* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download HKDSE CS – Physics Notes Waves Mechanics Mechanics Electricity
Survey
Document related concepts
Electromagnet wikipedia , lookup
Conservation of energy wikipedia , lookup
Thermal conduction wikipedia , lookup
Superconductivity wikipedia , lookup
Faster-than-light wikipedia , lookup
Lorentz force wikipedia , lookup
Speed of gravity wikipedia , lookup
Time in physics wikipedia , lookup
Diffraction wikipedia , lookup
Work (physics) wikipedia , lookup
Electrical resistance and conductance wikipedia , lookup
Centripetal force wikipedia , lookup
Wave–particle duality wikipedia , lookup
Matter wave wikipedia , lookup
Theoretical and experimental justification for the Schrödinger equation wikipedia , lookup
Transcript
St. Joseph’s Anglo-Chinese School HKDSE CS – Physics Notes H eat M echanics W aves E lectricity St. Joseph’s Anglo-Chinese School 0 Chapter 1 Temperature and Thermometers 1.1 Temperature Lower fixed point (ice point) Temperature of pure melting ice at normal atmospheric pressure Upper fixed point (steam point) Temperature of steam over pure boiling water at normal atmospheric pressure Celsius temperature scale Divide 100 equal divisions between the lower and upper fixed point. Each division is 1oC 1.2 Kinetic theory (a) All matter is made up of very tiny particles. (b) These particles are constantly in motion. (c) Forces between particles: (i) When particles are close together, they attract/repel each other strongly. (ii) When particles are far apart, they hardly attract/repel each other. Solid Liquid Gas Particle arrangement (1) close together (2) arranged in regular pattern (1) close together (2) not in fixed position (1) very far apart Particle motion Vibrate in fixed positions Can move freely from one place to another Move at random at very high speed (~500 ms-1) 1.3 Heat and Internal energy * Heat is the energy transferred from one body to another due to a temperature difference. heat warmer cooler internal energy = kinetic energy (K.E.) + potential energy (P.E.) of all particles K.E (depends on temperature) T ↑ ⇔ K .E. ↑ P.E. (depends on the state of matter) solid liquid gas (particles vibrate more rapidly at higher temperature) P.E. of particles increases * Temperature is a measure of average kinetic energy of the particles. St. Joseph’s Anglo-Chinese School 1 Chapter 2 Heat Capacity and Specific Heat Capacity Heat capacity (C) Definition Specific heat capacity (c) Energy required to raise the Energy required to raise the temperature o Formula temperature of a substance by 1 C of a 1 kg substance by 1oC E = C∆T E = mc∆T o -1 Unit J kg-1 oC-1 J C C = mc Example 1 After absorbing 1000 J of energy, the temperature of a substance increases by 4oC. If the mass of the substance is 2 kg, find (a) the heat capacity, and (b) the specific heat capacity of the substance. Solution (a) For heat capacity, E = C∆T 1000 = C(4) C = 250 J oC-1 (b) For specific heat capacity, E = mc∆T 1000 = (2)(c)(4) c = 125 J kg-1 oC-1 Power = rate of energy transferred P= E t or E = Pt Example 2 2 kg of water is heated by a heater of power 1500 W. Find the time it takes for the temperature of water to increase from 20oC to 98oC. Given: the specific heat capacity of the water = 4200 J kg-1 oC-1 Solution ∵ E = Pt ∴ Pt = mc∆T and E = mc∆T 1500t = (2)(4200)(98 – 20) t = 436.8 s St. Joseph’s Anglo-Chinese School 2 Example 3 The figure below shows the variation of temperature of an object with time. Temperature / oC 60 20 0 5 time / minute If the power of the heater is 800 W, find the heat capacity of the object. Solution ∵ E = Pt ∴ Pt = C∆Τ and E = C∆T (800)(5 × 60) = C(60 – 20) t = 6000 J oC-1 [5 minutes = 5 × 60 s] Example 4 A piece of 0.1 kg hot copper is put into a pond of water of 2 kg. If the initial temperatures of the copper and water are 500oC and 20oC respectively, find the final temperature of the copper. Given: the specific heat capacity of the water = 4200 J kg-1 oC-1 the specific heat capacity of the copper = 370 J kg-1 oC-1 Solution Copper: 500 oC → T Water: 20 oC → T Assume no heat loss to the surroundings Energy lost by hot object (copper) = Energy gained by cold object (water) (0.1)(370)(500 – T) = (2)(4200)(T – 20) 18500 – 37 T = 8400 T – 16800 T = 22.1oC Remark Since the specific heat capacity of water is large, water can absorb a large amount of energy with only a small temperature rise. (∆T = 22.1 – 20 = 2.1 oC) Uses of high specific heat capacity of water (1) coolant (2) body temperature regulation St. Joseph’s Anglo-Chinese School 3 Example 5 In an experiment to find the specific heat capacity of aluminium, the following results are obtained. Mass of aluminium block = 1 kg Initial joulemeter reading = 98 300 J Final joulemeter reading = 104 900J Initial temperature of aluminium block = 28.5 oC Final temperature of aluminium block = 35.0 oC (a) Find the specific heat capacity of aluminium. (b) The standard value of the specific heat capacity of aluminium is 900 J kg-1 oC-1, find the percentage error of the experiment. (c) How to improve the accurate of the experiment. Solution (a) By E = mc∆T 104900 – 98300 = (1)(c)(35 – 28.5) c = 1020 J kg-1 oC-1 (b) Percentage error = 1015.4 − 900 × 100% = 12.8% 900 (c) (1) Wrap the aluminium block with cotton to reduce heat loss to the surroundings. (2) Add a few drop of oil to the holes in the aluminium block to ensure a good thermal contact between the heater, thermometer and the block. St. Joseph’s Anglo-Chinese School 4 Chapter 3 Change of State 3.1 Latent heat Heating No change in state E = mc∆T Change in state E = ml * E = ml, where l: specific latent heat * Unit: J kg-1 Gas Release latent heat (P.E.↓) ∆T = 0 (K.E. remains unchanged) Absorb latent heat (P.E.↑) ∆T = 0 (K.E. remains unchanged) Liquid Release latent heat (P.E.↓) ∆T = 0 (K.E. remains unchanged) Absorb latent heat (P.E.↑) ∆T = 0 (K.E. remains unchanged) Solid Example 1 How much energy is required to melt 2 kg of ice at 0oC and to raise the temperature to 30oC? Given that: the latent heat of fusion of ice = 3.34 × 105J kg-1, and the specific heat capacity of water = 4200 J kg-1oC-1. Solution E = mc∆T E = ml Ice (0oC) Water (0oC) Water (30oC) = ml + mc∆T = ( 2)(3.34 × 10 5 ) + ( 2)( 4200)(30 − 0) Energy required = 9.2 × 10 5 J Example 2 A coffee machine injects 0.03 kg of steam at 100oC into a cup of cold coffee of mass 0.17 kg at 20oC. Find the final temperature of the coffee. Given that: the latent heat of vaporization of ice = 2.26 × 106 J kg-1, and the specific heat capacity of coffee = 5800 J kg-1oC-1. Solution 100oC Steam (0.03 kg): E = ml 20 C coffee E = mc∆T water o Coffee (0.17 kg): 100oC water T (final temperature) E = ml coffee T (final temperature) Energy lost by steam = Energy gained by coffee 6 (0.03)(2.26 × 10 ) + (0.03)(4200)(100 – T) = (0.17)(5800)(T – 20) T = 90.0oC St. Joseph’s Anglo-Chinese School 5 3.2 Measure the specific latent heat of fusion of ice Procedures (1) Set up the apparatus as shown. (2) Fill both funnels with roughly equal amounts of crushed melting ice. (3) Record the initial joulemeter reading (E1). (4) Switch on the heater for a period of time. (5) Find the mass of ice (m) melted by the heater and record the final joulemeter reading (E2). (6) Calculate the specific latent heat of fusion of ice (lf) by E2 – E1 = mlf. Precautions Ice should be crushed to increase the contact area with the heater. Melting ice is used so that it is at 0 °C. Before switching on the heater, pack the crushed ice in the two funnels so that the drip rates are steady and about the same. After switching off the heater, do not remove the beakers; wait until the drip rates have become steady and about the same. - A small piece of wire gauze or steel wool at the neck of the funnels can prevent the crushed ice from dropping into the beakers directly. Example 3 The following results are obtained from the above experiment: Mass of water in experimental cup = 0.050 kg Mass of water in control cup = 0.014 kg Initial joulemeter reading = 15 000 J Final joulemeter reading = 29 200 J (a) Find the specific latent heat of fusion of ice. (b) Calculate the experiment percentage error. Account for any difference of the value obtained from the standard value, 3.34 × 105 J kg-1. Solution (a) By E = ml 29200 − 15000 = (0.050 − 0.014)l f l f = 3.94 × 10 5 J kg-1 3.9444 × 10 5 − 3.34 × 10 5 × 100% = 18.1% 3.34 × 10 5 Possible sources of error include: (1) Water dripping down the two funnels at different rates. (b) Percentage error = (2) Energy is lost to the surroundings. [Since energy is lost to the surroundings, less amount of ice is melted by the heater. E By l f = , the measured lf is greater than the standard value.] m St. Joseph’s Anglo-Chinese School 6 3.3 Measure the specific latent heat of vaporization of water Procedures (1) Set up the apparatus as shown. (2) Take the reading from the electronic balance (m1) heater and the kilowatt-hour meter (E1) after the water boils. (3) Boil the water for a few minutes and turn off the heater. (4) Wait until the water becomes steady and take the kilowatt-hour meter electronic balance final reading of the balance (m2) and the kilowatt-hour meter (E2). (5) Calculate the specific latent heat of vaporization of water by E 2 − E1 = (m1 − m 2 )l v from the results. Precaution Do not switch on the heater unless the heating part is totally immersed in water. Possible sources of error (1) Steam condensing on the heater and drips back into the cup ⇒ Larger experimental lv (2) Energy is lost to the surroundings ⇒ Larger experimental lv (3) Some water ‘bubbles’ out of the cup ⇒ Smaller experimental lv Example 4 The following results are obtained from the above experiment: Mass of water boiled away = 0.10 kg Energy supplied to the heater = 246 000 J (a) Find the specific latent heat of fusion of ice. (b) Account for any difference of the value obtained from the standard value, 2.26 × 106 J kg-1. Solution (a) By E = ml 246000 = 0.1l v l f = 2.46 × 10 6 J kg-1 (b) Possible sources of error include: (1) Steam condensing on the heater and drips back into the cup. (2) Energy is lost to the surroundings. [These will cause a smaller amount of water boiled away. E By l v = , the measured lv is greater than the standard value.] m St. Joseph’s Anglo-Chinese School 7 3.4 Evaporation and Boiling (a) Common: absorb latent heat (E = ml) to change from liquid state to gas state (b) Difference: Evaporation Boiling Occurs at any temperature Occurs at a definite temperature – the boiling point Occurs at surface Occurs with liquid No bubbles formed Bubbles appear (c) Evaporation and particle motion (i) Some of the particles in a liquid have greater K.E. while the other have smaller K.E. (ii) Some of the particles at the liquid surface may gain enough KE to escape into the space above the liquid and become particles of vapour. (iii) As fast-moving particles fly away, the average KE of the remaining particles is lowered; so the liquid becomes colder. (d) Ways to increase the rate of evaporation (i) increase temperature of the liquid (ii) increase the surface area of the liquid (iii) decrease the humidity of air (iv) increase the movement of air Example 5 A person is wearing a wet shirt. There is 0.1 kg of water on the shirt in total. (a) How much energy is required to evaporate the water? (b) Where is the energy required taken from? Specific latent heat of vaporization of water = 2.26 × 106 J kg–1 Solution (a) Energy required evaporating 0.1 kg of water E = ml = (0.1)(2.26 × 106) = 2.26 × 105 J (b) The energy required is taken from the surroundings (or the skin of the person). St. Joseph’s Anglo-Chinese School 8 Chapter 4 Heat Transfer Conduction Convection Radiation Description Hot water rises Cold water falls – Particles at the hot end vibrate faster. – The fast vibrating – Fluid (gas or liquid) expands, rises and is particles bump into the replaced by the The solar energy is slower neighbouring surrounding cooler fluid. transferred to the earth by particles and make – Such movement of fluid them vibrate more is called convection. radiation rapidly. Medium required Solid, liquid or gas Fluid (liquid or gas) No medium is required Rate of heat transfer Conductor – faster Dark colour – good Insulator – slower absorber and radiator Daily examples A cotton jacket traps air (Vacuum ) (air is a poor conductor) – Air conditioners are – installed high on the wall – Heating element is fixed are black in colour – near the bottom of an electric kettle Most transformers Car engines are painted black. – Fuel storage tanks are painted silvery. Example 1 Solar heater glass (a) What is the function of the glass? (b) The temperature of the 0.3 kg water increases to 70oC after 30 minutes. If the initial temperature of water is 30oC, find the power of the solar heater. water bowl (c) Give a suggestion to improve the design of the solar container heater in order to obtain a greater temperature rise of water. Explain your answer briefly. Solution (a) The glass traps warm air to reduce heat loss by convection. (b) ∵ E = Pt and E = mc∆T ∴ Pt = mc∆T P(30 × 60) = (0.3)(4200)(70 – 30) P = 28 W (c) (1) Paint the bowl black in colour since dark colour object is a good radiation absorber. (2) Stick slivery paper (aluminium paper) onto the inner wall of the container so that more radiation is reflected to the bowl. St. Joseph’s Anglo-Chinese School 9 Vacuum flask Plastic or cork stopper reduces heat loss by conduction and convection. cap outer case Silvery glass or stainless steel walls reduce heat loss by radiation. A vacuum between the double walls reduces heat loss by conduction and convection. insulated support Vacuum cooker Inner pot Outer pot Inner and outer lids reduce heat loss by conduction and Silvery steel walls reduce convection heat loss by radiation Outer vacuum insulated pot (the vacuum Inner pot A thin film of air between inner pot and outer pot. Reduce heat between the double steel walls of the outer pot reduces heat loss by conduction and convection) loss by conduction and Food is kept at a high convection. temperature for a long time and is cooked without a fire. St. Joseph’s Anglo-Chinese School 10 Chapter 1 – 2 (a) Distance and Displacement Motion Speed and Velocity B Example 1 A person takes 4 s to walk from A to B along the following path. Find the distance, displacement, speed and velocity 5m A 40o Solution Distance = length of the path Displacement = the length of straight line AB = 10 m (N50oE) 10π = = 15.7m 2 Speed = Distance time = 15.7 = 3.93 ms-1 4 Velocity = = Scalar: Distance, Speed Displacement time 10 = 2.5 ms-1 (N50oE) 4 Vector: Displacement, velocity (b) Acceleration accelerati on = change in velocity time a= v−u t Unit: ms-2 Example 2 It takes 6 s for a racing car to attain a speed of 100 km h-1 from rest. Find the average acceleration of the car. Solution Initial velocity u = 0 ms-1 (start from rest) Final velocity v = 100 km h-1 = 100 -1 ms = 27.78 ms-1 3 .6 Time t = 6 s Acceleration a = = v−u t 27.78 − 0 = 4.63 ms-2 6 St. Joseph’s Anglo-Chinese School 11 (c) Motion graphs (s – t, v – t, a – t graph) (i) Uniform velocity s v 0 t a 0 t Displacement changes at a constant rate 0 Velocity remains constant. t No acceleration. a = 0 ms-2 (ii) Uniform acceleration s v 0 t 0 t Displacement: Velocity increase at a constant rate. Beginning: increases a little. End: increases significantly. (iii) Slope and area under graph s 0 slope = velocity a s t 0 t Acceleration remains constant. v 0 t slope = acceleration 0 t Area under v-t graph = displacement (d) Equations of uniformly accelerated motion (1) v = u + at u = initial velocity (2) v 2 − u 2 = 2as a = acceleration (3) 1 s = ut + at 2 2 (4) s= u+v t 2 St. Joseph’s Anglo-Chinese School v = final velocity s = displacement t = time 12 (e) Motion of free fall In the absence of air resistance, all objects fall with an acceleration of g (9.81 ms-2) Example 3 A stone is thrown vertically upwards from the ground with a speed of 20 m s-1. (a) How high does the stone rise? (b) How long does it take the stone to reach the ground? (c) Draw a v-t graph for the stone unit it returns to the ground. Solution (a) Consider the motion of the stone when it is thrown upward until it reaches the highest point. u = 20 ms-1 v = 0 ms-1 (highest point) a = -9.81 ms-2 v = 0 ms -1 (uniform deceleration) (momentarily at rest at the highest point) 20 ms-1 s=? By v 2 − u 2 = 2as 0 2 − 20 2 = 2( −9.81) s ∴ s = 20.4 m greatest height is 20.4 m (b) Method 1 Method 2 Consider the motion of the stone when it is Consider the motion of the stone when it is thrown upward until it reaches the highest thrown upward until it returns to the ground. point. u = 20 ms-1 u = 20 ms-1 a = -9.81 ms-2 (uniform deceleration) a = -9.81 ms-2 (uniform deceleration) v = 0 ms-1 (momentarily at rest) t=? s=0 v = 0 ms By v = u + at 0 = 20 + ( −9.81)t 20 ms-1 t = 2.04 s Time required for stone to reach the ground = 2 × 2.04 (return to ground) t=? 1 By s = ut + at 2 2 20 ms-1 s=0m 1 ( −9.81)t 2 2 t ( 4.905t − 20) = 0 0 = 20t + t = 0 (rejected) or 4.905t − 20 = 0 ∴ t = 4.08 s = 4.08 s (c) -1 v / ms-1 20 0 2.04 4.08 t/s -20 St. Joseph’s Anglo-Chinese School 13 Remark: Take upward direction as positive Stage 1: upward motion Stage 2: at highest point decelerate uniformly Momentarily at rest a = -9.81 ms-2 a = -9.81 ms-2 Stage 3: downward motion accelerate downward uniformly a = -9.81 ms-2 Misconception (1) The acceleration of the object at the highest point a = 0 ms-2. (2) We are certain that at the highest point, v = 0 ms-1. (3) If at the highest point a = 0 ms-2 and v = 0 ms-1, the object will remain at rest forever and will not fall down. Example 4 v / ms-1 The v-t graph of a car is shown. 50 (a) Describe the motion of the car from t = 0 s to t = 80 s. 0 (b) Find the deceleration of the car from t = 40 s to t = 70 s. -25 (c) Find the total displacement of the car. 40 70 80 t/s Solution (a) From t = 0 s to 40 s, the car moves with a uniform velocity. From t = 40 s to 70 s, the car decelerates uniformly and becomes momentarily at rest at t = 70 s. From t = 70 s to 80 s, the car accelerates uniformly in opposite direction. (b) by a = Method 1 v−u t From t = 40 s to 70 s -1 Method 2 a = slope of v-t graph From t = 40 s to 70 s -1 u = 50 ms , v = 0 ms , t = 30 s 0 − 50 = −1.67 ms-2 30 ∴ Deceleration is 1.67 ms-2 = slope of v-t graph a a= 50 − 0 = −1.67 ms-2 70 − 40 Deceleration is 1.67 ms-2 = − ∴ (c) Total displacement of the car = area under v – t graph = ( 40 + 70)(50) 1 − (10)( 25) 2 2 = 2625 m St. Joseph’s Anglo-Chinese School 14 Chapter 3 Force and Motion 3.1 Newton’s law of motion Newton’s 1st law Every object remains in a state of rest or uniform motion (i.e. constant velocity) unless acted on by a net force, or an unbalanced force. nd Newton’s 2 law The acceleration of an object is directly proportional to, and in the same direction as, the net force acting on it, and inversely proportional to the mass of the object. [ a ∝ F , a ∝ Newton’s 3rd law 1 ⇒ F = ma] m Action and reaction pair (1) Equal in magnitude (2) Opposite in direction (3) Acting on different bodies. 3.2 Forces in daily life (a) Weight (W) (b) Normal reaction (R) (c) Friction (f) (d) Tension (T) 3.3 Mass m and Weight W W = mg Mass m Weight W Scalar Vector A measure of inertia Gravitational force acting on the object Remain unchanged Vary in different planets (g is different) 3.4 Addition and resolution of forces Addition: parallelogram of forces method Resolution: Fx and Fy are components of F y FR (resultant force) F2 F Fy θ F1 Fx = F cos θ Fx x F y = F sin θ St. Joseph’s Anglo-Chinese School 15 3.5 Typical exam questions (a) Inclined plane Resolve W into two components: W sin θ θ θ along the plane W sin θ or mg sin θ perpendicular to the plane W cos θ or mg cos θ W cos θ W Example 1 A block is placed on the inclined plane and remains at rest. (a) (i) Draw a free diagram of the object. (ii) If the mass of the block is 5 kg, find the friction 30ο acting on the block (iii) Find the normal reaction acting on the block (b) Now, a force of 120 N is applied on the block and pulls the block upwards along the plane. If the friction between the block and the plane becomes 60 N, find the acceleration of the block. Solution (a) (i) Normal reaction (R) Friction (f) Weight (W = mg) (ii) Do not accept W, mg, f, R R W sin θ f = W sin θ = mg sin θ = (5)(9.81) sin 30° f = 24.5 N θ W cos θ W R = W cos θ = mg cos θ = (5)(9.81) cos 30° (iii) = 42.5 N (b) Consider the direction along the plane 120 N 120 – W sin θ – f = (5)(a) W sin θ θ By F = ma f = 60 N St. Joseph’s Anglo-Chinese School 120 – (5)(9.81)sin θ – 60 = 5a a = 7.10 ms-2 16 (b) Apparent weight in a lift Reading of weighing machine = normal reaction = R When the lift R a at rest or moves with uniform velocity R=W Feel normal weight accelerate upward R>W Feel heavier accelerate downward R<W Feel lighter Weight is always constant (W = mg) Weighing machine R – W = ma (Always take upward direction as +ve) W Example 2 (a) A boy of mass 55 kg stands on a weighing machine in a lift. Find the reading on the weighing machine when (i) the lift moves upwards with an acceleration of 1 ms-2, and (ii) the lift moves downward with a deceleration of 1.5 ms-2. (b) When the lift moves upward and the reading on the weighing machine is 480 N, find the acceleration of the lift. Solution (a) (i) The lift moves upward and accelerate, a = + 1 ms-2 R − W = ma R − (55)(9.81) = (55)(1) R = 595 N [The boy feels heavier] (ii) The lift moves downward and decelerate, a = + 1.5 ms-2 R − W = ma R − (55)(9.81) = (55)(1.5) R = 622 N (b) [The boy feels heavier] R − W = ma 480 − (55)(9.81) = (55)a a = −1.08 ms-2 ∴ The deceleration of the lift is 1.08 ms-2 Remark (1) When applying R − W = ma The lift moves (2) upward with acceleration a = +ve upward with deceleration a = – ve downward with acceleration a = – ve downward with deceleration a = +ve When the cable of the lift break, R * W = mg * a = -9.81 ms-2 * R=0 W St. Joseph’s Anglo-Chinese School 17 3.6 The turning effect of a force (a) Moment = Fs Unit: Nm 5 sin 30o 5N 2m 30o 2m pivot 5N 5 cos 30o pivot Moment = Fs = (5 sin 30o)(2) = 5 Ns Moment = Fs = (5)(2) = 10 Ns (b) Equilibrium of a rigid body No net force: all forces acting on the body are balanced No net torque: clockwise moment = anticlockwise moment Example 3 The mass of the meter rule is 0.5 kg. W1 School bag 4 kg R Block 10 kg 0.2 m 0.3 m W2 W3 d (a) Find the distance between the school bag and the pivot to maintain balance. (b) Find the reaction from the pivot. Solution (a) Clockwise moment = anticlockwise moment (W2)(0.3) + (W3)(d) = (W1)(0.2) (0.5 × 9.81)(0.3) + (4 × 9.81)(d) = (10 × 9.81)(0.2) d = 0.463 m (b) R = W1 + W 2 + W3 = (10)(9.81) + (0.5)(9.81) + (4)(9.81) = 142 N Remark (1) Moment = Force × Distance (distance is usually measured from the pivot) (2) Must label all forces clearly on the figure. St. Joseph’s Anglo-Chinese School 18 Chapter 4 Work, Energy and Power 4.1 Work (W) Definition: W = F// s Unit: J Example 1 Calculate the work done by F in each of the following case. (a) (b) F=2N 30o F=2N 5m 5m Solution (a) W = (2)(5) = 10 J (b) W = (2 cos 30o)(5) = 8.66 J Consider only the component of F along the direction of displacement. i.e. 2 cos 30o 4.2 Energy Mechanic energy: 1 mv 2 2 (2) P.E. = mgh (1) K.E. = (3) Elastic potential energy 4.3 Energy change Conservation of energy: Energy can be changed from one form to another, but it cannot be created or destroyed. Example 2 Find the speed of the particle at B. Assume that the track is smooth. A 2 ms-1 10 m B 3m Solution Total mechanical energy at A = Total mechanical energy at B P.E.A + K.E.A = P.E.B + K.E.B 2 Common mistake 2 m(9.81)(10) + ½ m(2) = m(9.81)(3) + ½ m(v) 98.1 + 2 = 29.43 + ½v2 v = 11.9 ms-1 St. Joseph’s Anglo-Chinese School cannot use the formula v 2 − u 2 = 2as reason: not a uniformly accelerated motion 19 Example 3 30o A block slides down for 8 m along the surface of a 30o inclined plane. It is given that the mass of the block is 3 kg and the friction acting on the block is 5 N (a) Find the P.E. loss of the block. (b) Find the work down against friction. (c) If the initial speed of the block is 4 ms-1, find its final speed. Solution (a) P.E. loss = mgh h = 8 sin 30o = (3)(9.81)(8 sin 30o) = 117.72 J ≈ 118 J (b) work done against friction W = fs = (5)(8) = 40 J (c) K.E. gained = 117.72 – 40 = 77.72 J 8m 30o 5N 8m 30o ½mv2 –½mu2 = 77.72 ½(3)(v2 – 42) = 77.72 v = 8.23 ms-1 Remark: Description of energy change K.E (77.72 J) P.E. (117.72 J) Heat (40 J) P.E. changes into kinetic energy and heat (internal energy) St. Joseph’s Anglo-Chinese School 20 Example 4 High diving A diver of 60 kg dives off a platform 10 m above the water surface. (a) If she falls directly downwards, what is her speed when she reaches the water? (b) If the water provides an average water resistance of 2500 N, how deep can she reach under the water surface? Solution (a) By conservation of energy, mgh = ½ mv2 v2 = 2gh v = 2gh = 2(9.81)(10) = 14.0 ms-1 A → B, the diver loses P.E. and K.E. Total energy loss = work done against water resistance (b) K.E. + P.E. = fs ½ mv2 + mgh = fs ½ (60)(14.0)2 + (60)(9.81)(d) = (2500)(d) d = 3.08 m Water resistance (2500 N) depth 4.4 Power P = Fv Unit: W Conclusion Work Energy Scalar Power St. Joseph’s Anglo-Chinese School 21 St. Joseph’s Anglo-Chinese School 22 Chapter 5 Momentum 5.1 Momentum Definition: mv Unit: kg ms-1 5.2 Force and change in momentum Net force = change in momentum time F= mv − mu t Example 1 20 ms-1 A tennis ball of mass 0.05 kg flies along a horizontal speed of 20 ms-1 towards a player. The player hits the ball, which leaves the racket at 30 ms -1 30 ms-1 in the opposite direction. If the time of impact is 0.005s, calculate the average force acting on the ball. Solution u = -20 ms-1 v = 30 ms-1 F= (towards the left) (0.05)(30) − (0.05)( −20) = 500 N 0.005 5.3 Impulse Definition: Ft Impulse = Ft = mv – mu = change in momentum F/N Area under F – t graph = Ft = mv – mu 0.02 0.04 t/s Example 2 If the area under the above F – t graph is 0.52 Ns, find (a) the change in momentum (b) The average force of impact. Solution (a) Change in momentum = area under F – t graph = 0.52 Ns (b) Average force of impact F= mv − mu 0.52 = = 26 N t 0.04 − 0.02 St. Joseph’s Anglo-Chinese School 23 5.4 Conservation of momentum: m1u1 + m 2 u 2 = m1 v1 + m 2 v 2 (for no external force) Example 3 Find the final velocity for the following completely inelastic collision. 10 ms-1 1 ms-1 2 kg 4 kg 2 kg Before collision 4 kg v After collision Solution By conservation of momentum m1u1 + m 2 u 2 = m1 v1 + m 2 v 2 ( 2)(10) + ( 4)( − 1) = ( 2 + 4)v (opposite direction) v = 2.67 ms-1 Example 4 2-dimensional impact (resolution of velocities required) Find the final velocities of the 5 kg and 1 kg objects. 5 kg 2 ms-1 At rest 10 v1 o 30o 1 kg 5 kg Before collision 1 kg After collision v2 Solution Along direction of x-axis (5)( 2) = (5)v1 cos 10° + (1)v 2 cos 30° 5v1 cos 10° + v 2 cos 30° = 10 ---- (1) Along direction of y-axis 0 = (5)v1 sin 10° − (1)v 2 sin 30° v 2 sin 30° ---- (2) 5 sin 10° Sub. (2) into (1) v sin 30° 5 2 cos 10° + v 2 cos 30° = 10 5 sin 10° v1 = v 2 = 2.70 ms-1 Sub. v 2 = 2.70 ms-1 into (2): v1 = 0.576 ms-1 St. Joseph’s Anglo-Chinese School 24 Example 5 Conservation of energy and energy changes A bullet of 4 g shoots into a piece of 200 g plasticine. Find the initial velocity of the bullet if the plasticine rises by 3 cm. u 200 g 3 cm 4g Solution Stage (1) Stage (2) impact: m1u1 + m 2 u 2 = m1 v1 + m 2 v 2 u v energy change: 1 mv 2 = mgh 2 v M = 200 g 3 cm m=4g Before impact K.E. → P.E. After impact Consider stage (2): K.E. → P.E. 1 ( M + m )v 2 = ( M + m ) gh 2 v 2 = 2 gh = 2(9.81)(0.03) [3 cm = 0.03 m] v = 0.7672 ms-1 Consider stage (1): m1u1 + m 2 u 2 = m1 v1 + m 2 v 2 mu = ( M + m )v 0.004u = (0.2 + 0.004)(0.7672) ∴ [4 g = 0.004 kg] u = 39.1 ms-1 the initial velocity of the bullet is 39.1 ms-1 St. Joseph’s Anglo-Chinese School 25 5.5 Different types of collisions Total momentum conserved? Elastic collision Inelastic collision Total energy conserved? Example 6 Identify the type of the following collisions 5 ms-1 4 ms-1 (a) 2 kg 1 kg 1 ms-1 1 kg 2 kg Before collision 4 ms-1 After collision Solution Total K.E. before collision = 1 1 (2)(5) 2 + (1)( 4) 2 = 33 J 2 2 1 1 (2)(1) 2 + (1)( 4) 2 = 9 J 2 2 there is K.E. loss during collision (24 J) Total K.E. after collision = ∵ ∴ the collision is inelastic. (b) 2 ms-1 At rest 2 kg 1 kg Before collision 2 kg 2 -1 ms 3 1 kg 8 -1 ms 3 After collision Solution Total K.E. before collision = 1 1 (2)( 2) 2 + (1)(0) 2 = 4 J 2 2 1 2 1 8 ( 2)( ) 2 + (1)( ) 2 = 4 J 2 3 2 3 there is no K.E. loss during collision Total K.E. after collision = ∵ ∴ the collision is elastic. St. Joseph’s Anglo-Chinese School 26 Chapter 6 Projectile motion Main point: Resolve projectile motion into horizontal motion and vertical motion y u u sin θ H θ u cos θ x Range sx Horizontal motion Uniform motion Velocity Displacement vx = ux (velocity is constant) s x = ux t vy = uy – gt sy = uyt – ½ gt2 Vertical motion Free fall motion (a = -g = -9.81 ms-2) Other useful formulae 1. u x = u cos θ u y = u sin θ 2. Time of flight T = 3. u 2 sin 2θ Range s x = g 4. Maximum height H = 5. Equation of trajectory: s y = (tan θ ) s x − 2 v = vx + v y 2u sin θ g u2 Maximum Range s x = when θ = 45o g u 2 sin 2 θ g g 2 sx 2 2u cos θ 2 Example 1 A small object is thrown horizontally towards a vertical wall 1.2 m away. If the initial speed of the ball is 8 ms-1, find the vertical displacement of the ball when it hits the wall Solution: Given 2 wall 1.2 m Horizontal motion ux = 8 ms-1 sx = 1.2 m, Vertical motion uy = 0 (thrown horizontally) sy = ? For horizontal motion: a = -9.81 ms-2 s x = ux t 1.2 = 8t ⇒ t = 0.15 s For vertical motion: s = ut + ½ at2 = 0 + ½ (-9.81)(0.15)2 = -0.110 m Vertical displacement sy = 0.110 m (downward) St. Joseph’s Anglo-Chinese School 27 Example 2 A ball is projected upward with velocity 10 ms-1 at 30o with the horizontal from a cliff which is 100 m above the sea level as shown. Find 10 ms-1 30 o 100 m (a) the time of flight of the ball, (b) the horizontal distance moved when the ball reaches the sea, (c) the direction of the ball when it reaches the sea, and (d) the speed of the ball when it reaches the sea. Solution: (a) Given Horizontal motion ux = 10 cos 30o Vertical motion uy = 10 sin 30o sy = -100 a = -9.81 ms-2 t =? sy = uyt + ½ at2 For vertical motion: -100 = 10sin 30ot + ½ (-9.81)t2 4.905t2 – 5t – 100 = 0 t = 5.05 s or t = -4.03 s (rejected) (b) horizontal distance = uxt = (10 cos 30o)(5.05) = 43.8 m (c) 10 ms-1 30 o vx 100 m vy θ v Horizontal motion : vx = ux = 10 cos 30º = 8.66 ms-1 Vertical motion tan θ = vy vx = : vy = uy + at = 10 sin 30º + (-9.81)(5.05) = -44.6 ms-1 (downward) 44.6 8.66 θ = 79.0o The direction makes 79.0o with the horizontal. (d) speed v = v x + v y = 8.66 2 + 44.6 2 = 45.4 ms-1 2 2 St. Joseph’s Anglo-Chinese School 28 Chapter 1 Reflection of Light 1.1 Laws of reflection (a) angle of incidence i = angle of reflection r (b) incident ray, reflected ray and the normal all lie on the same plane. Example 1 (a) Find the angle of incidence. (b) Find the angle between the incident ray and the reflected ray. Solution: (a) i = 90o – 25o Common mistake: = 65o i = 25o (b) r = i = 65o ∴ angle between the incident ray and the reflected ray =i+r = 130o 1.2 Images formed by a plane mirror * (a) Steps (i) image (image distance = object distance) (ii) reflected ray (iii) incident ray Note: (1) appropriate solid lines and dotted lines (2) arrows for light rays to show direction (b) Properties of image (i) Virtual (ii) Laterally inverted (iii) Same size as the object (iv) Image distance = object distance St. Joseph’s Anglo-Chinese School 29 Example 2 St. Joseph’s Anglo-Chinese School 30 Chapter 2 Refraction of Light 2.1 Laws of refraction (a) sin i and sin r are in direction proportion (b) incident ray, refracted ray and the normal all lie on the same plane. Note: Light bends towards the normal when it travels from a less dense medium into a denser medium. (e.g. from air into water) 2.2 Snell’s law n1sin θ1 = n2sin θ2 * Example1: Find the angle of refraction r. Glass (ng = 1.5) Water (nw = 1.33) Solution: angle of incidence i = 90o – 30o = 60o (Common mistake: i = 30o) By n1sin θ1 = n2sin θ2 1.5 sin 60o = 1.33 sin r sin r = 1.5 sin 60° 1.33 r ≈ 77.6o 2.3 Refractive index Glass block ** Slope of graph of sin i against sin r = sin i = n = refractive index of the glass block sin r St. Joseph’s Anglo-Chinese School 31 **Example 2 Do exercise: page 63 Q 13 St. Joseph’s Anglo-Chinese School 32 2.4 Refractive index and light speed n = c v Example 3 The speed of light in air is 3 × 108 ms-1. If the refractive index of water is 1.33, find the speed of light in water. Solution: n= c v 1.33 = v= 3 × 108 v 3 × 10 8 = 2.26 × 10 8 ms −1 1.33 2.5 Image formed by refraction * (a) Steps: (i) image (nearer to water surface) (ii) refracted ray (iii) incident ray Note: (1) appropriate solid lines and dotted lines (2) arrows for light rays to show direction (b) image property – virtual 2.6 Total internal reflection (a) Critical angle C: (i) when angle of refraction r = 90o, i = C n 1 (ii) C = sin −1 or C = sin −1 2 (n1 > n2) n n1 water [n1: denser medium] (b) Conditions: * (i) light enters from a denser medium (ii) angle of incidence i > critical angle C **Example 4 Determine whether total internal reflection would occur in each of the following cases: (a) (b) 60o Glass (ng = 1.7) water (nw = 1.33) water (na = 1.33) diamond (ng = 2.42) Solution: (a) ∵ Light enter from a denser medium, 1.33 = 51.5° < 60° 1 .7 total internal reflection would occur. C = sin −1 ∴ 60o St. Joseph’s Anglo-Chinese School (b) ∵ ∴ Light enters from a less dense medium total internal reflection would not occur. (No need to calculate the critical angle) 33 (c) Natural phenomena (自然現象) (i) mirage (海市蜃樓) (ii) sparkling of diamonds (d) Applications (i) optical fibres (transmit telephone signals) (ii) cat’s eyes found on highways (iii) using prisms as mirrors in cameras St. Joseph’s Anglo-Chinese School 34 Chapter 3 Lenses ** 3.1 Images formed by a convex lens The nature of image depends on the distance between the object and the lens. St. Joseph’s Anglo-Chinese School 35 ** 3.2 Images formed by a concave lens The images formed by a concave lens are always virtual, erect and diminished. 3.3 Magnification image size image distance m= = object size object distance m>1 m=1 m= hi v = ho u m<1 Magnified Same size Diminished *3.4 Real image and Virtual image Real image Virtual image Can be seen by eyes? Can be projected on screens? Remark We can see virtual images formed by plane mirrors Virtual images cannot be formed on screens – Plane mirrors Examples Convex lens (in – Convex lens projectors) (magnifying glasses) – Convex lens St. Joseph’s Anglo-Chinese School 36 * Example 1 Some letters on a single-lined paper are observed by using a lens as shown below. J J J J J J J J J J J J (a) What kind of lens is used? Explain briefly. (b) If the distance between the lens and the paper is 10 cm and the magnification is 0.4, find the image distance. (c) Draw a ray diagram to find the focal length of the lens. Solution: (a) Concave lens is used because the image formed is erect and diminished. (b) Object distance = 10 cm, u = 10 cm By m = v u 0.4 = v 10 v = 4 cm (c) 1.2 1.1 5 1.1 1.0 5 F C 1 2 cm 0.9 5 0 1 2 3 4 5 6 7 8 9 10 Focal length f ≈ 6.8 cm (measure the distance between F and the optical centre) St. Joseph’s Anglo-Chinese School 37 St. Joseph’s Anglo-Chinese School 38 Chapter 4 The Electromagnetic Spectrum 4.1 The EM spectrum Frequency f, Energy Ε Radio Waves Infra-red Visible Microwaves light Ultra-violet X – ray γ – ray Wavelength λ 4.2 Properties of EM waves (a) show reflection, refraction, diffraction and interference (b) are transverse waves (c) can travel through vacuum (d) travel at speed of light in air / vacuum (3 × 108 ms-1) (e) can apply the equation v = fλ * 4.3 Values commonly used in calculations f 1kHz = 103 Hz 1MHz = 106 Hz 1GHz = 109 Hz λ 1mm = 10-3 m 1µm = 10-6 m 1nm = 10-9 m t 1ms = 10-3 s 1µs = 10-6 s 1ns = 10-9 s Example 1 The wavelengths of visible light lie between 4 × 10-7 m to 7 × 10-7 m. Find the frequency range of light. Solution: ∵ v = fλ ∴ f=v/λ Minimum frequency = 3 × 10 8 = 4.29 × 1014 Hz −7 7 × 10 3 × 108 = 7.5 × 1014 Hz −7 4 × 10 ∴ frequency range of visible light is 4.29 × 1014 Hz to 7.5 × 1014 Hz. Maximum frequency = * Example 2: Many auto-focus cameras use infra-red radiation to judge the distance of an object from the camera. If the time between emitting an infra-red pulse and receiving the pulse is 25 ns, find the distance of the object from the camera. Solution: Distance = speed × time = (3 × 108) × (25 × 10-9 ÷ 2) [25 ns = 25 × 10-9 s] = 3.75 m St. Joseph’s Anglo-Chinese School 39 4.4 Useful information EM wave Detector Application Radio waves Radio & TV receiver Radio communication Microwaves Microwave receiver Microwave oven Satellite telecommunication Radar Infra-red Thermometer Auto-focus camera IR photographic film IR telescopes Skin Visible light Eyes -7 λ: red (4 × 10 m) violet (7 × 10-7 m) Photographic film Ultra-violet Fluorescent paper Solar cells Sterilizing (消毒) drinking water Checking fake banknotes X – ray Photographic film Medical diagnosis γ – ray Photographic film Radiothery Sterilization (消毒) St. Joseph’s Anglo-Chinese School 40 Chapter 5 Nature of Waves 5.1 Transverse and longitudinal waves Transverse Wave Waveform crest trough Longitudinal wave crest compression trough rarefaction * Definition Vibrations are at right angle to the direction of travel of the wave Vibrations are parallel to the direction of travel of the wave Wavelength Distance between 2 adjacent crests Distance between 2 adjacent compressions or troughs or rarefactions EM waves, water waves Sound waves Examples 5.2 Equations: (a) v = fλ 1 (b) f = T (c) d = vt frequency f: no. of waves produced in 1 s. If f = 5 Hz, 5 waves are generated in 1s. * period T: time required to produce 1 wave, OR time for a particle to make a complete vibration If T = 0.2 s, it takes 0.2 s to generate 1 wave, OR it takes 0.2 s for a particle to make 1 complete vibration. Example 1: It takes 10s for a slinky to produce 20 waves. (a) Find the frequency of the wave. (b) If the wavelength of the wave is 0.2 m, find the speed of the wave. Solution: (a) frequency f = no. of waves produced in 1 s f = 20 = 2 Hz (2 waves are produced in 1 s.) 10 (b) By v = fλ v = (2)(0.2) = 0.4 ms-1 St. Joseph’s Anglo-Chinese School 41 5.3 Vibrations of particles * Example 2: The following figure shows a transverse wave travelling to the right. direction of travel of the wave a e bI d f c (a) Which particles are (i) moving upwards, (ii) moving downwards, (iii) momentarily at rest? (b) Which particles are (i) vibrating in-phase, (ii) vibrating anti-phase? ** (c) If the frequency of the wave is 2 Hz, draw the shape of the waves after 0.125 s. Solution: (a) a e ib d f c (i) moving upards: b, f (ii) moving downwards: d (iii) momentarily at rest: a, c, e (b) (i) a,e or b,f are in-phase (ii) a,c or b,d or c,e or d,f are anti-phase Remark In phase Separation of 2 particles is λ, 2λ, 3λ, … Anti-phase Separation of 2 particles is 0.5λ, 1.5λ, 2.5λ, … a (c) Steps: (1) f = 2 Hz ⇒ T = 1/2 = 0.5 s (2) 0.125s = ¼ T e b d ¼λ f (3) In ¼ T, the wave travels by ¼ λ. c Note: particles only move up and down while the wave travels forward. St. Joseph’s Anglo-Chinese School 42 5.4 Displacement-time (s – t) graph and displacement-distance (s – d) graph s – t graph s – d graph Shows the displacement of ONE particle at Shows the displacement of ALL particle at a different times certain time s s Time for one complete vibration t d Τ λ * Example 3: An s – t graph and s – d graph of a wave are shown below. s / cm s / cm 5 5 0 0 0.1 0.2 0.3 0.4 t/s -5 -5 (a) P 0.1 0.2 d/m Q Find the speed of the wave. ** (b) Draw an s – t graph for particle Q (see the s – d graph.) Solution: (a) From s – t graph, we have T = 0.2 s ⇒ f = 1/0.2 = 5 Hz. From s – d graph, we have l = 0.2 m ∴ wave speed v = fλ = (5)(0.2) = 1 ms-1 (b) s / cm 5 0 0.1 0.2 0.3 0.4 t/s -5 St. Joseph’s Anglo-Chinese School 43 St. Joseph’s Anglo-Chinese School 44 Chapter 6 Wave Phenomena 6.1 Ripple tank * Crest: bright fringe Trough: dark fringe λ: distance between two adjacent bright fringes or dark fringes 6.2 Reflection i=r Frequency f No change Wavelength λ No change Wave speed v No change 6.3 Refraction (a) Deep region → Shallow region Deep region Shallow region (Put a glass block) Frequency f No change Wavelength λ Decrease Wave speed v Decrease Direction Bend towards normal Frequency f No change Wavelength λ Increase Wave speed v Increase Direction Bend away from normal (b) Shallow region → Deep region Shallow region Deep region St. Joseph’s Anglo-Chinese School 45 6.4 Diffraction (a) Diffraction is bending or spreading of waves around the edge of an obstacle. (b) Degree of diffraction depends on the relative size of the gap. Significant diffraction Insignificant diffraction Gap size is relatively small (d ≈ λ) Gap size is relatively large (d >> λ) (c) Ways to increase the degree of diffraction (i) increase wavelength (lower the frequency or increase the depth of water) (ii) decrease the size of gap 6.5 Interference (a) Condition: two sets of waves of the same frequency meet each other ** (b) Stable interference pattern: two sources MUST be coherent. Coherent sources are sources with (i) same frequency (ii) constant phase difference (e.g. constantly in-phase or antiphase) – amplitudes are not necessarily the same. ** (c) Constructive interference, destructive interference and path difference Path difference Type of interference A 4λ – 3λ = λ Constructive B 4λ – 3.5λ = ½ λ Destructive C 3.5λ – 2.5λ = λ Constructive ** Constructive interference: path difference = 0λ, λ, 2λ, 3λ, … ** Destructive interference: path difference = ½λ, 1½λ, 2½λ, 3½λ, … St. Joseph’s Anglo-Chinese School 46 *Example 1 If λ = 0.25 cm, determine what kind of interference occurs at P. Solution: P Path difference = S1P – S2P 3.2 cm = 3.2 – 2.7 2.7 cm = 0.5 cm S1 S2 ∴ = 2λ Constructive interference occurs at P. Common mistake 1: Common mistake 2: Did not express the path difference in terms of Wrong presentation λ. Path difference = 3.2 - 2.7 ∴ Path difference = 3.2 - 2.7 = 0.5 Destructive interference occurs at P. = 0.5 = [Note: 0 .5 = 2λ 0.25 0 .5 = 2 ≠ 2 λ] 0.25 **(d) Factors affecting interference pattern Increase source separation Result: Nodal lines and antinodal lines become closer together Decrease wavelength St. Joseph’s Anglo-Chinese School 47 **Example 2: St. Joseph’s Anglo-Chinese School 48 6.6 Stationary Waves A stationary wave is formed by the superposition of two waves with the same frequency and amplitude travelling in opposite directions. N: Node (amplitude = 0) A: Antinode (maximum amplitude) (a) Waveforms of a stationary wave (i) a and b (c and d) are in-phase but with different amplitudes (ii) a and c, a and d, b and c, b and d are anti-phase (iii) when t = 0 s, all particles (a, b, c and d) are momentarily at rest. *(b) Comparing travelling waves and stationary waves 1 Traveling waves Stationary waves Energy is transmitted from one place to Energy is localized. another 2 3 All particles vibrate with the same Different particles vibrate with different amplitude. amplitudes. Neighbouring particles always vibrate Particles in the same loop vibrate in phase. out of phase. Particles in two adjacent loops vibrate in anti-phase. 4 Different particles reach their All particles reach their maximum maximum displacement at different displacement at the same times. times. St. Joseph’s Anglo-Chinese School 49 * Example 3: The figure below shows a stationary. At the instant shown, all particles reach their maximum displacement. It is given that the frequency of the vibrator is 25 Hz. (a) Find particles which are (i) vibrator in-phase (ii) anti-phase (iii) momentarily at rest (b) Find the wave speed. * (c) Draw the shape of the wave after 0.01 s. (d) What should be the frequency of the vibrator in order to produce 4 vibrating loops? Solution: (a) (i) in-phase: c and d (ii) anti-phase: b and c, b and d (iii) b, c and d (a is always at rest) (b) λ = 1.2 × 2 / 3 = 0.8 m Wave speed (c) Steps: v = fλ = (25)(0.8) = 20 ms-1 (1) f = 25 Hz ⇒ T = 1/25 = 0.04 s (2) 0.01 s = ¼ T After 0.01 s (¼ T), the waveform is: (d) for 1 vibrating loop: f1 = 25 = 8.33 Hz 3 For 4 vibrating loops: f4 = 4f1 = 4 × St. Joseph’s Anglo-Chinese School 25 = 33.3 Hz 3 50 Chapter 7 Light Waves 7.1 Diffraction of light (a) Laser passes through a narrow slit and spreads into the shadow of the slit. Red Green Blue (b) λred > λblue ⇔ diffraction of red light > diffraction of blue light (c) λred = 7 ×10-7 m. To show the diffraction of light, the slit must be very narrow. 7.2 Interference of light Young’s double-slit experiment *(a) Precautions: (i) use strong light source (or black out the laboratory) (ii) use monochromatic light (單色光) a ∆x (iii) all slits should be as narrow as possible (iv) slit separation should be very small (~ 0.5 mm) D (v) screen should be placed 1 – 2 m (b) Ways to increase fringe separation ∆x behind the double-slit (i) use light of longer wavelength λ (ii) decrease slit separation a (iii) increase the distance between the double-slit and the screen D. 7.3 Values commonly used in calculations f 1kHz = 103 Hz 1MHz = 106 Hz 1GHz = 109 Hz λ 1mm = 10-3 m 1µm = 10-6 m 1nm = 10-9 m St. Joseph’s Anglo-Chinese School 51 Example 1 A yellow light of frequency 5 × 1014 Hz is used in Young’s double-slit experiment (a) Find the wavelength of light. (b) What is the path difference at C? What can be observed at C? (c) If successive bright fringes are formed at P, Q, R and S, what are the path differences at P, Q, R and S respectively? (d) If the yellow source is replaced by a violet light source, what changes would be observed on the screen? Solution (a) v = fλ λ= 3 × 108 v = = 6 × 10 −7 m 14 f 5 × 10 (b) Path difference at C = 0 (or 0λ). Constructive interference occurs at C. Hence, a bright fringe is observed at C. (c) P Path difference Q -6 2λ = 1.2 × 10 m R -7 λ = 6 × 10 m S -7 λ = 6 × 10 m 2λ = 1.2 × 10-6 m (d) Violet fringes will be observed. The fringe separation is smaller because the wavelength of violet light is shorter than that of yellow light. 7.4 Diffraction Grating 2nd order 1st order 0th order 1st order Diffraction grating 2nd order Screen (a) bright fringes – constructive interference (b) Fringe separation can be increased by (i) decreasing the grating spacing (using fine grating) (ii) using light of longer wavelength St. Joseph’s Anglo-Chinese School 52 7.5 Electromagnetic Waves (a) show reflection, refraction, diffraction and interference (b) are transverse waves (c) can travel through vacuum (d) travel at speed of light in air / vacuum (3 × 108 ms-1) (e) can apply the equation v = fλ Example 3 Microwaves are used to study interference as shown below. The amplitude of signal collected by the received is showed in the following graph. amplitude A P (a) Explain why the amplitude varies as the receiver moves from A to B? (b) If XP = 23 cm, YP = 30 cm, find (i) the path difference at P, (ii) the wavelength of the microwave (c) Now the separation between the slits is increased. Sketch the graph of amplitude of microwave received along AB. Solution: (a) Interference occurs. Maximum amplitude is obtained because of constructive interference and minimum signal is due to destructive interference. (b) (i) Path difference at P = YP – XP = 30 – 23 = 7 cm (ii) Path difference at P = 2λ ∴ 2λ = 7 cm (constructive interference) λ = 3.5 cm (c) amplitude Separation between positions of maximum amplitudes is reduced. Increase source separation St. Joseph’s Anglo-Chinese School Nodal lines and antinode lines become closer together 53 St. Joseph’s Anglo-Chinese School 54 Chapter 8 Sound 8.1 Wave nature of sound (a) Reflection of sound – echoes heard from an obstacle. (b) Refraction microphone speaker CO2 Refraction of light Sound waves are focused by the balloon filled with carbon dioxide. A large sound is received by the microphone. (c) Diffraction frequency of human speech f ~ 100 Hz – 300 Hz wavelength λ ~ 1.1 m – 3.3 m [λ = v / f, v = 330 ms-1] ∵ λ > width of doorway ∴ significant diffraction of sound Door (d) Interference * Experiment procedures: Signal generator Speaker Speaker 1. Connect two speakers to a signal generator. 2. Set the frequency of the signal generator at 2000 Hz. 3. λ = v / f = 330 / 2000 = 0.165 m. 4. Place the two speakers 3λ apart. i.e. 3 × 0.165 ≈ 0.5 m apart. Walk across in front of the loud speakers. Detect any change in the loudness of the sound. 5. Connect a microphone to a CRO and move it across in front of the two loudspeakers. Detect any change in the amplitude on the waveform on the CRO. CRO amplitude O ** A B C mic. position – Constructive interference: O, B (maximum amplitude) – Destructive interference: A, C (minimum amplitude) – Due to background noise, the amplitude at destructive interference ≠ 0. St. Joseph’s Anglo-Chinese School 55 8.2 Properties of sound (a) Longitudinal waves (b) Need a medium to transmit sound. (sound cannot travel through vacuum) (c) vsolid > vliquid > vgas (sound travels fastest in solid) (d) Sounds are produced by vibrations (e) Audible frequency range: 20 Hz – 20 kHz 8.3 Musical notes and Noise Musical note Noise Waveform on CRO (regular in shape) Waveform on CRO (irregular) Measured in decibel (dB) (分貝) Waveform of a tuning fork * Pitch ------------- frequency Loudness ------------- amplitude Sound quality ------------- waveform on CRO 8.4 Ultrasound (a) sound waves of frequency greater than 20 kHz (b) properties are the same as sound waves (c) applications (i) ultrasonic scan of the foetus (胎兒) (ii) detecting shoal of fish (魚群) (iii) ultrasound flaw (裂紋) detector Example 1: 30 kHz ultrasound is used to detect shoal of fish. It is given that the speed of ultrasound in sea water is 1500 ms-1. (a) Find the wavelength of ultrasound. * (b) Explain ultrasound is used instead of audible sound. Solution: (a) λ = v / f = 1500 / (30 × 103) = 0.05 m * (b) Since wavelength of ultrasound is shorter, the degree of diffraction of ultrasound is smaller. St. Joseph’s Anglo-Chinese School 56 Chapter 1 Electrostatics 1.1 Electric charge (a) Like charges repel; unlike charges attract. (b) Charging methods (i) Rub an acetate strip with a duster. The rubbed acetate strip becomes positively charged. (Electrons are transferred from the acetate strip to the duster) (ii) Rub a polystyrene strip with a duster. The rubbed ac polystyrene strips become negatively charged. (Electrons are transferred from the duster to the polystyrene strip) (iii) Charging by Van de Graaff generator (c) Charge Q unit: C Example 1 The charge of an electron is e = -1.6 ×10-19 C. How many extra electrons are gained by a negatively charged rod of 2 ×10-8 C? Solution No. of extra electrons gained = 2 × 10 −8 = 1.25 × 1011 −19 1.6 × 10 (d) Attraction of a neutral object by a charged object (i) When a positively charged rod is put near to a neutral object, negative charges are induced on the side near the rod and positive charges are induced on the other side. (ii) The attraction between the rod and the negative charges is greater than the repulsion between the rod and the positive charges. (iii) Therefore, the neutral object is attracted by the charged rod. St. Joseph’s Anglo-Chinese School 57 1.2 Coulomb’s law Coulomb’s law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. F= Q1Q2 4πε 0 r 2 ε0 = 8.85 x 10-12 C2 N-1 m-2 Example 2 There are three charges Q1 (–2 µC), Q2 (1 µC) and Q3 (–3 µC) as shown below. Find the resultant force acting on Q2. Solution Q2 is attracted by Q1 and Q3. By F = Q1Q2 4πε 0 r 2 (2 × 10 )(1 × 10 ) = 1.798 N = 4π (8.85 × 10 )(0.1 ) (3 × 10 )(1 × 10 ) = 2.698 N = 4π (8.85 × 10 )(0.1 ) −6 F1 −6 −12 −6 F3 2 [Note: 10 cm = 0.1 m] −6 −12 2 Resultant force = F3 − F1 = 0.899 N (toward the right) 1.3 Electric Field (a) Electric field pattern Direction of field lines Meaning Density of field lines St. Joseph’s Anglo-Chinese School Go from a positive charge to a negative charge Show the direction of electric force acting on a positive test charge Strength of electric field 58 (b) Electric field strength and gravitational field strength Electric field strength E Definition Electric force per unit charge F E= q Isolated point charge: E= Parallel plates: E= Equation Unit Q 4πε 0 r 2 σ ε0 N C-1 Vector Example 3 Two point charges A (3 × 10–8 C) and B (–2 × 10–8 C) are placed 20 cm apart. Find the magnitude of the electric field strength at B due to A. Solution Electric field strength due to A: 3 × 10 −8 E= = = 6740 NC-1 2 −12 2 4πε 0 r 4π (8.85 × 10 )(0.2) QA Remark: Electric field strength due to A is calculated from the charge of A. Example 4 Resultant electric field Three point charges A (–3 × 10–8 C), B (2 × 10–8 C) and C (2 × 10–8 C) are shown. Find the electric field strength due to B and C at the position of A. Solution Electric field strength due to B 2 × 10 −8 EB = = = 4496 NC-1 4πε 0 r 2 4π (8.85 × 10 −12 )(0.2) 2 QB Electric field strength due to C 2 × 10 −8 = 4496 NC-1 4πε 0 r 2 4π (8.85 × 10 −12 )(0.2) 2 Eresultant = E B cos 30° + EC cos 30° = 7790 NC-1 EC = QC = St. Joseph’s Anglo-Chinese School Eresultant 30o 30o EC EB EC EC cos 30o EB cos 30 60° EB o 30o 30o EC sin 30o EB sin 30o 59 Chapter 2 Electric Circuits 2.1 Simple circuit electron flow battery conventional current bulb bulb The conventional current goes round the circuit from the positive terminal of the battery to the negative Current is the same at every part of a simple circuit terminal. 2.2 Current Definition: current is the rate of flow of charge current = charge time Q I = t or Q = It Unit: A 2.3 Potential difference, electromotive force, voltage Electromotive force (e.m.f) Potential difference (p.d.) Energy gained by a unit charge when the charge is the amount of electrical energy which passes through a battery changes into other forms of energy when a unit charge passes between two points in a circuit. e.m.f = 4V ⇔ 4J of electrical energy is gained p.d. = 1V ⇔ 1 J of electrical energy is by 1C of charge changed into heat and light energy when 1 C of charge passes through the bulb e.m.f = 4V Suppose 6C of charge passes through the circuit. Total amount of energy gained by the charge from the battery = (4)(6) = 24 J For 1st bulb, Total amount of charge converted into heat and light energy = (1)(6) = 6 J p.d. = 1V e.m.f p.d. = 3V ε = V1 + V2 V = For 2nd bulb, Total amount of charge converted into heat and light energy = (3)(6) = 18 J E Q St. Joseph’s Anglo-Chinese School 60 2.4 Resistance Definition: Resistance = voltage current R= V or I V= IR Unit: Ω Example 1 A hairdryer has a resistance of 24 Ω. Find the current passing through it when it is connected to a 220-V power supply. Solution By V= IR 220 = I(24) I = 9.17 A 2.5 Ohm’s law The voltage across a conductor is directly proportional to the current flowing through it, provided the temperature and other physical conditions are unchanged. R= [V ∝ I] V V = slope of V – I graph I Higher R Lower R I Example 2 The V–I graph of conductor X is shown. (a) Find the resistance of conductor X. (b) Conductor Y obeys Ohm’s law and its resistance is half that of conductor X. Sketch the V–I graph of conductor Y. Solution (a) Resistance = slope of V – I graph = 5 = 0.5Ω 10 (b) Resistance of Y = ½ (0.5) = 0.25 Ω i.e. when V = 5V, I = 20 A St. Joseph’s Anglo-Chinese School Conductor Y (Lower resistance, smaller slope) 61 Example 3 Remark: 1. Ammeters must be connected in series. 2. Voltmeters must be connected in parallel. St. Joseph’s Anglo-Chinese School 62 Factors affecting resistance Temperature T Length l and thickness A For metals: T↑ ⇒ R↑ For semi-conductors: T↑ ⇒ R↓ R∝ l A R=ρ Conductor Resistivity (Ωm) Copper: 1.7 × 10 -8 longer wire ⇒ R↑ l A Thicker wire ⇒ R↓ ρ = resistivity Semi-conductor Insulator Silicon: 2300 Polystyrene 1015 Example 4 Find the resistance of a copper wire if its length and diameter are 70 cm and 1 mm respectively. Given that the resistivity of copper is 1.7 × 10-8 Ωm. Solution By R = ρ l 0.7 = (1.7 × 10 −8 ) = 0.0152 Ω 2 A 10 − 3 π 2 [Α = πr2 = π(d/2)2] 2.6 Resistors in Series and Parallel Resistors in Series Resistors in parallel ε ε (e.m.f of a cell) I1 R1 I2 V1 R2 I V2 I1 R1 I2 IV1 1 R2 V2 I1 = I2 (Common) I = I1 + I2 ε = V1 + V2 ε = V1 = V2 (Same) 1 1 −1 R eq = ( + ) R1 R 2 Req = R1 + R2 If R1 = R2, then Req = ½ R1 = ½R2 St. Joseph’s Anglo-Chinese School 63 Example 5 Circuit analysis 1 Find I6, I8, I12; V6, V8, V12. Example 6 Circuit analysis 2 Figure 1 Find the current passing through each resistor and voltage across each resistor. Solution: simplify the above circuit Solution 1. 2. 1 1 R eq = 8 + ( + ) −1 = 8 + 4 = 12 Ω 6 12 12 12 I = = =1A Req 12 3. ∵ ∴ 4. V6= V12 = 12 – V8 = 4 V 5. I6 = I8 = I = 1A V8 = I8(8) = (1)(8) = 8V Figure 2 V6 4 = = 0.667 A 6 6 V12 4 = = 0.333 A 12 12 (OR I12 = I8 – I6 = 1 – 0.667 = 0.333 A) I 12 = Figure 3 In Figure 2 St. Joseph’s Anglo-Chinese School 64 2.8 Internal Resistance in Ammeters, Voltmeters and Cells * Addition of an ammeter will increase the total resistance of the circuit ⇒ I ≠ I’ * An ideal ammeter should have small resistance. ⇒ I ≈ I’ * Addition of a voltmeter will affect the voltage across 4Ω resistor. ⇒ V4 ≠ Voltmeter reading * An ideal voltmeter should have large resistance. ⇒ V4 ≈ Voltmeter reading St. Joseph’s Anglo-Chinese School 65 St. Joseph’s Anglo-Chinese School 66 Chapter 3 Domestic Electricity 3.1 Electric power and energy E V2 = VI = I 2 R = t R Electric power: P= Electrical energy: E = Pt = VIt = I 2 Rt = V2 t R Example 1 Find the total amount of energy stored by the rechargeable battery when it is fully charged. Solution 2000 mAh: (1) I = 2000 mA = 2A (2) t = 1 h = 3600 s Energy stored E = VIt = (1.2)(2)(3600) = 8640 J 3.2 Pay for electricity Kilowatt-hour meter measures the electrical energy consumed. Unit: kWh (1 度電) 1 kWh is the amount of electrical energy consumed by an appliance of 1000 W for 1 hour Example 2 Find the cost of electricity to operate five lamps of power 60 W for 8 hours. It is given that electricity costs $1.02 per kWh. Solution 60 E = Pt = (8) × 5 = 2.4 kWh 1000 Cost of electricity = $1.02 × 2.4 = $2.448 3.3 Mains electricity and household wiring (a) Electric socket and plug Earth wire Earth hole Switch Live hole Fuse Neutral wire Live wire Neutral hole Remark: The earth pin is designed longer to open ‘shutters’ on the live and the neutral holes. This ensures the earth wire is connected before the live wire. St. Joseph’s Anglo-Chinese School 67 (b) Cable Live wire Neutral wire Earth wire +220 V , – 220 V 0V To ground (0V) Brown Blue Yellow and green * Connects the metal body of an electrical appliance to the Earth. * In case of a fault, a large current will flow through the earth wire to earth. This prevents the user from getting an electric shock. (c) Switch (i) Must be fitted in the live wire. (ii) This makes sure that no part of the electrical appliance is at high voltage when the switch is turned off. (d) Fuse (i) Must be fitted in the live wire. (ii) If an excess current flows through the circuit or the circuit overloads, the fuse blows and breaks the circuit before the cable overheats and causes a fire. St. Joseph’s Anglo-Chinese School 68 Chapter 4 Electromagnetism 4.1 Magnetic field due to permanent magnets Magnetic field lines are used to show the strength and direction of a magnetic field. (a) Field lines run from the N-pole round to the S-pole. (b) When field lines are closely-spaced, field is strong and vice-versa. neutral point N S x N N uniform field 4.2 Magnetic field due to current carrying conductors Current-carrying conductor Magnetic field pattern current direction up out of page Long straight wire wire (top view) solenoid N S current direction Solenoid Inside the solenoid: Uniform field Outside the solenoid: Similar to that around a bar magnet Ways to increase magnetic field strength of a solenoid (1) increase the current (2) increase the number of turns of solenoid (for the same length of solenoid) (3) insert a soft-iron core through the solenoid St. Joseph’s Anglo-Chinese School 69 4.3 Magnetic force acting on a current-carrying conductor Magnetic force F = BIl Fleming’s left hand rule The magnetic force is increased if (1) the current is increased, ( F ∝ I ) (2) the magnetic field is increased, ( F ∝ B ) (3) there is a greater length of wire inside the magnetic field. ( F ∝ l ) Example 1 The figure below shows two straight parallel wires X and Y . Wire X carries current 3A and wire Y carries current 6 A in the opposite direction. (a) Draw the magnetic field produced by Y at the position of wire X. (b) Determine the direction of magnetic force acting on X. (c) Mary says, ‘Since the current through Y is two times that through X, the magnetic force acting on Y is two times that acting on X.’ Comment on Mary’s statement. Solution (a) BY IX (b) By Fleming’s left-hand rule, the direction of magnetic force acting on X is towards the right. BY F (c) Her statement is incorrect. According to Newton’s third law, the forces acting on both wires are an action and reaction pair. The magnetic force acting on X is equal to that acting on Y. 4.4 Motor Structure (1) magnets (3) carbon brushes (2) a coil (4) commutator Function of a commutator: The commutator reverses the current every half turn to keep the coil rotating continuously in one direction. Ways to increase the turning speed of the coil (1) increase the current (2) increase the magnetic field (3) increase the number of turns of the coil (4) increase the area of the coil within the magnetic field St. Joseph’s Anglo-Chinese School 70 Chapter 5 Electromagnetic Induction 5.1 Induced e.m.f and induced current S N N An e.m.f. is induced because S the coil An e.m.f is induced because the conductor experiences a changing magnetic field. cuts across magnetic field lines. A current is induced because the circuit is A current is NOT induced because the circuit is complete. not complete. Ways to increase induced e.m.f or induced current: (1) move the magnet or the wire faster (2) use a stronger magnet (3) use a coil of more turns or increase the length of wire within the magnetic field (4) insert a soft iron core in the coil Working out the direction of induced current (a) Lenz’s law An induced current always flows to oppose the change which started it in a magnetic field. S N N S induced current S N S N induced current (b) Fleming’s right-hand rule (i) motion or force F (ii) magnetic field B (iii) induced current I St. Joseph’s Anglo-Chinese School 71 5.2 Generator An a.c. generator A d.c. generator coil rotated carbon brushes carbon brushes commutator slip rings (i) The coil cuts through the field lines most rapidly. (max. induced current I) (ii) The coil does not cut through any field line momentarily. (no induced current I) An a.c. generator becomes a d.c. generator if the slip rings are replaced by a commutator. * The commutator reverses the connections of the coil to the outside circuit every half turn. Therefore, the current in the outside circuit always flows in the same direction. Ways to increase induced e.m.f or induced current: (1) increase the number of turns of the coil (2) use a stronger magnet (1) Increase the area of coil within the magnetic field (2) winding the coil on a soft-iron core (3) rotate the coil at a higher speed St. Joseph’s Anglo-Chinese School 72 Example 1 St. Joseph’s Anglo-Chinese School 73 Chapter 6 6.1 Transformers Vp N p (a) = Vs Ns Transmission of Electricity Soft iron core Primary coil Secondary coil step-up transformer step-down transformer N p < N s ⇔ Vp < Vs N s < N p ⇔ Vs < Vp (b) Efficiency η Ip Is Vp Vs Power input = VpIp Power output = VsIs Efficiency η Power output = × 100% Power input = Vs I s × 100% Vp I p If efficiency η = 100% (ideal transformer / no power loss in transformer), then VpIp = VsIs (c) Power loss in transformers Cause Improvement Resistance of coils Use thick wire for the coil Magnetization and demagnetization of the core Use soft-iron core Induced currents in the core Use laminated core Example 1 If the transformer has 40 turns in its secondary coil and both bulbs can operate at their rated values, assume the transformer is ideal, find (a) the number of turns in the primary coil. (b) the currents in the primary coil and secondary coil. 220 V a.c. 10 V, 20 W Ns = 40 Solution Vp N p N p 220 (a) = ⇒ = ⇒ N p = 880 Vs Ns 40 10 (b) Power output = 40 W (two light bulbs) V s I s = 40 ⇒ 20 I s = 40 ⇒ I s = 2 A Power input = Power output (ideal transformer) V p I p = 40 ⇒ 220 I p = 40 ⇒ I p = 0.182 A St. Joseph’s Anglo-Chinese School 74 6.2 High voltage power transmission * Power loss in transmitting electricity * Reason: resistance in power transmission cable Analysis: consider the following power transmission system. I Cable 10 Ω Power station Consumer unit 200 V, 1200 W Cable 10 Ω Procedures: (1) Current in cable I cable = P 1200 = =6A V 200 (2) Power loss (in cable) 2 Ploss = I cable R = (6) 2 (20) = 720 W (3) Efficiency of the power transmission system useful power output η= × 100% power input = 1200 − 720 × 100% = 40% 1200 (4) Voltage drop (due to cable) V drop = I cable R = (6)( 20) = 120 V (5) Voltage available to consumer unit Vo = 1200 − Vdrop = 200 − 120 = 80 V Conclusion (a) low efficiency (η = 40 %) large amount of power loss (b) insufficient terminal voltage (200V → 80V) St. Joseph’s Anglo-Chinese School 75 Solution: use high voltage to transmit electrical power ⇒ reduce current in transmission cable ⇒ reduce power loss in cable C Cable 10 Ω E A Cable 10 W G 200 V Consumer 1200 W Unit power station B 1:20 Cable 10 Ω 20:1 Cable 10 W F D H Assumption: ideal transformers are used (1) ∵ VAB = 200 V ∴ VCD = 200 × 20 = 4000 V (step-up transformer) (2) Current in transmission cable IcableVCD = 1200 (no power loss in transformer) Icable(4000) = 1200 Icable = 0.3 A (3) Power loss (in cable) (high voltage ⇒ low current) 2 Ploss = I cable R = (0.3) 2 ( 20) = 1.8 W (4) Efficiency of the power transmission system useful power output η= × 100% power input = 1200 − 1.8 × 100% = 99.85% 1200 (5) Voltage drop (due to cable) V drop = I cable R = (0.3)( 20) = 6 V VEF = VCD − Vdrop = 4000 − 6 = 3994 V (6) Voltage available to consumer unit Vo = VEF × 1 3994 = = 199.7 V 20 20 Conclusion (a) high efficiency η = 99.85% (nearly perfect!) (b) terminal voltage (Vo = 199.7 V ≈ 200V) St. Joseph’s Anglo-Chinese School 76