Cambridge IGCSE™ Physics Answers to the Student Book * Supplement questions are indicated with an asterisk Cambridge Assessment International Education bears no responsibility for the example answers to questions taken from its past question papers which are contained in this publication. Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded may be different. References to assessment and/or assessment preparation are the publisher’s interpretation of the syllabus requirements and may not fully reflect the approach of Cambridge Assessment International Education. Section 1 Motion, forces and energy 1.1 Physical quantities and measurement techniques Test yourself questions 1 2 3 4 5 6 7 8 a 10 b 40 c 5 d 67 e 1000 a 3.00 b 5.50 c 8.70 d 0.43 e 0.1 a 1 × 105; 3.5 × 103; 4.28 × 108; 5.04 × 102; 2.7056 × 104 b 1000; 2 000 000; 69 200; 134; 1 000 000 000 a 1 × 10–3; 7 ×10–5; 1 × 10–7; 5 × 10–5 b 5 × 10–1; 8.4 × 10–2; 3.6 × 10–4; 1.04 × 10–3 Thickness of 100 pages = 100 × 0.10 mm = 10 mm Thickness of two covers = 2 × 0.20 mm = 0.40 mm Total thickness of book = 10 mm + 0.4 mm = 10.4 mm = 10 mm (correct to 2 s.f.) a two b three c four d two Volume of rectangular block = length × breadth × height = 4.1 cm × 2.8 cm × 2.1 cm = 24 cm3 Stopwatch; at least 5 Now put this into practice questions (Page 4) 1 Area of triangle = × base × height = × 8 cm × 12 cm = 48 cm2 2 Circumference 2πr = 2π × 6 cm = 38 cm (Page 5) Volume V = length × breadth × height = 30 cm × 25 cm × 15 cm = 11 250 cm3 = 11 250 × 10–6 m3 = 1.13 × 10–2 m3 2 Volume of cylinder V = πr2h = π × (5.0 cm)2 × 25cm = 2000 cm3 = 2 × 10–3 m3 (Page 9) 1 1 2 a a 10 b 0.577 b 8.6 1.00 c c 9.2 1.73 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 1 Cambridge IGCSE Physics Student Book answers 3 Let Fx = 5.0 and Fy = 7.0 then: and 4 𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �5.02 + 7.02 = √25 + 49 = √74 = 8.6 N so 𝜃𝜃 = 54º Resultant force is 8.6 N acting at 54º to the 5.0 N force Let Fx = 6.0 and Fy = 8.0 then: 𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �6.02 + 8.02 = √36 + 64 = √100 = 10 m/s and so 𝜃𝜃 = 53º Resultant velocity is 10 m/s acting at 53º to the horizontal Practical work questions (Page 6) 1 2 3 4 5 6 7 Students’ own responses based on their results Students’ own responses based on their results Students’ own responses based on their results Students’ own plans Record the time for at least 5 complete oscillations with a stopwatch; to determine the period divide the time by the number of oscillations BOAOB Length of pendulum is the distance from the lower end of the metal plates to the centre of the bob Exam-style questions (Page 11) 1 a b c 2 a b 3 a b 4 a b c Volume of chocolate bar, V1 = length × breadth × height = 10 cm × 2 cm × 2 cm = 40 cm3; Volume of chocolate bar, V2 = 2 cm × 2 cm × 2 cm = 8 cm3 Number of bars with same volume = V1 / V2 = 40 / 8 = 5 Time period = 8 s / 10 = 0.8 s [1] [2] [1] [2] [2] [Total: 8] Average thickness = 6 mm / 60 = 0.1 mm [2] Number of blocks = (40 cm × 40 cm × 20 cm) / (10 cm × 10 cm × 4 cm) = 80 [5] [Total: 7] Volume of water = 6 cm × 6 cm × 7 cm = 252 cm3 [3] 3 Volume of stone = 6 cm × 6 cm × (9 – 7) cm = 6 cm × 6 cm × 2 cm = 72 cm [4] [Total: 7] Metre, second [2] 3 [1] 2 2 i πr ii 2πr iii πr h [4] [Total: 7] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 2 Cambridge IGCSE Physics Student Book answers [Going further] 5 a 2.31 mm b 14.97 mm 6 a b C 13 N at 67º to 5 N force [2] [2] [Total: 4] [1] [7] [Total: 8] 1.2 Motion Test yourself questions 1 a Average speed = distance moved / time taken = 400 m / 20 s = 20 m/s b Distance moved / time taken = 1500 m / (4 × 60) s = 6.25 m/s 2 a Average speed = (10 m/s + 20 m/s) / 2 = 15 m/s b Distance = v t = 15 m/s × 60 s = 900 m *3 a Acceleration = change of speed / time taken = 6 m/s / 3 s = 2 m/s2 b Acceleration = –6 m/s / 2 s = –3 m/s2 *4 Time taken = change of speed / acceleration = 500 km/h / 10 km/h/s = 50 s 5 a Straight-line graph through the origin b Positive, constant *c Acceleration = slope of graph = 16 m/s / 4 s = 4 m/s2 d Area of triangle = base × height / 2 = 4 s × 16 m/s / 2 = 32 m e 32 m 6 a Straight line through the origin b Speed = gradient of graph = constant c Speed = distance/time = 18 m / 6 s = 3 m/s 7 Acceleration = change of speed / time taken, so Time = change of speed / acceleration = (30 – 0) m/s / 10 m/s2 = 3 s Distance = average speed × time = 30 m/s / 2 × 3 s = 45 m Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 3 Cambridge IGCSE Physics Student Book answers v = at = 9.8 m/s2 × 2 s = 20 m/s Distance = average velocity × time = (0 + 20) m/s / 2 × 2 s = 20 m 8 a b 9 Let Fx = 12.0 and Fy = 5.0 then: 𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = �12.02 + 5.02 = √144 + 25 = √169 = 13 m/s and so 𝜃𝜃 = 23º Resultant velocity is 13 m/s acting at 23º to the horizontal Now put this into practice questions (Page 17) 1 2 v = u + at = 0 + 0.8 m/s2 × 4 s = 3.2 m/s s = (u + v)t / 2 = (10 + 20) m/s × 5 / 2 = 75 m 3 s = ut + at2 = 0 × 5 s + (+2 m/s2) 52 s2 = 25 m Practical work questions (Page 19) 1 2 3 4 The speed increases because the mass falls further in each 1/50 s 33 × 1/50 = 0.66 s Reaction times will be longer than the small time interval to be measured. Also the stopwatch would only give an average speed for the fall; changes in speed and acceleration could not then be evaluated No difference Exam-style questions (Page 23) *1 a b c Average speed = (0 + 8) / 2 = 4 m/s s = v × t = 4 m/s × 4 s = 16 m Acceleration = 2 m/s2 *2 Change in speed = at = 1 m/s2 × 15 s = 15 m/s; final speed = (10 + 15) m/s = 25 m/s 3 a i 60 km ii (6pm – 1pm) = 5 hours [Total: 4] [1] [1] iii v = distance / time = 60 km / 5 h = 12 km/h [1] iv 2 (flat regions of graph) [1] v [1] (0.5 + 1.0) hours = 1.5 hours vi v = distance / time = 60 km / (5 – 1.5) h = 60 km / 3.5 h = 17 km/h 4 [2] [3] [2] [Total: 7] b Steepest line: EF a b c 100 m v = distance / time = 100 m / 5 s = 20 m/s Slows down (slope of graph decreases) Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [2] [2] [Total: 9] [1] [1] [2] 4 Cambridge IGCSE Physics Student Book answers d *5 a b c *6 a b c d e f [3] 2 Acceleration = change in speed / time taken = (5 – 0) m/s / 4 s = 1.25 m/s i Distance = average velocity × time = 5.0 m/s / 2 × 4 s = 10 m (or area under slope = (base × height of triangle) / 2 =10 m) ii Distance = average velocity × time = 5.0 m/s × 9 s = 45 m (= area under horizontal line between 4 s and 13 s) Remaining distance = (100 – 45 – 10) m = 45 m Time taken for remaining distance = distance / velocity = 45 m / 5 m/s = 9 s Total time for 100 m = (13 + 9) s = 22 s Graph needs to be extended horizontally to 22 s i Accelerating: OA, BC ii Decelerating: DE iii Constant speed: AB, CD OA: a = change in speed / time = +80 km/h2 AB: v = 80 km/h BC: a = (100 – 80) km/h / 0.5 h = +40 km/h2 CD: v = 100 km/h DE: a = (0 – 100) km/h / 0.5 h = –200 km/h2 Distance = average speed × time OA: distance = 80 km/h / 2 × 1 h = 40 km AB: distance = 80 km/h × 2 h = 160 km BC: distance = (80 + 100) km/h / 2 × 0.5 h = 90 km/h × 0.5 h = 45 km CD: distance = 100 km/h ×1 h = 100 km DE: distance = 100 km/h / 2 × 0.5 h = 25 km (40 + 160 + 45 + 100 + 25) km = 370 km Average speed = distance / time = 370 km / 5 h = 74 km/h Zero and 5 hours Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [Total: 7] [2] [2] [2] [1] [1] [1] [1] [Total: 10] [1] [1] [1] [3] [3] [1] [1] [1] [Total: 12] 5 Cambridge IGCSE Physics Student Book answers 7 a b c Constant speed Distance = 600 m Speed = distance / time = 600 m / 30 s = 20 m/s 8 a i ii i ii v = at = 9.8 m/s2 × 1 s = 9.8 m/s v = at = 9.8 m/s2 × 3 s = 29 m/s Distance = average velocity × time = (0 + 9.8) / 2 m/s × 1 s = 4.9 m Distance = average velocity × time = (0 + 29) / 2 m/s × 3 s = 44 m i ii Weight Air resistance b *9 a b Falls at constant velocity (terminal velocity) [2] [1] [2] [Total: 5] [2] [2] [2] [2] [Total: 8] [1] [1] [2] [Total: 4] 1.3 Mass and weight Test yourself questions 1 2 3 a a b a b 1N b 50 N c 0.50 N Weight on Earth = mass × g = 12 kg × 9.8 m/s2 = 120 N Weight on Moon = mass × gmoon = 12 kg × (10 / 6) m/s2 = 20 N Weight on Moon = mg = 80 kg × 1.6 N/kg = 128 N i 0 N/kg ii 0 N Exam-style questions (Page 28) 1 a i ii Mass is a measure of the quantity of matter in an object (at rest relative to an observer) Weight is a gravitational force on an object that has mass iii Using a balance [1] b c d C, mass × g B, mass i N ii 2 a b Gravitational field strength is the force per unit mass; g = W/m i W = mg so m = W/g = 1850 N / 9.8 m/s2 = 189 kg ii W = mg = 189 kg × 3.7 m/s2 = 700 N 3 a A gravitational field exerts a gravitational force on any mass in the field *b Determines the weight of a mass. c Force per unit mass; g = W/m d W = mg = 200 kg × 8.8 N/kg = 1760 N m/s2 [1] [2] iii N/kg Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [3] [Total: 9] [2] [2] [2] [Total: 6] [2] [1] [2] [2] [Total: 7] 6 Cambridge IGCSE Physics Student Book answers 1.4 Density Test yourself questions 1 a b i ρ = m/V = 100 g / 10 cm3 = 10 g/cm3 ii V = m/ρ = 38 g / 19 g/cm3 = 2.0 cm3 iii V = m/ρ = 95 g / 19 g/cm3 = 5.0 cm3 ii ρ = m/V = 9 kg / 3 m3 = 3 kg/m3 Volume of bar = 4 cm × 3 cm × 1 cm = 12 cm3 ρ = m/V = 96 g /12 cm3 = 8.0 g/cm3 b ρ = m/V = 96 × 10–3 kg / 12 × 10–6 m3 = 8.0 × 103 kg/m3 b ρ = m/V = 60 g / 10 cm3 = 6 g/cm3 3 a V = (60 – 50) cm3 = 10 cm3 *4 Liquid A 2 a Now put this into practice questions (Page 30) 1 2 𝜌𝜌 = 𝑚𝑚 𝑉𝑉 = 2.7 g/cm3 a m = V × ρ = 4 cm3 × 11 g/cm3 = 44 g b 𝑉𝑉 = 𝑚𝑚 𝜌𝜌 = 55 g 11 g/cm3 = 5 cm3 Exam-style questions (Page 32) 1 a b 2 a 3 A i m = V × ρ = 5 m3 × 3000 kg/m3 = 15000 kg ii m = V × ρ = 10 m × 5.0 m × 2.0 m × 1.3 kg/m3 = 130 kg Measure the mass m1 of an empty beaker on a balance. Transfer a known volume V of the liquid from a burette or a measuring cylinder into the beaker. Measure the mass m2 of the beaker plus liquid. Mass of liquid m = (m2 – m1) and so ρ = m/V can be calculated b Mass of liquid = (170 –130) g = 40 g Density of liquid ρ = m/V = 40 g / 50 cm3 = 0.8 g/cm3 c The density of ice is less than the density of water *d The density of oil is less than the density of water a b i V = 10 cm × 8 cm × 20 cm = 1600 cm3 = 1.6 × 10–3 m3 ii ρ = m/V = 1.2 kg / (1.6 × 10–3 m3) = 750 kg/m3 ρ = m/V = 33 g / 30 cm3 = 1100 kg/m3 [1] [3] [3] [Total: 7] [4] [4] [1] [1] [Total: 10] [2] [3] [3] [Total: 8] 1.5 Forces Test yourself questions *1 OE 2 May cause an object to move / change speed / change direction; change the shape/size of an object Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 7 Cambridge IGCSE Physics Student Book answers 3 Extension is proportional to stretching force over OE *4 k = F/x = 200 × 10–3 kg × 9.8 N/kg / (4 × 10–2) m = 49 N/m *5 The limit of proportionality can be defined as the point at which the load–extension graph becomes non-linear because the extension is no longer proportional to the stretching force. 6 If the ring does not move there is no net force on the ring: 140 N = 100 N + FH, so FH = 140 N – 100 N = 40 N 7 Resulting forward force = 100 N – 50 N = 50 N 8 Total upward force = 2 F Total downward force = 50 N In equilibrium, TUF = TDF so 2 F = 50 N and F = 25 N *9 Resultant force is 50 N at an angle of 53o to the 30 N force 10 D; Total force = 20 N + 30 N = 50 N is larger than any other case 11 Since acceleration = 0, there is no resultant force and F = P = 20 N *12 a Resultant force = mass × acceleration = 1000 kg × 5 m/s2 = 5000 N b Acceleration = resultant force / mass = 30 N / 2 kg = 15 m/s2 *13 a Acceleration = change in speed / time = 8 m/s / 2 s = 4 m/s2 b F = ma = (500/1000) kg × 4 m/s2 = 2 N 14 a Friction occurs when there is motion between two surfaces b impedes motion / reduces speed; causes heating 15 Air resistance between the air and the moving car acts to reduce speed; friction between the tyres and the road acts to reduce the speed; the motor acts to increase the speed. When the forces in each direction are equal and opposite, the resultant force is zero and the car maintains a constant speed *16 The force is greater than the stalk can resist *17 Less 18 Yes, moments equal. Clockwise moment = 40 N × 1 m = 40 N m; anticlockwise moment = 20 N × 2 m = 40 N m 19 Clockwise moment = W × (45 – 25) cm = W × 20 cm Anticlockwise moment = 20 N × (25 – 10) cm = 300 N cm Equating clockwise and anticlockwise moments gives W × 20 cm = 300 N cm so W = 300 N cm / 20 cm = 15 N 20 C; moment of force = force × perpendicular distance from pivot Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 8 Cambridge IGCSE Physics Student Book answers 21 Yes; clockwise moment = 60 N × 0.5 m = 30 N m; anticlockwise moment = 20 N × 1.5 m = 30 N m so clockwise and anticlockwise moments equal *22 200 N; clockwise moment = W × 3 m; Anticlockwise moment 160 N × 3 m + 120 N × 1 m = 480 N m + 120 N m = 600 N m Equating moments gives W = 600 N m / 3 m = 200 N 23 Balancing a beam experiment: Pivot a ruler at its centre, add a weight either side of pivot so that beam is balanced. Record the position of each weight and calculate its moment about the pivot. Check that the clockwise moment equals the anticlockwise moment when the ruler is in equilibrium. Repeat by placing weights at different distances along the ruler. 24 a Centre of ruler b Centre of sphere 25 a When the vertical line through its centre of gravity lies outside its base b By lowering its centre of gravity and increasing the area of its base 26 i unstable equilibrium ii stable equilibrium iii neutral equilibrium Now put this into practice questions (Page 35) *1 k = F/x = 4 N / (2 × 10–3) m = 2000 N/m *2 F = kx so x = F/k = (2 N × 9.8 N/kg) / 250 N/m = 78 × 10–3 m = 7.8 mm (Page 39) *1 a resultant force = 10 N – 8 N = 2 N. b a = F/m = 2 N / 5 kg = 0.4 m/s2. *2 F = ma = 7 kg × 0.5 m/s2 = 3.5 N Practical work questions (Page 34) 1 2 3 A straight line Extension proportional to stretching force Use a set square to ensure the ruler is vertical; avoid parallax errors in readings; add pointer at lower edge of spring to ensure readings taken from same place each time; repeat readings 4 Calculate force/extension for each pair of readings to test if a constant value is obtained or plot a graph of stretching force against extension. 5 Two springs, 2 hooks to support springs, 1 kg weight, string, paper, board, pins, protractor, ruler 6 Mass or angle POQ 7 Students’ own responses based on their results 8 Students’ own responses based on their results 9 Students’ own responses based on their results 10 Force and mass 11 Test independent variables separately: (a) calculate F/a for each pair of readings to see if a constant value is obtained. (b) calculate ma for each pair of readings to see if a constant value is obtained 12 Distances d1 and d2 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 9 Cambridge IGCSE Physics Student Book answers 13 a c 14 a b Moment = 5 N × 10 cm = 50 N cm b Moment = 5 N × 15 cm = 75 N cm Moment = 5 N × 30 cm = 150 N cm Tie a piece of string onto a weight used to find a vertical line; the string indicates the vertical if the plumb line is allowed to hang freely. 15 Vertically below the point of suspension Exam-style questions (Page 52) 1 a b *2 a b 3 a Hang a spring from a clamp and support a ruler vertically behind it. Mark the position of the lower end of the spring. Hang different weights from the bottom of the spring and record the scale readings in a table. Calculate the extension and plot a load–extension graph. Precautions: ensure ruler is vertical and no parallax occurs in readings, repeat readings. [6] The extension of the spring is proportional to the stretching force Mass/g Stretching force/N Scale reading/mm Extension/mm 0 0 20.0 0 100 0.98 20.2 0.2 200 1.96 20.4 0.4 300 2.94 20.6 0.6 400 3.92 20.8 0.8 500 4.90 21.0 1.0 k = F/x1 = 4 N / (22 – 10) cm = 0.33 N/cm; i x2 = F/k = 6 N / (4/12 N/cm) = 18 cm; ii Total length of spring = original length + extension = 10 cm + 18 cm = 28 cm i ii 12 N 5 N or 5 N 12N Resultant force = 12 N – 5 N = 7 N; 7 N or 7N [4] [Total:10] [3] [2] [1] [Total: 6] [2] [2] *b 𝐹𝐹 = �𝐹𝐹𝑋𝑋2 + 𝐹𝐹𝑌𝑌2 = √52 + 122 = √25 + 144 = √169 = 13 N and so 𝜃𝜃 = 67º Resultant force = 13 N at 67º to 5 N force *4 a b c a = change in velocity / time = 5 m/s / 10 s = 0.5 m/s2 Average speed = (0 + 5) m/s / 2 = 2.5 m/s Distance = average speed × time = 2.5 m/s × 10 s = 25 m Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [6] [Total: 10] [2] [2] [2] 10 Cambridge IGCSE Physics Student Book answers d *5 a b Backward frictional forces (due to air resistance and friction between road and tyres) are equal to the forward force exerted on the bicycle Rearranging F = ma gives a = F/m. If F is large and m is small (due to the use of lightweight material) the acceleration will be large, as is required for a racing car a = F/m; if the car has a small engine, F is small but m is also small. Low F reduces a, but a low m will increase a. If m is low enough, the acceleration can still be large. [2] [Total: 8] [3] [3] [Total: 6] 3 6 a Weight = mg = 500 kg × 9.8 N/kg = 4.9 × 10 N [1] b i Resultant force = 25 000 N – 4900 N = 20 100 N; [2] 2 [3] *ii From F = ma, initial acceleration a = F/m = 20 100 N / 500 kg = 40 m/s [Total: 6] *7 a Friction between tyres and road [2] b i larger ii smaller iii larger [3] c Slicks allow greater speed in dry conditions but in wet conditions treads provide frictional force to prevent skidding [2] [Total: 7] *8 a Circumference = 2 π r [1] b Speed, v = distance / time so T = 2 π r/v [2] c T = 2 π r/v = 2 × π × 6400 × 103 m / 8000 m/s = 5000 s (83 min) [4] [Total: 7] 9 a 0 [2] b between 0 and 5 N m [2] c 25 × 0.2 = 5 N m [2] [Total: 6] 10 Taking moments about pivot: a clockwise moment = 10 N × (50 – 40) cm = 10 N × 10 cm = 100 N cm [3] b anticlockwise moment = 3 N × (40 –10) cm = 3 N × 30 cm = 90 N cm [3] c clockwise moment is greater than anticlockwise moment so beam tips to right [2] [Total: 8] 11 a Suspend lamina so it can swing freely from a nail clamped in a stand. When the lamina comes to rest, locate and draw the vertical line from the point of suspension with a plumb line; repeat process using a different suspension point. Centre of gravity located where the two lines cross. [5] b i Stability increases because centre of gravity is lowered [2] ii Stability increases because the centre of gravity is lowered and the area of the base is increased Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [2] [Total: 9] 11 Cambridge IGCSE Physics Student Book answers Alternative to practical questions (Page 54) 12 a b 13 a Support a spring in a clamp and fix ruler vertically behind it. Draw up a table to record stretching force, scale reading and extension. Record scale reading opposite the lower end of the unweighted spring. Increase the load on the spring in 100g stages and record the scale reading at lower end of spring for each load. Calculate the extension for each load and plot a load–extension graph. Use set square to ensure the ruler is vertical; avoid parallax errors in readings; add pointer at lower edge of spring to ensure readings taken from same place each time; repeat readings. [1] [1] [1] [1] [1] [3] [Total: 8] i ii OL *iii L *b For example: gradient = 3 N / 6 mm = 0.5 N/mm *c 0.5 N/mm Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [4] [2] [1] [2] [1] [Total: 10] 12 Cambridge IGCSE Physics Student Book answers 14 a i ii iii iv [1] [2] [2] [2] Mass/g Force/N Ruler reading/cm d/cm F × d/N cm 50 0.49 5 20 9.8 A 50 0.49 10 15 7.4 B 50 0.49 15 10 4.9 C 50 0.49 20 5 2.5 D 100 0.98 30 5 4.9 E 100 0.98 35 10 9.8 F 100 0.98 40 15 14.7 G 150 1.47 20 5 7.4 H 150 1.47 35 10 14.7 I b AF, BH, CE [3] [Total: 10] 1.6 Momentum Test yourself questions 1 2 3 4 5 Momentum p = mv a p = 10 kg × 5 m/s = 50 kg m/s b p = 10 kg × 20 × 10–2 m/s = 2 kg m/s c p = 10 kg × 36 × 103 m/s / (60 × 60) = 100 kg m/s Momentum p = mv Total momentum before collision is (1 kg × 2 m/s) + (1 kg × 0 m/s) = 2 kg m/s Total momentum after collision is (1 kg × 0 m/s) + (1 kg × v) = 1 kg × v If momentum is conserved: (1 kg × v) = 2 kg m/s, so v = 2 m/s Momentum p = mv Total momentum before boy jumps on trolley is (50 kg × 5 m/s) + (20 kg × 1.5 m/s) = (250 + 30) kg m/s = 280 kg m/s Total momentum after collision is (50 + 20) kg × v = 70 kg × v If momentum is conserved: 70 kg × v = 280 kg m/s, so v = 280/70 = 4 m/s Forward momentum of girl = mv = 50 kg × 3 m/s = 150 kg m/s Backward momentum of boat = mv = 300 kg × v m/s If momentum is conserved: 300 kg × v = 150 kg m/s, so v = 150/300 = 0.5 m/s a Impulse = F Δt = 5 N × 0.02 s = 0.1 N s b Δp = F Δt = 0.1 N s Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 13 Cambridge IGCSE Physics Student Book answers 6 a b Δp = m Δv =1000 kg × 24 m/s = 2.4 × 104 kg m/s F = Δp/Δt = 2.4 × 104 kg m/s / 1.2 s = 2.0 ×104 N Now put this into practice questions (Page 56) 1 2 Total momentum before collision is (3 kg × 5 m/s) + (2 kg × 0 m/s) = 15 kg m/s Total momentum after collision is (3 kg + 2 kg) × v = 5 kg × v If momentum is conserved: 5 kg × v = 15 kg m/s, so v = 3 m/s Total momentum before collision is (5 kg × 5 m/s) + (2 kg × 0 m/s) = 25 kg m/s Total momentum after collision is (5 kg × 0 m/s) + 2 kg × v = 2 kg × v If momentum is conserved: 2 kg × v = 25 kg m/s, so v = 12.5 m/s Practical work questions (Page 56) 1 Momentum is conserved in a collision 2 To compensate for energy loss in friction between the trolley and the track 3 a b i mv = 2 kg × 0.2 m/s = 0.4 kg m/s iii mv = 2 kg × 5 × 10–2 m/s = 0.1 kg m/s i mv = 0.2 kg × 3 m/s = 0.6 kg m/s iii mv = 1 kg × 3 m/s = 3 kg m/s ii mv = 2 kg × 0.8 m/s = 1.6 kg m/s ii mv = 0.5 kg × 3 m/s = 1.5 kg m/s Exam-style questions (Page 59) 1 2 3 4 a b c d Momentum p = mv [1] Momentum of truck A before the collision is (500 kg × 4 m/s) = 2000 kg m/s [2] Momentum of truck B before the collision is (1500 kg × 2 m/s) = 3000 kg m/s [2] Total momentum after collision is (500 + 1500) kg × v = 2000 kg × v If momentum is conserved: 2000 kg × v = (2000 + 3000) kg m/s = 5000 kg m/s so v = 5000 kg m/s / 2000 kg = 2.5 m/s [4] [Total: 9] Momentum p = mv a Initial momentum (10 kg × 4 m/s) = 40 kg m/s [2] b Final momentum (10 kg × 8 m/s) = 80 kg m/s [2] c Total momentum gained in 2 s = (80 – 40) kg m/s = 40 kg m/s [2] 2 d F = rate of change of momentum = 40 kg m/s / 2s = 20 kg m/s [2] e Impulse = FΔt = 20 N × 2 s = 40 N s [2] [Total: 10] a Momentum = mass × velocity [1] b Backward momentum of ejected gas = 5 kg × 5000 m/s = 25 000 kg m/s [2] c Momentum is conserved in a collision if no external forces act [2] d If momentum is conserved forward momentum = 25 000 kg m/s = (10 000 – 5) kg × v so v = 25 000 kg m/s / 9995 kg = 2.5 m/s [3] [Total: 8] 2 a F = ma so a = F/m = 50 N / 0.03 kg = 1670 m/s [2] b Impulse = F Δt = 50 N × 0.001 s = 0.05 N s [2] c F Δt = Δ (mv) = 0.03 kg × v = 0.05 N s, so v = 0.05 N s / 0.03 kg = 1.7 m/s [2] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 14 Cambridge IGCSE Physics Student Book answers d Apply force for a longer time; increase the size of the force [2] [Total: 8] 1.7 Energy, work and power Test yourself questions 1 a b c 2 a c *3 a b c electric current mechanical working heating chemical energy b internal (thermal) energy kinetic energy d elastic (strain) energy 2 Ek = mv /2 = 1 kg × 2 m/s × 2 m/s / 2 = 2 J Ek = mv2/2 = 0.002 kg × 400 m/s × 400 m/s / 2 = 160 J Ek = mv2/2 = 500 kg × (72000 / (60 × 60)) m/s × (72000 / (60 × 60)) m/s / 2 = 100 000 = 105 J *4 a Rearranging Ek = mv2/2, gives v = √(2 × Ek/m) = √(2 × 200 J / 1 kg) = 20 m/s b i ∆Ep = mgΔh = 5 kg × 9.8 N/kg × 3 m = 147 J ii ∆Ep = mgΔh = 5 kg × 9.8 N/kg × 6 m = 294 J *5 ∆Ep of water flowing over falls/s = mgΔh/t = 7 × 106 kg/s × 9.8 N/kg × 50 m = 3.4 × 109 W = 3400 MW 6 W = F d = (3 × 9.8) N × 6 m =176 J 7 W = F d = (51 × 9.8) N × 300 m = 1.5 × 105 J 8 W = ∆Ep = mgΔh, so mg = W/Δh = 80 J /5 m = 16 N 9 Renewable, non-polluting (i.e. no CO2, SO2 or dangerous waste), low initial building cost of station to house energy converters, low running costs, high energy density, reliable, allows output to be readily adjusted to varying energy demands 10 a Heating, lighting, refrigeration, television, computers, sound systems... b i Install roof insulation, wall insulation, double-glazed windows, solar hot water system, solar generating panels, turn off lights and electrical devices when not in use, wear a jumper in cold weather so that home heating levels can be lowered, wait until washing machine and dishwater are full before switching on, collect materials for recycling... ii Develop energy efficient cars, buildings, power stations and power transmission systems, use waste energy from power stations (e.g. to heat local homes), encourage people to walk, cycle or use trains and buses to travel to work, turn off heating in offices at weekends and lights and computers at night, recycle materials such as metals, paper, glass *11 Efficiency = useful power output × 100% total power input = (9 J/s / 20 J/s) × 100% = 45% 12 P = W/t = Fd/t = 600N × 10 m / 12 s = 500 W 13 P = ΔE/t = 2400 J / 60 s = 40 W 14 Work done = F × d = (60 × 70 × 9.8) N × 5 m = 206 000 J P = work done / time taken = 206 000 J / 60 s = 3430 W = 3.4 kW Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 15 Cambridge IGCSE Physics Student Book answers Now put this into practice questions (Page 64) *1 Ek = mv2 = × 0.4 kg × (80 m/s)2 = 0.2 × 6400 kg m2/s2 = 1280 N m = 1280 J *2 Ek = mv2 = × (50 × 10–3) kg × (40 m/s)2 = 0.025 × 1600 kg m2/s2 = 40 N m = 40 J (Page 64) *1 ΔEp = mgΔh = 0.2 kg × 9.8 N/kg × 2 m = 4 N m = 4 J *2 ΔEp = mgΔh = 0.4kg × 9.8 N/kg × 3 m = 12 N m = 12 J (Page 65) *1 a Ek = mv2/2 = 2 kg × (10 m/s)2 / 2 = 100 kg m2/s2 = 100 N m = 100 J b ∆Ep = Ek = 100 J c ∆Ep = mgΔh so Δh = ∆Ep/mg = 100 J / 2 kg × 9.8 m/s2 = 5 m *2 ∆Ep = mgΔh = 0.4 kg × 9.8 m/s2. × 30 m = 120 J. By conservation of energy mv2 = 120 J and so v2 = 2 × 120 J / 0.4 kg = 600 m2/s2; v = 24.5 m/s ∆Ep = Ek = (Page 74) *1 W = Fd = 500 N × 12 m = 6000 J; efficiency = energy output / energy input × 100% = (6000 J / 8000 J) × 100% = 75% *2 Efficiency = useful power output × 100% total power input = (560 – 170) J/s / 560 J/s × 100% = (390 / 560) × 100% = 70 % Practical work questions (Page 64) 1 2 3 Students’ own responses based on their results a To reduce the effect of frictional losses b There is no longer a resultant force acting on the trolley ΔEp = mgΔh = 0.3kg × 10 N/kg × 0.80 m = 2.4 N m = 2.4 J 4 Ek = 5 6 Chemical energy stored in your muscles is transferred to gravitational potential energy Heat, sound, frictional losses mv2 = × 0.3 kg × (4.0 m/s)2 = 2.4 J Exam-style questions (Page 77) 1 a b c *2 a b c d e Electricity transferred to kinetic energy and thermal energy Electricity transferred to heat Electricity transferred to sound [2] [2] [2] [Total: 6] ∆Ep = mgΔh = 100/1000 kg × 9.8 N/kg × 1.8 m = 1.8 J [2] ∆Ep transfers to Ek = 1.8 J [1] 2 Rearranging Ek = mv /2, gives v = √(2 × Ek/m) = √( 2 × 1.8 J / 0.1 kg) = 6 m/s [3] ∆Ep = mgΔh at high point of rebound = 0.1 kg × 9.8 N/kg × 1.25 m = 1.23 J so Ek of rebound = 1.23 J [2] v = √(2 × Ek/m) = √(2 × 1.23 J / 0.1 kg) = 5 m/s [3] [Total: 11] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 16 Cambridge IGCSE Physics Student Book answers *3 a b c 4 a b c 5 a b c d e 6 a b *7 a b c Rearranging Ek = mv2/2, gives v = √(2 × Ek/m) = √( 2 × 100 J / 0.5 kg) = 20 m/s [3] [1] Ek = ∆Ep = 100 J 2 ∆Ep = mgΔh, so Δh = ∆Ep/mg = 100 J / (0.5 kg × 9.8 m/s ) = 20 m [3] [Total: 7] Work done = F d = 100 N × 1.5 m = 150 J [2] 150 J [1] Power = work done / time taken = 4 × 150 J / 60 s = 10 W [3] [Total: 6] 2% [1] Hydroelectric [1] Cannot be used up [1] Solar energy, wind energy [2] All energy ends up as thermal energy which is difficult to use and there is only a limited supply of non-renewable sources [2] [Total: 7] i Reliable, readily available at all times, high energy density [2] ii Polluting, non-renewable [2] i Renewable, non-polluting [2] ii Only available when the sun shines, needs large areas of land [2] [Total: 8] Efficiency = (useful energy output / total energy input) × 100% = (300 MJ / 1000 MJ) × 100 % = 30% [3] Energy lost = 1000 MJ – 300 MJ = 700 MJ; thermal energy [2] Warms surroundings; lost from cooling towers [2] [Total: 7] 1.8 Pressure Test yourself questions 1 To raise the level of the water supply above that in the reservoir and provide water pressure to the taps in the building 2 The pressure in a liquid increases with depth so the wall at the bottom of the dam must withstand a greater pressure than it does at the top of the dam *3 Δp = ρgΔh = 1000 kg/m3 × 9.8 m/s2 × 2 m = 2 × 104 Pa *4 Δp = ρgΔh so Δh = Δp/ρg = 3.0 × 106 Pa / (1.02 × 103 kg/m3 × 9.8 m/s2) = 300 m Now put this into practice questions (Page 79) 1 a b 2 a i ii i ii i ii The area of the base = 2 m × 2 m = 4 m2 Area of a side = 2 m × 5 m = 10 m2 Pressure on base = force / area = 80 N / 4 m2 = 20 Pa Pressure on side = force / area = 80 N / 10 m2 = 8 Pa Pressure = force / area = 50 N / 2.0 m2 = 25 Pa F/A = 50 N / 100 m2 = 0.50 Pa Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 17 Cambridge IGCSE Physics Student Book answers iii F/A = 50 N / 0.50 m2 = 100 Pa b F = pressure × area = 10 Pa × 3.0 m2 = 30 N (Page 80) 1 = 300 N a 0.1 m 2 = 70 N × =7N A 1.0 m 2 2 f= F × 3 Incompressibility Exam-style questions (Page 84) 1 2 A true, B true, C true, D false, E true, F false a i p = F/A = 2000 kN / 2 m2 = 1000 kN/m2 ii p = F/A = 200 kN / 0.2 m2 = 1000 kN/m2 iii p = F/A = 0.5 kN / 0.0002 m2 = 2500 kN/m2 b High heels; they produce a pressure greater than 2000 kN/m2 on the floor [Total: 6] [2] [2] [2] [2] [Total: 8] 2 [2] 3 a Pressure = force / area = 20 N / 0.20 m = 100 Pa b Force = pressure × area = 100 Pa × 2.0 m2 = 200 N [2] c A liquid is nearly incompressible [1] d A liquid transfers the pressure applied to it [1] [Total: 6] 4 a A true, B true, C false [3] 3 2 6 *b Δp = ρgΔh = 1150 kg/m × 9.8 N/kg × 100 m = 1 127 000 N/m = 1.13 × 10 Pa [4] [Total: 7] *5 a Δp = ρgΔh [2] b Pascal (Pa) [1] c Rearrange Δp = ρgΔh to give Δh = Δp/ρg = 7.5 × 106 Pa / (1.03 × 103 kg/m3 × 9.8 m/s2) = 740 m [3] [Total: 6] [Going further] 6 a Vacuum [1] b Atmospheric pressure [1] c 740 mm Hg [1] d It becomes less; atmospheric pressure lower (decreases with altitude) [2] [Total: 5] Section 2 Thermal physics 2.1 Kinetic particle model of matter Test yourself questions 1 a b Air is readily compressed Steel is not easily compressed Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 18 Cambridge IGCSE Physics Student Book answers 2 3 a melting b evaporation and boiling c solidification d condensation a Solid – ordered and densely packed particles vibrate about fixed positions b Liquid – less ordered, particles further apart than in a solid and can slide over each other c Gas – particles are wide apart and are free to move quickly in all directions 4 It is the temperature at which all particle motion ceases *5 If more high-speed molecules strike one side of the smoke particle than the other at a given instant, the particle will move in the direction in which there is a net force 6 When the temperature rises, the average speed of the gas particles increases leading to more frequent collisions with the surfaces of the container so the pressure increases if the volume is kept constant. 7 If the volume is reduced there will be more frequent collisions of the particles with the surfaces of the container so the pressure will increase if the temperature is kept constant. 8 a It is believed to be the lowest possible temperature b They are the same size Now put this into practice questions (Page 93) *1 p1V1 = p2V2 ; V2 = p1 × V1 / p2 = 1 × 105 Pa × 9 cm3 / 3 ×105 Pa = 3 cm3 *2 p1V1 = p2V2 ; p2 = p1 × V1 / V2 = 2 × 105 Pa × 40 cm3 / 20 cm3 = 4 × 105 Pa (Page 94) 1 T = 273 + θ = 273 + 80 = 353 K 2 θ = T – 273 = 100 – 273 = –173 ºC (Page 96) [Going further] 1 p1V1 / T1 = p2V2 / T2 so p2 = p1V1T2 / T1V2 = 1 × 105 Pa × 9 cm3 × 310 K / (300 K × 5 cm3) = 1.9 × 105 Pa [Going further] 2 p1V1/T1 = p2V2 /T2 so T2 = p2V2T1 / p1V1 = 3.2 × 105 Pa × 30 cm3 × 300 K / (2 × 105 Pa × 40 cm3) = 360 K = 87 ºC Practical work questions (Page 90) 1 Smoke particles 2 Haphazardly *3 Collisions with large numbers of light, fast moving air molecules. 4 a Temperature b Pressure 5 Only one variable should be changed in an experiment at a time 6 When taking a measurement, stop heating, stir the water and allow the gauge reading to become steady 7 a Pressure b Volume 8 Graph is a straight line through the origin [Going further] 9 a Temperature b Volume [Going further] 10 Volume is proportional to temperature measured in kelvin Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 19 Cambridge IGCSE Physics Student Book answers Exam-style questions (Page 98) 1 2 B a [Total: 1] [2] b [2] c [2] 3 a b c 4 Gas Particles strike the surfaces of the container in large numbers per second and cause a pressure on the surfaces The pressure increases When the temperature rises so does the average speed of the gas particles so there are more collisions per second and the pressure on the surfaces increases a Random *b Due to collisions with the air molecules in the box A smoke particle is huge compared to an air molecule but if there are a larger number of air molecules striking one side of the smoke particle at a given instant, it will move in the direction in which there is a net force The imbalance and hence the direction of the net force changes rapidly in a random manner Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [Total: 6] [1] [2] [1] [3] [Total: 7] [1] [1] [2] [1] [Total: 5] 20 Cambridge IGCSE Physics Student Book answers 5 a i ii iii b i ii iii True True False Absolute zero where particle motion ceases T = 273 + θ = 273 – 273 = 0 K. T = 273 + θ = 273 – 200= 73 K *6 a b p1V1 = p2V2 ; so p2 = p1V1 /V2 = 1 × 105 Pa × 10 cm / 40 cm = 2.5 × 104 Pa p1V1 = p3V3 ; so p3 = p1V1 /V2 = 1 × 105 Pa × 10 cm / 50 cm = 2.0 × 104 Pa *7 a p1V1 = p2V2; V2 = p1 × V1 / p2 = 1 × 105 Pa × 30 cm3 / 2 × 105 Pa = 15 cm3 (pressure doubled, volume halved) V2 = p1 × V1 / p2 = 1 × 105 Pa × 30 cm3 / 5 × 105 Pa = 6 cm3 b [1] [1] [1] [2] [1] [1] [Total: 7] [3] [3] [Total: 6] [3] [3] [Total: 6] Alternative to practical question (Page 99) 8 a [3] [1] b pressure/ 105 Pa volume/ cm3 1/volume / cm–3 24 1.0 1.00 12 2.0 0.50 8 3.0 0.33 6 4.0 0.25 4 6.0 0.17 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 21 Cambridge IGCSE Physics Student Book answers c [3] d Yes; values of pV all equal to 24 ×105 Pa cm3 [2] [Total: 9] 2.2 Thermal properties and temperature Test yourself questions 1 a b c 2 a b The metal lid expands more than the glass jar when it is held in hot water so becomes looser and easier to unscrew Wood and metal contract when the temperature falls, leading to creaking of the furniture Concrete expands when the temperature rises and would buckle if gaps are not left between sections; the pitch is easily squeezed out when expansion occurs Aluminium *3 The particles in a liquid are further apart and more mobile than in a solid so expansion is easier for liquids than for solids. In gases, particles are further apart than in liquids and can move about freely at high speeds; this means they are able to expand much more easily than liquids 4 a Water expands when it is cooled from 4 ºC to 0 ºC; most liquids contract when the temperature decreases b Water pipes burst when the water in them freezes; water-based liquids may burst their containers when they freeze; fish can survive below the frozen surface of a pond 5 Liquid in a glass bulb expands up a capillary tube when the bulb is heated; the temperature is marked in degrees on a scale next to the capillary tube 6 C *7 Heat needed = mass × temperature change × specific heat capacity = mΔθc = 5 kg × 10 ºC × 300 J/(kg ºC) = 15 000 J Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 22 Cambridge IGCSE Physics Student Book answers *8 Assume heat supplied ΔE = heat gained by water then ΔE = 3000 J/s × t = mcΔθ and t = mcΔθ / 3000 J/s = 5 kg × 4200 J/(kg ºC) × (50 – 30 )ºC / 3000 J/s = 140 s 9 a 1530 ºC b 19 ºC c 0 ºC d 100 ºC e 37 ºC 10 a Ice draws heat from the drink when it melts (it has a high specific latent heat of fusion) b Steam releases much heat when it condenses (it has a high specific latent heat of vaporisation) 11 a i Remain constant ii Absorbed b Remain constant 12 a A few energetic particles close to the surface of a liquid escape and become gas particles lowering the average kinetic energy of the remaining particles b Decreases *13 The water cools when evaporation occurs. Now put this into practice questions (Page 106) *1 Using c = 25 000 J / (2 kg × (35 − 10) ºC) = 25 000 J / 50 kg ºC = 500 J/(kg ºC) *2 Rearrange equation to give the heat equation: ΔE = mcΔθ = 3 kg × 500 J/kg ºC × 10 ºC = 15 000 J (Page 107) *1 Assume heat supplied to water = heat gained by water then 3000 J/s × t = mcΔθ and t = mcΔθ / 3000 J/s = 1 kg × 4200 J/(kg ºC) × (100 – 30 )ºC / 3000 J/s = 98 s *2 Heat lost by sphere = mcΔθ = 0.1 kg × c × (100 – 25) ºC = c × 7.5 kg ºC Heat gained by water = 0.2 kg × 4200 J/kg ºC × (25 – 20) ºC = 4200 J Equating heat lost to heat gained gives c = 4200 J / 7.5 kg ºC = 560 J/kg ºC Practical work questions (Page 106) *1 Heat used to raise temperature of aluminium pan as well as the water; heat is lost to the surroundings *2 Heat received by water = Power × t = 40 W × 300 s = 12 000 J So = 12 000 J / (1 kg × 2.5 ºC) = 4800 J/(kg ºC) *3 Reduce heat loss to surroundings by insulating the container and block *4 Heat received by aluminium cylinder = Power × t = 40 W × 300 s = 12 000 J So 5 6 7 = 12 000 J / (1 kg × 12.5 ºC) = 960 J/(kg ºC) Students’ own graphs bases on their results Temperature remains constant The liquid is solidifying Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 23 Cambridge IGCSE Physics Student Book answers 8 Faster [Going further] 9 Students’ own responses based on their results [Going further] 10 Collect the ice melted before the heater is switched on for the same time [Going further] 11 Wrap insulation around the funnel Exam-style questions (Page 113) 1 *a The particles in a gas are further apart than in a liquid and move around at higher speeds. They have less interaction with each other than the particles in a liquid and so a gas can expand more easily [3] b When water freezes to ice at 0 ºC, it expands and becomes less dense. The expansion causes metal pipes to burst [2] [Total: 5] 2 A correct [1] B correct [1] C correct [1] [Total: 3] *3 Heat supplied = ΔE = m × Δθ × c; so c = ΔE / (m × Δθ) A: cA = 2000 J / (1.0 kg × 1.0 ºC) = 2000 J/(kg ºC); [3] B: cB = 2000 J / (2.0 kg × 5.0 ºC) = 200 J/(kg ºC); [3] [3] C: cC = 2000 J / (0.5 kg × 4.0 ºC) = 1000 J/(kg ºC) [Total: 9] *4 a Specific heat capacity of jam is higher than that of pastry so it cools down more slowly [2] b Rearrange equation to give Δθ = ΔE / mc = 15 000 J / (3 kg × 500 J/kg ºC ) = 10 ºC *5 a b Heat supplied = ΔE = mΔθc = 10 g × 30 ºC × 4.0 J/(g ºC) = 1200 J. Heat is conducted from the milk to the water which cools when it evaporates [3] [Total: 5] [3] [3] [Total: 6] 6 a b c i ii iii i i The temperature at which a solid changes to a liquid The temperature at which a liquid changes to a gas The temperature at which a liquid changes to a solid 0 °C ii 100 °C released ii released *7 A incorrect [1] B incorrect [1] C incorrect [1] D correct [1] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 E incorrect [1] [3] [2] [2] [Total: 7] [Total: 5] 24 Cambridge IGCSE Physics Student Book answers Alternative to practical question (Page 114) 8 a b c 80 ºC; temperature constant They move nearer together and become more ordered [4] [2] [2] [Total: 8] 2.3 Transfer of thermal energy Test yourself questions *1 In conduction thermal energy is transferred through matter from places of higher temperatures to places of lower temperature without movement of the matter as a whole *2 Liquids and gases are less dense than solids; the atoms are further apart and they do not have a regular ordered structure to enable transfer of heat by lattice vibrations or free electrons 3 a If small amounts of hot water are to be drawn off frequently, it may not be necessary to heat the whole tank b If large amounts of hot water are needed, it will be necessary to heat the whole tank 4 Hot air is less dense than cool air 5 Black surfaces absorb radiation better than white ones; the ice on the black sections of the canopy melts faster than on the white sections 6 Infrared *7 a The Earth radiates energy back into space b Clouds reduce the amount of energy radiated into space, keeping the ground warmer 8 Metal is a better conductor of heat than rubber *9 a Conduction and radiation b Radiation and convection Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 25 Cambridge IGCSE Physics Student Book answers Practical work question (Page 121) a b i (80 – 42) ºC = 38 ºC ii c 80 ºC to 42 ºC d Exam-style questions (Page 126) (42 – 28) ºC = 14 ºC Higher temperatures 1 Fix a match to a metal rod with a little wax;[1] repeat with rods of identical dimensions but of different materials.[1] Support the rods on a tripod in a fan arrangement and heat where the ends are closest together.[1] The match will fall first from the best thermal conductor and last from the worst.[1] [Total: 4] *2 Atoms in the hot regions vibrate strongly and pass on some of their energy to colder neighbouring atoms through lattice vibrations.[2] Some materials such as metals also have large numbers of free electrons which when they gain kinetic energy in the hot regions can travel faster and further;[1] they can interact with atoms in cooler parts and transfer energy through the material quickly.[1] [Total: 4] 3 a Convection transfers thermal energy through a fluid by movement of the fluid from places of higher temperature to places of lower temperature;[2] this occurs because the density of the fluid is lower when it is hot than when it is cold.[1] b Drop a few crystals of potassium permanganate down a tube into the bottom of a beaker containing water.[1] When the tube is removed and the water is heated from below, purple streaks mark the motion of the water.[2] [Total: 6] 4 A true [1] B true [1] C false [1] D true [1] [Total: 4] 5 a Black surfaces are better emitters of radiation than white surfaces; dull surfaces are better emitters of radiation than shiny ones [2] b White surfaces are better reflectors of radiation than black surfaces; shiny surfaces are better reflectors of radiation than dull ones [2] c Black surfaces are better absorbers of radiation than white surfaces; dull surfaces are better absorbers of radiation than white ones [2] [Total: 6] *6 a Hold the back of your hands on either side of a hot copper sheet which has one side polished and the other side blackened; [3] your hand will feel warmer near the better emitter of radiation [1] b Attach a coin with wax to two different surfaces, one black and dull, the other shiny [2] Place each surface the same distance away from a heater; the wax will melt faster and the coin will fall from the surface which is the better absorber [2] [Total: 8] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 26 Cambridge IGCSE Physics Student Book answers 7 a b c 8 a b c d Newspaper is a poor conductor of heat The fur would trap more air, which is a good insulator, and so keep the wearer warmer The holes in a string vest trap air, which is a poor conductor of heat, next to the skin. Fibreglass traps air which is a poor conductor of heat; using it for roof insulation reduces heat loss by conduction Plastic foam-filled cavity walls trap air which is a good insulator; using it in the cavity walls reduces heat loss by conduction and convection Double glazed windows trap air, which is a poor conductor of heat, so reduce the loss of heat through windows by conduction Fibreglass and plastic foam are both good insulators; they trap air which is a poor conductor of heat Convection cannot occur in the plastic foam Implosion of glass could occur if glass is not strong enough [1] [2] [2] [Total: 5] [1] [2] [1] [2] [1] [1] [Total: 8] Alternative to practical question (Page 126) 9 For example: place two thermometers in a beaker of hot water until the temperature of each is about 80 ºC.[1] Remove each from the water and quickly wrap the bulb of one in one layer of fibreglass and the bulb of the other in two layers of fibreglass.[2] Record the temperature drop of each thermometer as a function of time.[2] The thermometer with two layers of fibreglass should take longer to cool over a given temperature range than the other, if the manufacturers’ claims are true.[1] [Total: 6] Section 3 Waves 3.1 General properties of waves Test yourself questions (Page 130) 1 2 3 4 5 a λ = 5 cm /5 = 1 cm b f = 5 cycles / 5 seconds = 1 cycle/s = 1 Hz c v = f λ = 1 Hz × 1 cm = 1 cm/s a Speed of ripple depends on the depth of the water b AB since ripples travel more slowly towards it so the water is shallower in this direction 35° B C Now put this into practice questions (Page 130) 1 2 3 λ = v/f = 5 cm/s / 50 Hz = 0.1 cm Rearrange v = f λ to give f = v/ λ = 10 m/s / 1 m = 10 Hz Rearrange v = f λ to give f = v/ λ = 10 m/s / 25 cm = 1000 cm/s / 25 cm = 40 Hz Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 27 Cambridge IGCSE Physics Student Book answers Practical work questions (Page 130) 1 2 Equally spaced parallel lines Distance between crests of ripples Exam-style questions (Page 135) 1 2 a b c Trough i λ = 3.0 mm (distance between consecutive wave-crests) ii v = d/t = 3.0 mm × 5 / 1s = 15 mm/s iii f = 5 cycles/s = 5 Hz Two small balls fitted to the bar of a ripple tank a b C B [1] [2] [3] [2] [2] [Total: 10] [1] [1] [Total: 2] *3 Wavelength longer after waves pass through gap [Total: 6] 3.2 Light Test yourself questions (Page 142) 1 2 3 Larger, less bright a Four images b Brighter but blurred C 4 Before; sound travels slower than light Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 28 Cambridge IGCSE Physics Student Book answers *5 A 6 7 8 B; the image is the same distance behind the plane mirror as the object is in front. D 9 Spear should be aimed below apparent position of the fish *10 Speed of light in medium = speed of light in air / refractive index = 300 000 km/s / (6/5) = 250 000 km/s 11 D; glass has a higher refractive index than water so the ray is refracted towards the normal 12 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 29 Cambridge IGCSE Physics Student Book answers 13 Parallel 14 Distance from lens: a Beyond 2F b 2F c Between F and 2F d Nearer than F 15 Towards 16 C 17 A; the ray is refracted towards the normal at the first surface and away from the normal at the second surface. Now put this into practice questions (Page 146) *1 n so sin r = sin i / n = sin 30° / 1.5 = 0.50 / 1.5 = 0.33 and r = 20° *2 refractive index, n = speed of light in air (or vacuum) speed of light in medium so speed of light in water = speed of light in air / n = 3.0 × 108 m/s / 1.3 = 2.3 × 108 m/s (Page 147) *1 sin 32° = 0.53 so n = 1 / 0.53 = 1.9 *2 Rearrange equation n = 1/sin c to give sin c = 1/n = 1/1.7 = 0.59 so c = 36° Practical work questions (Page 137) 1 2 3 4 5 Smaller, inverted, laterally inverted Larger image a brighter b less sharp c same size The angle of incidence equals the angle of reflection The reflected ray does not emerge from the front surface of the mirror at the same point that the incident ray strikes it. 6 Same size 7 Same points are perpendicular to glass 8 Same distance 9 Answer: image orientation changes from being parallel to 90° 10 Reflected and refracted 11 Towards 12 Away 13 They are parallel 14 Ray is undeviated 15 The ray strikes surface normally 16 Approximately 42° for glass 17 Incident, reflected and refracted rays 18 Parallel light from a distant object converges at principal focus of lens Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 30 Cambridge IGCSE Physics Student Book answers 19 A diverging lens produces a virtual image which cannot be formed on a screen 20 a Enlarged b Upright 21 Ray diagrams drawn as in Figures 3.2.39a–d. Values should roughly agree with measured values of image position Exam-style questions (Page 158) *1 a b c d 2 40° 40°, 50°, 50° Parallel, but turned through 180 ° (antiparallel) a b B Top half c She must stand 1m from the mirror if she is to be 2 m from her image (the image in a plane mirror is the same distance behind the mirror as the object is in front). She must walk 4 m towards mirror [2] [3] [3] [2] [Total: 10] [1] [2] [4] [Total: 7] Alternative to practical questions (Page 158) 3 a b c For example: Draw a straight line AOB on the paper and then a line OC perpendicular to AOB. Draw a further line OD at an angle of say 40°, to the normal OC; record the angle in a table. Set the mirror vertically on the line AOB. Press two of the pins, at least 5 cm apart along the line OD into the cork board through the paper and view their reflections in the mirror. Find the viewing position where the image of the pins appears to be in line (one behind the other) and mark the line, OE, by pressing two further pins into the board. Draw in line OE with a ruler and measure the angle of reflection. Repeat with line OD at different angles to the normal OC. The table should have headings ‘angle of incidence’ and ‘angle of reflection’ Check if the values of the angle of incidence and angle of reflection agree within experimental error Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [2] [2] 31 Cambridge IGCSE Physics Student Book answers d Ensure pins are vertical, ensure mirror is vertical and placed accurately on AOB, draw lines thinly 4 a b c d e Refraction POQ Towards 40° 90° – 65° = 25° 5 a b c Angle of incidence = 0° Angle of incidence is greater than the critical angle CB is refracted away from the normal *6 a b 7 i The ray passes into the air and is refracted away from the normal, since the angle of incidence is less than the critical angle ii Total internal reflection occurs in water, since the angle of incidence is greater than the critical angle n = 1/sin c, so sin c = 1/n = 3/4 = 0.75 so c = 49° a Converging [2] [Total: 10] [1] [1] [1] [1] [1] [Total: 5] [1] [2] [2] [Total: 5] [1] [2] [3] [3] [1] [Total: 10] [1] b [3] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 32 Cambridge IGCSE Physics Student Book answers c *8 a b correct ray diagram Image is 9 cm from the lens Image is 3 cm high [2] [2] [2] 4 cm Correct ray diagram Image is 8 cm behind lens, virtual, larger than object [Total: 10] [1] [2] [1] [1] [1] c Image height / object height = 8 cm / 4 cm = 2 9 Lens A: Object is near principal focus, F, of converging lens, so f = 10 cm Lens B: Object is at 2F of converging lens, so f = 5 cm [2] [Total: 8] [3] [3] [Total: 6] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 33 Cambridge IGCSE Physics Student Book answers 10 a b Dispersion Refraction in prism drawn correctly Refraction of emergent rays drawn correctly Red and blue rays in correct positions [1] [2] [2] [1] [Total: 6] 3.3 Electromagnetic spectrum Test yourself questions 1 a b 2 a b 3 a b c d e f 4 a b c d *5 a b c d 0.7 μm 0.4 μm B D Ultraviolet Microwave Gamma rays Infrared infrared/microwaves X-rays ultraviolet microwave infrared X-ray or gamma ray microwave radio light or infrared microwave Now put this into practice questions (Page 161) 1 2 Rearrange the equation v = f λ to give f = v/λ then f = 3 × 108 m/s / 6 × 10–7 m = 5 × 1014 Hz Rearrange the equation v = f λ to give λ = v/f then λ = 3 × 108 m/s / 4.0 × 1014 Hz = 7.5 × 10–7 m. Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 34 Cambridge IGCSE Physics Student Book answers Exam-style questions (Page 166) 1 a b 2 a b i ii i ii microwave ultraviolet gamma rays radio waves v = f λ so λ = v/f = 3 × 108 m/s / 100 × 106 Hz = 3 m v = s/t so t = s/v = 60 × 103 m / (3 × 108 m/s) = 2 × 10–4 s 3 D *4 a i b *5 a b c d smoothly and continuously varying; can have any value within a certain range ii has only two values: high or low Can be: transmitted at faster rate; sent over longer distances; less affected by noise [1] [1] [1] [1] [Total: 4] [4] [4] [Total: 8] [Total: 1] [3] [2] [3] [Total: 8] glass [1] infrared and visible light [2] glass or plastic is transparent to infrared and visible light [1] higher frequency than radio waves and so can transmit information at a higher rate [2] as a digital signal [1] data can be regenerated accurately; can be carried over long distances; is secure from electrical interference; signals can be transmitted at higher data rates; cannot be hacked easily are cheaper and easier to install [3] [Total: 10] 3.4 Sound Test yourself questions 1 2 Produced by a vibrating source such as a guitar string or loudspeaker Longitudinally; molecules vibrate about a fixed position in the direction in which the wave propagates *3 One wavelength 4 256 Hz 5 0.8 m (it has the highest frequency) 6 B 7 a 20 Hz to 20 000 Hz b 330 m/s to 350 m/s 8 v = s/t so s = v t = 330 m/s × 5 s = 1650 m (about 1 mile) 9 a Reflection, refraction, diffraction b In a transverse wave, vibrations are perpendicular to rather than along the direction of travel of the wave. Sound waves are longitudinal. Now put this into practice questions (Page 171) 1 Rearrange the equation v = f λ to give λ = v/f then λ = 340 m/s / 512 Hz = 0.66 m Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 35 Cambridge IGCSE Physics Student Book answers 2 Rearrange the equation v = f λ to give f = v/λ then f = 340 m/s / 1.0 m = 340 Hz (Page 172) *1 Using 2d = v t then d = = 1400 m/s × 0.5 s / 2 = 350 m *2 Using 2d = v t then d = = 1400 m/s × 2 s / 2 = 1400 m Practical work questions (Page 170) 1 2 Increase the distance between the microphones v = d/t = 1.2 m / (3.6 × 10–3) s = 330 m/s Exam-style questions (Page 174) 1 a b c Time for 1 clap, t = 60 s / 60 = 1 s, distance travelled to wall and back d = 2 × 160 m = 320 m Then v = d/t = 320 m / 1 s = 320 m/s t = 60 s / 80 = 0.75 s, d = 2 × 120 m = 240 m so v = d/t = 240 m / 0.75 s = 320 m/s t = 60 s / 30 = 2 s, d = v t = 320 m/s × 2 s = 640 m; distance to wall = d/2 = 320 m [1] [1] [1] [1] [1] [1] [1] [2] [1] [Total: 10] 2 a i [2] ii [2] b i ii v = f λ so λ = v/f = 340 m/s / 340 Hz = 1.0 m λ = v/f = 340 m/s / 170 Hz = 2.0 m *3 a i Compressions occur where the molecules of the medium transmitting a longitudinal sound wave are closer together than normal Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [3] [2] [Total: 9] [3] 36 Cambridge IGCSE Physics Student Book answers ii b c Rarefactions occur where the molecules of the transmitting medium are further apart than normal λ/2 v = f λ so λ = v/f = 330 m/s / 220 Hz = 1.5 m *4 a i solid ii gas b Time, t, for a sound wave to be reflected from an underwater surface and return to the transmitter is recorded. Knowing the speed of sound in the water, v, the depth of the reflecting object d can be calculated from the equation v = 2d/t c Medical imaging; non-destructive testing of materials [3] [1] [3] [Total: 10] [1] [1] [2] [2] [2] [Total: 8] Section 4 Electricity and Magnetism 4.1 Simple phenomena of magnetism Test yourself questions 1 2 C a NSNS b SNNS *3 Weaker: the magnetic field lines are further apart Practical work questions (Page 180) 1 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 37 Cambridge IGCSE Physics Student Book answers 2 3 4 5 Magnetic fields interact with each other Figure 4.1.9a N–S opposite N–S; Figure 4.1.9b N–S opposite S–N Current through the coil, number of turns on the coil The compass needle will point in the direction of the field lines emerging from the North pole of the current carrying coil Exam-style questions (Page 183) 1 2 a, b [1] [2] [1] [Total: 4] c Magnetic field line goes from the North to the South pole of the magnet. a Place a bar magnet on a piece of paper and a plotting compass near the N pole. Mark the position of the N and S poles of the compass on the paper;[1] then move the compass so that the S pole is at the point where the N pole was previously and mark the new position of the N pole.[1] Continue until compass is near the S pole then join up the points to give a field line.[1] Plot other field lines by repeating the process with the compass at different starting points.[1] Electromagnets are used where the strength of the magnetic field needs to be varied and turned on and off.[2] Permanent magnets do not require an electricity supply and are used when the magnetic field does not need to be varied.[2] Examples: compass, computer hard disk, electric motor or generator, microphone, loudspeaker, credit and debit cards. [2] [Total: 10] b c *3 a Through the interaction of magnetic fields; a magnet will experience a force in a magnetic field b Near the north and south poles c The magnetic field lines are i closest together ii furthest apart [2] [2] [2] [Total: 6] 4.2 Electrical quantities Test yourself questions 1 2 3 D Electrons are transferred from the cloth to the polythene a attracted b repelled *4 Radial field lines are perpendicular to the surface of the sphere; direction is towards the centre of the sphere Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 38 Cambridge IGCSE Physics Student Book answers 5 a b Ink-jet printer, photocopiers, paint and crop sprayers, flue ash precipitators… Damage to electronic equipment, explosion in presence of flammable vapour, lightning strikes… 6 By the movement of free electrons 7 Place ammeter in series with its positive terminal connected to the positive terminal of the supply *8 a I = Q / t = 10 C / 2 s = 5 A b I = Q / t = 20 C / 40 s = 0.5 A c I = Q / t = 240 C / 120 s = 2 A *9 Rearranging Q = I t gives t = Q/I = 5 C / 2 A = 2.5 s 10 a b c 11 Time for one cycle = 1/1000 = 10–3 s 12 a Electromotive force (e.m.f.) is the electrical work done by a source in moving a unit charge around a complete circuit *b Potential difference (p.d.) is the work done by a unit charge passing through a component *13 a W = Q V = 1 C × 12 V = 12 J b W = Q V = 5 C × 12 V = 60 J c W = Q V = I t V = 2 A × 10 s × 12 V = 240 J Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 39 Cambridge IGCSE Physics Student Book answers 14 R = V / I = 12 V / 4 A = 3 Ω 15 V = I R = 2 A × 10 Ω = 20 V 16 I = V / R = 6 V / 3 Ω = 2 A *17 I = V / R = 12 V / 4 Ω = 3 A; Q = I × t = 3 A × 1 s = 3 C *18 a b *19 a Calculate the gradient of the graph; resistance = 1/gradient b Resistance of filament increases as it heats up 20 a E = P t = 100 J/s × 1 s = 100 J b E = P t = 100 J/s × 5 s = 500 J c E = P t = 100 J/s × 60 s = 6000 J 21 a P = IV = 2 A × 12 V = 24 W b P = IV = 0.5 A × 6 V = 3 J/s Now put this into practice questions (Page 190) *1 Q = I × t = 2 A × 20 s = 40 C *2 I = Q / t = 3 C / 7 s = 0.4 A (Page 195) *1 p.d. across the lamp =W/Q =8 J / 4 C = 2 V *2 W = Q V = 2 C × 6 V = 12 J *3 Q = I t = 3A × 10 s = 30 C so W = Q V = 30 C × 6 V = 180 J (Page 196) 1 a 0.2 V (Page 198) 1 2 3 b upper scale c parallax error introduced R = 4.5 V / 0.3 A = 15 Ω V = IR = 0.2 A × 10 Ω = 2 V so I = 12.0 V / 24 Ω = 0.5 A (Page 200) *1 R = 60 Ω × 2 = 120 Ω since the length of wire is doubled *2 A1 / A2 = (0.20 mm)2 / (0.40 mm)2 = 0.25 so R2 = R1 × A1 /A2 = 60 Ω × 0.25 = 15 Ω Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 40 Cambridge IGCSE Physics Student Book answers (Page 203) 1 a p.d. V = IR = 1.0 A × 12 Ω = 12 V b P = IV = 1.0 A × 12 V = 12 W = 12 J/s c P = E/t so E = Pt = 12 J/s × 10 s = 120 J 2 a P = IV = 0.3 A × 12 V = 3.6 W b 3.6 J/s c P = E/t so E = Pt = 3.6 J/s × 60 s = 216 J (Page 205) 1 2 Electrical energy E = Pt = 6.4 kW × 2 h = 12.8 kWh Cost of using the oven = 12.8 kWh × 10 cents = 128 cents Electrical energy E = Pt = 0.150 kW × 12 h = 1.8 kWh Cost of using the refrigerator = 1.8 kWh × 10 cents = 18 cents Practical work questions (Page 186) 1 Rub the polythene with a cloth 2 Draw the rod firmly across the edge of the metal cap of the electroscope 3 The charge on the leaf has the same sign as that on the metal plate so is repelled from the plate 4 By connecting the metal cap to earth 5 When the lamp lights 6 a one b no 7 a two b yes 8 Students’ own responses based on their results 9 Students’ own responses based on their results 10 4.5 V / 3 = 1.5 V 11 a Connect voltmeter in parallel with the device; select appropriate range for the measurement b 1.5 V 12 Students’ own responses based on their results 13 a b 14 Coil of wire, ammeter, voltmeter, variable resistor/rheostat, battery/power supply, circuit board 15 R = V/I = 4.5 V / 0.15 A = 30 Ω 16 P = IV = 30 V × 0.5 A = 15 W 17 Po = mgh/t = 0.5 kg × 9.8 m/s2 × 0.8 m / 4 s = 0.98 W Exam-style questions (Page 206) 1 a b Rubbing with a cloth removes electrons from the cellulose acetate leaving it positively charged. repelled Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [3] [1] 41 Cambridge IGCSE Physics Student Book answers 2 c d attracted 2 a Charge a gold-leaf electroscope and then touch the cap with the test material. The gold leaf will fall quickly when the electroscope is discharged through a good conductor and only slowly or not at all when discharged through a bad conductor or insulator Any metal or carbon Plastics such as polythene and cellulose acetate, Perspex and nylon Conductors have some free electrons which can move through the material; the electrons in insulators are firmly bound to their atoms and are not free to move. b c d *3 a An electric field is a region in which an electric charge experiences a force b c *4 a coulomb (C) The direction of an electric field at a point is the direction of the force on a positive charge [1] [1] [Total: 6] [4] [2] [1] [3] [Total: 10] [2] [4] [1] [Total: 7] [3] b 5 6 7 B C a b c *8 a [4] [Total: 7] [Total: 1] [Total: 1] Electrons [1] In d.c. electrons flow in one direction only; in a.c. the direction of flow reverses regularly [2] Connect the ammeter in series in the circuit with the + of the ammeter closest to the + of the battery. For a digital ammeter, choose the d.c. setting. For either analogue or digital ammeters select a suitable range for the size of current. [3] [Total: 6] An electric current is the charge passing a point per unit time and is given by the equation I = Q/t [2] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 42 Cambridge IGCSE Physics Student Book answers b *9 a b i I = Q/t = 180 C / 60 s = 3 A [2] ii Q = I × t = 2 A × 60 s = 120 C i Q = I × t = 5 A × 10 s = 50 C ii Q = It = 5 A × 5 × 60 s = 1500 C Rearrange equation I = Q/t to give t = Q/I = 300 C / 5A = 60 s 10 b c d e f Very bright Normal brightness No light Brighter than normal Normal brightness *11 a b c V 2 = V – V1 = (18 – 12) V = 6 V W = QV1 = I × t ×V1 = 0.5 A × 60 s × 12 V = 360 J ammeter + and voltmeters + should be connected to the point nearest the + (left) of the battery 12 a b *13 a b c V = IR; graph is a straight line through the origin indicating V is proportional to I R = V / I = 6V / 3A = 2 Ω [3] [Total: 7] [2] [2] [3] [Total: 7] [1] [1] [1] [1] [1] [Total: 5] [2] [4] [4] [Total: 10] [4] [3] [Total: 7] The resistance R of a wire of a given material is directly proportional to its length l (R ∝ l), and inversely proportional to its cross-sectional area A (R ∝ 1/A ). Combining these two statements gives 𝑅𝑅 ∝ 𝑙𝑙 𝐴𝐴 R2 = R1 × l2/l1 = 70 Ω × 0.2 m / 1.0 m = 14 Ω A1/A2 = (1.0 mm)2 / (0.5 mm)2 = 4.0 R2 = R1 × A1/A2 = 70 Ω × 4 = 280 Ω [3] [3] [4] [Total: 10] *14 [Total: 9] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 43 Cambridge IGCSE Physics Student Book answers Alternative to practical questions (Page 208) 15 a b c 16 a b 17 a b [1] R = V/I Connect the wire to an ammeter, rheostat and battery in series; connect a voltmeter across the wire. Measure the current and voltage for different settings of the rheostat. Calculate R from the equation R = V/I or plot a graph of V versus I and determine R from the gradient of the graph [4] i straight line graph [3] ii with gradient 12 V / 0.24 A = 50 Ω = R [2] [Total: 10] E=Pt [1] = 6400 J/s × 30 × 60 s [1] 7 = 1.15 × 10 J [1] 80 minutes / 60 minutes = 4/3 hours; cost = 3 kW × 1.33 hours × 10 cents = 40 cents [3] [Total: 6] i 2 kW ii 60 W iii 850 W [3] P = I V so I = P / V = 920 W / 230 V = 4 A [4] [Total: 7] 4.3 Electric circuits Test yourself questions 1 All read 0.25 A (I1 = I2 + I4 and I2 = I3 = I1 / 2) 2 p.d. = 3 × 2 V = 6 V *3 a W = Q × V = 1 C × 2 V = 2 J b W=Q×V=1C×3×2V=6J 4 a *5 a R = R1 + R2 + R3 or b the same b larger *6 a b From the potential divider equation V1 / V2 = R1 / R2 = 12 Ω / 36 Ω = 1/3 Ratio of voltages is 1 : 3 c Dividing the supply voltage in the ratio 1 : 3 gives V1 = 1 × 20 V / 4 = 5 V and V2 = 3 × 20 V / 4 = 15 V 7 A thermistor B lamp C LDR D cell E resistor *8 A LED B semiconductor diode C relay D variable resistor Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 44 Cambridge IGCSE Physics Student Book answers 9 Now put this into practice questions (Page 212) 1 2 R = R1 + R2 + R3 = 4 Ω + 6 Ω + 8 Ω = 18 Ω a I = V/R = 4.5 V / (3 Ω + 6 Ω) = 4.5 V / 9 Ω = 0.5 A b V1 = IR1 = 0.5 A × 3 Ω = 1.5 V, V2 = IR2 = 0.5 A × 6 Ω = 3.0 V (Page 214) *1 a R = 1 Ω + 2 Ω +3 Ω = 6 Ω b current I = V/R = 12 V / 6 Ω = 2 A (same current in each resistor) c V1 = IR1 = 2 A × 1 Ω = 2 V, V2 = IR2 = 2 A × 2 Ω = 4 V, V3 = IR3 = 2 A × 3 Ω = 6 V *2 a 1/Ra = 1 / (2 Ω) + 1 / (3 Ω) = 5 / (6 Ω) so Ra = 6 / 5 Ω b 12 V, 12 V c i I = V/R = 12 V / 2 Ω = 6 A ii I = V/R = 12 V / 3 Ω = 4 A (Page 216) *1 V1 / V2 = R1 / R2 *2 a From the potential divider equation V1 / V2 = R1 / R2 = 9 Ω / 6 Ω = 3/2 Ratio of voltages is 3 : 2 b Dividing the supply voltage in the ratio 3 : 2 gives V1 = 3 × 30 V / 5 = 18 V and V2 = 2 × 30 V / 5 = 12 V Exam-style questions (Page 220) 1 x = V1 + V2 = 12 + 6 = 18 y = V – V1 = 6 – 4 = 2 z = V – V2 = 12 – 4 = 8 *2 Resistors in series: Rtotal = R1 + R2 + R3 + R4 = 4R1 I = V / R = 12 / (4 R1) = 3 / R1; VA = IR1 = 3 × R1 / R1 = 3 V; VB = IR2 = 3 × R2 / R1 = 3 V; VC = I(R3 + R4) = 3 × 2 × R1 / R1 = 6 V Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [2] [2] [2] [Total: 6] [1] [1] [2] [2] [2] [Total: 8] 45 Cambridge IGCSE Physics Student Book answers *3 a b 4 a b c 5 a b c 6 a b c *7 b c d Resistors in parallel: 1/R = 1/R1 + 1/R2 = 1/4 + 1/4 = 1/2 Ω–1; so R = 2 Ω For resistors in parallel: 1/R// = 1/R1 + 1/R2 = 1/6 + 1/2 = 4/6 Ω–1 so R// = 6/4 =1.5 Ω; then combined resistance = (1.5 + 6) Ω = 7.5 Ω [4] [6] [Total: 10] i R = (6 + 7 + 8) Ω = 21 Ω [2] ii Increased [1] The p.d. across each lamp is fixed (at the supply p.d.), so the lamp shines with the same brightness irrespective of how many other lamps are switched on. Each lamp can be turned on and off independently; if one lamp fails, the others can still be operated. [4] Less [1] [Total: 8] V 1 = V2 = 3 V [2] I = V/R = 6 V / (10 + 50) kΩ = 0.1 mA so V1 = IR1 = 0.1 mA × 10 kΩ = 1 V and V2 = IR2 = 0.1mA × 50 kΩ = 5 V [4] I = V/R = 6 V / (20 + 10) kΩ = 0.2 mA so V1 = IR1 = 0.2 mA × 20 kΩ = 4 V and V2 = IR2 = 0.2 mA × 10 kΩ = 2 V [4] [Total: 10] LDR and R in series with a battery [2] i I = V/Rtotal = 12 V / (20 + 28) Ω = 0.25 A ii V = IR = 0.25 A × 20 Ω = 5.0 V iii V = IR = 0.25 A × 28 Ω = 7.0 V i decreases ii increases iii increases L1 lights, L2 does not; no current flows through D2 as it is reverse biased L1 and L2 light; D forward biased and current splits between L2 and D L1 lights, L2 does not; D2 reverse biased [2] [2] [2] [3] [Total: 11] [3] [4] [3] [Total: 10] 4.4 Electrical safety Test yourself questions 1 2 3 C: P = IV = 5 A × 230 V = 1150 W; number of bulbs = 1150 W / 100 W = 11 P = IV = 13 A × 230 V = 2990 W = 2.99 kW a The metal case of the appliance b Double insulation; enclose all metal parts in an non-conducting plastic case Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 46 Cambridge IGCSE Physics Student Book answers Now put this into practice questions (Page 226) 1 2 3 a a a I = P/V = 1500W / 240 V = 6.3 A b I = P/V = 100 W / 240 V = 0.42 A b I = P/V = 6400 W / 240 V = 27 A b 13 A fuse should be used 3 A fuse should be used 30 A trip switch setting should be chosen Exam style questions (Page 228) 1 a b c d 2 3 Overheated cables, damaged insulation, overloading circuit by connecting too many appliances Damp conditions, faulty wiring, damaged insulation Disconnect appliance from electricity supply; look for any signs of a short circuit; check correct size of fuse is being used To prevent the user touching metal parts which could become live if a fault developed in the appliance [2] [2] [3] [3] [Total: 10] a It protects the circuit by ensuring the current-carrying capacity of the wiring is not exceeded. [2] b In a, the fuse is in the live wire, so when it blows and breaks the circuit, the lamp is disconnected from the live wire. In b, the fuse is in the neutral wire and when it breaks, the lamp is still connected to the live wire [4] [Total: 6] a i I = V/R = 240 V / 800 Ω = 0.3 A = 300 mA [2] ii Yes [1] iii By having dry hands or wearing shoes with rubber (insulating) soles [2] b i P = IV so I = P/V = 150 W / 230 V = 0.65 A; use a 3A fuse [2] ii I = P/V = 900 W / 230 V = 3.9 A; use a 13 A fuse [2] iii I = P/V = 2000 W / 230 V = 8.7 A; use a 13 A fuse [2] [Total: 11] 4.5 Electromagnetic effects Test yourself questions 1 D; the effect is called electromagnetic induction 2 *3 Slip rings (rotate with the coil) Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 47 Cambridge IGCSE Physics Student Book answers *4 a b coil i horizontal ii vertical 5 a North b East 6 S *7 a Inside the solenoid b The current; the number of turns on the solenoid 8 When a large current is to be controlled by a small current 9 The pull-on current I = V/R = 15 V / 300 Ω = 0.05 A = 50 mA 10 R = V/I = 12 V / 0.06 A = 200 Ω 11 Sound waves are produced by a paper cone attached to a coil vibrating with the a.c. signal *12 D (use Fleming’s left-hand rule) *13 Decrease, because force on the electrons is greater 14 a Increases b Reverses direction *15 Clockwise (use Fleming’s left-hand rule) *16They act to reverse the current through the coil every half turn so that the coil continues rotating in the same direction 17 B 18 B: Ns/Np = Vs/Vp so Ns = Np×Vs/Vp = 1000 × 46 / 230 = 200 Now put this into practice questions (Page 245) 1 a Np/Ns = Vp/Vs = 240 V / 12 V = 20 / 1 b Np/Ns = 20 so Np = 20 Ns = 20 × 80 = 1600 turns *2 a Turns ratio = Np/Ns = Vp/Vs = 240 V/ 960 V = 1 / 4 b Np/Ns = 1/4 so Ns = 4 Np = 4 × 500 = 2000 turns *3 IpVp = IsVs so Is = IpVp/Vs = 0.05 A × 240 / 12 = 1 A (Page 247) *1 a b *2 a i P = IV so I = P/V = 300 000 W / 20 000 V = 15 A ii P = IV so I = P/V = 300 000 W / 200 000 V = 1.5 A i Power lost in cables = I2R = (15 A)2 × 400 Ω = 9 × 104 W ii Power lost in cables = I2R = (1.5 A)2 × 400 Ω =9 × 102 W P = IV so I = P/V = 210 000 W / 700 000 V = 0.3 A; power lost in cables = I2R = (0.3 A)2 × 375 Ω = 34 W Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 48 Cambridge IGCSE Physics Student Book answers Practical work questions (Page 230) 1 2 3 4 5 6 When there is relative motion between the coil and the magnet Related by the equation V = IR Strength of magnet, number of turns on the coil, current through the coil Direction of rotation reversed No; there is no changing magnetic field to induce an e.m.f in the secondary coil Decrease Exam-style questions (Page 249) 1 a b 2 Connect a coil to a sensitive centre-zero ammeter in a complete circuit. Move a bar magnet towards and away from the coil; the meter shows a current is induced. Alternatively move the coil instead of the magnet Increase the strength of the magnet, the speed of the motion, the number of turns on the coil a The galvanometer needle swings alternately in one direction and then the other as the rod vibrates b This is due to an e.m.f. being induced in the metal rod when it cuts the magnetic field lines; current flows in alternate directions round the circuit as the rod moves in and out of the magnetic field *3 a b A: slip rings, B: brushes; slip rings connect the rotating coil to the brushes Increase the number of turns on the coil, the strength of the magnet and the speed of rotation of the coil. Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [4] [3] [Total: 7] [2] [4] [Total: 6] [1] [2] [3] 49 Cambridge IGCSE Physics Student Book answers c [4] [Total: 10] 4 a Pass a wire through a piece of card, support it vertically and pass a current through the wire; sprinkle iron filings on the card and tap gently so that they settle on concentric circles. Place a plotting compass at different points to find the direction of the magnetic field. [4] b [4] 5 c The direction of the magnetic field reverses a b To complete the circuits to the negative terminal of the battery A contains the relay contacts and starter; B contains the starter switch and relay coil Wire A carries a much larger current to the starter motor than wire B It allows the wires to the starter switch to be thin since they carry only the small current needed to energise the relay c d 6 7 A small coil placed between the poles of a magnet receives a varying current; the magnetic fields of the coil and the magnet interact causing the coil to vibrate with the frequency of the a.c. signal sound is produced by the vibration of a paper cone attached to the coil a b Support a wire horizontally between the poles of a magnet so that the direction of the wire is perpendicular to a horizontal magnetic field; the wire moves up or down when a current is passed through the wire; if the direction of the current or of the magnetic field is reversed, the wire moves in the opposite direction i increases [1] ii decreases [1] iii increases [1] [1] [Total: 9] [2] [2] [1] [2] [Total: 7] [2] [3] [1] [Total: 6] [2] [1] [2] [Total: 8] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 50 Cambridge IGCSE Physics Student Book answers 8 D *9 a [Total: 1] [3] b c The commutator rotates with the coil, so that the current through the coil reverses direction every half-turn [1] The forces on the coil then always act in the same direction, producing continuous rotation [1] If the current is as shown in the figure, Fleming’s left-hand rule gives an upward force on ab and a downward force on cd, producing clockwise rotation [1] Reversal of the battery connections would give anticlockwise rotation of the coil [1] i The motor would rotate in the opposite direction [1] ii The motor would rotate in the opposite direction [1] iii No change in the direction of rotation [1] [Total: 10] 10 a b c It is made up of two coils of wire, a primary and a secondary, wound on a complete soft iron core. It changes an a.c. voltage to an a.c. voltage of greater value Np/Ns = Vp/Vs so Ns = Np Vs/ Vp = 120 × 720 / 240 = 360 turns 11 a Ns/Np = Vs/Vp so Ns = Np × Vs /Vp = 460 × 12 V / 230 V = 24 *b Ip × Vp = Is × Vs so Is = Ip × Vp /Vs = 0.10 A × 230 V / 12 V = 1.9 A *12 a [4] [2] [4] [Total: 10] [4] [4] [Total: 8] When the switch is first closed, current will flow in coil A and a magnetic field will build up in A. Coil B will be cut by a changing magnetic field, a p.d. will be induced and a current will flow through the galvanometer deflecting the needle. Once the current and magnetic field become constant in coil A there will be no changing magnetic field linking it with coil B, no induced p.d. and the galvanometer reading will return to zero; this happens almost immediately after the switch is closed. When the switch is opened again the magnetic field in coil A falls, there is again a changing magnetic field linking it to coil B and a p.d. is induced in the opposite direction to previously; the galvanometer needle will swing briefly in the opposite direction before returning to zero again Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [2] [2] [3] 51 Cambridge IGCSE Physics Student Book answers b c 13 a b The deflection on the galvanometer would increase; the soft iron wires will become magnetised by the coil and will increase the magnetic field linking coil A with coil B. When the switch is closed or opened the induced p.d. will be larger The deflection on the galvanometer would increase Transformers step a.c. voltages up or down efficiently; p.d.s are stepped up at the power station before transmission and stepped down at sub-stations for local distribution. Energy lost as heat in cables is reduced; smaller currents result which allow thinner, cheaper wires to be used. [3] [1] [Total: 11] [3] [2] [Total: 5] Section 5 Nuclear physics 5.1 The nuclear model of the atom Test yourself questions 1 2 3 4 Positive charge and most of the mass are concentrated in a small dense nucleus; electrons with remaining mass carry the negative charge and orbit around the nucleus Removal of an orbital electron from the atom B a i 7 ii 3 iii (7 – 3) = 4 iv 3 b 5 a Z = 11, A = 23 b 11 *6 +2 *7 a 37 b mass of nucleus = total mass of the nucleons *8 a The decrease of total mass of the nuclei b It is converted into energy (E = mc2) *9 a Very high temperature needed so that the nuclei collide at speeds high enough to overcome the repulsive electrostatic forces between them b Large amounts of energy are released by the conversion of mass into energy *10 a Fission b Fusion Now put this into practice questions (Page 256) 1 2 Carbon-12: a Number of protons Z = 6 b number of nucleons A =12 c number of neutrons = A – Z = 6 Carbon-14: a Number of protons Z = 6 b number of nucleons A =14 c number of neutrons = A – Z = 8 A different B same C different Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 52 Cambridge IGCSE Physics Student Book answers Exam-style questions (Page 259) 1 a b c 2 a b c *3 a b *4 a b i Z is the number of protons in the nucleus of an atom ii A is the total number of protons and neutrons in the nucleus Isotopes of an element have the same number of protons in the nucleus but different number of neutrons. In there is one less neutron in the nucleus than in A = 40, Z = 20 i 20 ii 40 Positively charged. iii 20 [2] [2] [Total: 6] [2] [4] [1] [Total: 7] iv 20 The large deflection of a few alpha particles showed that most of the mass of the nucleus was concentrated at its centre and had a positive charge Most of the alpha particles were undeflected showing that the central nucleus was very small compared with the size of the whole atom i +1 ii 0 iii +2 i ii i ii [2] Nuclear fission is the break-up of a nucleus into smaller fragments In nuclear fusion light nuclei join up to form heavier nuclei A = 3 + 3 = 6, Z = 2 + 2 = 4 X = 4 – 1 – 1 = 2, Y is helium ( ) [3] [2] [3] [Total: 8] [2] [2] [2] [2] [Total: 8] 5.2 Radioactivity Test yourself questions 1 Any two from: cosmic rays, radon gas, rocks/buildings, food/drink, radioisotopes used in medicine 2 Ionising effect *3 a 300 counts / 60 s = 5 counts/s b (190 – 5) counts/s = 185 counts/s 4 a α b γ c α d β– *5 a β b γ c α d γ *6 neutron → proton + electron 7 a By alpha decay b By beta decay *8 a Corrected count rates for background radiation Time/s 0 30 60 90 120 150 180 Corrected counts/s 160 107 72 48 32 22 15 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 53 Cambridge IGCSE Physics Student Book answers b half-life = 52 s 9 D: 8 minutes = 4 half-lives so activity falls by 1/24 = 1/16 *10 a Gamma rays b Alpha-particles *11 If a radioisotope which emits gamma rays is placed on one side of a moving sheet of material and a GM tube on the other, the count rate increases if the thickness of the sheet decreases. Now put this into practice questions (Page 266) + *1 *2 + (Page 268) 1 a b The count rate drops by half between 15 and 20 minutes Half-life = 17 min 2 3 60 minute / 15 minutes = 4; after 4 half-lives fraction left = 1/2 × 1/2 × 1/2 × 1/2 = 1/16 After 1 half-life count rate =140 / 2 = 70 counts/minute, after 2 half-lives count rate = 70 / 2 = 35 counts/minute so 60 minutes = 2 half-lives. Half-life of the material = 60 / 2 = 30 minutes 4 After 1 × 5700 years the activity will be 80 / 2 = 40 counts per minute; after 2 × 5700 years the activity will be 40 / 2 = 20 counts per minute. Estimated age of the canoe is 2 × 5700 = 11 400 years Exam-style questions (Page 273) 1 a b B i C ii A iii B Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [3] [1] [1] [1] 54 Cambridge IGCSE Physics Student Book answers iv A [1] [Total: 7] 2 a b 3 a b The half-life of a radioisotope is the time taken for half the nuclei of that isotope in a particular sample to decay 2 minutes; count rate falls by a half every 2 minutes i ii iii iv After 1 × 1500 million years there will be N/2 atoms of argon left Number of potassium atoms formed = N − (N/2) = N/2 Ar : K ratio = N/2 : N/2 = 1 : 1 After 2 × 1500 = 3000 million years, there would be (N/2)/2 = N/4 argon atoms left v and N − (N/4) = 3N/4 potassium atoms formed vi Ar : K ratio of N/4 : 3N/4 = 1 : 3 Measured ratio is 1 : 3 so the rock must be about 3000 million years old *4 a i C b ii i 4 220 218 86Rn → 84Po + 2He ii 5 a B 234 234 0 90Th → 91Pa + −1e Can damage living cells and tissue leading to cell death, gene mutation, cancer, eye problems, radiation burns and sickness b i alpha ii gamma iii gamma iv alpha c Two from: reduce exposure time, increase distance between source and person/handle source with forceps, keep away from eyes, shield source, store source in a lead box when not in use [2] [4] [Total: 6] [1] [1] [1] [1] [1] [1] [1] [Total: 7] [1] [3] [1] [3] [Total: 8] [2] [4] [2] [Total: 8] *6 a b c A small amount of an α-particle emitter, such as Americium-241, causes ionisation of the air in an ionisation chamber and results in a current flow between two metal electrodes. When smoke enters the detector, it impedes the flow of ions and the current reduces. The fall in current is detected electronically and an alarm activated [3] An α-particle emitter is chosen because α-particles have only a short range in air; Americium-241 has a long half-life so its activity remains fairly constant over time [1] A radioisotope is placed on one side of a moving sheet of material and a GM tube on the opposite side; the count rate decreases when the thickness increases [2] β-emitters are suitable for thin sheets, but γ-rays are needed for thicker sheets because of their penetrating power; a long half-life source is preferred so that the activity of the source remains fairly constant over time [1] Gamma rays kill bacteria on the food without damaging the food itself [1] This leads to a longer shelf-life for the product [1] A γ-rays source is used for its high penetrating power; a long half-life is preferred so that the activity stays fairly constant over time [1] [Total: 10] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 55 Cambridge IGCSE Physics Student Book answers Section 6 Space physics 6.1 Earth and the Solar System Test yourself questions 1 2 D 3 a March and September; equinoxes b Summer *4 v = 2 π r / T = 2 π 380 000 km / (27 × 24 × 60 × 60) s = 1.02 km/s 5 Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto 6 C 7 A true B true C true *8 *9 a Circumference = 2 π r = 2 π × 228 × 106 km = 1.43 × 109 km b v = 2 π r / T = 1.43 × 109 km / (687 × 24 × 60 × 60) s = 24 km/s *10 a 23.1 N/kg b weight = mgJ = 50 kg × 23.1 N/kg = 1155 N *11 Higher; Jupiter is nearer the Sun than Saturn Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 56 Cambridge IGCSE Physics Student Book answers Now put this into practice questions (Page 281) 1 2 t = distance / speed = 5.8 × 1010 m / 3 × 108 m/s = 193 s t = distance / speed = 5.9 × 1012 m / 3 × 108 m/s = 2.0 × 104 s (over 5 hours) Exam-style questions (Page 285) 1 2 D [Total: 1] a The moon rotates on its axis in the same time it takes to orbit the Earth [2] b The Moon is orbiting the Earth. Different parts of the area of the Moon’s surface illuminated by the Sun are visible from Earth when the Moon moves to different positions in its orbit [3] c Because the Earth is rotating on its axis [1] [Total: 6] 3 Time = distance / speed of light = 228 × 109 m / 3 × 108 m/s = 760 s [Total: 3] 4 a The planets are thought to have formed from the remains of the cloud of hydrogen gas and dust from which the Sun evolved [1] In the region of space where the inner planets were condensing, the temperature close to the Sun would have been so high that only materials with a high melting point such as metals and silicates could exist in a solid form [2] Less than 1% of nebulae consists of heavy elements, so the inner planets only grew to a small size; they are small and rocky [1] b Further away from the Sun, the temperature would have been less, and light molecules with low melting points, such as hydrogen, helium, water and methane, condensed into solid ices [3] These molecules make up about 99% of the material in a nebula, so the outer planets grew to a size large enough to capture even the lightest molecule, hydrogen [1] [Total: 8] *5 a i increase ii increase [3] b i Greater as Venus is nearer the Sun [2] ii Less, as circumference of orbit is less and it travels faster [3] [Total: 8] *6 a i Lower; Jupiter is further from the Sun so receives less radiation/m2 [2] ii Larger [1] iii Lower [1] b i v = distance / time = 2 π r / T so vM/vE = (2 π RM / TM) / (2 π RE / TE) = (RM × TE) / (RE × TM) = 1.5 / 1.9 = 0.78 [3] ii Slower [1] [Total: 8] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 57 Cambridge IGCSE Physics Student Book answers 6.2 Stars and the Universe Test yourself questions 1 Infrared, visible light and ultraviolet *2 Nuclear fusion of hydrogen into helium 3 D 4 D *5 B *6 D *7 Redshift measurements show the universe is expanding and the further away galaxies are the faster their speed of recession from the Earth. The cosmic microwave background radiation left over from the Big Bang has been red-shifted into the microwave region. *8 B: Hubble’s law gives us v = Ho × d so d = v/Ho = 8500 / 2.2 ×10–18 km = 3.9 × 1021 km = 3.9 × 1021 km / 9.5 × 1012 light years = 400 million light years Exam-style questions (Page 294) 1 a b 2 C *3 a b *4 a b Light from glowing hydrogen and other gases in stars in distant galaxies has a longer wavelength than it does on Earth – the light is ‘shifted’ towards the red end of the spectrum Distant galaxies are moving away from us; the further away they are the faster the speed of recession Clouds of hydrogen gas collapse due to gravitational attraction and form a protostar As the protostar grows in size it becomes hotter and when the temperature is high enough in the core, nuclear fusion of hydrogen into helium starts; large amounts of energy are released which sustain the fusion process. The protostar has then become a star, powered by nuclear fusion When most of the hydrogen in the core has been used up in the nuclear fusion of hydrogen into helium Gravity acts inwards; thermal pressure, due to the high temperature of the star, acts outwards Low mass star → red giant → planetary nebula → white dwarf → black dwarf 1 million ly = 9.5 × 1012 × 106 km = 9.5 × 1018 km Ho = v/d = 11 000 km/s / (500 × 9.5 × 1018) km = 2.3 × 10–18 s–1 b It gives an estimate of the age of the Universe and is evidence that all matter in the Universe was created at a single point *5 a Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [3] [2] [Total: 5] [Total: 1] [1] [2] [2] [2] [Total: 7] [4] [5] [Total: 9] [5] [2] [Total: 7] 58 Cambridge IGCSE Physics Student Book answers Mathematics for physics (Page 295) 1 2 3 4 5 6 a b c d e f g h i a b c d e f g h a 3 5 8/3 20 12 6 2 3 8 f = v/λ λ = v/f I = V/R R = V/I m=d×V V = m/d s = vt t = s/v I2 = P/R b c d e f g h a b c d e I = √(P/R) f 3 × 108 a b c 2.0 × 105 10 8 d 2.0 × 108 e f a b c 20 300 4 2 5 a = 2s/t2 t2 = 2s/a t = √(2s/a) v = √(2gh) y = Dλ/a ρ = AR/l 10 34 2/3 1/10 10 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 59 Cambridge IGCSE Physics Student Book answers d 8 e 2/3 f –3/4 g 13/6 h –16 i 1 7 a = (v – u)/t a 5 b 60 c 75 8 a = (v2 – u2)/2s 9 a Students’ graphs of extension against mass b Extension ∝ mass because the graph is a straight line through the origin 10 a Students’ graphs of m against v b No: graph is a straight line but does not pass through the origin c 32 11 a Graph is a curve b Graph is a straight line through the origin, therefore s ∝ t2 or s/t2 = a constant = 2 Additional exam-style questions Motion, forces and energy (Page 299) 1 2 D a b 3 a b c 4 A 5 a b c 6 D *7 D 8 a b c d *9 a Average speed = s/t = 1600 m / (8 × 60) s = 3.3 m/s Increases mg = 90 kg × 9.8 m/s2 = 880 N 90 kg mg = 90 kg × 9.8 m/s2 / 6 = 147 N Yes, 1 mm = 0.001 m kg/m3 Density = m/V = 120 g / 15 cm3 = 8.0 g/cm3 = 8000 kg/m3 W = F d, where d is the distance moved by the force F in the direction of the force joules, J W = F d = 3 N × 5 m = 15 N m = 15 J 15 J i Electrical to heat and light ii Chemical to electrical Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 60 Cambridge IGCSE Physics Student Book answers iii Electrical to kinetic energy iv Kinetic to electrical b Efficiency is the ratio of the useful energy output to the total energy input expressed as a percentage 10 D *11 a Ep = mgΔh = 2 kg × 9.8 m/s2 × 4 m = 78 J b i Ek = mv2/2; rearrange to v2 = 2Ek/m = 2 × 100 J / 2 kg = 100 m2/s2 and v = 10 m/s ii Ek = Ep = mgΔh; rearrange to Δh = Ek/mg = 100 J / (2 kg × 9.8 m/s2) = 5.1 m *12 a W = F d = Ek; rearrange to give d = Ek /F = 10 J / 5 N = 2 m b F = ma = 6 kg × 3 m/s2 = 18 N c i Impulse = FΔt = 18 N × 3 s = 54 N s ii Δp = FΔt = 54 Ns Thermal physics (Page 300) *13 a Consists of equal lengths of two different metals of different expansivity welded together. When heated, one of the metals expands more than the other and the strip bends. It can be used in many applications from thermostats to fire alarms b i pV = constant ii p1V1 = p2V2 so p2 = p1V1 / V2 = p1 / 2; pressure halves c When the temperature of a gas increases the average speed of its molecules increase. If the volume of the gas remains constant, there will be more frequent collisions with the walls of the container and the pressure of the gas will increase 14 a Wood is a less good conductor of heat than metal, which conducts heat away from the hand b When a fluid is heated it expands and becomes less dense. Parts that are warm will rise above colder denser regions leading to a convection current being set up in the fluid c Conduction and convection require a medium to transfer thermal energy Waves (Page 300) *15 a b 16 a b c i Longitudinal ii Compression iii Rarefaction i Become circular ii No change iii No change 60° 30° The image is: as far behind the mirror as the object is in front with the line joining the object and image being perpendicular to the mirror; the same size as the object; virtual; laterally inverted Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 61 Cambridge IGCSE Physics Student Book answers 17 C 18 a b c 19 a b 20 a b Dispersion i Red ii Violet Violet, indigo, blue, green, yellow, orange, red i The line through the centre of a lens at right angles to the lens ii The point on the principal axis of the lens to which a parallel beam of light passing through the lens converges i Less than the focal length of the lens ii Upright iii Virtual i Infrared ii X-rays i Radio ii γ-rays Transverse Communications, microwave ovens for cooking c d 21 A 22 D 23 a Two from: electromagnetic (light, radio …), water, seismic S-waves, b Two from: sound, mechanical waves on a spring, seismic P-waves c Three from: reflected, refracted, diffracted, obey the wave equation, carry energy from place to place Electricity and magnetism (Page 302) 24 C 25 a b c *26a b c R = R1 + R2 = 2 Ω + 1 Ω = 3 Ω V = IR so I = V/R = 6 V / 3 Ω = 2 A (same in each resistor) V = IR: V1 = 2 A × 1 Ω = 2 V; V2 = 2 A × 2 Ω = 4 V resistors in parallel: 1/R = 1/R1 + 1/R2 = 1/2 + 1/2 = 1 Ω–1; so R = 1 Ω V = IR so I = V/R = 6 V / 2 Ω = 3 A (same in each resistor) 6V 27 a b i P = IV, rearrange to I = P/V = 3000 W / 230 V = 13 A Number of kWh = 3 × (100 × 10–3) × 10 = 3 Total cost = 3 ×10 cents = 30 cents ii D 28 B 29 a b c d e coulomb, C ampere, A volt, V joule, J watt, W Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 62 Cambridge IGCSE Physics Student Book answers Nuclear physics (Page 303) 30 a b After 1 half-life count rate will be 50 counts/s; after 2 half-lives count rate will be 25 counts/s. Time for two half-lives = 2 × 1 hour = 2 hours i alpha-particles < beta-particles < gamma rays ii alpha-particles > beta-particles > gamma rays Space physics (Page 303) 31 a b c d *32 a b Away At noon, around June 21 On or near to Dec 22 March and September distance = 2 π r = 2 × π × 385 000 km = 2.42 × 106 km average speed = distance / time so, time = distance / speed = 2.42 × 106 km / 1 km/s = 2.42 × 106 s. Time in days = 2.42 × 106 s / (60 × 60 × 24) = 28 days. 33 a 2.85 × 1014 km / (9.5 ×1012) km/ly = 30 light years b i B ii C iii A iv D *34 a Discovery of cosmic background radiation left over from the Big Bang; Red-shift of starlight shows that distant galaxies are moving away from us and that the further away a galaxy is, the higher the speed of recession; this means that the Universe is expanding. b v = Ho d, so d = v / Ho = 16 000 km/s / (2.2 × 10–18 s–1) = 7.3 × 1021 km = 7.3 × 1021 km / (9.5 × 1012 km/light year) = 7.70 × 108 light years = 770 million light years Theory past paper questions Motion, forces and energy Physical quantities and measurement techniques, Motion, Mass and weight, Density (Page 304) 1 a b c 2 a b Height of water/liquid i 3.10, 3.04, 3.16 ii Average time = 3.10 s Number of drops = 300 i 1.1: Time to swing from P to Q = 0. 55s, time for a complete oscillation = 1.1 s ii Any four from: use a fiducial mark, start watch as pendulum passes fiducial mark or when it is released, count at least 10 swings, time to centre of swing, stop watch as pendulum passes marker or starting point, divide total time by number of swings 1. 0.4 2. 0.0 Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 63 Cambridge IGCSE Physics Student Book answers 3 4 a b c a b *5 a b 6 a b *7 a b 8 9 a a b Deceleration Y Z, constant speed X Y Distance = area under YZ = 400 m WX has steeper gradient than YZ A uniform acceleration B constant speed C non-uniform deceleration D at rest Distance = 8.75 m i 1. Straight line through origin and (10, 50) 2. gradient/slope 2 ii 0.40 m/s i Δv = at = –35 m/s; straight line from (0, 50) to (100, 15) ii Distance = average speed × time = 3300 m Straight diagonal line passing from point (0, 0) to (10, 10); horizontal line drawn from (10, 10) to (70, 10); straight diagonal line drawn from (70, 10) to (85, 0) distance = speed × time = 1125 m change of speed / time = (v – u)/t i 1. acceleration 2. constant speed 3. deceleration ii 1. speed = distance / time = 7.5 m/s 2. change of distance / change of time = 12 m/s density b 2500 g c weight = mg = 40 N total weight of raft = mg = 5200 N i Volume of log 0.12 m3 Density = mass / volume = 550 kg/m3 ii Density of log is less than density of water Forces, Momentum, Energy, work and power and Pressure (Page 307) 10 a b c *11 a b 12 a b 13 a b c *14 a b i Gravity (or weight) ii 4.0 N Resultant force = 4.8 N upwards i Energy cannot be created or destroyed (but can be transferred) ii Air resistance causes friction with the moving spring/load and motion eventually ceases when all the kinetic energy has been transferred to thermal energy dissipated to the surroundings Accelerate or increase speed OR decelerate or decrease speed OR change speed, Change direction OR cause rotation Scale e.g. 30 N = 1 cm, vectors drawn at right angles and labelled, parallelogram or triangle completed, correct orientation with 360 N, T = 250 N Weight = mg =150 N i turning effect ii moment = Fd = 1063 N m iii move support rope nearer to P Distance = area under graph = (base × height) = 44 m C; cyclist is accelerating, so forward force greater than backward force Pressure = force / area = 7.5 N/cm2 mv – mu i Impulse = 2.4 N s ii Ft = m(v – u), u = 0, so v = 43 m/s Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 64 Cambridge IGCSE Physics Student Book answers *15 a b c d *16 a b c *17 a b *18 a b *19 a b iii 1. kinetic energy of racquet transfers to elastic/strain energy (in ball or strings) 2. elastic/strain energy (in ball or strings) transfers to kinetic energy (of ball) Change in momentum = 0.58 kg m/s F = rate of change in momentum = 0.096 N The truck accelerates backwards/to the right It accelerates faster due to truck having less mass (when empty); force is constant and F = ma Change in gravitational potential energy is the work done by the force Work done = force × distance moved = 8.3 × 107 J Work done by train = 270 × 107 J Efficiency = (work output/work input) × 100% = 3.1% Kinetic energy = = 2.7 × 107 J Work done = 1.8 × 105 N F, H, G, E i 1. 100 W 2. 500 W ii Less power OR energy used by LED, less CO2, OR greenhouse gases OR global warming i Pressure = 150 000 N/m2 ii Total pressure = 250 000 Pa iii p = F/A so F = 240 000 N Any two from: weight of lid, pressure inside box (or upthrust on lid), moment of force changes, friction at hinge, drag of water Thermal physics (Page 311) 20 a b *21 a b *22 a b c *23 a b 24 a Atoms in a more regular arrangement/fixed position, more tightly packed and slower moving The more energetic water molecules escape from the liquid surface and become gas molecules; the process is termed evaporation i showing straight lines with sudden change in direction ii Brownian motion – smoke particles bombarded by air molecules moving in random directions F = change in momentum/time = 1.4 N 1. solid to liquid 2. liquid to gas/vapour There are strong forces of attraction between molecules in a solid OR molecules in a gas are further apart so the attractive force between them is less V = 0.216 m3 Pressure increases, Any two from: Molecules travel shorter average distance between collisions with walls, Molecules hit walls more often OR more collisions (per unit area) with walls Greater force OR greater rate of change of momentum of molecules per unit area on walls 1st box: gas, 2nd box: solid Expansion/voltage/p.d./e.m.f./length/colour/pressure/volume/resistance Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 65 Cambridge IGCSE Physics Student Book answers b c *25 a b *26 a b i 8.30 pm ii 9.00 pm, gradient of graph decreases / rate at which temperature falls is slower Insulator, conduction, convection i Any good insulating material (such as fibre glass or newspaper) so that less heat is transferred to the surroundings and more energy is transferred to the block ii Insulation on top of block Heat supplied ΔE = 27 000 J So specific heat capacity = 450 J/(kg ℃) i Boiling ii Evaporation i Heat supplied ΔE = m c Δϴ = 44 000 J Power of heater ΔE/t = 292 J/s (290 W) ii Wrap insulation around the metal block to reduce heat lost to surroundings Waves (Page 314) 27 a b 28 a b 29 a b *30 a b *31 a b *32 a b c d 33 a i z ii water, other parts of electromagnetic spectrum i 3 curved waves after gap, evenly spaced, centred on gap ii diffraction i Normal ii Angle of incidence iii Doubles i Principal focus ii Diminished, inverted, real f = 5 cm i Straight line from top of object through F to lens and then parallel to principal axis ii Straight line from top of object through centre of lens iii Image drawn at point where lines cross; inverted image on RHS of lens i Dispersion ii A – red, B – violet iii Refractive index/speed different for different wavelengths i More reflection on top wall of fibre, between X and end of fibre and no reflections on lower wall of fibre and ray reaches end of fibre ii c = 43o iii Any two from: lighting, carry signals/communications, medicinal diagnosis/imaging i It passes through boundary and is refracted away from the normal ii Total internal reflection occurs i Total internal reflection at B with angle of incidence equal to angle of reflection. Refraction into air at right-hand face with angle of refraction greater than angle of incidence ii n = 1.5 Longitudinal, 35 000 Hz v = f λ, so λ = v/f = 3 × 108 m/s / (1.3 × 1017 Hz) = 2.3 × 10–9 m X-rays are ionising radiation which is harmful to humans Any one from: patient rarely exposed, dentist frequently exposed, total dose on dentist would be high if always stayed in room, low total dose on patient, benefit to patient outweighs danger Microwaves harmful to humans, microwaves could pass through open door Orange, yellow, green, blue, indigo Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 66 Cambridge IGCSE Physics Student Book answers b 34 a b 35 a b c i Detecting an intruder – infra-red, satellite communications – microwaves Detecting broken bones – X-rays ii Frequency i Vibrates ii Longitudinal iii Vacuum i 1000 ii Approximately 11 000 Hz iii Lowest frequency humans can hear is 20 Hz but elephants can hear frequencies down to 5 Hz iv ultrasound 20–20 000 Hz Sound waves with frequencies greater than 20 000 Hz P will sound quieter and have a lower pitch than Q because P has a smaller amplitude and lower frequency than Q Electricity and magnetism (Page 317) 36 a i Right hand end of bar marked N, left hand end of bar marked S ii Any two from; iron bar becomes induced magnet, a South pole is induced in the lefthand end of the iron bar and it is attracted to the North pole of the magnet, opposite poles attract b i Ends of coil connected to d.c. battery or power supply ii Can be switched on and off or easily magnetised/demagnetised *37 a Electrons must be removed b i + + + at top of sphere; – – – at bottom of sphere ii Earth sphere while rod in position (touch sphere with a conductor or finger); remove earth connection and then remove charged rod 38 a Friction causes electrons to be transferred from the cloth to the rod b Like charges repel so the suspended rod moves away 39 a i electrons ii R = 18.0 Ω iii V = 9 V b Current increases because combined resistance of resistors in parallel is less *40 a I = 4.2 A b i E = 3240 kJ ii Volume = 90 cm3 *41 a i Resistance is constant ii Resistance increases b R = 1.4 ohms c Current in lamp = 4.4 A, current in resistor = 4.0 A Total current = 8.4 A d p.d. across resistor = 6.0 V, p.d. across lamp = 4.9 V Total p.d. = 10.9 V 42 a Variable resistor or rheostat b 0.8 A c R = 5.6 Ω Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 67 Cambridge IGCSE Physics Student Book answers d *43 a b 44 a b c 45 a b c *46 a b 47 a b *48 a b A thicker wire has less resistance so current increases 1/Rparallel = 5/60 so Rparallel = 0.12 Ω Rcombined = 0.32 Ω The resistance of lamp increases because the temperature of the lamp increases V = IR so I = 0.33 A i Arrows from N to S ii force reverses direction, it acts upwards i Any two from: increase number of turns on coil, increase current in coil, increase strength of magnetic field ii Coil will rotate in the opposite direction because current is in the opposite direction Lamps can be switched on and off independently; each lamp has full battery voltage across it (so lamps brighter); if one lamp breaks, other still works i Useful energy output = 1.8 J ii The diagram should show smaller proportion of energy wastage OR reduced wasted energy output OR smaller energy input for same output Any two advantages from: renewable, non-polluting, conserves fossil fuels, does not contribute to global warming Any two disadvantages from: unreliable because wind not always available, may harm wildlife, large land area needed, unsightly, needs a windy location, expensive set-up costs, large number needed to replace a power station i Light-dependent resistor ii i Total R = 3.6 kΩ, I = 1.67 mA Voltmeter reads IR = 4.0 V ii Increase temperature of room OR replace 1.2 kΩ resistor with one of higher value Ammeter and coil of wire connected in series Move magnet towards or away from coil; ammeter shows current induced in the coil Any two from: speed of movement of magnet; strength of magnet; number of turns on the coil For transmission of power, IV, the current, will be low if V is high; if the current is low the thermal energy generated, and hence the power loss in transmission lines is low; at low currents thinner/lighter/cheaper transmission cables/pylons can be used i Ns/Np = Vs/Vp so Ns = Np Vs/Vp = 300 turns ii Iron Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 68 Cambridge IGCSE Physics Student Book answers Nuclear physics (Page 321) 49 Any four from: take background reading without source place a piece of paper between source and detector alpha rays are stopped by paper but beta rays will pass through reading on detector will be similar or unchanged if source does not emit alpha particles particles place a sheet of aluminium a few mm thick between source and detector reading on detector will then be similar to background reading because beta particles are stopped by a few mm aluminium *50 a i Two from: cosmic rays, the Sun, soil/rocks/buildings/ the Earth, medical sources, radon in air, food or drink ii Radioactivity is a random decay process b If background count rate stays constant, after 24 days source contribution is 10 counts/minute; after 4 half-lives expect source contribution to be 12.5 counts/minute. Half-life = 6 days c α-particle: most ionising – fire alarms β-particle: affected by small changes in amount of solid – aluminium foil manufacture γ -ray: highly penetrating – detect leaks in water pipes *51 a b c *52 a 231 90Th i Breakup of a nucleus into two or more parts ii To absorb ionising radiation emitted in nuclear fission iii Advantage: atmosphere not polluted with carbon or sulphur dioxide/greenhouse gases disadvantage: disposal of radioactive waste/leaks of radioactive material/radiation risk if an accident occurs After one half-life, 26 hours, 2.4 × 109 atoms will have decayed after two half-lives = 2 × 26 = 52 hours, a further = 1.2 × 109 atoms will have decayed Total number of atoms decayed after 52 hours = 3.6 × 109 atoms Background count rate = 19 ± 2 counts/s Time/hours Detector reading/ (counts/s) Corrected counts/ (counts/s) 0 1 2 3 4 5 6 7 8 9 10 324 96 39 23 21 17 21 20 19 20 18 305 77 20 4 2 –2 2 1 0 1 –1 In 1 half-life corrected count rate would fall to 153 counts/s In 2 half-lives corrected count rate would fall to 77 counts/s In 3 half-lives corrected count rate would fall to 39 counts/s In 4 half-lives corrected count rate would fall to 20 counts/s 2 half-lives occur in each hour, so one half-life is 1/2 hours = 30 minutes Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 69 Cambridge IGCSE Physics Student Book answers b 3 1 c Top track – path turned back to left; middle track – path goes right with downward deflection; bottom track – no deflection 86 protons, (220 – 86) neutrons = 134 neutrons, 86 electrons, *53 a b c 54 a b H = −01β + 23 X 220 86 Rn = 42 α + 216 84 Po 220/55 = 4; so after 4 half-lives count rate will be 45 counts/s Unpredictable Emission Relative ionising ability Relative penetrating ability alpha high low beta medium medium gamma low high c 2 protons, 2 neutrons, 0 electrons Practical Test past paper questions (Page 324) 1 Equipment clamp, boss and stand pendulum bobs of the same mass but different shapes lengths of thin string about 60 cm long metre rule, graduated in mm set-square stop-watch with a resolution of at least 0.1 s split cork, or similar to hold the string of the pendulum between the jaws of the clamp Notes The pendulum bobs can be made from modelling clay; one rolled into a spherical shape with a diameter about 2 cm and the other into a cylinder of length about 5.0 cm. The string should be embedded in the bob so that the pendulum can swing without the bob slipping from the string. The stand can be stabilised by putting a weight on the base. a b c d Correct use of set-square and vertical ruler i t1 = 29.5 ± 2.5 s ii Correct calculation of T1 = t1/20 s and correct units in i or ii i t2 = 27.7 = ± 2.5 s ii T2 = t2 /20 s, T2 less than T1 to 2 or 3 significant figures Statement to match readings Justification, e.g. beyond/close to/within experimental accuracy. Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [1] [1] [1] [1] 70 Cambridge IGCSE Physics Student Book answers e f 2 Method reduces the effect of errors when starting and stopping the stopwatch 4 or 5 correct, from top box V, V, V, V, P, P. [1] [1] [1] [Total: 11] Equipment Plastic or polystyrene drinks cup with base narrower than lip and a volume approximately 180 cm3–250 cm3 Ruler 30 cm long graduated in mm Water Measuring cylinder 250 cm3 or 100 cm3 Top-pan balance which can measure masses up to 200g to the nearest gram Supply of paper towels to mop up spills Note: beaker should be labelled W a b c d i 2 or more measurements DB correct within ± 2mm ii DT correct within ± 2mm iii D correct average i h sensible in cm ii 1. V correct and 2. V/2 correct i m recorded ii ρ recorded to 2 or 3 significant figures Value in range 0.9–1.3 g/cm3 Any one from: Drawn circle not exact / thickness of rim or cup Height is not length of side of cup / D2 increases the inaccuracy in D Volume measurements only to 1 or 2 cm3 / mass of cup has been ignored [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] 3 Equipment clamp, boss and stand spring metre rule, graduated in mm set-square masses of 100 g, 200 g, 300 g, 400 g and 500 g a b c d lo clearly marked on figure Correct value for lo Other values in table increasing Set square used to line up with scale or view perpendicularly/ scale close to spring On graph: Axes correctly labelled and not reversed Suitable scales chosen Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [1] [1] 71 Cambridge IGCSE Physics Student Book answers All points correctly plotted to 1/2 small square Good best fit line drawn with a thin continuous line. e No: line should not pass through the origin f Use of 2 × lo shown on graph. correct to 1/2 small square 4 5 Method: place truck on ramp and release measure distance travelled from bottom of ramp repeat with different masses loaded on the same truck Additional apparatus: metre ruler/measuring tape Variables: height/angle of ramp/number of supporting bricks Release position/height above bench Table: clear columns for mass, distance travelled with appropriate units in the headings of the table [1] [1] [1] [1] [1] [Total: 11] [1] [1] [1] [1] [1] [1] [1] [Total: 7] Equipment Thermometer –10 °C to 110 °C, graduated in 1 °C intervals Clamp, boss and stand 250 cm3 beaker Supply of hot water Stopclock or stopwatch showing seconds Paper towels to soak up any water spills Notes Apparatus should be set up as shown in Fig. P6 The bulb of the thermometer must be well below the 100 cm3 level of the beakers The hot water should be available at a constant temperature between 80°C and 100°C Students should be warned of the dangers of burns or scalds when using hot water. a b c d e f θ for 200 cm3 decreasing [1] 3 i θ for 100 cm decreasing more quickly [1] ii s, °C, °C, all correct [1] 30, 60, 90, 120, 150, 180 [1] Conclusion matching results [1] Justification matching conclusion with correct mention of comparative temperature change over 0 to 180 s [1] i unit °C/s [1] ii correct calculation of x1 and x2 – x1 [1] Statement matching results [1] with results used in explanation and reference to different starting temperature for x1 and x2 Experiment with lid and no insulation [1] Experiment with insulation and no lid [1] [Total: 11] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 72 Cambridge IGCSE Physics Student Book answers 6 Equipment Sheet of plain A4 paper Plain mirror mounted so that it is perpendicular to the bench Screen with slit (see note 1) Lamp, low voltage (24 W or greater) and power supply Protractor 30 cm ruler Note The screen can be a sheet of stiff card or thin wood (approximately 70 mm × 70 mm) fixed upright to a support. The slit in the screen should be a minimum of 35 mm long and 1mm to 2 mm wide. a b c d e f g h 7 AB, CD and normal correct θ = 5° ± 1° GN ≥ 5.0 cm All lines present, correct and neat 5 correct values of a increasing Graph: Axes correctly labelled with quantity and unit Suitable scales chosen Plots all correct to 1/2 small square and precise plots Well-judged line and a thin continuous line. Any suitable reason e.g.: ray has finite thickness, reflecting surface of mirror at rear, inaccuracies have more effect for smaller angles, small variations in mirror angle have significant effect on a Reflect ray below NL at same angles and take averages [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] Equipment Converging lens, focal length between 14.0 cm and 16.0 cm with a suitable holder Lamp, low voltage (24 W or greater) and power supply Object with a triangular hole of height 1.5 cm covered with thin translucent paper (e.g. tracing paper) Metre rule, graduated in mm Screen Note The screen can be a sheet of stiff card (approximately 15 cm × 15 cm) fixed to a wooden support a i ii u values: 20(.0), 22(.0), 25(.0), 30(.0), 35(.0) v values decreasing and all greater than 22.0 cm. Consistent 2 or consistent 3 significant figures for v b On graph: Axes correctly labelled and right way round Suitable scales chosen Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [1] 73 Cambridge IGCSE Physics Student Book answers All points correctly plotted to 1/2 small square Good best fit curve drawn with a thin continuous line c 8 9 i ii Correct straight line between points (25.0, 25.0) and (35.0, 35.0) u1 and v1 read correctly to 1/2 small square Correct calculation of f from values f value rounded to 14–16 cm Diagram: show power supply, ammeter, voltmeter and resistance wire correctly connected (variable resistor optional) Correct symbols for ammeter and voltmeter (and variable resistor if included) Method: measure p.d. (voltage) and current and calculate resistance Repeat with other types of wire Key variables: length and diameter stated Precautions: One of: Repeat with different voltages (or currents), repeat and take average of voltage and current readings Repeat entire experiment with different length or different diameter wire Use low current to prevent wire heating up Keep temperature of wire constant Use micrometer screw gauge to measure diameter/thickness of wire Table: Columns for type of wire, voltage, current, resistance with correct units (V, A and Ω) [1] [1] [1] [1] [1] [1] [Total: 11] [1] [1] [1] [1] [1] [1] [1] [Total: 7] Equipment Lamp X (2.5 V, 0.3 A) and Lamp Y (6 V, 0.4 A) or similar with terminals such that student can easily rearrange circuit Power supply of 2–3 V; if cells used they must remain fully charged for the experiment Switch (may be part of power supply) 10 Connecting leads 0–1 A ammeter with 0.05 A resolution (tape correct setting if variable) 0–3 V voltmeter with 0.1 V resolution (tape correct setting if variable) a b c i ii iii iv IS less than 1.00 A VX and VY both less than 3.00 V and VX less than VY VS within 10% of (VX + VY) Statement matches results Justification matching statement with comparative values used, e.g. beyond/close to/within experimental accuracy Correct calculation of R1 2 or 3 significant figures and unit Ω Lamps in parallel All circuit components in correct position and correct symbols used Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [1] [1] [1] [1] 74 Cambridge IGCSE Physics Student Book answers d i ii Ip and Vp present and V and A units correct R2 greater than R1 [1] [1] [Total: 11] Alternative to Practical past paper questions (Page 331) 1 2 a i ii l should be measured from lower surface of clamp to the centre of the bob [1] Any 2 from: use set square as horizontal reference to help line ruler up vertically, metre rule close to pendulum, measure from lower surface of clamp [2] b i T = t/20 = 20.22 / 20 = 1.01(1) s [1] ii Any two from: idea of averaging, reaction time/judgement of when to start/stop stopwatch, reduces effect of error / spreads error over 20 swings [2] c 1.02(212) With 2, 3, or 4 significant figures [1] Unit s2 [1] [Total: 8] Scale to print size. a i 2 or more measurements [1] DB = 4.8 ± 0.1 (cm) [1] ii D = 6.0 (cm) [1] b i h = 7.8 (cm) AND ii V = 220 cm3 [1] c ρ = 1/1.1/1.05 [1] 2 or 3 significant figures [1] 3 g/cm [1] d Any one from: [1] part a drawn circle not exact/thickness of rim or cup/thickness of pencil line part b difficult to measure the height/D2 increases inaccuracy in D part c Mass of cup has been ignored e Diagram showing: Line of sight perpendicularly to measuring cylinder [1] To bottom of meniscus [1] [Total: 10] 3 Measure length of band Suspend load, measure new length Repeat with different width bands Use same length of band each time Table with columns for thickness, load, length/extension with units Plot a graph of extension/length against thickness for the same load Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [1] [1] 75 Cambridge IGCSE Physics Student Book answers OR load against extension/length for different thicknesses OR compare using a table e.g. compare extensions/lengths of different thicknesses for the same load Mention one of the following: Use same load/same range of loads Use at least 5 thicknesses Show how to measure extension e.g. l – l0 Use same type/material of rubber band 4 a b c d e 5 a b c d e 6 23 (°C) s, °C, °C all correct 30, 60, 90, 120, 150, 180 Lid is more effective, correct mention of comparative temperature change over 180 s, supporting conclusion Additional experiment with both insulation and lid/neither insulation nor lid compare results of previous and additional experiments / only one factor changed in comparison i 0.081 °C/s ii Statement: cooling more rapid at higher temperatures explanation: comparison of temperature difference over first 30 s and last 30 s supporting statement Normal in centre of AB and CD and FE at 30º to normal P1 P2 distance at least 5 cm P3 P4 line and KE correctly drawn (to K) α in range 28 – 32 x in range 20 – 24 mm Yes; statement matches readings Justification: within limits of experimental accuracy Any one from: large pin separation ensure pins vertical/upright/erect view bases of pins use thin pencil lines/thin pins [1] [Total: 7] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 8] a Correct voltmeter symbol in parallel with lamp X [1] b IS = 0.34 (A) [1] [1] c i VX =1.2 (V) and VY 1.9 (V) ii VS given and correct units in b and c [1] iii Statement matching results [1] Justification matching statement with use of comparative values [1] (e.g. 3.1 (VX + VY) and 3.0 (VS) are within limits of experimental accuracy) d R1 = 3.5 Ω [1] Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 76 Cambridge IGCSE Physics Student Book answers e i ii 7 a b c d 8 a b c d e f Lamps in parallel all circuit elements in correct arrangement and all circuit symbols correct resistance increases with temperature R2 > R1 and brighter lamp has higher temperature i VT = 2.5(0) (V) IT = 0.18(0) (A) ii 0.45 (W) i PX = 0.23 W, PY = 0.22 W, ii Statement: yes, as within limits of experimental accuracy reason: e.g. close enough/very close/not too far apart Statement: no/disagree reason: low current not sufficient to make lamp glow/first lamp would not glow with no current/since there is a current (other lamp cannot be broken) Lamps and voltmeter in parallel correct symbols for lamps, ammeter and voltmeter variable resistor, correct symbol and position in correct circuit Correct voltmeter symbol shown in parallel V = 2.7 (V) I = 0.48 (A) Correct calculation of R: 5.63 Ω, 3.20 Ω, 2.59 Ω Consistent to 2 or 3 significant figures i Correct calculation of r: 6.26 Ω/m, 6.40 Ω/m, 6.48 Ω/m Ω/m given at least once and not contradicted ii Statement matching results. Justification matching statement and results e.g. within limits of experimental accuracy Arrow on wire between the inside edge of each crocodile clip Any suitable precaution: Reduce current/voltage, use longer/thinner resistance wires Cambridge IGCSE Physics 4th edition © Hodder & Stoughton Limited 2021 [1] [1] [1] [1] [Total: 11] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [Total: 11] 77