Pre-U STPM Physics Term 1 CC039242a

Physics PELANGI BESTSELLER STPMText TERM 1 Poh Liong Yong 1 COPIES SOLD MORE THAN PRE-U For the 21st Century Learner SCAN ME

Physics Poh Liong Yong PRE-U TERM 1 © Penerbitan Pelangi Sdn. Bhd. 2022 All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, photocopying, mechanical, recording or otherwise, without the prior permission of Penerbitan Pelangi Sdn. Bhd. ISBN: 978-967-2856-92-4 eISBN: 978-967-2878-05-6 (eBook) First Published 2022 Tel: 03-8922 3993 Fax: 03-8926 1223 / 8920 2366 E-mail: [email protected] Enquiry: [email protected] Printed in Malaysia by Percetakan Jiwabaru Sdn. Bhd. No. 2, Jalan P/8, Kawasan Miel Fasa 2, Bandar Baru Bangi, 43650 Bangi, Selangor Darul Ehsan. Please log on to https://plus.pelangibooks.com/errata/ for up-to-date adjustments to the contents of the book (where applicable). STPMText

Pre-U STPM Text Physics Term 1 is written based on the new syllabus made by the Malaysian Examinations Council (MEC) that has been implemented since 2012. The book is designed and well organised with the following features to help students to understand the concepts taught. STPM Scheme of Assessment Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 960/3 Physics Paper 3 Oscillations and Waves, Optics and Modern Physics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 960/5 Physics Paper 5 Written Physics Practical Written practical test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central First, assessment Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 To be scaled to 45 (20%) Throughout the three terms School-based assessment v Physics Term 1 STPM Chapter 1 Physical Quantities and Units 1 1 CHAPTER Bilingual Keywords Concept Map PHYSICAL QUANTITES 1 AND UNITS Base Quantities and SI Units – Base units – Derived units Scalars and Vectors – Sum, scalar product and vector product of coplanar vectors – Resolve a vector Dimensions of Physical Quantities – Determine dimensions of derived quantities – Check dimensional homogeneity – Construct equations in physics Uncertainties in Measurements – Uncertainty in derived quantity – Significant figures Physical Quantities and Units 1 Base quantity: Kuantiti asas Derived quantity: Kuantiti terbitan Dimension: Dimensi Error: Ralat Fractional uncertainty: Pecahan ketakpastian Physical quantity: Kuantiti fizik Random error: Ralat rawak Resolving vector: Meleraikan vektor Resultant vector: Vektor paduan Scalar product: Hasil darab skalar Scalar: Skalar Significant figures: Angka bererti Systematic error: Ralat bersistem Uncertainty: Ketakpastian Vector: Vektor 68 3 Physics Term 1 STPM Chapter 3 Dynamics 2009/P1/Q7, 2009/P2/Q2, 2010/P1/Q7, 2015/P1/Q18, 2016/P1/Q3 (a) What is the average extra force on the neck of the passenger if the mass of his head is 3.0 kg and remains firmly attached to the body. (b) Comment whether the force is big enough to cause serious injury to the passenger. 18. In order to stop a car of mass 2 000 kg travelling at 20 m s–1, the driver applies his brakes so that the stopping force F increases linearly to a maximum and then decreases to zero as shown in the figure. (a) Calculate (i) the change in momentum between t = 0 and t = 10 s. (ii) the value of Fmax. (iii) the maximum retardation of the car. (b) Sketch graphs to show the variation with time of (i) the velocity, (ii) the distance travelled. (c) Explain why this would seem to be a more gentle stop to a passenger compared to a constant force of Fmax is applied throughout the 10 second. 19. (a) State the relation between force and momentum. (b) A stream of sand falls at a constant rate from a height h on the pan of a balance marked in kg. After a time T, a total mass M has fallen and the sand stops falling. (i) Sketch a graph to show how the reading m of the balance changes with time t. Give the values of m at t = 0, t = T, and t = 2T. Explain the shape of the graph. (ii) How would the graph differ if the sand was allowed to fall from a height 2h at the same rate? 3.2 Friction Students should be able to: • explain the variation of frictional force with sliding force • define and use coefficient of static function and coefficient of kinetic friction Learning Outcomes 1. Friction between two solid surfaces is the force which opposes relative motion between the surfaces. 2. Figure 3.4 shows a block on a rough horizontal surface. When the block is pulled by a small force F, it does not move because static friction between the surfaces is in contact opposes the force F. Magnitude of static friction = magnitude of the force F. 3. When the pulling force is increased, and the block remains stationary, then the static friction too has increased. 4. The static friction between the surfaces in contact is maximum just before the block starts to move. 5. This maximum static friction is known as limiting static friction. mg R F Static friction Figure 3.4 Friction opposes motion INFO Normal Force and Friction Concept Map Physics 960/1 2015 2016 2017 Year A B C A B C A B C 1. Physical Quantities and Units 1 – – 1 – – 1 – – 2. Kinematics – – – 1 – 1 1 – – 3. Dynamics 2 – – 1 – – 2 – – 4. Work, Energy and Power 1 1 – 1 – – 1 – – 5. Circular Motion 2 – – 1 – – 1 – – 6. Gravitation 1 – 1 2 – 1 1 1 – 7. Statics 1 – – 1 1 – 1 – 1 8. Deformation of Solids 1 – 1 1 – – 1 – 1 9. Kinetic Theory of Gases 2 1 – 3 – – 2 – 1 10. Thermodynamics of Gases 2 – 1 2 1 – 2 1 – 11. Heat Transfer 2 – – 1 – 1 2 – – Total 15 2 3 15 2 3 15 2 3 Analysis of STPM Papers (2014 – 2017) iv PREFACE Bilingual Keywords STPM Scheme of Assessment 85 3 Physics Term 1 STPM Chapter 3 Dynamics (ii) From the two equations, obtain an equation relating (iii) Hence, deduce the condition v1 to u and v2. necessary for the mass m1 to stop after collision. Explain your answer. (iv) A trolley of mass 400 g moving with a velocity of 0.80 m s–1 collides elastically with another trolley of mass 600 g initially at rest. Calculate the velocities of the trolleys after collision. 17. Explain the terms force and linear momentum. (a) State how force is related to linear momentum. (b) In an elastic collision, kinetic energy is conserved. Discuss whether in a completely inelastic collision kinetic energy is lost completely. Use an example to illustrate your reasoning. (c) A mass m1 moving with a velocity collides elastically with a mass u stationary. After collision, m2 initially with velocity m1 and m2 move v1 and v2 respectively. (i) Write equations to represent the use of the principle of conservation of linear momentum and the principle of conservation of energy in the above collision. ANSWERS 1 1. C 2. B 3. B 4. B 5. B 6. B 7. C 8. A 9. D 10 11. B: Resultant force in the direction of . D a is T cos θ. Apply F = ma gives T cos θ = ma 12. C: s = (u + v 2 )t gives t = 2(0.75) 250 + 0 s = 6.0 ms F = m( v – u t ) = (0.050)( 0 – 250 6.0 × 10–3 ) N = 2.1 kN 13. C: Acceleration, a = gradient of Resultant force v-t graph F = ma 14. D 15. 55 N F – mg = ma s = ut + 1 2 at2 16. F = d dt (mv) F = 60 N 17. (a) 168 N F = ma, a = v – u t (b) Yes 18. (a) (i) 4.00 × 104 kg m s–1 Δp = m (v – u) (ii) Fmax = 8 000 N Δp = area under graph (iii) 4.00 m s–2 amax = Fmax (b) (i) m (ii) (c) Greater change of momentum Answers Analysis of STPM Papers Learning Outcomes QR code Info / Video Online Quick Quiz ii

34 2 Physics Term 1 STPM Chapter 2 Kinematics 2. For a body travelling along a straight line, if u = initial velocity, v = final velocity after a time t, and s = displacement in time t, then uniform acceleration, a = Change in velocity Time taken = v – u t v = u + at .............................. a Displacement, s = Average velocity × Time s = ( u + v 2 ) t .............................. b s = ( u + u + at 2 )t s = ut + 1 2 at 2 .............................. c From ①, v – u = at From ➁, v + u = 2s t (v – u)(v + u) = at × 2s t v 2 – u2 = 2as .............................. d ➃ Example 1 A motorist travelling at 72 km h–1 on a straight road approaches a traffic light which turns red when he is 55.0 m away from the stop line. The reaction time is 0.7 s. With the brakes applied fully, the vehicle decelerates at 5.0 m s–2. How far from the stop line will he stop and on which side of the stop line? Solution: 72 km h–1 = 72 000 m 3 600 s = 20 m s–1 The vehicle travels at constant speed of 20 m s–1 during the reaction time of 0.7 s. Distance travelled during the reaction time = 20 × 0.70 = 14.0 m When the vehicle stops, final velocity v = 0 Acceleration = –5.0 m s–2 (negative because of deceleration) Using v 2 = u2 + 2as Distance travelled when the brakes are applied, s = v2 – u2 2a = 0 – 202 2 (–5) = 40.0 m Total distance travelled = (14.0 + 40.0) m = 54.0 m Hence, the vehicle stops (55.0 – 54.0) m = 1.0 m before the stop line. Info Physics A380 Airbus take-off speed A wide-body aircraft such as the Airbus A380 has a takeoff speed of 280 km h–1 and requires a runway of length 3000 m. What then is its acceleration? Exam Tips There are five quantities, u, v, t, s, a, in the four equations for motion under constant acceleration. Each of the equations involves only four of the quantities. In a question, usually values for three of the quantities are provided, and you are asked to calculate the value of another of the quantities. Use the equation that contains the four quantities mentioned in the question. Physics Term 1 STPM Chapter 1 Physical Quantities and Units 2 1 1.1 Base Quantities and SI Units Students should be able to: • list base quantities and their SI units: mass (kg), length (m), time(s), current (A), temperature (K) and quantity of matter (mol) • deduce units for derived quantities Learning Outcomes Physical Quantities 1. In Physics, there is the need to make careful observations, precise and accurate measurements. 2. Physical quantities are quantities that can be measured. 3. Examples of physical quantities are length, mass, time, weight, electric current, force, velocity and energy. 4. To describe a physical quantity, two things must be specified. The first is a numerical value and the second is the unit. For example, a man’s height-1.70 m has the numerical value 1.70 and the unit is metre (m). SI Units 1. The unit of a physical quantity is the standard used to compare different magnitudes of the same physical quantity. For example, if the length of a table is 2 m, it means that the length of the table is twice the length of the standard metre (m). 2. Other units of length include foot and mile. 3. Although the foot can be converted into metres and vice versa, the SI (Le Systeme Internationale) is used to facilitate trade and communication between countries of the world. 4. The SI was established in 1960. 5. In SI, each physical quantity has only one unit. Prefixes are attached to SI units, as shown in Table 1.1. Table 1.1 Common SI prefixes Prefix Factor Symbol pico 10–12 p nano 10–9 n micro 10–6 μ milli 10–3 m centi 10–2 c deci 10–1 d kilo 103 k mega 106 M giga 109 G tera 1012 T 2008/P1/Q1, 2013/P1/Q1 Info Physics The systeme International d’Unites (International System of Units, SI) was adopted in 1960 at the 11th General Conference of Weights and Measures. 320 Paper 1 Kertas 1 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fifteen questions in Section A. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam Bahagian A. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. The following measurements are obtained in an experiment. L = (10.0 ± 0.1) cm and H = (2.0 ± 0.1) cm. Which expression has the greatest percentage uncertainty? Bacaan berikut diperolehi dalam suatu eksperimen. L = (10.0 ± 0.1) cm dan H = (2.0 ± 0.1) cm. Ungkapan yang manakah mempunyai peratus ketakpastian tertinggi? A (L – H) B C LH (L + H) D L H 2. A particle X moves from the point O to a point S via the point Q at constant speed. Another particle Y moves from point O to the point S via the point P with the same speed as shown in the diagram. Suatu zarah X bergerak dari titik O ke titik S melalui titik Q pada laju seragam. Suatu zarah yang lain Y bergerak dari titik O ke titik S melalui titik P pada laju yang sama seperti ditunjukkan dalam gambar rajah. Q S O Which statement is correct about the magnitude and direction of the average acceleration of P the particles X and Y? Pernyataan yang manakah yang betul tentang magnitud dan arah pecutan purata zarah X dan Y? Magnitude of acceleration Magnitud pecutan Direction of acceleration A Same / Arah pecutan Sama B Same / Same / Sama Sama Different / C Different / Berbeza Berbeza D Different / Same / Sama Berbeza Different / Berbeza STPM Model Paper (960/1) STPM Model Paper 98 4 Physics Term 1 STPM Chapter 4 Work, Energy and Power Quick Check 3 1. A toy car which is powered by a motor moves with a constant velocity v along a straight line. The figure shows the variation of the energy E supplied by the motor with time t. a If the constant force provided by the motor is F, the velocity v is given by A a Fb C Fa b B b Fa D Fb a 2. The power of a toy car of mass 0.400 kg varies with its velocity as shown in the figure. What is the acceleration of the toy car? Power / W Velocity / m s–1 8 2 2.0 0 A 1.60 m s–2 C 10.0 m s–2 B 7.50 m s–2 D 12.0 m s–2 3. A car of mass 1 000 kg is moving with a constant velocity of 20 m s–1 along a horizontal road. The output power of the car is 20 kW. The output power is then suddenly increased to 60 kW. What is the acceleration of the car? A 0.20 m s–2 C 2.0 m s–2 B 0.30 m s–2 D 3.0 m s–2 4. What is the power required by a body of mass m to produce an acceleration of a when it is moving with a velocity v up a plane inclined at an angle θ to the horizontal? (Neglect friction) A mavg sin θ B mav sin θ + mgv C mav + mgv sin θ D (mav + mgv) sin θ 5. A car of mass m has an engine which delivers a constant power P. What is the minimum time it takes to accelerate from rest to a speed v? A mv P C mv 2 2P B P mv D 2P mv 2 6. A machine requires 200 J of energy to do 120 J of work. What is its efficiency? 7. The efficiency of a car engine is 25%. The total resistance to motion is 1 000 N when the car is moving at a constant velocity. If the energy in one litre of petrol is 40 MJ, how far can the car travel with 1.0 litre of petrol? 8. The power of an electric motor is 200 W. It is used to raise a load of 100 N at a constant speed 0.6 m s–1. What is the efficiency of the motor? (Take g = 10 m s–2) 9. (a) The total resistance to the motion of a cyclist on a level road is 25 N. If his speed is 12 m s–1, what is his output power? (b) The cyclist managed to reduce the resistance to 15 N by bending low over the handle bar. If his output power is as calculated in (a) above, what would be his speed? Important Formulae 1. Work done by a constant force F, W = F.s = Fs cos θ 2. Work done by a variable force, W = ∫ 0 s Fdx = Area under force-displacement graph 3. Relation between force F and potential energy U F = – dU dx 4. Power = Work done Time taken = Fv 5. Efficiency = Work done by machine Energy input × 100% = Output power Input power × 100% Info Physics Physics Term 1 STPM Chapter 1 Physical Quantities and Units 25 1 (a) Calculate to the correct number of significant figures, (i) the average time for 20 oscillations, (ii) the period T of the simple pendulum. (b) Determine a value of g, with its uncertainty. Important Formulae 1. Resolving a vector X = R cos θ Y = R sin θ R = X 2 + Y 2 tan θ = Y X 1 STPM PRACTICE R Y X 1. The volume flow rate R of a fluid flows through a pipe of cross sectional area length A and l when the pressure difference between the two ends of the pipe, p is given by R = kA2 p ηl where c is a dimensionless constant, and the viscosity of the fluid. The unit of η is of base SI units is η in terms A kg m C kg m–1 s–1 B kg s–2 D kg m–3 s–1 2. The figure shows two dogs P and Q pulling a sledge S along the straight line O horizontal surface covered with snow. The x on a forces on the sledge by the dogs are in the directions shown. O x θ 20° 4000 N S 3600 N P Q What is the value of θ? A 16.0° C 67.7° B 22.3° D 73.9° 3. Which of the following is an example of random error? A Zero error of an ammeter B Error in the calibration of a thermometer C Error in assuming g, acceleration of free fall is 10 m s–2 D Error in recording the time shown on a stop-watch. 4. Three coplanar forces of 3.0 N, 5.0 N and 6.0 N acts on a point as shown in the figure below. O 3.0 N y x 6.0 N 5.0 N 30° What is the magnitude and direction of the resultant force? A 12.2 N at 48o below the x-axis B 12.2 N at 42o above the x-axis C 9.3 N at 76o below the x-axis D 9.3 N at 14o above the x-axis 5. What is the dimension for electric potential? A ML2 A–1T –3 C ML2 AT–2 B ML2 A–1T –2 D ML2 A2 T –2 2. P.Q = PQ cos θ 3. |P × Q| = PQ sin θ STPM Practices Exam Tips Important Formulae Quick Check Examples iii

Physics 960/1 2015 2016 2017 Year A B C A B C A B C 1. Physical Quantities and Units 1 – – 1 – – 1 – – 2. Kinematics – – – 1 – 1 1 – – 3. Dynamics 2 – – 1 – – 2 – – 4. Work, Energy and Power 1 1 – 1 – – 1 – – 5. Circular Motion 2 – – 1 – – 1 – – 6. Gravitation 1 – 1 2 – 1 1 1 – 7. Statics 1 – – 1 1 – 1 – 1 8. Deformation of Solids 1 – 1 1 – – 1 – 1 9. Kinetic Theory of Gases 2 1 – 3 – – 2 – 1 10. Thermodynamics of Gases 2 – 1 2 1 – 2 1 – 11. Heat Transfer 2 – – 1 – 1 2 – – Total 15 2 3 15 2 3 15 2 3 Analysis of STPM Papers (2014 – 2017) iv

STPM Scheme of Assessment Term of Study Paper Code and Name Theme/Title Type of Test Mark (Weighting) Duration Administration First Term 960/1 Physics Paper 1 Mechanics and Thermodynamics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 1 to 11. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Second Term 960/2 Physics Paper 2 Electricity and Magnetism Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 12 to 18. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment Third Term 960/3 Physics Paper 3 Oscillations and Waves, Optics and Modern Physics Written test Section A 15 compulsory multiple-choice questions to be answered. Section B 2 compulsory structured questions to be answered. Section C 2 questions to be answered out of 3 essay questions. All questions are based on topics 19 to 25. 60 (26.67%) 15 15 30 1—1 2 hours Central assessment 960/5 Physics Paper 5 Written Physics Practical Written practical test 3 compulsory structured questions to be answered. 45 (20%) 1—1 2 hours Central assessment First, Second and Third Terms 960/4 Physics Paper 4 Physics Practical School-based Assessment of Practical 13 compulsory experiments and one project to be carried out. 225 To be scaled to 45 (20%) Throughout the three terms School-based assessment v

Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 1 PHYSICAL QUANTITIES AND UNITS 1 1.1 Base Quantities and SI Units 2 1.2 Dimensions of Physical Quantities   4 1.3 Scalars and Vectors 13 1.4 Uncertainties in Measurements 19 STPM Practice 1 25 QQ1 28 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 2 KINEMATICS 32 2.1 Linear Motion 33 2.2 Projectile 45 STPM Practice 2 53 QQ2 56 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 3 DYNAMICS 59 3.1 Newton’s Laws of Motion 60 3.2 Friction 68 3.3 Conservation of Linear Momentum 70 3.4 Elastic and Inelastic Collisions 71 3.5 Centre of Mass 79 STPM Practice 3 82 QQ3 85 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 4 WORK, ENERGY AND POWER 88 4.1 Work 89 4.2 Energy 92 4.3 Power 96 STPM Practice 4 99 QQ4 101 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 5 CIRCULAR MOTION 104 5.1 Angular Displacement and Angular Velocity 105 5.2 Centripetal Acceleration 107 5.3 Centripetal Force 108 STPM Practice 5 120 QQ5 122 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 6 GRAVITATION 125 6.1 Newton’s Law of Universal Gravitation 126 6.2 Gravitational Field 127 6.3 Gravitational Potential 133 6.4 Satellite Motion in a Circular Orbit 138 6.5 Escape Velocity 148 STPM Practice 6 152 QQ6 155 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 7 STATICS 158 7.1 Centre of Gravity 159 7.2 Equilibrium of Particles 162 7.3 Equilibrium of Rigid Bodies 172 STPM Practice 7 179 QQ7 182 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 8 DEFORMATION OF SOLIDS 184 8.1 Stress and Strain 185 8.2 Force–extension Graph and Stress-strain Graph 187 8.3 Strain Energy 198 STPM Practice 8 202 QQ8 205 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 9 KINETIC THEORY OF GASES 208 9.1 Ideal Gas Equation 209 9.2 Pressure of a Gas 216 9.3 Molecular Kinetic Energy 219 9.4 The r.m.s Speed of Gas Molecules 223 9.5 Degree of Freedom and Law of Equipartition of Energy 226 9.6 Internal Energy of an Ideal Gas 229 STPM Practice 9 232 QQ9 234 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 10 THERMODYNAMICS OF GASES 237 10.1 Heat Capacities 238 10.2 Work Done by a Gas 240 10.3 First Law of Thermodynamics 244 10.4 Isothermal and Adiabatic Changes 255 STPM Practice 10 270 QQ10 274 Chapter • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • 11 HEAT TRANSFER 278 11.1 Conduction 279 11.2 Convection 304 11.3 Radiation 305 11.4 Global Warming 309 STPM Practice 11 311 QQ11 316 STPM Model Paper (960/1) 320 Summary of Key Quantities and Units 329 CONTENTS vi

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 1 1 CHAPTER Bilingual Keywords Concept Map PHYSICAL QUANTITES 1 AND UNITS Base Quantities and SI Units – Base units – Derived units Scalars and Vectors – Sum, scalar product and vector product of coplanar vectors – Resolve a vector Dimensions of Physical Quantities – Determine dimensions of derived quantities – Check dimensional homogeneity – Construct equations in physics Uncertainties in Measurements – Uncertainty in derived quantity – Significant figures Physical Quantities and Units 1 Base quantity: Kuantiti asas Derived quantity: Kuantiti terbitan Dimension: Dimensi Error: Ralat Fractional uncertainty: Pecahan ketakpastian Physical quantity: Kuantiti fizik Random error: Ralat rawak Resolving vector: Meleraikan vektor Resultant vector: Vektor paduan Scalar product: Hasil darab skalar Scalar: Skalar Significant figures: Angka bererti Systematic error: Ralat bersistem Uncertainty: Ketakpastian Vector: Vektor

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 2 1 1.1 Base Quantities and SI Units Students should be able to: • list base quantities and their SI units: mass (kg), length (m), time(s), current (A), temperature (K) and quantity of matter (mol) • deduce units for derived quantities Learning Outcomes Physical Quantities 1. In Physics, there is the need to make careful observations, precise and accurate measurements. 2. Physical quantities are quantities that can be measured. 3. Examples of physical quantities are length, mass, time, weight, electric current, force, velocity and energy. 4. To describe a physical quantity, two things must be specified. The first is a numerical value and the second is the unit. For example, a man’s height-1.70 m has the numerical value 1.70 and the unit is metre (m). SI Units 1. The unit of a physical quantity is the standard used to compare different magnitudes of the same physical quantity. For example, if the length of a table is 2 m, it means that the length of the table is twice the length of the standard metre (m). 2. Other units of length include foot and mile. 3. Although the foot can be converted into metres and vice versa, the SI (Le Systeme Internationale) is used to facilitate trade and communication between countries of the world. 4. The SI was established in 1960. 5. In SI, each physical quantity has only one unit. Prefixes are attached to SI units, as shown in Table 1.1. Table 1.1 Common SI prefixes Prefix Factor Symbol pico 10–12 p nano 10–9 n micro 10–6 μ milli 10–3 m centi 10–2 c deci 10–1 d kilo 103 k mega 106 M giga 109 G tera 1012 T 2008/P1/Q1, 2013/P1/Q1 Info Physics The systeme International d’Unites (International System of Units, SI) was adopted in 1960 at the 11th General Conference of Weights and Measures.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 3 1 Base Quantities and Base Units 1. In SI, seven physical quantities are chosen as base quantities with the base units, as shown in Table 1.2. Table 1.2 Base quantities and units Quantity SI Unit Symbol Length metre m Mass kilogram kg Time second s Electric current ampere A Thermodynamic temperature kelvin K Amount of substance mole mol Light intensity candela cd Derived Quantities and Derived Units 1. Physical quantities other than the base quantities are known as derived quantities. 2. A derived quantity is a combination of different base quantities. The unit for a derived quantity is known as derived unit. 3. The unit for a derived quantity is obtained by using the relation between the derived quantity and base quantities. Table 1.3 shows the units for some derived quantities. Table 1.3 Derived quantities and units Derived quantity Derived unit Area m2 Volume m3 Frequency Hz (hertz) Density kg m–3 Velocity m s–1 Acceleration m s–2 Force N (newton) or kg m s–2 Pressure Pa (pascal) or N m–2 Energy or work J (joule) or N m Power W (watt) or J s–1 Electric charge C (coulomb) or A s Electric potential V (volt) or J C–1 Electric intensity V m–1 Electric resistance Ω (ohm) or V A–1 Capacitance F (farad) or C V–1 Heat capacity J K–1 Specific heat capacity J kg–1 K–1 Info Physics The metre (1 m) is the length of the path travelled by light in vacuum during a time interval of 1/299 792 458 second.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 4 1 1.2 Dimensions of Physical Quantities Students should be able to: • use dimensional analysis to determine the dimensions of derived quantities • check the homogeneity of equations using dimensional analysis • construct empirical equations using dimensional analysis Learning Outcomes Dimensions of Derived Quantities 1. The dimension of a derived quantity is the relation between the physical quantities and the base physical quantities, such as mass (M), length (L), time (T), electric current (A),temperature (θ), and amount of matter (N). 2. The dimension of a physical quantity is represented by [physical quantity]. Example (a) Dimension of area = [Area] = [Length × Breadth] = L × L = L2 Unit of area = m2 (b) Dimension of velocity = [Velocity] = Displacement [ ] Time = L T = L T –1 Unit of velocity = m s–1 (c) Dimension of acceleration = [Acceleration] = Change of velocity [ ] Time = L T–1 T = L T –2 Unit of acceleration = m s–2 (d) Dimension of force = [Force] = [Mass × Acceleration] = M L T –2 Unit of force = kg m s–2 or N (e) Dimension of energy = [Energy] = [Force × Displacement] = M L T –2 L = M L2 T –2 Unit of energy = kg m2 s–2 or J (f) Dimension of electric charge = [Charge] = [Current × Time] = A T Unit of electric charge = A s or C 2010/P1/Q1, 2014/P1/Q1, 2017/P1/Q1

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 5 1 (g) Dimension of frequency = [Frequency] = 1 [ Period] = 1 T = T –1 Unit of frequency = s–1 or Hz (h) Dimension of strain = [Strain] = Extension [ Original length ] = L L which is dimensionless Uses of Dimensions Dimensional homogeneity of a physical equation 1. A physical equation is true in all systems of unit. Two physical quantities can only be equated, added or subtracted if they have the same dimension. 2. For a correct physical equation, the dimensions of all terms in the equation are the same. 3. The dimension on the left side of the equation and the dimension on the right side of the equation are the same. The equation is said to be dimensionally consistent or homogeneous. 4. For example, the equation of motion with uniform acceleration a, the initial velocity u, final velocity v and displacement s are related by the equation v2 = u2 + 2 as. [v2 ] = (L T –1)2 = L2 T–2 [u2 ] = (L T –1)2 = L2 T–2 [2 as] = L T –2 L = L2 T –2 Hence, every term in the equation has the same dimension, L2 T–2. The equation is dimensionally consistent. 5. On the other hand, if a student writes the equation v = u + 2as [v] = L T –1 [u] = L T –1 [2as] = L T –2 . L = L2 T –2 Hence [v] = [u]  [2as] The dimension is not consistent. We can conclude that the equation is incorrect. 6. An equation whose dimension is not homogeneous is incorrect. 7. On the other hand, a physical equation whose dimensions are consistent need not necessary be correct. There are two reasons an equation may be dimensionally consistent but incorrect. (a) The value of the constant of proportionality is wrong. For example, the correct equation that relates u, v, a and s in linear motion is v2 = u2 + 2as. If the relation has been wrongly written as v 2 = 2u2 + as where [v2 ] = [2u2 ] = [as] = L2 T–2. Hence, the dimension is consistent but the equation is incorrect. (b) The equation is incomplete or has extra terms. For example, the above relation between u, v, a and s when written as v 2 = u2 is obviously incomplete, but dimensionally consistent. On the other hand, the equation v2 = u2 + 2as + s2 t 2 is also incorrect because it has an extra term, s2 t 2 with the same dimension as the other terms.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 6 1 8. Experiments need to be carried out to confirm that a physical equation is correct. This is illustrated in Example 3. Example 1 The power P required to overcome external resistances when a vehicle is travelling at a speed v is given by the expression P = av + bv 2 where a and b are constants. Derive the dimensions for the constants a and b. Then deduce the units for a and b in terms of the base SI units. Solution: Unit of power, P is W = J s–1 = (N m) s–1 = (kg m s–2) m s–1 = kg m2 s–3 [P] = M L2 T–3 Dimension of av, [av] = [P] [a] = [P] [v] = M L2 T–3 L T–1 = M L T–2 Unit for a = kg m s–2 Example 2 The dependence of the heat capacity C of a solid on the temperature T is given by the equation C = αT + βT 3 What are the units of α and β in terms of the base units? Solution: The unit for the heat capacity C is J K–1 J K–1 = (N m) K–1 = (kg m s–2) m K–1 = kg m2 s–2 K–1 Unit for αT = Unit for C Unit for α = Unit for C Unit for T = kg m2 s–2 K–1 K = kg m2 s–2 K–2 Dimension of bv2 , [bv2 ] = [P] [b] = [P] [v 2 ] = M L2 T–3 L2 T–2 = M T–1 Unit for b = kg s–1 Unit for βT3 = Unit for C Unit for β = kg m2 s–2 K–1 K3 = kg m2 s–2 K–4

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 7 1 Example 3 The period of vibration t of a tuning fork depends on the density ρ, Young modulus E and length l of the tuning fork. Which of the following equations may be used to relate t with the quantities mentioned? (a) t = Aρ E gl3 (b) t = Al ρ E (c) t = AE ρ l g where A is a dimensionless constant and g is the acceleration due to gravity. The table below shows the data obtained from various tuning forks made of steel and are geometrically identical. Frequency/Hz 256 288 320 384 480 Length l/cm 12.0 10.6 9.6 8.0 6.4 Use the data above to confirm the choice of the right equation. Hence, determine the value of the constant A. For steel: density ρ = 8 500 kg m–3, E = 2.0 × 1011 N m–2 Solution: The unit for Young modulus E is N m–2 = (kg m s–2) m–2 = kg m–1 s–2 Hence dimension of [E] = M L–1 T –2 Dimension of t = [t] = T (a) Dimension of Aρ E gl3 = [ Aρ E gl3 ] = M L–3 M L–1T–2 (L T–2 L3 ) —1 2 = T = [t] Hence, the equation t = Aρ E gl3 is dimensionally consistent. (b) Dimension of [Al ρ E ] = L ( M L–3 M L–1 T–2 ) —1 2 = T = [t] Hence, the equation, t = Al ρ E is also dimensionally consistent. (c) Dimension of [ AE ρ l g ] = M L–1 T–2 M L–3 ( L L T–2 ) —1 2 = L2 T–1 ≠ [t] Hence, equation t = AE ρ l g is incorrect.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 8 1 Frequency f/Hz 256 288 320 384 480 Period t/s 3.91 × 10–3 3.47 × 10–3 3.13 × 10–3 2.60 × 10–3 2.08 × 10–3 Length l/cm 12.0 10.6 9.6 8.0 6.4 l —3 2 /cm—3 2 41.6 34.5 29.7 22.6 16.2 4.0 Graph (1) t against l 3.5 3.0 2.5 2.0 1.5 1.0 0.5 246 8 10 12 14 16 5 10 15 20 25 30 35 40 45 50 (1) (2) 0 18 20 l / cm 0 l / cm 2 3 – 2 3 – Graph (2) t against l 2 3 – t 10–3 + / s The graph of t against l is a straight line passing through the origin. Hence, the equation t = Al ρ E is correct. Although, the graph of t against l —3 2 is a straight line, it does not pass through the origin. Hence, the equation t = Aρ A gl3 is incorrect. From the correct equation t = Al ρ E = (A ρ E )l The gradient of the graph of t against l is (A ρ E ) 3.91 × 10–3 12 × 10–2 s m–1 = A 8 500 2 × 1011 A = 1.58 × 102

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 9 1. (a) Derive the dimensions of the following: 1 (i) force, F (ii) work, W (iii) power, P . (b) Hence express the following units in terms of the base SI units: (i) newton, N, (ii) joule, J and (iii) watt, W. 2. When a sphere of radius r moves with a velocity v under streamline conditions, it experiences a retarding viscous drag F given by F = krv where k is a constant. What is the dimension of k? 3. At low temperatures, the specific heat capacity of a certain metal is given by c = α T 3 , where T is the thermodynamic temperature and α is a constant characteristic of the solid. What is the unit for α? 4. The viscous drag F between two liquid layers of surface area of contact A in the region of velocity gradient dv dx is given by F = ηAdv dx where η is the coefficient of viscosity of the liquid. Find the dimension of η and write down its unit in SI base units. 5. Bernoulli equation which relates the pressure p with the velocity of flow v of liquid is p + hρg + 1 2 ρv 2 = k where h is height, ρ is density, g is acceleration due to gravity and k is a constant. Show that the equation is dimensionally consistent and give the SI unit for k. 6. The drag coefficient CD of a car travelling with a speed v through air of density ρ is given by CD = F 1 2 ρv 2 A ||||||||||||| where F is the drag force and A is the maximum cross sectional area of the car perpendicular to the direction of motion. (a) Show that CD is dimensionless. (b) When a certain car is travelling at 15 m s–1, it experiences the drag force of 300 N. What is the drag force when the car travels at 45 m s–1? 7. The following equations are suggested to describe the variation of pressure p with velocity v of a liquid flowing in a horizontal pipe. (a) p + Aρgv = X (b) p + Bρv 2 = Y (c) p + Cγgv–2 = Z Where A, B and C are dimensionless constants; X, Y and Z are constants with the same dimension as pressure. g = acceleration due to gravity ρ = density of liquid γ = surface tension of liquid, dimension of γ is M T –2 Identify the equations which are dimensionally consistent. Derivation of Physical Equations 2011/P1/Q1 1. From experimental observations, a physical quantity may be found to be dependent on a number of other physical quantities. 2. For example, the period T of a simple pendulum is found to be dependent on its length l and the acceleration due to gravity g. Using the dimensional method an equation that relates T to l and g can be derived. Suppose that period, T ∝ l x gy where x and y are dimensionless constants. Hence, T = kl x gy where k is the constant of proportionality Since the dimension of an equation must be consistent, [T] = [kl x gy ] T = Lx (L T–2)y = Lx+y T–2y Quick Check 1

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 10 1 Equating indices of T: 1 = –2y y = – 1 2 Hence T = kl —1 2 g – —1 2 = k l g The value of k, the constant of proportionality can be determined experimentally. Exam Tips T = kl x gy Identify x and y as dimensionless constants and k as constant of proportionality Example 4 The velocity v of waves in a liquid depends on the wavelength λ, surface tension γ and density ρ of the liquid. Derive an equation to relate v with λ, γ and ρ. In an experiment, the following values for the velocities and wavelengths are obtained. v/m s–1 2.30 2.70 3.40 4.50 5.90 λ/cm 8.50 6.20 3.90 2.20 1.30 The values of the density of the liquid and its surface tension are: ρ = 1 000 kg m–3, γ = 4.30 N m–1 Draw a suitable graph to deduce the value of the constant of proportionality in your equation. Solution: Velocity of waves, v ∝ λx γ y ρz where x, y and z are dimensionless constants. v = kλx γ y ρz k = constant of proportionality [γ] = [M L T–2 L–1 = M T2 ] Equating the dimension of both sides of the equation, [v] = [kλx γy ρz ] L T –1 = Lx (M T –2)y (M L–3)z = Lx–3z My+z T–2y Equating indices of T : –1 = –2y y = 1 2 Equating indices of M : 0 = y + z z = –y = – 1 2 Equating indices of L : 1 = x – 3z x = 1 + 3z = 1 – 3 2 = – 1 2 Hence, v = kλ– —1 2 γ—1 2 ρ– —1 2 = k γ lρ Equating indices of L: 0 = x + y x = –y = 1 2

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 11 1 To find the value of k, v = (k γ ρ ) 1 l which is in the form of the equation of a straight line y = mx Hence, a graph of v against 1 l should be plotted. The table below is first prepared. v / m s–1 2.30 2.70 3.40 4.50 5.90 1 l / cm– —1 2 0.34 0.40 0.51 0.67 0.88 cm 6.00 v / m s–1 Graph v against 1 5.50 5.00 4.50 4.00 3.50 3.00 2.50 2.00 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1 Gradient of the graph = k γ ρ k = (Gradient) 1 000 4.30 = 1 000 kg m–3 4.30 kg s–2 3.30 m s–1 0.5(10–2 m)– —1 2 = 10.1

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 12 1 Example 5 The rate of flow of a liquid through a horizontal tube under streamline condition depends on (i) a, radius of the tube (ii) η , viscosity of the liquid (iii) p l , the pressure gradient along the tube where p = pressure difference across the length of the tube l = length of the tube Derive an expression for the rate of the flow of a liquid through a horizontal tube in terms of a, η, p and l. Solution: The rate to flow, dV dt ∝ ax η y ( p l ) z dV dt = kax η y ( p l ) z k = constant of proportionality [ dV dt ] = [kax η y ( p l ) z ] [ dV dt ] = Lx (M L–1 T –1)y ( M L T–2 L2 . 1 L ) z L3 T–1 = My+z Lx–y–2z T–y–2z Equating indices of M : 0 = y + z .............................. ① Equating indices of T : –1 = –y –2z .............................. ② Equating indices of L : 3 = x – y – 2z .............................. ③ ③ – ② 4 = x ① + ② –1 = –z z = 1 From ① y = –z = –1 Hence, dV dt = ka4 η –1 ( p l ) 1 = k a4 p ηl Quick Check 2 1. The frequency f of the note produced by a stretched string depends on its length l, the tension T of the string and the mass per unit length m of the string. Use the method of dimension to derive an equation for f. 2. The viscous drag F on a sphere moving in a liquid depends on a, radius of the sphere, v its velocity and η the viscosity of the liquid. Use the method of dimension to derive an equation for F in terms of a, v and η.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 13 1 3. The velocity v of sound in a medium depends on the wavelength λ, Young modulus E and density ρ of the medium. Use the method of dimension to derive an equation for v in terms of λ, E, and ρ. 4. (a) How do you check the dimensional consistency of a formula? Why can’t this method give definite confirmation that an equation is correct? (b) Express the unit of force and charge in terms of the SI units. Hence, by reference to Coulomb’s law F = 1 4πε0 Q1 Q2 r2 where F is force, Q1 and Q2 are charges and r is distance between charges, express the unit of ε0 the permittivity of vacuum in terms of the SI base units. A unit for μ0 the permeability of vacuum, is kg m s–2A–2. Use the unit for µ0 and your unit for ε0 to decide which one of the following relation between ε0 , μ0 and c the speed of light in a vacuum is dimensionally consistent. ε0 μ0 = c2 ; ε0 μ0 = c; ε0 μ0 = c –1; ε0 μ0 = c –2 5. (a) Explain the meaning of the terms (i) base unit (ii) dimension of a physical quantity (b) How do you check a formula for dimensional consistency? State two ways in which a dimensionally consistent equation may be physically incorrect. (c) The current density J in a thermionic diode with parallel plane electrodes separated by a distance d depends on the potential difference V between the electrodes. Use the dimensions of the relevant quantities to find whether the equations J = Aμ0 –1 ( e me ) – —1 2 V —1 2 d–2 .................... |(i) J = Bε0 ( e me ) —1 2 V —3 2 d–2 .................... (ii) are dimensionally consistent. In these equations, μ0 and ε0 are the permeability and permittivity of vacuum and e me is the specific charge of the electron, A and B are dimensionless constants. The values of J and V obtained in an experiment using a diode where d = 2.0 mm are shown below. V/ V 40 60 80 120 160 J/A m–2 148 271 417 767 1180 Plot a graph of lg (J/A m–2) against lg (V/V) and use the graph to decide which of the equations (i) or (ii) is consistent with the obtained results. Find the value of the dimensionless constant in the correct equation. 1.3 Scalars and Vectors Students should be able to: • determine the sum, the scalar product and vector product of coplanar vectors • resolve a vector to two perpendicular components Learning Outcomes Scalar Quantities and Vector Quantities 1. A scalar quantity is a physical quantity which has only magnitude. For example, mass, length, time, speed, temperature, density and energy. 2. A vector quantity is a physical quantity which has magnitude and direction. For example, displacement, velocity, acceleration, force, momentum and electric field intensity. 2008/P1/Q2, 2009/P1/Q3, 2015/P1/Q1, 2016/P1/Q1

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 14 1 A vector quantity can be represented by a line segment where the length represents the magnitude and the direction is shown by an arrow. The length 2.0 cm represents the magnitude 20 m s–1. 3. Two vectors a and b are equal if (i) magnitude of a = magnitude of b (ii) direction of a = direction of b a = b v1  v2 although magnitude of v1 = magnitude of v2 because directions of v1 and v2 are different. Figure 1.2 Sum of Vectors 1. The sum of two vectors is another vector which is known as the resultant, (P + Q). 2. There are two methods to find the resultant vector. Method 1: Parallelogram of vectors The two vectors P and Q are represented by the adjacent sides AB and AD of parallelogram. The line DC is drawn parallel to AB and BC is drawn parallel to AD to complete the parallelogram ABCD. The resultant vector (P + Q) is represented by the diagonal AC in magnitude and direction. Exam Tips In the parallelogram method, the vectors P and Q together with the resultant (P + Q) should start from the same point. Method 2: Triangle of vectors P Q P Q C A B P + Q Figure 1.4 In the Figure 1.4, the line AB is drawn to represent the vector P in magnitude and direction. From B, the line BC is drawn to represent the vector Q. The resultant (P + Q) is represented by line AC in magnitude and direction. Exam Tips In the triangle method, direction of P and Q should follow one after the other. Direction of resultant (P + Q) is against the directions of P and Q in the triangle. velocity v = 20 m s–1 Figure 1.1 P Q D Q P P + Q C A B Figure 1.3

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 15 1 Polygon of Vectors P Q R Figure 1.5 1. To find the sum of three or more vectors, a polygon is drawn as shown in Figure 1.5. 2. The three vectors P, Q and R are represented in order by the three sides of a polygon. The resultant (P + Q + R) is represented by the line that completes the polygon as shown in Figure 1.5. Subtraction of Vectors The methods for the sum of two vectors discussed above can be used to find the subtraction of two vectors. The subtraction of two vectors P and Q can be written as the sum of two vectors. P – Q = P + (–Q) P Q P –Q P – Q P –Q P – Q Figure 1.6 Alternative Method B O A Q P P – Q O B A Q P Q – P B A (a) (b) Figure 1.7 1. In Figure 1.7, the lines OA and OB are drawn to represent the vectors P and Q respectively. 2. The side BA of the triangle represents (P – Q) both in magnitude and direction. The side AB represents (Q – P). Exam Tips • P and Q should start from the same point O. • Direction of (P – Q) points to the tip of the first vector P. VIDEO Resultant Vector

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 16 1 1. A scout starts from point O and walks 100 m east to point A. From A he walks another 100 m in the north-west direction to the point B. Draw a vector diagram to find the displacement of the scout from O when he is at B. 2. For each of the following figures, state the vector R in terms of the vectors P and Q. (a) Q P R (b) Q P R (c) Q P R (d) Q P R 3. Two vectors X and Y act at right angles. Draw labelled diagrams to show (a) (X + Y) (b) (X – Y) Give their magnitudes and state their directions with respect to the direction of vector X. 4. The maximum and minimum possible magnitudes of the sum of two forces are 12 N and 4 N respectively. What are the magnitudes of the two forces? Scalar Product of Vectors θ Q P Figure 1.8 1. The scalar product or dot product of two vectors P and Q is written as P.Q (read as P dot Q) is a scalar given by P.Q = PQ cos q where P is the magnitude of vector P, Q is the magnitude of vector Q and q is the angle between P and Q. 2. P.Q = PQ cos q = 0, when q = 90o ,that is when P and Q are at right angles. Maximum value of P.Q = 1, when q = 0, that is when P and Q are parallel. s F θ Figure 1.9 Quick Check 3

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 17 1 3. An example of scalar product of two vector is the work done W by a force F when the displacement of the point of application of the force F is s. Work done, W = F.s = Fs cos θ Vector Product of Vectors P P fi Q Q q Right hand Fingers curve from P to Q Thumb points in the direction of P fi Q (a) (b) Figure 1.10 1. The vector product or cross product of two vectors P and Q is written as P × Q (read as P cross Q) is a vector (Figure 1.10(a)) which has a magnitude given by P × Q = PQ sin θ where θ is the angle between P and Q. The direction of the vector P × Q is perpendicular to the plane that contains P and Q such that P, Q and P × Q form a right-hand system as shown in Figure 1.10(b). 2. Magnitude of P × Q is zero when θ = 0. That is when the vectors P and Q are parallel. Magnitude of P × Q is maximum when θ = 90o . That is when the vectors P and Q are at right-angles. Example 6 The force F on a charge q moving with a velocity v in a magnetic field B is given by the vector product F = q(v × B) In the figures below, a particle of charge 2.0 C enters a magnetic field B of 0.25 T with a velocity v of 350 m s-1. Show the direction and magnitude of the force F on the charge. (a) B v 30° (b) B v 30° (c) B v 90° Solution: Magnitude of the force, F = q(vB sin θ) In (a) and (b) F = (2.0)(350)(0.25) sin 30o = 87.5 N In (c) F = (2.0)(350)(0.25) sin 90o = 175 N

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 18 1 Directions of the force F are as shown in the figures below. (a) B F = q(v fi B) v 30° (b) B F = q(v fi B) v 30° (c) B v 90° F = q(v fi B) Quick Check 4 1. In each of the following figures, the vectors F and s are as shown. (a) Find the scalar product F.s. (b) Find the vector product F × s. Show clearly the direction of F × s. (i) F = 5 N s = 3 m (ii) F = 5 N s = 3 m (iii) F = 5 N 30° s = 3 m (iv) F = 5 N 120° s = 3 m Resolving a Vector 1. A vector R may be considered as the sum of two vectors. There are an infinite number of pairs of vectors whose resultant is R. 2. Figure 1.11 shows three possible combinations. The vector R is said to have two components. A B R C D R E F R (a) (b) (c) Figure 1.11 Components of the vector

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 19 1 3. Figure 1.12 shows a vector R with two mutually perpendicular components X and Y. θ R X = R cos θ Y = R sin θ R2 = X2 + Y2 R = X2 + Y2 tan θ = Y X Figure 1.12 Quick Check 5 1. Find the magnitudes of the components for the following vectors along the Ox and Oy directions. (a) (b) (c) (d) 30° y O x F = 10 N O 60° y O x s = 20 m 55° y O x a = 4.0 m s–2 1.4 Uncertainties in Measurements Students should be able to: • calculate the uncertainty in a derived quantity (a rigorous statistical treatment is not required) • write a derived quantity to an appropriate number of significant figures Learning Outcomes 1. Experiments in physics usually require us to make measurements. 2. Measurements of physical quantities consist of a number of readings. A reading is a specific value of the quantity being measured. 3. A certain measurement is only accurate up to a certain degree which is determined by the instrument used and physical constraints of the observer. 4. Any measurement of a physical quantity has errors or uncertainty. 5. When the length of a wire is measured by a metre rule and its length is recorded as 51.0 cm. It means that the measurement of the length is only accurate up to 1 mm or 0.1 cm. The uncertainty or error in the measurement can be shown by writing the length of the wire as (51.0 ± 0.1) cm.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 20 1 6. Similarly using a micrometer screw gauge the diameter of the wire is measured as 0.48 mm, we can record the diameter as (0.48 ± 0.01) mm to show the error. The value 0.01 mm represents the absolute error for the measurement of 0.48 mm. 7. The fractional error is represented by 0.01 0.48 = 0.021. Whereas the percentage error is 0.01 0.48 × 100% = 2.1% 8. A measurement is more accurate if its fractional error or percentage error is smaller. Types of Errors 2009/P1/Q1 1. Systematic Errors (a) Systematic errors are errors in the measurement of a physical quantity due to the instruments, the effects of surrounding conditions and physical constraints of the observer. (b) The main characteristic of systematic error is that its magnitude is almost constant. The value of the measurement is always greater, or is always less than the actual value. 2. Sources of systematic errors A Zero or end errors (a) Zero error occurs when the instrument gives a non-zero reading when in fact the actual reading is zero. Figure 1.13 Zero error = +0.01 mm (b) Figure 1.13 shows a micrometer screw gauge with its gap closed, but its reading is 0.01 mm, not zero. (c) The value of this error is constant, i.e. 0.01 mm and is positive. Obtained measurements using this micrometer would always be greater by 0.01 mm compared to the correct value. (d) The reading of an ammeter shows 0.2 A when the ammeter is not connected to any circuit. There is no current through the ammeter, the actual reading is zero. Hence, the zero error is 0.2 A. Readings obtained using this ammeter would always be bigger by 0.2 A. B Personal error of the observer (a) Physical constraints or limitations of the observer can cause systematic errors. An example is the reaction time. (b) The reaction time of an observer can be considered constant during an experiment. C Errors due to instruments (a) A stopwatch which is faster than normal would give readings which are always larger than the actual time. (b) Moving coil ammeter, which is used under different conditions from which it was calibrated. An ammeter manufactured in Japan had been calibrated under different temperature and earth’s magnetic field from Malaysia where the ammeter is used. D Errors due to wrong assumption We assumed that the value of the acceleration due to gravity g is 9.81 m s–2, but the actual value may be 9.79 m s–2. Hence, there is a positive error of 0.02 m s–2.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 21 1 3. Systematic errors cannot be reduced or eliminated by taking repeated readings using the same method, the same instrument and by the same observer. 4. Systematic errors can be eliminated or reduced by improving the procedure of taking the measurements, using a different instrument or getting somebody else to make the measurements. 5. Example of a good experimental procedure is to note the zero readings of instruments, such as micrometer screw gauge, stopwatches and ammeter before using the instruments. The measurement is then corrected by subtracting the zero reading from the obtained readings. 6. Random errors (a) The main source of random error is the observer. (b) The characteristics of random errors are: • it can be positive or negative. The obtained readings may be greater or less than the actual value. • its magnitude is not constant. (c) Examples of random errors. Parallax errors Figure 1.14 Parallax error (i) Figure 1.14 shows that position (2) is the correct position of the eye when reading a scale, such as a ruler. The line of vision from the eye should be perpendicular to the scale. (ii) When the eye is at position (1), the reading would be smaller than the actual value. (iii) When the eye in position (3), the reading would be larger than the actual value. (iv) Parallax errors occur when the eye is at position (1) or (3). (d) When a micrometer screw gauge is used to measure the diameter of a wire, random errors occur when different pressures are applied when closing the gap of the micrometer screw gauge. (e) Changes in the temperature during an experiment can result in the measurements being sometimes bigger or smaller than the actual value. (f) Reading the scale of the measuring instrument wrongly. This often occurs when using a dualscale electric meter. (g) Mistake in counting the oscillation of a simple pendulum. (h) To eliminate or reduce random errors as described in (f) and (g) above, repeated readings are taken. The wrong reading would be quite different from the rest. Uncertainties in Measurements 1. A measurement has an uncertainty that is determined by the precision or smallest division on the scale of the measuring instrument. 2. The uncertainty in a single reading from a measuring instrument is taken as half of the smallest division on the scale, if the half division can be easily read. When half of the smallest division on the scale cannot be read easily, the uncertainty equals the smallest division on the scale. Examples are readings of the vernier callipers and micrometer screw gauge.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 22 1 3. Table 1.4 shows the acceptable uncertainty of some common measuring instruments. Tabe 1.4 Instrument Smallest scale division Uncertainty Metre rule 0.1 cm  0.05 cm Vernier callipers 0.05 mm  0.05 mm Micrometer screw gauge 0.01 mm  0.01 mm Analog stopwatch 0.1 s  0.1 s Thermometer 1o C  0.5 oC Burette 0.1 cm3  0.05 cm3 Measuring cylinder 1 cm3  0.5 cm3 Voltmeter 0.2 V  0.1 V Ammeter 0.1 A  0.05 A 4. A single reading from a metre rule used to measure the length l of a rod is written as l  ∆l, where ∆l is the absolute uncertainty. Example Length of rod, l = (55.00  0.05) cm • The absolute uncertainty, ∆l = 0.05 cm • The fractional uncertainty in l is ∆l l = 0.05 55.00 = 9.1 × 10–4 • The percentage uncertainty in l is ∆l l × 100% = 0.05 55.00 × 100% = 0.091% 5. Uncertainty in a derived quantity is calculated as follows. • Absolute uncertainty for addition or subtraction of two readings equals the sum of the absolute uncertainty. Example if the two readings are (x  ∆x) and (y  ∆y) (i) A = x + y ∆A = ∆x + ∆y (ii) B = x – y ∆B = ∆x + ∆y • The derived quantity is expressed as the product and/or division of various readings. Example if A = 7 x2 y 1 2 z3 Fractional uncertainty, ∆A A = 2 ∆x x  + 1 2  ∆y y  + 3 ∆z z  Percentage uncertainty, ∆A A × 100% = 2 ∆x x  + 1 2  ∆y y  + 3 ∆z z  × 100% Significant Figure 1. The accuracy of a measurement is indicated by the number of significant figures. 2. When adding or subtracting readings, the number of decimal places of the answer should equal to the reading with the least decimal places. 3. If the answer is derived by multiplication, division, or trigonometric function, the number of significant figure in the answer equals to that of the data with the least number of significant figures. 4. Absolute uncertainty is calculated to only one significant figure.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 23 1 Example 7 A pipe has an external diameter D of (1.50  0.05) cm and also an internal diameter d =(1.10  0.05)cm. What is the thickness x of the wall of the pipe? State the uncertainty in the value of x. Solution: External diameter, D = d + 2x Thickness, x = 1 2 (D – d) = 1 2 (1.50 – 1.10) cm = 0.20 cm Uncertainty in (D – d), ∆(D – d) = ∆D + ∆d =  (0.05 + 0.05) cm =  0.10 cm x = 1 2 (D – d) ∆x = 1 2 ∆(D – d) = 1 2 (0.10) cm = 0.05 cm Thickness, x = (0.20  0.05) cm Example 8 The following readings were obtained by a student to determine the density of a metal cylinder. Mass of cylinder, M = (102.5  0.1) g Length of cylinder, L = (36.1  0.1) mm Diameter of cylinder, d = (20.0  0.1) mm (a) What is the percentage uncertainty in the value of the density? (b) Calculate the density of the metal and the uncertainty in its value. Solution: (a) Density of cylinder, ρ = mass volume = M  pd2 4 L Percentage uncertainty in ρ, ∆ρ ρ × 100% =  ∆M M + 2 ∆d d + ∆L L 100% =  0.1 102.5 + 2 0.1 20.0+ 0.1 36.1100% = 1.37% D d x (b) Density, ρ = M  pd2 4 L = 102.4 g  p 4(2.00)2 (3.61) cm3 = 9.04 g cm–3 ∆ρ ρ = 1.37 100 Uncertainty in ρ, ∆ρ =  1.37 100 (9.04) g cm–3 = 0.1 g cm–3 Density of metal ρ = (9.0  0.1) g cm–3 Note: ∆ρ is rounded up to 1 significant figure (0.1 g cm–3) ρ is rounded up to 0.1 g cm–3.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 24 1 Quick Check 6 1. Which of the following statements about errors is correct? A Zero error is random error. B Systematic errors can be due to instruments which are not sensitive C Random errors can be reduced by taking repeated readings. D Systematic errors cause the readings to be scattered on both sides of the actual value. 2. A student uses an ammeter to measure the current in a circuit. From the series of readings obtained, the results are as below: Minimum current = 3.4 A Mean current = 3.7 A Maximum current = 3.9 A The actual value of the current is 4.1 A. What in the most likely reason for the difference between the actual value of the current and the mean current? A The ammeter has zero error. B There are parallax errors. C The number of readings taken is not sufficient. D The ammeter is not sensitive. 3. Three characteristics of errors are listed below. Which corresponds to that of random errors? Error can Error is of Error will possibly be constant be reduced eliminated sign and by averaging magnitude repeated measurements A No No Yes B No Yes Yes C Yes No No D Yes Yes No 4. Which of the following experiment techniques can reduce systematic error of the quantity being measured? A Measuring the diameter of a wire at different points along the wire. B Adjusting an ammeter to read zero before measuring a current. C Timing a large number of oscillations to find the period of a pendulum. D Measuring the thickness of a large number of pieces of paper to find the thickness of one piece. 5. The length of a book is (268 ± 1) mm and its width is (194 ± 1) mm (a) What is the absolute error of the length? (b) What is the percentage error of the length? (c) What is the area of one page of the book? Give your answer to the correct number of significant number. 6. (a) What is meant by systematic error and random error? (b) Describe two ways of minimising each. 7. (a) Give two differences between systematic error and random error. (b) The period of a simple pendulum is determined by measuring the time for 20 oscillations using a stopwatch. Give a source of systematic error and another for random error in the determination of the period. (c) Suggest how each of the errors may be reduced or eliminated. 8. Two lengths x and y which are measured in an experiment are x = (10.0  0.1) cm and y = (5.0  0.1) cm Calculate the uncertainty in (a) (x + y) (b) (x – y) (c) xy (d) x y (e) 4x3 y 9. The period T of a simple pendulum of length l is given by the equation T = 2π l g where g is the acceleration due to gravity. A student recorded the time of 20 oscillations for a pendulum of length (50.0 ± 0.1) cm. Time for 20 oscillations ( t ± 0.1) s: 28.5 s, 28.6 s, 28.3 s

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 25 1 (a) Calculate to the correct number of significant figures, (i) the average time for 20 oscillations, (ii) the period T of the simple pendulum. (b) Determine a value of g, with its uncertainty. Important Formulae 1. Resolving a vector X = R cos θ Y = R sin θ R = X2 + Y2 tan θ = Y X STPM PRACTICE 1 R Y X 1. The volume flow rate R of a fluid flows through a pipe of cross sectional area A and length l when the pressure difference between the two ends of the pipe, p is given by R = kA2 p ηl where c is a dimensionless constant, and η is the viscosity of the fluid. The unit of η in terms of base SI units is A kg m C kg m–1 s–1 B kg s–2 D kg m–3 s–1 2. The figure shows two dogs P and Q pulling a sledge S along the straight line Ox on a horizontal surface covered with snow. The forces on the sledge by the dogs are in the directions shown. O x θ 20° 4000 N S 3600 N P Q What is the value of θ? A 16.0° C 67.7° B 22.3° D 73.9° 3. Which of the following is an example of random error? A Zero error of an ammeter B Error in the calibration of a thermometer C Error in assuming g, acceleration of free fall is 10 m s–2 D Error in recording the time shown on a stop-watch. 4. Three coplanar forces of 3.0 N, 5.0 N and 6.0 N acts on a point as shown in the figure below. O 3.0 N y x 6.0 N 5.0 N 30° What is the magnitude and direction of the resultant force? A 12.2 N at 48o below the x-axis B 12.2 N at 42o above the x-axis C 9.3 N at 76o below the x-axis D 9.3 N at 14o above the x-axis 5. What is the dimension for electric potential? A ML2 A–1T–3 C ML2 AT–2 B ML2 A–1T–2 D ML2 A2 T–2 2. P.Q = PQ cos q 3. |P × Q| = PQ sin q

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 26 1 6. The displacement s of a particle moving with a uniform acceleration a in time t is given by s = Camt n where C is a dimensionless constant. What is the value of m and n? m n A 1 1 B 1 2 C 2 1 D 2 0 7. A quantity D = 1 3 ρc 2 where ρ is density and c is speed. What is the dimension of the quantity D? A MLT–2 C ML–2T–1 B ML–1 T–2 D ML–2T–1 8. Three coplanar forces 155 N, 200 N and 300 N acting at a point O is shown in the diagram below. x y 200 N 300 N 155 N 53° 45° 30° O What are the net forces of the x and y components? x y A –132 N 188 N B –132 N 436 N C 188 N –132 N D 436 N 188 N 9. The speed v of a progressive wave in a stretched string of length l and mass m depends on l, m and the tension T in the string. If k is a dimensionaless constant, which dimension of equation is correct? A v = k Tm l C v = kl T m B v = k Tl m D v = km T l 10. A car moves with a velocity of 20 m s–1 towards the north, and then at 15 m s–1 eastward. What is the change in the velocity of the car? A 5.0 m s–1 in the direction 000o B 5.0 m s–1 in the direction 180o C 25 m s–1 in the direction 143o D 25 m s–1 in the direction 053o 11. The vectors X, Y and Z are related by the equations X = Y – Z. Which vector diagram correctly shows the relationship? A Z Y X C Z Y X B Z Y X D Z Y X 12. The following readings for the diameter of a bar were obtained in an experiment: 2.49 cm, 2.52 cm, 2.48 cm, 2.51 cm, 2.49 cm Which is the best way to express the diameter of the bar? A 2.5 cm B 2.495 cm C (2.495 ± 0.005) cm D (2.50 ± 0.01) cm 13. The dimensions of a rectangular piece of card are length, l = (8.0 ± 0.1) cm, breadth, b = (4.0 ± 0.1) cm The uncertainty in the perimeter of the card is A ± 0.1 cm C ± 0.4 cm B ± 0.2 cm D ± 1.0 cm 14. The air resistance F on a sphere falling through air is given by F = constant × cross sectional area × (velocity)2 The uncertainties in the values of the cross sectional area and velocity are ±2% and ±3% respectively. What is the uncertainty in the measurement of F? A ± 4% C ± 8% B ± 5% D ± 4% 15. Physical quantity X is related to the mass m, distance s, time t and acceleration due to gravity g by the equation X = mgst2 4π Deduce the dimension of X.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 27 1 16. The velocity v of a particle when its displacement x is given by the equation v2 = A + Bx2 . Determine the dimensions of the constant A and B. 17. The velocity of sound in a medium of density ρ is given by the equation v = k ρ where k is a constant. Deduce the dimension of k and write its unit in terms of SI base units. 18. The air resistance F on a car travelling with a velocity v in air of density ρ is given by F = kv x ρy Az where k is a dimensionless constant, and A the effective cross-sectional area of the car. Determine the values of x, y and z. 19. (a) The speed v of waves in the sea is related to its wavelength λ and g the acceleration due to gravity by the equation v = k gλ where k is a dimensionless constant. Show that the equation is dimensionally homogeneous. (b) The speed of a wave of wavelength 100 m is 9.0 m s–1. Calculate (i) the wave frequency, (ii) the value of k. (assume g = 10 m s–2) (c) The speed of another wave is 12.4 m s–1. Find its wavelength and frequency. 20. (a) Explain why the unit of work is a derived unit. (b) The density ρ and the pressure p of a gas is related by the expression c2 = γp ρ where c and γ are constants. Given that the constant γ has no unit, deduce the unit of c. From your answer above, suggest what quantity may be represented by c. 21. Discuss briefly whether in general the method of dimensional checking can positively confirm that an equation is correct. The flow of a liquid of density ρ, around an obstacle of width b changes from streamline to turbulence when the critical speed v is reached. If η is the viscosity of the liquid, which of the equations below relates v to ρ, b and η? (a) v = Aηb ρ (b) v = Bη bρ (c) V = Cρb η where A, B and C are dimensionless constants. 22. (a) Explain what is meant by the dimension of a physical quantity. (b) In an experiment to determine the viscosity η of a liquid of density ρ, the time t taken for the liquid level in an apparatus to fall a fixed distance is measured. The viscosity η is given by the equation η = Aρt – Bρ t where A and B are constants to be determined (i) The usual unit of η is N s m–2. Show that this unit is equivalent to kg m–1 s–1. (ii) Find the unit for A and B. (c) The results obtained in an experiment are as shown in the table below. θ /°C 10 20 30 50 80 η/N s m–2 1.30 × 10–3 1.00 × 10–3 8.0 × 10–4 5.5 × 10–4 3.6 × 10–4 ρ/kg m–3 1000 998 996 988 972 t/s 121 95 78 57 42 (i) Plot a graph ηt ρ against t 2 and use the graph to determine the values of the constants A and B. (ii) In an experiment, methanol of density 790 kg m–3 in the apparatus takes 73 s for its meniscus to drop through the fixed distance. Calculate the viscosity of methanol. 23. (a) What is (i) systematic error and (ii) random error? (b) In an experiment to determine the resistance R of a wire the ammeter and voltmeter readings were recorded by a student as: ammeter reading, I = (1.5 ± 0.1) A, voltmeter reading, V = (2.6 ± 0.2) V Calculate the resistance of the wire together with its uncertainty. (c) The length of the wire is recorded as l = (78.4 ± 0.1) cm. The diameter of the wire as measured by the student is given below. Diameter, (d ± 0.01) mm = 0.54 mm, 0.55 mm, 0.66 mm, 0.54 mm (i) There is an error in one of the readings for the diameter. Identify the wrong reading. Name the type of error and hence give a reason for taking repeat readings.

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 28 1 ANSWERS 1 1. (a) (i) [F] = [ma] = M L T–2 (ii) [W] = [Fs] = M L T–2 (L) = M L2 T–2 (iii) [P] = [ W t ] = M L2 T–3 (b) (i) N = kg m s–2 (ii) J = kg m2 s–2 (iii) W = kg m2 s–3 2. [k] = F rv = LMT–2 L(LT–1) = ML-1T-1 3. Unit for c is J kg–1 K–1 = (kg m2 s–2) kg–1 K–1 = m2 s–2 K–1 Unit for α = unit for c unit for T3 = m2 s–2K–1 K3 4. [η] = [ F A dv dx ] = MLT–2 (L2 )(LT–3 L ) = M L–1 T–1 , kg m–1 s–1 5. p + hρg + 1 2 ρv2 = k p = F A’ [p] = MLT–2 L2 = ML–1T–2 [hρg] = L(ML–3)(LT–2) = ML–1T–2 = [p] [ρv2 ] = (ML–3)(LT–1)2 = ML–1T–2 = [p] Hence dimension of equation is consistent [k] = [p] = ML–1T–2 Pa or kg m s–2 6. (a) [CD] = [ F 1 2 ρv2 ] = MLT–2 (ML–3)(LT–1)2 = MLT–2 (ML–2) which is dimensionless (b) 2700 N 7. Equation (a): [p] = ML-1T-2 (Refer answer 5) [Aρgv] = (ML-3)(LT-2)(LT-1) = ML-1T-3 [p] Hence dimension is not consistent. Equation (b): [Bρv2 ] = (ML-3)(LT-1)2 = ML-1T-2 = [p] Hence dimension is consistent. Equation (c): [Cρgv -2] = (MT-2)(LT-2)(LT-1)-2 = ML-1T-2 = [p] Hence dimension is consistent. Equation (b) and (c) dimensionally consistent. 2 1. f = klx Ty mz (m = M L ) [f] = [ 1 T ] = T-1 T-1 = Lx (MLT-2) y (ML-1) z = My + z Lx + y-zT-2y Equate indices of T : –1 = –2y y = 1 2 Equate indices of M: 0 = y + z z = – y = – 1 2 Equate indices of L: 0 = x + y – z x = z – y = – 1 2 – 1 2 = –1 Hence, f = k l-1 T 1/2 m-1/2 = k l T m 2. [η]=ML-1T-1(Refer no. 4 of Quick Check 1) [F] =[kax vy ηz ] MLT-2 = Lx (LT-1)y (ML-1T-1 )z = Mz Lx+y-zT-y-z (ii) The resistivity of the material ρ of a wire in terms of its resistance R, length l and cross sectional area A is given by the expression ρ = RA l . Discard the wrong reading of the diameter, and find the mean diameter of the wire. Use the value of the mean diameter to calculate a value for the resistivity, and state the uncertainty. (Resistance, R = V l )

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 29 Indices of M: 1 = z Indices of T: –2 = –y – z y = –1 + 2 = 1 Indices of L: 1 = x + y – z x = 1 - 1 + 1 = 1 F = kaηv 3. Unit for E =Pa = unit for pressure, p [E] = [p] = ML-1T-2 (Refer no.5 of Quick Check 1) [v] = [kEx ρy ] LT-1 = (ML-1T-2)x (ML-3)y = Mx+yL-x-3yT-2x Indices of T: –1 = –2x x = 1 2 Indices of M: 0 = x + y y = – x = – 1 2 Hence, v = k E ρ 4. (a) (Refer to page 5) (b) Charge, Q = current × time [Q] = AT Unit for charge is A s. Unit of force = N = Kg m s-2. F = 1 4πe0 Q1 Q2 r2 [e0 ] = [ 1 4πF Q1 Q2 r2 ] = (AT)(AT) (MLT–2)(L)2 = M-1L-3T4 A2 T4 Unit for e0 : kg–1m–3 A2 s4 Unit for e0 µ0 = (kg-1m-3 A2 s4 )(kgm A-2s-2) = m-2 s2 = (m s-1) -2 = (unit of c)-2 Hence e0 µ0 = c-2 5. (a) Refer to page 3 & 4 (b) Refer to page 5 (c) From answer above, [e0 ] = M-1L-3T4 A2 T4 [µ0 ] = MLA-2T-2 [J] = [ current area ] = AL-2 Volt, V = J C-1 = (kg m2 s–2)(As)–1 = kg m2 A–1 s–3 [V] = ML2 A–1T–3 [A µ0 –1( e me ) –1/2 V1/2 d–2 ] = (MLA–2T–2) –1( AT M ) –1/2(ML2 A–1T-3)1/2 L–2 = M–1 + 1 2 + 1 2 L –1 + 1 – 2A2 – 1 2 – 1 2T2 – 1 2 – 3 2 = L-2 A = [J] Dimension of equation (1) is consistent [B( e me ) 1/2 V3/2 d -2] = (M-1L-3T4 A2 T4 ) ( AT M ) 1/2(ML2 A-1T-3)3/2 L-2 = M–1 – 1 2 + 3 2 L –3 + 3 – 2A2 + 1 2 – 3 2T4 + 1 2 – 9 2 = L–2A = [J] Dimensions in both equations are consistent. 3 1. 67.5° 45° O A B s = s1 + s2 s1 = 100 m s2 = 100 m s = 76.5 m in the direction E 67.5°N 2. (a) R = P + Q (c) R = P – Q (b) R = Q – P (d) R = – (P + Q) 3. (a) (b) |X + Y| = X2 + Y2 |X – Y| = X2 + Y2 tan θ = Y X tan α = Y X θ = tan–1 ( Y X ) α = tan –1 ( Y X ) 4. 8 N and 4 N 4 1. (a) (i) F.s = Fs cos θ = (5)(3) cos 0o J = 15 J (ii) F.s = (5)(3) cos 90o J = 0 (iii) F.s = (5)(3) cos 30o J = 13 J (iv) F.s = (5)(3) cos 120o J = - 7.5 J (b) (i) |F × s| = Fs sin θ = (5)(3) sin 0o J = 0 (ii) |F × s| = (5)(3) sin 90o N m = 15 N m (iii) |F × s| = (5)(3) sin 30o N m = 7.5 N m (iv) |F × s| = (5)(3) sin 120o N m = 13 N m Direction of F × s is as shown in the figure below. (ii) F = 5 N F fi s s = 3 m (iii) F = 5 N 30° F fi s s = 3 m

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 30 1 (iv) F = 5 N 30° F fi s s = 3 m 5 1. (a) 8.66 N, 5.00 N (c) 10.0 m, 17.3 m (b) 3.21 m s–1, 3.83 m s–1 (d) 2.29 m s–2, 3.28 m s–2 6 1. C 2. A 3. A 4. B 5. (a) ± 1 mm (b) 0.373% (c) 5.20 × 104 mm2 6. (a) Systematic errors are errors in the measurement of a physical quantity due to the instruments, the effects of surrounding conditions and physical constraints of the observer. Random errors are errors in the measurement of a physical quantity due to observer. (b) Systematic error: (i) Improve the procedure of taking the readings (ii) Use different instruments to make the measurements Random error: (i) Take repeated readings (ii) Avoid parallax error 7. (a) Systematic error : same magnitude, same sign Random error : Different magnitudes, both positive and negative (b) Zero error of stopwatch – systematic error Wrong count of number of oscillation; mistake in noting down reading of stopwatch – random errors. (c) Systematic error : Subtract zero error from reading Random error : Repeat measurement of the time of oscillations 8. (a) ∆(x + y) = ∆x + ∆y = ± 0.2 cm (b) ∆(x – y) = ∆x + ∆y = ± 0.2 cm (c) ∆(xy) xy = ∆x x + ∆y y = 0.1 10.0 + 0.1 5.0 ∆(xy) =  0.1 10.0 + 0.1 5.0 (10.0 × 5.0) cm2 = 1.5 cm2 = 2 cm2 (to 1 significant figure) (d) ∆(x/y) x/y = ∆x x + ∆y y = 0.1 10.0 + 0.1 5.0 ∆ x y  =  0.1 10.0 + 0.1 5.0 10.0 5.0  = 0.06 (e) Let Z = 4x3 y  ∆Z Z  = 3 ∆x x  + 1 2  ∆y y  ∆Z = [3 ∆x x  + 1 2  ∆y y ]Z = [3 0.1 10.0 + 1 2  0.1 5.0 ] × 4(10.0)3 cm3 5.0 cm = 70 cm5/2 9. (a) (i) Average time t = 28.5 + 28.6 + 28.3 3 s = 28.5 s (To the same decimal place as data) (ii) Period T = t 20 = 28.5 20 s = 1.43 s (to the same number of significant figure as 28.5) (b) T = 2π l g g = 4π2 l T2 = 4π2 l (t/20)2 = 4π2 (0.5000) (1.43)2 m s–2 = 9.65 m s–2 ∆g =  ∆l l + 2 ∆t t g =  0.1 50.0 + 2 0.1 28.5(9.65) m s–2 = 0.09 m s–2 g = (9.65 ± 0.09) m s–2 STPM Practice 1 1. C : η = k A2 p RI Unit: (m2 )2 (kg m s–2/m2 ) (m3 s–1)(m) = kg m–1 s–1 2. B : 3600 sin θ = 4000 sin 20° θ = 22.3° 3. D : The rest are systematic errors. 4. A : x-component, Fx = 5.0 + 6.0 sin 30o N = 9.0 N y-component, Fy = -3.0 + 6.0 cos 30o N = 2.2 N

Physics Term 1 STPM Chapter 1 Physical Quantities and Units 31 1 Resultant force, F = 9.02 + 2.22 N = 9.3 N tan–1  2.2 9.0 2 = 14o above x-axis 5. A : Unit of potential difference is V = J C-1 = (N m)(A s)–1 = (kg m s–2 m)A–1 s–1 = kg m2 A–1 s–3 Dimension: ML2 A–1 T–3 6. B : Equate units on both sides of the equation: m = (m s–2)m sn = mm s–2m+n Hence m = 1, –2m + n = 0, n = 2m = 2 7. B: [ 1 3 ρc2 ] = (ML–3)(LT–1)2 = ML–1T–2 8. A: Fx = 200 cos 30o – 300 cos 45o – 155 cos 53o = –132 N Fy = 200 sin 30o + 300 sin 45o – 155 sin 53o = 188 N 9. B : v = Tx l y mz LT–1 = (MLT–2)x Ly Mz T: –1 = –2x gives x = 1 2 L: 1 = x + y, y = 1 – x = 1 2 M: 0 = x + z, z = –x = – 1 2 10. C v2 = 15 m s–1 v1 = 20 m s–1 v2 – v1 = 25 m s–1 θ Magnitude of v2 – v1 = 202 + 152 m s–1 = 25 m s–1 q = tan–1 15 20 = 37° Direction of v2 – v1 is (180° – 037°) = 0143°. 11. B: X = Y – Z gives Y = Z + X 12. D : Average value to the same decimal place as data. 13. C : Perimeter, p = (l + l +b + b) ∆p = (∆l + ∆l + ∆b + ∆b) 14. C : ∆F F × 100% = [∆A A + 2∆v v ] × 100% 15. M L2 16. [A] = L2 T–2 [B] = T–2 17. [k] = M L–1 T–2, kg m–1 s–2 18. x = 2, y = 1, z = 1 19. (a) [k] =  v gl  =  LT–1 (LT–2L)1/2  = LT–1 LT–1 (b) (i) 0.090 Hz (ii) k = 0.285 (c) λ = 189 m, f = 0.0656 Hz 20. Speed Unit of c: m s–1 21. Equation (b) 22. (a) The dimension of a physical quantity is the relation of the physical quantity to the basic physical quantities. (b) Unit of A: m2 s–2 Unit of B: m2 (c) (i) A = 1.1 × 10–8 m2 s–2 B = 4.0 × 10–6 m2 (ii) 5.9 × 10–4 N s m–2 23. (a) Refer to page 20 and 21 (b) R = V I = 2.6 1.5 Ω = 1.73 Ω ∆R R = ∆V V + ∆I I ∆R = ( 0.2 2.6 + 0.1 1.5 )(1.73) Ω = 0.2 Ω Hence R = (1.7 ± 0.2) Ω (c) (i) Wrong reading: 0.66 mm Random error Repeat readings to avoid wrong reading of the scale of the measuring instrument. (ii) ρ = RA l = (1.7)π (0.54 × 10–3)2 4(0.784) Ω m = 4.97 × 10–7 Ω m ∆ρ ρ = ∆R R + ∆l l + 2 ∆d d ∆ρ = ± 0.2 1.7 + 0.1 78.4 + 20.01 0.54(4.97 × 10–7)Ω m = ±0.8 × 10–7 Ω m = (5.0 ± 0.8) × 10–7 Ω m

CHAPTER Concept Map KINEMATICS 2 Bilingual Keywords Acceleration: Pecutan Displacement: Sesaran Kinematics: Kinematik Projectile: Luncuran Resistance: Rintangan Speed: Laju Velocity: Halaju Motion with Constant Acceleration • v = u + at • s = 1 2 (u + v)t • s = ut + 1 2 at2 • v2 = u2 + 2as Graphical Method 0 t s 0 t v Projectile u cos u sin u Effect of Air Resistance 0 t v 0 t a Linear Motion Kinematics 32

33 2 Physics Term 1 STPM Chapter 2 Kinematics 2.1 Linear Motion Students should be able to: • derive and use equations of motion with constant acceleration • sketch and use the graphs of displacement-time, velocity-time and acceleration-time for the motion of a body with constant acceleration Learning Outcomes 1. Kinematics is the study of motion without considering the causes of the motion. Linear motion is motion along a straight line. 2. The displacement s of a body from a point O to another point P is the distance moved by the body along the straight line OP. Displacement s has both magnitude and direction, hence it is a vector. 3. The velocity v of a body is its rate of change of displacement. Velocity, v = ds dt Velocity v is also a vector, its direction is along the direction of the change in displacement ds. (Figure 2.1). 4. The expression v = ds dt gives the instantaneous velocity. The instantaneous velocity of a car travelling along a straight road is shown by the speedometer. 5. If the time taken by a body to move through a displacement s is t, then average velocity = Final displacement, s Time taken, t Figure 2.2 The speedometer shows the instantaneous velocity 6. Acceleration a is the rate of change of velocity. Acceleration, a = dv dt Acceleration is also a vector. Its direction follows the direction of the change in velocity dv. When the velocity of a body changes either in magnitude, or in direction, or both, the body accelerates. Motion With Constant Acceleration 1. The acceleration of a body is uniform if the magnitude of the acceleration is constant and its direction remains unchanged. 2010/P1/Q3, 2016/P1/Q2,Q18 O ds P v Figure 2.1

34 2 Physics Term 1 STPM Chapter 2 Kinematics 2. For a body travelling along a straight line, if u = initial velocity, v = final velocity after a time t, and s = displacement in time t, then uniform acceleration, a = Change in velocity Time taken = v – u t v = u + at .............................. a Displacement, s = Average velocity × Time s = ( u + v 2 ) t .............................. b s = ( u + u + at 2 )t s = ut + 1 2 at 2 .............................. c From ①, v – u = at From ➁, v + u = 2s t (v – u)(v + u) = at × 2s t v 2 – u2 = 2as .............................. d ➃ Example 1 A motorist travelling at 72 km h–1 on a straight road approaches a traffic light which turns red when he is 55.0 m away from the stop line. The reaction time is 0.7 s. With the brakes applied fully, the vehicle decelerates at 5.0 m s–2. How far from the stop line will he stop and on which side of the stop line? Solution: 72 km h–1 = 72 000 m 3 600 s = 20 m s–1 The vehicle travels at constant speed of 20 m s–1 during the reaction time of 0.7 s. Distance travelled during the reaction time = 20 × 0.70 = 14.0 m When the vehicle stops, final velocity v = 0 Acceleration = –5.0 m s–2 (negative because of deceleration) Using v 2 = u2 + 2as Distance travelled when the brakes are applied, s = v2 – u2 2a = 0 – 202 2 (–5) = 40.0 m Total distance travelled = (14.0 + 40.0) m = 54.0 m Hence, the vehicle stops (55.0 – 54.0) m = 1.0 m before the stop line. Info Physics A380 Airbus take-off speed A wide-body aircraft such as the Airbus A380 has a takeoff speed of 280 km h–1 and requires a runway of length 3000 m. What then is its acceleration? Exam Tips There are five quantities, u, v, t, s, a, in the four equations for motion under constant acceleration. Each of the equations involves only four of the quantities. In a question, usually values for three of the quantities are provided, and you are asked to calculate the value of another of the quantities. Use the equation that contains the four quantities mentioned in the question.

35 2 Physics Term 1 STPM Chapter 2 Kinematics Example 2 The table below is from the handbook of a car. On a dry road, the car driven by an alert driver will stop in distances as shown below. Speed/ m s–1 Thinking distance/m Braking distance/m Overall stopping distance/m 5.0 3.0 1.9 4.9 10.0 6.0 7.5 13.5 15.0 9.0 17.0 26.0 20.0 12.0 30.0 42.0 25.0 15.0 47.0 62.0 30.0 18.0 68.0 86.0 35.0 21.0 92.0 113.00 The thinking distance is the distance travelled by the car during the driver’s reaction time. The braking distance is the distance travelled by the car before the car is stopped when the brakes are applied. (a) Explain why the thinking distance is directly proportional to the speed, whereas the braking distance is not. State the relationship between the braking distance and the speed. (b) What is the value of deceleration used in calculating the braking distance? (c) Calculate the overall stopping distance for a car travelling at 40 m s–1. Solution: (a) During the reaction time, t of the driver, the car travels at a constant speed, u. Hence, thinking distance, s1 = ut s1 ∝ u The reaction time t of a driver is constant, The final speed after the brakes are applied = 0 If deceleration = a, using v2 = u2 + 2as 0 = u2 – 2as2 Braking distance, s2 = u2 2a ∝ u2 (b) Using braking distance, s2 = u2 2a When u = 10 m s–1, s2 = 7.5 m Deceleration, a = u2 2s = 102 2 × 7.5 = 6.67 m s–2 (c) When u = 40 m s–1 Thinking distance, s1 = 40 × Reaction time = 40 × (3.0 5.0) = 24 m Overall distance travelled = (24 + 120) m = 144 m Braking distance, s2 = u2 2a = 402 2 × 6.67 = 120 m

36 2 Physics Term 1 STPM Chapter 2 Kinematics Quick Check 1 1. A car starts from rest and accelerates at 2.0 m s–2. (a) Find its speed at (i) 1.0 s, (ii) 2.0 s and (iii) 5.0 s. (b) What is the average speed in (i) the first second, (ii) 2.0 s and (iii) 5.0 s? (c) Calculate the distance travelled in (i) 1.0 s, (ii) 2.0 s and (iii) 5.0 s. 2. A car decelerates uniformly from 30 m s–1 to 15 m s–1 in a distance of 75 m. Calculate the further distance travelled before the car comes to a stop. 3. A car accelerates through three gear changes with the following speed: 20 m s–1 for 2.0 s 40 m s–1 for 2.0 s 60 m s–1 for 6.0 s What is the overall average speed of the car? Motion Under Gravity 2009/P1/Q2, 2011/P2/Q1, 2017/P1/Q2 1. When a body is released from rest, it falls towards the earth with an acceleration. 2. In free space where there is no air resistance, all objects irrespective of their mass, fall with the same acceleration, known as the acceleration of free fall or acceleration due to gravity, g. 3. Close to the earth, the value of g can be as assumed constant. In this book, the value of g is assumed to be 9.81 m s–2 unless otherwise stated. 4. Since the value of g is constant, equations of motion under uniform acceleration such as: v = u + at, s = ut + 1 2 at2 and v2 = u2 + 2as can be used for motion under gravity by substituting the acceleration a as +g, or as –g where applicable. 5. When the body is only falling downwards, it may be easier to use a = +g, since the direction of motion is in the direction of g, that is downwards. 6. When the body is projected vertically and upwards, if it is assumed that the velocity of the body is positive, it implies that the upwards direction is assumed positive. Since the direction of g is downwards, hence acceleration a = –g. 7. When the body moves in both directions, upwards then falls downwards or vice versa, it would be better to assume the upwards direction as positive. Hence, the acceleration a = –g. Upwards journey Displacement : + Velocity : + Acceleration : a = –g Downwards journey Displacement : + Velocity : – Acceleration : a = –g Below the level AI Displacement : – Velocity : – Acceleration : a = –g G H I J E D C B A F E Figure 2.3 INFO Free Fall

37 2 Physics Term 1 STPM Chapter 2 Kinematics 8. Figure 2.3 shows a particle being projected upwards from the point A. It reaches the highest point E and then falls down to a point below A. Example 3 A bullet is fired vertically upwards from the ground with an initial velocity of u and takes a time of t1 to reach a point P at a height h. From then the bullet takes a further time of t2 to move to the highest point and back to the ground. Find in terms of t1, t2 and g (a) the initial velocity u, (b) height h, and (c) the greatest height H reached. Solution: (a) Assume the upwards direction as positive, then a = –g. Total time taken for the bullet to move up and back to the ground, t = (t1 + t2) When the bullet returns to the ground, displacement, s = 0 Using s = ut + 1 2 at2 0 = u(t1 + t2) – 1 2 g(t1+ t2) 2 u = 1 2 g(t1 + t2) (b) When s = h, t = t1 Using s = ut + 1 2 at2 h = ut1 – 1 2 gt1 2 u = 1 2 g(t1 + t2) = 1 2 g(t1 + t2)t1 – 1 2 gt1 2 = 1 2 gt1 t2 Quick Check 2 H H h P v t 1 (t 1 + t 2) (c) At the highest point, s = H, velocity = 0 Using v2 = u2 + 2as 0 = u2 – 2gH H = u2 2g = 1 2g [ 1 2 g(t1 + t2) ] 2 = 1 8 g (t1 + t2)2 u = 1 2 g(t1 + t2) 1. A ball is thrown vertically upwards from the top of a building, 9.0 m above the ground with a speed of 8.0 m s–1. The ball hits the ground after a time T. Calculate (a) the time taken to reach the greatest height. (b) the greatest height reached by the ball. (c) the time, T taken by the ball to reach the ground. 2. A spaceship descends at a constant velocity of 10 m s–1 on the moon. When it is 120 m from the moon’s surface, an object falls off the spaceship. If the acceleration due to gravity on the moon is 1.6 m s–2, what is the velocity of the object when it reaches the surface of the moon?

38 2 Physics Term 1 STPM Chapter 2 Kinematics 3. The acceleration of free fall is determined by timing the fall of a ball bearing using photocells. The ball is released from the point P, and passes the point X and Y at time t1 and t2 after being released. Find an expression for the acceleration due to gravity in terms of h, t1 and t2. P Q X Y h Photo-cells Light beam Light beam Graphical Methods The motion of a body can be analysed by studying the various graphs for the motion: (a) Displacement-time graph (b) Velocity-time graph Displacement-time Graph 2010/P1/Q2, 2014/P1/Q2 1. Informations that can be deduced from the displacement-time graph: (a) The instantaneous displacement. The displacement s at any time t can be obtained off the graph. (b) The velocity = Gradient of the graph. (c) The manner that the velocity changes with time from the shape of the graph. 2. The displacement-time (s-t) graphs of a few types of motion are as shown in Figure 2.4. (a) s t 0 (b) s t 0 Constant velocity Increasing velocity Gradient = constant Gradient increasing (c) s t 0 (d) s t T 2T 0 Decreasing velocity Motion of a body projected vertically upwards and (deceleration) then falling back to the ground. When t = T, the body Gradient decreasing is at the highest point. When t = 2T, the body is back on the ground. Figure 2.4 Displacement-time graphs

39 2 Physics Term 1 STPM Chapter 2 Kinematics Exam Tips Note the difference between graph (a) and graph (b) when t = 0. (a) (b) Initial velocity u = 0 Initial velocity u ≠ 0 Gradient at t = 0 is zero. Gradient at t = 0 is not zero. 0 t v 0 t v Velocity-time Graph 1. Informations that can be deduced from the velocity-time graph: (a) The instantaneous velocity. (b) Between the time t = t1 and t = t2, displacement = ∫t1 t2 v dt = shaded area under the graph. (c) The acceleration, a = dv dt = Gradient of graph. (d) How the acceleration changes with time from the shape of the graph. 2. Figure 2.5 shows the velocity-time (v-t) graphs for various types of motion. (a) t 1 u t 2 0 t v (b) t 1 u t 2 0 t v (c) t 1 t 2 0 t v Uniform acceleration. Uniform deceleration. Increasing acceleration. Gradient = constant Constant negative gradient Gradient increasing (d) t 1 t 2 0 t v (e) t 1 t 2 0 t v Decreasing acceleration. Constant velocity Gradient decreasing Figure 2.5 Velocity-time graphs

40 2 Physics Term 1 STPM Chapter 2 Kinematics Example 4 The graph shows the speeds of two cars A and B which are travelling in the same direction over a period of 80 s. Car A travelling at a constant speed of 20 m s–1 overtakes car B at time t = 0. In order to catch up with car A, car B immediately accelerate uniformly for 30 s to reach a constant speed of 30 m s–1. (a) How far does car A travel during the first 30 s? (b) Calculate the acceleration of car B in the first 30 s. (c) What is the distance travelled by car B in this time? (d) What additional time will it take for car B to catch up with car A after car A passes car B? (e) How far would each car have travelled since t = 0? (f) What is the maximum distance between the cars before car B catches up with car A? Solution: (a) Distance travelled by car A during the first 30 s = 30 × 20 = 600 m (b) Acceleration of car B = Gradient of graph = 30 – 5 30 = 0.833 m s–2 (c) Distance travelled by car B during the first 30 s = Area under the graph from t = 0 to t = 30 s = 1 2 (5 + 30) × 30 = 525 m (d) After 30 s, car A is ahead of car B by (600 – 525) m = 75 m For car B to catch up with car A, Further distance travelled by car B – Further distance travelled by car A = 75 m 30t – 20t = 75 t = 7.5 s Hence, car B catches up with car A (30 + 7.5) s from when car A passes car B, i.e. 37.5 s. (e) Distance travelled by each car = 20 × 37.5 m = 750 m (f) The two cars are furthest apart when the two graphs intercept at P. At P, the difference (Area under graph A – Area under graph B) is maximum. Refering to the figure: x y = 10 15 x + y = 30 x = 2 3 y 2 3 y + y = 30 y = 3 5 × 30 = 18 s Hence, maximum separation = Shaded area = 1 2 × 18 × 15 = 135 m B A Time / s Speed / m s–1 0 30 30 20 5 80 B A Time / s Speed / m s–1 0 30 30 20 15 5 80 y P x

41 2 Physics Term 1 STPM Chapter 2 Kinematics Example 5 (a) A body accelerates uniformly from rest along a straight line. Sketch a graph to show the variation of displacement with time. How can the instantaneous velocity be deduced from the displacement-time graph? (b) A cricket player throws a ball vertically upwards and catches it 3.0 s later. Neglecting air resistance, calculate (i) speed of the ball when it leaves the player’s hand, (ii) the maximum height reached by the ball. (c) Sketch a graph to show how the velocity of the ball varies with time. Mark each of the following instants on the graph: (i) The ball leaves the player’s hand (t1). (ii) The ball at the maximum height (t2). (iii) The ball returns to the player’s hand again (t3). (d) You were told that the air resistance is negligible. In actual fact the ball experiences a retarding force during its motion. Without making any calculation, explain how the air resistance affects (i) the time taken to reach the highest point, (ii) the value of the maximum height reached. (e) Discuss, taking into account the air resistance, whether the time tu taken by the ball to reach the highest point is greater or smaller than td, the time taken for the ball to drop. (Assume g = 10 m s–2). Solution: (a) Instantaneous velocity = Gradient of the graph at time, t 0 t s Note that the gradient of the graph = 0 when t = 0 because initial velocity = 0 (b) (i) Suppose u = velocity of ball when it leaves the player’s hand. Acceleration a = –g = –10 m s–2 Time t = 3.0 s Displacement s = 0 Using s = ut + 1 2 at2 0 = 3.0 u – 1 2 × 10 × (3.0)2 u = 15 m s–1 (ii) At the maximum height H, velocity v = 0 Using v2 = u2 + 2as 0 = 152 – 2 × 10 × H H = 11.25 m

42 2 Physics Term 1 STPM Chapter 2 Kinematics (c) Velocity, v Time, t 0 + – t 1 t 2 t 3 (d) (i) Neglecting air resistance, the time taken to reach the highest point is given by v = u + at v = 0, a = –g 0 = u – gt Time taken, t = u g With air resistance, the retardation a1 > g using v = u + at 0 = u – a1 t1 Time taken, t1 = u a1 < u g (a1 > g) Hence, the time taken is shorter. (ii) Without air resistance, the greatest height reached = H At the greatest height, v = 0 Using v2 = u2 + 2as 0 = u2 – 2gH a = –g Greatest height, H = u2 2g With air resistance retardation a1 > g. Hence, the greatest height, H1 = u2 a1 < u2 g (a1 > g) Therefore, the greatest height reached is lower. (e) When the ball moves up, initial velocity = u, final velocity = 0, time = tu Using s = 1 2 (u + v)t H1 = 1 2 (u + 0)tu tu = 2H1 u Hence, the time taken to come down is greater than the time taken to move up. When the ball falls, initial velocity = 0, final velocity, v < u velocity of projection because of friction time = td using s = 1 2 (u + v)t H1 = 1 2 (0 + v)td td = 2H1 v > tu (v < u)

43 2 Physics Term 1 STPM Chapter 2 Kinematics Quick Check 3 1. A body initially at rest moves with uniform acceleration. Which graph best represents the variation of displacement s of the body with time t? A C s t 0 s t 0 B D s t 0 s 0 t 2. A brick is dislodged from a tall building and falls vertically under gravity. Which of the following curves represents the variation of its height h above the ground with time t if air resistance is negligible? A C h t 0 h t 0 B D h t 0 h t 0 3. The velocity-time (v-t) graph of a vehicle travelling along a straight line is shown below. Which of the following is the displacementtime (s-t) graph of the vehicle? A t 1 t 2 t 3 s 0 t C t 1 t 2 t 3 s 0 t B t 1 t 2 t 3 s 0 t D t 1 t 2 t 3 s 0 t 4. A tennis ball is released, it falls vertically to the floor and bounces back. Taking velocity upwards as positive. Which of the following is the velocity-time (v-t) graph of the ball? A v t + – 0 C v t + – 0 B v t + – 0 D v t + – 0