CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
aths Skills
M
Workbook answers
Chapter 1
1
a
i
ii
iii
6
a
s
m
kg
Line B
Arranging the numbers like this makes the
sequence more obvious.
distance _____
​​ = 0.8 m/s
Speed = ________
​​
​​ = ​​  1.6 m
time
2s
Cross-sectional area = 30 cm2
b
2
4
5
0
1
5
6
.
1
0
6
6
6
.
8
5
1
5
6
8
.
1
0
6
6
0
.
8
5
= 9000 cm3
6
8
6
.
5
0
1
b
Symbol
for the
variable
Name
of unit
power
P
watt
W
8
660.85 W 686.501 W Microwaves
cookers use significant amounts of energy.
c 6.0154 W 6.106 W Mobile phones use
a small amount of energy.
1
______
​​ th of a degree Celsius
To ​​
10 000
C (A shows 4 s.f., B shows 2 s.f., D shows 2 s.f.)
potential
difference
V
volt
V
9
a
3.0 × 108 m/s
current
I
amp (or A
ampere)
b
2.998 × 108 m/s
Variable
Symbol
for unit
7
Row A (v is the symbol for speed; centimetres
are a sensible unit of measurement when using
a ruler; d is a symbol for distance; t is the
symbol for time)
10 D (The number has been rounded using the
rounding rules shown in Figure 1.11.)
11 a
b
D
i
ii
iii
iv
v
12 a
b
c
d
31 560 000 s
1836.2
0.000 000 65 m
1 496 000 km
a
Variable Symbol for Name of unit
the variable
Symbol
of unit
power
P
watt
W
energy
E
joule
J
b
1
.
Volume = 30 cm2 × 300 cm
Yes you do need to change the units so that
they are consistent.
3
6
The expression for efficiency is a ratio of
two values of power, so there are no units
for efficiency. The % sign shows that the
answer is a number of parts in 100.
5.6752 × 104
2.533 12 × 102
1.0005 × 103
6 × 10−2
4.46 × 10−4
13 3.85 × 1026 W
14 1 μm = 10−6 m
1 nm = 10−9 m
Therefore there are 1000 or 103 nm in 1 μm.
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
15 a
b
c
d
e
0.7 kW = 7 × 102 W
14 ms = 1.4 × 10−2 s
23 MΩ = 2.3 × 107 Ω
1.8 µC = 1.8 × 10−6 C
475 nm = 4.75 × 10−7 m
16 a
b
1.0 × 10−3 m
19 B. (Picture each distance in your head and
imagine the cheetah travelling this distance
in one second. A cheetah can cover 30 metres
every second. The other values are unrealistic
for a maximum.)
3
[1]
b
2.7 × 106 N
c
1 cm2 = (1 cm)2 = (10−2 m)2 = 10−4 m2 [1]
540 000 cm2 = 540 000 × 10−4 m2 = 54 m2
[1]
6N
2.7
×
10
__________
Pressure = ​​
​​
[1]
54 m2
= 50 000 Pa
[1]
d
2
[1]
[1]
[1]
= 50 kPa
[1]
a
3 × 108 m/s
[1]
b
Distance = speed × time
= 3 × 108 m/s × 500 s
= 1.5 × 1011 m = 1.5 × 108 km
[1]
[1]
b
Q = I × t
t = 4 hours = 4 × 60 × 60 s
Q = 300 × 10−3 × 4 × 60 × 60
= 4.32 × 103 C
a
[1]
[1]
[1]
[1]
Round mass to: 1 × 104 kg
Convert mass to weight, giving: force
= 9.8 × 104 N
Round 1490 cm2 to 1500 cm2
1500 cm2 = 1.5 × 103 cm2
Convert 1.5 × 103 cm2 to m2
(because Pa = N/m2)
1 cm2 = 1.0 × 10−4 m2
Therefore 1.5 × 103 cm2
= 1.5 × 103 × 1.0 × 10−4 m2 = 0.15 m2
Applying this to the equation:
4
​​
Pressure = ________
​​  9.8 × 10
0.15
= 6.5 × 105 Pa
21 This is of order of magnitude 106 Pa.
Approximating the given values to 1 s.f.:
distance(km)
____________
​​
800 km/hour = ​​
20 hours
Distance = 800 km/h × 20 hours
= 16 000 km or 2 × 104 km (1 s.f.)
22 A
23 a
b
2
1 a 1st row: P, newton/(metre)2, N/m2
2nd row: F, newton, N
3rd row: A, (metre)2, m2
4.2 light-years = 4.2 × 9.5 × 1015 m
= 39.9 × 1015 m = 4.0 × 1016 km
17 6900 mA = 6.9 A
P = 2.4 × 102 × 6.9 = 16.56 × 102 W
P = 1.7 × 103 W
3.0 × 108 m/s
​​
18 Wavelength = ​​ ___________
500 × 106 Hz
3.0 × 108 m/s
​​
= ___________
​​
5.0 × 108 Hz
= 0.6 m
= 6.0 × 10−1 m
20
Exam-style questions
10 000 Ω + 200 Ω + 30 000 Ω = 40 200 Ω
Total resistance = 4.02 × 104 Ω
V = IR
V = 2 × 10−2 × 10−3 A × 4.02 × 104 Ω
V = 0.80 V
Chapter 2
Questions
1
Independent variable: distance –
continuous, quantitative
Dependent variable: time – continuous,
quantitative
Distance: a trundle wheel (a distance
measuring wheel) or a long tape measure
Time: a stopwatch
2
Independent variable: core material –
qualitative
Dependent variable: strength –
quantitative, continuous
Control variable: current
3
4
D
a R
ead the value when no weight added,
from the first diagram, then deduct this
value from the reading from the second
diagram.
b
1.6 N − 0.2 N = 1.4 N
c
Adjust the barrel of the newtonmeter
first, so that the reading on the scale is
zero with no force applied.
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5
a
b
c
d
6
a
0.72 V
0.04 V
0.16 V
0.96 V
Time / min Temperature of Temperature of
black can / °C
shiny can / °C
0
4
8
12
16
20
b
7
(Accept 2-minute intervals. Note
that maintaining a constant ambient
temperature while repeating the
experiment would be challenging
for this length of time, so repeating
this experiment would probably be
impracticable.)
The temperature of the surroundings is a
variable that needs to be controlled.
(The current value is only given to show that it was controlled, i.e. kept constant. If it is not shown in
the results table it needs to be recorded somewhere in the results section of the experimental write-up.)
Current / A Number
of coils
First reading Second reading Third reading
force / N
force / N
force / N
Mean (average)
reading force / N
0
5
10
15
20
(Number of coils could be 0, 4, 8, 12, 16, 20.)
8
p.d. V / V First current Second current Third current Mean (average) Resistance
reading I / A reading I / A
reading I / A current reading
V
R = __
​​  ​​ / Ω
I/A
I
1.0
0.16
0.17
0.17
0.17
5.9
2.0
0.34
0.36
0.34
0.35
5.7
3.0
0.54
0.55
0.54
0.54
5.6
The best value of the resistance is the mean value of
5.9 + 5.7 + 5.6
_____________
R =
​​
​​ = 5.7 Ω.
3
3
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9
Background count ____________
Net activity
​​      ​​
​​
​​ ______________
​​
counts / minute
counts / minute
Time /
hours
Activity
____________
​​
1
670
30
640
2
340
30
310
3
178
30
148
4
98
30
68
5
65
30
35
counts / minute
10 a
Velocity of A ______________
Velocity of B ______________
Mass of A ___________
Momentum of A _________
Mass of B ___________
Momentum of B __________________
Momentum of A + B
​ ​__________
​
​
​​
​​ ​​
​​ ​​
​​
​​
​​
​​
​​ ​​
​​
m/s
m/s
kg m/s
kg
kg m/s
kg m/s
kg
0.65
−2.30
–1.50
0.65
2.40
1.56
+0.06
0.65
−1.80
–1.17
0.65
1.70
1.11
–0.06
b
Within the tolerances of the readings taken: total momentum before the collision = total
momentum after the collision (= 0).
11 D (This statement is related to the timing
of the release of the ball. When the ruler is
not vertical the height is too large. Parallax
errors are caused when the eye is not directly
opposite the reading. Balls bounce very
quickly, making it difficult to judge precisely
the height reached. Therefore, in these
circumstances, the metre rule can only be used
to measure to the nearest centimetre.)
12 The half-life, the time taken for half of
the material to decay, should be consistent
throughout the curve. For the solid curve, the
half-life can be estimated as follows:
• time taken for 100 counts/minute to
become 50 is 24.5 − 18.5 = 6 hours
•
Possible reason: First readings should
always be treated with caution, as often the
experimenters have not practised taking
readings and so errors occur. This is especially
noticeable when timing is involved.
13 The first reading is anomalous and can be
ignored. No one has ever run 100 m in 8
seconds. It is likely that the student was late
pressing the start button on the stopwatch.
The mean (or average) value of the other
readings is 10.8 s.
14 B
Exam-style questions
1
• time taken for 25 counts/minute to become
12.5 is 36 − 30 = 6 hours
The mean (average) half-life is 6 hours.
Using the curve through point A, the time
taken for the count rate to change from 250
counts/minute to 125 counts/minute is 16 − 6
= 10 hours.
Using the curve through point B, the time
taken for the count rate to change from 200 to
100 is 18.5 – 12.0 = 6.5 hours.
Therefore, the curve through B is the correct
curve and point A is an outlier.
4
a
2
[1]
[1]
[1]
[1]
i
Current flowing through R
ii
p.d. across R
iii
Directly proportional
b
i
ii
Voltmeter
Voltmeter connected across R and
[1]
correct symbol for voltmeter
c
a
A2
i
Metre rule or measuring tape
time taken for 50 counts/minute to become
25 is 30 − 24.5 = 5.5 hours
b
ii
Stopwatch
i
t or time, cm or centimetre
ii
2, 4, 6, 8, 10
[1]
[1]
[1]
[1]
[1]
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
4
Chapter 3
60
Questions
50
Angle of refraction / °
a
Resistance / Ω
1
40
30
20
10
Distance of LDR from light source/ m
The distance from the light source is
the variable you are changing and so is
the independent variable and therefore
plotted on the x-axis. The resistance is the
dependent variable and hence plotted on
the y-axis. Units need to be added to each
variable.
A. Time is the independent variable.
2
3
120
0
20
40
60
80
Angle of incidence / °
100
4.0
2.0
0
10
20
–2.0
30
40 50
Time / ms
60
70
80
–4.0
100
6
80
(Small crosses or dots correctly plotted, using
a sharp pencil.)
Graph of temperature against time showing
cooling curve of 100 cm3 of water
60
100
40
90
20
0
0
2
4
6
8
10
Time / minutes
12
14
Temperature / °C
Temperature / °C
0
5
Voltage / volts
b
80
70
60
50
40
30
20
5
0
2
4
6
8 10 12 14 16 18 20
Time / minutes
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
8
Examples:
Should the line go through the origin?
Do the data points seem to form a curve or a
straight line?
Are there any outliers that I should ignore?
If I draw the line or curve like this, are there as
many points below the line as above it?
Are the points that are above and below the
line roughly evenly spaced along the line?
Am I sure this is the best estimate of the
trend?
a
B. (Although the curve does not touch
every data point this is the best estimate.
In A the drawn line is not smooth. Unless
there is a reason to believe otherwise, in
physics it is best to assume that changes
are gradual and curves should be
smoothly drawn. In C there is no evidence
for the rise at the end of the curve. In
D the broad line is wrong and leads to
uncertainty and so to inaccuracies.)
b
9
In either case, the point should be ignored
when determining the position of the
best-fit line.
10 (The line should be a curve, smoothly drawn,
with equal number of points above and below
the curve.)
Graph of acceleration against
mass for a truck
25.0
20.0
Acceleration
m / s2
7
15.0
10.0
5.0
0
0
1.0
2.0
3.0
4.0
5.0
6.0
Mass / kg
C. In this region (low volume, high
pressure) the data is changing rapidly as
shown by the steeply falling graph. More
readings would improve the accuracy of
the data plot.
11 a, b
Graph of period against length for a pendulum
2.50
a
2.00
Period / s
6.0
Extension / cm
5.0
4.0
1.50
1.00
0.50
3.0
0
2.0
0
0.5
1.0
Length of pendulum / m
1.0
1.5
12
0
1
2
3
4
Load / N
5
6
7
Yes, you need a ruler. (The graph is a
straight line and should pass through the
origin.)
b
The final point shows the spring has been
stretched beyond its elastic limit.
0.2
Amplitude / V
0
0.1
0
0
20
40
60
80
Time / ms
100 120
–0.1
–0.2
Or, the final point could be an outlier
and some repeat data should be taken to
check.
6
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13
b
40
Axes labelled [1]; suitable scale selected
[1]; all the points plotted up to half of the
small square [1]; best-fit line drawn [1].
10
30
8
20
Voltage / V
Orbital velocity km / s
50
10
0
0
40
80
120
160
200
240
6
4
2
280
Distance from the Sun / × 10 km
6
0
Exam-style questions
1
a
b
c
Axes labelled [1] suitable scale selected
graph starts at 50 or not 0 [1]; all the points
plotted up to half of the small square [1]; a
best-fit curve drawn [1].
Temperature/°C
d
[1]
Temperature [1]
x-axis: 10 small divisions = 30 seconds [1]
y-axis: 20 small divisions = 10
[1]
Time 3
b
4 g
3
4
Current / A
5
6
[1]
[1]
Chapter 4
Questions
2.9 s (allow 2.8 s or 3.0 s; this is the point at
which the graph becomes linear)
75
2
12 m/s
3
24.9 mph (accept 25 mph)
4
3800 Pa
5
The recession velocity is directly proportional
to the galaxy’s distance from Earth. (When the
recession velocity increases evenly, so does the
distance and the line goes through the origin.)
6
C
70
65
60
0
30
60
90
120
150
7
Time/s
7
2 cm3
2
1
50
a
a
1
80
55
2
0
[1]
Table completed
I/A
V/V
Resistance,
V
R = ​​ __​​ / Ω
I
1.0
2.0
2.0
2.0
4.0
2.0
3.0
6.0
2.0
4.0
7.8
1.95
5.0
9.6
1.92
8
(8 − 0 ) × 10−3
Gradient, m = _____________
​​
​​ A/V
2−0
8 × ​10​​  −3​
​​ A/V
= ​​  _______
2
= 4 × 10−3 A/V
b Intercept, c = 0 (the line goes through the
origin)
c I = 4 × 10−3V, where I is in mA and V is
in V.
300 − 100
​​ m/s
a Gradient, m = _________
​​
60 − 0
200
= ​​ ____ ​​ m/s
60
= 3.3 m/s
Intercept c = 100 m
a
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
b
c
9
a
b
Equation, using y = mx + c, is d = 3.3t +
100, where d is in m and t is in s
The value of the gradient tells us that the
object was travelling at a constant speed
of was 3.3 m/s.
The value of the intercept tells us the
object was 100 m from the start position
when timing started.
The new station needs to be 6.5 km from
station A.
1
The faster train is Q; it covers 20 km in __
​​   ​​
2
an hour
(speed 40 km/h), whereas train P covers
20 km in 1 hour (speed 20 km/h).
10 52 N
11
B (When braking the speed decreases slowly
at the beginning and then very quickly.)
ii
D (The graph shows that the speed is zero
at the start, as the car is parked. The final
speed is a horizontal line indicating a
constant speed has been reached.)
iii
A (Uniform deceleration means the speed
goes down evenly.)
iv
C (The horizontal line going through the
velocity axis shows at the start the rocket
is travelling at a constant speed. The
upward slope indicates acceleration.)
15 aA seen anywhere less than 7 mA and less
than 0.8 V.
b
80
bridge 2
Speed
m/s
60
40
0
0
1
2
3
Time / s
4
5
= 200 m
1
Area of triangle = __
​​   ​​ × 5 s × (80 − 40) m/s
2
1
= ​​ __ ​​ × 5 s × 40 m/s
2
= 100 m
Total distance between bridges = 200 m + 100 m
= 300 m
12 ‘Accelerates uniformly’ means that the line
on the speed–time graph is straight with a
constant gradient.
Distance travelled = area under line
1
Distance travelled = ​​ __ ​​× base × height
2
1
100 m = ​​ __ ​​ × 20 s × final speed
2
100 m = 10 s × final speed
Final speed = 10 m/s
b
c
increases, decreases; increases, increases
b
Yes, there is sufficient evidence because
graph 2 has a straight line which goes
through the origin.
c
Area of rectangle = 40 m/s × 5 s
13 a
B seen anywhere above 4 V and above
18 mA.
16 a
bridge 1
20
8
14 i
Power = 2700 × length, or P = 2700l,
where power P is in watts and length l is
in metres
1350 W (accept 1300 to 1400 W)
Any straight line with a lower gradient.
The gradient k of graph 2 is found by
taking two convenient numbers as far
apart as possible.
change in y
​​
Gradient
=___________
​​
change in x
7.8 − 2.6
= ​​ ________ ​​ Ω mm2
30 − 10
= 0.26 Ω mm2
Therefore by comparing to y = mx + c:
resistance = 0.26/cross-sectional area
where resistance is in Ω and crosssectional area is in mm2.
17 B
18 a
b
c
d
2.2 m
2.4 m
Point labelled F near 9 hours or 21 hours,
where the line is crossing the time axis
Point labelled S at any peak or trough
19 aA. The outside air has the biggest
temperature range.
b C. High thermal capacity materials either
limit the temperature range or cause the
biggest time lag. (Accept either.)
c 44° C – –6 °C = 50 °C
20 i
ii
75.5 TWh × (47.5/100) = 35.9 TWh to 3 s.f.
75.5 TWh × (23.6/100) = 17.8 TWh to 3 s.f.
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
21
c
Material
Density / Mass / g = density / g/cm3
g/cm3
× volume / cm3
mercury
13.6
4760
Acceleration = gradient of the speed–
time graph
[1]
(20
−
0)
m/s
Acceleration = __________
​​
[1]
​​
(10 − 0) s
= 2
[1]
copper
9.0
3150
iron
8.0
2800
Unit: m/s2
aluminium
2.6
910
d
22 a
Useful energy = total input energy −
wasted energy
Useful energy = 100 − 70 = 30 J
30
b Efficiency = ____
​​
​​× 100% = 30%
100
23 Wasted energy = 3400 − 600 = 2800 J
Exam-style questions
Distance between 15 and 20 s = area of
1
the triangle = __
​​   ​​(20 − 15) × 20 = 50 m [1]
2
Distance travelled during journey
= 100 + 100 + 50 = 250 and unit: m [1]
2
[1]
d
Extrapolate the graph [1] until it reaches
(50,25), so distance moved is 25 cm
a
b
50 °C
Yes, rate of cooling is higher for first
50 seconds compared to the last 50
seconds.
The gradient of the graph shows the
rate of cooling.
The gradient is greater for the first
50 seconds of the experiment. [1]
i
3 cm
ii
2 cm
[1]
[1]
[1]
c
3
a
b
c
4
9
5
aDistance moved is directly proportional to
the angle an inclined plane makes with the
floor, [1]; because the graph is a straight
line passing through the origin. [1]
distance moved
b Gradient = ______________
​​
[1]
​​
angle
(10 − 0) cm
c Gradient = __________
​​
[1]
​​
(20 − 0)°
Gradient = 0.5
[1]
Unit: cm/° a
20 m/s
b
10 s
[1]
[1]
[1]
Chapter 5
Questions
1
2
distance
Speed = ________
​​
​​
time
165 m × 2
​​
= _________
​​
1.0 s
= 330 m/s
a
[1]
[1]
Volume = 2 cm × 2 cm × 4 cm = 16 cm3
mass
Density = ​​ _______ ​​
volume
320 g
​​
= ______
​​
16 cm3
= 20 g/cm3
[1]
[1]
[1]
Total energy input = 1500 + 500 = 2000 J
1500 J
​​× 100% b Efficiency = ______
​​
2000 J
= 75%
b
3
a
Sound A
On the graph for sound A, a waveform
drawn with amplitude 4 cm [1]; with the
same time period as sound A. [1]
a
Distance travelled = area under the graph
[1]
Distance between 0 and 10 s = area of
1
triangle = __
​​   ​​× 10(20) = 100 m
[1]
2
Distance between 10 and 15 s = area of
the rectangle = (15 − 10) × 20 = 100 m [1]
24 Total output energy must be equal to total
input energy so at least one of the quantities is
wrong.
1
[1]
The object will sink, because it is denser
than water.
1
Kinetic energy = Ek = __
​​   ​​ mv2
2
1
= __
​​   ​​(6000 kg × 15 m/s × 15 m/s)
2
= 675 000 J
= 6.8 × 105 J to 2 s.f.
b
2.7 × 106 J (When the speed is doubled,
the kinetic energy is quadrupled.)
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CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
4
Percentage
5 m as a percentage ___
5
​​   ​​× 100 = 25%
of 20 m
20
Decimal
0.25
10 minutes as a
percentage of
5 minutes
10
​​ ___ ​​× 100 = 20%
50
650 cm3 of a liquid
650
_____
​​
​​× 100 = 52% 0.52
as a percentage of
1250 cm3 of liquid
0.2
1250
55 g as a percentage ____
55
​​× 100 = 6.9% 0.069
​​
of 800 g
800
5
6
7
75%
Find the % of useful energy first:
45
____
​​× 100 = 25%
​​
180
Subtract this value from 100% to find the
wasted percentage: 100% − 25% = 75%
The weight is the downward force: W = mg
W = 55 kg × 9.8 N/kg = 539 N
F
p = ​​ __ ​​
A
539N
p = ______
​​
​​
2.8 m2
p = 192.5 or 193 Pa (1 N/m2 = 1 Pa)
sin i
​​
10 Refractive index, n = ____
​​
sin r
For diamond, n = _______
​​  sin 20.0 ​​ = 2.398
sin 8.2
sin 10
If the crystal is diamond, sin A = ​​ _____​​
2.398
sin A = 0.0724
A = sin–1 0.0724
A = 4.15°
A = 4.2° to 2 s.f.
11 a
b W = 90 kg × 0.6 N/kg
= 54 N
c
750 MW
​​ × 100 = 29 %
​​ ___________
V
I
12 V
= _________
​​
​​
0.6 × 10−3
= 20 000 Ω or 20 kΩ
The information is correct.
13 Possible diagrams:
Acceleration
change in velocity
=
time taken
9
Vp N p
​​ ___ ​​= ​​ ___ ​​
Vs Ns
25 000V 5000
​​ ________ ​​ = _____
​​
​​
140 000V
10
Ek = 1 mv 2
2
ΔEp = mgh
Latent heat
ΔE = mL
Momentum
p = mv
Force
F = ma
Impulse
FΔt = Δ(mv)
Weight
W = mg
Pressure
p= F
A
Force
Moment
= force × ⊥ar distance
Density
ρ = mv
Work done = Fd
Spring constant
k = Fx
14 a
Ns
140
000
× 5000
Ns = _____________
​​ = 28 000
​​
25 000
Weight
W = mg
Change in energy
ΔE = mcΔT
750 MW × 100
​​ = 2586
Power input = ​​ _____________
29
= 2590 MW to 3 s.f.
Orbital period, T = 27.5 days
= 27.5 × 24 × 60 × 60
= 2 376 000 s
2πr 2π(3.82 × 108)
​​
Orbital speed of moon = ____
​​   ​​ =​​ _____________
T
2 376 000
= 1.01 × 103 m/s
Force
F = ma
Mass
Power input
8
1450 N = 90 kg × g
g = 16 N/kg
12 R = ​​  __ ​​
a
P = IV = 25 000 A × 30 000 V
= 750 000 000 W or 750 MW
Power
output
b​​ ____________
​​× 100 = efficiency as a
Power input
percentage
W = mg
= 90 kg × 9.8 N/kg
= 882 N
Impulse
FΔt = Δ(mv)
P = IV
0.5 × 10−4 W = I × 1.5 V
5.0 × 10−5 ​​
W
I = ​​ ___________
1.5 V
= 3.3 × 10−5 A
Cambridge IGCSE™ Physics: Maths Skills Workbook – Hamilton © Cambridge University Press 2022
CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
V
R = ​​  __​​
I
b
18 a
1.5 V
​​
= ​​ ____________
3.3 × 10−5 A
= 4.5 × 104 Ω
ΔE
P = ___
​​  t ​​
c
∆E
5.0 × 105 W = _________
​​
​​
1.6 × 10−7
∆E = 5.0 × 105 W × 1.6 × 10−7 s
= 8.0 × 10−12 J
1
1
1
15​
​ ___ ​​ = ​​ ___ ​​ + ​​ ___ ​​
R12
R1
R2
1  ​​ + ​​ ____
1  ​​ = ​​ ___
1  ​​
​​ ___
R12 5 Ω 10 Ω
3  ​​
1  ​​ = ​​ ____
​​ ___
R12 50 Ω
​​
R12 = ____
​​  10 Ω
3
R12 = 3.3 Ω
R3
b
= 21 kg × v1+2
145
v1+2 = ____
​​   ​​ m/s
21
v1+2 = 6.9 m/s
They move in the direction of the 10 kg mass.
19 F∆t = ∆(mv)
F × 4 s = 1.8 × 10−3 kg × 2.0 m/s
F = 9.0 × 10−4 N
KE lost = thermal energy gained,
1
so ​2 × ​ __ ​ m ​v​​  2​​ = mc∆T
2
v2 = c∆T and ∆T = 327 − 23 = 304
__________
v=√
​​ 130 × 304 ​​ m/s
v = 199 m/s
R4
1  ​​ + ​​ ___
1  ​​ = ​​ ____
1  ​​
​​ ___
R34 20 Ω 5 Ω
1
5  ​​
​​ ___ ​​ = ​​ ____
R34 20 Ω
R34 = 4 Ω
RT = R12 + R34 = 3.3 Ω + 4 Ω
= 7.3 Ω
21 FΔt = Δ(mv)
For 20 kg mass: F × 5 s = 30 Ns
So F = 6 N
16 D
V
RT = ​​  __ ​​
I
36 V
​​
= __________
​​
4 × ​10​​  −3​A
= 9000 Ω
1
1
1
​​ __ ​  = ​ ___ ​ + ​ __ ​​
​RT​  ​​
​RA​  ​​ ​RB​  ​​
1  ​ = ________
​  1  ​​
​  1  ​ + ___
​​ ______
90 000 Ω ​RB​ ​
9000 Ω
1  ​​ = _______
​​ ___
​​  1  ​​ − ________
​​  1  ​​ = ________
​​  9  ​​ = ________
​​  1  ​​
RB 9000 Ω 90 000 Ω 90 000 Ω 10 000 Ω
RB = 10 kΩ
17 Anticlockwise moment = clockwise moment
10 000 × 5 = force × 25
Force = 2000 N
This is the maximum weight that can be lifted
(because the balancing weight is at the end of
the jib).
11
10 kg × 20 m/s + 11 kg × −5 m/s
20 199 m/s
Two bullets each have the same kinetic energy
1
​​   ​​mv2
Total KE = Ek = 2 × __
2
Thermal energy = mc∆T
1
1
1
​​ ___ ​​ = ​​ ___ ​​ + ​​ ___ ​​
R34
Momentum before = momentum after
m1u1 + m2u2 = (m1 + m2)v1+2
For 10 kg mass, Δ(mv) = 6 N × 35 s = 210 Ns
218
4
22
​222
86​Rn ​→ ​ 84​ Po ​+ ​ 2​ ​  He​
14
14
0
23
​ 6​C ​→ ​ 7​N ​+ ​−1​e​
94
24 a​
​50​​S
b​
​99
51​​T
94
4
c​
​98
52​P ​→​50​S ​+​2​He​
99
d ​210
83​Bi ​→​51​T
99
​210
99​Po ​→​52​Q
25 C
26 The time between 7.00 a.m. and 7 p.m. is
6 half-lives. Each half-life going back from
7.00 p.m. to 7.00 a.m. doubles the count rate.
10 doubled 6 times is 640 counts/s.
27 C (Reduced by 50% after 7.5 h; reduced
by 50% of this, or by a further 25% of the
original, after 2 × 7.5 = 15 h)
Cambridge IGCSE™ Physics: Maths Skills Workbook – Hamilton © Cambridge University Press 2022
CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
Exam-style questions
1
a
W = m × g
= 0.1 × 9.8 = 0.98 N
b
Gain in kinetic energy of water = mgh [1]
[1]
[1]
= 0.1 × 9.8 × 0.1
= 0.098 J
c
2
a
Energy transferred in 1 second
= 0.098 and unit: J
5 × 10−3
​​ × 100%
Efficiency = ________
​​
0.098
= 5.1 %
W = 2 × 9.8 =19.6
Unit: N
b
Work = 19.6 × 12 = 235.2
Unit: J
3
c
235.2 and unit: J
a
P = V × I = 12 × 10 = 120
4
a
b
c
[1]
[1]
[1]
[1]
[1]
[1]
Clockwise moment = F1 × 0.5
[1]
[1]
F2 × 0.45 m = F1 × 0.5 m
or
F2 × 0.45 m = (15 N − F2) × 0.5 m
0.5 m × (15 N − 0.5F2)
___________________
​​
F2 =
​​
0.45 m
0.5 mF2
7.5 Nm _______
F2 = _______
​​ − ​​
​​
​​
0.45 m
0.45 m
F2 = 16.7 N − 1.11F2
[1]
2.11F2 = 16.7 N
5
C (Diagonals are straight lines joining two
opposite corners of a square, rectangle, or
other straight-sided shape.)
2
The picture shows a sideways version of the
wheel rolling. The dotted circle shows the
wheel after the first revolution. This diagram
is read from left to right. The total distance
for two circumferences is 12 − 1 cm = 11 cm.
Therefore one circumference is 5.5 cm.
3
B
A: area of contact = 2.0 cm × 4.0 cm
= 8.0 cm2
p = __
​​  F  ​​
A
240
p = ____
​​   ​​
8
p = 30 g/cm2
B: area of contact = 3.14 × 4.0 cm2
= 12.6 cm2
240
​​
p = ____
​​
12.6
p = 19 g/cm2
4
Volume of a cylinder = πr2h
Volume of space = 3.14 × (1.8)2 × 1.5 = 15.3 m3
5
Normal line added at 90° to mirror, labelled N;
angle of reflection drawn at 20°, correct to ±1°.
6
B
7
Physical quantity
Vector or scalar
Time taken for a ball to
bounce
Scalar
Distance travelled by a car
on a journey
Scalar
F2 = 7.9 N
15 = F1 + F2
[1]
The change in velocity of a
person on a zip wire
Vector
F1 = 7.1 N
[1]
a
b
v
v = H × d or d = ___
​​   ​​
[1]
The pressure inside a bicycle
inner tube
Scalar
17 × 106 × 9.5 × 1015 = 1.6 × 1019 km [1]
The acceleration of free fall
Vector
[1]
The kinetic energy store of
an aeroplane
Scalar
0
408 000
2.2 × 10
= 1.9 × 1020 km
d = __________
​​
​​
−18
12
1
[1]
In equilibrium,
F1 + F2 = 15 N or F1 = 15 N − F2
Anticlockwise moment = F2 × 0.45
Questions
[1]
Unit: coulomb/C
Q = I × t = 10 × 1 = 10
Chapter 6
[1]
[1]
[1]
[1]
[1]
Unit: watt/W
c
[1]
[1]
H0
[1]
Cambridge IGCSE™ Physics: Maths Skills Workbook – Hamilton © Cambridge University Press 2022
CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
Exam-style questions
1
a
Force, velocity, weight, acceleration: all
four underlined [2]; any two or three
underlined [1]
b
[1]
3400 ​​
d Density of oil = ​​ _____
4000
= 0.85
[1]
Unit: g/cm3
[1]
5
resultant
3
x
40 N
[1]
[1]
ray draw correctly
2
angle labelled r correctly
53.13°
1
A
b
B
air
0
1
2
3
4
Resultant force = 50 N at an angle
53.13° to the east
2
r
water
4
0
angle of refraction
air
C
5
a
[1]
a
A normal drawn at 90° to side AB
b
Angle measured 41° ± 1°
c
Refracted ray drawn, see diagram
x r
water
[1]
[1]
[1]
[1]
41°
F
angle of refraction
Applying more than
one skill
1
2
C
a
[1]
F
p = ​​  __​​
A
F = mg
F = (4 kg + 41 kg) × 9.8 N/kg
15°
F = 441 N
Area = 2.5 cm2 × 3
3
a
Q
b
Surface area = length × breadth
= 0.04 × 0.1 = 0.004 m2
c
4
13
F = 2.5 × 9.8 = 24.5 N
24.5 N
F
p = __
​​   ​​ = ________
​​
​​ = 6125 N/m2
A 0.004 m2
p = 6100 N/m2 to 2 s.f.
[1]
[1]
[1]
[1]
A
441 N  ​​
= ​​ _______
7.5 cm2
= 58.8 N/cm2
[1]
b
[1]
[1]
4000 g
[1]
b Mass of oil = 4000 − 600 = 3400 g
[1]
c Volume of cuboid = l × b × h
[1]
= 20.0 × 10.0 × 20.0 [1]
= 4000 cm2
[1]
a
= 7.5 cm2
F
p = ​​  __​​
New area = 2 × 2.5 cm2
= 5 cm2
New pressure = _______
​​  441 N2 ​​
5.0 cm
= 88.2 N/cm2
Mass = 39.2/9.8 N/kg = 4.0 kg
3
[1]
[1]
[1]
A is correct.
When balanced:
anticlockwise moment = clockwise moment
Cambridge IGCSE™ Physics: Maths Skills Workbook – Hamilton © Cambridge University Press 2022
CAMBRIDGE IGCSE™ PHYSICS: MATHS SKILLS WORKBOOK ANSWERS
Anticlockwise moment
= force × perpendicular distance from pivot
= 30 N × 0.4 m
= 12 Nm
Clockwise moment = 15 N × 0.2 m
= 3 Nm
Therefore an additional moment of 9 Nm is
needed in the clockwise direction to balance.
Answer A gives a 9 Nm anticlockwise moment
(20 N × 0.45 m)
4
5
14
c
d
E = P × t = 60 × 120 = 7.2 kJ
a
b
Work = 100 × 0.5 × 9.8 × 10 = 4900 J
Energy = 0.2 × 1000 × 40 = 8000 J
c
a, cNormal, incident ray and reflected ray
drawn and labelled.
b Angle of incidence = angle of reflection =
28°
8000 J
a
d
b
6
7
V
Total current, I = __
​​   ​​ = ____
​​  20  ​​ = 33.3 mA
R 600
So current through one resistor
= 33.3 ÷ 2 = 17 mA
Q = I × t = 17 × 10−3 × 10 = 170 mC
b
Total initial momentum
= (1500 × 5.0) + (1500 × 0) = 7500 Ns
After collision both wagons stick together
and move with same velocity.
Total initial momentum = T
otal final
momentum
7500 = (1500 + 1500)v
v = 2.5 m/s
1
__
c Kinetic energy = Ek = ​​   ​​ mv2
2
Total initial KE = 37.5 kJ
Total final KE = 18.75 kJ
Energy lost = 18.75 kJ
1
1
1
​​
​​ + ​​ ________​​
a​​ __ ​​ = ________
R 1.2 × 103 1.2 × 103
R = 600 Ω
4900 J
3100 J
8
9
useful energy output
Efficiency = __________________
​​
​​ × 100%
total energy input
= (4900/8000) × 100% = 61.3%
a
83, 210, α or He
b
Half-life = 20 minutes and work shown on
graph using horizontal line and vertical
line.
c
A curve drawn below the original curve
and starting at (0, 90).
a
2.54 × 106 years
2πr 2 × π × 2.28 × 108
v = ____
​​   ​​ = ​​ _________________
​​ = 24 km/s.
T
687 × 24 × 60 × 60
distance __________
2.28 × 1011
Time = ​​ ________
​​ = 760 s
​​ = ​​
speed
3.0 × 108
b
c
= 12.7 minutes
Cambridge IGCSE™ Physics: Maths Skills Workbook – Hamilton © Cambridge University Press 2022