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