Page 12 1 a) Between t = 30 and t = 45 mins b) 7.5 mins c) i) distance travelled = area under graph between t = 0 and t = 12½ mins ii) average speed = total distance travelled total time for journey = total area under graph 60 mins 2 a) ∆v = 32 m/s a =10 m/s² t = ∆v = 32 = 3.2 s a 10 b) 3 a) OP – constant acceleration PQ – constant acceleration (greater than OP) QR – constant speed RS – constant deceleration b) O and S c) 6 m/s d) 70 s e) Total distance travelled = area under graph 4 a) u = d = 25 = 12.5 s t 2 b) speed is decreasing c) time for tree 3 to tree 4 is greater than for time for tree 2 to tree 3 5 a) Stage 1: 20 s Stage 2: t = 4800 = 400 s 12 Stage 3: 80 s b) c) Total distance travelled = (½ × 20 × 12) + (400 × 12) + (½× 80 × 12) = 120 + 4800 + 480 = 5400m d) average speed = total distance = 5400 = 10.8 m/s time taken 500 Page 17 1 Weigh the car using the Newton meter. Find the mass from mass = weight 10 m/s² (for rest of question see description on p. 16) 2 (For first part see description on p.16) To find the density of cork the mass, M, could still be found using the balance. But the volume must be calculated by finding its dimensions and using cross-sectional area × length to calculate the volume. Area = area of a circle, πr² d = 2 × radius = 2r Area = πr² Volume = πr²l Density = πr²l m 3 a) Volume = Reading Q – Reading P b) Mass = Reading S – Reading R c) i) P = m v ii) p = 57.5 = 2.3 g/cm³ 25 Page 17 1 l = 3.6 cm 2 A = πr² - π(d)² = π(0.6)² = 0.28 cm² 2 2 3 V = πr²l = 1.0 cm³ Volume of bundle – 10 × 10 – 10.0 cm³ P = m = 59.1 – 5.91 g/cm³ v 10.0 Page 21 1 a) length = 17.5 cm b) load = 2.25 N c) extension = length – initial length = 15 – 11 = 4 cm 2 a) It has both size and direction b) f = m × a =5×3 = 15 N 3 a) 11 cm b) 19 cm c) 19 − 11 = 8 cm Page 22 4 a) Q = elastic limit b) The extension is directly proportional to the force applied c) In the region QR, the extension is greater for the same increase in applied force d) k = F/x = gradient of graph = 8/x = 4 N/m 5 c) Extension = 67 – 40 = 27 cm d) 2.5 N Practical Question 1 m/g θ/° 2 Graph is a straight line through (0,0) Page 26 1 i) 1: force 2: moment ii) F = F1 + w + F2 iii) F 2 a) Resultant force = 0 Resultant moment = 0 b) 6 × 40 = F × 30 F = 240 = 8.0 N 30 c) Force = 0.5 N downwards Page 27 1 Use a set square 2 1/d ÷ 1/m 1.11 1.18 1. 25 1.33 1.43 3 c) G = 0.53 Nm 4 w = 0.53 = 1.1 N 0.490 Page 29 1 a) Line of action of weight of box acts outside its base b) i) less than ii) The centre of mass is higher and so the line of action of the weight will act outside the base at a longer angle Page 30 2 a) Plastic will rotate clockwise b) Plastic will not move 10 N ←−−−−−− Page 34 1 a) chemical, gravitational PE, kinetic, internal b) kinetic c) gravitational PE d) chemical 2 i) kinetic + gravitational potential ii) kinetic energy is dissipated as heat due to frictional forces Page 37 1 a) gravitational b) kinetic c) kinetic d) electrical e) heat Page 39 1 a) WD = f × d = mgh b) i) WD = 100 × 8 = 800 J ii) power = WD = 800 = 160 W t 5 iii) Increasing internal energy 2 a) i) WD = high = 50 × 10 × 4 = 200 J ii) P = WD = 200 = 100 W t 20 b) gravitational potential → kinetic → heat 3 a) PE = mgh = 1×10×7 = 70J b) mgh = ½mv² 70 = 0.5v² V = [symbol] 140 = 1.2 m/s c) energy dissipated as heat Page 43 1 a) Skate, since area is smaller for the same weight so pressure is greater. b) The area of the vehicle is large and so for a constant wind pressure, the face is high and could blow the vehicle over 2 a) Pressure = pgh = 1000 x 10 x 2 = 20000 N/m² b) Area = force = 50 = 0.0025m² pressure 20000 c) gravitational potential to kinetic Page 44 1. distance; time; m/s; gradient; direction; magnitude; rate of change; m/s² 2. 2m/s 3. 3m/s 7. 4g/cm³ 9. 5cm 12. 4.9J 12. 2500J 14. 5m 15. 200W 16. 36000 N/m² 17. 0.6 N/cm² Page 46 1 a) Speed increases with decreasing acceleration between P and R. Between R and S acceleration is zero, speed is steady. b) c) forces are balanced; resultant force = 0 d) distance between l and T = 40x120 = 4800m distance between P and Q = ½x6x50 = 150 m Page 47 2 a) Time several complete oscillations. Divide time by number of oscillations and then divide by 2 to find the time between P and Q. b) i) weight and tension ii) Towards centre of circle c) PE = mgh = 0.2×10×0.05 = 0.1J 3 b) a = ∆v = 12 = 1.5 m/s² t 8 c) distance = ½×5×12 = 30m d) f=mxa = 4000 × 1.2 = 4800N e) For the same accelerating force, there is a greater mass due to the extra passengers and so acceleration is less. 4 a) i) ii) Force increases b) i) ii) The force of friction was not large enough to provide the centripetal force. c) i) steady speed ii) circumference = 2TTr = TTd = 20.4cm iii) increase in speed per second = 25 = 8.3 m/s² 3 5 a) When clockwise moment due force of steam > anticlockwise moment due to weight, the beam turns clockwise and W is lifted. b) i) 12Nm = F × 0/2 f = 60N ii) P = F = 60 = 200000 N/m² A 0.0003 Page 53 1 a) Constant random motion b)As the temperature increases, the velocity of the particles increases. Particles collide with walls of container more often and with greater force and so the pressure increases. 2 a) Evaporation b) Particles with greatest ICE escape from surface of water, reducing the volume of water. Page 54 3 a) The level of water in the tube becomes lower. b) The air in the flask expands as the temperature increases. c) Below 0°C, water would freeze. 4 a) The dust particle is constantly bombarded by the air molecules, which are in constant random motion. b) i) The ones with the greatest kinetic energy. ii) These have the energy required to overcome the attractive forces between molecules. Page 55 5 a) i) Constant, random motion ii) Collides with container walls putting force on walls. b) i) Increases ii) Remains constant c) i) Movement with greater KE in gas. ii) Separation of molecules in gas is greater. Practical Question 1 The temperature of thermometer 2 decreases more rapidly than that of thermometer 1, due to the evaporation of the water. 2 The temperature of thermometer 3 decreases more rapidly than that of thermometer 2, because acetone evaporates more rapidly than water. Page 58 1 a) Choose a different liquid, one with a larger range of temperature between its freezing and boiling point. b) Increase the number of scale divisions and have a thinner thread in the container. 2 a) The piston must move to the right. b) The heat supplied to the piston is used to do work in pushing the piston to the right. The temperature remains constant. 3 a) They vibrate with greater kinetic energy and move further apart. b) i) Thermometers ii) Railway tracks Page 64 1 a) Mass of aluminium on balance. Change in temperature with thermometer energy delivered to the aluminium. b) Higher. c) Some heat is lost to the surroundings. 2 a) Evaporation occurs over a range of temperatures. Boiling occurs at a specific temperature (the boiling point). Evaporation decreases the temperature of the liquid. The temperature of the liquid remains constant during boiling. b) Energy is required to break the bonds between the liquid particles. All the heat energy is used to do work against the surroundings, not to increase the internal energy of the liquid. c) Є = P × t = 100 × 20 × 60 = 120000J 120000 = ΔML = 0.05 L L = 2.4 × 106 J/kg = 2400 J/kg 3 a) i) 20°C, 15°C ii) As the temperature of the water increases, more heat is lost to the surroundings and so the temperature rises at a lower rate. b) C = Є/MΔT = 60 × 210/0.075 × 40 = 4200 J/kg/°C c) 4 a) Find the mass of the beaker as a balance. Start the stopclock and simultaneously switch on the heater. After most of the ice has melted, stop the stopclock. Find the mass of the beaker and the water. b) Є = ΔML 60 × t = 120 × 340 t = 120 × 340/60 = 680s Page 66 c) i) Heat from the surroundings melts the ice. ii) Insulate the ice from the surroundings. Practical question 1 24°C 2 Beaker A: 8°C, Beaker B: 5°C 3 The Bunsen burner loses more heat to the surroundings. The power of the fixed voltage source is greater than the Bunsen burner. Page 71 1 Take the starting temperature on the thermometers. Read the thermometers after a fixed interval of time. The best absorber of heat radiation will have a higher temperature. 2 a) Radiation b) Conduction Page 73 1 a) Copper is a good conductor of heat. b) Plastic is a poor conductor of heat to prevent burns. c) i) The metal base conducts heat through the increased vibration of the particles. ii) Convection. 2 a) i) Copper is a good conductor of heat. ii) Plastic is an insulator. b) i) Conduction ii) Radiation and convection. c) equal to 40W 3 a) i) Conduction ii) The particles at A vibrate faster and pass on the increased vibrations to neighbouring particles. b) 4 different coloured surfaces are heated at equal distances from a radiant heater (see diagram page 70). The initial and final temperatures of each of the surfaces are taken after equal time intervals. Practical question 1 t/s, θ/°C 2 Place insulation on the bottom of the beaker. Cover the top of the beakers. Page 76 Summary Questions on unit 2 1 2 melt, 1, bal, 1, temperature, 1 3 Structure – see diagrams page 50. Arrangement of Close together; vibrating Far part; in constant molecules in fixed positions random motion Compressible No No ?? No Yes Shape Fills the bottom of the Fills the whole container container 4 Arrows point clockwise. 5 Є = MCΔT C = E / MΔT = 4500 / 0.5 × 10 = 9000 J/kg/°C 6 Є = ΔML L = 66 000 / 0.2 = 330000 J/kg 7 The pressure increases since the volume decreases at constant temperature. Collisions between the particles and walls of the container occur more frequently. Page 78 1 reading 2 = mass of beaker + stirrer + thermometer + water a) reading 3 – mass of beaker + stirrer + thermometer + water + melted ice b) i) heat lost = mass of water × 20 × specific heat capacity of water ii) heat gained = mass of ice × specific latent heat of fusion c) L = 12800 / 0.030 = 4.3 × 105 J/kg = 430000 J/kg d) The ice may not be at 0°C. Heat may be taken in from the surroundings to melt the ice. 2 Temperature decreasing. Volume increasing. 3 a) Vibrate faster; move further apart. b) i) Thermometer ii) Railway tracks 4 a) i) Constant random motion ii) Collide with walls and exert a force. b) i) increases ii) remains constant c) i) gas – constant random motion in all directions solid – vibrating about fixed positions ii) separation of molecules in a gas is much higher than in a solid. Page 83 1 a) b) Count the number of waves passing a point in 30s. Divide number of waves by 30s. Frequency = number of waves per second. 2 a) b) Air molecules are vibrating, alternately moving closer together and then further apart. c) λ = v / f = 330 / 500 = 0.66m Page 84 3 a) Separation of waves: should be equal between one wavefront and the next. Shape of wavefronts should be semicircular, not straight. b) c) V = f × λ = 8 × 0.012 = 0.096 m/s Page 88 1 a) Angle of incidence – 0° i.e. along normal b) Speed decreases, frequency is constant, wavelength decreases. c) d) sin30° / sin θ = 0.67 θ = sin-1 (sin30° / 0.67) = 48° 2 a) b) i) 90° ii) 43° iii) n = 1 / sin43° = 1.5 c) speed = 3 × 108 / 1.5 = 2.0 × 108 m/s Page 89 3 a) b) Virtual, inverted, same size c) Angle of incidence = 0° d) speed = 3 × 108 / 1.5 = 2 × 108 m/s e) angle of incidence at b> critical angle Page 93 1 a) infra red, UV, X-ray, (symbol)-ray b) sand 2 a) i) (symbol) or x-ray ii) infra-red or radio b) f = v / λ - 3 × 108 / 1.0 × 10–14 = 3.0 × 1022 Hz c) 3 × 108 m/s 3 a) b) i) dispersion ii) red iii) violet 4 5 a) Page 97 1 a) t = d / u = 440 / 330 = 1.3s b) First clap = 220 / 330 = 0.67s after clap Second clap = 660 / 330 = 2s after clap Page 98 2 a) 2000 / 60 = 33 times per second b) Frequency = 33 Hz c) Yes, hearing range is typically 20 Hz – 20 kHz 3 a) i) A building/wall. ii) d = u × t = 320 × 1.5 = 480 m b) v = d / t = 6 / 5 = 1.2 m/s Page 99 1 longitudinal; transverse; transverse; longitudinal; vibrations; parallel; wavelength; time period; waves; second; V = f × λ; reflection; refraction; diffraction. 2 3 4 a) sin r = sin 56 / 1.5 r = 34° angle of reflection = 56° b) sin r = sin 35 / 0.67 r = 59° angle of reflection = 59° c) angle of reflection = 60° There is no refracted ray. 5 7 a) λ = v / f = 3 × 108 / 1 × 1016 = 3 × 10–8m UV b) f = v / λ = 3 × 108 / 0.1 = 3 × 109 Hz radio c) λ = v / f = 3 × 108 / 1 × 1020 = 3 × 10–12m (symbol) 9 Different wavelengths of light refract by different amounts in a denser medium, causing white light to split into a spectrum. Page 101 10 11 f = v / λ = 330 / 0.1 = 3300 Hz speed = 3300 × 0.45 = 1485 m/s 12 u = d / t – 2 / 0.00042 = 4760 m/s Page 102 1 The image is inverted, virtual, same size as object Page 103 2 a) t = d / u – 498 / 332 = 1.5s b) 0.75s, 2.25s 3 a) They are standing in the path of the reflected second wave. b) 400 Hz c) λ = v/f – 330/400 = 0.825m d) Vibrations of particles are parallel to direction of wave. Page 104 4 a) i) ii) same size, upright, virtual, laterally inverted 5 a) Stop clock, starting period b) Start stopclock when smoke from pistol is seen Stop stopclock when sound is heard. Measure distance between student with starting pistol and observer. Time of travel is reading on stopclock. c) speed = distance / time d) Repeat the experiment several times, eliminate anomalous results and average e) i) 100 m/s ii) 1000 m/s 6 a) b) sin r = sin 40° / 1.52 = 25° c) i) 3 × 108 m/s ii) 3 × 108 m/s Page 108 1 a) A south pole is a south seeking pole. It is attracted to a north seeking pole. b) repulsion c) i) ii) The induced magnetism is lost. 2 a) i) iron ii) 1 nothing, 2 nothing b) 3 Page 115 1 a) 6.3 V; 330 (symbol) c) iii) A best fit line allows values of time or resistance between the plotted values to be found with greater accuracy. d) Temperature is greatest for the lowest value of resistance on the graph. 2 a) i) conductor ii) metal iii) Put a potential difference across it. b) i) insulator ii) plastic iii) insulator Page 116 3 a) b) Vary resistance of variable resistor to obtain 5 different voltage readings, taken on voltmeter. Take 5 corresponding values of current on the ammeter. c) i) R = V / I – 6 / 0.5 = 12 Ώ unknown resistor = 9 Ώ ii) Q = It – 0.5 × 120 = 60C iii) P = IV – 0.5 × 1.5 = 0.75W 4 a) Attach plate to earth. Bring positively charged rod towards plate. Remove earth lead and then remove rod. b) i) Q = It = 0.02 × 15 = 0.3C ii) V = IR = 0.02 × 10000 = 200 V Page 117 1 AB 50 cm AC 75 cm AD 100 cm 3 AB R = V / I = 2.53 Ώ AC = 4.00 Ώ AD = 5.20 Ώ 5 R/Ώ 6 R = 5.2 × 1.5 = 7.8 Ώ Page 124 1 a) same b) 2Ώ c) 4Ώ d) i) 1 Ώ ii) greater than e) i) increases ii) increases Page 125 2 a) b) i) I = P / V = 36 / 12 = 3A ii) R = V / I = 12 / 3 = 4 Ώ iii) 2 Ώ iv) Є = Pt – 36 × 30 = 1080 J c) lamps A and B have a potential difference of 6V not 12V across each of them. 3 The resistance of the thermistor varies with temperature. a) As the temperature increases, it takes a smaller share of the voltage, switching on the transistor and lighting the lamp. b) Vary the size of resistor R. c) An indicator light to show the oven has reached the required temperature. Page 126 4 a) d) R = V / I = 12 / 3 = 4 Ώ e i) Parallel circuit ii) 4.0 A 5 a) A – resistor B – light dependent resistor C – transistor D – bulb b) transistor, C c) Resistance of LDR increases in the dark, increasing in share of voltage from the supply. The potential difference across the base is enough to switch on the collector-emitter current and so the bulb lights. 6 a) Region around an electrostatic charge where another charge experiences a force. b) c) I = θ / t = 0.060 / 30 = 0.002 A d) Є = VIt – 1500 × 0.0080 × 10 = 120 J 7 a) i) low ii) low i) ii) No effect Page 128 1 2 There is an ammeter reading the current. 3 Variable resistor. 4 Increase resistance of variable resistor. Reduce voltage of power source. 5 A since current has to pass through more coils of wire. Page 136 1 a) Primary turns > Secondary turns b) Output voltage < Input voltage c) 12V d) The bell would receive too high a pd, which could cause it to overhead and be damaged. Page 137 2 a) The wire has a magnetic field around it when a current flows through it. The field due to the magnetic poles interacts with this field and puts a force on the wire. b) Out of the page. c) Electrical to kinetic. d) i) commutator and brushes (see page 135 – dc motor) ii) When the current flows through the coil a force acts on each side of the coil in opposite directions, due to the interaction with the magnetic field. The couple causes a turning effect on the coil. The commutator allows the polarity of the power supply to swap every half turn to keep the coil turning. 3 a) b) i) For an emf to be induced across the secondary coil, there must be a changing magnetic field in the primary coil. ii) The changing magnetic field due to the ac current induces an emf across the secondary coil. c) i) P = IV – 1.5 × 12 = 18W ii) Є = 18 × 3 = 54J Page 138 4 a) An alternating current produces an alternating magnetic field. In the core an alternating magnetic field links the two coils. In the secondary coil, an emf is induced due to the changing magnetic field. b) More turns on secondary coil than on primary. c) The magnetic field linking the two coils would be too weak. d) 0.2A 5 a) Attach the millivoltmeter across the wire. Move the wire between the poles of the magnet. b) The millivoltmeter would give a reading c) generator Page 140 a) i) ii) The positive changes on the positive phase attract the negatively charged electron. b) i) horizontal deflection plates ii) vertical deflection plates c) Page 141 2 a) i) A and B ii) in the filament iii) electrons iv) straight line from left to right v) light is emitted b) c) Otherwise the electrons would collide with air molecules and lose kinetic energy. Page 142 2 3 a) Q = It = 3 × 6 = 18C b) I = Q/t = 0.006 / 100 = 60 μC c) t = Q / I – IJ / 0.1 = 150s 4 V = IR a) V = 100 × 0.5 = 50V b) I = V / R = 10 / 330 = 0.030 A c) R = V / I = 1.5 / 0.8 – 1.9 Ώ Page 142 5 a) P = IV – 3 × 12 – 36W b) I = P / V = 40 / 230 = 0.17A c) V = P / I = 100 / 0.5 = 200V 6 If the live wire touches the metal casing of an appliance, the current flows through the earth wire to ground. The current is large enough to melt the fuse and break the circuit, protecting the person from an electric shock. 7 8 See page 122 9 A small current in the coil produces a magnetic field which attracts the iron armature. The iron armature pivots to close the contacts in the high voltage circuit, switching on a large pd. 10 11 field; force; force; left hand; field; current; force; current; strength; magnetic field 12. See page 135-136 13 See page 132 14 See page 132 15 Spin the coil faster Stronger magnet More turns on coil 16 The transformer requires a changing current to produce a changing magnetic field. Page 148 2 a) When the switch is closed a current flows in the circuit, through the metal spokes. The magnetic field due to this current and due to the fixed magnets interact and put a force on the spokes, which makes the wheel turn. b) Anticlockwise. c) See page 135 d) The commutator keeps electrical contact with the brushes but the brushes swap sides of the commutator every half turn. This keeps the coil moving in the same direction. 3 a) I = P / V = 9 / 6 = 1.5A b) i) 8 Ώ ii) 6V c) i) brightness decreases ii) Resistance in the circuit increases and so the current decreases, which reduces the brightness. d) i) 4 Ώ ii) 4 Ώ 4 a) i) Prevents the current flowing in one direction. ii) b) i) ii) Inverts the input 5 a) i) Group 1 ii) Group 2 iii) Plastics b) c) Region around a changed particle where another changed particle experiences a force. 6 a) electrons b) A c) i) D ii) Produces a bright spot where the electrons hit the stream. d) Deflects the cathode rays vertically. e) Otherwise the electrons would collide with air molecules and not reach the screen. 7 a) i) Potential divider. ii) 12 Ώ iii) I = V / R = 6 / 12 = 0.5 A iv) 5V v) 5V b) i) 1:6V, 2:0V ii) Page 157 1 a) proton and neutron b) electron c) 4 d) +2 2 α - into page β - out of page (symbol) – net deflected 3 a) An electron is removed from the atom by collision with the ionising radiation. b) α radiation consisted of helium nuclei, which are much more massive than β radiation, which consists of electrons. Page 153 1 a) 0 b) –1 c) β 2 a) 4 b) 2 c) α 3 a) 0 b) –1 c) β Page 157 1 ½ life – 1.6 min 2 a) b) Take count over one minute without the α source present i.e. the background count. Take count over one minute with the α source very close to the detector. Take a count over one minute with the α source 3 cm from the detector. Repeat each reading two more times. 3 See page 154-156 Page 161 1 Protons 92; newtons: 146 [note printing error – 192 should be 92] 2 a) b) The vast majority of the α particles go straight through undetected, which is evidence that the atom is mainly space. Very few are deflected straight back, because they approach a gold nucleus head on. The gold nucleus must contain all of the positive charge of the atom and therefore almost all of the mass. 3 a) 2 protons and 2 neutrons b) 1 electron Page 162 1 alpha; βeta; gamma; nucleus; gamma; alpha; paper; millimetres; deflected; positive; deflected right Page 162 2 3 4 6 8 2 1 92 146 95 136 Page 164 1 a) b) Measure count with detector opposite beam of β particles with pd switched off. Switch on pd and take new count. Move detector until a high reading is obtained closer to the positive plate, indicating that the particles have been deflected. c) The reduction in count rate when the pd was switched on and the increase in count rate when the detector was moved closer to the positive plate. d) The negative β particles are attracted to the positive plate. 2 a) 16 min b) i) ii) Take count over 1 minute without the β source. Take count over 1 minute with β source. Take count over 1 minute with β source and aluminium sheet in place. Repeat each of the three readings twice more. 3 a) 92 b) in orbitals around the nucleus c) 146 d) in the nucleus e) decreases by 3 4 a) i) Time taken for half of the radioactive nucleis to decay. ii) The radiation in the surroundings due to man-made and natural sources. b) i) 80–25 = 55 counts/minute ii) 1:15 counts/minute, 2:40 counts/minute iii) 28 minutes iv) 25 counts/minute = background count 5 a) i) 2 protons, 2 neutrons ii) electron b) 11 protons, 11 electrons, 13 neutrons c) i) 0 ii) –1 iii) β 6 a) strontium –90 since it has the greatest half life.
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