Public Assessment of the HKDSE Physics Examination

Transcription

1 Public Assessment of the HKDSE Physics Examination 1. Public Assessment The public assessment of the Hong Kong Diploma of Secondary Education (HKDSE) Physics Exam consists of a public examination component and a school-based assessment component. A. Public Examination component In Physics, the mark of the public examination component contributes 80% of the final subject mark. Paper Section Question types Weighing Duration Curriculum Section A Multiple-choice Questions 21% 2 hours Paper 1 Short Questions and Structured Compulsory Part Section B 39% 30 minutes / Essay-type Questions Paper 2 Ð Multiple-choice Questions and Structured Questions 20% 1 hour Elective Part (2 out of 4) B. School-based Assessment component In Physics, students are assessed by their teachers on their performance of a wide range of skills involved in practical and non-practical related tasks throughout S.5 and S.6. The mark of the school-based assessment component contributes 20% of the final subject mark. Year of examination Assessment types Weighing 2012 and 2013 Practical related component 20% Starting from 2014 Practical related component 18% Non-practical related component 2% 2. Standards-referenced Reporting The HKDSE makes use of standards-referenced reporting, which means candidates levels of performance will be reported with reference to a set of standards as defined by cut scores on the variable or scale for a given subject. The following diagram represents the set of standards for a given subject: Cut scores U Variable/ scale Within the context of the HKDSE there will be five cut scores, which will be used to distinguish five levels of performance (1 5), with 5 being the highest. The Level 5 candidates with the best performance will have their results annotated with the symbols and the next top group with the symbol. A performance below the threshold cut score for Level 1 will be labelled as Unclassified (U). I

2 Exam Strategies Section A Multiple-choice Questions 1. The statements in the options may be misleading. Do not spend time in reading the 4 options only. To tackle the problem, follow the steps below: Read the question (not options) twice or more. Recall the related situation / knowledge. Organize the memory. Read the options. Cross-out the incorrect options if necessary. Choose the correct answer. Example: Which of the following statements is a correct definition of specific heat capacity? A. The heat required to rise the temperature of a substance by 1 C. B. The heat required to rise the temperature of a 1 kg substance by 1 C. C. The temperature change of a substance when it gains 1 J of heat. D. The temperature change of a 1 kg substance when it gains 1 J of heat. Ans: B 2. The given statements may be correct. However, each of them may or may not be a relevant explanation to the situation. To tackle this kind of question, think critically about the relationship between the situation and each of the options. Example: Sweating helps us to maintain our body temperature in hot weather. Which of the following statements correctly explain this phenomenon? (1) The boiling point of water is higher than our body temperature. (2) Water has a high specific latent heat. (3) Evaporation of water may occur at 37 C. A. (1) only B. (1) and (2) only C. (2) and (3) only D. (1), (2) and (3) Ans: C II

3 1 Heat Heat Temperature Heat, internal energy and power Temperature and thermometers Temperature is the degree of hotness. In the Celsius scale, the melting point of water is 0 C and the boiling point of water is 100 C. Examples of temperaturedependent properties: volume, electrical resistance, etc. Temperature is associated with the average kinetic energy of molecules due to random motion. Calibration of a thermometer: T = X - X C X X 0 Types of thermometers: Liquid-in-glass thermometer (Clinical thermometer) Thermistor thermometer Resistance Thermometer Infra-red thermometer Rotary thermometer Heat and internal energy Heat is the name given to the energy transferred between two objects at different temperatures. Internal energy is the name given to the energy stored in an object. Internal energy = Kinetic energy of molecules due to random motion + Potential energy of molecules in a system Average kinetic energy and potential energy of molecules are associated with the temperature and the physical state of the system respectively. Power Power P is the rate of energy transferred, i.e., Power = Energy Time Unit: watt, W or joule per second, J s -1 Heat capacity and specific heat capacity E Heat capacity C = (T final - T initial ) Unit: J C -1 where E: heat required T final : final temperature T initial : initial temperature E Specific heat capacity c = where m: mass of the object m(t final - T initial ) Unit: J C -1 kg -1 Energy transfer during heating and cooling: E = mcdt where DT = T final - T initial Practical importance of the high specific heat capacity of water Mild climate in coastal regions High water content in living organisms Coolant in power plants, car engines, etc. Fire extinguishing

4 Heat Transfer processes Change of state Means of heat transfer Conduction: Heat is transferred from a hot object to a cool object through the contact surface. Convection: Heat is transferred by motion of heated part in liquid or gas. A convection current is formed. Radiation: Heat is transferred by infra-red radiation emitted by a hot body. It is the only means to transfer heat through vacuum. Greenhouse effect Part of the solar energy emitted from the Sun is absorbed by the Earth. Gases like carbon dioxide (CO 2 ), chlorofluorocarbons (CFCs), methane (CH 4 ), nitrous oxide (N 2 O) and water vapour in the atmosphere absorb infra-red radiation emitted from the warmed Earth, and this warms up the atmosphere. The atmosphere radiates infra-red to the space and back to the lands and oceans. As a result, the Earth is kept warm. Enhanced greenhouse effect causes global warming. Between solid state and liquid state Melting is the process in which a substance changes from a solid to a liquid state and latent heat of fusion is gained. Freezing (solidifying) is the process in which a substance changing from liquid to solid state and latent heat of fusion is released. Between liquid state and gaseous state In vaporization, latent heat of vaporization is gained. In condensation, latent heat of vaporization is released. In evaporation, a substance changes from liquid to gaseous state below boiling point. Rate is increased by increasing temperature, surface area and air current over the liquid. Energetic molecules at liquid surface escape to form vapour. Latent heat and specific latent heat Latent heat is heat transferred when a substance changes state. Specific latent heat of fusion l f or vaporization l v : l f (or l v ) = E m where E: heat required m: mass of the substance which has changed state Unit: joule per kilogram, J kg -1 Energy transfer during change of state: E = ml where l: specific latent heat Latent heat is gained during a change of state due to an increase in the potential energy of molecules of a substance.

5 Physics: Key Notes & Exambuilder Do you know the meaning of temperature? How do we define the Celsius scale? Can we use substances with temperature-dependent properties to measure temperature and make thermometers? What are the conditions for a heating process? What is the meaning of the internal energy of a system? What are the factors affecting it? What is the relationship between temperature and the motion of molecules in a system? 1.1 Temperature, Heat and Internal Energy A Temperature Temperature is a quantity to describe the degree of hotness. It measures how hot a body is. The molecules in an object are in constant random motion. The hotter the object, the more vigorously the molecules move, i.e., they have a higher average kinetic energy. The temperature of a body increases with the average kinetic energy of molecules in the body. The concept of temperature is different from that of energy. Energy can be transferred between two bodies but temperature cannot be transferred. B Temperature scale Fixed points of the Celsius scale: Lower fixed point (0 C) Melting point of pure ice under normal atmospheric pressure Upper fixed point (100 C) Boiling point of pure water under normal atmospheric pressure All types of thermometers should give the same reading at these two fixed points. The temperature difference between the lower fixed point and the upper fixed point is divided into 100 equal divisions in Celsius scale. 4 Figure 1.1

6 Heat D Joulemeter and kilowatt-hour meter If the power of the electric heater is not known, either a joulemeter or a kilowatt-hour meter can be used to measure the energy supplied by the heater. The means of connection was asked in past exam questions. The meter should be connected between the heater and the power supply. Figure 1.7 A joulemeter measures the electrical energy passing through in the unit of joule (J). A kilowatt-hour meter measures the electrical energy passing through in the unit of kwh. 1 kwh = Energy consumed by a 1 kw appliance in 1 hour 1 kwh = 1 kw 1 h = J Measuring specific heat capacity of a solid (metal block) Procedure: Prepare the following set-up: Record the following quantities: Figure 1.8 m : mass of the block E initial and E final : Initial and final joulemeter readings T initial and T final : initial temperature and final temperature of the block Calculate the result: Specific heat capacity of the block, c = Q m DT, where Q = E final - E initial The sources of errors, effects on the result and methods of improvement are frequently asked in past exam questions. Students have to tackle the problem according to the given scenario. The reading of the meter is cumulative and it cannot be reset to zero. So the energy transferred is the difference between the final reading and the initial reading of the meter. 13

7 Physics: Key Notes & Exambuilder If three of the above quantities are known, the fourth one can be calculated by the relation: V 1 T 1 = V 2 T 2 V 1 and V 2 must be expressed in the same unit. T 1 and T 2 must be expressed in Kelvin scale. B Explanation in terms of molecular motion When the temperature of a gas is increased at a fixed external pressure, the average speed of the molecules increases. They collide more frequently with the wall of the container and produce a higher pressure. The instantaneous pressure difference causes the volume to increase. As the volume increases, each molecule takes a longer time to go from one wall to the opposite. Collisions are less frequent. Pressure drops until it is the same as the external pressure. Since the temperature is increased, the average speed of molecules increases. For details, refer to the kinetic theory in Section 2.6. Guided Example 9 A fixed mass of gas is heated from -30 C to 120 C at constant pressure. If initial volume of the gas is 3000 cm 3, what will be the final volume? Ans. Pressure is constant, apply CharlesÕ law: \ V 1 = V 2 T 1 T (-30) = V V 2 = 4850 cm 3 Convert temperature to the Kelvin scale. It is not necessary to convert the unit of volume into m 3. The gas law only relates the ratio of the initial and final volumes. Guided Example 10 Figure 2.24 A fixed mass of gas is trapped inside a tube by a little mercury thread as shown above. The mercury thread can slide smoothly along the glass tube. The volume of the gas at 25 C is 5 cm 3. What will be the volume if the gas is heated to 85 C? (You may assume that the atmospheric pressure remains unchanged during the heating process.) The pressure of the gas inside the tube is equal to the external pressure. If there is a pressure difference, the mercury thread will move to a new position to balance the two pressures. The external pressure is assumed constant. Hence the pressure of the gas in the tube is constant. 62

8 Physics: Key Notes & Exambuilder B Braking a Car Figure 3.60 When a car brakes hard, it leaves a skid mark. The length of the skid mark is the braking distance. Stopping distance = Thinking distance + Braking distance Figure 3.61 Note: The stopping distance is NOT directly proportional to the initial velocity. Doubling the value of u will be more than double of the stopping distance. Road Safety In a traffic accident, the skid mark left by a vehicle provides important information. The length of this skid mark is equal to the braking distance s 2. The vehicle retards to stop in this distance. The police will carry out some tests with the vehicle to find out the magnitude of the retardation. The Figure 3.62 vehicle moving with different speed is braked to stop on a road with similar conditions. The skid marks are recorded and analyzed. The magnitude of retardation a can be found. Thus the speed u of the vehicle just before the accident can be determined by the formula v 2 - u 2 = 2(-a)s 2, where v is zero. This is the evidence to decide whether the driver was driving at a speed higher than the speed limit of the road. The initial position of the vehicle may also be identified according to the statements of witnesses and the nature of environment. Thus the thinking distance s 1 can be found. The driver s reaction time t can then estimated by the formula t = s 1 u. 106

9 Work, Energy and Power What is gravitational potential energy? Do you know that gravitational potential energy is the energy possessed by an object due to its position under the action of gravity? How to calculate gravitational potential energy? How to solve problems involving gravitational potential energy? What is kinetic energy? Do you know that kinetic energy is the energy possessed by an object due to its motion? How to calculate kinetic energy? How to solve problems involving kinetic energy? The P.E. of an object far away from Earth is given by the formula - GMm. r The discussion of gravitational potential energy is in the elective part Astronomy and Space Science. 5.2 Mechanical Energy A Gravitational potential energy Gravitational potential energy (P.E.) is the energy possessed by a body due to its vertical position under the action of gravity. It is the mechanical work done by a force F to raise a body with uniform motion by a vertical displacement h: Magnitude of F = W = mg Mechanical work done by F = F h = mgh Figure 5.12 To raise a body with uniform motion, the net force is zero. Hence the magnitude of the force is equal to the weight of the object. where m is the mass in kg, g is the gravitational constant ( 9.81 N kg -1 near the Earth surface), h is the height from a reference level in m. Hence the gravitational potential energy is given by: P.E. = mgh In most cases, P.E. is taken to be zero at ground level and h is measured from ground level. g is also called the acceleration due to gravity. Its unit is m s -2 or N kg -1. The units of m, g and h must be correct to match the unit of P.E. (which is in J, N m or kg m 2 s -2 ). P.E. is negative if the body is below the ground level. Figure

10 Gravitation Similarly, y-component of the field strength 1.354g. 2 2 Hence the resultant gravitational field strength g = 1.91g or g 9.1 A. Determine whether each of the following statements is correct. Put a 3 in the appropriate box. (a) The gravitational forces between two bodies are always equal in magnitude. (b) The gravitational forces between two bodies form an action-reaction pair. (c) Gravitational field strength is a vector quantity. (d) The gravitational field strength on the Earth s surface is greater than that at an altitude of 1000 km. (e) The gravitational field strength at a certain point in space is equal to the gravitational force acting on a unit mass at that point. Correct Incorrect B. Fill in the blanks. (1) forces arise between any two bodies. The magnitude of each force is directly proportional to the product of the two (2) and inversely proportional to the square of the (3) between them. The directions of the gravitational forces between two bodies are (4) to each other. When the separation between two bodies increases, the gravitational forces between them (5). The unit of the gravitational filed strength is (6). The gravitational field established by an object can be represented by (7) lines. (For answers, see the bottom of the page.) Suggested Answers (Check Your Progress 9.1) (a) Correct (b) Correct (c) Correct (d) Correct (e) Correct (1) Gravitational (2) masses (3) distance (4) opposite (5) decreases (6) N kg -1 (7) field 277

11 Physics: Key Notes & Exambuilder Concept Check Topics Common errors Correct concepts Inelastic collision Elastic collision on a wall Ball rebounds on the ground Momentum in 2-D collision Total momentum decreases in an inelastic collision. When a ball rebounds elastically on a wall, its momentum remains unchanged. A ball rebounds on the ground. The average force F exerted by the ground on mv - mu the ball during impact is F =. t In an oblique collision between two bodies, the total momentum is the sum of the momenta of the two bodies. Total momentum still conserves in an inelastic collision. Only the total K.E. decreases in an inelastic collision. Momentum is a vector quantity. Although the magnitude of the momentum remains unchanged, the direction of the momentum changes in the collision. A ball rebounds on the ground. The force F in the mv - mu formula F = is the average unbalanced force t acting on the ball during impact. This is the vector sum of the average force R exerted by the ground on the ball and the weight W of the ball. In an oblique collision between two bodies, the total momentum should be the vector sum of the momenta of the two bodies. Section A Multiple-choice Questions 1. cart Figure 6.32 A cart of mass 200 kg is moving at a speed of 12 m s -1 on a level track. A man of mass 70 kg hanging on a rail slightly over the cart jumps into the cart when it passes. What is the final speed of the cart? Hint 1 A m s -1 B m s -1 C m s -1 D m s A head-on collision occurs between two private cars. Which of the following statements is/are correct? Hint 2 (1) Total energy is conserved in the collision. (2) Some K.E. is converted into heat and sound energy. (3) Some K.E. is used to twist the chassis (body) of the two cars. A. (1) only B. (2) only C. (1) and (2) only D. (1), (2) and (3) 226

12 Momentum 6. Figure 6.50a Figure 6.50b A 0.25 kg volley ball is falling vertically. The speed of the ball is 7 m s -1 just before it is hit by a player. Just after the smash, the ball leaves the player s hand at a velocity 18 m s -1 which makes an angle of 30 with the horizontal as shown in Fig 6.50b. (a) Find the initial momentum and final momentum of the ball. Hint 11 (b) Find the impulse exerted by the player. (c) If the time of impact is 0.15 s, what is the average force exerted by the player. You may ignore the effect of gravity during impact. 7. Figure 6.51 Two balls A and B of masses 1.8 kg and 1.2 kg respectively are moving on a smooth level surface. A collision occurs between them and they stick together after the collision. The above diagram shows their moving paths before and after the collision. The velocities of A and B before collision are respectively 2 m s -1 and 2.4 m s -1. (a) Find the total momentum of A and B just before the collision. Hint 12 (b) Find the size of θ. (c) How much kinetic energy is lost in the collision? 231

13 Physics: Key Notes & Exambuilder Section A Multiple-choice Questions Hint 1 The man s vertical motion does not affect the horizontal momentum of the system. Therefore, calculate the speed by considering the horizontal momentum only. Hint 2 Total energy in the collision is heat, sound and the kinetic energy of the cars. Hint 3 As momentum is a vector quantity, it involves direction. Use the correct +/- sign to represent its direction first. The magnitude of the ball will change while it rebounds. Hint 4 Total momentum in a system is conserved no matter if it is an elastic or an inelastic collision. Hint 5 Consider the vertical motions of objects under gravity. Hint 6 The momentum of the ball has two components p x and p y. p x is parallel to the side and p y is perpendicular to the side. Only py changes in the rebound and it is reversed in direction. Hint 7 Apply Newton s Third Law of motion. The action and reaction pairs of force act on X and Y respectively during impact. Section B Short Questions and Structured / Essay-type Questions Hint 8 Water resistance and tension acting in the chain may affect the momentum of the buoy. Make assumption on these. Hint 9 When the resident descends onto the cushion, his vertical position is still changing. Hint 10 Note that gravity does not exist in space. There is no downward force acting on any object. The mass of the object cannot be found by the supporting force. Hint 11 The final momentum of the ball is the vector sum of the initial momentum and the impulse received. Hint 12 The two balls move together after the collision. So the total momentum after collision can be represented by a single vector. Draw a vector diagram of the initial momenta of A and B. 232

14 Uniform Circular Motion Chapter 8 Uniform Circular Motion Section A Multiple-choice Questions 1. B Since v = wr, v v v P Q P 2r = r = 2 = 2v Q \ Ratio of their K.E. 1 1 = mvp : ( mv ) Q 1 1 = m( 2vQ ) : ( mv ) = 2 : Q 2. A When the puck starts to slide, 2 mv mg r 4 2 gr v 4 gr v 2 3. B For (1): The direction of motion is changing. For (2): Centripetal force is a vector. Its direction changes with time. For (3): Since the linear speed is constant, by v = wr, the corresponding angular speed is also constant. 4. C For (1): The weight of the cabin remains constant. For (2): Since the speed is constant, the unbalanced force (centripetal force) is constant in magnitude. For (3): The unbalanced force is the vector sum of the weight of the cabin and the force F exerted by the rod on the cabin. Since the unbalanced force is changing in direction and the weight is constant, F must be changing. Section B 1. (a) v = rω = ω 2 Short Questions and Structured/Essay-type Questions ω = rad s 2π (b) T = = 319 s Number of turns = 319 = (a) 2πr v = T 2 π = = 465 m s (b) a = 2 v r = = m s 2 3. (a) a = v r = = 136. m s (b) Friction f = ma = 70(. 136) = N