CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
Exam-style questions and sample answers have been written by the authors. In examinations, the way marks are awarded
may be different.
Coursebook answers
Chapter 7
b
Self-assessment questions
1
volume of cube = 3.0 × 3.0 × 3.0 = 27 cm3
density =
2
3
4
5
6
7
1
mass
240
=
= 8.89 g cm−3 = 8890
volume
27
kg m−3
4
4
volume of sphere, V = pr3 = × π × (0.15)3
3
3
= 0.0141 m3
mass
rearrange: density =
volume
so, mass = density × volume = 7850 × 0.0141 =
111 kg
80
F
pressure, p =
=
= 20 kPa
4 × 0.0010
A
Estimate weight = 600 N, area of feet = 500
cm2 = 0.05 m2
F
600
so, pressure p =
=
= 12 kPa
A 0.05
Pressure at depth 0.8 m is p = ρgh1 = 1000 ×
9.81 × 0.8 = 7.85 × 103 Pa
Pressure at depth 2.4 m is p = ρgh2 = 1000 ×
9.81 × 2.4 = 2.35 × 104 Pa
maximum total pressure, p = patm + pwater
= 1.01 × 105 + 2.35 × 104 = 1.25 × 105 Pa
p
Rearrange p = ρgh to give height h =
ρ
g
1.01× 105
=
= 7980 m ≈ 8000 m
1.29 × 9.81
This figure is too small because it assumes the
density of the air is constant. In fact, density
decreases with height. You may have sensibly
assumed a smaller value for the density of air,
say half the value quoted.
a
The ball displaces a lot of water and the
upthrust is larger than its weight.
8
I nitially, with the water inside the ballast
tanks, the upthrust was equal to the weight
of the submarine plus the water inside the
tanks. When the water is pushed out of the
tanks the upthrust is still the same but the
submarine without the water weighs less.
The upthrust is now larger than the weight.
ass of extra water displaced = 15 × 1200
m
= 18 000 kg
extra volume displaced = 18 000/1000 = 18 m3
extra depth = 18/750 = 0.024 m
9
easure the sides of the cube with the
M
micrometer. Multiply the three sides together
to obtain the volume of the cube. Use mass
= density of water × volume of cube to find
the mass of water displaced. The weight =
mass × g. Use the newton-meter to measure
the weight of the cube in air and when fully
submerged in water. The difference is the
upthrust. This should equal the weight of the
water displaced.
10 mass of hydrogen and fabric = 3000 × 0.09
+100 = 370 kg
upthrust in air = 3000 × 1.2 × 9.81 = 35300 N
g reatest mass it can lift = 35300 ÷ 9.81 − 370 =
3200 kg
11 a
B
b
C
12 a
Spring D has the greatest value of force
constant (the graph has the steepest
gradient).
b
pring A is the least stiff (it extends the
S
most for each unit of force applied).
c
pring C does not obey Hooke’s law:
S
there is no section of the graph that forms
a straight line.
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020
CAMBRIDGE INTERNATIONAL AS & A LEVEL PHYSICS: COURSEBOOK
13 Metals from stiffest to least stiff:
Most stiff
Metal
oung modulus /
Y
GPa
steel
210
iron (wrought) 200
Least stiff
copper
130
brass
90−110
aluminium
70
tin
50
lead
18
14 Stiffest non-metal is glass (Young modulus =
70−80 GPa)
15 For material A:
Young modulus, EA =
1.5 × 10 Pa = 15 GPa
σ 15 × 106
stress
=
=
=
ε
strain
0.001
10
For material B:
6
stress = σ = 12 × 10 = 5.0
Young modulus, EB = strain
ε
0.0024
× 109 Pa = 5.0 GPa
force
16 stress =
cross-sectional area
50
=
= 1.0 × 108 Pa
0.5 × 10 −6
(Remember that 0.5 mm2 = 0.5 × 10−6 m2)
extension
0.1
strain =
=
200.0
original length
strain = 5.0 × 10−4 (0.05%)
1.0 × 108
stress
Young modulus =
=
strain 5.0 × 10 −4
= 2.0 × 1011 Pa
stress
17 Young modulus =
strain
stress
Rearrange so strain =
Young modulus
x
Then insert formulae for stress and strain
L
F
πd 2
=
and cross-sectional area, A =
A× E
4
4 FL
This gives extension = 2
πd × E
4 × 10 × 1.00
=
2
π × ( 0.001) × 130 × 109
force
4F
=
cross-sectional area πd 2
4 × 1.00
=
= 8.0 × 106 Pa
π × ( 0.0004 )2
extension
0.001
strain =
=
original length 0.800
18 stress =
= 1.25 × 10−3 (at most)
8.0 × 106
stress
Young modulus =
=
strain 1.25 × 10 −3
= 6.4 × 109 Pa (but could be more, because
extension may be less than 1 mm)
stress 150 × 106
19 a
Young modulus =
=
=
0.003
strain
50 GPa
stress 100 × 106
b Young modulus =
=
=
0.001
strain
100 GPa
Note that the Young modulus is only
found for the straight portion of the
stress−strain graph.
stress 100 × 106
c Young modulus =
=
= 25
0.004
strain
GPa
20 elastic potential energy, E = Fx = 1 × 12 ×
1
2
0.18 = 1.08 J ≈ 1.1 J
2
he rubber band is assumed to obey Hooke’s
T
law; hence, the answer is an estimate.
21 elastic potential energy, E = 12 Fx = 12 kx2
= 12 × 4800 × (0.0020)2 = 9.6 × 10−3 J
22 a
A has greater stiffness (less extension per
unit force).
b
requires greater force to break (line
A
continues to higher force value).
c
requires greater amount of work done
B
to break (larger area under graph).
= 9.796 × 10−5 m
≈ 9.8 × 10−5 m
2
Cambridge International AS & A Level Physics – Sang, Jones, Chadha & Woodside
© Cambridge University Press 2020