PHYSICS PELANGI ONLINE TEST PELANGI ONLINE TEST 4 5 KSSM FORM Bonus AUDIO Chapter Reflection SPM QUICK REVISION Chong Chee Sian Problem Solving Numerical PELANGI
Form 4 1 CHAPTER Measurement 1 1.1 Physical Quantities 2 1.2 Scientific Investigation 3 SPM Practice 6 2 CHAPTER Force and Motion I 10 2.1 Linear Motion 11 2.2 Linear Motion Graphs 13 2.3 Free Fall Motion 16 2.4 Inertia 17 2.5 Momentum 19 2.6 Force 20 2.7 Impulse and Impulsive Force 22 2.8 Weight 23 SPM Practice 24 3 CHAPTER Gravitation 31 3.1 Newton’s Universal Law of Gravitation 32 3.2 Kepler’s Laws 37 3.3 Man-Made Satellites 39 SPM Practice 41 4 CHAPTER Force and Motion I 47 4.1 Thermal Equilibrium 48 4.2 Specific Heat Capacity 50 4.3 Specific Latent Heat 53 4.4 Gas Laws 57 SPM Practice 64 5 CHAPTER Waves 70 5.1 Fundamentals of Waves 71 5.2 Damping and Resonance 73 5.3 Reflection of Waves 74 5.4 Refraction of Waves 75 5.5 Diffraction of Waves 76 5.6 Interference of Waves 77 5.7 Electromagnetic Waves 82 SPM Practice 84 6 CHAPTER Light and Optics 92 6.1 Refraction of Light 93 6.2 Total Internal Reflection 98 6.3 Image Formation by Lenses 102 6.4 Thin Lens Formula 106 6.5 Optical Instruments 107 6.6 Image Formation by Spherical Mirrors 109 SPM Practice 114 Form 5 1 CHAPTER Force and Motion II 120 1.1 Resultant Force 121 1.2 Resolution of Forces 125 1.3 Forces in Equilibrium 126 1.4 Elasticity 127 SPM Practice 131 iii
2 CHAPTER Pressure 140 2.1 Pressure in Liquids 141 2.2 Atmospheric Pressure 143 2.3 Gas Pressure 147 2.4 Pascal’s Principle 148 2.5 Archimedes’ Principle 149 2.6 Bernoulli’s Principle 153 SPM Practice 158 3 CHAPTER Electricity 165 3.1 Current and Potential Difference 166 3.2 Resistance 170 3.3 Electromotive Force (e.m.f) and Internal Resistance 176 3.4 Electrical Energy and Power 179 SPM Practice 180 4 CHAPTER Electromagnetism 188 4.1 Force on a Currentcarrying Conductor in a Magnetic Field 189 4.2 Electromagnetic Induction 193 4.3 Transformer 197 SPM Practice 201 5 CHAPTER Electronics 210 5.1 Electron 211 5.2 Semiconductor Diode 213 5.3 Transistor 216 SPM Practice 220 6 CHAPTER Nuclear Physics 227 6.1 Radioactive Decay 228 6.2 Nuclear Energy 231 SPM Practice 235 7 CHAPTER Quantum Physics 240 7.1 Quantum Theory of Light 241 7.2 Photoelectric Effect 244 7.3 Einstein’s Photoelectric Theory 247 SPM Practice 251 Numerical Problem Solving 258 SPM Model Paper 312 iv
1 Physics SPM Chapter 1 Measurements 4 Form Concept Map Measurement 1 CHAPTER Measurement Physical Quantities Scalar Quantities Vector Quantities Base Quantities Interpreting graph of different shapes Writing a complete experimental report Derived Quantities • Length, l • Mass, m • Time, t • Electric current, I • Temperature, T • Luminous intensity, Iv • Amount of substance, n • Relationship between two physical quantities • Gradient of graph • Area under the graph • Interpolation of graph • Extrapolation of graph Analysing graph to summarise an investigation Example: Area, A = l × l Volume, V = l × l × l Velocity, v = l t Acceleration, a = v t = l t × t Force, F = ma = m × l t × t Scientific Investigation Chapter Outline 1.1 Physical Quantities 1.2 Scientific Investigation Chapter Reflection Theme 1 Elementary Physics 1
2 Physics SPM Chapter 1 Measurements 4 Form 1.1 Physical Quantities 1. Measurement is a method to determine the value of a physical quantity. 2. A physical quantity is a quantity that can be measured. 3. Physical quantities can be classified into base quantities and derived quantities as shown in Table 1.1. Table 1.1 Base quantity Derived quantity Length Volume Mass Density Time Acceleration Temperature Weight Electric current Momentum 4. A base quantity is a quantity that cannot be defined in terms of other base quantities. 5. Table 1.2 shows the seven base quantities with their respective symbols and S.I. units. Table 1.2 Base quantity Symbol of quantity S.I. unit Symbol of S.I. unit Length l metre m Mass m kilogram kg Time t second s Temperature T kelvin K Electric current l ampere A Luminous intensity l v candela cd Quantity of matter n mole mol 6. A derived quantity is a physical quantity derived from a combination of various base quantities. 7. Table 1.3 gives a summary of several derived quantities with their respective symbols and derived units. Table 1.3 Derived quantity Symbol of quantity Relationship with base quantities Symbol of derived unit Volume V l × l × l m3 Density ρ m l × l × l kg m–3 Force F m × l t × t kg m s–2 Example 1 Determine the S.I. unit for area. Solution Area = Length × Width = l × l = [m] × [m] = m2 Common mistake kg ms–2 ✗ kg m s–2 ✓ SPM Tips 8. Physical quantities can be classified into scalar quantities and vector quantities. 9. Scalar quantities are quantities that have magnitude but no direction. 10. Vector quantities are quantities that have both magnitude and direction.
3 Physics SPM Chapter 1 Measurements 4 Form 11. Some examples of scalar and vector quantities are shown in Table 1.4. Table 1.4 Scalar quantity Vector quantity Distance Displacement Speed Velocity Time Acceleration Mass Weight Temperature Momentum Energy Force Power Pressure 1.2 Scientific Investigation 1. Some graphs that are commonly found in experiments: (a) Directly proportional (y ∝ x) 0 y x The graph is a straight line through the origin. y = mx where m is the gradient. (b) Linearly proportional OR 0 m > 0 y x 0 m < 0 c c y x 0 m > 0 y x 0 m < 0 c c y x The graph is a straight line but does not pass through the origin. y = mx + c where m is the gradient and c is the y-intercept. (c) Inversely proportional y ∝ 1 x OR 0 y x 1 0 y x 0 y x 1 0 y x (i) (ii) y = m x where m is the gradient of the graph (i). 2. The process of scientific investigation follows a sequence as shown in the flow map below. Conclusion List and arrangement of apparatus Observation Analysis Discussion Problem statement Inference Hypothesis Variables Procedure / Method
4 Physics SPM Chapter 1 Measurements 4 Form 3. After doing some preliminary study of a problem, an investigation should be started by making an inference. 4. Making an inference is the process of formulating a possible solution to answer the problem that was raised, using known facts or observations gathered in the study. 5. This possible solution is called a hypothesis and its validity has to be tested via experiments. When proven to be valid, it will be incorporated into theories or laws. 6. When designing experiments, an aim is stated to identify the purpose of the experiment. There are three types of variables which need to be identified. 7. These three types of variables are: (i) Manipulated variable (ii) Responding variable (iii) Fixed variable 8. The apparatus and materials that are needed for the experiment are listed out. The arrangement of the apparatus is shown. 9. The procedure is a set of instructions for carrying out the experiment and is normally listed in steps. 10. An observation is the recording and tabulation of the data obtained from the experiment. 11. Analysis of the data is done by: (i) calculating the rates of change, (ii) plotting graphs, (iii) interpretation of data, (iv) comparisons 12. Discussion is carried out to find out whether the hypothesis is accepted or not. 13. Conclusion is a statement concerning the results of the experiment. It summarises whether the hypothesis is valid or not. Simple Pendulum Experiment 1. Diagram 1.1 shows two pendulums, A and B, using similar bobs with identical masses, m. The time taken for both pendulums to make one complete swing is however different. θ m θ m θ m θ m A B Diagram 1.1 2. Problem: Why does the time taken to make one complete swing for both the pendulums is different? 3. The only difference between the two pendulums is the length of the pendulum. 4. By inference, the length of the pendulum affects the time taken to make one complete swing. 5. Period is the time taken to complete one oscillation. What is the purpose of fixed variable? Quiz Quiz
5 Physics SPM Chapter 1 Measurements 4 Form Aim: To investigate the relationship between the length of a pendulum and the time taken for the pendulum to make one complete oscillation. Hypothesis: The longer the length of the pendulum, the longer the time taken for the pendulum to make one complete oscillation. Variables: (i) Manipulated variable: Length of pendulum (ii) Responding variable: Time taken for one complete oscillation. (iii) Constant variable: Number of oscillations / Mass of the pendulum bob / Angle of release Apparatus: Metre rule, stopwatch, protractor, pendulum bob, retort stand, 60 cm thread, 2 pieces of small plywood. Procedure: Plywood Retort Thread stand Metal bob Diagram 1.2 1. The apparatus was arranged as shown in Diagram 1.2. 2. A pendulum of length 10 cm was displaced and allowed to oscillate. 3. The time taken for 10 complete oscillations was recorded. The period of oscillation, T, of the pendulum is calculated. 4. The experiment was repeated with pendulums of length 20 cm, 30 cm, 40 cm and 50 cm. Results: Length of pendulum, x / cm Time taken for 10 complete oscillations / s Period of oscillation, T / s 10 20 30 40 50 Analysis: 0 Period, T / s Length, x / cm Diagram 1.3 Discussion: 1. The graph in Diagram 1.3 shows a curve with a positive gradient. 2. As x increases, T also increases. 3. Precautionary steps: The experiment is carried out in an enclosed room to reduce the effect of wind on the results of the experiment. Conclusion: The hypothesis is supported. The longer the length of the pendulum, the longer the time taken for the pendulum to make one complete oscillation. EXpERIMENT 1.1
6 Physics SPM Chapter 1 Measurements 4 Form Objective Questions 1. Which physical quantity has the correct S.I. unit? Physical quantity S.I. unit A Temperature degree Celcius B Time second C Mass newton D Length centimetre 2. Which of the following physical quantities is a base quantity? A Time C Weight B Momentum D Velocity 3. Which quantity is a scalar quantity? A Acceleration B Force C Mass D Momentum 4. Diagram 1 shows the relationship between physical quantities v and t. v 6 2 t Diagram 1 Which statement about the graph is correct? A If v = 6, then t = 1. B The gradient of the graphs is 3. C v is inversely proportional to t. D The equation of the graph is v = 3t + 6. 5 The mathematical expression for pressure is given as, Pressure, P = F A Where, F = Force A = Area What is the unit of pressure? A N m–1 C kg m–1 B N m–2 D kg m–2 6. Diagram 2 shows the relationship between s and t. s y x t 0 Diagram 2 The relationship between s and t can be expressed by the equation A s = y x t + x B s = x y t + y C s = – y x t + y D s = – x y t + x 7. Which of the following graphs obeys the equation F = ma, where m is a constant? A F a SPMPRACTICE
7 Physics SPM Chapter 1 Measurements 4 Form B F a C F a D F a 8. What is physical quantity? A Quantity that cannot be measured. B Quantity that can be measured. C Quantity that cannot be derived by others base quantities. D Quantity that has magnitude and direction. 9. Which of the following groups is the derived quantities? A Time, mass, length B Length, area, volume C Volume, weight, density D Electric current, voltage, power 10. Which pair of quantities is correct? Vector quantity Scalar quantity A Has magnitude only Has direction only B Has magnitude and direction Has magnitude only C Has direction only Has magnitude only D Has magnitude only Has magnitude and direction Subjective Questions Section A 1. A wooden block of mass 2.45 kg is released from a building and undergoes free fall. The wooden block experiences a gravitational pull of 19.6 N. The time taken for the block to reach the ground is 3 s. The velocity of the block just before it touches the ground is 29.4 m s–1. (a) Classify the physical quantities above into the following categories. (i) Scalar quantity (ii) Vector quantity [2 marks] (b) Convert the velocity of the block just before it touches the ground into km h–1. [2 marks]
8 Physics SPM Chapter 1 Measurements 4 Form 2. Diagram 2 shows an experiment to study the relationship between the length of the pendulum and the time taken for 10 complete oscillations. A pendulum of mass 0.1 kg Length of pendulum A complete oscillation Diagram 2 (a) Match the following physical quantities to the correct variables. Physical quantity Variable (i) The mass of the pendulum The constant variable (ii) The time taken for 10 complete oscillations The manipulated variable (iii) The length of the pendulum The responding variable [3 marks] (b) State the inference and hypothesis for the experiment. [2 marks] (c) The result of the experiment is tabulated by a student as shown below. Length of the pendulum / cm The time taken for 10 complete oscillations / s 30 6.3 35 7.5 40 10 45 14.4 50 19.5 What is the mistake made by the student in tabulating the readings? [1 mark]
9 Physics SPM Chapter 1 Measurements 4 Form 3. Diagram 3 shows the result in the form of graph obtained from an experiment. v 0 0 1 2 3 4 1 2 3 4 5 t Diagram 3 (a) State the relationship between the velocity, v and the time, t. [1 mark] (b) What is the velocity, v when time, t = 3.0 s? Show on the graph how you determine v. [2 marks] (c) Calculate the gradient of the graph. Show on the graph how you determine the gradient. [2 marks] (d) Based on the unit of the gradient in (c), what physical quantity is represented by the gradient? [1 mark] Answers Chapter 1
Physics SPM Chapter 1 Force and Motion II 5 Form Concept Map 1 CHAPTER Theme 1 Newtonian Mechanics Chapter Outline 1.1 Resultant Force 1.2 Resolution of Forces 1.3 Forces in Equilibrium 1.4 Elasticity Chapter Reflection Force and Motion II Resultant Force, F Resolution of Forces Elasticity Equilibrium of Forces Free body diagram Two perpendicular components Relationship between force, F and spring extension, x Three forces in equilibrium Triangle method Parallelogram method F = 0 Fx = F cos q Triangle of forces Hooke’s law F = kx Fy = F sin q Vector Force Table Kit Graph of F against x Area under the graph Elastic potential energy, Ep Ep = 1 2 Fx Ep = 1 2 kx2 Gradient Spring constant, k Stationary Uniform velocity F ≠ 0 Acceleration Force and Motion II 120
Physics SPM Chapter 1 Force and Motion II 5 Form 1.1 Resultant Force 1. Two or more forces can be combined to obtain the resultant force. 2. Resultant force is a single force that represents total vector of two or more forces acted on an object. Determine Resultant Force Combining forces (A) Algebraic Method (1-dimension) (i) Combining forces in the same direction: 3 N 2 N +ve Diagram 1.1 Resultant force, F = F1 + F2 = 2 + 3 = 5 N to the right (ii) Combining forces in the opposite direction: 3 N +ve 2 N Diagram 1.2 Resultant force, F = F1 + F2 = 3 + (–2) = 1 N to the right (iii) Combining forces which are perpendicular to each other θ Fx Fy F Diagram 1.3 Given that Fx = 3 N, Fy = 4 N. F is the combined force. Using Pythagoras’ theorem: F2 = Fx 2 + Fy 2 F2 = (3)2 + (4)2 F = (3)2 + (4)2 = 25 = 5 N 121
Physics SPM Chapter 1 Force and Motion II 5 Form Vector Force Table Kit 1. Diagram 1.6(a) shows the vector force table kit that can be used to determine the combined force of two or more than two forces. 180 170 160 150 140 130 120 110 100 80 70 60 50 40 30 20 10 0 90 350 340 330 320 310 300 290 280 250 260 270 240 230 220 210 200 190 F2 F1 R Metal ring Weight Hanger Slotted masses Lever Diagram 1.6(a) Vector force table kit 350 340 330 320 310 300 290 250 260 270 280 240 230 220 210 200 190 180 170 160 150 140 130 120 110 100 80 70 60 50 40 30 20 10 0 90 R F1 F2 F Diagram 1.6(b) Free body diagram 2. Diagram 1.6(b) shows the arrangement to find out the combined force of F2 and F2. (a) F1 and F2 can be manipulated by changing the slotted masses and angles. (b) R is loaded and adjusted gradually until the metal ring is moved to the centre point of the table. (B) Graphical Method of Forces (i) The Triangle Method F1 F2 F Diagram 1.4 Step 1 The forces are drawn in scale. Step 2 The forces are arranged with the head of the previous force, F1 connected to the tail of the successive force, F2 . Step 3 The resultant force, F is formed by connecting the tail of the first force, F1 to the head of the last force, F2 . (ii) The Parallelogram of Forces Method F2 F1 F Diagram 1.5 Step 1 All the forces (F1 and F2 ) are placed at the same initial point. Step 2 The parallelogram is completed. Step 3 The resultant force, F is the diagonal of the parallelogram. 122
Physics SPM Chapter 1 Force and Motion II 5 Form (c) The magnitude and direction of R is recorded. (d) The combined force, F is the counter force of R. Free Body Diagram 1. The diagrams below show the forces acting on an object in various situations. (a) At rest Normal reaction, R Weight, W (b) Moving with a uniform velocity (F = 0 N) Frictional force, F Thrust, T (c) Moving with a uniform acceleration (F ≠ 0 N) Air friction, R Gravitational force, F Diagram 1.7 Diagram 1.8 Diagram 1.9 Normal reaction, R = Weight, W Resultant force, SF = 0 Thrust, T = Frictional force, F Resultant force, SF = 0 Gravitational force, F Air resistance, R Resultant force, SF ≠ 0 Example 1 Diagram 1.10 shows loaded trolley which has an overall mass of 20 kg. The trolley accelerates with an acceleration of 1.5 m s-2 when it is pushed with a force of 50 N. Find out the frictional force acting upon the trolley. Frictional force = ? 50 N Diagram 1.10 Solution F – R = ma 50 – R = 20 (1.5) R = 50 – 20 (1.5) R = 20 N Example 2 A load of 20 kg is lift by using a smooth trolley as shown in Diagram 1.11. 2 kg 30 N a Diagram 1.11 123
Physics SPM Chapter 1 Force and Motion II 5 Form What is the acceleration of the load if a pull of 30 N is applied? [The acceleration due to gravity, g = 10 m s-2] Solution R is the weight of the load. F – R = ma (30) – 2(10) = 2a 10 = 2a a = 5 m s-2 2. The table below shows the forces acting upon a person in a lift in different situations. 3. R is the normal reaction, which is indicated by the reading of the weighing machine. Moving upwards R mg Constant Velocity Diagram 1.12 R mg Acceleration Diagram 1.13 Forces R = mg R – mg = ma Moving downwards R mg Constant velocity Diagram 1.14 R mg Acceleration Diagram 1.15 Forces R = mg mg – R = ma 124
Physics SPM Chapter 1 Force and Motion II 5 Form Example 3 a < 0 mg R Diagram 1.16 A lift is moving upwards with a deceleration of 0.2 m s-2. If a girl in the lift has a mass of 60 kg, what is the normal reaction, R acting upon her? [The acceleration due to gravity, g = 10 m s–2 ] Solution R – mg = ma R – 60(10) = 60 (–0.2) R – 600 = – 12 R = 600 – 12 = 588 N The apparent weight is smaller than her actual body weight. She feels lighter when the lift is decelerating upwards. Deceleration can be known by the negative sign (–). SPM Tips 1.2 Resolution of Forces 1. A force can be resolved into its components. Resolving a force into its components (a) Graphical Method θ Fx Fy F Y R O X Diagram 1.17 • A force, F, is represented by OR, with its length drawn proportional to the magnitude of the force. • Draw a rectangle/square with OR as the diagonal. • The two sides of the rectangle/ square represent FX and FY. • The magnitudes of FX and FY can be found by measuring their lengths and multiply by the scale factor. (b) Algebraic Method θ Fx Fy F Diagram 1.18 Using trigonometry: Fx = F cos q Fy = F sin q q = tan-1 Fy Fx 125
Physics SPM Chapter 1 Force and Motion II 5 Form 1.3 Forces in Equilibrium 1. An object at rest or moving at constant velocity is in equilibrium. 2. The resultant force acting on the object which is in equilibrium is zero. 3. Diagram 1.20 shows the vector triangle of forces. F3 F1 F2 Diagram 1.19 (a) The vectors of these three forces form a closed triangle. (b) The vector sum is zero. Hence, the net force is zero. (c) F1 , F2, and F3 are in equilibrium. 4. Diagram 1.20 shows an object at rest on an inclined plane with its corresponding free body diagram and triangle of forces. The resultant force acting on the system = 0. θ R F W A (a) Forces in equilibrium θ θ Horizontal axis Vertical axis R F W A (b) Free body diagram R F W (c) Triangle of forces Diagram 1.20 Example 4 A flower pot is hung by two wires over a horizontal beam. The angle between the wires are shown in Diagram 1.21. Weight = mg 30° 30° T T Diagram 1.21 126
Physics SPM Chapter 1 Force and Motion II 5 Aim: Form To study the relationship between force, F and extension of a spring, x. Hypothesis: The extension of a spring is directly proportional to the applied force. Variables: (i) Manipulated variable: Force (ii) Responding variable: Extension of the spring (iii) Fixed variable: Force constant of the spring Apparatus: Spring, retort stand with clamp, slotted weights, pin, metre rule, and plasticine EXPERIMENT 1.1 Procedure: Table Spring Half metre rule Slotted weight Retort stand Pin Diagram 1.22 1. The apparatus was arranged as shown in Diagram 1.22. 2. The initial position of the pin against the metre rule was recorded. Calculate the tension, T of each wire if the mass, m of the pot is 0.5 kg. [The acceleration due to gravity, g = 10 m s–2] Solution 2T cos 30° = mg 2T (0.866) = (0.5) (10) T = 2.886 N 1.4 Elasticity 1. Elasticity is the ability of an object to return to its original shape and size when distorting forces are removed. 2. Materials which able to return to its original shape and size are elastic. 3. Materials which do not behave in this way are plastic. The term is used technically to describe materials which undergo plastic deformation and not the commonly known plastic. 127
Physics SPM Chapter 1 Force and Motion II 5 Form 3. A 20 g weight was suspended on the spring. 4. The new position of the pin against the metre rule was recorded. 5. The extension of the spring was calculated. 6. The experiment was repeated with slotted weight of 40 g, 60 g, 80 g and 100g. Results: Mass of slotted weight, m (g) Weight of slotted weight, F (N) Extension of spring, x (cm) 20 40 60 80 100 Force, F (N) 0 Extension, x (cm) Diagram 1.23 Conclusion: The extension of a spring is directly proportional to the applied force. Hooke’s Law 1. Diagram 1.24 shows the forceextension graph of a spring. F 0 x Limit of propotionality, P Elastic limit, E Gradient = Force constant, k Diagram 1.24 2. The applied force is directly proportional to the extension of the spring. However, beyond point P (see Diagram 1.24) the graph is non-linear. 3. Beyond point E (see Diagram 1.24), the spring is said to have a permanent extension and not able to return to zero extension. 4. A material obeys Hooke’s law if the material produces a straight-line graph as shown in Diagram 1.25. Force, N 0 Extension, m Diagram 1.25 Hooke’s Law states that the force applied to maintain extension is directly proportional to the extension. F ∝ x where, F = Force x = Extension therefore, F = kx where, k = Force constant Physics SPM Chapter 1 Force and Motion II 128
Physics SPM Chapter 1 Force and Motion II 5 Form Force constant is the force that required to produce one unit of extension to a spring. k = F x The S.I. unit of force constant = N m–1 5. A spring which has a higher value of force constant, k is stiffer. 6. The table below summarises the factors that affect the elasticity of a spring. Factors Change in factor Elasticity Material of spring Different materials of spring yield different elasticity. Length of spring Longer More elastic Shorter Less elastic Diameter of spring wire Larger diameter Less elastic Smaller diameter More elastic Diameter of spring coil Larger diameter More elastic Smaller diameter Less elastic 7. Diagram 1.26 shows a load supported by the arrangement of springs A, B, and C which is made up of identical springs and their respective force-extension graph. M A B M C M Force, N 0 Extension, m C (k = 2 N m–1) B (k = 1 N m–1) A (k = N m–1) 1 2 Diagram 1.26 Why spring on a cradle have a high spring constant? Quiz Quiz 129
Physics SPM Chapter 1 Force and Motion II 5 Form Elastic Potential Energy 1. When a force is used to extend or compress an elastic material or object, work is done and energy is stored in the object as elastic potential energy. 2. The elastic potential energy stored in the object can easily be found from the force-extension graph as shown in Diagram 1.27. Force, F 0 Extension, x Gradient = k Area under the graph = Elastic potential energy Diagram 1.27 3. Elastic potential energy is given by: Ep = 1 2 Fx or Ep = 1 2 kx2 4. EP is in Joules if F is measured in Newton, x is in metres and k is in Newton per metre. Solving Problems Involving Force and Extension of a Spring Arrangement of Identical Springs in Series B (a) (b) Diagram 1.28 Tension = F (a) Extension = x (b) Extension = x Extension of spring system = x + x = 2x Arrangement of Identical Springs in Parallel A Diagram 1.29 Tension = F 2 Extension of spring system = x 2 Example 5 Diagram 1.29 shows two spring systems made using identical springs. A P M D Q Diagram 1.30 Which of the arrangement has the least extension? Solution Spring Q system has the least extension because three springs are arranged in parallel. 130
Physics SPM Chapter 1 Force and Motion II 5 Form Objective Questions 1. Diagram 1 shows two forces, F1 and F2 , acting upon a wooden block that is placed on the surface of a table. The frictional force between the wooden block and the table surface is 2 N. F1 F2 Diagram 1 Among the following pairs of F1 and F2 , which one will cause the wooden block to move with a uniform velocity? F1 F2 A 3 N 7 N B 4 N 5 N C 5 N 3 N D 7 N 4 N 2. Diagram 2 shows a car with a mass of 1000 kg moving with an acceleration of 2 m s-2. There is a frictional force of 900 N acting on the car. Frictional force, 900 N Acceleration Diagram 2 What is the force exerted by the engine of the car? A 900 N B 1100 N C 2000 N D 2900 N 3. Diagram 3 shows a box being pulled by a boy. 4.0 kg 60° 5.0 N 30 N Diagram 3 What is the acceleration of the box? A 2.50 m s–2 B 3.75 m s–2 C 5.00 m s–2 D 6.25 m s–2 4. Diagram 4 shows a picture frame is hung on the wall. Tension, T2 Tension, T1 Weight, W Diagram 4 The forces acting upon the frame are in equilibrium. Which of the following shows the correct vector diagram which describes the forces acting on the picture? SPMPRACTICE 131
Physics SPM Chapter 1 Force and Motion II 5 Form A T2 W T1 B T2 W T1 C T2 W T1 D T2 W T1 5. Diagram 5 shows a picture hanging on a wall. Tension, T1 Tension, T2 Weight, W Nail String Diagram 5 Which is correct about the forces? A T1 + T2 + W = 0 B T1 = T2 = W C T1 + T2 = W D T1 T2 > W 6. Which diagram shows the unbalanced forces acting on an object? A A stack of book on the surface of a table. Normal reaction Weight HOTS HOTS 132
Physics SPM Chapter 1 Force and Motion II 5 Form B A dropping durian. C A car moving with constant velocity. Thrust Frictional force D An airplane navigating at a fixed altitude Weight Lift Upthrust Drag 7. Diagram 6 shows a car on tow. Diagram 6 Which of the following vector diagrams represents the resultant force, F, which acts on the car? A F HOTS B F C F D F 8. Which variable below is fixed when a spring is extended? A Force constant B Elastic potential energy stored in the spring C Extension of the spring D Length of the spring 9. Diagram 7 shows a spring that is being compressed. Determine the weight of object T. 14 cm 10 cm 8 cm 40 N T Diagram 7 A 40 N B 50 N C 60 N D 70 N 133
Physics SPM Chapter 1 Force and Motion II 5 Form 10. Diagram 8 shows load M supported by the arrangement of springs, P, Q, and R. All the springs are identical. R M Q M P M Diagram 8 Which comparison is correct about the extension of P, Q, and R? A P Q R B Q R P C R Q P D Q P R Subjective Questions Section A 1. Diagram 1.1 and Diagram 1.2 show a boy standing on a weighing scale in two different situations. Constant Velocity 50 kg a = 2 m s-2 50 kg Diagram 1.1 Diagram 1.2 (a) (i) Label and name the forces acting on the boy in Diagram 1.1. [1 mark] HOTS (ii) State the relationship between the forces given in your answer in (a)(i). [1 mark] (b) What is the reading of the weighing scale in Diagram 1.2? [Assumed that g = 9.81 N kg–1] [1 mark] (c) (i) State the change in the scale’s reading in Diagram 1.1 when the lift accelerates upwards suddenly? [1 mark] (ii) Explain your answer in (c) (i). [1 mark] Section B 2. Diagram 2.1 shows two arrangement of spring system attached with the same weight of load. All spring used in both systems are the same in terms of force constant. Spring T System J Slotted weight Spring T System K Slotted weight Diagram 2.1 (a) What is meant by force constant? [1 mark] (b) Using graph of force, F against extension, x, show the comparison and difference of force constant for system J and system K. Give a relationship between force constant for both systems. [4 marks] 134
Physics SPM Chapter 1 Force and Motion II 5 Form (c) Table 1 shows arrangement of spring system S, T, U, and V with difference specification. Table 1 Spring arrangement 100 N S 100 N T 100 N U 100 N V Force applied 100 N 100 N 100 N 100 N Force constant 166.17 N m–1 666.67 N m–1 200 N m–1 1 000 N m–1 Density Low Low High High HOTS Diagram 2.2 shows the usage of spring in portable cradle that can withstands 10 kg baby. Diagram 2.2 You are assigned to study the design and characteristics of the spring system S, T, U, and V in Table 1 based on the following aspects: • Extension of spring when a baby’s weight is applied on the spring. • Density of the wire of the spring. Explain the suitability of each aspects and determine the most suitable arrangement to be used for the cradle in Diagram 2.2. [10 marks] (d) Diagram 2.3 shows spring with force constant of 20 N cm-1 that compress when a 4 kg block is placed on top of the spring. Extension, x 4 kg Diagram 2.3 Calculate: (i) the compression, x of the spring. [3 marks] (ii) elastic potential energy that stored in the spring. [2 marks] 135
Physics SPM Chapter 1 Force and Motion II 5 Form 3. (a) (i) What is meant by force constant? [1 mark] (ii) using the force-extension graph, explain the relationship between applied force on the spring with the spring extension. [4 marks] (b) Diagram 3.1 shows the arrangements of spring system S, T, U, and V. The springs are varying in force constant, rate of rusting, type of materials, and cost. 100 N 100 N 100 N 100 N S T U V Diagram 3.1 You are required to set up a baby cradle which enables a baby with a weight of 100 N to use it safely. Table 2 Force constant (N m–1) Rate of rusting Diameter of spring wire (cm) Cost (RM per metre) S 100 High 1.0 15.00 T 300 Low 2.0 5.00 U 50 High 1.5 3.00 V 25 Low 1.5 20.00 HOTS HOTS Using the information in Diagram 3.1 and Table 2: (i) Choose the most suitable design among the spring systems. Ex p l a i n t h e suitability of the design. (ii) Explain why the other designs are not suitable. [10 marks] (c) Diagram 3.2 shows a spring with a force constant of 20 N cm-1 is compressed when a block of mass 4 kg is placed on the spring. Extension, x 4 kg Diagram 3.2 136
Physics SPM Chapter 1 Force and Motion II 5 Form Calculate: (i) the compression, x of the spring. (ii) the elastic potential energy stored in the spring. [5 marks] Section C 4. (a) Diagram 4.1 shows two identical steel balls placed onto spring M and spring N. Both springs are elastic and fixed to a horizontal surface. The springs are pushed down until the lengths of spring M and spring N are the same. Diagram 4.2 shows the maximum displacement reached by the balls when the hands are released. [Assume spring M and spring N are of the same material, same coil diameter and same original length] Spring M Thin spring wire F1 Spring N F2 Thick spring wire Diagram 4.1(a) Diagram 4.1(b) Spring M v Diagram 4.2(a) v Spring N Diagram 4.2(b) 137
Physics SPM Chapter 1 Force and Motion II 5 Form (i) What is the meaning of elasticity? [1 mark] (ii) Based on Diagram 4.1 and Diagram 4.2, compare the thickness of the spring wire and the maximum displacement of the payload. Relate the thickness of the spring wire with the maximum displacement of the payload to make a deduction regarding the relationship between the thickness of the spring wire and the elastic potential energy of the spring. [5 marks] (b) The forces used to compress the springs in Diagram 4.1(a) and Diagram 4.1(b) are F1 and F2 respectively. (i) Compare F1 and F2 . Give one reason for the answer. [2 marks] (ii) Based on Diagram 4.1 and Diagram 4.2, state the energy changes that take place from the moment the spring is compressed until the ball reaches its maximum height. [2 marks] (c) Diagram 4.3 shows a gymnastic athlete is doing a jump. Diagram 4.3 Using the suitable physics concept, explain the use of the equipment and suitable technique to improve the technique of jumping safely. HOTS Your answer must include the following aspects. (i) The runway (ii) Vaulter’s attire (ii) Vaulter’s movement (iii) Elasticity of the pole used (iv) Thickness of the mat [10 marks] 5. Diagram 5.1 shows two identical steel balls on spring M and spring N. The springs are pushed until the length of spring M and spring N is the same. Diagram 5.2 shows the maximum displacement when hand is released. [Assume spring M and spring N are from the same material, the same coil diameter and the same original length.] Spring M Steel ball Spring N Steel ball Diagram 5.1 Spring M Steel ball Spring N Steel ball Diagram 5.2 138
Physics SPM Chapter 1 Force and Motion II 5 Form (a) What is meant by elasticity? [1 mark] (b) Using Diagram 5.1 and Diagram 5.2, compare the thickness of the spring wire with the maximum distance achieved by the balls. Relate the thickness of the spring wire with the maximum displacement of the balls to make a conclusion regarding the relationship between the thickness of the spring wire and elastic potential energy of the spring. [5 marks] (c) Forces applied to compress the spring M and spring N are F1 and F2 . (i) Compare F1 and F2 . Give a reason for your answer. [2 marks] (ii) Based on Diagram 5.1 and Diagram 5.2, state the changes in energy occurs from the moment the spring is compressed until the balls achieved the maximum displacement. [2 marks] (d) Diagram 5.3 shows a spring balance to weight daily use things. Diagram 5.3 Diagram 5.4 By using the suitable physics concept, state the modifications that can be made to change the spring balance into industrial balance as shown in Diagram 5.4. Your answer must include the following aspects: (i) Density of the materials. (ii) Rate of rusting of spring materials. (iii) Spring force constant. (iv) Characteristics of the reading scale to help ease the measurement. (v) Methods to put on and put off the load. [10 marks] Answers Chapter 1 139
SPM Model Paper 1. Which of the following physical quantities is a base quantity? A Time B Momentum C Weight D Velocity 2. Which of the following quantities is a scalar quantity? A Acceleration B Force C Mass D Momentum 3. Diagram 1 shows the relationship between physical quantities s and t. s t 6 2 Diagram 1 Which of the following statements is correct in describing the graph? A If s = 6, then t = 1 B s is inversely proportional to t C The gradient of the graph is 3 D The equation of the graph is s = 3t + 6 Paper 1 1 hour 15 minutes Question 1 to Question 40 are followed by four answer options, A, B, C, and D, or three answer options, A, B and C. Choose the best answer for each question. 4. Diagram 2 shows a strip of ticker tape that is dragged by trolley from a ticker timer with a frequency of 50 Hz. 1.0 cm 4.0 cm Direction of motion Diagram 2 Acceleration of the trolley is A 200.0 cm s-2 B –200.0 cm s-2 C 1500 cm s-2 D –1500.0 cm s-2 5. Which of the following velocity-time graphs shows the movement of an object with uniform acceleration? A v t C v t B v t D v t SPM MODEL PAPER 312
Physic SPM SPM Model Paper SPM Model Paper 37. The following equation represents the decay of an uranium nucleus. 238 92U → 230 90Th + x 0 –1e + y 4 2 He What are the values of x and y? x y A 0 1 B 1 1 C 1 2 D 2 2 38. In a nuclear fission, 0.005 u mass of a radioactive element is changed to nuclear energy. Calculate the energy released. [1 a.m.u. = 1.7 × 10-27 kg, speed of light = 3.0 × 108 m s-1] A 2.55 × 10 -21 J B 5.10 × 10 -19 J C 3.83 × 10 -13 J D 7.65 × 10 -13 J 39. Which of the following statements is true regarding the black-body radiation? A An ‘ideal’ radiator is also an ‘ideal’ absorber of radiation. B Black-body radiation does not depend on the temperature. C Black-body radiation can be well explained by both classical and quantum theory. D An infinite amount of energy is radiated at high frequency. 40. Metal X has a threshold frequency of 5.8 × 1014 Hz. Calculate the work function of metal X. [Planck’s constant, h = 6.63 × 10–34J s] A 3.85 × 10–19 J B 6.54 × 10–19 J C 7.53 × 10–19 J D 6.91 × 10–19 J Paper 2 2 hours 30 minutes Section A (60 marks) Answer all the questions in this section. 1. Diagram 1 shows a container of margarine. There is a label showing several physical quantities on the container. Minimum volume contained = 1.5 litre Vitamin D per serving: mass = 12 mg Keep at temperature < 25°C Diagram 1 Physical quantities can be classified as base quantity and derived quantity. (a) What is the meaning of base quantity? [1 mark] (b) Based on Diagram 1, (i) classify all the physical quantities into Table 1. Table 1 Base quantity Derived quantity [2 marks] 321
Physic SPM SPM Model Paper SPM Model Paper (c) Based on your answers in (b), choose the most suitable metal to be applied. [1 mark] 8. Mercury-in-glass thermometer is a tool that applies the thermometric property of mercury to determine the temperature. (a) State one thermometric property that is applied in the mercury-inglass thermometer. [1 mark] (b) An uncalibrated thermometer is attached to a centimetre scale. The length of the mercury column reads 5.0 cm in pure melting ice and 30.0 cm in steam. What is the length of mercury column when the thermometer is immersed in a liquid of temperature 40°C? [2 marks] (c) Diagram 8 shows a prototype of thermometer which is produced in a factory. 110 100 90 80 70 60 50 40 30 20 10 0 –10 –20 Rajah 8 Based on the following aspects, give suggestions on how the sensitivity of a thermometer can be increased. (i) Total mass of the mercury: Reason: [2 marks] (ii) Thickness of the capillary tube: Reason: [2 marks] (iii) Thickness of the bulb wall: Reason: [2 marks] Section B (20 marks) Choose any one question from this section. 9. During a hot daytime, a driver found that the road surface in front of him seems shiny and wet but the road is actually dry. Rajah 9.1 (a) Name the light phenomenon involved in the situation. [1 mark] (b) Diagram 9.2 shows a light signal travelling through an optical fibre. Core total internal reflection occurs here cladding cladding Rajah 9.2 Based your knowledge and understanding about total internal reflection, explain how the light ray can be retained in the core. [4 marks] 326
Physic SPM SPM Model Paper SPM Model Paper Explain the modification need to be done to the a.c. generator to change into a direct current (d.c.) generator that can produces stable direct current. Your answer should be based on the following aspects: (i) Modification of the external circuit to produce stable current. (ii) Modification to the slip ring. (iii) Number of turn of wire in the coil. (iv) Strength of magnet. (v) Shape and position of the magnet used. [10 marks] Answers SPM Model Paper 330
Bahasa Melayu English Matematik Mathematics Sains Science Sejarah Pendidikan Islam Biologi Biology Fizik Physics Kimia Chemistry Matematik Tambahan Additional Mathematics Ekonomi Perniagaan Prinsip Perakaunan Concise Notes HOTS & i-THINK SPM Practices Numerical Problem Solving SPM Model Paper Answers Audios, Videos & Quizzes QR code SPM QUICK REVISION PelangiPublishing PelangiBooks PelangiBooks How to Access MCQ Quiz W.M: RM19.95 / E.M: RM20.95 KC118544 ISBN: 978-629-470-518-0 The RANGER SPM series allows students to do express revision before the SPM examination. The content of this book is aligned with the Standard Curriculum for Secondary Schools (DSKP) for Form 4 and Form 5, textbooks, and the latest SPM assessment format. It also includes engaging digital resources to enhance understanding of the subject. Form 4 b*f6#P34 Form 5 mwCS5gN# Purchase eBook here! Scan, register and insert Enrolment Key