Cambridge IGCSE Physics Coursebook Answers [PDF]

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Answers to end-of-chapter questions Chapter 1 7 a volume = l × b × h

1 Mass

Length

Volume

Time

balance

metre rule

measuring cylinder

stopclock

scales

tape measure

= 80 × 40 × 15 = 48 000 m3 b mass = volume × density = 48 000 × 1.3 = 62 400 kg

electronic timer

[1] [1] [1]

8 a Half-fill a measuring cylinder with water;

vernier calipers

record volume. Place pebble in water, ensuring that it is submerged. Record new volume. Volume of pebble equals difference in recorded volumes.

micrometer screw gauge

2 a density = mass

volume

b

[1] [1] [1]

[1] [1] [1] [1]

Unit of mass

Unit of volume

Unit of density

kg

m3

kg/m3

9 a V1 = 70 cm3

[1]

g

cm3

g/cm3

V2 = 95 cm3

[1]

b V = 95 − 70 = 25 cm3

[1] [1]

mass c density = volume

[1]

b mass of pebble

3 a vernier callipers; micrometer screw gauge b Time, say, 40 drops and divide by 40.

4 a volume = l × b × h = 8.4 × 8.0 × 5.5 = 369.6 cm3 mass b density = volume 340 = 369.6 = 0.92 g/cm3

5 a 70 – 15 = 55 cm3 b 43 – 12 = 31 s

6 mass of liquid = 203 – 147 = 56 g mass density = volume 56 = 59 = 0.95 g/cm3 © Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

102 25 = 4.08 g/cm3

=

[1] [1]

[1] [1] [1] [1] [1] [1] [1]

[1] [1]

d Sample m / g

[1] [1] [1]

[1]

V2 [1] / V1 [1] / V / cm3 Density / cm3 [1] cm3 [1] [1] g/cm3 [1]

B

144

80

44

36 [1]

4.0 [2]

C

166

124

71

53 [1]

3.1 [2]

10 a 30.98 − 30.72 = 0.26 g b density = mass volume

[1] [1] [1]

= 0 26 200

[1]

= 0.0013 g/cm3

[1] [1]

Answers to end-of-chapter questions: Chapter 1

1

11 a water

[1]

b volume (of water) or water level

[1]

c the stone

[1]

d volume (of water)

[1]

e subtract or calculate the difference between [1] first volume from (or and) second volume [1]

© Cambridge University Press 2014 IGCSE Physics

Answers to end-of-chapter questions: Chapter 1

2

Answers to end-of-chapter questions Chapter 2 7 a

1 a average speed = distance travelled

800

time taken

Distance / m

b m/s c Graph is a horizontal straight line, showing that speed does not change. d distance travelled = area under graph

600 400 200

2 a graph A; speed = gradient (slope) of graph

0 0

b Graph is a straight line. c graph B; acceleration = gradient (slope) of graph

Description

Examples

scalar

has magnitude only

speed, distance

vector

has magnitude and direction

velocity, acceleration, weight

4 average speed = distance time 400 50 = 8.0 m/s =

5 Speed is uniform (constant) in both. The bus travels faster during B than A.

6 distance = speed × time = 15 × 30 = 450 m

20 30 Time / s

40

suitable scales chosen horizontal axis and scale correct vertical axis and scale correct five points correctly plotted and straight line drawn

3 a, b Quantity

10

b Graph is straight line, so constant speed.

[1] [1]

graph is horizontal

[1] [1] [1] [1] [1] [1] [1]

b point 3, graph is steepest

[1] [1]

c point 2, graph is becoming steeper (gradient is increasing)

[1] [1]

d point 4, graph is becoming less steep (gradient is decreasing)

[1]

e point 6, distance is decreasing

[1] [1]

[1]

9 a speed of light

[1]

b distance = speed × time

10

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

8 a point 1 or 5, [1]

[1] [1] [1]

[1]

Object

Distance travelled

Time taken

Speed

bus

20 km

0.8 h

25 km/h

taxi

6 km

200 s [1] 30 m/s

aircraft

4950 km

snail

3 mm

[1] 5.5 h 10 s

[1]

900 km/h 0.3 mm/s [1]

Answers to end-of-chapter questions: Chapter 2

1

11 speed is constant

[1] [1]

acceleration = 0

Speed

12 a

horizontal axis showing time vertical axis showing speed rising straight-line graph starting at origin

[1] [1] [1]

Speed

b

Time

horizontal and vertical axes showing time and speed horizontal straight-line graph above axis, then decreases down to zero change in velocity 13 acceleration = time taken =

b acceleration = gradient of graph

[1]

8.0 2.0

[1]

14 initial speed = 0 m/s

[1] [1] [1] [1]

change in speed = acceleration × time = 2.3 × 4.0 = 9.2 m/s speed acceleration 24 = 5.6 = 4.3 s

15 time =

[1]

c distance = area under speed against time graph = area of triangle + area of rectangle

[1] [1]

= 12 × 30 × 27 + 20 × 27

[1]

= 405 + 540 = 945 m

[1] [1]

17 a B, D

[2]

b A, E

[2]

c Acceleration is changing in the other section, C.

[2]

distance time 1425 = 75 = 19 m/s

[1]

b the direction of its motion

[1]

[1]

30

10 0 0

10

20 30 Time / s

© Cambridge University Press 2014 IGCSE Physics

40

50

[1]

[1]

b i

[1]

accelerating or increasing speed

ii steady or constant speed

[1]

iii decelerating or slowing down

[1]

c less than

20 a i

[1]

constant/steady/uniform speed or velocity or speed or velocity = 2.5 (m/s) [1] speed or velocity = 2.5 m/s [1]

ii shape curving upward but not to vertical [1] b horizontal (straight) line (parallel to time / x-axis) c i

20

[1]

19 a 25 km

[1] [1]

[1]

= 0.9 m/s2

[1] [1] [1]

[1]

= 4.0 m/s

27 30

18 a speed =

[1]

2

Speed / m/s

[1] [1] [1] [1]

=

Time

16 a

horizontal axis and scale correct vertical axis and scale correct six points correctly plotted graph drawn through points

[1]

horizontal straight line at 2.5 m/s from 0 to 2 s

[1]

ii straight line rising to the right as far as the edge of the graph area Δv = 4 m/s or gradient clearly 2 m/s2

[1] [1]

d horizontal straight line at 0 m/s

Answers to end-of-chapter questions: Chapter 2

[1] [1]

2

Answers to end-of-chapter questions Chapter 3 11 a weight = mass × g

1 A force can make an object change direction,

= 80 × 10 = 800 N

decelerate, or accelerate.

2 resultant force

[1] [1] [1]

3 weight

b the same

[1]

4 a force = mass × acceleration

c less

[1]

12 a the two 5000 N forces

b, c Quantity

Unit

Scalar or vector?

mass

kg

scalar

acceleration

m/s2

vector

force

N

vector

They are equal in size but act in opposite directions.

[1] [1] [1]

c The lorry will speed up (accelerate).

[1]

= 20 × 5 = 100 N

6 a weight downwards, air resistance upwards

c The resultant force on it is zero, so it does not accelerate.

force mass 1 400 000 = 800 000

d terminal velocity

= 1.75 m/s2

14 acceleration =

b zero

7 a impulse of force = change of momentum b F = force, t = time, m = mass, v = final velocity, u = initial velocity

15 acceleration = =

c Momentum is a vector quantity.

8 a kilogram (kg) or gram (g)

[1]

b newton (N)

[1]

c metre per second per second (m/s2)

[1]

direction (like an arrow). b Mass is not a vector quantity (it is a scalar), so it does not have direction.

10 the bigger force acting on the smaller mass, that is, the 10 N force acting on the 5 kg mass

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1] [1]

[1] [1] [1] [1] [1] [1]

change in speed time

[1]

(20 – 12) 6.4

[1]

= 1.25 m/s2 force = mass × acceleration = 1200 × 1.25 = 1500 N

16 weight = mass × g

9 a Force is a vector quantity, that is, it has

[1]

b resultant force = 1300 – 1200 = 100 N forwards (to the left)

13 force = mass × acceleration

5 90°

[1]

[1] [1] [1] [1] [1] [1] [1]

= 50 × 1.6 = 80 N

17 a resultant force = 680 – 600 = 80 N upwards b He will accelerate upwards.

Answers to end-of-chapter questions: Chapter 3

[1] [1] [1] [1]

1

18 a i (engine) thrust and (air) friction ii force shown vertically upwards, anywhere on plane b i speed = distance in any form time

[1] [1] [1]

= 2200 2 75

[1]

= 800 (km/h)

[1]

ii idea of headwind on outward journey or tailwind on return journey or routes of different lengths or less friction or less weight

© Cambridge University Press 2014 IGCSE Physics

(v − u) or v or 8 3 t t = 2.7 m/s2

[1]

ii F = ma or 42 × 8/3 = 112 N

[1] [1]

19 a i

iii distance in first 3 s = 12 m so distance in last 11.2 s = 88 m so final speed = 88 = 7.9 m/s 11.2 b Any two from: lower top speed, longer total time, less steep slope at first, etc.

[1]

[1] [1] [1] [2]

[1]

Answers to end-of-chapter questions: Chapter 3

2

Answers to end-of-chapter questions Chapter 4 1 a increase b increase

5 moment

[1]

6 no resultant force (forces balanced)

[1] [1]

no resultant moment

2 a resultant

7 See Activity 4.3.

b zero

Make three small pinholes around the edge of the lamina. Suspend the lamina freely from a pin through one hole. Mark a vertical line below the pin using a plumb-line. Repeat this process for the other two pinholes. The centre of mass is where the three lines intersect.

3 a, b for example centre of mass

stable object

8

contact force A

for example

1m

0.9 m

centre of mass

[1] [1] [1] [1] [1]

B

1.5 m

weight of beam centre of mass unstable object

x

4 a pivot

F

b moment = force × distance from pivot c

Quantity

Unit

force

N

distance

m

moment of force

Nm

© Cambridge University Press 2014 IGCSE Physics

a centre of mass correctly marked, as in diagram

[1]

b arrows and labels added correctly

[2]

c moment of weight = force × distance = 200 N × 0.5 m = 100 N m moment of force F is F × 1.0 = 100 N m so F = 100 N

[1] [1] [1] [1] [1]

d upward contact force = sum of downward forces = 200 N + 100 N = 300 N

[1] [1] [1]

Answers to end-of-chapter questions: Chapter 4

1

9 a force and perpendicular distance (of force) from the point b i

downward force arrow at centre of bar

ii 0.50 m or 50 cm

[1]

© Cambridge University Press 2014 IGCSE Physics

[1]

[1]

30 = 0.60 kg

[1]

[1]

b (1.5 + 0.6) × 10 = 21 N

[1]

c i stays in position

[1]

iii moment of force = 40 × 1.2 = 48 N m [1] moment of weight = + 30 × 0.5 = 15 N m [1] total clockwise moment = 48 + 15 = 63 N m [1] iv F × 0.2 = 63 63 = 315 N F= 0.2 v make bar longer or move pivot/stone to the left or move pivot to left or increase mass of bar

10 a mass = 1 5 × 12

[1]

ii as the parrot is rotated, both distances change in proportion so clockwise moment = anticlockwise moment

[1]

Answers to end-of-chapter questions: Chapter 4

2

[1]

[1]

[1]

Answers to end-of-chapter questions Chapter 5 1 a extension = length when stretched − original length b graph B

2 a The extension of a spring is proportional to the load, provided the limit of proportionality is not exceeded.

Student must measure: ◆ length of spring when weights added ◆ unstretched length of spring ◆ repeated for at least five different weights.

8 a

Load / N

Length / cm

Extension / cm

0.0

83.0

0.0

5.0

87.0

4.0 [1]

10.0

91.0

8.0 [1]

3 a increases

15.0

95.0

12.0 [1]

b decreases

20.0

99.0

16.0 [1]

b load = stiffness × extension c See Figure 5.7a.

[1] [1] [1]

c increases b

d decreases

4 a pressure = force

1.5 Extension / cm

area

b P= F A c

Quantity

Unit

force

N

area

m2

pressure

Pa

2.0

1.0

0.5

0 0

5

10 Load / N

15

20

horizontal axis and scale correct vertical axis and scale correct five points correctly plotted and graph drawn through points

5 a A fluid is a liquid or a gas; any substance that can flow. b g = 10 m/s2 c P = hρg

[1] [1] [1]

9 a See Figure 5.14.

6 extension = change in length = 66 – 58 = 8.0 cm

[1] [1] [1]

7 See Activity 5.1. Diagram or list indicating: ◆ spring hanging vertically from clamp ◆ weights hanging from end of spring ◆ ruler.

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

Diagram showing: ◆ vertical tube with closed upper end ◆ open end submerged in mercury reservoir ◆ mercury in tube continuous from reservoir up to empty (vacuum) space near top

Answers to end-of-chapter questions: Chapter 5

[1] [1]

[1]

1

b An increase in atmospheric pressure causes the level of mercury in the tube to rise / the length of the mercury column to increase. [1]

10 If you stand upright, your weight is pressing down on a small area. [1] This gives a high pressure. [1] If you use a ladder, the pressure is less because your weight is spread over a greater area. [1]

11 extension for 5 N is 15 – 12 = 3.0 cm

[1] [1] [1]

extension for 15 N is 3 × 3 cm = 9.0 cm length is 12 + 9 = 21 cm

12 a

Load / N

0

Length / m

3.200

3.207

Extension / mm

0

7

Extension / mm

b

10

20

30

3.215 15

40

3.222 22

50

3.230 30

3.242 42

60

70

3.255 55

80

14 pressure = height × density × g

60

= 0.760 × 13 600 × 10 = 103 400 N/m2

3.270 70

[4]

[1] [1] [1]

15 a extension indicated between two broken lines

40

b i 20

0 0

20

40 Load / N

60

horizontal axis and scale correct vertical axis and scale correct eight points correctly plotted graph drawn through points c Draw up from 25 N to intersect graph line, from this intersection, go across to axis, 19 mm d the point where the graph line ceases to be straight 40 N approximately

13 a force = pressure × area = 100 000 × 2.0 × 1.25 = 250 000 N b There are equal forces on both sides of the window.

© Cambridge University Press 2014 IGCSE Physics

80

[1] [1] [1] [1] [1] [1]

[1]

four points correctly plotted straight line through points and origin

[2] [1]

ii proportional

[1]

iii 1 newton(s) 2 extension = 25 − 26 mm length = 75 − 76 mm

[1] [1] [1]

16 a wall A has bigger area so lower pressure (on soil) b i two from ◆ depth/height of air/atmosphere ◆ density of air/atmosphere ◆ acceleration due to gravity or weight of air ii 1 the same as 2 greater than or four times

[1] [1]

[2] [1] [2]

[1] [1] [1] [1] [1] [1]

Answers to end-of-chapter questions: Chapter 5

2

Answers to end-of-chapter questions Chapter 6 6 energy supplied = 100 J

1 Name

Description

kinetic energy

energy of a moving object

internal energy

energy stored in a hot object

chemical energy

energy stored in a fuel

light energy

energy that we can see

sound energy

energy that we can hear

strain (elastic) energy

energy stored in a squashed spring

electrical energy

energy released = 93 + 7 = 100 J Energy before is equal to energy after, so energy is conserved.

[1] [1] [1]

7 a gravitational potential energy → kinetic energy b kinetic energy → gravitational potential energy

[2] [2]

energy carried by an electric current

c Some energy is lost as heat due to friction and/or air resistance, so the final g.p.e. cannot equal the original g.p.e.

[1]

nuclear energy

energy stored in the nucleus of an atom

d She needs to supply energy, by jumping up as she starts off.

[1] [1]

heat thermal energy

energy escaping from a hot object

8 a, b

2 a heat energy b efficiency c conservation

3 a k.e. = 1 mv 2 (m = mass, v = speed) 2

b g.p.e. = mgh (m = mass, g = acceleration due to gravity, h = height)

4 a waste energy = energy input − useful energy output b efficiency =

[1]

Low-energy bulb

Filament bulbs

cost of one bulb

400 p

50 p

number of bulbs required for 10 000 hours

1

10

cost of electricity for 1 hour

0.2 p

1.0 p

total cost of electricity for 10 000 hours

2 000 p [1]

10 000 p [1]

total cost of bulbs and electricity

2 400 p [1]

10 500 p [1]

useful energy gy output p × 100% energy input

5 a chemical energy → light + heat

[2]

b electrical energy → kinetic energy

[2]

c kinetic energy → electrical energy

[2]

d kinetic energy → thermal (heat) energy

[2]

© Cambridge University Press 2014 IGCSE Physics

c money saved = 10 500 – 2400 = 8100 p

[1] [1]

d initial cost is high difficult to dispose of

[1] [1]

Answers to end-of-chapter questions: Chapter 6

1

9 a thermal (heat) energy, electrical energy

[2]

b thermal (heat) energy

[1]

c Yes, because 90% of the energy is used, and only 10% is wasted.

[1] [1]

10 a k.e. of moving air → electrical energy electrical energy → k.e. of car b Process is less than 100% efficient, so car will not gain speed.

11 a weight = mass × g = 180 × 1.6 = 288 N

[1] [1] [1] [1] [1] [1] [1]

b change in g.p.e. = weight × change in height = 288 × 100 = 28 800 J

[1] [1] [1]

c g.p.e. increases

[1]

© Cambridge University Press 2014 IGCSE Physics

12 work potential / gravitational / p.e. / g.p.e. / position kinetic / k.e. / movement constant / the same / uniform joule(s) or J

13 a mgh = 0.5 × 10 × 1.1 = 5.5 J b i 1.5 (J)

[1] [1] [1] [1] [1] [1] [1] [1]

ii energy used to deform ball/ground or strain energy stored in (deformed) ball/ground or heat generated in deformed ball/ground [1] c 9 + 5.5 = 14.5 J

[1]

k.e. = 1 mv 2 2 v = 7.6 m/s

[1]

Answers to end-of-chapter questions: Chapter 6

[1]

2

Answers to end-of-chapter questions Chapter 7 1 a resource

6 a i

fission ii uranium (or plutonium)

b Sun

b i fusion ii hydrogen iii helium

c renewable d fossil fuels; non-renewable e wind, electricity

b nuclear fission using sunlight as their energy source. b Sunlight causes evaporation, producing clouds; rain falls, and finally enters rivers, whose water is trapped behind a dam.

[1] [1] [1]

7 Renewable: two from hydroelectricity, solar,

2 a nuclear fusion 3 a Trees and plants grow

[1] [1]

[1] [1]

tidal, wind [2] Non-renewable: two from coal, oil, nuclear [2] (At least two correct in each column for 4 marks; deduct 1 mark for any in incorrect column.)

8 a oil nuclear fission

[1] [1] [1]

4 a Sunlight is always available in space, and

b i

gas lamp

[1] [1] [1]

ii electric motor or loudspeaker

[1]

iii microphone

[1]

not much power is needed on a spacecraft. [1] But in cities, there are large numbers of people in a small area, [1] so there is not enough roof space for all the solar cells that would be needed to generate enough power. [1] b (for example) In a desert for roadside phones, because there is then no need to connect the phone to the mains electricity supply. c A rechargeable battery can store the energy produced by solar cells, and can therefore supply electricity when the sun is not shining.

[1] [1] [1] [1]

5 a g.p.e.

[1]

b k.e.

[1]

c g.p.e. → k.e. → electrical energy

[2]

© Cambridge University Press 2014 IGCSE Physics

Answers to end-of-chapter questions: Chapter 7

1

Answers to end-of-chapter questions Chapter 8 8 a work done = force × distance moved

1 a more

= 250 × 12.0 = 3000 J

b more

2 a energy

b gain in g.p.e. = weight × increase in height = 700 × 2.5 = 1750 J

b work

3 work done = energy transferred

9 a weight = mass × g

4 a work done = force × distance moved (in the

= 45 × 10 = 450 N

direction of the force) b

Quantity

Unit

ΔW

joule, J

F

newton, N

x

metre, m

b gain in g.p.e. = weight × increase in height = 450 × 0.20 × 36 = 3240 J work done time taken 3240 = 4.2

c power =

c ΔE = energy transferred

5 a power = work done

= 770 W = 0.77 kW

time taken

b

Quantity

Unit

P

watt, W

ΔW

joule, J

t

second, s

6 Ahmed He lifts them to a greater height.

7 a Millie: speed = Lily: speed =

25 = 0.50 m/s 50

100 = 0.40 m/s 250

10 a work done = force × distance moved

[1] [1] [1] [1]

b Millie [1] Because they are identical, the one with the greater speed has the greater power. [1]

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1] [1]

= 780 × 100 = 78 000 J

[1] [1] [1]

b work done = force × distance moved = 240 × 100 = 24 000 J

[1] [1] [1]

c k.e. = 12 × 750 × 122

[1]

= 54 000 J d work done by engine = work done against friction + k.e. 78 000 = 24 000 + 54 000 so energy is conserved

Answers to end-of-chapter questions: Chapter 8

[1] [1] [1] [1]

1

11 a i gravitational potential energy (g.p.e.) ii force/mass/weight of (basket of) rocks and height of cliff

[1] [1] [1]

b chemical energy

[1]

c time taken to raise basket up cliff

[1] [1]

12 a M = V × D = 103 × 10−3 = 1.0 kg b mgh = 1 × 10 × 0.8 = 8.0 J (or 8.0 N m)

[1] [1]

c P = E = 8 × 90 t 60

[1]

= 12 W (or 12 J/s or 12 N m/s) d P = ρgh 8000 Pa (or 8000 N/m2)

© Cambridge University Press 2014 IGCSE Physics

[1] [1]

Answers to end-of-chapter questions: Chapter 8

[1] [1] [1]

2

Answers to end-of-chapter questions Chapter 9 1 See Figure 9.2.

11 a Molecules of ethanol leave the surface of the liquid so that its mass decreases.

2 See Figure 9.5. 3 a energy

b The more energetic molecules of ethanol are more likely to leave the liquid, so the average energy of the molecules remaining decreases. Hence its temperature decreases.

b temperature

4 a evaporation b faster-moving or more energetic; decrease or fall/drop

5 a smoke particles b molecules of the air

b The pressure will decrease.

[1]

b In the liquid, forces between the particles hold them together. If it is to become a gas, energy must be supplied to overcome these forces and separate the particles.

c quickly

7 a pressure × volume = constant

[1]

9 a solid b The particles are well separated and can move about within the volume of their container, colliding with its walls and with each other.

10 a particles of smoke b The smoke particles are moving because the particles of the air are continually colliding with them, changing their speed and direction of motion.

© Cambridge University Press 2014 IGCSE Physics

[1]

[1] [1]

15

a

Solid Liquid

[1]

[1]

[1]

120 000 × 20 = 160 000 × V2 V2 = 15 m3

b

[1]

[1]

[1]

[1]

c liquid

[1]

p1V1 = p2V2

14

[1]

[1] [1] [1]

b quickly

b solid

[1]

12 a The pressure will increase. 13 a evaporation (or vaporisation)

6 a slowly

b pV = constant p1V1 = p2V2 p∝ 1 V 8 a gas

[1] [1]

c

Gas

Shape

Molecules

fixed shape

vibrate about a fixed position

[2]

shape fills the container from the bottom

move around, close together

[1]

completely fills the container

move around, far apart

[1]

[1] [1]

Answers to end-of-chapter questions: Chapter 9

1

b i increases

16 a i bombardment/collisions with air molecules/particles

[1]

ii any two from lighter / very small / smaller than smoke particles / too small to be seen fast-moving / high kinetic energy random movement / movement in all directions [2]

© Cambridge University Press 2014 IGCSE Physics

[1]

ii air molecules/particles/atoms bombard/hit walls molecules faster / higher energy when temperature raised (not vibrate faster) greater force (per unit area) or more collisions per second

[1]

Answers to end-of-chapter questions: Chapter 9

2

[1]

[1]

Answers to end-of-chapter questions Chapter 10 1 a Liquid in bulb absorbs energy; gets hotter;

7 a Mercury expands as its temperature

expands; pushes up tube. b melting point of pure ice (0 °C); boiling point of pure water (100 °C)

2 a A has greater range (120 °C, from −10 °C to +110 °C). (B’s range is only 60 °C, from −10 °C to +50 °C.) b B is more sensitive. Each degree is a wider interval on the scale, so smaller changes can be measured.

3 solids, liquids, gases 4 more, greater or less, smaller

5 a specific heat capacity – the energy required per kilogram and per degree celsius to raise the temperature of a substance. b specific latent heat of fusion – the energy per kilogram required to cause a substance to change state from solid to liquid at its melting point. c specific latent heat of vaporisation – the energy per kilogram required to cause a substance to change state from liquid to gas at its boiling point.

6 a energy = mass × specific heat capacity × change in temperature energy in J, mass in kg, s.h.c. in J/(kg °C), change in temperature in °C b energy = mass × specific latent heat energy in J, mass in kg, specific latent heat in J/kg

© Cambridge University Press 2014 IGCSE Physics

increases.

[1]

b

Definition lower fixed point

melting point of pure ice

upper fixed point

boiling point of pure water [1]

Value 0 °C [1] 100 °C

c (for example) the resistance of a resistor or thermistor

8 a internal energy b the steel block It takes more energy to raise the temperature of the steel block by a certain amount than that of the copper block.

9 a the thermocouple thermometer

[1] [1] [1]

[1] [1]

b 100 °C This is a fixed point on the Celsius scale.

[1] [1]

c the liquid-in-glass thermometer It can measure to 0.5 °C (or better); the other measures to the nearest 1 °C.

[1]

d The properties of the two materials used in the thermometers do not vary linearly with temperature. The voltage of the thermocouple does not increase at a steady rate as the temperature goes up.

10 a the final temperature of the block the mass of the block

[1]

[1]

[1] [1] [1]

b If poorly insulated, some energy will be lost.

[1]

c too high (because the heater will have to supply more energy to make the temperature rise by 1 °C)

[1]

Answers to end-of-chapter questions: Chapter 10

1

11 a 0 and 100 (°C)

[1]

b i expands

[1]

ii moves along the tube/up/to the right stops at/near 100 mark c arrow slightly to left of −10 mark

[1] [1] [1]

b i Q = mcθ = 100.6 − 12.0 = 88.6 = 0.80 × 3900 × 88.6 = 276 432 J ii Q = Wt so t = 276 432 620 = 446 s

[1] [1] [1] [1] [1] [1]

12 a i electrical method – 3 marks for all 6 points (deduct 1 mark for each point omitted) lagged container + lid liquid (allow water) heater in liquid heater connected to electrical supply voltmeter and ammeter appropriately connected thermometer [3] or mixtures method – 3 marks for all 6 points (deduct 1 mark for each point omitted) lagged container liquid hot solid/hot liquid means of heating hot solid/liquid (seen or stated) means of weighing hot solid/liquid/use of known mass (seen or stated) thermometer [3] ii electrical method – 3 marks for all 5 points (deduct 1 mark for each point omitted) initial and final temperatures of liquid or temperature rise voltmeter reading ammeter reading heating time mass of liquid [3] or mixtures method – 3 marks for all 5 points (deduct 1 mark for each point omitted) initial and final temperatures of liquid or temperature rise initial and final temperatures of added solid/ liquid or temperature drop mass of added solid/liquid mass of liquid s.h.c. of added solid/liquid [3]

© Cambridge University Press 2014 IGCSE Physics

Answers to end-of-chapter questions: Chapter 10

2

Answers to end-of-chapter questions Chapter 11 1 a temperature; higher; lower

7 a As the air is heated, it expands. Its density decreases. It is lighter than the surrounding air, so it floats upwards.

b metal; non-metal

2 convection

Warm fluid moves, carrying energy with it.

radiation

Energy travels as infrared waves.

conduction

Energy travels through a material without the material moving.

b The surrounding air is cooler and so less dense. It sinks and replaces the warm air rising above the flame.

with cooler neighbours and share energy; these vibrate more, pass energy on to their neighbours; and so on. Electrons collide with particles in hotter region, gain energy; move randomly to cooler region, collide with particles there, give them energy.

4 expands; greater; less; lighter; rises; more; gravity; convection

energy, so vibrate more. They collide with neighbours, sharing energy with them. Energy is thus transferred from the hot end to the cold end.

Good absorber

Good emitter

Good reflector

matt

matt

shiny

black

black

white

6 a Air is a good insulator, so less heat

c The glass wool prevents the movement of air in the gap, so it is difficult for a convection current to be set up, which would transfer energy from the inner wall to the outer wall.

© Cambridge University Press 2014 IGCSE Physics

[1] [1]

[1]

[1] [1] [1]

[1]

[1] [1] [1]

b The temperature of the cold end of the rod would rise more rapidly, because metals are better conductors than plastics.

[1]

c electrons

[1]

9 a walls made of glass – poor conductor

5

b Infrared (heat) radiation from below is reflected back into the house, so that less escapes from the house.

[1]

8 a Particles at the hot end have greater

3 Particles in hotter region vibrate more; collide

is lost by conduction. Cold air from the window cannot flow into the room, so convection current losses are reduced.

[1] [1]

vacuum between walls – no conduction or convection silvering – reflects away infrared radiation lid – prevents convection losses (but see part b) b A liquid that is colder than its surroundings does not heat the air above it, so no convection current rises above it. Hence a lid is not essential.

[1]

[2] [2] [2] [2]

[1] [1]

10 a i conduction

[1]

ii convection

[1]

b heat lost at same rate as heat supplied

[1]

c i boiling

[1]

ii steam

[1]

[1]

Answers to end-of-chapter questions: Chapter 11

1

11 a i conduction ii atoms/free electrons at hot end vibrate more/have more energy share energy with others by collisions b copper is a better conductor or iron is a worse conductor c iron conducts heat slowly so gas above gauze is hot enough to burn copper conducts heat rapidly so gas above gauze is not hot enough to burn

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1] [1] [1] [1] [1] [1]

Answers to end-of-chapter questions: Chapter 11

2

Answers to end-of-chapter questions Chapter 12 1 a source

7 a trace A The amplitude of trace A is the greatest.

b vibrations

b trace C The frequency of trace C is the greatest (because more waves are contained in the same time interval).

c echo d frequency; second e hertz

8 You need a source of sound,

f gases; vacuum

and two detectors in line with the sound. You need to measure the distance between the two detectors, and the time interval between the sound reaching them. distance Then use speed = time to calculate the speed of sound.

2 a greater frequency b greater amplitude

3 a

B A

b

D

9 a solid

C

4 a shaded from 20 Hz to 20 kHz b region beyond 20 kHz

5

rarefaction

where particles of the medium are spread out

compression

where particles of the medium are squashed together

6 a the air inside the instrument

[1]

b the strings of the instrument

[1]

c The vibrations of the instrument cause the air near the instrument to vibrate. Compressions and rarefactions are formed, and these propagate through the air to the listener’s ear.

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

[1] [1] [1] [1] [1]

[1] [1]

b (for example) Place ear against table, tap table at a distance and hear the sound through the wood.

[1] [1]

c distance travelled = 2 times length of rod = 800 m

[1]

distance time

[1]

800 m 0.16 s

[1]

= 5000 m/s

[1]

speed = =

10 a i

reflection or wave bounces back from large object/sea bed

[1] [1]

[1]

ii distance = speed × time = 1500 × 0.80 = 1200 (m)

[1] [1] [1]

[1]

iii 1200/2 = 600 (m)

[1]

b graph should show uniformly sloping line with positive gradient

[1] [1]

[1]

Answers to end-of-chapter questions: Chapter 12

1

11 a any large surface, e.g. wall/cliff/mountain b i when hears bang/sees flash ii when hears echo c i reading = 2.25 s speed = distance time

[1] [1] [1] [1] [1]

= 720 2.25

[1]

= 320 (m/s)

[1]

ii one from inaccurate distance from firework reaction time wind

© Cambridge University Press 2014 IGCSE Physics

[1]

Answers to end-of-chapter questions: Chapter 12

2

Answers to end-of-chapter questions Chapter 13 1 a See Figure 13.5.

9 a Ray diagram correctly drawn showing

b angle of incidence = angle of reflection i=r

2 a virtual

that the ray passes through both surfaces undeflected, that is, the ray remains straight.

[2]

ray bends towards normal then away again so that it ends up parallel to original path

[1] [1] [1]

b

b the same size as c object d left–right inverted

3 See Figure 13.9a. 4 a n=

speed of light in a vacuum speed of light in the material

b n = sin i sin r

c Parallel rays remain parallel.

n = refractive index, i = angle of incidence, r = angle of refraction

10 a

5 See Figure 13.12a.

30° 30° A

[1]

50° 50° B

6 See Figure 13.16a. 7 a converging b closer than c virtual; magnified

8

normal

mirror angle of incidence i

angle of reflection r

incident ray

reflected ray

incident and reflected rays correctly drawn normal correctly drawn angle of incidence correctly marked angle of reflection correctly marked

© Cambridge University Press 2014 IGCSE Physics

In block A, reflected ray at equal angle and refracted ray bent away from normal. In block B, reflected ray only, at equal angle.

[1] [1] [1] [1]

b When the angle of incidence is greater than the critical angle, there is only an internally reflected ray; all of the ray is totally internally reflected.

[1] [1] [1]

[1] [1] [1] [1]

Answers to end-of-chapter questions: Chapter 13

1

11 a

13 a i image behind mirror

image

image same distance from mirror, along line perpendicular to mirror ii reflected ray reaching eye direction of reflected ray coming from image

mirror

object

Two rays in different directions from a single point on the lamp reflect off the mirror correctly and are extrapolated back behind the mirror, so that the image is at the point where they cross.

[1] [1] [1] [1]

b Each ray is reflected so that angle of incidence equals angle of reflection.

[1] [1]

c Light appears to come from a point behind the mirror but no light actually travels behind the mirror.

12 a

[1] [1]

ray 2 I

F

ray 1 F

O

[1] [1] [1]

c both rays straight on at first surface 30° prism ray refracted down in air at second surface 45° prism ray reflected down in glass at second surface 90° reflection straight on at third surface

[1] [1] [1] [1] [1]

14 a i any two of these three rays from top of object: parallel to axis to lens and on through focal point undeviated through centre of lens as if from focal point to lens and then parallel to axis traced back to locate image

[2] [1]

ii any two of: virtual / upright / magnified / further from lens / dimmer

[1]

ii magnifying glass

[1] [1]

[1] [1] [1] [1]

b The image is real because it is formed where rays of light meet.

[1]

c The image is diminished because it is shorter than the object.

[1] [1]

d The image is inverted because it is below the axis.

[1]

© Cambridge University Press 2014 IGCSE Physics

[1]

b HIS, because S is not its own mirror image

b i 3.4–3.6 cm Ray 1 continues straight through the centre of the lens, ray 2 bends at the lens and passes through the principal focus F, so that the image is at the point where they cross.

[1]

[1]

Answers to end-of-chapter questions: Chapter 13

2

Answers to end-of-chapter questions Chapter 14 1 energy; matter

7

2 transverse

describes a wave that varies from side to side, at right angles to the direction of travel

longitudinal

describes a wave that varies back and forth along the direction of travel

3

Symbol

Quantity

Unit

v

speed

m/s

f

frequency

Hz

λ

wavelength

m

waves correctly reflected at barrier separation remains as before

8 a decreases

4 a bounces off

[1]

b stays the same

[1]

c decreases

[1]

9 a speed = frequency × wavelength, v = f λ

b speed

5 reflection, refraction, diffraction (in any order) 6 a 4.0 cm

[1]

b 3.0 cm

[1]

c one wave = 4 cm so 10 cm = 2.5 waves so 2.5 waves pass in 1 s frequency = 2.5 Hz

[1] [1] [1]

[1] [1]

[1]

b v = f λ = 6 × 1014 × 3.75 × 10−7 = 2.25 × 108 m/s

[1] [1]

waves are curved in space beyond barrier separation remains as before

[1] [1]

10

d 2

y / cm

1 0 1

2

3

4

5

–1

6

7

8 9 x / cm

–2

correct value of amplitude correct value of wavelength

© Cambridge University Press 2014 IGCSE Physics

[1] [1]

Answers to end-of-chapter questions: Chapter 14

1

11 a i amplitude ii wavelength b i string moves air backwards and forwards or up and down or produces compressions and rarefactions

[1] [1] [1]

[1]

ii gets quieter/softer/less loud

[1]

12 a i R in correct position, by eye

[1]

ii three reflected waves correctly meeting mirror three reflected waves equidistant and centred on R b first ray plus reflection correct second ray plus reflection correct reflected rays projected back, to meet behind mirror or labelled I and in correct position

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1] [1]

[1]

Answers to end-of-chapter questions: Chapter 14

2

Answers to end-of-chapter questions Chapter 15 1 a red, orange, yellow, green, blue, indigo, violet

8 White light is dispersed because it is a

b red = lowest frequency, longest wavelength violet = highest frequency, shortest wavelength

2 a radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, gamma rays b radio waves = lowest frequency, longest wavelength gamma rays = highest frequency, shortest wavelength

3 300 000 000 m/s or 3.0 × 108 m/s 4 monochromatic; frequency (or wavelength) 5 a

prism

ray broadens on entering prism, and again on leaving red and violet ends of spectrum correctly indicated

ii dispersion

[1]

b i A (red, at top) ii C (yellow, third from top)

[1] [1]

c any two from: gamma, cosmic, X-rays, UV, IR, microwaves, radio, TV

[2]

b violet c Different colours travel at different speeds in glass. The slowest colour is the most strongly dispersed.

b film or photographic film or electronic detector or charge-coupled device (CCD)

[1] [1] [1] [1]

[1] [1]

short

red orange yellow green blue violet indigo

[1] [1]

9 a i refraction

10 a electromagnetic

spectrum white light

mixture of different colours/wavelengths, which travel at different speeds in glass. Laser light is a single wavelength (monochromatic) and so cannot be dispersed.

[1] [1] [1]

c absorbed/stopped by bone (not deflected/ reflected) [1] less absorption by flesh or penetrates/passes through flesh [1] d any one of: photographic film badges, behind screen when operating X-ray machine, protective clothing, minimise exposure

[1]

[1]

6 a false b true c true

7 300 000 000 m/s = 3.0 × 108 m/s

© Cambridge University Press 2014 IGCSE Physics

[1]

Answers to end-of-chapter questions: Chapter 15

1

Answers to end-of-chapter questions Chapter 16 b

1 a i repel

N

ii attract b See Figure 16.3.

S

S

N

N

S

2 a i soft ii hard b i for example: steel ii for example: soft iron

S

3 stroking with one pole of a permanent magnet

magnets in a square arranged N–S–N–S–N–S–N–S with attractive forces shown

place in electromagnet connected to d.c. supply

4 hammer it place in electromagnet connected to a.c. supply

6 a one of the following:

2

1

attract N

N

S attract

3

S N

S

[1]

9 a A soft magnetic material is easy

b one of the following: N and S poles at opposite ends field lines have same pattern

S

[1] [1]

b (for example) in a scrapyard crane or an electromagnetic door bolt

electromagnet can be switched on and off strength can be varied by changing current poles can be reversed by reversing current

N

[1]

more turns of wire or turns of wire closer together add an iron core

b See Figure 16.7b.

repel

[1] [1]

8 a bigger current

5 a See Figure 16.7a.

7 a

N

to magnetise and to demagnetise. A hard magnetic material is difficult to magnetise and demagnetise.

[1] [1]

b A hard material because it retains its magnetisation well.

[1] [1]

c A soft material because its magnetisation can change easily.

[1]

[1]

[1]

10

repel

4

each correct pair of attractive or repulsive forces

© Cambridge University Press 2014 IGCSE Physics

N

N

S

S

[4]

Answers to end-of-chapter questions: Chapter 16

1

a left-hand end of solenoid N right-hand end of solenoid S b lines of force out of N poles and into S poles lines close together at poles, farther apart elsewhere similar pattern for both magnet and solenoid repulsion indicated by distortion of pattern

11 a (S) N S N b i

Switch closed

[1] [1]

ii attractive force

[1]

iii with soft iron core

[1]

iv can be switched on and off (or can be stronger)

[1]

12 a can be switched off can vary the strength

[1]

b i

[1] [1]

1000 turns

[1]

ii iron

[1]

iii 3.0 A

[2]

Switch open

Soft iron

magnetised

loses its magnetism

Steel

magnetised

keeps its magnetism

© Cambridge University Press 2014 IGCSE Physics

[1] [1]

[2]

Answers to end-of-chapter questions: Chapter 16

2

Answers to end-of-chapter questions Chapter 17 1 a rubbed, friction, opposite Because the electrons have negative charge, this gives the balloon a negative charge.

b repel, attract

2 a electrons b negative

b The balloon repels electrons in the paper, so that there is a positive charge on the area of paper closest to the balloon. This positive charge and the negative charge of the balloon attract each other.

c positively

3 charged, attract, attract, attract, induction 4 a electric force

8 a i iron or ferromagnetic

+

unmagnetised (before being brought near magnet) (not non-magnetic)

b The field is in same direction as the force on a positive charge.

5

Symbol for unit

Quantity

Unit

force

newton

N

electric charge

coulomb

C

6 a positive

[1] [1]

c Suspend one so that it can turn freely. Bring the other close to one end and observe repulsion.

[1]

causes electrons to be transferred from atoms in the wool to atoms in the balloon.

© Cambridge University Press 2014 IGCSE Physics

b attracts (at first) repels after touching or angle of thread increases as XY decreases

9 a rub/rubbing

b They are equal.

7 a The force of friction

ii magnet

[1]

[1] [1] [1] [1] [1] [1] [1] [1] [1]

with dry cloth

[1] [1]

b i negative (−)

[1]

ii opposite charges attract

[1]

c horizontal arrow to L, starting or ending on sphere

[1]

d swings to R / moves away / is repelled

[1]

[1]

[1]

Answers to end-of-chapter questions: Chapter 17

1

Answers to end-of-chapter questions Chapter 18 1 a charge

7

cell

b positive, negative

2 a ammeter, series

current

switch

b voltmeter, parallel

3 a lamp A

V

b R=V I

a series circuit correctly drawn correct symbols with labels

[1] [3]

b at least two arrows around circuit from positive of cell

[1] [1]

c voltmeter

[1]

d volt (V)

[1]

8

4 Unit

+

–

Symbol for unit

Potential difference

volt

V

Current

ampere

A

Resistance

ohm

Ω

A

R

5 a positive, charge b negative, positive V

6 Equation

In words

In units

Q = It

charge = current × time

coulomb = ampere × second (C = A s)

R=V I

resistance =

P = IV

power = current × p.d.

watt = ampere × volt (W = A V)

E = IVt

energy = current × p.d. × time

joule = ampere × volt × second (J = A V s)

© Cambridge University Press 2014 IGCSE Physics

p.d. current

ohm = volt/ampere (Ω = V/A)

Answers to end-of-chapter questions: Chapter 18

1

a correct symbols for resistor, ammeter and power supply connected in series with voltmeter in parallel with resistor

[3] [1] [1]

b current

[1]

c potential difference (p.d.) V d R= I 6.5 = 1.25 = 5.2 Ω

[1] [1] [1] [1]

9 a light

[1]

b heat

[1]

c power = 36 W

[1]

d energy = power × time = 36 × 60 = 2160 J P e I= V 36 = 12

[1] [1] [1]

= 3A

[1] [1] [1]

Q 10 a I = t 30 = 20 = 1.5 A

[1]

b E = Pt

battery/cell, ammeter, coil in series voltmeter in parallel with coil standard symbols used for battery/cell, voltmeter and ammeter ii R = V I iii any two of: length (of wire) diameter/cross-section/area (of wire) resistivity/type of material temperature b R=60 15 = 4.0 Ω resistance of AB = 1.0 Ω resistance per metre = 0.50 Ω/m

12 a increases as current increases at an increasing rate b i

25 Ω

[1] [1] [1] [1]

[2] [1] [1] [1] [1] [1] [1] [1]

ii V = I R = 0.070 × 25 = 1.8 V iii P = I V (or P = I 2 R etc.) = 0.12 W c i 1.8 V (same as answer to b ii)

[1] [1] [1] [1] [1]

[1]

ii 1 = 1 + 1 R R1 R2

[1]

[1]

R = 12.5 Ω

[1]

[1]

= IVt

[1]

= 1.5 × 10 × 20

[1]

= 300 J

[1]

11 a i

A coil of wire

V

© Cambridge University Press 2014 IGCSE Physics

Answers to end-of-chapter questions: Chapter 18

2

Answers to end-of-chapter questions Chapter 19 1

Name of device

b i Circuit symbol

Description

lightdependent resistor (LDR)

resistance decreases when light falls on it

thermistor

resistance changes when temperature changes

relay

See Figure 19.15c.

ii See Figure 19.15a. iii See Figure 19.15b. iv See Figure 19.18a. v

an electromagnetic switch

See Figure 19.18b.

6 a melting, burning, fumes b wire melts, breaks circuit c circuit breaker d Fuse will not break for normal current, but will break when current rises above this value.

7

2 a current b sum

3 a voltage (or p.d.) b shared

each symbol correctly drawn

8 a

[4]

6V

c more (greater)

4 a series b parallel c series d parallel

5 a i

series circuit correct symbols for resistor, switch and power supply

[1] [3]

ii

b 10 + 40 = 50 Ω

[1] [1]

iii

c 0.12 A

[1]

d 0.12 A

[1]

iv

v

© Cambridge University Press 2014 IGCSE Physics

9 a capacitor

[1]

b light-dependent resistor

[1]

c relay

[1]

Answers to end-of-chapter questions: Chapter 19

1

10 a wires overheat (risk of fire)

[1]

b fuse, trip switch

[2]

c Use thicker wires, which have lower resistance, so there is less heating.

[1] [1] [1]

11 a in parallel

[1]

b 6.0 V

[1]

across each resistor

[1]

14 a

[1]

1 1 1 d = + R 2 3 3 2 5 = + = 6 6 6

[1] [1]

R=

6 = 1.2 Ω 5

[1]

I=

V 6 = R 1.2

[1]

= 5.0 A

[1] A

B electrons

diode

a diode correctly labelled

[1]

b cell correct way round

[1]

c arrow in correct direction

[1]

13 a AND gate Output is ON only if both inputs are ON. Input 1

d OFF (0)

[1]

because the resistance is lower.

b

[1] [1]

[1]

c The 2 Ω resistor,

12

c AND gate NOT gate

Input 2

[1] [1]

[1] [1] [1]

c L2 has blown or is missing

[1]

d i

[1]

blows

0

0

0

[1]

1

0

1

[1]

0

1

1

[1]

1

1

0

[1]

[1] [1] [1] [1]

ii nothing / does not light / off (not turns off )

[1]

iii nothing / does not light / off (not turns off )

[1]

15 a i 4.0 V

[1]

ii 12 V

[1]

b i 6Ω ii 1 = 1 + 1 R 3 6 R = 2Ω c I=V R = 6.0 A d i stays the same ii decreases

Output

© Cambridge University Press 2014 IGCSE Physics

circuit diagram with two lamps in parallel switch alongside power supply correct symbols for lamps and switch used b R=V I = 12 16 = 7.5 ohm(s) or Ω

Answers to end-of-chapter questions: Chapter 19

[1] [1] [1] [1] [1] [1] [1]

2

Answers to end-of-chapter questions Chapter 20 1 relays, motors, electric bells (any order)

9 a i

magnetised ii attracted or magnetised iii close

2 current, magnetic, turning, rotate 3 a force (motion)

b any two from: armature becomes permanently magnetised would not release from core contacts always closed [2]

b magnetic field c current

4 a charged, field, force

10 a i

b electrons, cathode ray (or television)

5 Someone presses the bell push.

[1] A current flows through the electromagnet. [1] The electromagnet attracts the iron armature. [1] The hammer strikes the gong. [1] At the same time, the circuit is broken at point A. [1] The springy metal pulls the hammer back. [1] The circuit is completed again at A. [1]

6 a The wire will swing the other way.

[1]

b The wire will swing the other way.

[1]

7 a downwards b to the right, by Fleming’s left-hand rule

current clockwise when viewed from top [1]

ii anticlockwise or down on left and/or up on right [1] b i

faster or greater turning effect

[1]

ii faster or greater turning effect

[1]

iii faster or greater turning effect

[1]

[1] [1] [1]

8 a downwards

[1]

b upwards

[1]

c The forces are unbalanced, and so provide a turning effect.

[1] [1]

d The force is zero, because the current does not cut across the magnetic field (it is parallel to the field).

[1]

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

[1]

Answers to end-of-chapter questions: Chapter 20

1

Answers to end-of-chapter questions Chapter 21 1 conductor, magnetic, induced, circuit, current 2 (answers from the top) d.c.; a.c.; a.c.; d.c.; a.c.; d.c.; a.c.

c Use the primary coil as the secondary and the secondary as the primary.

11 a i

X: coil ii Y: slip rings iii Z: brushes

3 coil, rotate/turn, magnetic, e.m.f., current 4 a movement b field

[1] [1] [1]

b A.c. flows back and forth, changing direction. D.c. flows in one direction only.

c current

5 high, smaller, less

[1] [1]

+

6 a primary, core, secondary

d.c. Current

b step-up, e.m.f./voltage c step-down, e.m.f./voltage

0

Time

7 Vp = p.d. across primary coil

a.c.

Vs = p.d. across secondary coil Np = number of turns on primary coil Ns = number of turns on secondary coil

–

correct (labelled) diagram

8 Ip = current in primary coil

12 a more turns

Vp = p.d. across primary coil Is = current in secondary coil Vs = p.d. across secondary coil

bigger area b stronger magnetic field turn the coil faster

9 a The magnetic field around the wire is changing (it is cutting across field lines).

[1]

b It will change sign / direction (from positive to negative, or the other way round). [1] c She should move the wire more quickly.

[1]

d No, because it is not cutting across the field lines / the magnetic field is not changing.

[1]

13 a

Vp Np = Vs Ns Ns =

Ip =

Ns Np 3 × 200 = = 60 V 10

Vs =Vp ×

© Cambridge University Press 2014 IGCSE Physics

[1]

14 a i

[1] [1] [1] [1]

[1] [1]

0.40 × 12 230

= 0.021 A

[1] [1]

5000 × 12 230

b Ip × Vp = Is × Vs

[1]

[1]

[1]

= 261

10 a So that less energy is lost during transmission. Vp Np b = Vs Ns

[1]

[1] [1] [1]

deflection to one side then goes back to zero again

[1] [1]

ii same as i but opposite direction

[1]

b larger

[1]

[1]

Answers to end-of-chapter questions: Chapter 21

1

c smaller d nothing (or small oscillations about zero position or blurred light spot)

15 a i

no deflection on voltmeter

ii deflection on voltmeter

© Cambridge University Press 2014 IGCSE Physics

[1] [1] [1]

iii deflection on voltmeter (in same direction as in ii) b use a stronger magnet move coil or magnet faster add more turns to coil

[1] [1] [1] [1]

[1]

Answers to end-of-chapter questions: Chapter 21

2

Answers to end-of-chapter questions Chapter 22 1

9 a i alpha, positive electron

gold nucleus, positive electron, negative

nucleus proton

neutron 4 2

2

[3]

b Electrons (the ‘plums’) are distributed through a sphere of positive charge (the ‘pudding’).

[1] [1]

What it tells us

X

chemical symbol

name of element

Z

proton number

number of protons in nucleus number of nucleons in nucleus

3 proton number + neutron number = nucleon number

4 a different numbers

alpha particle smaller than gold nucleus, with positive charges marked track of alpha particle correctly shown d Most of the gold atom is empty space / the nucleus makes up a small fraction of the volume of the atom, so the chance of a head-on collision between an alpha particle and a gold nucleus is very small.

10 a i

b the same number c different numbers

5 alpha, deflected, thin, mass, positive, centre 6 a 6 protons

[1]

b 6 neutrons

[1]

c 6 electrons

[1]

7 a 79 + 118

gold nucleus alpha

Name

nucleon number

ii electron, alpha particle, gold nucleus

c

He

Symbol

A

[1] [1] [1]

ii 3

[1]

iii 4

[1]

iv 3 + 4 = 7

[1] [1]

Particle

Charge

Mass m

[1] [1]

electron

−1

neutron

0

[1]

2000m

[1]

197 79

[2]

proton

+1

[1]

2000m

[1]

8 a 19

[1]

b 39

[1]

c

Au

40 19

K

© Cambridge University Press 2014 IGCSE Physics

[2]

[1] [1]

= 197 b

[1]

3

b 73 Li

11 a

[1] [1]

b i 92 ii 146 iii 92

[1] [1] [1]

Answers to end-of-chapter questions: Chapter 22

1

Answers to end-of-chapter questions Chapter 23 1 Radiation

2

Symbol

Type of particle or electromagnetic radiation

Mass

alpha

α

2 protons + 2 neutrons (He nucleus)

beta

β

electron

gamma

γ

electromagnetic radiation

Charge

1

+2

small

−1

0

0

radiation in the environment

background

detectors of ionising radiation

Geiger counter, photographic film

three types of ionising radiation from radioactive substances

alpha, beta, gamma

3 a it has negative charge b charged; Fleming’s left-hand rule c it is uncharged

4

Radiation

Penetration

Absorption

Absorbed by

alpha

least penetrating

most easily absorbed

thin paper, a few cm of air

beta

in between

in between

thin metal foil

gamma

most penetrating

least easily absorbed

thick lead or concrete

5 a (average) time, half, decay b See Figure 23.10a.

6

Use

… because …

Finding the age of an object

radioactive substances decay at a known rate.

Seeing through solid objects

radiation can penetrate matter.

Sterilising medical equipment

radiation can destroy living cells.

Tracing the movement of hazardous substances

small amounts of radiation can be detected.

© Cambridge University Press 2014 IGCSE Physics

Answers to end-of-chapter questions: Chapter 23

1

7 β is more penetrating than α.

[1] [1]

Detect using Geiger counter. Place thin paper over sources − α does not pass through. Place thin aluminium foil over sources − neither passes through.

[1] [1]

8 a 15

[1] [1]

counts per minute b 65 − 15 = 50 counts per minute c

[1] [1]

Count rate / counts per minute

65

40

11 a i background or any of the following: contaminated surfaces other radioactive material nearby radiation from rocks/soil cosmic rays/radiation from space radon gas from ground ii count rate = 136 4 = 34 counts/min b i alpha or α ii 876 4 − 34 = 185 counts/min

12 a i proton

15 0 half- 2 4 6 Time / hours life

8

correct graph drawn After one half-life, the measured count rate will be down to 25 + 15 = 40 Reading across from 40 on the graph, and then down, half-life = 1.3 h approximately.

9 a The formation of an ion by the removal of one or more electrons from an atom. b X-rays

10 a (for example) A patient with cancer is exposed to γ-radiation. This damages the cancerous cells, which then die. b (for example) During the manufacture of cardboard, β-radiation is passed through the card. If the card is too thick (too thin), the amount of radiation detected will be too low (too high). The machinery is automatically adjusted to give the correct thickness.

© Cambridge University Press 2014 IGCSE Physics

[1] [1]

[1] [1] [1] [1] [1] [1] [1]

ii proton and neutron

0

[1]

[1]

b number of protons = 47 number of neutrons = 107 − 47 = 60

[1] [1]

c i 8 h ± 0.25 h

[1]

ii Choose two points on the graph; for each, halve the value and add 8 h to the time.

[2]

[1] [1] [1] [1] [1] [1] [1]

[1]

[1] [1]

Answers to end-of-chapter questions: Chapter 23

2