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GCSE Physics for CCEA second edition

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Chapter 1 Motion In-text questions 1. a) average speed = total distance ÷ total time = 800 ÷ 40s = 20m/s

b) Car may be overtaking or braking.

2. a) distance = speed x time = 10m/s x 9s = 90m

b) time = distance/speed = 220m ÷ 10 m/s = 22s

3. a) Displacement is distance in a given direction i.e., it is a vector quantity.

b) Velocity is speed in a given direction i.e., it is a vector quantity.

4. a) Runner: average speed = 400m ÷ 44s = 9.09 m/s

b) Car: average speed = 175 miles ÷ 3 hour = 5 miles per hour

c) Shuttle: average speed = 43 750 000m ÷ 2.5 x 60 x 60s = 4860 m/s

5. a) An acceleration of 3m/s2 means that the train’s velocity increases by 3m/s in each and every subsequent second.

b) The velocity of the bus decreases by 2m/s in each and every subsequent second.

6. acceleration = change in velocity/time = (30 – 3) m/s ÷ 8s = 27 ÷ 8 = 3.375 m/s2

7. u = 25 m/s, v = 0 m/s (comes to a halt), t = 5 s, a = ?

t

u - v a

5

25 - 0

5

25- = -5m/s2 or deceleration = 5m/s2

8. a) u = 0 m/s, v = ?, t = 5 s, a = 4 m/s2

t

u - v a

5

0 - v 4 v = 20m/s

b) u = 28 m/s, v = ?, t = 8 s, a = 4 m/s2 v = u + at

= 28 + (4 x 8) = 28 + 32 = 60 m/s


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9.

10. a) & b)

c) i) Paul

ii) Paul takes 3.5s, Jim takes 5.5s.

iii) 3m.

iv) No, gradient changes after 6s.

v) Average speed = 14m ÷ 7s = 2m/s.

11. Speed increases then becomes steady. Speed is not zero at time t = 0s.


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12. a) Acceleration = gradient = 25 ÷ 5 = 5 m/s2

b) total displacement = area of triangle + area of rectangle = (125 ÷ 2 ) + (25 x 10) = 312.5m

13. a) i) OA = Uniform acceleration

ii) AB = Constant velocity

iii) BC = Uniform deceleration

b) i) Acceleration = gradient = 5m/s ÷ 2s = 2.5m/s2

ii) total distance = area of triangle + area of rectangle + area of triangle = 5m + 5m + 10m = 20m

iii) average speed = total distance travelled ÷ time taken = 20m ÷ 7s = 2.86m/s

14. u = 20m/s, v = ?, t = 10s, a = 3 m/s2 v = u + at = 20 + (3 x 10) = 20 + 30 = 50 m/s

15. u = 0m/s, a = 0.3m/s2, t = 10s, s = ? v = u + at = 0 + 10 x 0.3 = 3m/s s = ut + ½ at2 = 0 + ½ x 0.3 x 102 = 0.3 x 50 = 15m

16. u = 20m/s, v = 0, a = -10m/s2, t = ?, s = ? v = u + at 0 = 20 – (10 x t) 10t = 20 t = 25 s s = ut + ½ at2 = (20 x 2) – (½ x 10 x 22) = 40 - 20 = 20m

17. u = 0m/s, v = ?, a = 10 m/s2, t = 3s, s = ? a) v = u + at v = 0 + 10 x 3 v = 30m/s


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b) average speed = (30 + 0) ÷ 2 = 15m/s

c) s = ut + ½ at2 = 0 + ½ x 10 x 32 = 45 m

18. s = 500m, u = 0m/s, v = ?, a = 10 m/s2, t = ? a) v2 = u2 + 2as = 2 x 10 x 500 = 10 000 v = 100m/s

b) v = u + at 100 = 0 + 10t 10t = 100 t = 10s

19. u = 50m/s, v= 0, a = -10m/s2, t = ?, s = ? a) v = u + at 0 = 50 – (10 x t) 10t = 50 t = 5s

b) s = ut + ½ at2 = (50 x 5)-(½x10x25) = 250 – 125 = 125m

20. a) u = 0 m/s, v = ?, a = 3 m/s2, t = 5s v = u + at = 0 + 3 x 5 = 15m/s

b) u = 15, v = 0, t = ?, a = -0.5 v = u + at 0 = 15 – (0.5 x t) 0.5t = 15 t = 30s

area of triangle = ½ x b x h = ½ x 35 x 15 = 262.5m

c)


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Exam questions (pages 11–12) 1. a) distance = 3km = 3000m, time = 5 mins = 300s, average speed = ?

timetotal

distance total speed average =

s300

3000m = 10m/s (6 marks)

b) i) acceleration = gradient = 14m/s ÷ 20s = 0.85 m/s2 (5 marks)

ii) RF = m x a = 1500kg x 0.85m/s = 1275N RF = driving force – drag drag = 3000 – 1275 = 1725N (7 marks)

iii) Distance travelled from 80s to 200s is the area of the trapezium: = ½ (a + b) x h = ½ (120 + 80) x 20 = 100 x 20 = 2000m (6 marks)

c) i) Constant velocity

ii) Deceleration

iii) Stationary (3 marks)

d) mass = 1500kg, velocity = 14m/s, momentum = ? momentum = mass x velocity = 1500kg x 14m/s = 21 000kgm/s (5 marks)

2. a) mass = 10 000kg, g = 10m/s2, weight = ? weight = mass x g = 10 000kg x 10m/s2

= 100 000N (2 marks)

b) 100 000N (1 mark)

c) area = 20m x 5m = 100m2, upward force = 100 000N, pressure =?

area

force pressure

2m 100

N 000 100 = 1000 Pa (4 marks)

d) Arrow to the right. (1 mark)

e) RF = 20000 – 15000 = 5000N, mass = 10000kg, acceleration = ? RF = m x a 5000 = 10000 x a

000 10

5000 a = 0.5 m/s2 (4 marks)

3. a) The graph does not go below time axis. (1 mark)

b) acceleration = gradient = 10m/s ÷ 50s = 0.2m/s2 (3 marks)

c) t = 100s (1 mark)

d) i) Stage BC = area of rectangle = 50 x 10 = 500m (2 marks)


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ii) Stage CD = area of triangle = ½ x b x h = ½ 125 x 10 = 625 m (3 marks)

e) timetotal

distance total speed average =

225

625 500 250 = 6.11 m/s (3 marks)

f) work done = force x distance = 5000N x 500m = 2500 000J (3 marks)

4. a) i) speed = gradient = 14m ÷ 7s = 2m/s (2 marks)

ii) (2 marks)

b) i) potential energy = mass x g x height 320 000J = 80 kg x 10 m/s2 x height height = 320 000 ÷ 800 = 400m (3 marks)

ii) kinetic energy = ½ x m x v2 = ½ x 80 x 142 = 7840J (3 marks)


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Chapter 2 Forces In-text questions 1. a) No acceleration means balanced forces, so there must be a backwards force of 40N acting on

bicycle and rider; friction and air resistance.

b) RF = 70 – 40 = 30N, m = 90kg, a = ? RF = m x a 30 = 90 x a a = 30 ÷ 90 = 0.33m/s2

2. m = 1200 kg, a = 2 m/s2, RF = (3000 – frictional force) RF = m x a 3000 – frictional force = 1200 x 2 3000 – frictional force = 2400 3000 – 2400 = frictional force 600 = frictional force or frictional force = 600N

3. a) Forwards.

b) Consider horizontal forces only. RF = (2000 – 400) = 1600N, m = 800kg, a = ? RF = m x a 1600 = 800 x a a = 1600 ÷ 800 = 2m/s2

4. a) weight = mass x acceleration due to gravity = 2000 kg x 10 m/s2 = 20000N

b) Consider vertical forces. RF = (25000 – 20000) = 5000N, m = 2000kg, a = ? RF = m x a 5000 = 2000 x a a = 5000 ÷ 2000 = 2.5m/s2

5. 300N forward thrust. Frictional force due to water must also be 300N backwards, if it travels at constant speed. RF = (? - 300), m = 500kg, a = 2 m/s2

RF = m x a (? -300) = 500 x 2 (? -300) = 1000 ? = 1000 + 300 = 1300N

6. RF = (? - 400) N, m = 1200kg, a = 3m/s2 RF = m x a (? -400) = 1200 x 3 (? -400) = 3600 ? = 3600 + 400 = 4000N

7. a) u = 24m/s, v = 0m/s (brings to rest), t = 8s, a = ?


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t

u - v a

8

24 - 0

8

24 = -3m/s2 or deceleration = 3m/s

b) RF = ?, a = 3 m/s, m = 1200 kg RF = m x a = 1200 x 3 = 3600N NB: No need to substitute a = -3 m/s into formula, as we are only asked for the size of the

unbalanced force!

8. a) Forces acting on girl must be balanced i.e., the backwards frictional force must also be 120N if the girl is travelling at a steady speed.

b) RF = 300 – 120 = 180N, m = 60kg, a = ? RF = m x a 180 = 60 x a a = 180 ÷ 60 = 3m/s2

9. a) u = 50m/s, v = 0m/s, t = 5s, a = ?

t

u - v a

5

50 - 0 = -10m/s2

b) a = 10 m/s2, m = ?, RF = 18000N RF = m x a 18000 = m x 10 m = 18000 ÷ 10 = 1800kg

10. a) u = 20m/s, v = 0m/s, t = 0.1s, a = ?

t

u - v a

0.1

20 - 0 = -200m/s2 or deceleration = 200m/s2

b) F = ?, m = 1000kg, a = 200m/s2 F = m x a = 1000 x 200 = 200000N

11. a) 30km/hr = 30000m in 3600 seconds = 8.33m/s

b) u = 8.33m/s, v = 0m/s, t = 0.03s, a = ?

t

u - v a

0.03

8.33 - 0 = -277.7m/s2

c) a = 277.7m/s2, m = 70kg, RF = ? RF = m x a = 70 x 277.7 = 19 444 N

12. Weight is measured in Newtons, so Julie should have said ‘My weight is 350 N’ or ‘My mass is 35 kg’.

13. As soon as it enters the oil, the ball will decelerate until its weight is equal to the viscous drag. Then the ball will fall with constant (terminal) velocity.


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14. There is no atmosphere on the Moon. The hammer and feather will experience no drag forces, so they will both accelerate at 1.6 m/s2 and hit the surface of the moon simultaneously.

15. When the parachute opens, the vertical drag force increases, decreasing the velocity of the parachutist.

16. AB – Parachutist is decelerating, the faster he falls the greater the air resistance he experiences until at BC – air resistance equals weight, balanced forces, so that parachutist falls at constant (terminal) velocity of roughly 60 m/s. CD – at C the parachute opens whereupon the air resistance increases, deceleration occurs, until at DE – the parachutist reaches a smaller and safer terminal velocity.

17. volume = 3 m3, mass = 57.9 g, density = ?

volume

mass density

3

57.9 = 19.3 g/cm3

Consulting table: substance = gold

18. a) volume = 20 cm3, mass = ?, density = 2.7 g/cm3 mass = density x volume = 2.7 x 20 = 54 g

b) volume = ?, mass = 54 g, density = 2.7 g/cm3

density

mass volume

2.7

54 = 20 cm3

19. volume = 15 cm3, mass = 120 g, density = ?

density

mass volume

15

120 = 8 cm3

20. volume = 10 m x 5 m x 3 m = 150 m3, mass = ?, density = 1.26 kg/m3 mass = density x volume = 1.26 x 150 = 189 g

21. volume = (35 -15) = 20 cm3, mass = 60 g, density = ?

volume

mass density

20

60 = 3 g/cm3

22. volume = 0.8 m3, mass = ?, density = 800 kg/m3 mass = density x volume = 800 x 0.08 = 64 g

23. volume of air = 100 cm3, mass of air = (351.2 – 350) = 1.2 g, density = ?

volume

mass density

100

1.2 = 0.012 g/cm3


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24. a) volume of rivets = 70 – 50 = 20 cm3

b) volume of 1 rivet = 20/100 = 0.2 cm3

c) volume of air = 20 cm3, mass of air = 180 g, density = ?

volume

mass density

20

180 = 9 g/cm3

25. a) clockwise moment = force x perpendicular distance to pivot = 150 x 0.9 = 135 Nm

b) clockwise moment = anticlockwise moment 135 = F x 1.35

1.35

135 F = 100 N

26. clockwise moment = anticlockwise moment 3 x X = 4 x 36

X = 3

36 x 4 = 48 cm

27. a) moment exerted by boy = force x perpendicular distance to pivot = 100 N x 4 m = 400 Nm

b) If in equilibrium: moment exerted by man = moment exerted by boy = 400 Nm

c) moment exerted by man = force x distance to pivot 400 = force x (4 – 3) = force x 1 force = 400 N

28. a) clockwise moment of 5 N force = 5N x 40 cm = 200 Ncm

b) anticlockwise moment = clockwise moment weight x 25 cm = 200 Ncm

25

200 weight

= 8N

29. clockwise moment = anticlockwise moment F x 225 = 600 x 75

225

75 x 600 F

= 200 N


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30. clockwise moment = anticlockwise moment F x 0.6 = 100 x 0.3

= 0.6

0.3 x 100

= 50 N

31. a) Where medians of triangle meet, middle of rectangle.

b) In the centre of the cut-out circle, ie., not in the plastic sheet itself.

32. a) It is a point through which the whole weight of the object appears to act.

b) i) The centre of gravity of the pencil and the penknife is on a line below the point of the pencil.

ii) The penknife and pencil would topple.

33. a)

b) Low centre of gravity and wide wheel base.

34. a)

b) The whiskey glass.

c) Lower centre of gravity and wider base.

35. a)

i) unstable ii) neutral


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b) i)

ii) B.

iii) Lower centre of gravity and wider base.

36. a)

b)

Exam questions (page 35) 1. a) Density is the ratio of the mass to volume. (2 marks)

b) See Figure 12a page 23. (4 marks)

c) Volume = 2.4 cm3, mass = 46g, density = ?

volume

mass density

4.2

46 = 19.2g/cm3 (3 marks)

d) Gold. (1 mark)

2. a) i) volume = l x b x d = 1.8 x 1.2 x 0.1 = 0.216 m3 (4 marks)

X Y Z

centre of gravity outside lamina


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ii) volume

mass density

216.0

520 = 2407.4kg/m3

This density is greater than 2350 kg/m, therefore this slab will be suitable. (4 marks) NB: mass of slab = 520kg, not 520g as in question.

b) i) The centre of gravity is a point through which the whole weight of the body appears to act. (1 mark)

ii) moment = force x perpendicular distance to pivot = 1500N x 0.4m = 600Nm (5 marks)

iii) anticlockwise moment = clockwise moment effort x (0.4 + 0.7) = 600 effort x 1.1 = 600

effort 1.1

600 = 545N (4 marks)

iv) upward forces = downward forces reaction + 545N = 1500N reaction = 1500 – 545 = 955N (2 marks)


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Chapter 3 Energy In-text questions 1. Any three from: coal, oil, (natural) gas, lignite, peat (turf).

2. Sound, electricity and heat are energy forms. The others are not forms of energy.

3. Hydroelectricity, wind and tides are renewable. Gas, oil and coal are non-renewable.

4. The solar cells change light energy into electrical energy. The battery stores chemical energy. As the propellers turn they change electrical energy into useful kinetic energy. As the model aircraft gains height, it gains gravitational potential energy. The model aircraft crashes into the ground. As it does so, it produces wasted heat and sound energy.

5. Renewable resources are in limitless supply because they are replaced by nature in less than a human lifetime.

6. Both use water to produce steam which drives a turbine. A nuclear power station uses fission of uranium to produce heat and the waste products are

dangerously radioactive. A fossil fuel power station burns fossil fuels to produce heat. One of the waste products,

carbon dioxide, is a major contributor to global warming.

7. The waste will be dangerously radioactive for a very long time and there can be no guarantee that it will not leak. There is also the possibility of seismic activity (earthquakes) bringing it to the surface.

8. Carbon dioxide.

9. Prevailing winds blow sulphur dioxide from Britain to Norway.

10. See table and text on pages 39–40.

11. See table and text on pages 39–40.

12. Open-cast mining for lignite would ruin the beauty of a naturally very attractive area.

13. See text on pages 37 and 40.

14. Conserve fossil fuels, wind is renewable, less atmospheric pollution leading to global warming.

15. The materials used to generate electricity (fossil fuels and uranium) themselves produce very polluting waste products.


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16. Device/situation Input energy form Useful output

energy form Microphone sound energy → electrical energy Electric smoothing iron electrical energy → heat energy Loudspeaker electrical energy → sound energy Coal burning in an open fire chemical energy → heat energy A weight falling towards the ground

gravitational potential energy

→ kinetic energy

A candle flame chemical energy → heat energy and light energy

Battery-powered electric drill chemical energy → electrical energy →

kinetic energy

17. work (in J) = Force (in N) x distance (in m) = (100 x 10)N x 5.5m = 5500J

18. work (in J) = Force (in N) x distance (in m) = 60N x 20m = 1200J

19. a) inputenergy total

outputenergy useful efficiency

1000

750 = 0.75

b) Heat is lost to the metal of the boiler, to the surroundings and in the hot smoke through the chimney. Sound is lost to the environment.

20. inputenergy total


Since, by the Law of Conservation of Energy, energy is neither created nor destroyed, the useful energy output can never be greater than the total energy input.

21. inputenergy total


inputenergy total

000 140 0.28

0.28

000 140 input energy (chemical) total = 500 000kJ

22. energy (in J) = power (in W) x time (in s) = 3600W x (5 x 60)s = 1 080 000 J

23. a) weight = mg = 1500kg x 10N/kg = 15 000N

b) work = force x distance = 15 000N x 12m = 180 000J

c) (in W)power

J)(in work s)(in time

3000

180000 = 60s

d) s)(in time

m)(in distance m/s)(in speed

60

12 = 0.2m/s


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24. a) work (in J) = force (in N) x distance (in m) = 1000N x 0.4m = 400J

b) inputenergy total


1200

400 = 0.33

25. KE = ½ mv2 = ½ x 120 x 30002 = 540 000 000 J

26. GPE of rubber = mgh = 0.050 kg x 10 N/kg x 280 m = 140 J KE of shell = ½ mv2 = 0.5 x 0.010 x 1502 = 112.5 J Comment: if energy losses are ignored, KE of shell is less than that of rubber as it hits the

ground.

27. KE of tanker = ½ mv2 200 000 000 = 0.5 x 100 000 000 x v2 v2 = 4 v = 2 m/s

28. Height above ground in m

Gravitational potential energy in J

Kinetic energy in J

Total energy in J

Speed in m/s

5.0 100 0 100 0 4.0 80 20 100 4.47 3.2 64 36 100 6.0 1.8 36 64 100 8.0 0.0 0 100 100 10.0

29. KE of car = ½ mv2 160 000 = 0.5 x 800 x v2 v2 = 400 v = 20 m/s = (20 x 60) metres/min = 1.2 km/min = 1.2 x 60 km/hr = 72 km/hr

30. GPE = mass x gravitational field strength x height 176 = 2 x g x 10 g = 176 ÷ 20 = 8.8 N/kg By inspection of the table, the planet was Venus.

31. a) KE (in J) as ball rises from ground = GPE of ball at its maximum height 10 J = 0.2 kg x 10 N/kg x height in metres Height = 10 ÷ 2 = 5 metres

b) In practice, energy is lost as heat and sound against air resistance as the ball rises. So, not all of the KE of the ball is converted into GPE.

32. a) inputenergy total


25

)2025( = 0.2

b) The principle of Conservation of Energy.

33. a) useful energy output = efficiency x total energy input = 0.3 x 2000 = 600 J

b) total wasted energy = 2000 – 600 = 1400 J wasted heat energy = 0.9 x total wasted energy = 0.9 x 1400 = 1260 J percentage of the input energy lost as heat = (1260 ÷ 2000) x 100% = 63%


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Exam questions (pages 55–57) 1. a) Kinetic energy and gravitational potential energy. (2 marks)

b) i) Input kinetic energy of wind → Useful output electrical energy (2 marks)

ii) Wind energy is in almost limitless supply because it is replaced so quickly by nature. (1 mark)

iii) Other renewables are wave energy and solar energy. (2 marks)

c) Gravitational potential energy → Kinetic energy → Electrical output energy stored in upper lake in moving water from the power station (3 marks)

2. Advantage: tidal energy has low running costs (but huge set-up costs). Disadvantage: while the tides (unlike wind and waves) are predictable, they vary from day to day

and month to month. This makes them unsuitable for producing a constant daily amount of electrical energy. (2 marks)

3. a) i) Oil is a non-renewable fuel because it is not being replaced by nature. (2 marks)

ii) Wind energy is renewable because it is in almost limitless supply because it is replaced so quickly by nature. (2 marks)

iii) Non-renewable resources, such as coal, oil and gas are generally more reliable than renewable energy sources like wind, waves and solar energy. (1 mark)

b) i) GPE lost per second = mgh = 100 000 000 x 10 x 50 = 50 000 000 000 J/s or 5 x 1010 J/s (3 marks)

ii) Maximum power output = (0.8 / 100) x 50 000 000 000 = 400 000 000 W (3 marks)

iii) The water flowing over the falls comes from rainfall. Rainfall occurs as part of the water cycle in which water in the oceans is evaporated by the sun, rises and condenses to form clouds. The clouds are driven by the wind over land and the water falls as precipitation to form rivers and lakes. Some of this water passes over the falls. (2 marks)

c) i) useful power output = efficiency x electrical power input = 0.6 x 500 = 300 W (3 marks)

ii) 300 J (1 mark)

iii) power = force in rope x distance moved per second 300 = 1200 x distance moved per second constant speed of vehicle = distance moved per second = 300 ÷ 1200 = 0.25 m/s

(3 marks)

4. a) work = force x distance = 8000N x 1.8m = 14 400J (3 marks)

b) power

work time

5200

26000 = 5s (3 marks)

c) efficiency

outputpower useful input power

0.26

5200 = 20 000 W = 20 kW (3 marks)


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5. work = force x distance = 550N x 3m = 1650J (3 marks)

6. a) work = force x distance = 24 000N x 40m = 960 000J = 960kJ (3 marks)

b) time

work power

20

960 = 48kW (3 marks)

c) inputenergy total


1200

960 = 0.8 (3 marks)

d)

Energy Increases/decreases/ unchanged

Potential energy of the top tramcar

DECREASES

Kinetic energy of the top tramcar

UNCHANGED

Kinetic energy of the bottom tramcar

UNCHANGED

Potential energy of the bottom tramcar

INCREASES

Heat energy INCREASES (5 marks)

7. Quantity Increases Decreases Constant Speed of ball Potential energy of ball Total energy of ball Kinetic energy of ball

(4 marks)

8. a) Grade A dishwasher costs less to run than D. (1 mark)

b) Electrical energy is converted into USEFUL HEAT energy and USEFUL KINETIC energy and WASTED SOUND energy. (4 marks)

c) A shopper might not always buy the most efficient appliance which a shop has for sale because the most efficient appliance might be the most expensive. (2 marks)


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Chapter 4 Radioactivity In-text questions 1. 1. 6 protons, 8 neutrons, 0 electrons (electrons are not in the nucleus).

2. Na2311

3. The nuclei of isotopes have the same atomic number (same number of protons), but they have a different mass number (a different number of protons + neutrons).

4. a) β decay (no change in mass number, but change in name of nucleus)

b) α decay (reduction in mass number of 4)

c) α decay (reduction in mass number of 4)

5. a) 210

b) 212

c) 210

6. a) Decreases by 4

b) Decreases by 2

c) Unchanged

d) Increases by 1

Exam questions (pages 72–3) 1. a)

Particle Mass Charge Number Location Electron 1/1840 –1 6 Orbiting nucleus Neutron 1 0 6 In the nucleus Proton 1 +1 6 In the nucleus

(7 marks)

b) i) A nucleus is radioactive if it decays by emitting α particles, β particles or γ radiation. (2 marks)

ii) Radon is radioactive and it emits radiation which can cause serious damage to lung tissues. The decay products might also lodge permanently in the lungs. These decay products are also radioactive and will go on causing damage to tissues. This damage may lead to cancers. (2 marks)

ii) Nuclei with the same number of protons but a different number of neutrons are called isotopes. (2 marks)

c) i) Points plotted at: (0, 800), (20, 400), (40, 200), (60, 100), (80, 50) (5 marks)

ii) Curve passing through these points. (1 mark)


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d) i) A beam of beta particles will all be absorbed by a sheet of aluminium several mm thick placed in their path. (1 mark)

ii) (3 marks)

iii) First measure the background count rate. Then measure the count rate for increasing thicknesses of aluminium. (3 marks)

iv) Increase the thickness of the aluminium between the source and the Geiger tube and measure the corresponding count rate until the count rate is equal to the background count. Then measure the thickness of the aluminium – this thickness is the range of beta particles in aluminium. (2 marks)

v) (2 marks)

2. a) β N C 01-

147

146 (4 marks)

b) Firstly, alpha radiation is so ionisng that it is likely to cause serious harm to tissues within the body. Secondly, alpha radiation has such a small range that it cannot leave the body and be detected by the camera. (2 marks)

c) i) Background activity comes from rocks containing uranium and other radioactive elements, the waste products from nuclear power stations and Nuclear Medicine Departments in hospitals and cosmic rays from space. To obtain the corrected count rate the experimentalist subtracts the background count rate from the measured count rate. Corrected count rate = Observed count rate less Background count rate. (2 marks)


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ii) (5 marks)

iii) Half life = time taken for activity to fall from original value to half of original value. From graph, the half life is 29 seconds. (2 marks)

3. a) i) In alpha decay the mass number decreases by 4 and the atomic number decreases by 2. (2 marks)

ii) In beta alpha decay the mass number remains the same and the atomic number increases by 1. (2 marks)

b) (3 marks)

c) Student’s graph. (5 marks)

d) Gamma emission gives no change to atomic number and mass number. (1 mark)

e) U23892 and U234

92 are isotopes; Th23490 and Th230

90 are isotopes. (2 marks)

4. a) i) The half-life of an isotope is defined as the time taken for its radioactivity to fall by half. (1 mark)

ii) Nuclei with the same number of protons but a different number of neutrons isotopes. (1 mark)

b) Student’s graph. (4 marks)

Element (symbol)

Atomic number

Mass number

Decays by emitting

Leavingelement

U 92 238 Th Th 90 234 Pa Pa 91 234 U U 92 234 Th Th 90 230 Ra Ra 88 226 Rn Rn 86 222 Po Po 84 218 Pb Pb 82 214 Bi Bi 83 214 Po


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c) Half-life = 21 days (2 marks)

d) Three weeks is 1 half-life. So one half-life BEFORE arrival, the mass of the uranium present in the sample would be double what it was when it arrived; so the mass of uranium present would have been 2 x 100 = 200 grams. (2 marks)

5. One hour is 60/12 = 5 half lives, so proportion left after 1 hour is 1/25 = 1/32 of material. (3 marks)

6. Radioactivity is a spontaneous, random process governed by the laws of probability. Just as one is unlikely to obtain exactly 3 heads from 6 tosses of a fair coin, so the count from a long half-life source is unlikely to be exactly the same in each of four 10 second intervals. (2 marks)


23

Chapter 5 Waves, sound and light In-text questions 1. a) Energy flows from A towards B.

b) In a longitudinal wave, X vibrates parallel to the axis of the slinky.

c) In a transverse wave, X vibrates perpendicular to the axis of the slinky.

d) Another example of a transverse wave is a water wave.

2. a) = 60 ÷ 4 = 15 cm

b) number of waves pass the marker in 1 second = 10 ÷ 4 = 2.5

c) 2.5 Hz

d) v = f = 2.5 Hz x 15 cm = 37.5 cm/s

3. a) A lightning flash is generally seen before thunder is heard because light travels faster in air than sound.

b) distance = speed x time = 330 m/s x 4 s = 1320 m

4. total distance travelled by sound = speed x total time = 330 m/s x 6 s = 1980 m

distance to cliff = ½ x total distance = ½ x 1980 = 990 m

5. a)

Frequency (Hz) 680 440 170 136 85 68 Wavelength (m) 0.5 1.0 2.0 2.5 4.0 5.0

1/Wavelength (m-

1) 2.0 1.0 0.50 0.40 0.25

0.20

b) Graph: line of best fit is a straight line through the origin.

c) Incorrect point is (1.0, 440)

d) directly proportional

e) The value of f for all table values (except the second) is 340 m/s.

6. a) wavelength = 14 cm ÷ 6 = 2.33 cm

b) distance between ripple A and ripple F = 2.33 x 5 = 11.65 cm

time

distance wavesof speed

10

11.65 = 1.165 cm/s

wavelength

speed frequency

2.33

1.165 = 0.5 Hz


24

7.

On reflection from deep water into shallow water

On reflection from a barrier

Property Increases/Decreases/Does not change Increases/Decreases/Does not change

Wavelength Decreases Does not change

Frequency Does not change Does not change

Speed Decreases Does not change

8. 0°

9. 50°

10. 65°

11.

So that motorists will see the word “AMBULANCE” when they see the ambulance in their rear-view mirrors.

12. Angle of reflection at M1 is 30o, so angle between reflected ray and M1 is 60o. Angle between incident ray to M2 and mirror M2 is 30o, so angle of incidence at M2 is 60o.

Angle of reflection at M2 is 60o. The ray incident on M1 is parallel to the ray reflected from M2.

13. 0°. The ray is reflected back along the path of the incident ray.

14. i) Light is fastest in air.

ii) Light is slowest in B. (Light bends towards normal when it leaves air and enters B, so light is slower in B than in air. Light bends away from the normal when it leaves B and enters A, so light is faster in A than in B. Light bends even further away from the normal when it leaves A and enters air, so light is faster in air than in A.)

15. See Figure 29 on page 92.

16. a) Ray diagram similar to first diagram in table on page 96.

b) Image is 6 cm from lens, but on same side as the object.

c) Image is enlarged, erect and virtual.

17. Focal length is 12 cm.

18. a) Minimise exposure to sunlight by remaining indoors when sun is hottest. Use high sun protection cream (high SPF cream) on skin when in sunshine.

b) To detect forgeries of banknotes and to kill bacteria in water chillers.

19. a) Infra red light is used in toasters in kitchens and in PIR (passive infra red) systems to detect intruders and burglars.

b) Infra red light can burn the skin.


25

20. Electromagnetic waves can travel in a vacuum; in a vacuum all electromagnetic waves travel at the same speed.

21. a) radio waves

b) X-rays

22. speed = f = 200 000 Hz x 1500 m = 300 000 000 m/s = 300 000 km/s

speed

distance time

300000

900 = 0.003 s

Exam questions (pages 105–7) 1. a) i) The diagram shows 3 complete wavelengths cover 60 m. Each wavelength is therefore

60/3 = 20 m. (1 mark)

ii) Frequency is the number of waves passing a fixed point in 1 second. (1 mark)

iii) frequency = number of waves passing in 1 second = 10/25 = 0.4 Hz (1 mark)

iv) v = f = 0.4 x 20 = 8 m/s (1 mark)

b) i) See page 79. (1 mark)

ii) See page 79. (2 marks)

iii) The reflected waves have been refracted as they pass from deep water into shallow water. In shallower water they have a smaller speed, but an unchanged frequency. They therefore have a smaller wavelength, so their wavefronts are closer together. (2 marks)

2. a) i) Vibrating (1 mark)

ii) The skin is vibrating with a larger amplitude. (1 mark)

iii) The cone is making more vibrations per second. (1 mark)

b) i) He could become deaf (or become hard of hearing). (1 mark)

ii) They could fit double (or triple) glazed windows in their homes. (1 mark)

c) Statement Waves Can cause sunburn and skin cancer Ultraviolet Can heat food Infrared or microwaves Have the greatest wavelength Radio wave

(2 marks) 3. a) i) Dispersion (1 mark)

ii) See Figure 30 on page 92. (2 marks)

iii) Each colour in white light travels at a different speed in glass and therefore each colour bends by a different amount. Red slows down least and violet slows down most. (2 marks)

b) i) See page 99. (2 marks)


26

ii) See page 99. (1 mark)

iii) f = 2.7 cm (1 mark)

iv) Magnified and upside down (3 marks)

4. a) The first sound Tom heard came directly from Sean 200 m away. The second sound was the echo from the cliff. In this case the sound travelled 350 m to the cliff and the reflected echo traveled a further 150 m back to Tom. (2 marks)

b) speed

distance time

Tom hears the first sound after 200/340 = 0.588 seconds. Tom hears the second sound after 500/340 = 1.471 seconds. Time interval = 1.471 – 0.588 = 0.883 seconds. (5 marks)

5. a) A: Transverse; B: Longitudinal. (2 marks)

b) Energy (1 mark)

c) X vibrates vertically, at 90° to the direction in which the wave is travelling. (2 marks)

d) 0.3 m (1 mark)

e) 0.4 m (1 mark)

f) 3 Hz (1 mark)

g) v = f = 3 x 0.4 = 1.2 m/s (3 marks)

6. a) A transverse wave is one in which the particles vibrate at 90° to the direction in which the wave is travelling. (2 marks)

b) X-rays – Detects broken bones Ultra-violet light – Detects banknote forgeries Infra-red light – Night-time photography Radio waves – Long distance communications Microwaves – Fast food preparation (5 marks)

7. a) See text. (1 mark)

b) angle of incidence = angle of reflection = 63° (2 marks)

c) Reflected ray from M2 is parallel to incident ray on M1 (2 marks)


27

Chapter 6 Electricity In-text questions 1. a) Static electricity is the study of stationary charges.

b) Positive and negative charges.

c) Like charges repel, unlike charges attract.

2. a) Free electrons move from the cloth (which is now deficient of negative charge and so becomes positively charged) to the polythene cloth (which now has an excess of free electrons and so becomes negatively charged).

b) Charging by friction.

3.

4. a) Useful: photocopier; paint spraying of cars.

b) Nuisance: clothes stick together when removed from tumble dryer; ‘shocks’ from walking across a nylon carpet and then touching door handle.

c) Dangerous: being struck by lightning; refuelling aircraft.

5. a) 3.0 A = 3000 mA

b) 0.2 A = 200 mA

6. a) 400 m A = 0.4 A

b) 1500 m A = 1.5 A

7. a) current = 6 A, time = 10 s, charge = ? charge = current x time = 6 A x 10 s = 60 C

b) current = 300 m A = 0.3A, time = 1minute = 60 s, charge = ? charge = current x time = 0.3 A x 60 s = 18 C

8. a) current = ?, time = 5 s, charge = 100 C charge = current x time 100 C = current x 5 s current = 100 ÷ 5 = 20 A

Material of rod Bead movement Charge on rod Positive Negative Uncharged Perspex Repelled Cellulose acetate Repelled Polythene Attracted Steel None


28

b) charge = 500 mC = 0.5 C, time = 50 s, current = ? charge = current x time 0.5 C = current x 50 s current = 0.5 ÷ 50 = 0.01 A

c) charge = 60 μC = 0.000006 C, time = 200 s, current = ? charge = current x time 0.000006 C = current x 200 s current = 0.000006 ÷ 200 = 0.3 x 10-6 A = 0.3 μA

9. Two volts means 2 joules of energy are supplied to each coulomb of charge that the cell supplies.

10. energy = 9 J, charge = 6 C, voltage = ?

charge

energy voltage

6

9 = 1.5V

11. energy = ?, charge = 3 C, voltage = 18 V

charge

energy voltage

3

energy 18

energy = 18 x 3 = 54 J

12. energy = 150 J, charge = 25 C, voltage = ?

charge

energy voltage

25

150 = 6 V

13. a) Voltmeter, since it is in parallel.

b) 12 V = 5V + ? ? = 12 – 5 = 7 V

c) 12 V means 12 joules of energy are supplied to each coulomb of charge.

d) 7 volts are dropped across motor, therefore 7 joules of energy are deposited by each coulomb of charge that passes through the motor.

14. a)

b)

15. a) Light Dependent Resistor

b) c) The resistance of an LDR decreases when light is shone on it.


29

16. current = ?, resistance = 10 Ω, voltage = 20V V = I x R 20V = I x 10 Ω I = 20V ÷ 10 Ω = 2 A

17. current = 3 A, resistance = ?, voltage = 15 V V = I x R 15V = 3 A x ? R = 15 V ÷ 3 A = 5 Ω

18. current = 2 A, resistance = 25 Ω, voltage = ? V = I x R = 2A x 25 Ω = 50 V

19. a) current = 2.5 A, resistance = ?, voltage = 15 V V = I x R 15V = 2.5 A x R R = 15 V ÷ 2.5 A = 6 Ω

b) current = 2 A, resistance = 6 Ω, voltage = ? V = I x R V = 2A x 6 Ω = 12 V

20. a) current = ?, resistance = 12 Ω, voltage = 6V V = I x R 6V = I x 12 Ω I = 6V ÷ 12 Ω = 0.5 A

b) current = 15 A, resistance = 12 Ω, voltage = ? V = I x R V = 1.5A x 12 Ω = 18 V

21. current = 100mA = 0.1 A, resistance = ?, voltage = 6 V V = I x R 6V = 0.1 A x R R = 6 V ÷ 0.1 A = 60 Ω


30

22. a)

b) No. Graph is non-linear which is necessary for Ohm’s Law.

23. a)


31

b) No. Graph is non-linear.

c) Voltage in V 0 0.1 0.2 0.3 0.4 0.5 0.6 0.9 Current in mA 0 0.1 0.1 0.1 0.1 1.5 73 250 Resistance in Ω 0 1000 2000 3000 4000 333.3 8.219 2.8

24. a) Simple series circuit R = R1 + R2 = 6 Ω+ 5 Ω = 10 Ω

b) Simple parallel circuit

21T R

1

R

1

R

1

6

1

6

1

6

2

3

1

RT = 3 Ω

c) Simple series circuit RT = R1 + R2 + R3

= 2 Ω + 4 Ω + 6 Ω = 12 Ω

d) Simple parallel circuit

21T R

1

R

1

R

1

6

1

12

1

12

3

4

1

RT = 4 Ω

e) 3T R sum

product R

= 2 + 1 = 3 Ω

f) 321

T R R

1

R

1 R

5 4

4

4

4 R T

= 2 + 5 = 7 Ω


32

g)

3

sum

product R T

3 2

2

2

2 R T

= 7 Ω

h) We must do top branch of circuit first: 2 Ω + 2Ω = 4 Ω

sum

product R T

4

4

4

4 R T

= 2 Ω

i) sum

product R T

6

6

3

3 R T

= 2 Ω

j) In effect, there are three series resistors when we reduce the two parallel networks

6

6

4

4 R T

+ 1.6 Ω +

18

18

9

9

= 2.4 + 1.6 + 6.0 = 10 Ω

25. The 3 resistors in series = 1 + 2 + 3 = 6Ω The 3 resistors in parallel:

321T R

1

R

1

R

1

R

1

3

1

2

1

1

1

6

2 3 6

6

11

Therefore RT = 6/11 Ω 1 Ω and 2 Ω in parallel with 3 Ω in series = 3 ⅔ Ω 1 Ω and 3 Ω in parallel with 2 Ω in series = 2 ¾ Ω 2 Ω and 3 Ω in parallel with 1 Ω in series = 2 1/5 Ω 1 Ω and 2 Ω in series and in parallel with 3 Ω = 1.5 Ω 1 Ω and 3 Ω in series and in parallel with 2 Ω = 1.33 Ω etc., etc

26. a) i) total resistance = 2 + 4 + 6 = 12 Ω total voltage = 3V

A25.012

3

R

V drawn current total


33

ii) 4 16

64

sum

product resistance total

total voltage = 8 V

A24

8

R

V drawn current total

b) The 8Ω resistors are in parallel, so the voltage dropped will be the same voltage ie., 8V.

27. a) i)

6 12 12

12 12 resistance total

total voltage = 9 V

A .51 6

9 current total

A1 = 1.5A A2 = 0.75A A3 = 0.75A

ii) 4 4

4 4

66

6 6 resistance total

= 3 + 2 = 5 Ω total voltage = 10 V

A25

10

R

V current total

A3 = 2A A1 = 1A A2 = 1A

b) Since each 6 Ω resistor is carrying 1 A, there must be the same (V = IR) voltage of 6 Ω x 1A = 6V across each 6 Ω resistor.

Similarly, since each 4 Ω resistor is carrying 1 A, there must be the same voltage of 4 Ω x 1A = 4V across each of the 4 Ω resistors.

28. energy = ?, power = 1000 W, time = 1 hr = 3600s energy = power x time = 1000 x 3600 = 3 600 000 J

29. a) energy = 15 000 J, power = ?, time = 10 s

time

energy power

10

15000 = 1500 W

b) 1500 W = 1.5 kW

30. current = 0.25 A, voltage = 240 V, time = 60 s, energy = ? E = V x I x t = 240V x 0.25A x 60s = 3600 J


34

31. a) energy = ?, power = 960 W, time = 60 s energy = power x time = 960 W x 60 s = 57 600 J

b) current = ?, voltage = 12 V, power = 960 W

P = V x I 960 W = 12 V x I

12

960 I = 80 A

32. current = 4 A, voltage = 240 V, power = ? P = V x I = 240 V x 4A = 960 W

33. current = ?, voltage = 12 V, power = 48W P = V x I 48 W = 12 V x I

12

48 I = 4 A

34.

35. resistance = 48, voltage = 240 V, power = ?

R

V P

2

48

402

2

= 1200 W

36. a) i) Brown.

ii) Blue.

iii) Green and yellow.

b) i) Current will flow from the live wire, through the metal case to the person touching the case, and then to earth. The person touching the case would be electrocuted.

Name of appliance Power rating Current

V

P I Resistance

I

V R

Bulb of study lamp 60 W A 0.25

240

60 960

0.25

240

Television 80 W A 0.33

240

80 720

0.333

240

Toaster 120 W A 0.5

240

120 480

0.5

240

Convector heater 2 kW A 8.33

240

2000 28.8

8.33

240

Shower 3 kW A 12.5

240

3000 192

12.5

240


35

ii) Since the earth wire is now properly connected, the current will flow through the earth wire to the ground, melting the fuse, cutting off the current. The person touching the case would not be electrocuted.

37. a) Direct current – always flows in the same direction, from a fixed positive terminal to the fixed negative terminal of a supply. See Figure 39 b, page 133.

b) Alternating current – the voltage (and hence the current) change size and direction in a regular and repetitive way. See Figure 40, page 133.

38. Circuit breakers react much more quickly than fuses so protecting the user.

39. a) current = ?, voltage = 230 V, power = 60W P = V x I 60 W = 230 V x I

230

60 I = 0.26 A

b) A fuse rated greater than 3 A would pass too high a current which would be dangerous.

40. a) current = ?, voltage = 240 V, power = 3000W P = V x I 3000 W = 240 V x I

240

3000 I = 12.5 A i.e. the fuse would ‘blow’.

b) current = ?, voltage = 240 V, power = 60 W P = V x I 60 W = 240 V x I

240

60 I = 0.25 A

If a 13 A fuse was used, the fuse would not ‘blow’ until the current exceeded 13 A. So a faulty study lamp would electrocute user.

c) current = ?, voltage = 240 V, power = 800 W P = V x I 800 W = 240 V x I

240

800 I = 3.33 A

So a 5A fuse would be preferable.

41. Consider 1 bulb: current = ?, voltage = 240 V, power = 60 W P = V x I 60 W = 240 V x I

240

60 I = 0.25 A

So you should use less than 20 bulbs so that the fuse would not be overloaded.


36

42. a)

b) Total current drawn (32.06 A) is much greater than the 13 A fuse in extension lead, so the fuse would ‘blow’ and none of the appliances would work.

43. energy (in kW h) = power (in kW) x time (in hours) a) energy = 0.1 kW x 12 hr = 1.2 kWh b) energy = 0.25kW x 4 hr = 1 kWh c) energy = 2.4 kW x 1/12 hr = 0.2 kWh

44. Lamp: 1.2 kWh @ 12p per unit = 14.4p Television: 1 kWh @ 12p per unit = 12p Kettle: 0.2 kWh @ 12p per unit = 24p

45. a) current = 15 A, voltage = 230 V, power = ? P = V x I = 230 V x 15 A = 3450 W

b) number of units used = power (in kW) x time (in hr) = 3.45 kW x 0.166 hr = 0.575 units @ 12p per unit = 6.9p

Exam questions (pages 139–42) 1. a) Lightning strike would mean a large current would flow through building to earth. Fire or

destruction of buildings could happen. (2 marks)

b) i) Friction between balloon and cloth cause electrons to move from the balloon to the cloth. The balloon would have a deficiency of electrons and hence would become positively charged. (3 marks)

ii) The charges on the balloons are both positive hence the balloon repel each other. Like charges repel. (4 marks)

Device Power (W) Current drawn (A) Kettle 2400

10.43 230

2400

Washing Machine 3000 12.5

230

2300

Television 800 3.48

230

800

Toaster 1300 5.65

230

1300

So total current drawn = 32.06 A


37

c) i) (3 marks)

ii) The currents decreased. (1 mark)

d) i)

3 6

3 6 8 R Total

= 8 + 2 = 10 Ω (4 marks)

ii) R

V I

10

5 = 0.5 A (3 marks)

iii) V = IR = 0.5 x 8 = 4 V (3 marks)

iv) power = V x I = 4 V x 0.5 A = 2 W (4 marks)

v) Since there are 4 V across the 8 Ω resistor then there must be (5 – 4) V = 1 V across the 6Ω resistor. (2 marks)

vi) One volt, since resistors in parallel have the same voltage across them. (1 mark)

2. a) i) (2 marks)

ii) sum

product R r

0.3 0.6

0.3 0.6

0.9

0.18 = 0.2 Ω (3 marks)

iii) Brightness remains the same because the voltage across bulb has not changed. (2 marks)

X

X


38

b) i) (3 marks)

ii)

(4 marks)

iii) 70 mA (1 mark)

iv) I

V R

50

0.4 = 8 Ω (4 marks)

v) Yes, gradient is constant. (2 marks)

vi) To keep temperature of wire as low as possible, otherwise resistance would increase. (1 mark)

3. a) i) (4 marks)

A

A

V

X

X


39

ii) Parallel. (1 mark)

iii) Bulbs in parallel will have 1.5 V across each of them i.e. 1.5V in this case. So when one bulb is removed, the second bulb will still have 1.5 V across it, so its brightness will not change. (2 marks)

iv) voltage = 1.5 V, current = 0.3 A, resistance = ? V = IR 1.5 = 0.3 x R

0.3

1.5 R = 5 Ω (4 marks)

b) i) 3 V (2 marks)

ii) voltage = 3 V, current = ?, resistance = 5 Ω V = IR 3 = I x 5

5

3 I = 0.6 A (4 marks)

iii) Bulb will be brighter. (2 marks)

c) i) The resistance increases because length of resistance wire increases. (2 marks)

ii) Current decreases as the resistance has increased. (2 marks)

4. a) i) (2 marks)

ii) I

V R

0.1

0.5 = 5 Ω

I

V R

0.4

3.2 = 8 Ω (3 marks)

X

A

V


40

iii)

(2 marks)

iv) The gradient of the graph increases as the current increases because the resistance of bulb filament increases as the current increases. The reason for this is that temperature of the filament wire increases. (2 marks)

b) i) power = 2000 W, voltage = 240 V, current = ? P = V x I 2000 W = 240 V x I

240

2000 I = 8.33 A

Fuse size = 13 A (4 marks)

ii) live = 8.33 A; neutral = 8.33 A; earth = 0 A (1 mark)

iii) A current of up to 13 A would flow through person touching casing of fire. (1 mark)

iv) Earth wire should be connected to metal body of electric fire. (1 mark)

v) The fuse must be connected to live wire. (1 mark)

vi) The 13 A current would flow through fuse and earth wire to ground. The fuse will blow isolating the device from the high voltage. (1 mark)

5. a) i) 12107 – 11847 = 210 units (1 mark)

ii) 1 unit = 12 p 210 units = 210 x 12 = 2520 p = £25.20 (2 marks)

b) i) A = fuse (1 mark)

ii) 2 (1 mark)

iii) 1 = blue; 2 = brown (2 marks)

c) i) As thickness increases, resistance decreases. (2 marks)

ii) As length increases, resistance increases. (2 marks)

d) i) Draw lines to connect points 1 to 2 and 4 to 6; or draw lines to connect points 1 to 3 and 4 to 5. (2 marks)


41

ii) The current flows from positive terminal to 1 to 2 to 6 to 4 to the negative terminal. (2 marks)

e) i) Parallel. (1 mark)

ii) sum

product R r

5 5

5 5

10

25 = 2.5 Ω (2 marks)

iii) (1 mark)

6. a) i) 240 V means 240 J deposited by each coulomb of charge that passes through TV. 80 W means 80 J of energy consumed by the TV in each and every second. (2 marks)

ii) power = 80 W, voltage = 240 V, current = ? P = V x I 80 W = 240 V x I

240

80 I = 0.33 A (3 marks)

iii) 1 A fuse would be the safest size of fuse to select. The other fuses will allow currents which would be too large. (2 marks)

iv) voltage = 240 V, current = 0.33 A, resistance = ? V = IR 240 = 0.33 x R

0.33

240 R = 720 Ω (3 marks)

v) Blue wire to the left pin and brown wire to the right pin. (3 marks)

vi) The live wire. (1 mark)

vii) To isolate the appliance and user from high voltage. (1 mark)

viii)All the live electrical wires and circuits inside the TV are insulated by the non-metallic casing of the TV. (3 marks)

b) i) energy = power x time = 8 x 2 = 16 units @ 12p per unit = 192 p = £1.92 (3 marks)


42

ii) 15m means resistance will be 3 times bigger. Similarly the cross-sectional area will be 3 times bigger, therefore the resistance will be 3 times smaller. The combined effect is that the resistance will be the same as before i.e., 0.045 Ω. (6 marks)

iii) The current is small in the cables so that the (I2 R) heating effect is smaller. (2 marks)


43

Chapter 7 Electromagnetism

In-text questions 1. a) Use a larger current; use more turns of wire; put a soft iron rod into the middle of the

solenoid.

b) Lifting and dropping magnetic materials like iron and steel in scrap yards.

2. a) The direction of the magnetic lines of force are from the N-pole of the magnet to the S-pole.

b) The brass rod will move to the right, parallel to the brass rails.

c) The brass rod will move to the left.

d) Assuming the magnet is also in its original position i.e., as in part (a), the brass rod will move to the left.

3. a) ‘d.c.’ means direct current.

b) The direction of the force on AB is vertically downwards.

c) The direction of the force on CD is vertically upwards.

d) The forces form a couple which results in a clockwise moment about the axis, causing the coil to rotate.

e) The force on side BC is zero, since the current in BC is parallel to the magnetic field lines.

4. a) There is a rate of change of flux (magnetic field lines) cutting which induces an e.m.f. which results in an induced current.

b) The induced current will flow in the opposite direction.

c) Julie should move the wire faster.

d) No current will be induced because the wire is moving parallel to the magnetic field lines and is no longer cutting the field lines.

5. Power is transmitted at high voltage to minimise the current in the cables which reduces heat losses, sometimes referred to as ohmic losses, in the transmission system.

6. a) Primary coil; soft-iron laminated core; secondary coil.

b) The core transfers energy, in the form of an oscillating magnetic field, from the primary coil to the secondary coil.

c) Soft iron is the best material for magnetic field lines to pass through.

7. If d.c. was used there would be no changing magnetic field produced by the primary coil, consequently there would be no induced e.m.f. in the secondary coil.


44

8. VP = 240 V, Vs = 48 V, NP = 2000 turns, Ns = ?

S

P

S

P

N

N

V

V

SN

2000

48

240

240

48 2000 NS

= 400 turns

9. VP = 4 V, Vs = ?, NP = 20 turns, Ns = 80 turns

S

P

S

P

N

N

V

V

80

20

V

4

S

20

80 4 VS

= 16 V

10. a) Step-down transformer.

b) All electrical appliances in the home operate at 240 V, 24000 V would seriously damage appliances and would be a health hazard to the occupants.

c) Current in the supply cables would be very large causing a lot of electrical energy to be wasted as heat energy.

d) VP = 24 000 V, Vs = 240 V, NP = 5000 turns, Ns = ?

S

P

S

P

N

N

V

V

SN

5000

240

000 24

000 24

240 5000 NS

= 50 turns

11. a) Step-up transformer.

b) VP = 25 000 V, Vs = 400 000 V, NP = 1000 turns, Ns = ?

S

P

S

P

N

N

V

V

SN

000 10

000 400

000 25

000 25

000 400 000 10 NS

= 160 000 turns

c) The voltage is 400 000 V to reduce the current in the grid. This reduces the energy losses in the form of heat so that more electrical energy reaches factories and homes.

Exam questions (pages 157–9) 1. a) There is a rate of change of flux linkage between the magnetic field of the magnet and the

coil of copper wire. (2 marks)

b) Current is alternating. The flux linkage increases and decreases inducing a voltage which changes in size and direction as long as the magnet is oscillating up and down. (2 marks)

c) See Figure 39 b, page 133. (2 marks)

2. a) i) a.c. stands for alternating current. Alternating means that the polarity of the supply changes sign periodically whereas direct means that the polarity remains constant. (3 marks)


45

ii) Soft iron. (1 mark)

iii) VP = 16 V, Vs = ?, NP = 100 turns, Ns = 50 turns

S

P

S

P

N

N

V

V

50

100

V

16

S

50

100 16 VS

= 32 V (4 marks)

b) i) 10 steps = 20 V 1 step = 2 V So voltage will increase or decrease by 2 V. (3 marks)

ii) 1 step = 50 turns 10 steps = 500 turns (1 mark)

iii) VP = 240 V, Vs = 20 V, NP = ?, Ns = 500 turns

S

P

S

P

N

N

V

V

500

N

20

240 P

20

240 500 N P

= 6000 turns (3 marks)

iv) If a current flows which is greater than the fuse rating, the fuse will ‘blow’ protecting the device to which it is connected. (2 marks)

c) i) Kinetic energy of fission fragments is converted into heat energy. (2 marks)

ii) X = generator which converts rotational kinetic energy into electrical energy. (2 marks)

d) i) Step-up; step-down. (2 marks)

ii) High; low. (2 marks)

iii) High voltages are used to minimise current in the grid. This reduces electrical energy lost as heat energy. (2 marks)

3. a) i) See Figure 6, page 144. (2 marks)

ii) See Figure 6, page 144. (4 marks)

b) i) See Figure 24, page 155. (7 marks)

ii) a.c. voltages. (1 mark)

iii) VP = 6 V, Vs = 12 V, NP = 200 turns, Ns = ?

S

P

S

P

N

N

V

V

SN

200

12

6

6

12 200 NS

= 400 turns (4 marks)

c) The output of a power station is a high current and relatively low voltage. If this form of energy was transmitted directly, without a step-up transformer, the large current would generate a large amount of electrical energy in the form of heat, lost to the atmosphere. Hence there would be less energy transmitted to users. (4 marks)


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4. a) i) 25 kJ of energy per coulomb. (1 mark)

ii) Stepping up the voltage, decreases the current in the grid. This has the effect of decreasing the electrical energy lost as heat due to the resistance of the cables in the grid. (2 marks)

iii) Step-down transformers must also be used before electrical energy is put into factories and homes. (1 mark)

iv) For safety. (1 mark)

b) i) energy (in kWh) = power (in kW) x time (in hours) = 0.8 kW x 0.25 hr = 0.2 kWh (4 marks)

ii) 1 unit costs 12p, 0.25 units costs 3p (1 mark)

c) i) 1) Momentary deflection in current meter. (1 mark) 2) Momentary deflection in the opposite direction in current meter. (2 marks)

ii) The needle in the current meter will oscillate at 1 Hz. (2 marks)


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Chapter 8 The Earth and Universe In-text questions 1. asteroid, planet, star, galaxy, Universe.

2. a) The light from distant galaxies has a longer wavelength than we would have expected. Having a longer wavelength means that it is closer to the red end of the visible spectrum. This is called ‘red-shift’.

b) These galaxies are moving away from our galaxy (the Milky Way).

3. Mercury, Venus, Earth, Mars, Jupiter, S aturn, Uranus, Neptune.

4. The heliocentric (Sun-centred) model explains: the apparent ‘looping’ of planets was due to the combined motion of the Earth and the planet

itself; why Venus is sometimes closer to Earth than Mars so it appears brighter, but at other times

Venus is further away than Mars and appears less bright; why Venus and Mercury should show phases, just as our Moon does.

5. The Big Bang occurred between 13 and 14 billion years ago. It was then that the Universe (matter, energy and time) came into existence. The Big Bang (or explosion) came from a tiny point that physicists call a singularity.

6. In order to explain the expansion of the Universe, Steady State theory suggests that matter is being created continually. Physicists are uncomfortable with this idea because it suggests that our thoughts about energy conservation might not be correct. In addition, the theory fails to explain why the Universe is expanding or why there is cosmic microwave background radiation.

7. At temperatures around 15 million Celsius, hydrogen nuclei collide with each other at unbelievably high speeds. At such speeds they can form new, heavier nuclei such as helium-3 and helium-4. This process is called nuclear fusion and results in the production of vast quantities of heat and light energy. It occurs in stars.

8. Clouds of gas and dust come together as a result of mutual gravitational attraction, but do not have enough material for the temperature to reach 15 million °C to form a star. Such gas and dust clouds are called planetary nebulae and eventually they become planets as a result of accretion due to gravitational attraction. The presence of a massive star may cause them to become trapped in its orbit. Since the gas and dust clouds originally spiralled in the same direction, so the planets would orbit the Sun in same direction and in the same plane.

9. Currently astrophysicists estimate the age of the Universe to be 13.75 billion years.

10. When rock strata of similar ages are studied in various countries, they show remarkable similarity – suggesting that these countries were joined together when the rocks formed. The most compelling evidence in favour of continental drift comes from the magnetic orientation of the rocks in under-ocean ridges. As the liquid magma erupts out of the gap, the iron particles in the rocks tend to align themselves with the Earth’s magnetic field, and as it cools they set in that position. Every half million years or so, the Earth’s magnetic fi eld tends to swap direction. This means the rocks on either side of the ridge have bands of alternate magnetic polarity.


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11. Earthquakes are caused when tectonic plates slide past each other. This happens where the plates meet.

12. The Earth’s lithosphere is the lower part of the crust and the upper, solid part of the upper mantle.

13. Every century the continents move apart by 1.2 cm x 100 = 120 cm =1.2 m. In 50 000 centuries they move apart by 50 000 x 1.2 m = 60 000 m = 60 km

Exam questions (pages 175–6) 1. a) i) Nebula. (1 mark)

ii) Hydrogen and dust. (1 mark)

iii) It compresses under its own gravity and its temperature rises appreciably. (1 mark)

iv) A star. (1 mark)

b) i) Nuclear fusion. (1 mark)

ii) Heat. (1 mark)

2. Planet Distance from Sun in millions of kilometres Earth 150 Jupiter 778 Mercury 58 Venus 108 Uranus 2870 Neptune 4497 Mars 227 Saturn 1247

(4 marks)

3. a) Universe. (1 mark)

b) i) Neptune. (1 mark)

ii) C. (1 mark)

iii) D and E. (1 mark)

iv) Jupiter, Saturn, Uranus, Neptune. (2 marks)

c) Nuclear fusion. (1 mark)

d) The journey would take a very long time. There are logistics problems in carrying enough fuel, oxygen, food and water for the journey. (2 marks)

e) Steady State theory. (1 mark)

f) i) The Big Bang occurred between 13 and 14 billion years ago. It was then that the Universe (matter, energy and time) came into existence. The Big Bang (or explosion) came from a tiny point that physicists call a singularity. (1 mark)


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ii) The Universe is continuing to expand. (1 mark)

4. a) Our present model is heliocentric. This means the planets orbit the Sun. The ancient theory was geocentric. This means that the planets and the Sun orbited the Earth. In our present model there are 8 planets. The ancient theory had only 6 planets because Uranus and Neptune had not yet been discovered. (2 marks)

b) Stars form when clouds of hydrogen, known as a stellar nebula, come together because of gravity. As these clouds become more and more compressed, they start to spiral inwards and the temperature rises enormously. Gravity eventually compresses the hydrogen so much that the temperature reaches about 15 million °C. At this temperature, nuclear fusion reactions start and a star is born. Clouds of gas and dust, called planetary nebulae, come together as a result of mutual gravitational attraction, Such gas and dust clouds eventually become planets as a result of accretion due to gravitational attraction. (2 marks)

c) i) The arrows all point from points 1, 2 and 3 towards the centre of the Sun. (3 marks)

ii) Gravitational attraction between the Sun and the comet. (1 mark)

iii) Points 1 and 2 are equidistant from the centre of the Sun and therefore the size of the force on the comet at these two positions is the same. Point 3 is further from the Sun than points 1 and 2. At point 3, the force on the comet is less than at points 1 and 2. This is because gravitational force decreases with distance. (2 marks)

iv) Kinetic energy. (1 mark)

v) As the comet approaches the Sun it loses gravitational potential energy. This GPE is converted into kinetic energy. As it moves away from the Sun it loses KE. The lost KE is converted into gravitational potential energy. (1 mark)

vi) The comet is fastest when it is closest to the Sun and slowest when it is furthest away from the Sun. This is because it has most KE when it is closest the Sun. (2 marks)

5. a) The light from distant galaxies has a longer wavelength than we would have expected. Having a longer wavelength means that it is closer to the red end of the visible spectrum. This is called ‘red-shift’. (1 mark)

b) Red shift tells us that these galaxies are moving away from our galaxy (the Milky Way). (2 marks)

6. a) i) Red shift. (1 mark)

ii) Away from our galaxy. (1 mark)

b) Big Bang Theory. (1 mark)

c) See answer to question 4(b). (4 marks)

d) The temperature near the centre of the Sun is so high that hydrogen nuclei collide with each other at unbelievably high speeds. When they collide at such speeds they can form new, heavier nuclei such as helium-3 and helium-4. This process is called nuclear fusion and results in the production of vast quantities of heat and light energy. (3 marks)

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