑒 As are in parallel, the potential difference is the same: 3 As the three resistors are in parallel, the potential difference is the same: As the 𝑅𝑅 three resistors are in parallel, the potential difference is the same: = 𝑉𝑉1=resistors V 𝑉𝑉1𝑒𝑒𝑒𝑒= As the are in parallel, the potential difference is the same: 𝑅𝑅 =𝑉𝑉23three 𝑉𝑉1 = 𝑉𝑉2 = 𝑉𝑉1= V Commented [U168]: Pse complete (theory?) To calculate current, must apply Ohm’s the law current through each one is the same. As the the resistance, 𝑉𝑉2resistors = 𝑉𝑉1=resistors V havewe 𝑉𝑉1 = As the three are same in parallel, the potential difference is the same: As the 𝑉𝑉2 resistors have the same resistance, the current through each one is the same. 𝐼𝐼Solving: 2 =the Commented [U169]: Perhaps solution (i.e.noun – solving is a 𝑅𝑅2 resistors have the same resistance, the current through each one is the same. As 𝑉𝑉2 = 𝑉𝑉1= V 𝑉𝑉1 = verb). Commented [U169]: Perhaps solution (i.e.noun – solving is a Solving: As the resistors have the same resistance, the current through each one is the same. 𝑅𝑅 verb). The resistors are connected in parallel: Commented [U169]: Perhaps solution (i.e.noun – solving is a Solving: 𝑉𝑉 = 𝐼𝐼𝐼𝐼 = 𝐼𝐼 As have the same resistance, the current through each one is the same. 2 the resistors 𝑒𝑒𝑒𝑒 3 𝑅𝑅 verb). 𝑉𝑉21 = 𝐼𝐼𝐼𝐼1𝑒𝑒𝑒𝑒 =1 𝐼𝐼 1 3𝑅𝑅 𝑅𝑅 + = 𝐼𝐼+ = 𝑉𝑉 = 𝐼𝐼𝐼𝐼 Commented [U169]: Perhaps solution (i.e.noun – solving is a Solving: 2 𝑒𝑒𝑒𝑒 𝐼𝐼 Solving: 𝑅𝑅 𝑅𝑅 𝐼𝐼 𝑅𝑅 3 𝑅𝑅 verb). 𝐼𝐼2𝑒𝑒𝑒𝑒= 𝐼𝐼𝑅𝑅𝑅𝑅3 1= 3 2 3 𝐼𝐼 3 𝑅𝑅 𝑅𝑅𝑅𝑅 = 3= 𝐼𝐼𝑉𝑉2 == 𝑅𝑅𝐼𝐼𝐼𝐼𝐼𝐼 𝐼𝐼 𝐼𝐼 2 𝑒𝑒𝑒𝑒 3 3 𝐼𝐼𝑅𝑅2𝑒𝑒𝑒𝑒==3𝑅𝑅 3= (6) 𝐼𝐼2 = 3 = 13 𝐴𝐴 3 (6) 𝐼𝐼2 = 3𝐼𝐼𝑅𝑅= 1𝐼𝐼𝐴𝐴 As the 33 three resistors are in parallel, the potential difference is the same: 𝐼𝐼22 = 𝑉𝑉𝑅𝑅 ==1= 𝐴𝐴 (6) 2.4. 3𝑒𝑒𝑒𝑒𝑒𝑒 3 𝜀𝜀 − 𝑉𝑉𝑖𝑖 = 𝜀𝜀 − 𝑉𝑉 2.4. 𝑉𝑉 𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉1 = 3𝑉𝑉 2 = 𝑉𝑉1 = V 𝑖𝑖 =𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉 2.4. = 𝑉𝑉 =12 1=− 𝐴𝐴𝜀𝜀3−×𝑉𝑉0,4 (6) 𝐼𝐼2𝑒𝑒𝑒𝑒𝑒𝑒 𝑖𝑖 3 = 12 − 3 × 0,4 𝑉𝑉 𝑒𝑒𝑒𝑒𝑒𝑒the As resistors have the same resistance, the current through each one is the same. 𝑉𝑉 10,8 (3) = 12 −𝑉𝑉3 × 0,4 𝑒𝑒𝑒𝑒𝑒𝑒 𝑉𝑉 2.4. 𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 𝑉𝑉 (3) Commented [U169]: Perhaps solution (i.e.noun – solving is a Solving: 𝑒𝑒𝑒𝑒𝑒𝑒 = 10,8 𝑉𝑉 verb). 2.5 𝑉𝑉 = 10,8 𝑉𝑉 (3) 𝑒𝑒𝑒𝑒𝑒𝑒Increases 𝑉𝑉 𝑒𝑒𝑒𝑒𝑒𝑒 = 12 − 3 × 0,4 𝑅𝑅 2.5 𝑉𝑉2 =Increases 𝐼𝐼𝐼𝐼𝑒𝑒𝑒𝑒 = 𝐼𝐼 3 TheIncreases resistors are now connected in series then total resistance increases, emf is constant then(3) current 2.5 𝑉𝑉 𝑒𝑒𝑒𝑒𝑒𝑒 = 10,8 𝑉𝑉 The resistors are connected in series total resistance increases, emf=is𝜀𝜀constant then current decreases, dropnow of potential in the batterythen decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 − 𝑉𝑉𝑖𝑖 terminal 𝑅𝑅 𝐼𝐼 𝐼𝐼 resistors are now connected in series total resistance increases, emf then(4) current 3 decreases, of potential in the batterythen decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = is𝜀𝜀 constant − 𝑉𝑉𝑖𝑖 terminal Commented [U170]: Pse check / clarify. potential 2.5 = increases. drop 𝐼𝐼The 2 =Increases 𝑅𝑅 3 (6) [U170]: Pse check / clarify. decreases, drop of potential in the battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal (4) Commented potential increases. Exercise 2 are now connected in series then total resistance increases, emf is constant then(4) 3 The resistors current Commented [U170]: Pse check / clarify. potential increases. (6) 𝐼𝐼2 = 3 = 1 𝐴𝐴 Exercise 2 drop of potential in the battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal 2.4. decreases, 3.1. What is the relationship between potential difference and the strength of the electric current in a circuit Exercise 2 Commented [U170]: Pse check / clarify. potential increases. 3.1. What is 𝜀𝜀the relationship between potential difference and the strength of the electric current in(4) a circuit = −with 𝑉𝑉 2.4. 𝑖𝑖 constant resistance? with 𝑉𝑉 a𝑒𝑒𝑒𝑒𝑒𝑒 resistor 3.1. What is the relationship between potential difference and the strength of the electric current in a circuit with a resistor with constant resistance? Exercise 3.2. When increases, potential difference also increases. 𝑉𝑉 12 2−current 3× 0,4 𝑒𝑒𝑒𝑒𝑒𝑒 = with a resistor with constant resistance? 3.2. When current increases, potentialpotential difference also increases. 3.1. is the relationship difference and the strength of the electric current in a circuit 3.3. What Independent variable – between current 3.2. When current increases, potential difference also increases. 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 Independent = 10,8 𝑉𝑉 variable – current (3) 3.3. with aDependent resistor with constant resistance? variable – potential difference 3.3. Independent variable – current Dependent variable – potential difference 2.5 3.2. When current increases, potential difference also increases. 3.4.Increases Dependent variable – potential difference 3.4. 3.3. Independent – current (3) The resistors are variable now connected in series then total resistance increases, emf is constant then current 3.4. decreases, drop of potential in the difference battery decreases and according to 𝑉𝑉𝑒𝑒𝑒𝑒𝑒𝑒 = 𝜀𝜀 − 𝑉𝑉𝑖𝑖 terminal Dependent variable – potential Commented [U170]: Pse check / clarify. potential increases. (4) 3.4. 2.5 Increases Exercise 2 3.1. What is the relationship potential difference and total the strength of the electric current in aemf circuit The resistors are now between connected in series then resistance increases, is constant then current 94 with a resistor with constant resistance? decreases, drop of potential in the battery decreases and according to terminal 94 potential increases. 3.2. When current increases, potential difference also increases. (4) 3.3. Independent variable – current Dependent variable – potential difference 94 94 3.4. 86 94 Exercise 2 3.1. What is the relationship between potential difference and the strength of the electric current in a circuit with a resistor with constant resistance? 3.2. When current increases, potential difference also increases. 3.3. Independent variable – current Dependent variable – potential difference V (V) V (V) 3.4. I (A) I (A) 2.5 ∆YY ∆X Gradient == 2.5 gradient ∆ VYV = gradient X = I 2.5 ∆XI 24 6 24 6 0 , 08 Gradient== 0,02 Gradient V I Gradient = 0,08 0,02 Gradient = 300 Gradient==of 300 2.6. Resistance the resistor. Gradient 300 2.7. The is correct: the potential difference is directly proportional to the strength of the 2.6.hypothesis Resistance of the resistor. electric when resistance is constant. 2.7.current The hypothesis is correct: the potential difference is directly proportional to the strength of the 2.6. Resistance of thethe resistor. electric current when the resistance is constant. 2.7. The is correct: the potential difference is directly proportional to the strength of the electric 7.8 hypothesis Electrodynamics current when1 the resistance is constant. Exercise 7.8 Electrodynamics 7.5 Electrodynamics Exercise 1 1.1 DC generator Exercise 1 Mechanical energy to electrical energy 1.1 1.2 1.1 (2) DC generator To make the direction of the (induced) current to be the same in every half cycle/half DC generator turn Commented [U171]: Pse check – not understood. Mechanical energy to electrical energy (2) OR Mechanical energy to electrical energy 1.2 (2) cycle/half To make the direction of the (induced) current to be the same in every half keepturn the (induced) unidirectional. To To make the directioncurrent of the (induced) current to be the same in every half cycle/half turn 1.2 1.3 Commented [U171] Graph A OR OR (2) The DC generator becomes an AC generator. keep the (induced) current unidirectional. To keepTo the (induced) current unidirectional. 1.3 (2) Voltage changes the polarity with every half cycle. Graph A A 1.3 Graph (2) (3) The DC generator becomes an AC generator. The DC generator becomes an AC generator. (3) (3) Voltage changes the polarity every halfcycle. cycle. Voltage changes the polarity withwith every half 95 95 87 1.4 1.4 1.4 Pav g/ gem V 2 rms / wgk R Pav g/ gem Commented [U172]: Pse c V 2 rms / wgk R Commented [U172]: Pse check items (languages). 7.9 Commented [U172]: Pse check items (languages). Photoelectric Effect IrmsEffect Total/wgk totaal = 2x(0,71) Photoelectric Exercise 1 Exercise 1 1.1 Alternative solution: Alternative solution: V 2 maks 2 rms / wgk ) ( Vmax/ Pav g/ gem R2 V Pavg/gem Vmax/ maks 2 ( max/ maks ) 2 ) ( R 2 2 Pavg Vmax/ maks 2 Pavg/gem ) ( R R V 2 rms / wgk solution: 12 Alternative ( )2 2 1.4 Pav g/ gem 2 Pavg R R2 rms / wgk 6 12 12 2 V ( )2 ( ) R Pav g/ gem 2 2 R V 6 6 12 ( max/ maks ) 2 R= 12 Ω ( )2 R R 2 2 V Pavg/gem 6 ( max/ maks ) 2 R R= 12 Ω R= 12 Ω Vrms / wgk R 2 Pavg Irms / wgk R 12 Ω Rrms/wgk = IrmsTot/wgk tot RT 12 V R= Vrms / wgk ( ) 2 Irms / wgk 2 6R 12 V max/ maks tot2RT Vrms/wgk =( V IrmsTot/wgk R ) ( ) 2 rms / wgk I R ) rms Total/wgk tot ( 2 2 2 2 Irms/wgk V maks 2 6 R Vrms / wgk ) ( max/ R R= 12 Ω Irms Total/wgk tot ( )R 2 2 2 Irms/wgk 12 12 R R= 12 Ω Irms / wgk ( ) 12 2 Vrms / wgk ) ( 2 12 Irms / wgk 2 12 2 V R Irms rms/wgk tot RT ( = )IrmsTot/wgk 12 2 / wgk ( ) 2 2 12 Irms Total/wgk tot = 1,42 A Vmax/ maks 2 Vrms / wgk 2 R ) ( Irms/wgk = 0,71tot = 1,42 A Irms Total/wgk tot ( ) 12 Irms Total/wgk 2 2 2 Irms/wgk R Irms/wgk = 0,71 Irms Total/wgk totaal = 2x(0,71) 12 12 Irms / wgk ( ) 12 Irms Total/wgk totaal = 2x(0,71) ( )2 =1,42 A2 2 2 =1,42 A 12 Irms Total/wgk tot = 1,42 A Irms/wgk = 0,71 7.9 V 2 rms / wgk Pav g/ gem R Alternative solution: =1,42 A 1.1 Minimum energy needed to eject an electron from a metal surface (substance). 7.6 Photoelectric Effect Minimum energy needed1.2 to ejectThe an electron from a metal of surface (substance). extraction (ejection) electrons from a metal surface by radiation (when light strikes its surface). Photoelectric 1.2 The7.9extraction (ejection)Effect of electrons from a metal surface by radiation (when light strikes its Exercise 1 surface). 1.3.1 𝐸𝐸 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘 Exercise 1 1.3.1 1.1 1.1 = 𝑊𝑊𝑜𝑜an + 𝐸𝐸electron 𝐸𝐸 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘 𝑘𝑘 Minimum energy needed toℎ𝑓𝑓eject from a metal surface (substance). Minimum energy needed to eject an electron from a metal surface (substance). (6,63 × 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19 ℎ𝑓𝑓 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘 The extraction (ejection) of electrons from a metal by surface radiation (when light strikes its light strikes its surface). 1.21.2 The extraction (ejection) of electrons fromsurface a metal by radiation (when (6,63 × 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19 𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻 surface). 1.3.1 1.3.1 𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻 𝐸𝐸 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘 ℎ𝑓𝑓 = 𝑊𝑊𝑜𝑜 + 𝐸𝐸𝑘𝑘 96 96 (6,63 × 10−34 )𝑓𝑓 = 3,52 × 10−19 + 4,20 × 10−19 𝑓𝑓 = 1,6 × 1015 𝐻𝐻𝐻𝐻 1.3.2 1.3.2 𝑐𝑐 = 𝑓𝑓𝑓𝑓 3 × 108 = (1,16 × 1015 )𝜆𝜆 𝜆𝜆 = 2,59 × 10−7 𝑚𝑚 88 96 chemistry 1. How to use this book This book serves as a guide to understanding the Grade 12 Physical Sciences. However, it does not replace your textbook. The book focuses more on the challenges that have been observed from learners’ responses in the Grades 12 National Examinations over the past few years. The authors have used their experience to focus their attention on the areas that learners seem to struggle with. 2. Key subject concepts The two areas of study in Chemistry are: 1. materials 2. chemical change. In demonstrating concepts, laws and theories in this book, examples are used; this however does not mean that the laws, concepts and principles applies only to those examples used. It is therefore important to understand the concept itself and where and when it applies. Some theories/laws/concepts apply throughout the content in both organic and inorganic chemistry e.g. atomic theory, The Kinetic Molecular Theory, the mole concept, stoichiometry, rules of oxidation, inter-molecular forces etc. These theories and concepts are applied throughout the chemistry content. There are aslo laws that apply only to a particular group or class of compounds; for example gas laws apply only to gases and not to solids or aqueous substances. It is important to always know where the laws/ rules can be applied. 3. The Representation model This model employs a strategy of three stages in understanding Chemistry Education. These stages should be applied almost simultaneously when studying chemistry: 1. The macro level: This is what you see as an observer of either a material or chemical change (state, colour, size, etc.) 2. The Sub-micro level: This is what happens at the sub-atomic level, e.g. with electrons, protons in atoms or molecules. 3. The Symbolic level: This is when you represent what happens at the sub-atomic level and macro level using symbols, e.g. instead of saying ice, you write H2O (s). Learners of Chemistry must be able to move through these three levels with ease when they discuss chemistry. When you deal with a chemistry question, always think about it in terms of these three levels. Macroscopic Symbolic Sub-micro representation model 89 4. Tips on specific topics Organic chemistry: Study how to name organic molecules for all groups. always remember that the carbon atom always takes a maximum of 4 bonds in all organic compounds, not more, not less. Electrochemistry: Pay attention to half reaction and their respective oxidation states. use the Table of Reduction Potentials as discussed in the text. Read about the similarities and differences between the two types of cells: Galvanic and Electrolytic! Fertilizers: The processes involved in the manufacture of fertilisers is clearly indicated in the text. Study the molecules and their reactions, including their reaction conditions (these would include temperature, pressure, concentrations of reacting species and in some cases catalysts). 5. Study and examination tips 6. message to Grade 12 learners from the writers As you prepare to write your examination, it is important to carefully understand the rules governing certain aspects of your work, and the laws and concepts. Ensure you understand these rules/laws/ concepts properly. Understand what they mean, where they apply and when they apply - and also when and where they do not apply. Always: 1. 2. 3. 4. 5. Read the question that you are working on carefully. Understand what it says and what is required of you. Write down the information that you have. Write down the information that you do not have. Use the information you have to derive the information you need to answer the question. 6. Check your work by going through these steps again! 90 7. The mole concept The Mole: The quantity of a substance that contains 6,02 x 1023 particles. This number is also called Avogadro’s number. A mole is a very, very BIG number and if 12g of Carbon C12 contains this number of Carbon atoms, you can imagine how small atoms are. Imagine this many atoms contained in only 12g! This also demonstrates how small atoms are! All Chemistry calculations require an understanding of the mole concept. When we do calculations on reactions in chemistry we always use the moles that reacted NEVER the mass that reacted. If you are given the mass, you must always work out the number of moles contained in the mass! The masses that reacts in a chemical reaction will almost always have one or more of the reactant(s) having a larger than necessary number of moles or fewer that necessary number of moles and the reactions will almost always leave some unreacted reactant(s) that we call an excess reactant and the reactant(s) that gets used up while there is still an excess reactant is called a limiting reactant(s). 8. Stoichiometry 91 8.1 Excess and Limiting Reagents See how many tyres you are able to use in this example. The remaining tyres are in excess. The cars are said to be limiting. Which of the two reactants is in excess and which is the limiting reagent? Look at the reactants in the jar as they form products in the jar on the right of the arrow. Study this reaction and decide which of the two molecules that are involved in this reaction is limiting H2 or N2 is the limiting reagent? 92 9. Practice activities from Grade 10 Reminder: One MOLE has 6.022 x 1023 items or there are 6.022 x 1023 items/mole. Show your work! 1. How many atoms of potassium make up one MOLE? _____ 2. How many atoms of potassium make up 2 MOLES? _____ 3. How many formula units of salt make up 10 MOLES? _____ 4. How many molecules of water make up 1 MOLE? _____ 5. How many molecules of water make up 5 MOLES? _____ 6. How many moles are 6.022 x 1023 atoms of sodium? _____ 7. How many moles are 12.04 x 1023 atoms of carbon? _____ 8. How many moles are 18.06 x 1023 atoms of sodium? _____ 9. How many moles are 60.22 x 1023 atoms of sodium? _____ 10. How many moles are 6.022 x 1023 molecules of water? _____ 11. How many moles are 12.04 x 1023 molecules of water? _____ 12. How many moles are 30.10 x 1023 molecules of water? _____ 10. dynamic chemical Equilibrium Graphs 93 11. cONcENTRaTION – TImE GRaPHS STudy TIP 1: Concentration, pressure and temperature changes made to a gaseous chemical equilibrium are identified as follows: concentration change: The graph of one of the substances spikes upward or downward Pressure change: The graph spikes upward or downward for each substance in the equation. Temperature change: The graph does not spike but shows gradual upward and downward changes that are opposite for the forward and reverse reactions STudy TIP 2: Upward spiking and gradual upward changes show increase in concentration as well as reaction rate. Downward spiking and gradual downward changes show decrease in concentration as well as reaction rate. STudy TIP 3: Pressure is increased by decreasing the volume of the container containing the reaction mixture. Conversely, pressure is decreased by increasing the volume of the closed container. Therefore pressure changes are really concentration changes. STudy TIP 4: Once the change is made Le Chatelier’s Principle (LCP) can be used to explain the graph shape. ExamPLE In the Haber Process, the reaction N2(g) + 3H2(g) ⇌ 2NH3(g) closed container. ΔH < 0 takes place in a Changes are made to the reaction at times t1, t2 and t3. The graphs below show the situation. 1. What do the shapes of the graphs between the times 0 and t1 indicate about the reaction? 2. State the change that was made to the reaction at: 2.1 t1 2.2 t2 2.3 t3. Give a reason for each answer. 3. Use Le Chatteliers Principle (LCP) to explain the shapes of the graphs between: 3.1 t1 and t2 3.2 t2 and t3 aNSwERS 1. The reaction is in equilibrium 2.1 [N2] was increased 94 Reason: There is an upward spike only for N2 ANSWERS 1. The reaction is in equilibrium. 2.1 [N2] was increased. Reason: There is an upward spike only for N2. 2.2 Pressure was increased. Reason: There is an upward spike for N2, H2 and NH3. 2.3 Temperature was increased. Reason: All changes in the graph line shapes are gradual. The changes are upward for the reverse reaction and downward for the forward reaction. 3.1 STEP 1: Change made: [N2] increases. STEP 2: According to LCP, the forward reaction is favoured (i.e. the reaction that reduces the [N2] is favoured). STEP 3: Therefore, the [H2] decreases and the [NH3] increases until a new equilibrium is established. 3.2 STEP 1: Change made: Pressure was increased. STEP 2: According to LCP, the forward reaction is favoured (i.e. the reaction that produces less molecules is favoured). STEP 3: Therefore the [NH3] increases, but the [N2] and [H2] decrease until a new equilibrium is established. 95 FERTILIzERS 12. FERTILISERS NH4NO3 and (NH4)2SO4 3 Industrial processes Fractional distillation of liquid air gives N2. Ostwald Process 4NH3 + 5O2 2NO + O2 4NO + 6H2O 2NO2 Haber Process Contact Process N2 + 3h2 2NH3 catalyst: Iron Oxide + SO2 2SO2 + O2 2SO3 SO3 + H2SO4 H2S2O7 (Oleum) Catalyst in Step2:V2O5 3NO2 + H2O 2HNO3 + NO Catalyst in step 1: Pt: HNO3 S + O2 NH3 + H2SO4 FERTILISERS (ONLY 2 Fertilisers to study) NH4NO3 Ammonium Nitrate 96 (NH4) 2SO4 Ammonium Sulphate Primary nutrients (N P K) Essential Nutrients (c H and O) NPK ratio TIP 1: Make a CHART showing the OSTWALD, HABER and CONTACT processes and place it in your ROOM where you can always see it. Practice writing out the equations on your own. TIP 2: Play this game: Prepare the following cards for your game: 4 cards marked O2; 2 cards marked NO; 2 cards marked NO2; 2 cards marked SO2; 2 cards marked SO3; 2 cards marked H2O; 1 card for each of the following - S, H2SO4, H2S2O7 and HNO3. Use THese CArDs To DePICT THe reACTIoNs For THe TWo ProCesses oN A WALL. ComPeTe WITH YoUr CLAssmATe To see WHo FINIsHes FAsTer. activities QuESTION 1 The flow diagram below represents two industrial processes that are used to make fertilizer Z. The letters (a) to (d) represent steps in one of the industrial processes. The letter J represents the other industrial process. S + O2 X (a (b) SO3 (c) Gas P + H2 R J (d) Y NH3 Fertilizer Z 1.1 Write down the NAME or FORMULA of: 1.1.1 Product X formed in step (a). (1) 1.1.2 Product Y formed in step (d). (1) 1.1.3 Process J, by which NH3 is produced. 97 1.2 Write down a balanced chemical equation for the formation of fertilizer Z. 1.3 A farmer has two labelled bags of fertilizer - x and y -as shown below. (3) X 50 kg 20 kg 6:2:1 (20) 3:5:2 (30) She mixes the contents of x and y together in another bag - z. Calculate the total mass of phosphorus that z now contains. (7) 1.3 The bag of fertiliser shown below was prepared by mixing 25 kg of pure ammonium sulphate (NH4)2SO4 with potassium salts, phosphates and sand. 3:5:1 (30) x kg Calculate the mass of the bag of fertiliser. (4) (ASSUME THAT AMMONIUM SULPHATE IS THE ONLY SOURCE OF NITROGEN. QuESTION 2 2.1 The flow diagram below shows the processes involved in industrial preparation of fertiliser y. Process a: Proces A H2 + Gas X NH3 Fertiliser Y Process B: Gas B + O2 Step 1 98 NO + H2O + O2 Step 2 Gas C H2O Step 3 HNO3 2.1.1 Write down the NAME of: a. The RAW material for gas x in process a. (1) b. Process a. (1) c. Process B, represented by Steps 1 to 3. (1) 2.1.2 Write down the balanced equation for: a. Process a (3) b. Step 1 (3) c. Step 2 (3) d. The reaction that leads to the formation of fertiliser y. (3) 2.2 Two 50 kg bags contain fertiliser P and Q. Fertiliser P: 5:2:3 (25) Fertiliser Q: 1:3:4 (20) 2.2.1 What do the numbers 25 and 20 represent? (1) 2.2.2 Do a calculation to determine which fertiliser (P or Q) contains the greater mass of potassium. (4) SOLuTIONS QuESTION 1 1.1.1 Sulphur dioxide (or SO2) - give the NAME or FORMULA, not both. 1.1.2 Sulphuric acid (or H2SO4) 1.1.3 Haber process 1.2 2NH3 + H2SO4 (NH4)2SO4 1.3 Approach: Calculate the mass of phosphorus in each bag and add these figures together. Bag X; m Phosphorus = 2/9 x 30/100 x 50 = 2.22 kg Bag Y: m Phosphorus = 5 /10 x 30 x 20 = 3 mTOTAL = 3 + 2.22 = 5.22 kg 99 1.4 Approach: First find mass N in (NH4)2SO4 m Nitrogen = % N in (NH4)2SO4 x 25k = M nitrogen/M (NH4)2 SO4 X 25 Kg = (2 x 14 ) / (14 x 2) + (2 X 4 X 1)+ 32 + (4 x 16) x 25kg = 28+3 +32+64+25 = 28/132 x 25 Kg = 5.3 Kg Mass nitrogen in bag = 5.3 = 3/9 x 30/100 x 5.3 = 0.1 x 53 Kg = x QuESTION 2 2.1.1 a. Air b. Haber c. Ostwald 2.1.2 a. N2 + 3H2 b. 4 NH3 + 5 O2 c. 2 NO + O2 d. NH3 + HNO3 2 NH3 4 NO + 6 H2O 2NO2 NH4NO3 2.2.1 Percentage of fertiliser in the bag. 2.2.2 P mK = 3/10 x 25/100 x 50 = 3.75 kg Q mK = 4/8 x 20 /100 x 50 = 5 100 Q contains more potassium 13. ORGaNIc cHEmISTRy ORGaNIc cHEmISTRy BaSIc ORGaNIc cHEmISTRy PHySIcaL PROPERTIES ORGaNIc REacTIONS 13.1 BASIC ORGANIC CHEMISTRY Structural Formulae Condensed Molecular Functional groups 9 Empirical General formulae Hydrocarbons Basic IuPac naming Organic chemistry saturated/Unsaturated Terminology Homologous series Functional group Isomers Chain Structural isomers Positional Functional 101 13.2 NOMENCLATURE OF ORGANIC COMPOUNDS NOTE: organic compounds in these notes are named according to the 1993 IUPAC RULES. STRaIGHT cHaIN aLKaNES IuPac NOmENcLaTuRE The IUPAC names of the first eight are: Name condensed structural formula Name condensed structural formula Methane CH4 Pentane CH3CH2CH2CH2CH3 Ethane CH3CH3 Hexane CH3CH2CH2CH2CH2CH3 Propane CH3CH2CH3 Heptane CH3CH2CH2CH2CH2CH2CH3 Butane CH3CH2CH2CH3 Octane CH3CH2CH2CH2CH2CH2CH2CH3 IuPac NOmENcLaTuRE OF STRaIGHT cHaIN aLKaNES cONTaINING aN aLKyL In organic chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many possible substitutions. ... The smallest alkyl group is methyl, with the formula CH3− In general, alkyl substituents are named by changing the ending “ane” in an alkane’s name to “yl”. EXAMPLE: CH3 is methyl; it is derived from CH4 (methane) by changing the “ane” ending in methane to “yl”. IUPAC RULES should be followed when naming these alkanes: The following compound will be used to illustrate the IUPAC Rules: STEP 1: 102 Select the longest continuous chain for the parent name. In this case it has six carbon atoms and is thus hexane. STEP 2: Number the chain from either end, so that the substituents are attached at the lower numbers. In the example, numbering from left to right gives substituents at carbon 2 and 4, whereas numbering from right to left gives substituents at carbon 3 and 5. Thus the lower numbers occur when numbering from left to right. STEP 3: Substituent groups are assigned the number of the carbon to which they are attached. In the example, the two substituents are called the 2-methyl and 4-ethyl substituent groups. Note that the number is written in front of the substituent group and a hyphen (-) separates the number and the alkyl group’s name. STEP 4: The name of the compound is now composed of the parent name preceded by the names of the substituent groups in alphabetical order. The correct IUPAC name for the example is 4-ethyl-2-methylhexane. NOTES: In IuPac nomenclature, a hyphen (-) separates numbers from words, and a comma separates numbers from numbers. Otherwise there is no space between the words that make up the name. STEP 5: If the same substituent occurs more than once in the molecule, the prefixes “di”, “tri”, “tetra” and so on, are used to indicate how many times it occurs. If the substituent occurs more than once on the same carbon, the number is repeated. The following examples illustrate this step. NOTES: In IUPAC nomenclature, the prefixes “di”, “tri”, “tetra” and so on, are not alphabetised. EXAMPLES: 2,3 - dimethylpentane 3,3 – dimethylhexane NOTE: The IUPAC nomenclature of the following homologous series is similar to that of alkanes. The only differences in the IUPAC nomenclature of these series is with precedence, position of the functional group and parent name. 103 HOmOLOGOuS SERIES: aLKENES IuPac NOmENcLaTuRE Precedence Position Parent name The longest chain is numbered so that the C=C has the lowest number in the chain. The first carbon in C=C is used to locate the double bond. Change the “ane” ending in the alkane to “ene” for the alkene. EXAMPLES (a) pent-2-ene (b) 4, 4-dimethylpent-2-ene H H C H H H H C C C H H H C H C C H H H H HOmOLOGOuS SERIES: aLKyNES IuPac NOmENcLaTuRE Precedence Position The longest chain is numbered so that the C=C has the lowest number in the chain. The first carbon in C=C is used to locate the triple bond. EXAMPLES Parent name Change the “ane” ending in the alkane to “yne” for the alkyne. (a) 4,4-dimethylpent-2-yne (b) 2-methylhex-3-yne H H H H C H C C C C H C H 104 H H C H H H H H H C C H H C H H C H C H C C H H H HOmOLOGOuS SERIES: HaLOaLKaNES (aLKyL HaLIdES) IuPac NOmENcLaTuRE The halogen substituents that occur on the alkane chains are named as follows: cl – chloro Br- bromo Precedence Position The longest chain is numbered to give the halogen substituents the lowest numbers in the chain. The carbon atoms to which halogen substituents are bonded are used to locate their positions. Examples: (a) 1-bromo-4-methylpentane Br H Cl H H C C C C C H H H 2 3 4 H C Parent name There is no change made to the parent alkane chain. It remains an alkane. (b) 2-bromo-1-chloropropane H 1 I – iodo 5 Cl H H H H HOmOLOGOuS SERIES: aLcOHOLS C H H H H C C Br H H IuPac NOmENcLaTuRE NOTES: Just like there are primary, secondary and tertiary alcohols, there are primary, secondary and tertiary alkyl halides. PRImaRy aLcOHOL: The carbon atom attached to the –OH group is attached to one other carbon atom in the molecule. Example: CH3CH2CH2CH2OH (butan-1-ol). SEcONdaRy aLcOHOL: The carbon atom attached to the –OH group is attached to two other carbon atoms in the molecule. Example: TERTIaRy aLcOHOL: The carbon atom attached to the –OH group is attached to three other carbon atoms in the molecule. Example: 105 IuPac NOmENcLaTuRE OF aLcOHOLS Precedence Position Parent name Longest chain is numbered so that the carbon atom bonded to the OH group has the lowest number in the chain. The carbon atom to which the OH group is bonded is used to locate it. Change the “e” ending in the alkane to “ol” for the alcohol. Examples 2-methylbutan-1-ol The table below provides information on the structural formula, condensed structural formula, molecular formula, isomer(s) and functional group of some alcohols when their IUPAC name is given. IuPac Name Structural Formula H Pentan-1-ol condensed Structural Formula H H H H H C C C C C H H H H H molecular Formula Functional / Structural Isomers Pentan-2-ol O H CH3CH2CH2CH2CH2OH C5H12O Pentan-3-ol 3-methylbutan-2ol Butan-1-ol 106 H H H H H C C C C H H H H Functional Group (structural formula) -O-H Butan-2-ol O H CH3CH2CH2CH2OH C4H10O 2-methylpropan2-ol -O-H HOmOLOGOuS SERIES: caRBOxyLIc acIdS Precedence IuPac NOmENcLaTuRE Position The longest chain is numbered so that the carbon atom in the COOH group has the lowest number in the chain. It will always be 1, but 1 is omitted in the IUPAC name. Parent name The carbon atom in COOH is used to locate the COOH group. Examples: Change the “e” ending in the alkane name to “oic acid” for the carboxylic acid. CH3 (a) HCOOH methanoic acid (b) CH3COOH (c) CH3CH2COOH (d) CH3CH2CH3CH2CH2C1OOH ethanoic acid propanoic acid 4-methylhexanoic acid Draw all the molecules above (a-d). The table below provides information on the structural formula, condensed structural formula, molecular formula, isomer and functional group of some carboxylic acids when their IUPAC name is given. IuPac Name Structural Formula H Ethanoic acid H condensed Structural Formula molecular Formula Functional / Structural Isomers O O C C O H CH3COOH C2H4O2 Methyl methanoate H H Propanoic acid H H C C H H Functional Group (structural formula) Methyl ethanoate O C - C –O-H O O H CH3CH2COOH C3H6O2 - C –O-H Ethyl methanoate 107 In general, an isomer of a carboxylic acid is usually an ester, and vice versa. HOmOLOGOuS SERIES: ESTERS Precedence The alkyl group from the alcohol is named first, followed by the name of the acid, with the “ic” changed to “ate”. Examples: A) butyl ethanoate B) ethyl butanoate C) ethyl ethanoate 108 Position Not applicable. Parent name Not applicable. D) ethyl methanoate HOmOLOGOuS SERIES: aLdEHydES IuPac NOmENcLaTuRE Precedence Position Parent name The longest chain is numbered so that the carbon atom in the CHO group has the lowest number in the chain. It will always be 1, but 1 is omitted in the IUPAC name. The carbon atom in CHO is used to locate the CHO group. Change the “e” ending in the alkane to “al” for the aldehyde. a) B) (a) HCHO (b) CH3CH2CHO methanal propanal HOmOLOGOuS SERIES: KETONES IuPac NOmENcLaTuRE Precedence Position Parent name The longest chain is numbered so that the carbon atom in the C=O group has the lowest number in the chain. The carbon atom of the C=O group is used to locate C=O. Change the “e” ending in the alkane name to “one” for the ketone. In general, the functional isomer of a ketone is an aldehyde, and vice versa. Examples: CH3 (a) HCOH methanone (b) CH3COCH3 Propan-2-one (c) CH3CH2CH2CH2COCH3 5-methylhexan-2-one Draw structural formulae for all the molecules above. 109 PHySIcaL PROPERTIES OF ORGaNIc mOLEcuLES definitions (REFER TO yOuR Exam GuIdELINES) - Boiling point - Vapour pressure - Melting point Types of inter-molecular forces (NOTE: ALL inter-molecular forces are CALLED as Van der Waals forces; therefore, Hydrogen bonds are also Van der Waals forces, according to caPS) REmEmBER THE FOLLOwING: The relationship between inter-molecular forces and boiling point: As the strength of inter-molecular forces increases, boiling point increases. The relationship between inter-molecular forces and vapour pressure: As the strength of inter-molecular forces increases, vapour pressure decreases. What type of inter-molecular forces are present between molecules of HydROcaRBONS Alkanes Alkenes Only London Forces Alkynes (Haloalkanes/aldehydes/Ketones contain BOTH London Forces and dipoledipole forces (organic molecules where there is a halogen atom or oxygen atom (more electro-negative atoms than C there are dipole dipole forces) dipole–dipole forces are stronger than London Forces. In Haloalkanes/aldehydes/Ketones, the stronger force is the Dipole- dipole force. (Note: London Forces are also present.) SummaRy Alkanes Hydrocarbons Alkenes have ONLy London forces between molecules Alkynes Haloalkane/aldehydes and Ketones have: 110 BOTH London forces and dipole-dipole forces (i.e. a compound with Br or cl or O will have dipole–dipole forces). dipole–dipole forces are stronger than London Forces. alcohols and carboxylic acids have: London Forces, Dipole-dipole forces and Hydrogen bonds (the strongest of the inter-molecular forces) between their molecules. SummaRy OF TyPES OF BONdS IN ORGaNIc mOLEcuLES ONLy London Forces Alkane Alkene BOTH London forces and dipole –dipole forces Haloalkane Aldehyde Alkyne Ketone aLL THREE London forces dipole-dipole forces Hydrogen bonding Alcohol Carboxylic acid Summary of bonds Hydrogen Bonds dipole dipole Hydrogen bonding occurs in molecules containing the highly electronegative elements F, O, or N that are directly bound to a hydrogen atom in a molecule. Since H has an electronegativity of 2.2 the bonds are not as polarized as purely ionic bonds and possess some covalent character. However, the bond to hydrogen will still be polarized and possess a dipole. Van der waals dipole-dipole interactions. Other groups of molecules beside hydrogen can be involved in polar covalent bonding with strongly electronegative atoms due to their electro negativities. The unequal distribution of charge gives the molecule a dipole nature. For instance, each of these molecules contains a dipole: The dipole of one molecule can align with the dipole from another molecule, leading to an attractive interaction that we call hydrogen bonding. Owing to rapid molecular motion in solution, these bonds are temporary (shortlived) but have significant bond strengths ranging from (9 kJ/mol (2 kcal/mol) (for NH) to about 30 kJ/mol These dipoles can interact with each other in an attractive fashion, which will also increase the boiling point. However since the electronegativity difference between carbon (electronegativity = 2.5) and the dispersion Forces/ London Forces The weakest intermolecular forces of all are called dispersion forces or London forces. These represent the attraction between instantaneous dipoles in a molecule. If you cool a gas like Argon to –186 °C, you can actually condense it into liquid argon. The fact that it forms a liquid it means that something is holding it together. That “something” is dispersion forces. Think about. on average the electrons in the valence shell, are evenly dispersed. But at any given instant, there might be a mismatch between how many electrons are on one side and how many are on the other, which can lead to an instantaneous difference in charge. 111 (7 kcal) and higher for HF. As you might expect, the strength of the bond increases as the electronegativity of the group bound to hydrogen is increased. So in a sense, HO, and NH are “sticky” – molecules containing these functional groups will tend to have higher boiling points than you would expect based on their molecular weight. electronegative atom (such as oxygen or nitrogen) is not as large as it is for hydrogen (electronegativity = 2.2), the polar interaction is not as strong. So on average these forces tend to be weaker than in hydrogen bonding. For hydrocarbons and other nonpolar molecules which lack strong dipoles, these dispersion forces are really the only attractive forces between molecules. The dipoles are weak and transient, they depend on contact between molecules – which means that the forces increase with surface area. A small molecule like methane has very weak intermolecular forces, and has a low boiling point. However, as molecular weight increases, boiling point also goes up. That’s because the surface over which these forces can operate has increased. Therefore, dispersion forces increase with increasing molecular weight. Individually, each interaction isn’t worth much, but if collectively, these forces can be extremely significant. COMPARE THE STRENGTH OF INTER-MOLECULAR FORCES London Forces – WEAKEST Dipole-dipole - STRONGER than London Forces, but WEAKER than Hydrogen bonds Hydrogen bonds - STRONGEST ExamPLES 1.1 Consider the three compounds: a: CH3OH B:CH3CH2OH and c:CH3CH2CH2OH 1.1.1 Which of the three compounds has the HIGHEST boiling point? 1.1.2 Explain your answer to QUESTION 1.1.1 above Answer: Approach: Using the summary above, check where the molecules in question belong and check the forces that can be found in that group. The compounds belong to the same homologous series. Mention the trend with regard to chain length (surface area). Mention the trend with regard to London Forces. 112 More energy is required to break forces. Generally: If the compounds belong to the same homologous series, the explanation must mention London Forces, as they are the oNLY forces affected by an increase/decrease in surface area. A common ERROR is using Hydrogen bonds in the answer, since we are dealing with alcohols - but hydrogen bonds/dipole-dipole forces are not affected by chain length. aNSwER: 1.1.1 c 1.1.2 From a to c Chain length (or surface area or molecular size) increases. The strength of London forces increases. More energy is needed to break inter-molecular forces. (DO NOT SAY “break bonds”. SAY “Break inter-molecular forces”.) CH3 1.2 Consider the TWO compounds, d: CH3COCH3 and E:CH3CH2CH3 1.2.1 Which compound has the HIGHEST boiling point? 1.2.2 Explain your answer to QUESTION 1.1 above. Approach: The two compounds belong to different homologous series Step 1: Identify the type of inter-molecular force in each Step2: Compare the strength of the forces. Step 3: More energy is needed to break inter-molecular forces. Answer: 1.2.1 d 1.2.2 STEP 1: Between the molecules of d there are dipole–dipole forces. STEP 2: Between the molecules of E, there are London forces. Dipole-dipole forces are stronger than London forces. STEP 3: More energy is required to break inter-molecular forces in d. 113 ExERcISE : PHySIcaL PROPERTIES Consider compounds a to c in the table below: compound a B c IuPac name ethane propane butane 1.1 To what homologous series do compounds a to c belong? 1.2 Is compound a SATURATED or UNSATURATED? Give a reason. 1.3 Which ONE of the compounds has the HIGHEST vapour pressure? 1.4 Explain your answer to QUESTION 1.3 ABOVE. 1.5 Which compound has the highest boiling point? ExcERcISE 2: PHySIcaL PROPERTIES 2.1 The table shows five organic compounds represented by the letters A to D. a B c d E CH3CH2OH CH3CO2H CH3CHO CH3CH2CH3 CH3CH3 2.1.1 Which compound in the table is an aldehyde? Consider the boiling points of compounds a to E, given in random order below, and use them where applicable to answer the questions that follow. 118 oC -89 oC 78 oC 2.1.2 Write down the boiling point of: a. Compound B b. Compound c 2.1.3 Explain the difference in the boiling point of: a. a and B 114 -42oC 21 oC b. B and c c. c and d d. D and E 2.1.4 Describe the trend in vapour pressure from c to E 2.2 The relationship between boiling point and the number of carbon atoms in straight chain molecules of alkanes, carboxylic acids and aldehydes is investigated. Curves P, Q and R are obtained. GRAPH OF BOILING POINT VERSUS NUMBER OF C ATOMS 500 Boiling point (K) P Q 300 R 100 2 3 4 5 6 7 8 Number of C atoms 2.2.1 Define the following terms: a. Boiling point b. Vapour pressure (2) (2) 2.2.2 For curves a, P, Q, R write down a conclusion that can be drawn from these results. (2) 115 2.2.3 Identify the curve that represents each of the following (a ,B or c): a. Aldehydes (1) b. Carboxylic acids (1) 2.2.4 Explain your answer to question 2.2.3 above. (5) SOLuTIONS: PHySIcaL PROPERTIES ExcERcISE 1: PHySIcaL PROPERTIES 1.1 Alkanes 1.2 Saturated - it contains only single bonds. 1.3 Ethane 1.4 From a to c: Surface area (chain length) increases. Strength of London forces increases. More energy is needed to break the inter-molecular forces. 1.5 C ExcERcISE 2: PHySIcaL PROPERTIES 2.1.1. c 2.1.2. a.118 oC b. 21 oC 2.1.3 a. Both compound A and B have Hydrogen bonds. Compound B has more sites for Hydrogen bonds than A (therefore the bonds are stronger in B). More energy is needed to break inter-molecular forces in B. b. B has strong Hydrogen bonds. C has weak dipole–dipole forces. More energy is needed to break the bonds in B. c. c has dipole dipole forces. d has weak London forces. c needs more energy to break forces. 116 d. From E to d Surface area (chain length) increases. The strength of London forces increases. More energy is needed to break inter-molecular forces. GENERaLLy, when explaining trends in boiling point and vapour pressure: IF the compounds differ in chain length: Mention the trend in surface area. Mention the trend in strength of London forces. (Remember ALL compounds have London forces, including alcohols and carboxylic acids.) Lastly, more energy is needed to break forces in the compound where forces are strongest. IF the compounds belong to different homologous series: Identify the type of force in each compound. Compare the strength of the forces - mention which one is stronger. More energy is needed to break inter-molecular forces in the compound where forces are stronger. (NOTE: IT IS NOT ACCEPTABLE TO WRITE, “TO BREAK BONDS, AS THESE ARE INTER-MOLECULAR FORCES, NOT INTER-ATOMIC BONDS, LIKE COVALENT BONDS, ETC.”) 117 ORGaNIc REacTIONS addITION, SuBSTITuTION, ELImINaTION ESTERIFIcaTION, cOmBuSTION, POLymERISaTION Draw contrasts between substitution and elimination addition and elimination SUBSTITUTION ELIMINATION (NO alkene in products or reactants) Alkene is a Product Reaction Conditions Reaction conditions -mild heat for in all Substitution Strong Heat in all Elimination For reaction of Haloalkane to For Hydrohalogenation Produce Alcohol (hydrolysis) - Use concentrated strong base KOH - Use H2O OR or NaOH –Dilute strong base KOH or NaOH ORGANIC REACTIONS ADDITION ELIMINATION Alkene is a reactant (On left of arrow) Alkene is a product Hydrogenation Alkene to Alkane: Add H2 (Catalyst Pt or Pd) Hydrohalogenation: Add HBr or HCl Alkene to Haloalkane (Product will have ONE halogen atom) Halogenation (Add Br2 or Cl2) Alkene to haloalkane Two halogen atoms on Product (Bromine test) Hydration (Add H2O) Catalyst: H2SO4 .Alkene to alcohol (add H2O catalyst H2SO4) 118 Reaction Conditions Strong Heat for all Elimination reactions Cracking e.g. C15H32 2C2H4 + C3H6 + C8H18 Catalytic cracking Low Temp/Low Pressure Thermal cracking High Temp/High Pressure Dehydrohalogenation (Removal of HBr or HCl) Conditions: Strong heat/Conc KOH or NaOH Dehydration (Removal of H2O) Conditions: Strong heat and excess conc H2SO4) TIPS In addition, reactions involving compounds where there is no symmetry across the double bond remember H “likes many Hs”) OH in substitution in addition reaction and substitution reactions. Think of water H E.g. Write down the structural formula of the Major product using structural formulae. 1 2 3 CH3CH=CH2 + H2O CH3 Approach: Think of H2O as H OH HO H H Answer: H C comes here C H H C H H C H H Look at the Carbon atoms at the double bond: Carbon 2 has ONE H; carbon 3 has 2. Therefore, there is NO symmetry across double bond H will add where there are many Hs, at Carbon 3 and OH- adds on Carbon 2 on to form the major product H Answer: H H O H C C C H H H H C H H For substitution the negative part (written second in formula undergoes substitution e.g. when using dilute KOH it is the OH- that substitutes, not the K+ ExERcISE 1: ORGaNIc REacTIONS 119 The flow diagram below shows how 2-bromopropane can be used to prepare other organic compounds. Alkene a (Major Product) B 2-bromobutane c Alcohol d 1.1 Classify 2-bromobutane as a PRIMARY, SECONDARY or TERTIARY haloalkane. (1) 1.2 Write down the STRUCTURAL FORMULA of a positional isomer of 2-bromobutane. (1) 1.3 Write down the type of reaction represented by: 1.3.1 a 1.3.2 c 1.3.3 d (1x3) 1.4 Write down the: 1.4.1 NAME or FORMULA of the inorganic reagent in a. reaction a b. reaction c c. reaction d (1x 3) 1.5 Write down the NAME or FORMULA of the inorganic product(s) in a. reaction a, if KOH is used as a reagent (1) b. reaction c, if KOH is used as a reagent (1) c. reaction d (1) 1.6 Write down the FORMULA of the catalyst used in reaction B. (1) ExERcISE 2: ORGaNIc REacTIONS Consider the following compounds: a: CH4 B: Ethanol C: Propanoic acid 2.1 Compound a reacts with unknown compound x in the presence of sunlight to produce bromomethane. Write down a balanced equation for the reaction between: 120 2.1.1 compound a and x (3) 2.1.2 compound a with excess oxygen (3) 2.2 Compound B and c react in the presence of an inorganic acid. Write down the: 2.2.1 name of the inorganic acid (1) 2.2.2 the equation for the reaction between B and c, using structural formulae for the organic reagents (1) 2.2.3 reaction condition (1) ExERcISE 3: ORGaNIc REacTIONS 3.1 A mixture of 4 grams of ethanol (C2H5OH) and 25 dm3 of air is ignited at 25 ºC and standard pressure. A reaction occurs according to the balanced equation: C2H5OH(ℓ) + 3O2(g) → 2 CO2(g) + 3 H2O(g) 3.1.1Determine, with the aid of a calculation, whether all of Compound d will be used up at the end of the reaction, if the molar volume of a gas at 25 ºC and standard pressure is 24,47 dm3. (ASSUME THAT AIR CONTAINS 21% OXYGEN (O2).) 3.1.2 If the ethanol is not used up, calculate the mass in grams of ethanol that is not used. (3) (5) SOLuTIONS ExERcISE 1: ORGaNIc REacTIONS 1.1 Secondary H Carbon 2 H H Br H c c c c H Look at carbon 2 (middle one) bonded to ONE H; therefore H H H H SEcONdaRy If there were TwO Hs on the carbon bonded to Br (carbon 2) then PRImaRy. If there were NO Hs on carbon 2, then TERTIaRy. 1.2 Change position of Br. (Change position of functional group, halogen or side chain to make a positional isomer.) 121 H H H H Br c c c c H H H H H 1.3.1 A Elimination OR Dehydrohalogenation 1.3.2 c Substitution or Hydrolysis 1.3.3 d Substitution 1.4.1 a. concentrated KOH b. KOH or H2O c. HBr 1.5 a. NaBr + H2O b. KBr c. H2O B H2SO4 1.6 ExERcISE 2: ORGaNIc REacTIONS H 2.1.1 H C H H H + Br2 C Br + HBr O H H C C C H H O H H C C H H H H H 2.1.2 CH4 + 2 O2 2CO2 + H2O 2.2.1 Sulphuric acid 2.2.2 H H H C C H H H 122 O H + H H C C H H H O O C H + H2O 2.2.3 Heat mildly over a water bath. ExERcISE 3: ORGaNIc REacTIONS 3.1 n ethanol = m/M = 4/46 = 0.087 mol Vo2 in 25 dm3 of air = 21/100 x 25 = 5.25 dm3 n O2 = V/Vm = 5.25/24.47 = 0,215 mol n ethanol = m/M = 4 /46 = 0.087 Ratio from equation Ethanol: Oxygen 1: 3 0.087 mol ethanol too much for 0.215 oxygen (Ratio 0.087:0.215 = 1:2,5 required ratio 1:3) O2 is the limiting reagent; Ethanol is in excess 3.2 n ethanol reacting = 1/3 x (0.215) = 0.072 mol n ethanol in Excess = n initial – n final = 0.087 - .072 = 0.015 mol m ethanol = nM= 0.015 x 46 = 0.69 g 123 14. ELEcTROcHEmISTRy ELEcTROcHEmISTRy E ELECTROLYTIC CELL GALVANIC CELL ELEcTROLyTIc cELLS (There are 5 electrolytic cells that you must understand.) 124 Convert electrical energy to chemical energy. (Note there is NO mention of mEcHaNIcaL energy in chemistry.) Reactions at electrodes are ENdOTHERmIc (reactions not self – sustaining.) In the diagram for electrolytic cell, there will be ONE CONTAINER and a power source or battery. ELECTROLYSIS OF EELECTROLYSIS OF saturated NaCl (aq) concentrated CuCl2(aq) Cℓ ℓ ℓ EXTRACTION OF ALUMINIUM Need for cryolite ELECTROLYTIC CELL (To lower melting poit saving on operational costs) Reactions at electrodes At CATHODE: Al3+ + 3e ELECTROPLATING Al At ANODE: C + O2 CO2 The substance that is plated forms the CATHODE. The electrolyte must have ions of the metal that are plated. ELECTROREFINING of Copper (Or Purification) The substance that is refined forms the ANODE. The electrolyte must contain ions of the metal that is refined. ELEcTROLySIS ELEcTROLySIS OF cucl2 and Nacl solutions “Second part of formula” goes to anode for oxidation. “First part” may go to cathode for reduction if it is a STRONGER OXIDISING agent. Positive ion “First part of formula” may NOT be reduced at ALL but H2O will undergo reduction 125 if Positive ion (Metal ion) “First part of formula” is a weaker oxidising agent than H2O e.g. in NaCl electrolysis Na+ dOES NOT undergo REducTION Take CuCl2 (Cu2+ Cℓ- Cℓ- ). The chloride ion Cl- “Second part” always migrates towards +ve ANODE (Positive attracted to negative) to undergo OXIDATION (Cℓ - is the reducing agent) Compare oxidising ability of Cu2+ the “first part” of the formula with that of water H2O in order to decide which species will undergo reduction (H2O or Cu2+). Cu2+ is a stronger oxidising agent than H2O therefore Cu2+ will undergo reduction Consider electrolysis of NaCl (aq) 5.1 Write down the following: 5.1.1 oxidation half reaction 5.1.2 formula of the reducing agent 5.1.3 reduction half reactio5.1.4 formula of the oxidising agent approach: 5.1.1 Negative part of formula (second part) Cl- goes to anode for oxidation Therefore, Oxidation half reaction 2 Clreversed) Cl2 + 2 e (From table and 5.1.2 The reducing agent is found in the oxidation half reaction on left of arrow) 2 Cl- Cl2 + 2 e Therefore Cl- 5.1.3 Compare the oxidising ability of Na+ with that of water H2O. H2O is a stronger oxidising agent than Na+, therefore H2O undergoes reduction. Therefore: H2O(l) + 2e H2(g) + OH- 5.1.4 The oxidising agent is at the left in the reduction half reaction H2O(l) + 2e H2(g) + OH- Therefore: H2O is the oxidising agent. 126 TIPS: You are NOT permitted to put double arrows in equations - always use ONE arrow. Reducing agent found on the left of arrow in OXIDATION half reaction The oxidising agent is ALWAYS on the left of the REDUCTION half reaction. There will be NO REACTION involving Na+ but H2O in electrolytic cells Know the properties of ANODE then know those of CATHODE as the opposite of anode Reverse the OXIDATION half reaction when taking it from the table Oxidising agents are bbon the FAR LEFT on TABLE of REDUCTION POTENTIALS Reducing agents immediately on the right of arrow when reading from left to right Smaller EO value for ANODE Write the formula as it is given DATA SHEET e.g. Eocell = Eocathode - Eoanode (No abbreviations are acceptable, e.g. E = Eocat - Eoan - NO marks for this.) To get marks for the formula, you must substitute values, even if they are INCORRECT. In Chemistry, there is no MECHANICAL energy - only CHEMICAL energy QuESTION 1 Consider the electrolytic cells a and B given below. X, Y, W and Z are carbon electrodes. w X X z Y Y CuCl2(aq) NaCl(aq) cELL a cELL B 1.1 Write down the: a. energy conversion in these cells b. the reason why a DC power source is used in place of an AC power source 127 1.2 Write down THREE observable changes in cell B. 1.3 Which electrode is the ANODE in: 1.3.1 Cell a 1.3.2 Cell B 1.4 Write down the half reaction taking place at electrode: 1.4.1 x and w 1.4.2 y 1.4.3 z 1.5 Write down the FORMULA and NAME of the reducing agent in cell: 1.5.1 a 1.5.2 B 1.6 Write down the FORMULA and NAME of the oxidising agent in cell: 1.6.1 a 1.6.2 B 1.7 Write down the name of the gas formed at: 1.7.1 electrode x 1.7.2 electrode y 1.8 In cell a, the Na+ ions do not form sodium Na. Explain be referring to the strength of oxidising agents involved. QuESTION 2 A chemist wants to determine the PERCENTAGE of copper by mass in an IMPURE copper sample containing platinum and silver as the only impurities. He first sets up an electrolytic cell to purify copper. Power Source ImPuRE cOPPER P Q P cuSO4(aq) 2.1 Is electrode P the POSITIVE or NEGATIVE electrode? (1) 2.2 Which electrode (P or Q) will show an increase in mass? (1) 2.3 Write down the following: 128 PuRE cOPPER 2.3.1 The SYMBOL or NAME of the product formed at electrode P. (1) 2.3.2 The equation for the half reaction taking place at electrode Q. (2) 2.4 Use relative strengths of reducing agents involved to explain why the metals platinum and silver do not form ions during this process. (2) [12] QuESTION 3 The electrolytic cell below is used for electro refining (purification) an IMPURE copper sample. Electrode Q contains copper, which contains 3% IMPURITIES by mass. Power Source Pure Copper IMPURE Copper P Q CuCℓ2(aq) 3.1 Write down the observable changes at the: 3.1.1 Anode 3.1.2 Cathode 3.2 Write down the half reaction taking place at the: 3.2.1 Anode 3.2.2 Cathode 3.3 CuCl2 solution is light green. How will the intensity of the colour of the solution change as the cell is working? Give a reason. 3.4 Calculate the mass of the IMPURE copper sample if 2 moles of electrons are transferred when ALL the copper is extracted from the IMPURE sample. 129 3.5 The IMPURE copper sample contains metals like silver. Explain why silver does undergo oxidation in this cell. 3.6 What happens to the electrolyte when the cell is in operation? Explain. GaLVaNIc cELLS Questions on galvanic cells may include: dIaGRam OF a GaLVaNIc cELL REacTION EQuaTIONS (increase in oxidation number for aNOdE) cELL NOTaTION (anode is on the left) KNOw THE PROPERTIES OF aN aNOdE. aNOdE OXIDATION OCCURS (ELECTRONS MOVE AWAY FROM THE ANODE) SMALLER EO-values NEGATIVE ELECTRODE (CONNECTED TO NEGATIVE TERMINAL OF VOLTMETER) DECREASE IN MASS [POSITIVE IONS] INCREASE ANIONS MOVE TOWARDS ANODE OXIDATION NUMBER INCREASES (Based on the notes above, you can deduce what happens at the caTHOdE. No need to memorise both.) cOmmON mIScONcEPTION The CATHODE DOES NOT undergo REDUCTION. 130 What undergoes reduction (i.e. oxidising agent) is an ION in solution at the CATHODE half cell. QuESTION 4 STRucTuREd QuESTIONS The diagram below shows a galvanic cell operating under standard conditions. The cell reaction taking place when the cell is functioning is: 6Cℓ - (aq) + 2Au3+(aq) → 3Cℓ2(g) + 2Au(s) With switch S OPEN, the initial reading on the voltmeter is 0,14 V. Write down the following: 4.1 The NAME or FORMULA of the oxidising agent. (1) 4.2 The half-reaction that takes place at the anode. (2) 4.3 The cell notation for this cell. (3) 4.4 Calculate the standard reduction potential of Au. (4) 4.5 The concentration of the Au3+ is increased. How will this change affect the initial EMF of the cell. Write down INCREASES, DECREASES or REMAINS THE SAME. (1) Switch S is now closed and the bulb lights up. 131 4.5 How will the reading on the voltmeter now compare to the INITIAL reading of 0,14 V? Write down only LARGER THAN, SMALLER THAN or EQUAL TO. Give a reason for your answer. QuESTION 5 A learner sets up a galvanic cell as shown below. The cell functions under standard conditions. Nitrates are used as electrolytes. 5.1 Define the following terms: 5.1.1 Electrolyte (2) 5.1.2 Galvanic cell (2) 5.2 Write down the initial concentration of the Copper (II) nitrate (Cu (NO3)2 Solution. (1) 5.3 Calculate the mass of Copper (II) nitrate needed to prepare the solution if the volume of the solution is 250 cm3. (4) 5.4 In which direction (a or B) will ANIONS move in the salt bridge? (1) 5.5 Calculate the emf of the above cell under standard conditions. (4) 5.6 Write down the balanced equation for the net cell reaction that takes place in this cell. 132 (3) 5.7 Write down the cell notation for this cell. (3) QuESTION 6 A standard electrochemical cell is set up using a standard hydrogen half-cell and a standard X|X2+ half-cell, as shown below. The standard reduction potential of the X|X2+ half-cell is - 0.74 V. 6.1 Write down the following: 6.1.1 The function of component Q. 6.1.2 The initial reading on the voltmeter. (1) (1) 6.1.3 The half-reaction that takes place at the cathode of this cell. (2) 6. 2 The hydrogen half-cell is now replaced by a m|m2+ half-cell. The cell notation of this cell is: X(s)| X2+(aq) || m2+(aq) | m(s). The initial reading on the voltmeter is now 0.61 V. 6.2.1 Identify metal m. Show how you arrived at the answer. (5) 6.2.2 Is the cell reaction EXOTHERMIC or ENDOTHERMIC? (1) 6.3 The reading on the voltmeter becomes zero after using this cell for several hours. Give a reason for this reading by referring to the cell reaction. (1) 133 QuESTION 7 Consider the following reaction: Ce3+ (aq) + Ti(s) Ti3+ (aq) + Ce(s) Standard electrode potentials are given in the table below. Redox pair Standard Electrode Potential Ti3+/Ti -1.63 V Ce3+/ Ce -2.48 V 7.1 Determine by calculation if the reaction is spontaneous. 7.2 According to the data in the given table, which substance is the strongest reducing agent? SOLuTIONS ELEcTROcHEmISTRy: GaLVaNIc aNd ELEcTROLyTIc cELLS QuESTION 1 1.1 a. Chemical energy to electrical energy. b. To ensure the polarities of the anode and cathode do not change. 1.2 Bubbles of chlorine gas at Z (Anode) Brown coating on Y (Copper) Intensity of blue colour decreases. (The concentration of Cu2+ decreases as Cu2+ reduced to Cu) 1.3.1 x 1.4.1 2Cl- 134 Cl2 + 2e 1.3.2 z 1.4.2 2 H2O(l) + 2e 1.5.1 Cl 1.5.2 Cl- 1.6.1 H2O 1.6.2 Cu2+ 1.7.1 Chlorine 1.7.2 hydrogen H2O + OH- 1.8 H2O is a stronger oxidising agent than Na+. Therefore, H2O is reduced. QuESTION 2 2.1 POSITIVE 2.2 Q 2.3.1 Cu2+ or Copper (II) ions 2.3.2 Cu2+ + 2e (From oxidation Cu Cu2+ + 2e ) Cu 2.4 Cu is a stronger reducing agent than both silver and platinum. QuESTION 3 3.1.1 Mass decreases (ANODE) 3.2.1 Cu Cu2+ + 2e 3.3 nCu = ½ x 2 = 1 mol 3.1.2 Mass increases (CATHODE) 3.2.2 Cu2+ + 2e Cu (Ratio of mol Cu : mol electrons 1:2) M pure Cu = n .M = 1 x 63.5 = 63.5 g Therefore: 97/100 x IMPURE Copper = Pure Copper (3% impurities; therefore 100% - 3% = 97% is pure copper) 3.4 Cu is a stronger reducing agent than Ag. QuESTION 4 4.1 Au3+ (From oxidation numbers Oxidation number of Au DECREASES from +3 to 0) 4.2 2 Cl- Cl2 + 2e 4.3 Pt Cl- Cl2 Au3+ Au 4.4 INCREASE 4.5 SMALLER THAN Lost volts due to internal resistance (when current flows through the cell). QuESTION 5 5.1.1 Electrolyte – a solution or liquid that conducts electricity because it contains ions. 135 5.1.2 Galvanic cell: converts chemical energy to electrical energy. 5.2 c = 1 mol.dm-3 (standard conditions) 5.3 mCu = ? nCu = c.V = 1 x 250/1000 = 0,25 mol mCu = n.M= 0.25 x 63.5 = 15.88 g (15.875 g) 5.4 a 5.5 Eocell = Eocathode - Eo anode = 0.80 (Formula) – 0.34 (Substitution) = 0.46 V (Answer with UNIT) Cu2+ + 2e 5.6 Cu Ag+ + e electrons.) Ag 2 Ag+ + 2 e 5.7 (Multiply by 2 to balance the number of 2 Ag Cu/Cu2+// Ag+/Ag QuESTION 6 6.1.1 Completes the circuit/ensures electrical neutrality of the solutions. 6.1.2 0,74 V ( an use Eocell = Eocathode - Eo anode = 0 – (- 0.74) = 0.74 V Remember: Eocell cannot be negative 6.1.3 2 H+ + 2e H2(g) 6.2.1 m is the cathode (M is on the right-hand side in cell notation.) Eocell = Eocathode - Eo anode 0.61 = Eocathode - (- 0.74) 0.61 = Eocathode + 0.74 0.61 – 0.74 = Eocathode - 0.31 V = Eocathode m is Pb 136 QuESTION 7 7.1 According to the equation, Ce is the CATHODE (Reason: Oxidation number decreases from +3 to 0; OR Ti is the ANODE - Oxidation number increases from 0 to +3.) Approach: Calculate EMF. If EmF is POSITIVE it means Reaction is SPONTaNEOuS Eocell = Eocathode - Eo anode Eocell = -2.48 – (- 1.63) = - 0.85 V Reaction NOT SPONTANEOUS, Eocell is NEGATIVE 7.2 Ce INdIREcT TRaNSFER OF ELEcTRONS e. g. A zinc rod is placed in a copper(II) sulphate solution, as shown below. Zn CuSO4(aq) ExamPLE 1 Write down THREE observable changes that will occur. 1.1 Write down the: 1.1.1 oxidation half reaction 1.1.2 reduction half reaction 1.1.3 name of the oxidising agent 1.1.4 name of the reducing agent 1.1 Oxidation half: Zn Zn2+ + 2e 137 1.2 Reduction half reaction: Cu2+ + 2e Cu 1.3 Oxidising agent (on the left of the Reduction half reaction) Cu2+ + 2e Cu Oxidising agent: Cu2+ 1.4 Reducing agent (on the left of the Oxidation half reaction) Zn Zn2+ + 2e Zn is the Reducing agent NOTE: If the reducing ability of the METAL is lower than that of the metal of the positive ion in solution, no reaction will occur. EXAMPLE 2: A piece of silver is placed in magnesium sulphate, e.g.: Ag MgSO4 2.1 Will a spontaneous reaction occur? 2.2 Can we store a magnesium sulphate solution in a container made of silver, without the container leaking? Explain. 138 Approach: Consult the Table of Reduction potentials. Extract form Table 4B OXIDISING AGENTS + e ⇌ Li Li+ ……………………… ……………………… Ba Ba2+ + 2e ⇌ ……………………… ………………………. Mg2+ + 2e ⇌ Mg ……………………… ……………………… ……………………… Cu2+ + 2e ⇌ Cu ……………………. ……………………. + e ⇌ Ag Ag+ …………………….. Cl2 (g) + 2e ⇌ 2ClF2(g) + 2e ⇌ Li - strongest reducing agent REDUCING AGENTS 2F- F2 - strongest oxidising agent RULE: Top right reduces bottom left. Consider example 1 above: Zn + CuSO4 Find Zn and Cu2+ in the table. Remember Zn is the reducing agent in Example 1. Zn2+ + 2e ⇌ Cu2+ + 2e ⇌ Zn Zn is a stronger reducing agent than Cu. Cu Zn reduces Cu2+ to Cu. 139 Example 2: If a reaction occurs, Ag will be the reducing agent. (It is the METAL.) Mg2+ Ag+ + + 2e ⇌ Mg e ⇌ Ag THIS CAN NOT TAKE PLACE SPONTANEOUSLY. In Example 2: 2.1 There will be NO reaction. Reason: Ag is a weaker reducing agent than Mg. Ag cannot reduce Mg2+ to Mg 2.2 To be able to store a solution in a container, the SOLUTION AND THE METAL USED IN THE CONTAINER MUST NOT REACT. Yes, we can store MgSO4 in Ag container. Reason: Reaction will not occur spontaneously, as Ag is a weaker reducing agent than Mg. Ag cannot reduce Mg2+ to Mg The table of reduction potentials only gives values of redox reactions under standard conditions. This means that if the experiment is carried out when the conditions are not standard, we cannot rely on the usual equation to give us the. We will need a different type of equation called the Nernst Equation. The Nernst equation however is not prescribed in our current curriculum. Our calculations should therefore confine themselves to Standard Conditions only. 140 PRACTICE QUESTIONS QuESTION 1 An aluminium rod is placed in a copper (II) sulphate solution as shown below. Al CuSO4 (aq) ExamPLE 1 1.1 Write down THREE observable changes that will occur. 1.2 Write down the following: 1.2.1 oxidation half reaction 1.2.2 reduction half reaction 1.2.3 name of the oxidising agent 1.2.4 name of the reducing agent ExamPLE 2 2.1 Can we store: 2.1.1 CuSO4 in a container made of silver metal? Explain. 2.1.2 MgSO4 in a container made of zinc? Explain. 2.1.3 AgNO3 in a container made of copper? . 141 142 143 144 Published in 2017 by the Department of Basic Education 222 Struben Street Private Bag X895, Pretoria, 0001 Telephone: 012 357 3000 Fax: 012 323 0601 © Department of Basic Education website facebook twitter www.education.gov.za www.facebook.com/BasicEd www.twitter.com/dbe_sa
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