ii Contents Must Know iii – x Chapter 1 Force and Motion II 1 – 20 NOTES 1 Paper 1 3 Paper 2 11 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 19 Chapter 2 Pressure 21 – 42 NOTES 21 Paper 1 23 Paper 2 33 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 41 Chapter 3 Electricity 43 – 61 NOTES 43 Paper 1 46 Paper 2 52 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 60 Chapter 4 Electromagnetism 62 – 79 NOTES 62 Paper 1 65 Paper 2 72 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 78 Chapter 5 Electronics 80 – 101 NOTES 80 Paper 1 84 Paper 2 92 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 100 Chapter 6 Nuclear Physics 102 – 115 NOTES 102 Paper 1 104 Paper 2 108 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 114 Chapter 7 Quantum Physics 116 – 132 NOTES 116 Paper 1 119 Paper 2 124 Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 131 Answers 133 – 150 Content_1202 Physics F5.indd 2 10/01/2022 12:27 PM
Mnemonics (Chapter 1) 1 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 3) 3 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 5) 5 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 4) 7 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 4) 9 @ Pan Asia Publications Sdn. Bhd. Mnemonics (Chapter 5) 11 @ Pan Asia Publications Sdn. Bhd. Triangle of Forces Triangle of forces: • is a vector diagram whose sides represent three forces in equilibrium. • is a closed circle with forces in clockwise or in anticlockwise direction. a b F3 F2 F1 c (q) Current and Potential Difference Electrons flow from the Negative to the Positive terminal ˜ ENP Current flow from the Positive to the Negative terminal ˜ CPN npn transistor Not Pointing IN pnp transistor Pointing IN Factors Affecting Resistance of Wire Mnemonic Factors Last Length of wire, L Rocket Resistivity of wire, r Cross Cross-section of wire, A Tower Temperature of wire, q Fleming’s Left-hand Rule The direction of the force can be determined by using Fleming’s left-hand rule. Force, F Current, I Magnetic field, B Fleming’s left-hand rule → FBI Fleming’s Right-hand Rule The direction of the induced current in a conductor can be determined by using Fleming’s right-hand rule. F1 sin a = F2 sin b = F3 sin c F2 3 = F2 1 + F2 2 + 2F1 F2 cos q Thumb Movement Current is affected Magnetic field Second finger B C C n P n E Collector Emitter B Base E B C C P n P E Collector Emitter B Base E npn Transistor and pnp Transistor First finger KNOW Mnemonics MUST Must know_1202 Physic F5.indd 1 07/01/2022 11:41 AM
Important Facts (Chapter 4) 20 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 5) 22 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 6) 24 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 1) 14 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 2) 16 @ Pan Asia Publications Sdn. Bhd. Important Facts (Chapter 3) 18 @ Pan Asia Publications Sdn. Bhd. Direction of Force on a Current-carrying Conductor in a Magnetic Field • The direction of the magnetic field of the current-carrying wire can be determined by using right-hand grip rule. When the wire is placed in a magnetic field, a force acts on it, F. N S I F • The direction of the force can be determined by using Fleming’s left-hand rule. • Equation for Alpha Decay (α or)4 2 He Parent nucleus has nucleon number, A and proton number, Z. After a α-decay, the daughter nucleus has a nucleon number, A – 4 and a proton number, Z – 2. A ZX ˜ A–4 Z–2X + 4 2 He + Energy • Equation for Beta Decay (β or) 0 –1e Parent nucleus with nucleon number, A and proton number, Z. After a β-decay, the daughter nucleus has a nucleon number, A (no change) and a proton number, Z + 1. A ZX ˜ A–0 Z+1X + 0 –1e + Energy • Equation for Gamma Decay (γ) There is no changes in nucleon number and proton number. The nucleus is less energetic after gamma decay. A ZX ˜ A ZX + γ + Energy Deflection of Cathode Rays in an Electric Field Filament Cathode + – Anode Maltese cross Shadow P Q If P and Q are not connected to the power supply, cathode rays will move straight and shadow of the maltase cross is at the center position of the screen. + – If P and Q are connected to the power supply, cathode rays is deflected upwards and shadow of the maltase cross is deflected upwards. Zero Resultant Force 1. Forces are balanced when the resultant force is zero. 2. When resultant force is nil, acceleration is nil and the body will be: (a) Stationary or Ffriction F F = Ffriction (b) Moves with uniform velocity Engine oil Steel sphere Fbuoyant Fdrag mg mg = Fbuoyant + Fdrag Buoyant Force Buoyant force = Weight of fluid displaced Air V1 V2 Vobject = V1 + V2 Liquid with density, ρ FB = rV2 g where V2 is the volume of the part of the body submerged in the liquid Air V Liquid FB = rVg where V is the volume of the body Internal Resistance The internal resistance of a dry cell is the obstruction or resistance to the flow of charge by chemicals electrolyte in the dry cell. Rheostat V A Ԑ = IR + Ir that is, V = IR Ԑ = V + Ir KNOW Important Facts MUST Must know_1202 Physic F5.indd 4 07/01/2022 11:41 AM
Resolution of Forces Wrong Correct 30° Vertical component 100 N Fx = 100 cos 30° Fy = 100 sin 30° Fx = 100 sin 30° Fy = 100 cos 30° or find the angle with the horizontal line first: 60° Horizontal line 100 N Fx = 100 sin 30° or 100 cos 60° Fy = 100 cos 30° or 100 sin 60° Total Pressure What is the ratio of the pressure at X to the pressure at Y? Air Water X Y 20 m 30 m Wrong Correct Px Py = 20 m H2 O 30 m H2 O = 2 : 3 Px Py = (10 + 20) m H2 O (10 + 30) m H2 O = 3 : 4 Note: Total pressure = Atmospheric pressure + pressure due to water Magnetic Field Wrong Correct • The electric field lines – incorrectly labelled. • Positively charged particles • Negatively charged particles • The electric field lines are always directed away from positive charges. • The electric field lines are always directed toward negative charges. Common Mistakes (Chapter 1) 37 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 2) 39 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 3) 41 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 4) 43 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 5) 45 @ Pan Asia Publications Sdn. Bhd. Common Mistakes (Chapter 7) 47 @ Pan Asia Publications Sdn. Bhd. Effect of Magnetic Field on Cathode Rays • The diagram below shows a Maltese cross tube. Filament Cathode + – Anode Maltese cross Shadow • What will happen to the velocity of the cathode ray if the potential difference, V between the cathode and the anode is increased? Wrong Correct No change Cathode rays velocity, v = 2eV m , thus the velocity increases as V increases. Current-carrying Conductor • The diagram below shows a current-carrying conductor placed in a magnetic field. I N S • State the direction of the force on the conductor. Wrong Correct Fleming’s right-hand rule is used to determine the direction of force. The current-carrying conductor is not an induced current, so need to use Fleming’s left-hand rule to determine the direction of the force. Working Functions for Photoelectric Effects If the threshold wavelength is 420 nm, calculate the metal working function in units of eV. Wrong Correct Students solve this question by using equation: hc l = 1 2 mv2 + W Students need to simplify the photoelectric equation: W = hfo + – + – KNOW Common Mistakes MUST Must know_1202 Physic F5.indd 7 07/01/2022 11:41 AM
Important Diagrams (Chapter 4) 44 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 5) 46 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 7) 48 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 1) 38 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 2) 40 @ Pan Asia Publications Sdn. Bhd. Important Diagrams (Chapter 3) 42 @ Pan Asia Publications Sdn. Bhd. Pressure • Liquid pressure is due to weight of liquid above. A h W Surface of liquid Container filled with liquid • Atmospheric pressure is due to weight of atmosphere above. 0 kPa 31 kPa 101 kPa Electric Field Electric field pattern between two charged spheres and a charged sphere with a plate: + + – + – – – Calculation of Potential Difference Through Resistors R1 R2 Vin Vout V1 = ( R1 R1 + R2 ) × Vin or V2 = ( R2 R1 + R2 ) × Vin Spring F Q Q P P x Copper Steel F P Q Q P x Short Long Steel spring stiffer Short spring stiffer F P Q Q P x Small Large F Q Q P P x Thin Thick Spring with smaller coil diameter stiffer Spring with thicker wire stiffer Magnetic Field Field ines Right hand Current-carrying conductor Other fingers show the direction of the magnetic field Electron Interference Experiment Electron beam gun Electron Doubleslit Screen Interference pattern KNOW Important Diagrams MUST Must know_1202 Physic F5.indd 8 07/01/2022 11:41 AM
1 Chapter 1 Force and Motion II 1.1 Resultant Force 1. A resultant force is a single force that produces the same effect when it replaces two or more forces acting on the object. F1 m a = 3 m s–2 a = 3 m s–2 = F2 FR m FR = ma 2. Simple rules: (a) Forces in same direction F1 = 3 N F2 = 4 N FR = 7 N = FR = F1 + F2 (b) Forces in opposite direction F1 = 3 N F2 = 4 N FR = 1 N = FR = F1 – F2 (c) Two perpendicular forces F1 F2 F2 FR θ FR = F1 2 + F2 2 (d) Two non-perpendicular forces F1 F2 θ FR 2 = F1 2 + F2 2 + 2F1 F2 cos q 3. By drawing a scaled diagram: A C B O FR θ Use scale = 1 cm : p N (a) Draw OA and OB in the same direction as F1 and F2 respectively. (b) Draw a parallelogram OACB. (c) Measure the length of OC = q cm. FR = p µ q N 4. By vice versa, if F = 0, object is stationary / moving with uniform velocity. If F ≠ 0, object moves with uniform acceleration. Stationary v – u a = –––– t Uniform velocity Uniform acceleration u v 1.2 Resolution of Forces 1. A force can be resolved into two components, usually vertical component, Fy and horizontal component, Fx. F F sin θ F cos θ θ Angle with horizontal line 2. On a smooth plane, where mass of object = m and R = normal reaction force. R = mg cos θ a = g sin θ Mass, m θ NOTES u = v = 0 a = 0 u = v ≠ 0 a = 0 u ≠ v a ≠ 0 F = ma = 0 F = ma ≠ 0 123 Chp 1_1202 QB Physics F5.indd 1 10/01/2022 12:33 PM
2 1.3 Forces in Equilibrium 1. Forces are in equilibrium when the resultant force is equal to zero. 2. Methods of solving numerical problems for three forces in equilibrium. (a) By resolution of forces β α F2 F1 F3 F3 F2 cos β F1 cos α F1 sin α Vertically F1 sin α + F2 sin β = F3 Horizontally F1 cos α = F2 cos β F2 sin β (b) By drawing a scaled triangle of force F3 F2 F1 β α (i) Use suitable scale. (ii) Measure the corresponding sides. (iii) Calculate the required force using the scale used. (c) If the triangle is right-angled, use simple trigonometry. F3 F3 F2 F2 F1 F1 β β β α α Example: F1 = F3 sin α F2 = F3 cos α F3 F3 F2 F2 F1 F1 β β β α α Example: F1 = F3 sin α F2 = F3 cos α 3. Alternative method: (a) Using sine rule F3 F1 F2 a b c sin a –––– = F1 sin b –––– = F2 sin c –––– F3 sin a F1 = sin b F2 = sin c F3 (b) Using cosine rule F3 F3 2 = F1 2 + F2 2 – 2F1 F2 cos θ F1 F2 θ F3 2 = F1 2 + F2 2 – 2F1 F2 cos q 1.4 Elasticity 1. Hooke’s law states that the extension of spring is proportional to the applied force. (a) F ∝ x F = kx where k = the force constant of the spring Gradient of graph = k 1 0.1 0.2 0.3 Extension/ m 0.4 0 2 3 4 Force/ N (b) Gradient of X > Gradient of Y Steeper slope ˜ Greater value of k ˜ Stiffer spring. x/ cm F/ N X Y 2. Area under force-extension graph: Extension Force Area = = Elastic potential energy x F O 1 – Fx 2 Elastic potential energy = Area under the graph EP = 1 2 Fx = 1 2 (kx)(x) = 1 2 kx2 Vertically F1 sin a + F2 sin b = F3 Horizontally F1 cos a = F2 cos b Chp 1_1202 QB Physics F5.indd 2 10/01/2022 12:33 PM
23SOS TIP Answer all questions. Take g = 10 N kg– 1, density of Hg, r = 13 600 kg m–3 and density of air, r = 1 000 kg m–3 unless stated otherwise. PAPER 1 2.1 Pressure in Liquid 1. Diagram 1 shows the water supply system to a house. h X Diagram 1 If the pressure meter reading at X is 6.2 × 105 Pa, what is the value of h? A 6.2 m C 620 m B 62 m D 6 200 m 2. Diagram 2 shows that the speed of water draining out can be increased by tilting the water container. Wooden block Water container Diagram 2 Which factor produces such an effect? A The density of water B Water depth C Gravity D Volume of water 3. Liquid pressure depends on I the depth of the liquid II the density of the liquid III volume of liquid IV the acceleration due to gravity there A I and II C I, II and III B II and IV D I, II and IV CLONE SPM CLONE SPM 4. Diagram 3 shows a tube filled with liquid P. The density of the liquid is 1.2 g cm-3. The angle between the tube and the horizontal line is 50o . Liquid P 50° X Diagram 3 If the pressure at X due to liquid P is 7 200 Pa, what is the length of the column of liquid in the tube? A 78 cm C 100 cm B 90 cm D 110 cm 5. Diagram 4 shows three containers with mercury, water and cooking oil. Liquid pressures at the bases are PX, PY dan PZ respectively. X Water Cooking oil Mercury Y Z Diagram 4 Which comparison is correct? A PX = PY PZ C PX PZ PY B PX PY PZ D PZ PY PX 6. Diagram 5 shows a metal cylinder immersed in a liquid. The upper surface of the cylinder, X is h m from the surface of the liquid. Liquid Cylinder Y surface h Diagram 5 Question 4: P = hrg h = depth from the liquid surface Chp 2_1202 QB Physics F5.indd 23 07/01/2022 12:08 PM
24SOS TIP The pressure on surface X depends on the following except A the density of a liquid B the acceleration due to gravity C value of h D area of X 7. The pressure due to a liquid at the bottom of the container containing the liquid is 3 600 Pa. What is the pressure when the container is on a planet where the gravitational field strength is 5.6 N kg– I? HOTS Applying A 640 Pa C 2 000 Pa B 1 800 Pa D 3 600 Pa 8. Diagram 6 shows two liquids, P and Q with densities of 1 200 kg m–3 and 1 600 kg m–3 respectively. Liquid P X 20 cm Liquid Q Diagram 6 What is the depth in liquid Q where the pressure is equal to the pressure at 20 cm below the surface of liquid P? A 12.0 cm C 18.0 cm B 15.0 cm D 26.7 cm 9. Diagram 7 shows two holes, P and Q (same level) of different sizes on the wall of a tall cylinder. The water from the holes P and Q crosses a horizon distance of x (in the diagram) and x’ respectively. Small hole P Big hole Q x Diagram 7 Which comparison is true? A x < x’ B x = x’ C x > x’ 10. Diagram 8 shows water ejected out at same velocity at points normally perforated to the surface of a can filled with water. Hole at the same position Diagram 8 The following statements are conclusions from observations except A liquid pressure acts normally to the surface B liquid pressure at same level is equal C liquid pressure acts in all directions D liquid pressure depends on shape of container 11. Diagram 9 shows a container with water. Water Diagram 9 Which of the following is a correct vector diagram of water pressure acting on the walls? A C B D Question 7: P = hrg P ∝ g Question 9: Liquid pressure does not depend on area. Chp 2_1202 QB Physics F5.indd 24 07/01/2022 12:08 PM
25SOS TIP 12. Diagram 10 shows a container with water. X Y Z Diagram 10 Which comparison of water pressure at points X, Y and Z is true? A PX PY PZ B PX = PY PZ C PX = PY = PZ D PX PY = PZ 13. Dams are built with the bottom being thicker than the top. What is the reason for doing so? A Water pressure increases with depth B The deeper the water, the colder it is C The density of water is directly proportional to the depth of water D The size of water molecules increases with the depth of water 14. What is the force due to water pressure on a fish with a body surface area of 360 cm2 at a depth of 25 m below the surface of sea water where the density of seawater is 1 025 kg m–3. A 900 N C 9 000 N B 8 600 N D 9 200 N 15. A cylinder with diameter D contains mercury. The pressure due to mercury is P. If the mercury is poured into another cylinder with diameter 2D, what is the new pressure P’ due to mercury? HOTS Applying A P C 1 4P B 1 2P D 1 8P 2.2 Atmospheric Pressure 16. Mercury is used in Fortin barometers because A it is opaque B it is shiny C it does not stick to the glass tube D it has very high density 17. The equipment commonly used to read altitude above sea level is basically a A Aneroid barometer B Fortin barometer C Bourdon gauge D mercury manometer 18. Diagram 11 shows a tube with mercury at sea level. h mm Vacuum Mercury Patm Diagram 11 What is the value of h? A 760 mm B 800 mm C 100 cm D 120 cm 19. Diagram 12 shows two turtles. Small turtle Big turtle Diagram 12 The small turtle is 10 m below the surface of the water. What is the ratio of the total pressure on the large turtle to the small turtle? A 1:1 C 2:1 B 1:2 D 4:1 20. Consider the density of air is 1.29 kg m–3 and does not change with height. What is the difference in pressure between the ground and the top of a hill with height 300 m? A 890 Pa B 1 927 Pa C 3 870 Pa D 5 600 Pa Question 12: The length of the arrow shows the magnitude of the pressure. Question 15: Base area ∝ diameter2 . Question 20: ∆pressure = hr∆h Chp 2_1202 QB Physics F5.indd 25 07/01/2022 12:08 PM
92SOS TIP Section A Answer all questions. 1. Diagram 1 shows the thermal emission and production of cathode ray generation. – + e X Y E.H.T. e Diagram 1 (a) What is meant by the term thermionic emission? [1 mark] (b) State the name of X and Y. [2 marks] (c) Why vacuum is created in discharge tube? [1 mark] (d) State two factors which affects the rate of thermionic emission. [2 marks] (e) Potential difference, V = 2 500 V is supplied between X and Y, HOTS Applying (i) calculate the velocity of electron. [2 marks] (ii) state any change in speed of electron if the potential difference between X and Y increased? [1 mark] CLONE SPM PAPER 2 Question 1: (b) Using the concept of conservation of energy, the work done (W = eV) is equal to the increase in kinetic energy ( 1 2 mv2 ). Discharge tube Chp 5_1202 QB Physics F5.indd 92 07/01/2022 2:09 PM
97SOS TIP Section B Answer all questions. 6. (a) Diagram 6.1 shows the transistor symbol. E C B Diagram 6.1 (i) State the type of transistor above. [1 mark] (ii) Write an equation to relate IB, IC and IE. [1 mark] (iii) Explain the principle of operation of transistor above using a simple circuit. [4 marks] (b) Diagram 6.2 shows a transistor circuit for activating the fire alarm. R Switch Bell Extra high tension 1 kΩ 100 kΩ Diode 5V Thermistor Diagram 6.2 Explain the operation of the transistor circuit as an auto switch for a fire alarm system. [4 marks] (c) Diagram 6.3 shows a circuit with transistor used as a sound amplifier. R Q P S Diagram 6.3 Question 6: (c) On all four electronic devices, the loudspeaker is an important electronic device for the sound amplifier. Chp 5_1202 QB Physics F5.indd 97 07/01/2022 2:09 PM
99SOS TIP Section C Answer all questions. 7. Diagram 7.1(a) and 7.1(b) show the effects on cathode rays using a deflection tube. Power supply Vacuum Cathode Anode 0.5 kV Potential difference Spotted light Power supply Vacuum 2.5 kV Spotted light Cathode Anode Potential difference Diagram 7.1(a) Diagram 7.1(b) (a) What is meant by thermionic emission? [1 mark] (b) (i) Observe Diagram 7.1(a) and 7.1(b). Compare the size of the light spot seen on the screen with the given potential difference. [2 marks] (ii) State the relationship between size of the light spot and the potential difference to make a deduction about the relationship the size of light spot and kinetic energy of cathode ray. [2 marks] (iii) State the physical quantity to be conserved in 7(b)(ii). [1 mark] (c) Diagram 7.2 shows the cathode rays entering the region of the electric and magnetic field in the deflection tube. Electron beam O 1 cm – N N + Diagram 7.2 State and explain in which direction the cathode ray will be deflected. [4 marks] (d) You are given a task to modify the design of cathode ray deflection tube so that it can increase the rate of thermionic emission and brightness of light spot on the screen. State and explain the modification based on the following aspects: [10 marks] (i) Type of metal used as cathode (ii) Type of the wire as a filament (iii) Surface area of cathode (iv) Potential difference charged in cathode gun (v) Type of the screen Question 7: (b) (ii) Direction of magnetic force exerted on cathode rays can be determined by using Fleming left-hand rule. Chp 5_1202 QB Physics F5.indd 99 07/01/2022 2:10 PM
100SOS TIP Reinforcement and Assessment of Science Process Skills for Paper 3 (Practical Test) 1. In this experiment, you will investigate the relationship between the base current, IB with the collector current, IC for npn transistor. (a) Set up the apparatus as shown in Diagram 1. I B R1 = 1 kΩ R3 = 1 kΩ R4 = 0 – 1 kΩ R2 = 50 kΩ I C A1 6 V A2 Diagram 1 (i) Close the switch and adjust the rheostat until the reading on ammeter A1 is 10 mA. (ii) Record the reading on ammeter A2 in the table. (iii) Repeat the above procedure by adjusting the rheostat until the readings on ammeter A2 are 20 mA, 30 mA 40 mA and 50 mA. [6 marks] IB / mA IC / mA (b) State the variables for this experiment; (i) Manipulated variables [1 mark] (ii) Responding variable [1 mark] (c) (i) Plot a graph of collector current, IC (y-axis) against base current, IB (x-axis). (ii) Calculate the gradient of the graph. [3 marks] (d) (i) The equation relating the collector current, IC to the base current, IB is IC = kIB where k is a constant Using answer from (c)(ii), calculate the value of the constant k. (ii) Calculate the magnitude of the collector current, IC when the base current, IB is 85 mA. [3 marks] Question 1: (d) Linearise the equation and establish a relationship with y = mx + c. Chp 5_1202 QB Physics F5.indd 100 07/01/2022 2:10 PM
101SOS TIP (e) State one precaution that needs to be taken to improve the accuracy of the readings in this experiment. [1 mark] If you are unable to carry out the experiment described, you can answer this question by using the data obtained as shown in Table 1. Table 1 IB/A IC/A µA 0 10 20 40 30 50 µA 0 5 10 20 15 25 µA 0 10 20 40 30 50 µA 0 5 10 20 15 25 µA 0 10 20 40 30 50 µA 0 5 10 20 15 25 µA 0 10 20 40 30 50 µA 0 5 10 20 15 25 µA 0 10 20 40 30 50 µA 0 5 10 20 15 25 Question 1: (e) Avoid parallax errors when taking readings from measuring apparatus. Chp 5_1202 QB Physics F5.indd 101 07/01/2022 2:10 PM
133 CHAPTER 1 Paper 1 1. D 2. C 3. C 4. B 5. C 6. A 7. A 8. A When the resultant force is zero, the acceleration is zero. Velocity is zero or uniform. 9. D Reading = Weight + ma 10. C Upward resultant force = Balancing weight 11. A FR = ma 500 – 120 = (60 + 10)a a = 5.4 m s–1 12. D 13. D 14. C Friction always opposes motion 15. B Apply formula: Fp 2 = F1 2 + F2 2 + 2F1 + 2F1 F2 cos q 16. C FR = ma F – 20 = 5(2) F = 30 N 17. A 18. B 19. C 20. D Fg = 30 – 12 sin 30° = 24 N 21. B Ff is the frictional force while R is the normal reaction force. 22. D 23. C 24. B 25. B Magnitude of resultant force is doubled and toward left. 26. A Weight down the track = mg sin q 27. D Moving along OX means the vertical forces are balanced. 28. A 29. B 30. D 31. A 32. A 33. B 34. B Resolve the forces vertically. 35. D What is the meaning of balanced forces? 36. A 37. A 38. B 39. B 40. D 41. C Weight down the track is balanced by friction. 42. B 43. B 44. D 45. B Hooke’s Law is true for compression and stretching. 46. D 47. A 48. B 49. B The elastic energy is directly proportional to the square of the elongation. 50. A The addition of a load of 150 g causes a compression of 6 cm. 51. A 52. A Note the half -length spring can still only stand 10 kg (even it is stiffer now). 53. C Paper 2 Section A 1. (a) 6 cm 8 cm 120o R (i) Use a scale of 1 cm: 10 N. Draw a parallelogram Diagonal length = 7.0 cm Magnitude of resultant force = 7 µ 10 N = 70 N Direction = 39o counter clock wise from 80 N or N51o E (by measurement using protractor) (ii) Fp 2 = 602 + 802 + 2(60)(80) cos 120° Fp = 72 N (b) F-friction = ma 70 − 30 = 20a a = 2.0 m s–2 (c) Magnitude of opposite force = 40 N direction = S51o W 40 + 30 = 70 N (d) The new acceleration exceeds 4.0 m s–2 because the resultant force = 2(70) − 30 = 110 > 2 times the previous resultant force (40 N) (e) Fʹ = 70 + 30 = 100 N because now the frictional force acts in the opposite direction. 2. (a) Weight = mg = 5 µ 10 = 50 N (b) 50 – 30 = 20 N (c) (i) F = msystem µ a 20 = (50 + 30)a a = 0.25 m s–2 (ii) Consider 5 kg mass only T 5 kg 50 N a = 0.25 m s-2 F = m µ a 50 – T = 5(0.25) T = 48.75 N (d) (i) a’ a (ii) Because now, the tension in the rope is 50 N, exceeding the tension before. 3. (a) (i) When the resultant is nil. (ii) 10 N (b) Fnet = ma 30 – 10 = 5a a = 4 m s–2 Chapter 1 Answers Answer_1202 Physics F5.indd 133 10/01/2022 12:50 PM
134 (c) Fnet = ma 30 – 10 – friction = ma 20 – 10 = 5a a = 2 m s–2 (d) Fnet = ma 2(30) cos q 2 – 10 = 0 q 2 = 80.4° q = 160.8° (e) (i) Trolley will decelerate (ii) Trolley will accelerate (f) Because the component to the right for both 30 N will decrease when angle increases (F cos q 2 will decrease when q increase). 4. (a) R mg (b) Wx = mg sin q (c) Resultant force (d) a = g sin q a / m s-2 9.8 q/° 90° (e) S u = 0 h 30o t = 1.2 s a = g sin q = 9.8 sin 30° = 4.9 m s–2 s = ut + 1 2 at 2 h sin 30° = 0 + 1 2 (4.9)(1.2)2 h = 1.8 m 5. (a) Vector: force Scalar: extention, energy, work, length Direction is the one that differentiates vector from scalar. (b) (i) Directly proportional (ii) x / m F / N (a) (b) 0.5 1600 (c) F = kx k = 1 600 0.5 = 3 200 N m–2 kʹ = 3k = 3 µ 3 200 = 9 600 N m–2 (d) W = 1 2 kʹk2 = 1 2 (9 600)(0.15)2 = 108 J Section B 6. (a) • The length of the board should be long so that the component of weight down the board is less. The workers need less force to do their work. • The board should be strong and rigid so that it won’t be broken by the heavy load. • The surface of the board should be smooth in order to reduce the frictional force resisting the movement of the load. • Angle q = 0° so that the rope is parallel to the board to ensure pulling force needed is minimal. • The arrangement for Q is chosen because the board is long, rigid, smooth and the pulling force is parallel to the board. (b) Let frictional force = Fg , Fnet = ma 5 – Fg = 2(1) Fg = 3 N Apply Fnet = ma again 20 – 10 – 3 = 2a a = 3.5 m s–2 (c) (i) Bouyant force,FB Dragging force, FV W = mg (ii) Initially, the weight of the sphere exceeds the buoyant force and dragging force. The sphere accelerates. However, as the velocity of the sphere increases, the dragging force also increases until finally, the sum of the buoyant force and dragging force is equal in magnitude to the weight of the sphere. Then and after, the forces are balanced and the sphere moves with a uniform downward terminal velocity. Section C 7. (a) (i) Stiffness is a measure of the resistance offered by an elastic body to deformation. (ii) • The thickness of the coil winding for Y is more. • The elongation for Y is less. • The time for 20 oscillations for Y is less, thus the oscillation period of the load is less. • The stiffer the spring the less the swing period. (b) • The spring constant should be high so that it is hard and can support and withstand the weight of the child including a fat one. • The material for the spring and seat should be durable and able to resist wear due to the weather so that it can last longer. • The diameter of the spring should be large to add stability and prevent the child from slipping down. • The center of gravity of the spring rider is on the axis of the spring to add stability while the boy uses it. • The seat base should be rough so that the child does not slip while using it. • The rider’s spring should be mounted on a coarse fabric material to prevent injury in case the child falls. (c) l o x Chapter 1 Answer_1202 Physics F5.indd 134 10/01/2022 12:50 PM
137 (e) h1 r1 = h2 r2 900 µ 1.2 = rY µ 0.85 rY = 1.27 µ 103 kg m–3 6. (a) (i) Vacuum (Torricelli) (ii) Same (b) h1 r1 = h2 r2 0.760 µ 13.6 = h2 µ 1 hwater = 10.3 m (c) h1 r1 = h2 Vr2 0.760 µ 13 600 = h2 µ 1.2 hair = 8 613 m (d) The mercury column on the right will drop less because the larger space there will produce less trapped air pressure than the space on the left. (e) 2 µ 13.6 cm = 27.2 cm Section B 7. (a) • The cross-sectional area of the output piston should be large to produce a large output force. • The boiling point of the liquid should be high so that it does not boil or evaporate easily. If bubbles are produced in such a way in a liquid, the efficiency of the system will be reduced. • The compressibility of the fluid should be low so that no work is used to compress it. • The material for the transmission pipe should be steel so that the corrosion rate is slow and the transmission pipe is durable. • System S was chosen because it has a large output piston, high boiling point and hydraulic fluid of low compressibility, and its transmission pipe is made of steel. (b) F1 F2 = A1 A2 80 + W 6 + 2 = p 4 (3.5d) 2 p 4 d2 W = 18 N (c) (i) Same (ii) The mercury column of the barometer Q rises higher. The area for the container in Q is less, so the height of the water in the container is more for the same volume of water poured in. As pressure of liquid increases with depth, the pressure exerted by the water on mercury in container Q is more and pushes the mercury column up higher. Section C 8. (a) (i) The pressure of a liquid acts in all directions, acts normally to the surface and increases with the depth of the liquid. (ii) h is more for Diagram 8.1. The horizontal distance traversed by the water before it touches the basin is further away for Diagram 8.1. (iii) The deeper the water, the faster the water gushes out. Deduction: The water pressure increases with the depth of water. (b) • The shape of the boat should be streamlined to reduce the dragging force by water. • The overall density of the boat should be less and enable it to accommodate more tourists. • The bottom of the boat should be of glass material to facilitate viewing of the coral reef. It is also better because it can eliminate while viewing directly by the side of the boat reflection of sunlight. • The glass used must be of high quality so that it does not break and is easily scratched. • The glass should be slightly convex to magnify the image seen. (c) • The weight of a liquid is equal to the weight of a floating cube. Thus, the weight or mass of water displaced is equal to the weight or mass of oil displaced. • Mass is equal to the product of density and volume. • Since the density of the oil is less, the volume to be displaced must be more. So, the wooden cube sinks more to do so. Paper 3 1. (a) (i) Manipulated variable: Height of turpentine column, h1 (ii) Responding variable: Height of water column, h2 (b) h1 / cm h2 / cm 3.1 2.6 6.0 5.2 9.2 7.6 11.8 10.2 15.0 12.8 (c) 0 2 4 6 8 10 12 14 16 2 4 6 8 10 12 h2 / cm Graph of h2 against h1 h1 / cm (d) k = 11.0 – 3.4 13.0 – 4.0 = 0.844 (e) The gradient remains the same CHAPTER 3 Paper 1 1. C Based of meaning of electric field. 2. D Direction of electric field must be outward. 3. D Magnitude of the electric field at P and Q is same. Since electric field is vector quantity, net electric field at X become zero. 4. B Based on definition, E = F q . 5. C Based on definition, I = Q t . 6. C Heavier positive ions are pulled towards the negative plate slowly. 7. B Q = It = 0.85(8 × 60) = 408 C 8. C Work, W done to move one coulomb charge, Q between two points in electric field. 9. B Based on definition, V = W Q. 10. B W = QV V = 85 000 × 10–3 18 = 4.72 V Chapter 2 – Chapter 3 Answer_1202 Physics F5.indd 137 10/01/2022 12:50 PM