1QA_Physics_AnyflipWM

PENERBIT ILMU BAKTI SDN. BHD.

Contents FORM 4 Theme 1: Elementary Physics Chapter 1 Measurement 1 Paper 1 Objective Questions 1 Paper 2 Section A – Structured Questions 5 Section B & C – Essay Questions 7 Theme 2: Newtonian Mechanics Chapter 2 Force and Motion I 10 Paper 1 Objective Questions 10 Paper 2 Section A – Structured Questions 18 Section B & C – Essay Questions 22 Chapter 3 Gravitation 26 Paper 1 Objective Questions 26 Paper 2 Section A – Structured Questions 32 Section B & C – Essay Questions 36 Theme 3: Heat Chapter 4 Heat 38 Paper 1 Objective Questions 38 Paper 2 Section A – Structured Questions 44 Section B & C – Essay Questions 51 Theme 4: Waves, Light and Optics Chapter 5 Waves 58 Paper 1 Objective Questions 58 Paper 2 Section A – Structured Questions 65 Section B & C – Essay Questions 72 Chapter 6 Light and Optics 76 Paper 1 Objective Questions 76 Paper 2 Section A – Structured Questions 83 Section B & C – Essay Questions 89 FORM 5 Theme 1: Newtonian Mechanics Chapter 1 Force and Motion II 93 Paper 1 Objective Questions 93 Paper 2 Section A – Structured Questions 98 Section B & C – Essay Questions 101 Q&A SPM Phyics Contents 3rd.indd 2 14-Jan-23 1:11:02 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter 2 Pressure 106 Paper 1 Objective Questions 106 Paper 2 Section A – Structured Questions 113 Section B & C – Essay Questions 118 Theme 2: Electricity and Electromagnetism Chapter 3 Electricity 122 Paper 1 Objective Questions 122 Paper 2 Section A – Structured Questions 128 Section B & C – Essay Questions 135 Chapter 4 Electromagnetism 143 Paper 1 Objective Questions 143 Paper 2 Section A – Structured Questions 149 Section B & C – Essay Questions 152 Theme 3: Applied Physics Chapter 5 Electronics 157 Paper 1 Objective Questions 157 Paper 2 Section A – Structured Questions 163 Section B & C – Essay Questions 167 Theme 4: Modern Physics Chapter 6 Nuclear Physics 172 Paper 1 Objective Questions 172 Paper 2 Section A – Structured Questions 178 Section B & C – Essay Questions 184 Chapter 7 Quantum Physics 189 Paper 1 Objective Questions 189 Paper 2 Section A – Structured Questions 197 Section B & C – Essay Questions 201 SPM Model Test 207 Answers 231 Q&A SPM Phyics Contents 3rd.indd 3 14-Jan-23 1:11:03 PM PENERBIT ILMU BAKTI SDN. BHD.

1 Paper 1 Objective Questions 1.1 Physical Quantities 1 Which physical quantity is not a base quantity? A Time B Weight C Temperature D Electric current Answer: B There are only seven physical quantities in base quantities: mass, time, length, thermodynamic temperature, electric current, luminous intensity and amount of substance. 2 Which of the following is an S.I. unit for base quantity? A N m C °C B cd D J Answer: B N m is a unit for work, °C is a unit for temperature, cd is an S.I. unit for luminous intensity and J is a unit for energy. 3 Which of the following shows a scalar quantity and a vector quantity correctly? Scalar quantity Vector quantity A Mass Amount of substance B Luminous intensity Mass C Weight Luminous intensity D Amount of substance Weight Answer: D Weight is a vector quantity and is directed towards the centre of the Earth. The other quantities do not have direction, so they are scalar quantities. 4 Which quantity is a vector quantity? A Pressure C Power B Energy D Force Answer: D A vector quantity has both a magnitude and a direction. Force is a push or a pull. Therefore, it is pulling or pushing an object in a specific direction. 5 Which quantity is a scalar quantity? A Velocity C Acceleration B Distance D Displacement Answer: B A scalar quantity has only a magnitude but no direction. Distance is a scalar quantity because it has only a magnitude. 6 Which physical quantity is equal to displacement time ? A Speed B Velocity C Distance D Acceleration Answer: B The formula for velocity is ν = displacement time Measurement 1 CHAPTER Theme 1: Elementary Physics Textbook Pages: 2 – 22 Q&A SPM Phyics F4 C1 3rd.indd 1 14-Jan-23 1:13:28 PM PENERBIT ILMU BAKTI SDN. BHD.

2 Chapter Form 4 1 7 Which of the following is a derived quantity? A Mass C Weight B Length D Temperature Answer: C There are only seven physical quantities in base quantities: mass, time, length, thermodynamic temperature, electric current, luminous intensity and amount of substance. All other quantities are derived quantities. Weight, W = mg 8 Which measuring instrument measures a derived quantity? A C B D Answer: B Thermometer is used to measure temperature. Ammeter is used to measure electric current. Voltmeter is used to measure potential difference. Triple beam balance is used to measure mass. Potential difference is a derived quantity. 9 Force depends on mass and acceleration. Which of the following is a unit for force? A kg m s2 C kg m s–2 B kg m s–1 D kg m s–3 Answer: C Force is given by, F = ma. The S.I. units for mass and acceleration are kg and m s–2 respectively. Therefore, the unit for force is kg m s–2. 10 What is the S.I. unit for density? A g cm3 C kg m3 B g cm–3 D kg m–3 Answer: D Density is given by ρ = mass, m volume, V The S.I. units for mass and volume are kg and m3 respectively. Therefore, the unit for density is kg m–3. 1.2 Scientific Investigation 11 Diagram 1 shows a displacementtime graph. Displacement (m) Time (s) 0 Diagram 1 What is the physical quantity represented by the gradient of the graph? A Speed C Acceleration B Velocity D Displacement Answer: B To calculate the gradient of the graph, divide the change in y by the change in x. Gradient = displacement time = velocity 12 Diagram 2 shows the graph of acceleration, a against the reciprocal of mass, 1 m. a (m s–2) 0 1 m (kg) Diagram 2 Q&A SPM Phyics F4 C1 3rd.indd 2 14-Jan-23 1:13:28 PM PENERBIT ILMU BAKTI SDN. BHD.

3 Chapter Form 4 131 A X → Y → Z B X → Y → Z → Y C X → Y → Z → Y → X D X → Y → Z → Y → X → Y → Z Answer: C One complete oscillation occurs when the pendulum moves from a starting point to the next point and comes back to the starting point. 15 Diagram 4 shows the relationship between two inversely proportional variables. 0 x y 2 2 4 Diagram 4 What is the value of y when x = 1? A 0 C 1.5 B 1 D 2 Answer: B Students must draw the interpolation line at x = 1. Then, from the interpolation line, y = 1. 16 In general, there are five important aspects in analysing graphs. Which of the following is not an aspect in analysing graphs? A State the relationship between two variables B Calculate the gradient of the graph C Calculate the area under the graph D Calculate the volume of the graph Based on the graph, which relationship is correct? A a is inversely proportional to m B a is directly proportional to m C a is decreasing linearly with m D a is increasing linearly with m Answer: A The graph is a straight line with a positive gradient passing through the origin. Then, a is inversely proportional to m. 13 Which graph shows the relationship between two physical quantities, M and N increases linearly? A M N 0 C 0 M N B M 0 N D 0 M N Answer: B M increases linearly with N when the straight line with a positive gradient cuts the y-axis. 14 Diagram 3 shows the oscillation of a simple pendulum. The oscillation starts at point X. X Z Y O Diagram 3 Which of the following is the correct sequence for one complete oscillation? Q&A SPM Phyics F4 C1 3rd.indd 3 14-Jan-23 1:13:28 PM PENERBIT ILMU BAKTI SDN. BHD.

4 Chapter Form 4 1 Answer: D Five important aspects in analysing graphs: relationship of two variables, gradient of the graph, area under the graph, interpolation and extrapolation of the graph. 17 When planning an experiment, how many types of variables must be stated in the report? A 1 C 3 B 2 D 4 Answer: C There are three types of variables in an experimental report: manipulated variable, responding variable and constant variable. 18 Diagram 5 shows the arrangement of apparatus for a simple pendulum experiment. Retort stand Thread Plywood Length of pendulum, I Pendulum bob G-clamp Diagram 5 Which statement is the correct inference for the experiment? A The length of a simple pendulum depends on its period of oscillation B The period of oscillation of a simple pendulum depends on its length C The length of a simple pendulum depends on the time of all its oscillations D The time of all oscillations of a simple pendulum depends on its period of oscillations Answer: B To state the inference of an experiment, the responding variable depends on the manipulated variable. 19 The relationship between variables p and q is p = m√q + n. m is the gradient of the graph and n is the y-intercept of the graph. Which graph must be plotted to obtain the values of m and n? A Graph of p against q B Graph of p against √q C Graph of p against 1 q D Graph of q against √p Answer: B The graph needs to be drawn based on a straight line equation, y = mx + c. So, y-axis is p and x-axis is √ q. 20 A student carries out an experiment to investigate the relationship between the mass and extension of a spring. What measuring instruments should be used to obtain data from the experiment? Mass Extension of the spring A Vernier calliper Triple beam balance B Triple beam balance Metre rule C Triple beam balance Vernier calliper D Metre rule Triple beam balance Answer: B In the school laboratory, usually the mass of an object is measured using a triple beam balance or an electronic balance, while the extension of the spring is measured using a metre rule. Q&A SPM Phyics F4 C1 3rd.indd 4 14-Jan-23 1:13:29 PM PENERBIT ILMU BAKTI SDN. BHD.

5 Chapter Form 4 131 Paper 2 (Section A – Structured Questions) 1 (a) What is the meaning of base quantities? Base quantities are physical quantities that cannot be derived from other physical quantities. [1 mark] (b) State two base quantities with their S.I. units. Mass (kilogram)//Length (metre)//Time (second)//Absolute temperature (kelvin)//Electric current (ampere)//Luminous intensity (candela)// Amount of substance (mole) [1 mark] (c) Work done by an object, W is given by the formula W = Fs where F is a force and s is the displacement of the object. Derive the unit for W in terms of S.I. base units. W = Fs Formula for force, F = ma = m1 v – u t 2 = m1 s t t 2 = m1 s t × t2 Expressed in base quantities: m1 l t × t 2 Expressed in S.I. base units: kg1 m s2 2 = kg m s–2 Therefore, unit for work done, W = [Unit for F] × [Unit for s] = kg m s–2 × m = kg m2 s–2 [2 marks] 2 Diagram 1 shows the labels on a banana bread packaging. There are several physical quantities stated on the labels. Nutritional Information BANANA BANANA Per 100g Ingredients: Flour, banana, brown sugar, eggs, milk, olive oil, walnut, baking powder Serving suggestions: Consume warm, heat in microwave for 30 seconds Storage: This product is best stored in the fridge at 4°C. Energy 1460 kJ Protein 5.9g Fat- Total 15.7g Saturated 2.5g Trans <0.1g Monounsaturated 9.3g Polyunsaturated 4.0g Carbohydrate 44.9g Sugars 24.5g Sodium 140mg Dietary Fibre 2.3g PRICE : $15 Diagram 1 Q&A SPM Phyics F4 C1 3rd.indd 5 14-Jan-23 1:13:29 PM PENERBIT ILMU BAKTI SDN. BHD.

6 Chapter Form 4 1 Physical quantities are classified into two quantities: base quantities and derived quantities. (a) What is the meaning of derived quantities? Derived quantities are physical quantities which can be derived from other physical quantities. [1 mark] Derived quantities can be derived using multiplication, division or combination of multiplication and division of other physical quantities. Exam Tips (b) Based on Diagram 1, state one base quantity and one derived quantity that stated on the labels. (i) Base quantity: Mass//Temperature//Time (ii) Derived quantity: Energy [2 marks] (c) The nutritional information for sodium is 140 mg where m in the unit is a prefix. State the value of m. 10–3 [1 mark] Tera = 1012, Giga = 109 , Mega = 106 , Kilo = 103 , Deci = 10–1, Centi = 10–2, Milli = 10–3, Micro = 10–6, Nano = 10–9, Pico = 10–12 Exam Tips 3 Adam Zikri carried out an experiment to determine the relationship between the length of a pendulum and the period of oscillation. Diagram 2 shows the arrangement of apparatus for the experiment. Table 1 shows the data obtained from the experiment. 60 5 10 15 20 25 30 35 40 45 50 55 Thread Split cork P Q Pendulum bob Pin Diagram 2 Q&A SPM Phyics F4 C1 3rd.indd 6 14-Jan-23 1:13:30 PM PENERBIT ILMU BAKTI SDN. BHD.

7 Chapter Form 4 131 Length, l (cm) Time for 10 oscillations, t (s) Period, T (s) T2 (s2) t1 t2 taverage 15.0 16.1 16.3 16.2 1.62 2.62 20.0 18.3 18.7 18.5 1.85 3.42 25.0 22.0 22.2 22.1 2.21 4.88 30.0 24.8 25.0 24.9 2.49 6.20 35.0 26.4 26.4 26.4 2.64 6.97 Table 1 (a) Name the labelled apparatus and state the physical quantities measured. Name of the apparatus Physical quantity measured P: Metre rule Length Q: Stopwatch Time [2 marks] (b) Calculate the value of taverage, T and T2 , for each length of the pendulum. Complete Table 1. [3 marks] Period of oscillation is time taken for one complete oscillation. Exam Tips Paper 2 (Section B & C – Essay Questions) 4 Diagram 3.1 shows the label on a tub of chocolate flavoured ice cream. Diagram 3.2 shows the label on a tub of durian flavoured ice cream. There are several physical quantities stated on the labels. CHOCOLATE FLAVOUR KEEP FROZEN AT TEMPERATURE BELOW -18°C Serving size: One scoop (mass 50 g) VOLUME IN CONTAINER: 1 000 mL DURIAN FLAVOUR KEEP FROZEN AT TEMPERATURE BELOW -18°C Serving size: One scoop (mass 60 g) VOLUME IN CONTAINER: 1 000 mL Diagram 3.1 Diagram 3.2 Q&A SPM Phyics F4 C1 3rd.indd 7 14-Jan-23 1:13:30 PM PENERBIT ILMU BAKTI SDN. BHD.

8 Chapter Form 4 1 (a) What is the meaning of physical quantities? [1 mark] (b) Based on Diagram 3.1 and Diagram 3.2, HOTS Analysing (i) compare the masses of serving sizes, the volumes of ice cream in the tubs and the temperatures to keep the ice cream frozen. [3 marks] (ii) state the base quantities and derived quantities that are stated on the labels. [2 marks] (c) Today, an ice cream shop near Zayed’s house is holding a promotional sale for both ice cream flavours. Zayed wants to go to the ice cream shop to buy them. The distance from his house to the ice cream shop is 0.8 km. Zayed rides his bicycle at a speed of 30 km h–1. He wants to reach the shop in 10 minutes. Therefore, he increases his speed by accelerating at 1.5 m s–2. Identify the scalar quantities and the vector quantities involved in the above situation. State the similarity and the difference between both quantities. [4 marks] (d) Diagram 3.3 shows four measuring instruments used to determine the measurement of physical quantities. R T S U Diagram 3.3 Based on Diagram 3.3, you are given a set of measuring instruments that can be used to measure the lengths of different objects. State the name of each measuring instrument and explain the characteristics based on its sensitivities and functions. Choose the measuring instrument that is most appropriate for measuring the diameter of a marble accurately. Give reasons for your choice. [10 marks] Q&A SPM Phyics F4 C1 3rd.indd 8 14-Jan-23 1:13:30 PM PENERBIT ILMU BAKTI SDN. BHD.

9 Chapter Form 4 131 4 (a) Physical quantities are quantities that can be measured. (b) (i) – The mass of serving size in Diagram 3.1  Diagram 3.2. – The volume of ice cream in the tub in Diagram 3.1 = Diagram 3.2. – The temperature to keep the ice cream frozen in Diagram 3.1 = Diagram 3.2. (ii) Base quantities: Mass and temperature Derived quantity: Volume (c) Scalar quantities: Speed//Time//Distance Vector quantity: Acceleration Similarity: Both quantities have magnitude. Difference: Scalar quantity has no direction, while vector quantity has direction. (d) Measuring instruments Sensitivity Functions R: Metre rule 0.1 cm Used to measure the length of objects such as books, pencils, etc. S: Measuring tape 0.1 cm Used to measure the length of bigger objects such as tables, rooms, etc. T: Micrometre screw gauge 0.01 mm Used to measure smaller lengths such as the thickness of a coin, the diameter of a ping-pong ball, etc. U: Vernier calliper 0.01 cm – Used to measure smaller lengths such as the outer and inner diameters of a beaker – Used to measure the depth of a shallow hole using depth rod T is chosen because it can measure the smallest lengths and is accurate up to two decimal places. 4 (b) (i) – Candidates can answer the question using symbols:  (less than),  (more than) or = (equal). – The comparison must be done separately for each quantity. Exam Tips Answers Q&A SPM Phyics F4 C1 3rd.indd 9 14-Jan-23 1:13:30 PM PENERBIT ILMU BAKTI SDN. BHD.

10 Paper 1 Objective Questions 2.1 Linear Motion 1 Diagram 1 shows a part of Shah’s neighbourhood. 2 km 1 km Shah’s house School Office Kindergarden 400 m Diagram 1 What is the shortest distance from Shah’s house to the school? A 1.08 km C 2.24 km B 1.40 km D 3.00 km Answer: C Use Theorem Pythagoras to solve the problem. School (B) Shah’s house (A) 1 km 2 km C (AB)2 = (AC)2 + (BC)2 = (1)2 + (2)2 AB = √ 1 + 4 = 2.24 km The shortest distance refers to the displacement. Exam Tips 2 Fadhlie plans to cycle with his friends. He plans to finish cycling 50 km in 2 hours. If he rides with constant acceleration, what is his final velocity? A 6.94 m s–1 C 11.04 m s–1 B 7.68 m s–1 D 13.89 m s–1 Answer: D Since the acceleration is constant, then the linear motion equation is used. s = 1 2 (u + v)t (50 × 1 000) = 1 2 (0 + v)(2 × 60 × 60) v = 13.89 m s–1 3 Diagram 2 shows part of a ticker tape that is pulled by a moving toy car. 1.2 cm 5.6 cm Direction of motion Diagram 2 The ticker timer is connected to a 50 Hz alternating current power supply. What is the acceleration of the toy car? A –22 m s–2 C 11 m s–2 B –11 m s–2 D 22 m s–2 Answer: B 1 tick = 50 Hz = 0.02 s u = s t = 0.056 0.02 = 2.8 m s–1 v = s t = 0.012 0.02 = 0.6 m s–1 t = (number of ticks – 1) × 0.02 s = (11 – 1) × 0.02 s = 0.2 s a = v – u t = 0.6 – 2.8 0.2 = –11 m s–2 Force and Motion I 2 CHAPTER Theme 2: Newtonian Mechanics Textbook Pages: 24 – 75 Q&A SPM Phyics F4 C2 2nd.indd 10 14-Jan-23 1:15:54 PM PENERBIT ILMU BAKTI SDN. BHD.

11 Chapter Form 4 213 The negative sign indicates that the toy car experiences deceleration. Exam Tips 4 Diagram 3 shows the arrangement of photogate system and electronic timer that is used to study linear motion more accurately. Shutter plate Height of runway is adjusted to 15 cm Runway Trolley First photogate Second photogate Electronic timer Diagram 3 The width of the shutter plate is 5 cm. If the time interval is 0.095 s, what is the final velocity of the trolley? A 0.195 m s–1 C 0.385 m s–1 B 0.256 m s–1 D 0.526 m s–1 Answer: D Final velocity of the trolley, v = "Width of shutter plate, s" "Time interval, t" = 0.05 0.095 = 0.526 m s–1 5 Diagram 4 shows two ticker tapes produced by a toy car and a toy lorry respectively. Toy car Toy lorry Diagram 4 Which statement is correct? HOTS Analysing A The time taken for the toy car and the toy lorry are the same B The acceleration of the toy car and the toy lorry are the same C The average velocity of the toy car and the toy lorry are the same D The distance travelled by the toy car and the toy lorry are the same Answer: D The time taken, acceleration and average velocity for both toys are different. 6 Diagram 5 shows a ticker tape chart of an object accelerating down on a smooth track. 20 Time (s) Length (cm) 2 Diagram 5 Calculate the acceleration of the object. A 350 cm s–1 B 400 cm s–1 C 450 cm s–1 D 500 cm s–1 Answer: C 1 tick = 50 Hz = 0.02 s u = 2 (5 × 0.02) = 20 cm s–1 v = 20 (5 × 0.02) = 200 cm s–1 a = 200 – 20 (4 × 0.1) = 450 cm s–1 Q&A SPM Phyics F4 C2 2nd.indd 11 14-Jan-23 1:15:55 PM PENERBIT ILMU BAKTI SDN. BHD.

12 Chapter Form 4 2 2.2 Linear Motion Graphs 7 Diagram 6 shows a velocity-time graph for the motion of a car driven by Mr Ryyan. Velocity, v (m s–1) Time, t(s) 10 20 30 20 4 6 8 10 12 Diagram 6 Mr Ryyan was driving at a velocity of 30 m s–1. When a cat suddenly jumps onto the middle of the road, he stepped on the brakes immediately and the velocity started to decrease uniformly. Calculate the displacement of the car after he stepped on the brakes until the car stopped completely. HOTS Applying A 18 m C 150 m B 60 m D 300 m Answer: B For the velocity-time graph, displacement is the area under the graph. Mr Ryyan starts stepping on the brakes at the 8th second and the car stops completely at the 12th second. It takes 4 seconds for the car to come to a stop. Displacement = Area under the graph = 1 2 (30)(12 – 8) = 60 m 8 Which displacement-time graph shows a man riding a motorcycle with a constant velocity and then returning to his start position? A C Displacement, s (m) 0 Time, t (s) Displacement, s (m) 0 Time, t (s) B D Displacement, s (m) 0 Time, t (s) Displacement, s (m) 0 Time, t (s) Answer: C The first part is constant velocity. It means that there is an increase in displacement. When he turns back to return to his start position, the gradient of the graph becomes negative. 9 Diagram 7 shows a velocity-time graph for a remote controlled car. Velocity, v (m s–1) Time, t (s) 20 40 60 100 20 30 40 50 60 Diagram 7 Determine the displacement of the remote controlled car when it is moving with constant velocity. A 1 800 m C 3 000 m B 2 700 m D 3 300 m Answer: A The remote controlled car moves with constant velocity from 10th second to 20th second. The displacement is equal to the area under the graph. Displacement = Area under the graph = 60 × (40 – 10) = 1 800 m Q&A SPM Phyics F4 C2 2nd.indd 12 14-Jan-23 1:15:55 PM PENERBIT ILMU BAKTI SDN. BHD.

13 Chapter Form 4 213 2.3 Free Fall Motion 10 Diagram 8 shows a stone and a feather in a vacuum cylinder. The feather and the stone are released simultaneously and reach the bottom of the vacuum cylinder at the same time. Feather Stone Vacuum Diagram 8 Which statement best describes the situation? HOTS Analysing A The air resistance is high in a vacuum cylinder B The mass of the rock and the feather are the same C The velocity of the rock is higher than the feather D The acceleration of the feather is equal to the gravitational acceleration Answer: D In vacuum, there is no air resistance. Therefore, both objects are in free fall motion. During the free fall, the accelerations of both objects are solely due to gravitational force, and thus, are equal to the gravitational acceleration. 11 A rock that is thrown vertically downwards experiences free fall. Which physical quantity remains unchanged? A Displacement C Distance B Acceleration D Velocity Answer: B Velocity, distance and displacement increase as an object falls. Only acceleration remains unchanged. 12 Diagram 9 shows Nuh jumping up vertically from a trampoline with a velocity of 5 m s–1. Diagram 9 What is the maximum displacement reached by Nuh? A 0.54 m C 1.27 m B 0.98 m D 3.15 m Answer: C In this case, the maximum displacement is the maximum height that Nuh can reach. At the maximum height, the velocity is 0. Thus, the final velocity is 0. Magnitude of acceleration, a is equal to the magnitude of gravitational acceleration, g. Using linear motion equation, v2 = u2 + 2as (0)2 = (5)2 + (2)(–9.81)s s = 1.27 m 13 Diagram 10 shows a man jumping off an 8 m height cliff. 8 m Diagram 10 Q&A SPM Phyics F4 C2 2nd.indd 13 14-Jan-23 1:15:55 PM PENERBIT ILMU BAKTI SDN. BHD.

14 Chapter Form 4 2 If the man takes 5 s to reach the surface of the water, find the velocity just before he entering the water. [g = –9.81 m s–2] A 49.05 m s–1 C –25.95 m s–1 B 25.95 m s–1 D –49.05 m s–1 Answer: D Initial velocity, u is 0. v = u + at = 0 + (–9.81)(5) = –49.05 m s–1 2.4 Inertia 14 Which statement best describes inertia? A Inertia is affected by the weight of an object B Moving objects with larger inertia are easier to stop C Stationary objects with smaller mass are easier to push D Fast-moving objects have larger inertia compared with slow-moving objects Answer: C Inertia is a measure of the tendency of an object to remain in motion or at rest, which depends on the mass of the body or the amount of matter it has. 15 Diagram 11 shows a wet umbrella that is being rotated back and forth vigorously to dry it. Diagram 11 How can the umbrella dry by doing so? A Rotating the umbrella will generate heat B The mass of the umbrella is greater than the water droplets C Rotating the umbrella vigorously will increase the speed of motion D Changing the direction of rotation will keep the water droplets in motion Answer: D Changing the direction of rotation causes the umbrella to stop abruptly but the inertia of the water droplets will resist the motion and the water droplets are spun off the umbrella. 16 A student observes that the blades of a fan continue to rotate for a few moments even after the fan is switched off. What is the concept involved in the above situation? A Inertia B Free fall C Momentum D Gravitational acceleration Answer: A Inertia is the tendency of an object to resist the change in its state of motion. The object at rest will remain at rest or, if moving, will continue its motion in a straight line at uniform velocity. The fan blades continue to move because of their tendency to remain in motion. They finally stop due to an external force acting on them i.e., the frictional force. Q&A SPM Phyics F4 C2 2nd.indd 14 14-Jan-23 1:15:56 PM PENERBIT ILMU BAKTI SDN. BHD.

15 Chapter Form 4 213 2.5 Momentum 17 Diagram 12 shows a bullet is fired horizontally from a pistol. v Diagram 12 The momentum of the bullet that is moving forward is 4 kg m s–1. What is the velocity of the pistol if its mass is 2.0 kg? A 2.0 m s–1 C 6.0 m s–1 B 4.0 m s–1 D 8.0 m s–1 Answer: A In an explosion, m1 v1 = –m2 v2 4 = –2.0v2 v2 = –2.0 m s–1 18 A 11 000-kg bus that was moving with a velocity of 90 km h–1 had collided with a 4 500-kg small truck that was moving at 50 km h-1 in the same direction. Both the vehicles move together after the collision. Which statement is correct about the collision? A Total momentum after the collision is increased B Both vehicles move in opposite directions after the collision C Total kinetic energy before and after the collision is not conserved D The momentum of the bus before collision is equal to its momentum after collision Answer: C Total kinetic energy before and after an inelastic collision is not conserved because some of the energy is converted to other forms of energy. Total kinetic energy is conserved during an elastic collision. Exam Tips 19 Diagram 13 shows a pool player hitting a white ball to strike a black ball. Each ball has a mass of 0.2 kg. Diagram 13 The white ball moves at 0.5 m s–1 and strikes the black ball which is at rest. After hitting the black ball, the white ball moves at 0.2 m s–1 in the opposite direction from its initial motion. Find the final velocity of the black ball after the collision. A 0.3 m s–1 B 0.5 m s–1 C 0.7 m s–1 D 0.9 m s–1 Answer: C Principle of Conservation of Momentum states that the total momentum before collision is equal to the total momentum after collision. The black ball was initially at rest, thus u1 is 0 m s–1. The white ball moves at 0.2 m s–1 in the opposite direction from its initial motion. If we consider the initial direction to be positive, thus v1 is –0.2 m s–1. m1 u1 + m2 u2 = m1 v1 + m2 v2 (0.2)(0.5) + (0.2)(0) = (0.2)(–0.2) + (0.2)v2 v2 = 0.7 m s–1 Q&A SPM Phyics F4 C2 2nd.indd 15 14-Jan-23 1:15:56 PM PENERBIT ILMU BAKTI SDN. BHD.

16 Chapter Form 4 2 2.6 Force 20 Diagram 14 shows a man pushing a-50 kg cart at a supermarket with an acceleration of 2 m s–2. 25 N Diagram 14 If the frictional force of 25 N acting on the wheels of the cart, calculate the force exerted on the cart. A 50 kg m s–2 C 100 kg m s–2 B 75 kg m s–2 D 125 kg m s–2 Answer: D F – Frictional force = ma F – 25 = (50)(2) F = 100 + 25 = 125 kg m s–2 21 Which statement is related to Newton’s Second Law? A The rate of change of momentum is directly proportional to the force B An object tends to resist any change in its initial state of motion C The total momentum before a collision remains unchanged after the collision D An object in motion will move forever unless it is acted on by an external force Answer: A Newton’s Second Law of Motion states that the rate of change of momentum is directly proportional to the force and acts in the direction of the applied force. From the relationship F ∝ ma, F = kma, where k is a constant. 22 A car stops at a traffic light when it turns red. After a few moments, the traffic light turns green and the driver steps on the accelerator to start moving. The car accelerates for 5 s until it reaches the uniform velocity of 20 m s–1. Find the force exerted by the engine if the mass of the car is 1 200 kg. HOTS Applying A 3 600 N C 12 000 N B 4 800 N D 24 000 N Answer: B Find the acceleration using formula, a = v – u t = 20 – 0 5 = 4 m s–1 Substitute the magnitude of acceleration, a in F = ma F = (1 200)(4) = 4 800 N 2.7 Impulse and Impulsive Force 23 Diagram 15 shows a boy throwing a 0.05-kg dart onto a target board with a velocity of 5 m s–1. The dart stuck on the target board. Diagram 15 Calculate the magnitude of impulse produced. A 0.25 kg m s–1 C 0.75 kg m s–1 B 0.50 kg m s–1 D 1.00 kg m s–1 Answer: A Impulse is a change in momentum. J = mv – mu = (0.05)(0) – (0.05)(5) = –0.25 kg m s–1 Q&A SPM Phyics F4 C2 2nd.indd 16 14-Jan-23 1:15:57 PM PENERBIT ILMU BAKTI SDN. BHD.

17 Chapter Form 4 213 24 Diagram 16 shows a tennis ball moves towards a tennis racket. 55 m s-1 Diagram 16 If the mass of the tennis ball is 55 g and the impact time is 5 ms, find the impulsive force if the tennis ball moves at 55 m s–1 after hitting the racket. A 605 N B 1 210 N C 3 025 N D 6 050 N Answer: B Impulsive force, F is a rate of change of momentum. F = mv – mu t = (0.055)(55) – (0.055)(–55) 0.005 = 1 210 N 25 Jane has an online business selling a souvenir made of glass. She makes sure to wrap the box with a thick air bubble wrap before delivering it to the customer. Why does she do that? A To decrease the impulse B To decrease the impact time C To decrease the momentum D To decrease the rate of change of momentum Answer: D In the event that the box is dropped, the air bubble wrap will increase the impact time. This will decrease the rate of change of momentum or the impulsive force. 2.8 Weight 26 A watermelon has a mass of 1.5 kg. What is its weight if the gravitational acceleration is 9.81 m s–2? A 14.7 N C 17.6 N B 16.5 N D 18.7 N Answer: A W = mg = (1.5)(9.81) = 14.715 N 27 An astronaut discovered a new planet. He noticed that his weight on the new planet is less than his weight on the Earth as shown in Diagram 17. Earth New planet Weight = 800 N g = 9.81 N kg-1 Weight = 660 N Diagram 17 What is the gravitational field strength, g on the new planet? A 8.09 N kg–1 C 9.09 N kg-1 B 8.91 N kg–1 D 9.91 N kg-1 Answer: A Use formula W = mg to find the mass of the astronaut: Weight on the Earth, We = mge 800 = (m)(9.81) m = 81.55 kg Weight on the new planet, Wx = mgx 660 = (81.55)(gx ) gx = 8.09 N kg–1 Q&A SPM Phyics F4 C2 2nd.indd 17 14-Jan-23 1:15:57 PM PENERBIT ILMU BAKTI SDN. BHD.

18 Chapter Form 4 2 Paper 2 (Section A – Structured Questions) 1 (a) Diagram 1 shows an astronaut floating freely in outer space. The astronaut continues moving and is unable to stop due to the absence of any external force such as frictional force. Diagram 1 (i) Which physics law supports the above statement? Newton’s First Law [1 mark] (ii) Explain your answer in 1(a)(i). Newton’s First Law states that an object will remain at rest or continue to move at uniform velocity unless acted upon by an external force. [1 mark] (iii) Name one physics concept involved in the above situation. Inertia [1 mark] (b) Tick (✓) the vehicle with the largest inertia. ✓ [1 mark] Q&A SPM Phyics F4 C2 2nd.indd 18 14-Jan-23 1:15:58 PM PENERBIT ILMU BAKTI SDN. BHD.

19 Chapter Form 4 213 2 Diagram 2 shows a displacement-time graph that is plotted for the motion of a boy who is riding a bicycle on a road. 25 100 Displacement, s (m) 10 A E B C D Time, t (s) 20 30 40 Diagram 2 (a) State the type of motion at (i) AB:Uniform velocity//Zero acceleration (ii) BC:At rest [2 marks] (b) What is the distance travelled after 20 s? 100 m [1 mark] (c) What is the total displacement travelled by the boy? Total displacement = (100 – 0) + (100 – 100) + (–25 – 100) = –25 m * A negative sign indicates that the final position is 25 m from the original position in the direction opposite from the original direction of movement. [2 marks] 3 Diagram 3 shows a boy in a moving car driven by his father. The boy is seated without a child restraint system (CRS). The CRS can reduce the risk of injury or death during a collision. Diagram 3 Q&A SPM Phyics F4 C2 2nd.indd 19 14-Jan-23 1:15:58 PM PENERBIT ILMU BAKTI SDN. BHD.

20 Chapter Form 4 2 (a) The car was moving at 90 km h–1 when the father applied the brakes abruptly. (i) Predict what will happen to the boy without the CRS. HOTS Analysing The boy will be thrown forward. [1 mark] (ii) Give a reason for your answer in 3(a)(i). Inertia [1 mark] (b) One of the properties of the material used in the seat belt is the ability to stretch when a large force is exerted. (i) Explain how the property of the material can help to prevent injury. HOTS Analysing The stretch will increase the impact time to reduce the impulsive force. [1 mark] (ii) Calculate the impulsive force if the impact time is 0.2 s and the mass of the boy is 35 kg. u = 90 km h–1 = 25 m s–1 Impulsive force, F = mv – mu t = (35)(0) – (35)(25) 0.2 = –4 375 N [2 marks] Candidates should show clearly the values substituted in the formula used. Exam Tips 4 Diagram 4 shows a cannonball fired from a cannon. When it is fired, the cannon recoils backwards. Diagram 4 (a) Define momentum. Momentum is product of mass and velocity. [1 mark] Q&A SPM Phyics F4 C2 2nd.indd 20 14-Jan-23 1:15:59 PM PENERBIT ILMU BAKTI SDN. BHD.

21 Chapter Form 4 213 (b) The masses of the cannonball and the cannon are 4.0 kg and 96 kg respectively. The cannonball is fired at 120 m s–1. Calculate (i) the total momentum before the cannon is fired, p = m1 u1 + m2 u2 = (96)(0) + (4)(0) = 0 [2 marks] (ii) velocity of the cannon after the cannon is fired. 0 = m1 v1 + m2 v2 0 = (96)v1 + (4)(120) v1 = –5 m s–1 [2 marks] The total momentum before explosion is equal to the total momentum after explosion. Exam Tips (c) State the principle of physics involved in 4(b). Principle of Conservation of Momentum [1 mark] 5 Diagram 5.1 shows an astronaut moving a heavy equipment easily in outer space. Diagram 5.2 shows the same astronaut trying to lift the same equipment on the Earth. Diagram 5.1 Diagram 5.2 (a) Based on Diagram 5.1 and Diagram 5.2, compare the HOTS Analysing (i) mass of the equipment, The mass of the equipment in outer space = mass on the Earth [1 mark] Q&A SPM Phyics F4 C2 2nd.indd 21 14-Jan-23 1:15:59 PM PENERBIT ILMU BAKTI SDN. BHD.

22 Chapter Form 4 2 (ii) gravitational field strength, The gravitational field strength in outer space  on the Earth [1 mark] (iii) weight of the equipment. The weight of the equipment in outer space  on the Earth [1 mark] (b) State the relationship between the weight and the gravitational field strength. The greater the gravitational field strength, the greater the weight. [1 mark] (c) The mass of the equipment is 70 kg. If the gravitational field strength in outer space, gs is 1 5 from gravitational field strength on the Earth, ge , calculate the weight of the equipment in outer space. [ge = 9.81 N kg–1] W = mgs gs = 1 5 (9.81) = (70)(1.962) = 1.962 N kg–1 = 137.34 N [2 marks] Give the final answer correct to at least two decimal places. Exam Tips Paper 2 (Section B & C – Essay Questions) 6 (a) Diagram 6.1 shows air is pumped into the water rocket. Diagram 6.2 shows the water rocket is launched when the valve is released. The water rocket accelerates upwards, while a jet of water expelled downwards. Water rocket Valve Pump Water Water rocket Water jet expelled Diagram 6.1 Diagram 6.2 Q&A SPM Phyics F4 C2 2nd.indd 22 14-Jan-23 1:16:00 PM PENERBIT ILMU BAKTI SDN. BHD.

23 Chapter Form 4 213 (i) What is meant by momentum? [1 mark] (ii) Explain how a water rocket can be launched just by using water. HOTS Analysing [5 marks] (iii) The mass of the water rocket with air pumped is 0.3 kg and the mass of water is 0.5 kg. What is the velocity of the jet of water expelled if the water rocket moves with a speed of 25 m s–1? [2 marks] (iv) What is the maximum height reached by the water rocket? [2 marks] (b) Diagram 6.3 shows the structure of a water rocket model. Nose Body Fin Nozzle Diagram 6.3 Your school will be participating in a water rocket launching competition. Table 1 shows the characteristics of four water rockets, P, Q, R and S. Model Nose Body Fins Nozzle P Pointy Large diameter Arranged asymmetrically Large diameter Q Hemisphere Small diameter Arranged symmetrically Large diameter R Pointy Small diameter Arranged symmetrically Small diameter S Hemisphere Large diameter Arranged asymmetrically Small diameter Table 1 Explain the suitability of each characteristic of the water rocket model that helps it to hit the target precisely. Determine the best water rocket model and give reasons for your choice. [10 marks] Q&A SPM Phyics F4 C2 2nd.indd 23 14-Jan-23 1:16:00 PM PENERBIT ILMU BAKTI SDN. BHD.

24 Chapter Form 4 2 Answers 6 (a) (i) Product of mass and velocity (ii) The air pumped into the water rocket through the nozzle increases the air pressure inside the water rocket. When the valve is released, the water jet expelled through the nozzle at high speed, creating an upwards momentum that is equal to the backwards momentum. (iii) Total momentum before lauching = Total momentum after lauching 0 = m1 v1 + m2 v2 0 = (0.3)(25) + (0.5)(v2 ) v2 = –15 m s–1 (iv) v2 = u2 + 2as (0)2 = (25)2 + 2(–9.81)s s = 31.86 m (b) Characteristics Explanations Pointy nose Reduces air resistance//Aerodynamic shape Small diameter body Reduces air resistance Fins are arranged symmetrically Creates more stability Small diameter nozzle Produces high velocity R is chosen because it has a pointy nose, small diameter, fins that are arranged symmetrically and a small nozzle. 7 Diagram 7.1 shows the water balloon toss game in a telematch between Dhani and his father. The balloon filled with water bursts when Dhani catches it. Diagram 7.2 shows that the balloon filled with water does not burst when Dhani uses a follow-through action when catching the balloon. Father Father Dhani Dhani Diagram 7.1 Diagram 7.2 (a) What is meant by impulse? [1 mark] (b) Based on Diagrams 7.1 and 7.2, HOTS Evaluating (i) compare the impact time, force acting on the balloon filled with water, and the change in momentum if the mass of the balloon and velocity of each toss are identical, (ii) state the relationship between the impact time and the change in momentum, (iii) relate the impact time and the force produced. [5 marks] Q&A SPM Phyics F4 C2 2nd.indd 24 14-Jan-23 1:16:00 PM PENERBIT ILMU BAKTI SDN. BHD.

25 Chapter Form 4 213 (c) Explain how the follow-through action can reduce the risk of the balloon bursting. [4 marks] (d) Diagram 7.3 shows a child is wearing a 3-point harness safety seat belt in a car without a proper child restraint system. Diagram 7.4 shows an incomplete child restraint system. Headrest Latch Seat-belt Diagram 7.3 Diagram 7.4 As an engineer, you are required to build a child restraint system to reduce the risk of injury during a collision and offer comfort to the child. Your design should include the aspects of material, safety features and comfortable design. HOTS Creating [10 marks] 7 (a) Impulse is the change in momentum. (b) (i) The impact time in Diagram 7.1  Diagram 7.2. The force produced in Diagram 7.1  Diagram 7.2. The change in momentum in Diagram 7.1  Diagram 7.2. (ii) The longer the impact time, the greater the change in momentum. (iii) The longer the impact time, the smaller the force produced. (c) When doing the follow-through action, the impact time increases. The change in momentum will also increase. Thus, the force produced will decrease. The balloon can withstand the small force, therefore it will not burst. (d) Specifications Reasons Mesh fabric for material of seat cover Breathable fabric//Less heat trapped Webbing seat-belt material Able to elongate to increase the impact time and reduce the magnitude of impulsive forces Seat-belt using nylon material Durable Isofix//Attachment point// Latch//Clip Secure the child and seat more tightly to the car 5-point harness Forces due to impact distributed to rigid points//To secure child more tightly, to prevent from being thrown out Side head support//Headrest Protect neck and prevent head injury during collision from the side Answers Q&A SPM Phyics F4 C2 2nd.indd 25 14-Jan-23 1:16:00 PM PENERBIT ILMU BAKTI SDN. BHD.

93 Paper 1 Objective Questions 1.1 Resultant Force 1 Diagram 1 shows two forces P and Q acting on an object. The acceleration of object is a. a P Q Diagram 1 Which of the following shows the correct relationship between P and Q? A P = Q B P  Q C P  Q Answer: C The object accelerates when acted on by an unbalanced force. The direction of acceleration of the object is the same as the direction of the resultant force. Resultant force, F = P – Q 2 Diagram 2 shows a car with a mass of 1 200 kg moving with a constant acceleration. The frictional force and thrust acting on the car are 1 000 N and 4 500 N respectively. Frictional force Thrust Diagram 2 What is the acceleration of the car? A 1.67 m s–2 B 2.92 m s–2 C 7.75 m s–2 D 4.58 m s–2 Answer: B F = ma, where F is the resultant force. Resultant force, F = 4 500 – 1 000 = 3 500 N a = F m = 3 500 1 200 = 2.92 m s–2 3 Diagram 3 shows two forces, R and S acting on an object. R S Diagram 3 Which of the following shows the correct vector addition? A C R S F R S F B D R S F R S F Answer: C Use the triangle of forces method. The resultant force is the vector connecting the tail of R to the tip of S. → → → F = R + S Force and Motion II 1 CHAPTER Theme 1: Newtonian Mechanics Textbook Pages: 1 – 37 Q&A SPM Phyics F5 C1 3rd.indd 93 14-Jan-23 1:23:11 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 94 4 Diagram 4 shows a man standing on a weighing scale in an elevator. The reading of the weighing scale is 800 N. Diagram 4 What happens to the reading of the weighing scale when the elevator moves upwards with constant acceleration? A Increases B Decreases C Unchanged Answer: A Resultant force, F = R – mg R = F + mg = m(a + g) Thus, the reading of weighing scale increases. 5 Two forces of magnitude 8 N and 15 N act on an object. Which of the following are possible magnitudes of the resultant force acting on the object? I 7 N II 15 N III 23 N A I and II only B I and III only C II and III only Answer: B The resultant force is a single force which is equal to the vector sum of all forces acting on an object. 1.2 Resolution of Forces 6 Diagram 5 shows a box at rest on a rough inclined plane. The weight of the box is 25 N. 30° W G Diagram 5 What is the frictional force, G acting on the box? A 12.5 N C 21.7 N B 16.1 N D 50.0 N Answer: A Wx = W sin 30° = (25)(0.5) = 12.5 N When the box is at rest, the horizontal component, Wx = frictional force, G Exam Tips 7 Diagram 6 shows a wooden block being pulled by a pulling force, F. The angle between the force and the surface of the floor is 30°. F 30° Diagram 6 Which of the following relationships of the perpendicular components Fx and Fy is true? A Fx = F sin 30° B Fy = F sin 30° C Fy = F cos 30° D Fx = F cos 60° Answer: B sin θ = Fy F Fy = F sin 30° θ F Fy Fx Q&A SPM Phyics F5 C1 3rd.indd 94 14-Jan-23 1:23:12 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 95 8 Diagram 7 shows a box with a weight of 20 N moving down an inclined plane. The acceleration of the box is 0.5 m s–2. G W 45° Diagram 7 What is the frictional force, G acting on the box? HOTS Applying A 13.12 C 15.16 B 14.14 D 16.14 Answer: A Wx = 20 × sin 45° = 14.14 Mass of the box, m = W g = 20 9.81 = 2.04 kg Resultant force, F = ma = (2.04)(0.5) = 1.02 N G = Wx – F = 14.14 – 1.02 = 13.12 N 1.3 Forces in Equilibrium 9 Diagram 8 shows a pail with a mass of 3 kg is hung by two inelastic strings. The pail is in equilibrium. 3 kg 110° T T Diagram 8 What is the tension of the string, T? HOTS Applying A 17.96 N C 35.93 N B 25.65 N D 51.30 N Answer: B When the pail is in equilibrium, Weight, W = 2Ty; 2Ty = (3)(9.81) 2Ty = 29.43 N Ty = 14.715 N (Ty = T cos 55°) Tension of the string, T = Ty cos 55° = 14.715 cos 55° = 25.65 N 10 Diagram 9 shows a pail of water hanging from the end of an inelastic string. The mass of the pail and water is 6 kg. Diagram 9 What is the tension of the string? A 6.0 N C 49.1 N B 15.2 N D 58.9 N Answer: D Tension of the string = Weight of the pail + water W = mg = (6)(9.81) = 58.9 N 11 Diagram 10 shows a photo frame that is hung on the wall with a nail. The tensions of the strings are T1 and T2 respectively. T1 T2 Diagram 10 Q&A SPM Phyics F5 C1 3rd.indd 95 14-Jan-23 1:23:13 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 96 Which of the following shows the correct triangle of forces in equilibrium? A T2 T1 W C T2 T1 W B T2 T1 W D T2 T1 W Answer: C The three forces, T1 , T2 and W acting on the frame are in equilibrium. The direction of the forces follows the sequence in the triangle. 12 Which situation shows forces are in equilibrium? A A motorcycle moving with a constant acceleration B A car moving at a constant velocity C A bus slowing down at bus stop D A runner accelerating from rest Answer: B When the forces acting on an object are in equilibrium, the object is either at rest or moving with uniform velocity. 13 Diagram 11 shows three forces acting on an object. The object is in equilibrium. F1 F3 F2 Diagram 11 Which relationship is correct? A F1 + F2  F3 B F1 + F2 = F3 C F1 + F2 + F3 = 0 D F1 + F2 + F3  0 Answer: C F1 , F2 and F3 are in equilibrium. The resultant force produced is zero. 1.4 Elasticity 14 Which graph shows the correct relationship between the applied force, F, and the extension of a spring, x? A F 0 x C F 0 x B F 0 x D F 0 x Answer: C Hooke’s law states that the extension of the spring is directly proportional to the applied force. 15 Which of the following arrangements, A, B, C or D produces the shortest extension? A B C D 200 g 200 g 200 g 200 g Answer: A The extension of the springs is x 2. Q&A SPM Phyics F5 C1 3rd.indd 96 14-Jan-23 1:23:14 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 97 For springs in parallel, the total extension is x 2. For springs in series, the total extension is nx, where n is the number of springs and x is the extension of one spring. Exam Tips 16 Diagram 12 shows a graph of force, F against extension of a spring, x. 15 0 5 Elastic limit x (cm) F (N) Diagram 12 What is the elastic potential energy stored in the spring? A 37.5 N cm–1 B 40.5 N cm–1 C 75.0 N cm–1 D 80.0 N cm–1 Answer: A Elastic potential energy, Ep = 1 2 Fx Ep = 1 2 (15 × 5) = 37.5 N cm–1 17 Diagram 13 shows a load of mass 100 g, placed on a spring. The initial length of the spring is 12 cm. 100 g 10 cm 12 cm Diagram 13 What is the spring constant of the spring? [Gravitational acceleration, g = 9.81 m s–2] A 9.81 N m–1 B 19.62 N m–1 C 25.50 N m–1 D 49.05 N m–1 Answer: D Spring constant, k = F x , where F is weight of load k = F x = (0.1)(9.81) 0.02 = 49.05 N m–1 18 Diagram 14 shows a graph of force, F against the extension, x of two springs, P and Q. F (N) P Q 0 x (cm) Diagram 14 Which comparison of springs P and Q is correct? A Spring Q is stiffer than spring P B Spring Q is shorter than spring P C The diameter of spring P is smaller than spring Q Answer: C The stiffer the spring, the greater the gradient of graph. When the diameter of spring decreases, the spring constant increases. Gradient of the graph = spring constant Spring constant P  spring constant Q Exam Tips Q&A SPM Phyics F5 C1 3rd.indd 97 14-Jan-23 1:23:15 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 98 Paper 2 (Section A – Structured Questions) 1 Diagram 1 shows three forces, P, Q and R which are in equilibrium. P R Q Diagram 1 (a) What is the meaning of forces in equilibrium? Resultant force is zero [1 mark] (b) In the space below, draw a triangle of forces in equilibrium. Q P R [3 marks] The forces form a closed triangle. The direction of forces follows the sequence in the triangle. The resultant force produced is zero. Exam Tips 2 Diagram 2.1 shows a man pushing a wooden box. The wooden box moves with uniform velocity. The mass of the wooden box is 55 kg. R (b) F W G Diagram 2.1 (a) What is the resultant force acting on the wooden box? 0 N [1 mark] Q&A SPM Phyics F5 C1 3rd.indd 98 14-Jan-23 1:23:15 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 99 (b) On Diagram 2.1, draw and label all the forces acting on the wooden box. [2 marks] (c) Diagram 2.2 shows a wooden box is pulled by two forces, F1 and F2 . F1 = 100 N F2 = 80 N 60° Diagram 2.2 In the space below, draw a triangle of forces to determine the resultant force acting on the wooden box. HOTS Analysing (Scale = 1 cm : 20 N) F1 = 5 cm F2 = 4 cm F = 7.7 × 20 = 154 N F = 7.7 cm 120° [3 marks] 3 Diagram 3 shows a box on an inclined plane. The weight of the box is 80 N. 45° 56.57 N Wy W (b) Wx Diagram 3 (a) What is the meaning of weight? Gravitational force // mass × gravitational acceleration [1 mark] (b) On Diagram 3, label the horizontal and vertical components of the weight of the box, Wx and Wy . [2 marks] (c) (i) Calculate the component of weight parallel to the inclined plane. Wx = W sin 45° = 80 sin 45° = 56.57 N [2 marks] Q&A SPM Phyics F5 C1 3rd.indd 99 14-Jan-23 1:23:16 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 100 (ii) Calculate the resultant force acting on the box. Resultant force, F = Wx – G = 56.57 – 56.57 = 0 N [1 mark] 4 Diagram 4.1 shows a stone suspended on a stretched spring. The initial length of the spring is 10 cm. The spring constant of the spring is 1.5 N cm–1. 12 cm Stone Diagram 4.1 (a) What is the meaning of spring constant? Force required to produce a unit of extension [1 mark] (b) Calculate the mass of the stone. [Gravitational acceleration, g = 9.81 m s–2] Extension of spring, x = 12.0 – 10.0 = 2.0 cm F = kx; F = (1.5)(2.0) = 3.0 N Mass of the stone, m = F g = 3.0 9.81 = 0.306 kg [3 marks] (c) Diagram 4.2 shows two springs, P and Q. Diagram 4.3 shows a graph of the force applied, F against the extension of spring, x of springs P and Q. P Q P F (N) 0 x (cm) Q Diagram 4.2 Diagram 4.3 Q&A SPM Phyics F5 C1 3rd.indd 100 14-Jan-23 1:23:17 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 101 Based on Diagram 4.2 and Diagram 4.3, HOTS Analysing (i) compare the lengths of the springs. Length of spring P  spring Q [1 mark] (ii) compare the gradients of the graphs of springs P and Q. Gradient of spring P  spring Q [1 mark] (iii) compare the spring constants of the springs. Spring constant of spring P  spring Q [1 mark] (iv) state the relationship between the gradient of graph and the spring constant. The gradient of graph increases, the spring constant also increases. [1 mark] (v) relate the spring constant with the length of the spring. HOTS Evaluating The spring constant increases when the length of the spring decreases. [1 mark] Paper 2 (Section B & C – Essay Questions) 5 (a) Diagram 5.1 shows a rubber band that has been stretched. Rubber band Diagram 5.1 (i) What is the meaning of elasticity? [1 mark] (ii) Based on kinetic theory, explain why the rubber band is elastic. [4 marks] Q&A SPM Phyics F5 C1 3rd.indd 101 14-Jan-23 1:23:17 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 102 (b) Diagram 5.2 shows a swing chair. Base Stand Spring Diagram 5.2 Table 1 shows four models of swing chairs W, X, Y and Z. You are required to investigate the characteristics of the swing chairs as shown in Table 1. Swing chair Material of the stand Spring constant of spring (N m–1) Thickness of the spring wire Diameter of base W Iron 1 500 Thin Big X Iron 650 Thick Small Y Steel 1 600 Thick Big Z Steel 800 Thin Small Table 1 Explain the suitability of each characteristic of the swing chairs that makes them safe for use during rest and leisure activities. Determine the most suitable swing chair and give the reasons for your answers. HOTS Evaluating [10 marks] (c) Diagram 5.3 shows 300 g of load hung on a spring. The original length of the spring is 15 cm. 20 cm Diagram 5.3 (i) Name the energy stored in the spring. (ii) Calculate the spring constant of the spring. (iii) Calculate the energy stored in the spring. [5 marks] Q&A SPM Phyics F5 C1 3rd.indd 102 14-Jan-23 1:23:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 103 6 (a) Diagram 6.1 shows a worker pulling a large wooden box on a rough surface. The angle between the rope and the surface is 30°. The frictional force acting on the wooden box is 200 N. 30° Diagram 6.1 (i) What is the meaning of frictional force? [1 mark] (ii) Based on Diagram 6.1, draw and label two perpendicular forces produced. [2 marks] 5 (a) (i) The ability of an object to return to its original shape or size when the applied force is removed. (ii) – There is a repulsive and attractive force acting between the molecules. – When the rubber band is stretched, the distance between molecules increases. – The attractive force increases. – The attractive force pulls the molecules back to their original positions. (b) Characteristics Reasons Material of the stand is steel Stronger//Does not break easily Spring constant of the spring is high Stiffer//Stronger Thickness of the spring wire is thick Stiffer//Stronger Diameter of the base is big More stable Y is chosen because it is made of steel, has a high spring constant, thick spring wire and a base with a larger diameter. (c) (i) Elastic potential energy (ii) k = F x ; k = (0.3)(9.81) 0.05 = 58.86 N m–1 (iii) Ep = 1 2 Fx = 1 2 (2.943)(0.05) = 7.36 × 10–2 J Answers Q&A SPM Phyics F5 C1 3rd.indd 103 14-Jan-23 1:23:18 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 104 (iii) The worker pulled the wooden box with a force of 250 N. Calculate the magnitude of the perpendicular forces produced. [3 marks] (b) Diagram 6.2 and Diagram 6.3 show two identical boxes pulled by a force of 50 N. 50 N = 30° Diagram 6.2 = 45° 50 N Diagram 6.3 Based on Diagram 6.2 and Diagram 6.3, (i) calculate the horizontal force, Fx produced, [2 marks] (ii) compare the angles between the force exerted and the horizontal surface, [1 mark] (iii) state the relationship between the angle between the force exerted and the horizontal surface, θ and the horizontal force, Fx . HOTS Evaluating [1 mark] (c) Diagram 6.4 shows a man pushing a lawn mower. Diagram 6.5 shows the bottom view of the lawn mower. Handle Wheel Blade Bottom view of lawn mower Diagram 6.4 Diagram 6.5 Using appropriate physics concepts, explain the modifications that can be made to the lawn mower that will help the man cut more grass in a short time. HOTS Creating [10 marks] Q&A SPM Phyics F5 C1 3rd.indd 104 14-Jan-23 1:23:19 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 131 105 6 (a) (i) Force that opposes the direction of motion. (ii) Fy Fx F 30° (iii) Fx = F cos 30° = (250)(0.866) = 216.5 N Fy = F sin 30° = (250)(0.5) = 125 N (b) (i) Diagram 6.2: Fx = F cos 30° = (50)(0.866) = 43.3 N Diagram 6.3: Fx = F cos 45° = (50)(0.7071) = 35.36 N (ii) The angle between the force exerted and the horizontal surface in Diagram 6.3 is greater than that of in Diagram 6.2. (iii) When the angle between the force exerted and the horizontal surface increases, the horizontal force decreases. (c) Modifications Reasons Material of the body is low density Lighter and easy to move Diameter of the wheel is bigger Moves faster Diameter of the blade is bigger Cuts more grass Higher power motor Blade rotates faster Higher handle To control the lawn mower easily Answers Q&A SPM Phyics F5 C1 3rd.indd 105 14-Jan-23 1:23:19 PM PENERBIT ILMU BAKTI SDN. BHD.

106 Paper 1 Objective Questions 2.1 Pressure in Liquids 1 Diagram 1 shows a container containing water. 12.0 cm 5.0 cm P Diagram 1 What is the pressure exerted by the water at point P? [Density of water = 1 000 kg m–3, gravitational acceleration = 9.81 m s–2] A 490.5 Pa B 686.7 Pa C 1 177.2 Pa D 1 200.0 Pa Answer: B h is measured from the water surface (12 – 5 = 7 cm = 0.07 m) P = hρg = (0.07)(1 000)(9.81) = 686.7 Pa Common Error Candidates often make a mistake by not changing the unit of height, h from centimetre to metre. 2 Diagram 2 shows a mercury barometer. A B Mercury Trapped air C D Diagram 2 Which point, A, B, C or D has the highest pressure? Answer: D Pressure at D, P = Patm + Pmercury Pressure increases when depth increases. 3 Diagram 3 shows three containers filled with cooking oil. P R Q Diagram 3 Which of the following comparisons is correct about the pressure at points P, Q and R? A P = Q = R C P  Q  R B P  Q  R D Q  R  P Answer: A Pressure at the same depth is equal. Factors affecting pressure in a liquid are depth, h, density, ρ and gravitational acceleration, g. Pressure 2 CHAPTER Theme 1: Newtonian Mechanics Textbook Pages: 38 – 89 Q&A SPM Phyics F5 C2 2nd.indd 106 14-Jan-23 1:24:48 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 213 107 4 Diagram 4 shows a dam. Wall 15.0 m Diagram 4 What is the water pressure at the base of the dam? [Density of water = 1 000 kg m–3, gravitational acceleration = 9.81 m s–2] A 14 715 Pa C 147 150 Pa B 15 700 Pa D 157 000 Pa Answer: C P = ρgh = (15)(1 000)(9.81) = 147 150 Pa 5 Diagram 5 shows a fish in an aquarium. Diagram 5 Which action will increase the pressure acting on the fish? A Remove some water from the aquarium B Place the aquarium at a higher place C Add more water into the aquarium D Use a bigger aquarium Answer: C The depth of the water increases. Pressure is directly proportional to depth of water. 2.2 Atmospheric Pressure 6 Which of the following uses the concept of atmospheric pressure? A C B D Answer: D Higher atmospheric pressure pushes water into the straw. The hydraulic jack uses Pascal's principle. The foot pump uses gas pressure.The hydrometer uses Archimedes' principle. 7 Diagram 6 shows a mercury barometer. Vacuum 75 cm Diagram 6 What is the atmospheric pressure in Pascal? [Density of mercury = 13 600 kg m–3, gravitational acceleration = 9.81 m s–2] A 98 727 C 101 396 B 100 062 D 102 000 Answer: B P = hρg = (0.75)(13 600)( 9.81) = 100 062 Pa Q&A SPM Phyics F5 C2 2nd.indd 107 14-Jan-23 1:24:49 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 132 108 8 Which instrument is used to measure atmospheric pressure? A Aneroid barometer B Bourdon gauge C Hydrometer D Manometer Answer: A Instruments for measuring atmospheric pressure are Fortin barometer, Aneroid barometer and mercury barometer. 9 Diagram 7 shows a syphon being used to drain the water out of a container. Diagram 7 Why does the water continuously flow out of the container? A Higher atmospheric pressure B Difference in pressure C High water pressure D High speed of water Answer: B The difference in pressure at different levels causes the water to flow out of the tube. The water pressure at the lower end of the tube is higher than the atmospheric pressure. 10 A reading of an Aneroid barometer at the top of the mountain is 9.94 × 104 Pa. What is the atmospheric pressure in cm Hg? A 78.0 cm Hg B 76.0 cm Hg C 75.5 cm Hg D 74.5 cm Hg Answer: D P = hρg 9.94 × 104 = h(13 600)(9.81) h = 0.745 m ∴Atmospheric pressure = 74.5 cm Hg 2.3 Gas Pressure 11 Diagram 8 shows gas molecules in a closed metal container. Diagram 8 What causes the gas pressure in the container? A Gas molecules that move randomly B Gas molecules that move with equal speed C Gas molecules that collide with each other D Gas molecules that collide with the inner wall of the container Answer: D Collision between gas molecules and the inner wall of the container will exert pressure. 12 Diagram 9 shows a manometer used to measure the gas pressure in a closed container. Gas 10 cm Mercury Diagram 9 Q&A SPM Phyics F5 C2 2nd.indd 108 14-Jan-23 1:24:50 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 213 109 What is the gas pressure in Pascal? [Atmospheric pressure = 76 cm Hg] A 8.8 × 104 C 1.15 × 105 B 1.01 × 105 D 1.20 × 105 Answer: C P = hρg = (0.76 + 0.1)(13 600)(9.81) = 1.15 × 105 Pa At the same level, pressure is equal. Gas pressure = Atmospheric pressure + Pressure exerted by the mercury column Exam Tips 13 Diagram 10 shows a mercury manometer used to measure gas pressure. X D C B Gas A Diagram 10 At which point, A, B, C or D is the gas pressure equal to the gas pressure at X? Answer: C At the same level, the pressure is equal. Gas pressure at X = gas pressure at C because X and C are at the same level. 14 Which of the following is not factor that affects gas pressure? A Temperature C Speed B Volume D Mass Answer: D The number of gas molecules in a closed container is constant. Therefore, the mass of gas molecules in the closed container is also constant. 2.4 Pascal’s Principle 15 Diagram 11 shows a perforated flask filled with water. The water spurts out from all the holes with equal speed when a compression force is applied to the piston. Piston Diagram 11 Which principle explains the above phenomenon? A Pascal’s principle B Bernoulli’s principle C Archimedes’ principle D Principle of conservation of momentum Answer: A Pressure is transmitted equally in all directions. Water spurts out from the holes at the same speed. 16 Diagram 12 shows a simple hydraulic system. Input force, F1 is exerted on a small piston. The areas of the small piston and big piston are 5.0 cm2 and 30.0 cm2 respectively. Small piston Big piston F1 F2 Diagram 12 What is the output force, F2 when the input force F1 = 10 N? A 60 N C 80 N B 70 N D 100 N Answer: A F2 = A2 A1 × F1 = 30 5 × 10 = 60 N Q&A SPM Phyics F5 C2 2nd.indd 109 14-Jan-23 1:24:50 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 132 110 17 Diagram 13 shows a simple hydraulic jack. The input force, F, is able to support a load of weight, W. The areas of the small piston and big piston are A1 and A2 respectively. Big piston F Small piston W Diagram 13 Which of the following shows the correct relationship between F, W, A1 and A2 ? HOTS Applying A W F = A1 A2 C W = A1 A2 × F B W A1 = F A2 D W = A2 A1 × F Answer: D Pascal’s principle formula: F1 A1 = F2 A2 W = F2 , F = F1 Therefore, F A1 = W A2 W = A2 A1 × F 18 Diagram 14 shows a hydraulic jack that is used to lift a heavy load. Large piston Hydraulic oil Small piston Diagram 14 What is a characteristic of the hydraulic oil used? A High density B Low boiling point C Difficult to compress D High rate of oxidation Answer: C Characteristics of hydraulic oil: difficult to compress, high boiling point, low rate of oxidation and low density. 19 Which action shows an application of Pascal's principle? A Squeezing a toothpaste tube B Squeezing an empty bottle C Lifting rocks in the water D Hitting a tennis ball Answer: A When toothpaste tube is being squeezed, the pressure produced is transmitted to every part of the toothpaste. 2.5 Archimedes’ Principle 20 Diagram 15 shows a submarine moving at a constant depth with uniform velocity. Diagram 15 Which of the following statements is correct? A Weight of the submarine  buoyant force B Weight of the submarine  buoyant force C Weight of the submarine = buoyant force Answer: C At a constant depth, the weight of the submarine is equal to buoyant force because the submarine is in equilibrium. Q&A SPM Phyics F5 C2 2nd.indd 110 14-Jan-23 1:24:51 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 213 111 21 Diagram 16 shows two identical barrels floating on the surface of a river and the sea respectively. River Sea Diagram 16 Which statement is correct about the above observation? A Density of river water = density of sea water B Buoyant force in the river  buoyant force in the sea C Weight of river water displaced  weight of sea water displaced D Volume of river water displaced  volume of sea water displaced Answer: D Volume of water displaced = volume of the object that is submerged 22 Diagram 17 shows a ship floating on the surface of the sea. The weight of the ship and containers is 250 000 N. Containers Diagram 17 What is the volume of sea water displaced? [Density of sea water = 1 020 kg m–3, gravitational acceleration = 9.81 m s–2] A 10 m3 C 50 m3 B 25 m3 D 65 m3 Answer: B V = F ρg = 250 000 (1 020)(9.81) = 25 m3 23 Diagram 18 shows a stone immersed in water. The volume of water displaced is 125 cm3 . Water displaced Stone Diagram 18 What is the buoyant force on the stone? [Density of water = 1 000 kg m–3, gravitational acceleration = 9.81 m s–2] A 1.00 N C 1.23 N B 1.15 N D 12.3 N Answer: C Buoyant force, FB = ρVg = (1 000)(125 × 10–6)(9.81) = 1.23 N 24 Diagram 19 shows a hydrometer floating in a liquid. The readings are marked on the stem of the hydrometer. Diagram 19 Which of the following relationships is correct? A Volume of water displaced = volume of hydrometer Q&A SPM Phyics F5 C2 2nd.indd 111 14-Jan-23 1:24:52 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 132 112 B Weight of the hydrometer  weight of water displaced C Weight of the hydrometer  buoyant force D Weight of water displaced = buoyant force Answer: D Archimedes’ principle states that the buoyant force is equal to the weight of the fluid displaced. 2.6 Bernoulli’s Principle 25 Diagram 20 shows a Bunsen burner. Air is pushed into the Bunsen burner due to atmospheric pressure. Air flow Gas Diagram 20 What is the principle that explains this phenomenon? A Pascal’s principle B Bernoulli’s principle C Archimedes’ principle D Principle of conservation of momentum Answer: B The pressure inside the Bunsen burner is lower because gas flows out of the jet at high speed. Bernoulli's principle states that in a steady flow of a fluid, the pressure of the fluid decreases when the velocity of the fluid increases. 26 Diagram 21 shows a Formula One car. Rear spoiler Diagram 21 What is the function of the rear spoiler? A To produce upwards force B To produce downwards force C To increase the mass of the car D To increase the speed of the car Answer: B The rear spoiler is an inverted aerofoil. The pressure above the spoiler is higher than the pressure below the spoiler. A downwards force is produced to make the car more stable. 27 Diagram 22 shows the aerofoil shape of an aeroplane wing. Air flow Diagram 22 Which of the following statements is correct? A The atmospheric pressure below the wing is higher than above B The speed of air below the wing is higher than above C The pressures above and below the wing are equal D The resultant force on the wing is zero Answer: A The speed of air above the aerofoil is higher than the air below the aerofoil. When the velocity of the air increases, the pressure decreases. Difference in pressure produces a resultant force on the aerofoil. F = PA where P = pressure difference A = cross-sectional area of the aerofoil Exam Tips Q&A SPM Phyics F5 C2 2nd.indd 112 14-Jan-23 1:24:53 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 213 113 28 Which situation shows an application of Bernoulli's principle? A A ship that is floating at sea B A spinning ping-pong ball C A car moving at constant velocity D A hot air balloon that is moving upwards Answer: B A spinning ping-pong ball causes a resultant force to act on the ball. The ball will move in a curved path. Paper 2 (Section A – Structured Questions) 1 Diagram 1 shows an instrument that is used to measure atmospheric pressure. Mercury Vacuum 75.5 cm Diagram 1 (a) Name the instrument. Mercury barometer [1 mark] (b) What is the reading of the atmospheric pressure? 75.5 cm Hg [1 mark] (c) State two applications of atmospheric pressure in daily life. Dropper//Rubber sucker//Drinking straw [2 marks] Q&A SPM Phyics F5 C2 2nd.indd 113 14-Jan-23 1:24:53 PM PENERBIT ILMU BAKTI SDN. BHD.

Chapter Form 5 132 114 2 Diagram 2 shows the apparatus and materials set up for an experiment to determine buoyant force, FB. Stone Water displaced FB (a) Diagram 2 (a) On Diagram 2, draw an arrow to show the direction of the buoyant force acting on the stone. [1 mark] (b) Name the physics principle involved. Archimedes’ principle [1 mark] (c) The stone displaced 250 cm3 of water. Calculate HOTS Applying (i) the weight of water displaced, [Density of water = 1 000 kg m–3, gravitational acceleration = 9.81 m s–2] W = ρVg = (1 000)(250 × 10–6)(9.81) = 2.45 N [2 marks] (ii) the buoyant force, FB. FB = W = 2.45 N [1 mark] Q&A SPM Phyics F5 C2 2nd.indd 114 14-Jan-23 1:24:54 PM PENERBIT ILMU BAKTI SDN. BHD.