8 Marking scheme Worksheet (A2.pdf

AS and A Level Physics Original material © Cambridge University Press 2010 1
18 Marking scheme: Worksheet (A2)
1 a θ =
360
30
× 2π =
6
π
≈ 0.52 rad [1]
b θ =
360
210
× 2π ≈ 3.7 rad [1]
c θ =
360
05
.
0
× 2π ≈ 8.7 × 10−4
rad [1]
2 a θ =
π
2
0
.
1
× 360 = 57.3° ≈ 57° [1]
b θ =
π
2
0
.
4
× 360 ≈ 230° [1]
c θ =
π
2
15
.
0
× 360 ≈ 8.6° [1]
3 a 88 days is equivalent to 2π radians.
θ =
88
44
× 2π = π rad [1]
b θ =
88
1
× 2π ≈ 0.071 rad (4.1°) [1]
4 a Friction between the tyres and the road. [1]
b Gravitational force acting on the planet due to the Sun. [1]
c Electrical force acting on the electron due to the positive nucleus. [1]
d The (inward) contact force between the clothes and the rotating drum. [1]
5 a ω =
r
v
=
20000
150
[1]
ω = 7.5 × 105
rad s−1
[1]
b a =
r
v2
[1]
a =
000
20
1502
[1]
a = 1.125 m s–2
≈ 1.1 m s−2
[1]
c F = ma = 80 × 1.125 [1]
F = 90 N [1]
6 a i Time =
10
2
.
8
= 0.82 s [1]
ii Distance = circumference of circle = 2π × 0.80 = 5.03 m ≈ 5.0 m [1]
iii speed =
time
distance
=
82
.
0
03
.
5
[1]
speed, v = 6.13 m s−1
≈ 6.1 m s−1
[1]
iv a =
r
v2
[1]
a =
80
.
0
13
.
6 2
[1]
a = 47 m s–2
[1]

v F = ma = 0.090 × 47 [1]
F ≈ 4.2 N [1]
b The tension in the string. [1]
c The stone describes a circle, therefore the angle between the velocity and the acceleration
(or centripetal force) must be 90°. [1]
7 a i speed =
time
distance
speed v =
1.6
0.12
2π
2π ×
=
T
r
[1]
v = 0.471 m s−1
≈ 0.47 m s−1
[1]
ii
r
mv
ma
F
2
=
= [1]
12
.
0
471
.
0
300
.
0 2
×
=
F [1]
frictional force ≈ 0.55 N [1]
b Frictional force = 0.7mg [1]
r
mv
mg
2
7
.
0 = [1]
12
.
0
81
.
9
7
.
0
7
.
0 ×
×
=
= gr
v [1]
speed = 0.908 m s−1
≈ 0.91 m s−1
[1]
8 a Kinetic energy at B = loss of gravitational potential energy from A to B
2
1
mv2
= mgh or v = gh
2 [1]
v = 2
.
5
81
.
9
2 ×
× =10.1 m s–1
≈ 10 m s−1
[1]
b i a =
r
v2
[1]
a =
16
1
.
10 2
[1]
a = 6.38 m s−2
≈ 6.4 m s−2
[1]
ii Net force = ma
R − mg = ma [1]
R = mg + ma = m(a + g) = 70(6.38 + 9.81) [1]
R ≈ 1.1 × 103
N [1]

9 a R cos 20° = W = 840 × 9.8 [1]
R =
°
×
20
cos
8
.
9
840
= 8760 N [1]
b R sin 20° = centripetal force =
r
mv2
[1]
r =
R
mv2
[1]
r =
°
×
20
sin
8760
32
840 2
[1]
r = 287 m ≈ 290 m [1]
10 net force =
r
mv2
net force =
80
.
0
0
.
4
120
.
0 2
×
= 2.4 N [2]
weight W of stone = mg = 0.120 × 9.81 = 1.18 N ≈ 1.2 N [1]
At the top: W + TB = 2.4 so TB = 2.4 − 1.2 = 1.2 N [1]
At the bottom: TA − W = 2.4 so TA = 2.4 + 1.2 = 3.6 N [1]
ratio =
B
A
T
T
=
2
.
1
6
.
3
= 3.0 [1]

1 Gravitational field strength at a point, g, is the force experienced per unit mass at that point. [1]
2 F = 2
r
GMm
− [1]
Therefore: G =
Mm
Fr2
→ ⎥
⎦
⎤
⎢
⎣
⎡
2
2
kg
m
N
→ [N m2
kg–2
] [1]
3 g =
m
F
F = mg = 80 × 1.6 [1]
F = 128 N ≈ 130 N (F is the ‘weight’ of the astronaut.) [1]
4 a F = 2
r
GMm
[1]
F =
( )2
14
27
27
11
10
0
5
10
7
1
10
7
1
10
67
6
−
−
−
−
×
×
×
×
×
×
.
.
.
.
[1]
F ≈ 7.7 × 10–38
N [1]
b F =
( )2
12
28
28
11
10
0
8
10
0
5
10
0
5
10
67
6
×
×
×
×
×
× −
.
.
.
.
[1]
F = 2.61 × 1021
N ≈ 2.6 × 1021
N [1]
c F = 2
2
11
0
.
2
1500
10
67
.
6 ×
× −
[1]
F = 6.00 × 10–6
N ≈ 6.00 × 10–6
N [1]
5 a g = 2
r
GM
− [1]
b The field strength obeys an inverse square law with distance (g ∝ 2
1
r
). [1]
Doubling the distance decreases the field strength by a factor of four. [1]
c ratio =
( )
( )2
2
59
5
R
GM
R
GM
[1]
ratio = 2
2
5
59
=
2
5
59
⎟
⎠
⎞
⎜
⎝
⎛
[1]
ratio ≈ 140 [1]
6 g = 2
r
GM
− [1]
g =
( )2
7
26
11
10
2
.
2
10
0
.
1
10
67
.
6
×
×
×
× −
(magnitude only) [1]
g = 13.8 N kg–1
≈ 14 N kg–1
[1]
7 g = 2
r
GM
− [1]
r2
=
g
GM
=
0
.
4
10
0
.
5
10
67
.
6 23
11
×
×
× −
= 8.34 × 1012
m2
[1]
r = 12
10
34
.
8 × ≈ 2.9 × 106
m [1]

8 a F = 2
r
GMm
− [1]
F =
( )2
10
24
11
10
9
.
3
10
0
.
6
1800
10
67
.
6
×
×
×
×
× −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
=
×
= m
10
9
.
3
2
10
8
.
7 10
10
r [1]
F = 4.74 × 10–4
N ≈ 4.7 × 10–4
N [1]
b F = 2
r
GMm
−
F =
( )2
10
23
11
10
9
.
3
10
4
.
6
1800
10
67
.
6
×
×
×
×
× −
[1]
F = 5.05 × 10–5
N ≈ 5.1 × 10–5
N [1]
c a =
m
F
(F is the net force.) [1]
a =
4 5
4.74 10 5.05 10
1800
− −
× − ×
[1]
a ≈ 2.4 × 10–7
m s–2
(towards the centre of the Earth) [1]
9 a F = 2
r
GMm
− [1]
F =
( )2
3
24
11
10
6800
10
0
.
6
5000
10
67
.
6
×
×
×
×
× −
(r = 6400 + 400 = 6800 km) [1]
F = 4.33 × 104
N ≈ 4.3 × 104
N [1]
b a =
m
F
=
5000
10
33
.
4 4
×
[1]
a = 8.66 ≈ 8.7 m s–2
[1]
c a =
r
v2
[1]
v2
= ar = 8.66 × 6800 × 103
[1]
v = 7.67 × 103
m s–1
≈ 7.7 km s–1
[1]
10 a The work done in bringing unit mass [1]
from infinity to the point [1]
b 0 J [1]
c Ep =− 6
24
11
10
4
6
10
0
6
10
67
6
×
×
×
× −
.
.
.
([1] mark only if minus sign missed) [2]
= −6.25 × 106
J [1]
d 6.25 × 106
J [1]
11 a Gravitational force on planet = 2
r
GMm
[1]
Centripetal force =
r
mv2
[1]
Equating these two forces, we have: 2
r
GMm
=
r
mv2
[1]
Therefore: v2
=
r
GM
or v =
r
GM
[1]

b v =
r
GM
= 11
30
11
10
5
.
1
10
0
.
2
10
67
.
6
×
×
×
× −
[1]
v = 2.98 × 104
m s–1
≈ 30 km s–1
[1]
12 The field strengths are the same at point P.
2
M
x
GM
=
( )2
E
x
R
GM
−
[1]
R − x = x ×
M
E
M
M
[1]
R − x = x × 81 so R − x = 9x [1]
10x = R so x =
10
R
[1]

1 a The period of an oscillator is the time for one complete oscillation. [1]
b The frequency of an oscillator is the number of oscillations completed per unit time
(or per second). [1]
2 a The gradient of a displacement against time graph is equal to velocity. [1]
The magnitude of the velocity (speed) is a maximum at 0 s or 0.4 s or 0.8 s. [1]
b For s.h.m., acceleration ∝ –displacement.
The magnitude of the acceleration is maximum when the displacement is equal to the
amplitude of the motion. [1]
The magnitude of the acceleration is a maximum at 0.2 s or 0.6 s or 1.0 s. [1]
3 a T =
12
2
.
13
[1]
T = 1.1 s [1]
b f =
T
1
=
1
.
1
1
[1]
f = 0.909 ≈ 0.91 Hz [1]
4 a Amplitude = 0.10 m [1]
b Period = 4.0 × 10–2
s [1]
c f =
T
1
=
04
.
0
1
[1]
f = 25 Hz [1]
d ω = 2πf = 2π × 25 [1]
ω = 157 rad s–1
≈ 160 rad s–1
[1]
e Maximum speed = 10
.
0
157×
=
A
ω [1]
maximum speed = 15.7 m s−1
≈ 16 m s−1
[1]
5 a Phase difference = 2π × ⎟
⎠
⎞
⎜
⎝
⎛
T
t
where T is the period and t is the time lag between the motions of the two objects.
phase difference = 2π × ⎟
⎠
⎞
⎜
⎝
⎛
T
t
= 2π × ⎟
⎠
⎞
⎜
⎝
⎛
10
5
.
2
[1]
phase difference =
2
π
≈ 1.6 rad [1]
b Phase difference = 2π × ⎟
⎠
⎞
⎜
⎝
⎛
T
t
= 2π × ⎟
⎠
⎞
⎜
⎝
⎛
10
0
.
5
[1]
phase difference = π ≈ 3.1 rad [1]
6 a A = 16 cm [1]
b ω = 2πf =
T
π
2
=
2.8
π
2
[1]
ω = 2.24 rads–1
≈ 2.2 rads–1
[1]
c a = (2πf )2
x (magnitude only) [1]
For maximum acceleration, the displacement x must be 16 cm.
a = 2
2
10
16
8
.
2
1
π
2 −
×
×
⎟
⎠
⎞
⎜
⎝
⎛
× [1]
a = 0.806 m s–2
≈ 0.81 m s–2
[1]
d Maximum speed = 16
.
0
24
.
2 ×
=
A
ω [1]
maximum speed = 0.358 m s−1
≈ 0.36 m s−1
[1]

7 a ω = 2πf =
T
π
2
=
2.0
π
2
[1]
ω = 3.14 rad s–1
≈ 3.1 rad s–1
[1]
b a = –(2πf )2
x or a = –ω2
x [1]
a = 3.142
× 3.0 × 10–2
[1]
a ≈ 0.30 m s–2
[1]
c x = A cos (2πft) = A cos (ωt) [1]
x = 3.0 × 10–2
cos (3.14 × 6.7) [1]
x ≈ –1.7 × 10–2
m [1]
8 a Gradient of x–t graph = velocity
[2]
b Gradient of v–t graph = acceleration
(for s.h.m. acceleration ∝ −displacement)
[2]
c Kinetic energy =
2
1
mv2
∝ v2
[2]
d Potential energy = total energy − kinetic energy
[2]

9 a a = −(2πf )2
x [1]
Therefore (2πf )2
= 6.4 × 105
[1]
f =
π
×
2
10
4
.
6 5
= 127 Hz ≈ 130 Hz [1]
b F = ma
Acceleration is maximum at maximum displacement, so magnitude of maximum force
is given by:
F = ma = 0.700 × (6.4 × 105
× 0.08) [1]
F = 3.58 × 104
N ≈ 3.6 × 104
N [1]
10 a According to Hooke’s law, F = –kx [1]
(The minus sign shows that the force is directed towards the equilibrium position.)
From Newton’s second law: F = ma [1]
Equating, we have: ma = –kx [1]
Hence: a = x
m
k
⎟
⎠
⎞
⎜
⎝
⎛
−
b For s.h.m. we have a = –(2πf )2
x [1]
Hence (2πf )2
=
m
k
[1]
Therefore f =
m
k
π
2
1
c f =
T
1
=
4
.
0
1
= 2.5 Hz [1]
2.5 =
850
2
1 k
π
[1]
k = (2π × 2.5)2
× 850 ≈ 2.1 × 105
N m–1
[1]

1 a The atoms in a solid are arranged in a three-dimensional structure. [1]
There are strong attractive forces between the atoms. [1]
The atoms vibrate about their equilibrium positions. [1]
b The atoms in a liquid are more disordered than those in a solid. [1]
There are still attractive electrical forces between molecules but these are weaker than
those between similar atoms in a solid. [1]
The atoms in a liquid are free to move around. [1]
c The atoms in a gas move around randomly. [1]
There are virtually no forces between the molecules (except during collisions) because they
are much further apart than similar molecules in a liquid. [1]
The atoms of a gas move at high speeds (but no faster than those in a liquid at the same
temperature). [1]
2 The atoms move faster [1]
because their mean kinetic energy increases as the temperature is increased. [1]
The atoms still have a random motion. [1]
3 a The internal energy of a substance is the sum (of the random distribution) of the kinetic
and potential energies of its particles (atoms or molecules). [1]
b There is an increase in the average kinetic energy of the aluminium atoms as they vibrate
with larger amplitudes about their equilibrium positions. [1]
The potential energy remains the same because the mean separation between the atoms
does not change significantly. [1]
Hence, the internal energy increases because there is an increase in the kinetic energy of
the atoms. [1]
c As the metal melts, the mean separation between the atoms increases. [1]
Hence, the electrical potential energy of the atoms increases. [1]
There is no change in the kinetic energy of the atoms because the temperature remains
the same. [1]
The internal energy of the metal increases because there is an increase in the electrical
potential energy of the atoms. [1]
4 Change in thermal energy = mass × specific heat capacity × change in temperature [1]
5 The specific heat capacity refers to the energy required to change the temperature of a substance. [1]
Specific latent heat of fusion is the energy required to melt a substance; there is no change in
temperature as the substance melts. [1]
6 E = mc∆θ [1]
E = 6.0 × 105
× 4200 × (24 – 21) [1]
E = 7.56 × 109
J ≈ 7.6 × 109
J [1]
7 E = mc∆θ [1]
E = 300 × 10–3
× 490 × (20 – 300) [1]
E = –4.1 × 104
J (The minus sign implies energy is released by the cooling metal.) [1]
8 E = mLf = 200 × 10−3
×3.4 × 105
[1]
= 6.8 × 104
J [1]

9 a i T = 273 + 0 = 273 K [1]
ii T = 273 + 80 = 353 K [1]
iii T = 273 – 120 = 153 K [1]
b i θ = 400 – 273 = 127 °C [1]
ii θ = 272 – 273 = –1 °C [1]
iii θ = 3 – 273 = –270 °C [1]
10 a The thermal energy E supplied and the specific heat capacity c remain constant.
The mass m is larger by a factor of 3. [1]
Since E = mc∆θ, we have:
∆θ =
mc
E
; ∆θ ∝
m
1
[1]
Therefore ∆θ =
3
15
= 5.0 °C [1]
b The thermal energy E supplied is halved but the specific heat capacity c and the mass m
remain constant. [1]
Since E = mc∆θ, we have:
∆θ =
mc
E
; ∆θ ∝ E [1]
Therefore ∆θ =
2
15
= 7.5 °C [1]
11 a Melting point = 600 °C [1]
(There is no change in temperature during change of state.)
b The lead is being heated at a steady rate and therefore the temperature also increases
at a steady rate. [1]
c The energy supplied to the lead is used to break the atomic bonds and increase the
separation between the atoms of lead (and hence their potential energy increases). [1]
d E = mc∆θ [1]
E = 200 × 10–3
× 130 × (600 – 0) [1]
E = 1.56 × 104
J ≈ 1.6 × 104
J [1]
e In a time of 300 s, 1.56 × 104
J of energy is supplied to the lead.
Rate of heating = power
power =
300
10
56
.
1 4
×
[1]
power = 52 W [1]
f Energy supplied = 52 × 100 = 5200 J [1]
Lf
2
.
0
5200
=
∆
∆
=
m
E
[1]
= 26 000 J kg−1
[1]
12 The energy supplied per second is equal to the power of the heater.
In a time of 1 s, water of mass 0.015 kg has its temperature changed from 15 °C to 42 °C. [1]
E = mc∆θ (where E is the energy supplied in 1 s) [1]
E = 0.015 × 4200 × (42 – 15) [1]
E = 1.7 × 103
J [1]
The power of the heater is therefore 1.7 kW. [1]
(You may use P = (
t
m
)c∆θ )
13 The gas does work against atmospheric pressure. [1]
Energy to do this work is taken from the internal energy of the gas. [1]

14 Heat ‘lost’ by hot water = heat ‘gained’ by cold water. [1]
0.3 × c × (90 – θ) = 0.2 × c × (θ – 10) [1]
where c is the specific heat capacity of the water and θ is the final temperature.
The actual value of c is not required, since it cancels on both sides of the equation.
Hence:
0.3 × (90 – θ) = 0.2 × (θ – 10) [1]
27 – 0.3θ = 0.2θ – 2.0 [1]
0.5θ = 29 so θ = 58 °C [1]
15 Heat ‘lost’ by metal = heat ‘gained’ by cold water [1]
0.075 × 500 × (θ – 48) = 0.2 × 4200 × (48 – 18) [1]
(θ is the initial temperature of the metal.)
θ – 48 =
500
075
.
0
30
4200
2
.
0
×
×
×
[1]
θ – 48 = 672 [1]
θ = 720 °C [1]

1 a number of atoms = number of moles × NA
number of atoms = 1.0 × 6.02 × 1023
≈ 6.0 × 1023
[1]
b Number of molecules = 3.6 × 6.02 × 1023
≈ 2.2 × 1024
[1]
c Number of atoms = 0.26 × 6.02 × 1023
≈ 1.6 × 1023
[1]
2 There are 6.02 × 1023
atoms in 4.0 g of helium. [1]
mass of atom = 23
10
02
.
6
004
.
0
×
= 6.645 × 10–27
kg ≈ 6.6 × 10–27
kg [1]
3 a There are 6.02 × 1023
atoms in 0.238 kg of uranium. [1]
mass of atom = 23
10
02
.
6
238
.
0
×
= 3.95 × 10−25
kg ≈ 4.0 × 10−25
kg [1]
b i number of moles =
uranium
of
mass
molar
uranium
of
mass
[1]
number of moles =
0.12
238
= 5.04 × 10–4
≈ 5.0 × 10–4
[1]
ii number of atoms = number of moles × NA
number of atoms = 5.04 × 10–4
× 6.02 × 1023
= 3.06 × 1020
≈ 3.1 × 1020
[1]
4 The absolute zero of temperature is –273.15 °C or 0 K. [1]
This is the lowest temperature any substance can have. [1]
At absolute zero of temperature, the substance has minimum internal energy. [1]
5 a Pressure × volume
= number of moles × universal gas constant × thermodynamic temperature [1]
b PV = nRT [1]
P =
V
nRT
=
020
.
0
293
31
.
8
0
.
1 ×
×
[1]
P = 1.22 × 105
Pa ≈ 1.2 × 105
Pa (120 kPa) [1]
6 a PV = nRT [1]
Comparing this equation with y = mx, we have:
y = PV, x = T, gradient, m = nR [1]
A graph of PV against T is a straight line through the origin.
Correct graph [1]
n =
R
gradient
[1]
b PV = nRT [1]
At a constant temperature, the product
PV is a constant. [1]
Hence a graph of PV against P is a
straight horizontal line. [1]

7 a PV = nRT [1]
n =
29
0
.
4
= 0.138 moles [1]
P =
V
nRT
=
030
.
0
)
34
273
(
31
.
8
138
.
0 +
×
×
[1]
P = 1.17 × 104
Pa ≈ 1.2 × 104
Pa (12 kPa) [1]
b
T
P
is constant when the volume of the gas is constant. [1]
The pressure is doubled, hence the absolute temperature of the gas is also doubled. [1]
Therefore:
temperature = 2 × (273 + 34) = 614 K [1]
temperature in °C = 614 – 273 = 341 °C ≈ 340 °C [1]
8 a n =
RT
PV
[1]
n =
)
13
273
(
31
.
8
10
0
.
2
10
180 2
3
−
×
×
×
× −
+
)
13
273
(
31
.
8
10
0
.
2
10
300 2
3
−
×
×
×
× −
[1]
n = 4.44 moles ≈ 4.4 moles [1]
b Total volume, V = 4.0 × 10–2
m3
, T = 273 – 13 = 260 K
P =
V
nRT
[1]
P = 2
10
0
.
4
260
31
.
8
44
.
4
−
×
×
×
[1]
P ≈ 2.4 × 105
Pa (240 kPa) [1]
9 a P =
A
F
= 3
10
6
.
1
400
−
×
[1]
P = 2.5 × 105
Pa [1]
b n =
RT
PV
[1]
n =
)
0
.
5
273
(
31
.
8
10
4
.
2
10
5
.
2 4
5
+
×
×
×
× −
[1]
n = 2.6 × 10–2
moles [1]
c i mass = number of moles × molar mass
mass = 2.6 × 10–2
× 29 = 0.754 g ≈ 0.75 g [1]
ii density =
volume
mass
density = 4
3
10
4
.
2
10
754
.
0
−
−
×
×
[1]
density = 3.14 kg m–3
≈ 3.1 kg m–3
[1]

10 Mean kinetic energy of atom ∝ absolute temperature [1]
2
2
1 mv ∝ T or v2
∝
m
T
2
[1]
Since the mass m of the atom is constant, we have: v ∝ T [1]
The temperature of 0 °C in kelvin is T = 273 K
The absolute temperature increases by a factor of
273
000
10
(= 36.6) [1]
Hence the speed will increase by a factor of
273
000
10
= 6.05 [1]
The speed of the atoms at 10 000 K = 1.3 × 6.05 ≈ 7.9 km s–1
[1]
11 a The particles have a range of speeds and travel in different directions. [1]
b i Mean kinetic energy = 5400
10
38
.
1
2
3
2
3 23
×
×
×
= −
kT [1]
= 1.118 × 10–19
J ≈ 1.1 × 10–19
J [1]
ii kT
mv
2
3
2
1 2
= [1]
27
23
10
7
.
1
5400
10
38
.
1
3
3
−
−
×
×
×
×
=
=
m
kT
v [1]
speed = 1.147 × 104
m s–1
≈ 11 km s–1
[1]
12 a Mean kinetic energy = 273
10
38
.
1
2
3
2
3 23
×
×
×
= −
kT [1]
= 5.65 × 10–21
J ≈ 5.7 × 10–21
J [1]
b There are 6.02 × 1023
molecules of carbon dioxide. [1]
mass of molecule = 26
23
10
31
.
7
10
02
.
6
044
.
0 −
×
=
×
kg [1]
21
2
10
65
.
5
2
1 −
×
=
mv J [1]
26
21
10
31
.
7
10
65
.
5
2
−
−
×
×
×
=
v [1]
speed = 393 m s–1
≈ 390 m s–1
[1]
c Total kinetic energy of one mole of gas = ×
kT
2
3
NA = RT
2
3
(Note: R = k × NA) [1]
For an ideal gas, the change in internal energy is entirely kinetic energy.
Change in internal energy = 100
31
.
8
2
3
)
273
373
(
2
3
×
×
=
−
×
R [1]
change in internal energy = 1.2465 kJ ≈ 1.2 kJ [1]
13 a i The molecule has 3 degrees of freedom for translational motion and 2 degrees of freedom
for rotation – making a total of 5. [1]
Therefore, mean energy = 5 × kT
2
1
= kT
2
5
[1]
ii The molecule has 3 degrees of freedom for translational motion and 3 degrees of freedom
for rotation – making a total of 6. [1]
Therefore, mean energy = 6 × kT
2
1
= 3kT [1]
b Internal energy = ×
kT
3 NA = RT
3 (Note: R = k × NA) [1]
internal energy per unit kelvin = 3R [1]
= 3 × 8.31 ≈ 25 J K–1
[1]

1 Electric field strength is the force per unit charge at that point. [1]
The potential at a point is the work that must be done to bring unit charge
from infinity to that point. [1]
2 a E =
d
V
, d =
E
V
[1]
d =
000
400
5000
= 1.25 × 10–2
m ≈ 1.3 cm [1]
b F = EQ = 400 000 × 1.6 × 10–19
N [1]
F = 6.4 × 10–14
N [1]
3 E = 2
0
π
4 r
Q
ε
so k = 12
0 10
85
.
8
4
1
π
4
1
−
×
×
π
=
ε
k = 8.99 × 109
m F–1
≈ 9.0 × 109
m F–1
[1]
4 The force between the charges obeys an inverse square law with distance; that is, F ∝ 2
1
r
[1]
Point B: The distance is the same. The force between the charges is F. [1]
Point C: The distance is doubled, so the force decreases by a factor of 4. [1]
The force between the charges is
4
F
. [1]
Point D: The distance is trebled, so the force decreases by a factor of 32
= 9. [1]
9
F
. [1]
Point E: The distance between the charges is 8 R. [1]
The force between the charges decreases by a factor
of 2
)
8
( = 8 [1]
8
F
. [1]
5 a E = 2
0
π
4 r
Q
ε
[1]
E = 2
12
6
15
.
0
10
85
.
8
π
4
10
5
.
2
×
×
×
×
−
−
[1]
E = 9.99 × 105
V m–1
≈ 1.0 × 106
V m–1
[1]
b The distance from the centre of the dome increases by a factor of 3.
The electric field strength decreases by a factor of 32
= 9. [1]
Therefore E =
9
10
0
.
1 6
×
= 1.1 × 104
V m–1
[1]

6 a i E = 2
0
π
4 r
Q
ε
[1]
E = 2
12
6
40
.
0
10
85
.
8
π
4
10
20
×
×
×
×
−
−
(r =
2
80
= 40 cm) [1]
E = 1.124 × 106
V m–1
≈ 1.1 × 106
V m–1
[1]
ii E = 2
12
6
40
.
0
10
85
.
8
π
4
10
40
×
×
×
×
−
−
[1]
E = 2.248 × 106
V m–1
≈ 2.2 × 106
V m–1
[1]
(The electric field doubles because the charge is doubled, E ∝ Q.)
b Net field strength, E = 2.2 × 106
– 1.1 × 106
= 1.1 × 106
V m–1
[1]
The direction of the electric field at X is to the left. [1]
7 a Q = V × 4πεor = 20 000 × 4 × π × 8.85 × 10−12
× 0.15 [1]
Q = 3.3 × 10–7
C [1]
b E = 2
r
kQ
= 2
7
9
15
.
0
10
3
.
3
10
0
.
9 −
×
×
×
[1]
= 1.32 × 105
V m–1
≈ 1.3 × 105
V m–1
[1]
c F = eV = 1.6 × 10−19
× 1.32 × 105
[1]
F = 2.11 × 10−14
N ≈ 2.1 × 10−14
N [1]
8 a V =
r
Q
0
π
4 ε
=
r
kQ
= 2
9
9
10
5
10
2000
10
9.0
−
−
×
×
−
×
×
([1] mark only if minus sign omitted) [2]
V = 3.6 × 105
J C–1
=360 kV [1]
9 Similarities
• Both produce radial fields. [1]
• Both obey an inverse square law with distance; that is, F ∝ 2
1
r
. [1]
• The field strengths are defined as force per unit (positive) charge or mass. [1]
• Both produce action at a distance. [1]
Differences
• Electrical forces can be either attractive or repulsive, whereas gravitational forces are
always attractive. [1]
• Gravitational forces act between masses, whereas electrical forces act between charges. [1]

10 The electric field strength due to the charge +Q is equal in magnitude but opposite in direction
to the electric field strength due to the charge +3Q. [1]
Therefore:
2
0
π
4 x
Q
ε
= 2
0 )
(
π
4
3
x
R
Q
−
ε
(where R is the distance between the charges = 10 cm) [1]
2
1
x
= 2
)
(
3
x
R −
[1]
so
x
x
R −
= 3 [1]
x(1 + 3 ) = R so x =
3
1+
R
= 0.37 R [1]
x = 0.37 × 10 = 3.7 cm [1]
11 ratio = 2
2
2
0
2
π
4
r
Gm
r
e ε
(where m = mass of proton and r = separation) [2]
ratio = 2
0
2
π
4 Gm
e
ε
[1]
The r2
terms cancel and so this ratio is independent of the separation. [1]
ratio = 2
27
11
12
2
19
)
10
7
.
1
(
10
67
.
6
10
85
.
8
π
4
)
10
6
.
1
(
−
−
−
−
×
×
×
×
×
×
×
[1]
ratio ≈ 1.2 × 1036
[1]
12 a F = 2
0
2
1
π
4 r
Q
Q
ε
= 9 × 109
× 2
15
19
19
)
10
(
10
6
.
1
10
6
.
1
−
−
−
×
×
×
[1]
= 230 N [1]
b V =
r
Q
0
π
4 ε
= 9 × 109
× 15
19
10
10
6
1
−
−
×
.
[1]
= 1.44 × 106
V [1]
c W = VQ = 1.44 × 106
× 1.6 × 10–19
[1]
= 2.3 × 10–13
J [1]
d
2
1
mv2
= W ⇒ v2
=
m
W
2
[1]
v2
= 27
13
10
7
1
10
3
2
2
−
−
×
×
×
.
.
= 2.7 × 10–14
[1]
v = 14
10
7
2 −
×
. = 1.6 × 107
m s–1
[1]

1 a Q = VC = 9.0 × 30 × 10–6
[1]
Q = 2.7 × 10–4
C (270 µC) [1]
b number of excess electrons =
e
Q
= 19
4
10
6
.
1
10
7
.
2
−
−
×
×
[1]
number = 1.69 × 1015
≈ 1.7 × 1015
[1]
2 a i The charge is directly proportional to the voltage across the capacitor.
Hence doubling the voltage will double the charge. [1]
charge = 2 × 150 = 300 nC [1]
ii Since Q ∝ V for a given capacitor, increasing the voltage by a factor of three
will increase the charge by the same factor. [1]
charge = 3 × 150 = 450 nC [1]
b C =
V
Q
=
0
3
10
150 9
.
−
×
[1]
C = 5.0 × 10–8
F [1]
3 a E =
2
1
V2
C =
2
1
× 9.02
× 1000 × 10–6
[1]
E = 4.05 × 10–2
J ≈ 4.1 × 10–2
J [1]
b For a given capacitor, energy stored ∝ voltage2
. [1]
energy = 22
× 4.05 × 10–2
≈ 0.16 J [1]
4 a Ctotal = C1 + C2 [1]
Ctotal = 20 + 40 = 60 nF [1]
b
total
1
C
=
1
1
C
+
2
1
C
[1]
total
1
C
=
100
1
+
500
1
= 0.012 µF−1
[1]
Ctotal =
012
.
0
1
≈ 83 µF [1]
c
total
1
C
=
1
1
C
+
2
1
C
+
3
1
C
[1]
total
1
C
=
10
1
+
50
1
+
100
1
= 0.13 µF−1
[1]
Ctotal =
13
.
0
1
≈ 7.7 µF [1]
d Total capacitance of the two capacitors in parallel = 50 + 50 = 100 µF. [1]
total
1
C
=
50
1
+
100
1
= 0.03 µF−1
[1]
Ctotal =
03
.
0
1
≈ 33 µF [1]
e Total capacitance of the two capacitors in series is 83 µF (from b). [1]
Ctotal = 83 + 50 = 133 µF ≈ 130 µF [1]
5 a Ctotal = C1 + C2 [1]
Ctotal = 100 + 500 = 600 µF [1]
b The potential difference across parallel components is the same and equal to 1.5 V. [1]

c Q = VC = 1.5 × 600 × 10–6
[1]
Q = 9.0 × 10–4
C (900 µC) [1]
d E =
2
1
QV =
2
1
× 9.0 × 10–4
× 1.5 [1]
E = 6.75 × 10–4
J ≈ 6.8 × 10–4
J [1]
6 a E =
2
1
V2
C =
2
1
× 322
× 10 000 × 10–6
[1]
E = 5.12 J ≈ 5.1 J [1]
b P =
t
E
=
300
.
0
12
.
5
[1]
P ≈ 17 W [1]
7 a Q = VC = 12 × 1000 × 10–6
[1]
Q = 1.2 × 10–2
C (12 mC) [1]
b i Ctotal = C1 + C2 [1]
Ctotal = 1000 + 500 = 1500 µF [1]
ii V =
C
Q
(The charge Q is conserved and C is the total capacitance.) [1]
V = 6
2
10
1500
10
2
.
1
−
−
×
×
= 8.0 V [1]
8 a ∆Q = I∆t =
50
10
225 3
−
×
=
f
I
= 4.5 × 10–3
C [1]
b C =
0
.
9
10
5
.
4 3
−
×
=
V
Q
[1]
= 5.0 × 10–4
F = 500 µF [1]
c i The capacitors are in parallel, so the total capacitance = 2C and charge stored = 2Q;
current =2I = 2 × 225 = 450 mA [1]
ii The capacitors are in series, so the total capacitance =
2
1
C and charge stored =
2
1
Q;
current =
2
1
I =
2
1
× 225 = 113 mA [1]
9 a Q = CV = 200 × 10–6
× 200 = 0.040 C [1]
b E =
2
1
CV2
=
2
1
× 200 × 10–6
× 2002
= 4.0 J [1]
c The two capacitors now make a pair of capacitors in parallel of total capacitance
= 100 µF + 200 µF = 300 µF [1]
The charge is conserved, therefore V = 6
10
300
040
0
−
×
=
.
C
Q
[1]
= 133 V [1]
d E =
2
1
CV2
=
2
1
× 300 × 10–6
× 1332
[1]
= 2.67 J [1]
e The energy is lost as the connected wires are heated as the current passes through them. [1]

10 The capacitors are in parallel, so the total capacitance = 3C. [1]
The total charge Q remains constant. [1]
The energy stored by a capacitor is given by E =
C
Q
2
2
. [1]
Einitial =
C
Q
2
2
and Efinal =
)
3
(
2
2
C
Q
[1]
Fraction of energy stored =
initial
final
E
E
=
C
Q
C
Q
2
)
3
(
2
2
2
=
3
1
[1]
Fraction of energy ‘lost’ as heat in resistor = 1 –
3
1
=
3
2
. [1]
The resistance governs how long it takes for the capacitor to discharge. The final voltage
across each capacitor is independent of the resistance. Hence, the energy lost as heat is
independent of the actual resistance of the resistor. [1]

1 The direction of the magnetic field should be clockwise (and not anticlockwise). [1]
The separation between adjacent circular field lines should increase further away from
the wire (see below). [1]
2 a The conductor is pushed to the left. [1]
b The conductor is pushed to the left. [1]
c The conductor is pushed out of the plane of the paper. [1]
3 B =
Il
F
[1]
[B] =
m
A
N
= N A−1
m−1
[1]
4 F = BIl [1]
F = 0.12 × 3.5 × 0.01 (length = 1.0 cm) [1]
F = 4.2 × 10–3
N [1]
5 a F = BIl sin θ
F = 0.050 × 3.0 × 0.04 × sin 90° [1]
F = 6.0 × 10–3
N [1]
b F = 0.050 × 3.0 × 0.04 × sin 30° [1]
F = 3.0 × 10–3
N [1]
c F = 0.050 × 3.0 × 0.04 × sin 65° [1]
F = 5.44 × 10–3
N ≈ 5.4 × 10–3
N [1]
6 Force experienced by PQ = force experienced by RS (but in opposite direction). [1]
No force experienced by QR and PS (since current is parallel to the field). [1]
torque = one of the forces × perpendicular distance between forces = (BIL)x [1]
torque = BI(Lx), Lx = area of loop = A [1]
torque = BIA ∝ A [1]
The torque is directly proportional to the area of the loop.
7 a Current is at right angles to magnetic field. [1]
Left-hand rule produces force on AB towards the right. [1]
Wire leaves mercury and breaks contact/current stops/force stops. [1]
Weight causes AB to fall back/return and make contact again. [1]
b i Moment = Fd [1]
F =
035
0
10
5
3 5
.
. −
×
= 1.0 × 10−3
N [1]

ii F = BIl ⇒ I =
Bl
F
[1]
I =
)
07
.
0
10
0
.
6
(
10
0
.
1
3
3
×
×
×
−
−
= 2.38 A ≈ 2.4 A [1]
8 a F = BIl × number of turns [1]
F = 0.19 × 2.8 × 0.07 × 25 = 0.93 N [1]
b Torque= Fd = 0.93 × 0.03 [1]
torque = 0.028 N m [1]
c The longest side always stays at 90º to the magnetic field as the coil turns
so the force is constant. [1]
The perpendicular distance between the forces changes as the coil turns
so the torque (moment) changes. [1]
9 a F = mg = (103.14 – 102.00) × 10–3
× 9.81 so F = 1.12 × 10–2
N ≈ 1.1 × 10–2
N [1]
b B =
Il
F
[1]
B =
05
.
0
2
.
8
10
12
.
1 2
×
× −
[1]
B = 2.73 × 10−2
T (27 mT) [1]

1 F = EQ = 5.0 × 105
× 3.2 × 10−19
[1]
F = 1.6 × 10−13
N [1]
2 a E =
d
V
= 2
10
0
.
3
600
−
×
[1]
E = 2.0 × 104
V m−1
[1]
The field acts towards the negative plate. [1]
b The electric field is uniform between the plates (except at the ‘edges’). [1]
The electric field is at right angles to the plate. [1]
c i Since the droplet is stationary,
the electric force on the droplet
must be equal and opposite to its weight. [1]
The electric force must act upwards,
so the charge on the droplet must
be negative. [1]
ii E =
Q
F
Q =
E
F
= 4
15
10
0
.
2
10
4
.
6
×
× −
[1]
Q = 3.2 × 10−19
C [1]
3 F = BQv [1]
F = 0.18 × 1.6 × 10–19
× 4.0 × 106
[1]
F = 1.15 × 10–13
N ≈ 1.2 × 10−13
N [1]
4 a F = BQv [1]
F = 0.004 × 1.6 × 10–19
× 8.0 × 106
[1]
F = 5.12 × 10−15
N ≈ 5.1 × 10−15
N [1]
b a =
m
F
=
15
31
5.12 10
9.11 10
−
−
×
×
[1]
a = 5.63 × 1015
m s−2
≈ 5.6 × 1015
m s−2
[1]
c From circular motion, the centripetal acceleration a is given by:
a =
r
v2
r =
a
v2
= 15
2
6
10
63
.
5
)
10
0
.
8
(
×
×
[1]
r = 1.14 × 10−2
m ≈ 1.1 × 10−2
m (1.1 cm) [1]

5 a
Both arrows at A and B are towards the centre of the circle. [1]
b The force on the electron is at 90° to the velocity. Hence the path described by the
electron is a circle. [1]
c The magnetic force provides the centripetal force. [1]
Therefore: BQv =
r
mv2
[1]
BQ =
r
mv
or v =
m
BQr
[1]
v = 31
2
19
3
10
1
.
9
10
0
.
5
10
6
.
1
10
0
.
2
−
−
−
−
×
×
×
×
×
×
[1]
v = 1.76 × 107
m s–1
≈ 1.8 × 107
m s−1
[1]
d v =
m
BQr
, so the speed v is directly proportional to the radius r. [1]
Radius is halved, so v =
2
10
76
.
1 7
×
= 8.8 × 106
m s−1
[1]
6 a Ek = 15 × 103
× 1.6 × 10−19
= 2.4 × 10−15
J (1 eV = 1.6 × 10−19
J) [1]
2
1
mv2
= 2.4 × 10−15
v = 27
15
10
7
.
1
10
4
.
2
2
−
−
×
×
×
[1]
v = 1.68 × 106
m s−1
≈ 1.7 × 106
m s−1
[1]
b F = ma =
r
mv2
[1]
F =
05
.
0
)
10
68
.
1
(
10
7
.
1 2
6
27
×
×
× −
[1]
F = 9.60 × 10−14
N ≈ 9.6 × 10−14
N [1]
c F = BQv [1]
B =
Qv
F
= 6
19
14
10
68
.
1
10
6
.
1
10
60
.
9
×
×
×
×
−
−
[1]
B ≈ 0.36 T [1]
d speed =
time
distance
time =
speed
nce
circumfere
= 6
10
68
.
1
05
.
0
π
2
×
×
[1]
time = 1.87 × 10−7
s ≈ 1.9 × 10−7
s [1]

7 a In order for the positively charged ions to emerge from the slit,
the net force perpendicular to the velocity must be zero. [1]
electrical force on ion = magnetic force on ion [1]
EQ = BQv [1]
The charge Q cancels.
E = Bv [1]
The electric field strength is E =
d
V
. Therefore, the magnetic flux density is:
B =
v
E
=
v
d
V
= 6
3
10
0
.
6
024
.
0
/
)
10
0
.
5
(
×
×
[1]
B = 3.47 × 10–2
T ≈ 35 mT [1]
b v =
m
BQr
so r =
BQ
mv
[1]
∆r =
BQ
v
m
m )
( 2
1 −
[1]
8 a The centripetal force is provided by the magnetic force. [1]
Therefore: Bev =
r
mv2
[1]
Be =
r
mv
or v =
m
Ber
[1]
T =
speed
nce
circumfere
=
m
Ber
r
π
2
[1]
The radius r of the orbit cancels. Hence: T =
Be
m
π
2
The time T is independent of both the radius of the orbit r and the speed v. [1]
b The faster electron travels in a circle of larger radius. [1]

1 a i Magnetic flux, Φ = BA [1]
ii Flux linkage = NΦ = NBA [1]
b Flux linkage = N(B cos θ)A = NBA cos θ [1]
(The component of B normal to the plane of the coil is B cos θ.)
2 Area A of coil = x2
[1]
flux linkage = NBA = NBx2
[1]
3 a There is no change to the magnetic flux linking the coil, hence according to Faraday’s law,
there is no induced e.m.f. [1]
b The magnetic flux density B increases as the magnet moves towards the coil. [1]
There is an increase in the magnetic flux linking the coil, hence an e.m.f. is induced across
the ends of the coil. [1]
4 a flux linkage = NBA so B =
NA
linkage
flux
[1]
B = 4
4
10
0
.
4
70
10
4
.
1
−
−
×
×
×
[1]
B = 5.0 × 10−3
T [1]
b flux linkage = NBA cos θ [1]
flux linkage = 70 × 0.050 × 4.0 × 10−4
× cos 60° [1]
flux linkage = 7.0 × 10−4
Wb [1]
5 a i magnetic flux, Φ = BA = 40 × 10–3
× (0.03 × 0.03) [1]
Φ = 3.6 × 10−5
Wb [1]
ii flux linkage = NΦ = 200 × 3.6 × 10−5
[1]
flux linkage = 7.2 × 10−3
Wb [1]
b Final flux linkage = 0, initial flux linkage = 7.2 × 10−3
Wb. [1]
Hence, change in magnetic flux linkage is 7.2 × 10−3
Wb. [1]
6 a Initial magnetic flux = BA = 0.15 × (π × [8.0 × 10−3
]2
) [1]
initial magnetic flux = 3.02 × 10−5
Wb [1]
final magnetic flux = 0 [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage
E =
t
Φ
N
∆
∆
, so E = 1200 ×
020
.
0
10
02
.
3
0 5
−
×
−
[1]
E = 1.81 V ≈ 1.8 V (magnitude only) [1]
b average current =
resistance
e.m.f.
=
3
.
6
81
.
1
[1]
I = 0.287 A ≈ 0.29 A [1]

7 a distance = speed × time = 2.0 × 1.0 = 2.0 m [1]
b Area swept = length × distance travelled = 0.10 × 2.0 = 0.20 m2
[1]
c Change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = 0.20 × 0.050 = 1.0 × 10−2
Wb [1]
d Magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
E =
t
NΦ
∆
∆ )
(
(N = 1)
E =
0
.
1
10
0
.
1 2
−
×
= 1.0 × 10−2
V (1 Wb s−1
= 1 V) [1]
e E = Blv = 0.050 × 2.0 × 0.10 = 1.0 × 10−2
V [1]
8 Initial magnetic flux = BA = 0.060 × π × (1.2 × 10−2
)2
[1]
initial magnetic flux = 2.72 × 10−5
Wb [1]
final magnetic flux = –2.72 × 10−5
Wb (since the field is reversed) [1]
average magnitude of induced e.m.f. = rate of change of magnetic flux linkage
E =
t
Φ
N
∆
∆
, so E = 2000 ×
030
.
0
10
72
.
2
10
72
.
2 5
5 −
−
×
−
×
−
[1]
E = 3.62 V ≈ 3.6 V (magnitude only) [1]
9 a There is a current in the primary coil when the switch is closed. This current creates a
magnetic flux in the primary coil. [1]
Due to the soft iron ring, the magnetic flux created by the primary coil also links the
secondary coil. With the switch closed, there is no change in the magnetic flux linkage
at the secondary and hence the lamp is not lit. [1]
When the switch is opened, the magnetic flux decreases to zero in a short period.
The rapid change in magnetic flux at the secondary coil creates an e.m.f. and the lamp
illuminates for a short period. [1]
Eventually there is no magnetic flux at either the primary or the secondary coil and
hence there is no e.m.f. induced – the lamp stays off. [1]
b Change the cell to an a.c. supply (or keep switching on and off very fast). [1]

10
distance = speed × time = vt [1]
area swept = length × distance travelled = Lvt [1]
change in magnetic flux = area swept × magnetic flux density [1]
change in magnetic flux = (Lvt) × B = BLvt [1]
magnitude of e.m.f. = rate of change of magnetic flux linkage [1]
E =
t
BLvt
= BLv [1]
E = BLv = 40 × 10−6
× 0.20 × 0.30 [1]
E = 2.4 × 10−6
V (2.4 µV) [1]
11 a The magnitude of the induced e.m.f. in a circuit is directly proportional to the rate of
change of magnetic flux linkage. [1]
b Lenz’s law expresses the principle of conservation of energy. [1]
c i magnetic flux = magnetic flux density × cross-sectional area of coil
or Φ = BA [1]
ii Flux linkage = NBA [1]
Coil X: flux linkage = NBA = 200 × 0.10 × (π × 0.022
) ≈ 2.5 × 10−2
Wb [1]
Coil Y: flux linkage = NBA = 4000 × 0.01 × (π × 0.032
) ≈ 1.1 × 10−1
Wb [1]
The coil Y has greater flux linkage. [1]

1 a T =
f
1
[1]
0.1 s [1]
b i t = 0, 0.05 and 0.15 [1]
ii t =
4
1
of a period [1]
t = 0.025 s [1]
iii t = 0.075 s [1]
iv t =
2
1
= sin (20πt) [1]
t = 0.0125 s [1]
2 a Corresponding parts of each wave occur at the same time. [1]
b Power P always positive. [1]
Two peaks for P in time taken for one peak of I and V. [1]
Sinusoidal shape above axis. [1]
3 a i I =
V
P
=
230
690
[1]
Irms = 3.0 A [1]
ii Ipeak = 3 2 = 4.2 A [1]
iii Vpeak = 230 2 = 325 V [1]
b Correct shape all above axis. [1]
Two cycles shown (i.e. one cycle of current waveform). [1]
Peak and average power marked. [1]

4 a a.c. [1]
Current is positive and negative OR current flows one way, then the opposite way. [1]
b Voltage switches between +2 and −2 V. Power is
R
V 2
, so it has the same value for both
positive and negative values of the voltage. [1]
c i 2 V [1]
ii 2 V [1]
5 Average power = ⎟
⎠
⎞
⎜
⎝
⎛
2
5
.
2
× ⎟
⎠
⎞
⎜
⎝
⎛
2
6
[1]
= 7.5 W [1]
6 a i Period = 80 ms = 0.080 s [1]
f =
T
1
= 12. 5 Hz [1]
ii Peak voltage = 4.5 V [1]
r.m.s. voltage =
2
5
.
4
= 3.2 V [1]
b V0 = 4.5 V [1]
ω = 2πf = 2π × 12.5 = 78.5 s−1
≈ 79 s−1
[1]
7 a I =
V
P
=
6
24
[1]
I = 4.0 A [1]
b Vp = 6 ×
30
1150
[1]
Vp = 230 V [1]
c Ip =
230
24
or Ip =
1150
4
30×
[1]
Ip = 0.10 A [1]
d Maximum p.d. = 6 2 [1]
maximum p.d. = 8.5 V [1]
e Heat is still produced inside the wires even if it cannot be conducted to the
outside. The wires may melt if they cannot lose the heat. [1]
8 a Number of turns on the primary =
230
115
× 500 = 1000 turns [1]
b Is =
R
Vs
=
5000
115
= 0.023 A r.m.s. [1]
c Ip = 0.023 ×
1000
500
= 0.0125 A ≈ 0.013 A r.m.s. [1]
d Peak voltage = 115 2 = 162 V [1]
so the cables will break down. [1]
9 a I =
P
V
=
1000
100
= 10 A r.m.s. [1]
P = I 2
R = 102
× 5 = 500 W [1]
b At high voltages the current is less for the same power. [1]
Power lost in cable = I2
R so power lost is less. [1]

1 A ‘packet’, or quantum, of electromagnetic energy. [1]
2 The energy of a photon is proportional to the frequency of the radiation. [1]
Hence a γ-ray photon has greater energy than a photon of visible light (and therefore
is more harmful). [1]
3 a Electromagnetic radiation travels through space as waves and, as such, shows diffraction
and interference effects. [1]
b Electromagnetic radiation interacts with matter as ‘particles’. The photoelectric effect
provides strong evidence for the particle-like (photon) behaviour of electromagnetic
radiation. [1]
4 a c = fλ so f = 7
8
10
4
.
6
10
0
.
3
−
×
×
=
λ
c
[1]
f = 4.69 × 1014
Hz ≈ 4.7 × 1014
Hz [1]
b E = hf [1]
E = 6.63 × 10−34
× 4.69 × 1014
[1]
E = 3.1 × 10−19
J [1]
5 For an electron to escape from the surface of the metal, it must absorb energy from the photon
that is greater than the work function. [1]
The work function is the minimum energy required by the electron to escape from the surface
of the metal. [1]
The photon of visible light has energy less than the work function of the metal, whereas
the photon of ultraviolet radiation has energy greater than the work function. [1]
6 a The electron loses energy. [1]
This energy appears as a photon of electromagnetic radiation. [1]
b Energy of photon = E1 − E2 [1]
Therefore:
hf = E1 − E2 or
h
E
E
f 2
1 −
= [1]
c The change in energy ∆E is greater. [1]
Hence the frequency of the radiation is greater (f ∝ ∆E). [1]
The spectral line will be the right side of the line shown on the spectrum diagram.
7 There are six spectral lines. [1]
Correct transitions shown on the energy level diagram. [1]

8
λ
hc
hf
E =
=
∆ [1]
9
8
34
10
670
10
0
.
3
10
63
.
6
−
−
×
×
×
×
=
∆E [1]
∆E = 2.97 × 10−19
J ≈ 3.0 × 10−19
J [1]
9 a Continuous spectrum [1]
b Emission spectrum [1]
c Absorption spectrum [1]
10 Electrons travel through space as waves. Evidence for this is provided by the diffraction of
electrons by matter (e.g. graphite). [1]
11 The electronvolt is the energy gained by an electron travelling through a potential difference
of one volt. [1]
12 The kinetic energy Ee of the electron is:
Ee = VQ = 6.0 × 1.6 × 10–19
[1]
Ee = 9.6 × 10–19
J [1]
The energy EUV of the ultraviolet photon is:
EUV = hf =
λ
hc
[1]
EUV = 7
8
34
10
5
.
2
10
0
.
3
10
63
.
6
−
−
×
×
×
×
[1]
EUV = 7.96 × 10−19
J ≈ 8.0 × 10−19
J [1]
The energy of the ultraviolet photon is less than the kinetic energy of the electron.
(The student is correct.)
13 a The threshold frequency is the minimum frequency of electromagnetic radiation that just
removes electrons from the surface of the metal. [1]
b At the threshold frequency, the energy of the photon is equal to the work function φ of the
metal. Hence:
φ = hf0 (f0 = threshold frequency) [1]
f0 = 34
19
10
63
.
6
10
6
.
1
9
.
1
−
−
×
×
×
[1]
f0 = 4.6 × 1014
Hz [1]
14 a E = hf =
λ
hc
[1]
E = 9
8
34
10
550
10
0
.
3
10
63
.
6
−
−
×
×
×
×
[1]
E = 3.62 × 10–19
J ≈ 3.6 × 10–19
J [1]
b Power emitted as light = 0.05 × 60 = 3.0 W [1]
Number of photons emitted per second = 19
10
62
.
3
0
.
3
−
×
[1]
= 8.3 × 1018
[1]
15 φ = 4.3 × 1.6 × 10–19
= 6.88 × 10–19
J [1]
Energy of photon =
λ
hc
= 7
8
34
10
1
.
2
10
0
.
3
10
63
.
6
−
−
×
×
×
×
= 9.47 × 10−19
J [1]
energy of photon = work function + maximum kinetic energy of electron [1]
maximum kinetic energy of electron = (9.47 − 6.88) × 10−19
≈ 2.6 × 10−19
J [1]

16 λ =
mv
h
[1]
ν =
λ
m
h
= 11
27
34
10
0
.
2
10
7
.
1
10
63
.
6
−
−
−
×
×
×
×
[1]
v = 1.95 × 104
m s−1
(20 km s−1
) [1]
17 Energy lost by a single electron = energy of photon [1]
(The energy lost by a single electron travelling through the LED reappears as the energy of a
single photon.)
Therefore:
λ
hc
eV = [1]
19
7
8
34
10
6
.
1
10
8
.
5
10
0
.
3
10
63
.
6
−
−
−
×
×
×
×
×
×
=
=
e
hc
V
λ
[1]
V = 2.14 V ≈ 2.1 V [1]
18 a Kinetic energy of electron = VQ = Ve
2
1
mev2
= Ve or
e
2
2m
p
= Ve (where p = mev) [1]
p = Ve
me
2 [1]
λ =
p
h
v
m
h
=
e
(de Broglie equation) [1]
Therefore, λ =
Ve
m
h
e
2
b λ =
Ve
m
h
e
2
or V =
e
m
h
2
e
2
2 λ
[1]
V = 19
2
11
31
2
34
10
6
.
1
)
10
0
.
4
(
10
1
.
9
2
)
10
63
.
6
(
−
−
−
−
×
×
×
×
×
×
×
[1]
V ≈ 940 V [1]
19 Using f =
λ
c
and Einstein’s photoelectric equation (hf = φ + )
2
1 2
max
mv : [1]
Red light ⇒ 9
8
10
640
10
0
.
3
−
×
×
×
h
= φ + (0.9 × 1.6 × 10−19
)
⇒ 4.688 × 1014
h = φ + 1.440 × 10−19
(equation 1) [1]
Blue light ⇒ 9
8
10
420
10
0
.
3
−
×
×
×
h
= φ + (1.9 × 1.6 × 10−19
)
⇒ 7.143 × 1014
h = φ + 3.040 × 10−19
(equation 2) [1]
Equations 1 and 2 are two simultaneous equations.
(7.143 – 4.688) × 1014
h = (3.040 − 1.440) × 10−19
[1]
h = 14
19
10
)
688
.
4
143
.
7
(
10
)
440
.
1
040
.
3
(
×
−
×
− −
≈ 6.5 × 10−34
J s [1]

20 a External energy has to be supplied to excite or free an electron. [1]
(Allow: The electrons are trapped in an energy well.)
b An energy level of 0 eV means the electron is free from the atom.
The minimum energy is equal to 3.00 eV. [1]
Energy needed to free electron = 3.00 × 1.6 × 10−19
[1]
Energy needed to free electron = 4.80 × 10−19
J [1]
c i The difference between the energy levels −3.00 eV and −1.59 eV is equal to 1.41 eV. [1]
Hence, an electron jumps from −3.00 eV energy level to −1.59 eV energy level. [1]
ii
λ
hc
hf
E =
=
∆ [1]
19
8
34
10
6
.
1
41
.
1
10
0
.
3
10
63
.
6
−
−
×
×
×
×
×
=
∆
=
E
hc
λ [1]
λ = 8.82 × 10−7
m [1]
21 a This is the lowest energy level occupied by an electron in an atom. [1]
b The shortest wavelength corresponds to the change in energy between the two most widely
separated energy levels.
Hence, ∆E = 10.43 eV [1]
λ
hc
hf
E =
=
∆ [1]
19
8
34
10
6
.
1
43
.
10
10
0
.
3
10
63
.
6
−
−
×
×
×
×
×
=
∆
=
E
hc
λ [1]
λ = 1.19 × 10−7
m [1]
22 a E1 = 2
18
1
10
18
.
2 −
×
− = –2.18 × 10−18
J [1]
E2 = 2
18
2
10
18
.
2 −
×
− = –5.45 × 10−19
J [1]
b E2 − E1 = ∆E =
λ
hc
[1]
19
8
34
10
)
45
.
5
8
.
21
(
10
0
.
3
10
63
.
6
−
−
×
−
×
×
×
=
∆
=
E
hc
λ [1]
λ = 1.22 × 10−7
m [1]
This spectral line lies in the ultraviolet region of the spectrum. [1]
c ∆E = 2.18 × 10−18
⎟
⎠
⎞
⎜
⎝
⎛
− 2
2
7
1
6
1
= 1.607 × 10−20
J [1]
20
8
34
10
607
1
10
0
3
10
63
6
−
−
×
×
×
×
=
∆
=
.
.
.
E
hc
λ = 1.238 × 10−5
m [1]
λ ≈ 1.24 × 10−5
m (12.4 µm) [1]
This spectral line lies in the infrared region of the electromagnetic spectrum. [1]

1 a change in energy = change in mass × (speed of light)2
or ∆E = ∆mc2
[1]
b i ∆E = ∆mc2
∆E = 0.001 × (3.0 × 108
)2
[1]
∆E = 9.0 × 1013
J [1]
ii ∆E = ∆mc2
∆E = 9.1 × 10−31
× (3.0 × 108
)2
= 8.19 × 10−14
J [1]
∆E ≈ 8.2 × 10−14
J [1]
2 a i Mass = 27
27
10
66
.
1
10
65
.
6
−
−
×
×
= 4.01 u [1]
ii Mass = 27
26
10
66
.
1
10
16
.
2
−
−
×
×
= 13.01 u [1]
b i Mass = 1.01 × 1.66 × 10−27
= 1.68 × 10−27
kg [1]
ii Mass = 234.99 × 1.66 × 10−27
= 3.90 × 10−25
kg [1]
3 In all nuclear reactions the following quantities are conserved:
• charge (or proton number)
• nucleon number
• mass–energy
• momentum.
Any three of the above. [3]
4 a The nucleons within the nucleus are held tightly together by the strong nuclear force. [1]
b The binding energy of a nucleus is the minimum energy required to separate the nucleus
into its constituent protons and neutrons. [1]
c binding energy per nucleon =
nucleons
of
number
energy
binding
binding energy per nucleon =
16
128
[1]
binding energy per nucleon = 8.0 MeV [1]
5 a The half-life of a radioactive isotope is the mean time taken for the number of nuclei of the
isotope to decrease to half the initial number. [1]
b i 20 minutes is 1 half-life, so number of nuclei left =
2
0
N
[1]
ii 1.0 hour is 3 half-lives. [1]
Number of nuclei left =
8
2
1 0
0
3
N
N =
⎟
⎠
⎞
⎜
⎝
⎛
[1]
6 a i Activity is equal to the number of emissions (or decays of nuclei) per second.
Hence, there are 540 α-particles emitted in 1 second. [1]
ii Number of α-particles emitted in 1 h = 540 × 3600 ≈ 1.9 × 106
[1]
b Energy released in 1 s = number of α-particles emitted in 1 s × energy of each α-particle [1]
energy released in 1 s = 540 × 8.6 × 10−14
[1]
energy released in 1 s = 4.64 × 10−11
J ≈ 4.6 × 10−11
J [1]
c Rate of emission of energy = energy released per second
rate of emission of energy = 4.6 × 10−11
J s−1
= 4.6 × 10−11
W [1]

7 a The nuclide Fe
56
26 is the most stable. [1]
It has the maximum value for the binding energy per nucleon. [1]
b Binding energy = binding energy per nucleon × number of nucleons
binding energy ≈ 12.3 × 10−13
× 12 [1]
binding energy ≈ 1.5 × 10−11
J [1]
c From the graph, the binding energies per nucleon of H
2
1 and He
4
2 are approximately
1.0 × 10−13
J and 11.2 × 10−13
J. [1]
energy released = difference in binding energy per nucleon × number of nucleons [1]
energy released = [11.2 × 10−13
– 1.0 × 10−13
] × 4 [1]
energy released = 4.08 × 10−12
J ≈ 4.1 × 10−12
J [1]
d High temperatures (~108
K) and pressures. [2]
8 uranium
neutron
143
proton
92 235
92
1
0
1
1 →
+ [1]
mass defect = [(143 × 1.009) + (92 × 1.007)]u – (234.992)u [1]
mass defect = 1.939 u = 1.939 × 1.66 × 10−27
kg [1]
mass defect = 3.219 × 10−27
kg [1]
binding energy = mass defect × (speed of light)2
[1]
binding energy = 3.219 × 10−27
× (3.0 × 108
)2
= 2.897 × 10−10
J [1]
nucleons
of
number
energy
binding
235
10
897
.
2 10
−
×
= 1.233 × 10−12
≈ 1.2 × 10−12
J [1]
9 a Fission is the splitting of a heavy nucleus like U
235
92 into two approximately equal fragments.
The splitting occurs when the heavy nucleus absorbs a neutron. [1]
b i All particles identified on the diagram. [1]
ii In the reaction above, there is a decrease in the mass of the particles. [1]
According to ∆E = ∆mc2
, a decrease in mass implies that energy is released
in the process. [1]
iii The change in mass is ∆m given by:
∆m = [1.575 × 10−25
+ 2.306 × 10−25
+ 2(1.675 × 10−27
)] – [3.902 × 10−25
+ 1.675 × 10−27
][1]
∆m = –4.250 × 10−28
kg [1]
(The minus sign means a decrease in mass and hence energy is released in this reaction.)
∆E = ∆mc2
[1]
∆E = 4.250 × 10−28
× (3.0 × 108
)2
[1]
∆E = 3.83 × 10−11
J ≈ 3.8 × 10−11
J [1]

10 Binding energy of ‘reactant’ = 236 × 7.59 = 1791 MeV (binding energy of neutron = 0) [1]
Total binding energy of ‘products’ = (146 × 8.41) + (87 × 8.59) ≈ 1975 MeV [1]
Therefore energy released = 1975 – 1791 = 184 MeV [1]
11 a t1/2 =
λ
693
.
0
so λ =
1/2
693
.
0
t
[1]
λ =
56
693
.
0
[1]
λ = 1.238 × 10−2
s−1
≈ 1.2 × 10−2
s−1
[1]
b A = λN [1]
A =
56
693
.
0
× 6.0 × 1010
[1]
A ≈ 7.4 × 108
Bq [1]
12 a A = λN so λ =
N
A
[1]
λ = 14
9
10
0
.
8
10
0
.
5
×
×
[1]
λ = 6.25 × 10−6
s−1
≈ 6.3 × 10−6
s−1
[1]
b t1/2 =
λ
693
.
0
[1]
t1/2 = 6
10
25
.
6
693
.
0
−
×
[1]
t1/2 = 1.11 × 105
s ≈ 1.1 × 105
s [1]
c N = N0 e−λt
[1]
N = 8.0 × 1014 )
3600
40
10
(6.25 6
e ×
×
×
− −
[1]
N = 3.25 × 1014
≈ 3.3 × 1014
[1]
13 a The decay constant is the probability that an individual nucleus will decay per unit time. [1]
b i t1/2 =
λ
693
.
0
so λ =
1/2
693
.
0
t
[1]
λ =
3600
24
18
693
.
0
×
×
[1]
λ = 4.46 × 10−7
s−1
≈ 4.5 × 10−7
s−1
[1]
ii A = λN [1]
A = 4.46 × 10−7
× 4.0 × 1012
[1]
A = 1.78 × 106
Bq ≈ 1.8 × 106
Bq [1]
iii 36 days is equal to 2 half-lives. [1]
activity =
2
2
1
⎟
⎠
⎞
⎜
⎝
⎛
× 1.78 × 106
= 4.45 × 105
Bq ≈ 4.5 × 105
Bq [1]
14 a number of nuclei = number of moles × NA
number of nuclei =
226
10
0
.
1 6
−
×
× 6.02 × 1023
[1]
number of nuclei = 2.66 × 1015
≈ 2.7 × 1015
[1]
b A = λN [1]
A =
3600
24
365
1600
10
66
.
2
693
.
0
693
.
0 15
1/2 ×
×
×
×
×
=
×
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
N
t
[1]
A ≈ 3.7 × 104
Bq [1]

15 According to Einstein’s equation: ∆E = ∆mc2
[1]
In this case, ∆E is the energy of two photons and ∆m is the mass of two protons. [1]
Hence:
2 ×
λ
hc
= (2 × mp) c2
[1]
8
27
34
p
2
p 10
0
.
3
10
7
.
1
10
63
.
6
×
×
×
×
=
=
= −
−
c
m
h
c
m
hc
λ [1]
λ = 1.3 × 10−15
m [1]
16 For fusion, we have:
energy released per kg = number of ‘pairs’ of H
2
1 in 1 kg × 4.08 × 10−12
J (from 7 c) [1]
energy per kg = ⎟
⎠
⎞
⎜
⎝
⎛
×
×
× 23
10
02
.
6
2
1000
2
1
× 4.08 × 10−12
[1]
energy per kg = 6.14 × 1014
J ≈ 6.1 × 1014
J [1]
For fission, we have:
energy released per kg = number of nuclei in 1 kg × 3.83 × 10−11
J (from 9 b) [1]
energy per kg = ⎟
⎠
⎞
⎜
⎝
⎛
×
× 23
10
02
.
6
235
1000
× 3.83 × 10−11
[1]
energy per kg = 9.8 × 1013
J [1]
There is less energy released per fusion than per fission. However, there are many more nuclei
per kg for fusion. Hence fusion produces more energy per kg than fission. [1]
17 N = N0 e−λt
and λ =
1/2
693
.
0
t
[1]
fraction left = )
/
693
.
0
(
0
1/2
e
e t
t
t
N
N −
−
=
= λ
[1]
fraction left = )
10
5
.
4
/
10
0
.
5
693
.
0
( 9
9
e ×
×
×
−
[1]
fraction left = 0.463 ≈ 0.46 [1]

1 Processor (or processing device) [1]
Output device [1]
2 a Axis labels [1]
Graph correct shape [1]
b Thermistor, strain gauge, microphone, etc. [1]
3 a Length of fine/thin wire [1]
sealed in a plastic case [1]
b i
A
l
R
ρ
= so
R
l
A
ρ
= =
150
15
.
0
10
0
.
5 7
×
× −
[1]
A = 5.0 × 10−10
m2
[1]
ii Extends by
150
1
so resistance increases by a factor of
150
1
, which is 1 Ω. [1]
4 a Fraction of the output of a device is fed back to the input [1]
reducing any changes in the input to the system. [1]
b Increased bandwidth, less distortion, greater stability in gain (less affected by temperature
changes). [2]
5 a Inverting amplifier [1]
When Vin is positive then Vout is negative, or vice versa. [1]
b The output voltage reaches the power supply voltage and cannot exceed this value. [1]
c 10 V [1]
d i Gain becomes −10 rather than −5, i.e. sloping line has twice the gradient. [1]
Vout is +10 V for Vin below −1 V or Vout is −10 V for Vin above +1 V. [1]
ii Gain becomes −2.5 rather than −5, i.e. sloping line has half the gradient. [1]
iii The sloping line increases in length before flattening off. [1]
e Output voltage peaks at ±5 V. [1]
Input and output voltage out of phase. [1]
f When the input signal reaches above 2 V or below −2 V [1]
the output voltage remains at −10 V or +10 V. [1]

6 a The output is not saturated so the potential at both the inverting and non-inverting inputs
is very nearly the same. [1]
Since the non-inverting input is earthed, this value is 0 V. [1]
b 1.0 V [1]
c I = =
=
2000
1
R
V
5.0 × 10−4
A [1]
d Negligible current goes into the (–) terminal of the op-amp since it has a very high input
resistance (impedance). [1]
e Size of gain =
in
f
R
R
= 5 so Rf = 5Rin
Rf = 5 × 2 = 10 kΩ [1]
7 a Any two from:
infinite open-loop voltage gain, infinite input resistance (or impedance),
infinite bandwidth, zero output resistance (or impedance) [2]
b i I = 4
3
10
10−
=
R
V
[1]
I = 1.0 × 10−7
A [1]
ii V = IR = −1.0 × 10−7
× 100 × 103
= −1.0 × 10−2
V [1]
iii Gain = 3
2
in
out
10
0
.
1
10
0
.
1
−
−
×
×
−
=
V
V
= −10 [1]
iv 1.0 × 10−2
V (with Q being more positive) [1]
8 a For [2] marks, any two from the following points. [2]
With an inverting amplifier the output is half a cycle out of phase with the input,
whereas with a non-inverting amplifier the input and output are in phase.
The input goes to the (−) terminal on the op-amp for an inverting amplifier and
to the (+) terminal for the non-inverting amplifier.
The input resistance (impedance ) is higher for an op-amp used as a non-inverting amplifier.
For an inverting amplifier the gain is
in
f
R
R
−
but for an inverting amplifier
the gain is
2
1
1
R
R
+ .
b G =
2
1
1
R
R
+ =
20
40
1+ [1]
G = 3 [1]
c
in
out
V
V
G = ⇒ =
=
=
3
0
.
8
out
in
G
V
V 2.67 V ≈ 2.7 V [1]
d
( ) 3
10
40
20
0
.
8
×
+
=
=
R
V
I [1]
I = 1.33 × 10−4
A ≈ 1.3 × 10−4
A [1]
e The voltage is the same as for part c; voltage across the 20 kΩ resistor = 2.7 V [1]

green
red
output of
op-amp
9 a The output is either +9 V (+Vs) or −9 V (−Vs) [1]
depending on whether V +
> V –
or V –
> V +
[1]
b −9 V (−Vs) [1]
c i If the resistances are all the same, then V –
and V +
are equal, so it is unclear which of the
two is larger. The comparator is at its switching point and could go either way. [1]
ii V–
stays the same (at 4.5 V). [1]
V+
falls (due to larger resistance of X and larger potential drop across X). [1]
Vout is −9 V (it may have already been −9 V). [1]
10 a Two LEDs and series resistor shown [1]
Correct direction shown for LEDs [1]
b When output of op-amp is +9 V then the green LED is forward biased with enough
p.d. across it. The red LED is reverse biased. [1]
When output of op-amp is −9 V then the red LED is forward biased with enough
p.d. across it. The green LED is reverse biased. [1]

1 It is an electromagnetic wave. [1]
Wavelength in the range 10−8
m to 10−13
m [1]
2 a Energy = 50 × 103
× 1.6 × 10−19
[1]
= 8.0 × 10−15
J [1]
b
λ
hc
E =
15
8
34
10
0
.
8
10
0
.
3
10
63
.
6
−
−
×
×
×
×
=
λ [1]
wavelength = 2.49 × 10−11
m ≈ 2.5 × 10−11
m [1]
3 Compton scattering and pair production. [2]
4 A CAT scan produces a three-dimensional image of the body. [1]
5 A technique that does not involve cutting the body, e.g. CAT scan. [1]
6 It is a wave similar to sound (a longitudinal wave) [1]
but with a frequency above the audible range/above about 20 kHz. [1]
7 a λ
f
v =
wavelength = 6
3
10
8
.
1
10
5
.
1
×
×
[1]
= 8.33 × 10−4
m ≈ 0.83 mm [1]
b High-frequency ultrasound implies shorter wavelengths. Hence, smaller details can be
seen on an ultrasound scan. [1]
8 Acoustic impedance = density of medium × speed of ultrasound (Z = ρc) [1]
9 a Z = ρc
6.40 × 106
= ρ × 4000 [1]
1600
4000
10
40
.
6 6
=
×
=
ρ kg m−3
[1]
b fraction of intensity reflected =
2
1
2
1
2
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
−
Z
Z
Z
Z
[1]
= 5
2
10
3
.
8
63
.
1
66
.
1
63
.
1
66
.
1 −
×
=
⎟
⎠
⎞
⎜
⎝
⎛
+
−
[1]
percentage of intensity reflected = 0.0083% [1]
c There is only a very small amount of ultrasound reflected at the boundary between these
two materials. [1]
This is because their acoustic impedances are very similar. [1]
10 Superconducting magnet, RF transmission coil, RF receiver coil, gradient coils and computer. [5]
11 a The rotation of the spin/magnetic axis about the direction of the external magnetic field. [1]
b 40 MHz [1]
c wavelength = 6
8
10
40
10
0
.
3
×
×
=
f
c
[1]
= 7.5 m [1]

12 High-energy electrons made to bombard a metal target. [1]
When the electrons are accelerated and/or slowed down, electromagnetic radiation is produced. [1]
For large accelerations, radiation is in the X-ray region. [1]
Many different values of acceleration so many different wavelengths [1]
giving a continuous spectrum. [1]
Incident electron may excite an orbiting electron in a target atom. [1]
De-excitation of these electrons gives rise to an emission line spectrum. [1]
13 a intensity =
area
sectional
cross
power
-
[1]
b Power = 250 × π(2.0 × 10−3
)2
[1]
≈ 3.1 × 10−3
W [1]
14 A material (e.g. barium) with high-atomic number/attenuation coefficient used to
absorb X-rays. [1]
It is used to image the edges of soft tissues, such as the intestines (following a barium meal). [1]
15 Maximum energy of X-ray photon = 80 keV [1]
19
3
min
10
6
.
1
10
80 −
×
×
×
=
λ
hc
[1]
m
10
6
.
1
m
10
554
.
1
10
6
.
1
10
80
10
0
.
3
10
63
.
6 11
11
19
3
8
34
min
−
−
−
−
×
≈
×
=
×
×
×
×
×
×
=
λ [1]
16 The energy of the incident X-ray photon is used to remove an electron close to the nucleus
of an atom. [1]
The attenuation coefficient is proportional to the cube of the atomic (proton) number (µ ∝ Z3
). [1]
The attenuation coefficient is inversely proportional to the cube of the photon energy (µ ∝ E −3
).[1]
17 a
Sketch graph showing: labelled axes [1]
correct shape of line [1]
intensity showing an
exponential decrease [1]
b I = I0e−µx
[1]
where µ is the (linear) attenuation/absorption coefficient. [1]
c 0.10 = e−0.693x
[1]
ln 0.10 = –0.693x
–2.3 = −0.693x [1]
x = 3.32 mm ≈ 3.3 mm [1]
18 a X-ray image produced of target area. [1]
Image is stored in a computer. [1]
X-ray beam rotated about target so that many images produced from different angles. [1]
Computer stores and processes images to give image of slice through target. [1]
Process repeated for different slices. [1]
b X-ray image is flat shadow image of whole structure. [1]
CAT scan gives two-dimensional image of a slice through structure. [1]
Images of slices can be processed to give image of structure in any plane. [1]

19 Pulse of ultrasound transmitted into body [1]
where it is reflected at the boundary between different tissues. [1]
Reflected wave is detected and processed. [1]
Time for echo to reach detector indicates depth of tissue boundary. [1]
Intensity of echo gives information about tissue boundary. [1]
20 Amount of reflection of a wave at a boundary depends on the difference in acoustic
impedance of the two media. [1]
Acoustic impedance is speed of wave × density of medium. [1]
Acoustic impedance of air is very different from that for skin [1]
so very little ultrasound energy would pass into skin at the skin–air boundary. [1]
Coupling medium has acoustic impedance close to that of skin (and probe) [1]
so greatly reduces reflection at skin surface. [1]
21 a 2 × thickness = 4000 × (13 × 10−6
) [1]
2 × thickness = 5.2 × 10−2
[1]
thickness = 2
2
10
6
.
2
2
10
2
.
5 −
−
×
=
×
m (2.6 cm) [1]
b In a B-scan, the transducer is moved around the body. [1]
This produces a two-dimensional outline of an organ. [1]
22 a Nuclei of hydrogen (and certain other nuclides) have spin/have a magnetic axis. [1]
In a strong magnetic field, the nuclei precess about the direction of the field [1]
with a frequency known as the Larmor frequency. [1]
Radio wave pulse at Larmor frequency causes resonance [1]
and nuclei precess in high-energy state. [1]
After pulse has ceased, nuclei relax, emitting an RF signal. [1]
b
Diagram showing: magnet producing large field [1]
magnets producing non-uniform field [1]
RF coils. [1]
Patient subject to large magnetic field and calibrated non-uniform field. [1]
RF pulse at Larmor frequency transmitted to patient. [1]
RF emissions from patient are detected and processed. [1]
Hydrogen nuclei within patient [1]
have Larmor frequency dependent on magnetic field strength [1]
so that location (and concentration) of hydrogen nuclei can be detected. [1]
Total image built up by varying the non-uniform field to give specific field strength at
different positions within the patient. [1]

23 X-rays: diagnosis of broken bones [1]
relatively cheap, can detect flaws within the bone structure. [1]
Ultrasound: fetal development [1]
can distinguish between soft tissues; it is immediate and relatively safe. [1]
MRI: spinal/back problems [1]
fine detail provided of soft and hard tissues. [1]

1 a The audio signals have the same frequency. [1]
The amplitude of the signal/trace/carrier rises and falls with the same number of cycles
(two in the time of the trace). [1]
The audio signals have different volumes/loudness/amplitudes. [1]
The amplitude of the trace rises and falls more in the bottom trace. [1]
b The time (or horizontal distance along each trace) for one rapid variation
is the same in both traces. [1]
2 a In amplitude modulation the amplitude of the carrier wave is altered to carry the signal
(the frequency remains the same). [1]
In frequency modulation the frequency of the carrier wave is altered to carry the signal
(the amplitude remains the same). [1]
b i 30 × 2 = 60 kHz [1]
ii 800 − 60 = 740 kHz [1]
iii Alters from 740 kHz to 860 kHz 6000 times a second. [1]
c More transmitters may be needed as the range of FM is less than that of AM. [1]
Equipment to transmit and receive FM is more expensive. [1]
3 a i Carrier wave [1]
ii Sidebands [1]
iii 5 kHz [1]
b i 2.5 × 10−5
s [1]
ii 2.0 × 10−4
s [1]
iii Correct amplitude-modulated shape [1]
8 carrier wave oscillations per oscillation of the amplitude [1]
Correct times marked [1]
c Correct shape [1]
Sidebands extending from 40 to 55 kHz
and from 40 to 25 kHz [1]
time for one wave of audio signal, b ii
time for one wave of carrier, b i
25 40 55 f / kHz

4 a Analogue signal can have any value (within limits). [1]
Diagram to show analogue signal. [1]
Digital can have only a few values, e.g. two, and nothing in-between these values. [1]
Diagram to show digital signal. [1]
b The value of the signal is measured at regular intervals of time. [1]
The value obtained is converted into a binary number (with a certain number of bits). [1]
The binary numbers obtained are placed one after the other and form the digital signal. [1]
5 a 0101 and 1000 [2]
b Values shown as horizontal lines of 0.5 ms duration [1]
Values plotted: 5, 8, 8, 6, 2, 1, 2, 6, 8, 8 [1]
Graph correct with labels [1]
c i Any variation in the signal that occurs between sampling is not detected [1]
Increasing sampling frequency decreases the time between samples [1]
Frequency at least 2 × signal, i.e. 600 Hz (frequency of signal ≈ 300 Hz) [1]
ii The variation in voltage can use more voltage levels (28
levels rather than 24
levels). [1]
The signal voltage at every sample is closer to the actual value. [1]
6 Signal-to-noise ratio (in dB) = 10 lg ⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−
−
19
3
10
0
2
10
0
6
.
.
= 164.7 dB ≈ 165 dB [1]
7 a Signal-to-noise ratio = 10 lg 3
3
10
0.001
10
1.0
−
−
×
×
= 30 dB [1]
b Signal becomes 0.001 mW or signal-to-noise ratio is 1 [1]
0 dB [1]
8
6
4
2
0
0 1 2 3 4 5 t / ms
V / mV

8 Any two reasons and explanation.
Less attenuation so fewer repeater/regeneration amplifiers needed. [2]
More bandwidth so more data can be sent per second/more telephone calls made at once. [2]
Less interference/noise so fewer regeneration amplifiers needed. [2]
Lower diameter/weight so easier to handle/cheaper. [2]
9 a i Any value less than 10 m (and more than 1 mm) [1]
ii Any value between 10 and 100 m [1]
b The sky wave uses reflection by the ionosphere for transmission. [1]
The ionosphere fluctuates in its ability to reflect. [1]
10 a Orbits around the Earth’s equator. [1]
Takes one day for a complete orbit. [1]
Stays over one point on the Earth OR height of orbit 36 000 km above Earth’s surface. [1]
b First satellites used wavelengths of about 5 cm; typically now between 1 mm and 1 cm. [1]
c Advantage: permanent link with ground station/dishes do not have to be moved. [1]
Disadvantage: greater time delay for signal OR further away so signal weaker. [1]
11 The public switched telephone network connects every telephone through exchanges. [1]
Without exchanges too many telephones and interconnecting wires are needed. [1]
One cable can handle many telephone conversations at once. [1]
Sampling places a series of digital bits from many telephone conversations on one cable. [1]

18 Worksheet (A2)
Data needed to answer questions can be found in the Data, formulae and relationships sheet.
1 Convert the following angles into radians.
a 30° [1]
b 210° [1]
c 0.05° [1]
2 Convert the following angles from radians into degrees.
a 1.0 rad [1]
b 4.0 rad [1]
c 0.15 rad [1]
3 The planet Mercury takes 88 days to orbit once round the Sun.
Calculate its angular displacement in radians during a time interval of:
a 44 days [1]
b 1 day. [1]
4 In each case below, state what provides the centripetal force on the object.
a A car travels at a high speed round a sharp corner. [1]
b A planet orbits the Sun. [1]
c An electron orbits the positive nucleus of an atom. [1]
d Clothes spin round in the drum of a washing machine. [1]
5 An aeroplane is circling in the sky at a speed of 150 m s−1
.
The aeroplane describes a circle of radius 20 km.
For a passenger of mass 80 kg inside this aeroplane, calculate:
a her angular velocity [2]
b her centripetal acceleration [3]
c the centripetal force acting on her. [2]
6 The diagram shows a stone tied to the end of a length of string.
It is whirled round in a horizontal circle of radius 80 cm.
The stone has a mass of 90 g and it completes 10 revolutions in a time of 8.2 s.
a Calculate:
i the time taken for one revolution [1]
ii the distance travelled by the stone during one revolution (this distance is equal to the
circumference of the circle) [1]
iii the speed of the stone as it travels in the circle [2]
iv the centripetal acceleration of the stone [3]
v the centripetal force on the stone. [2]
b What provides the centripetal force on the stone? [1]
c What is the angle between the acceleration of the stone and its velocity? [1]

18 Worksheet (A2)
7 A lump of clay of mass 300 g is placed
close to the edge of a spinning turntable.
The centre of mass of the lump of clay travels
in a circle of radius 12 cm.
a The lump of clay takes 1.6 s to complete one revolution.
i Calculate the rotational speed of the clay. [2]
ii Calculate the frictional force between the clay and the turntable. [3]
b The maximum magnitude of the frictional force F between the clay and the turntable is
70% of the weight of the clay. The speed of rotation of clay is slowly increased.
Determine the speed of the clay when it just starts to slip off the turntable. [4]
8 The diagram shows a skateboarder of mass 70 kg who drops through a vertical height
of 5.2 m.
The dip has a radius of curvature of 16 m.
a Assuming no energy losses due to air resistance or friction, calculate the speed of the
skateboarder at the bottom of the dip at point B.
You may assume that the speed of the skateboarder at point A is zero. [2]
b i Calculate the centripetal acceleration of the skateboarder at point B. [3]
ii Calculate the contact force R acting on the skateboarder at point B. [3]
9 A car of mass 820 kg travels at a constant speed
of 32 m s−1
along a banked track.
The track is banked at an angle of 20°
to the horizontal.
a The net vertical force on the car is zero.
Use this to show that the contact force R on the car is 8.56 kN. [2]
b Use the answer from a to calculate the radius of the circle described by the car. [4]

18 Worksheet (A2)
10 A stone of mass 120 g is fixed to one end of a light rigid rod.
The stone is whirled at a constant speed of 4.0 m s−1
in a vertical circle of radius 80 cm.
Calculate the ratio:
B
A
at
rod
in the
tension
at
rod
in the
tension
[6]
Total:
59
Score: %

19 Worksheet (A2)
1 Define gravitational field strength at a point in space. [1]
2 Show that the gravitational constant G has the unit N m2
kg−2
. [2]
3 The gravitational field strength on the surface of the Moon is 1.6 N kg−1
.
What is the weight of an astronaut of mass 80 kg standing on the surface of the Moon? [2]
4 Calculate the magnitude of the gravitational force between the objects described below.
You may assume that the objects are ‘point masses’.
a two protons separated by a distance of 5.0 × 10−14
m
(mass of a proton = 1.7 × 10−27
kg) [3]
b two binary stars, each of mass 5.0 × 1028
kg,
with a separation of 8.0 × 1012
m [2]
c two 1500 kg elephants separated by a distance of 5.0 m [2]
5 The diagram shows the Moon and an artificial satellite orbiting round the Earth.
The radius of the Earth is R.
a Write an equation for the gravitational field strength g at a distance r from the centre of
an isolated object of mass M. [1]
b By what factor would the gravitational field decrease if the distance from the centre of
the mass were doubled? [2]
c The satellite orbits at a distance of 5R from the Earth’s centre and the Moon is at a
distance of 59R. Calculate the ratio:
Moon
of
position
at
strength
field
nal
gravitatio
satellite
of
position
at
strength
field
nal
gravitatio
[3]
6 The planet Neptune has a mass of 1.0 × 1026
kg and a radius of 2.2 × 107
m.
Calculate the surface gravitational field strength of Neptune. [3]

19 Worksheet (A2)
7 Calculate the radius of Pluto, given its mass is 5.0 × 1023
kg and its surface
gravitational field strength has been estimated to be 4.0 N kg−1
. [3]
8 A space probe of mass 1800 kg is travelling from Earth to the planet Mars.
The space probe is midway between the planets. Use the data given to calculate:
a the gravitational force on the space probe due to the Earth [3]
b the gravitational force on the space probe due to Mars [2]
c the acceleration of the probe due to the gravitational force acting on it. [3]
Data
separation between Earth and Mars = 7.8 × 1010
m
mass of Earth = 6.0 × 1024
kg mass of Mars = 6.4 × 1023
kg
9 An artificial satellite orbits the Earth at a height of 400 km above its surface.
The satellite has a mass 5000 kg, the radius of the Earth is 6400 km and the mass of the
Earth is 6.0 × 1024
kg. For this satellite, calculate:
a the gravitational force experienced [3]
b its centripetal acceleration [2]
c its orbital speed. [3]
10 a Explain what is meant by the term gravitational potential at a point. [2]
b Write down the gravitational potential energy of a body of mass 1 kg when it is at
an infinite distance from another body. [1]
c The radius of the Earth is 6.4 × 106
m and the mass of the Earth = 6.0 × 1024
kg.
Calculate the potential energy of the 1 kg mass at the Earth’s surface. [3]
d Write down the minimum energy required to remove the body totally from the Earth’s
gravitational field. [1]
11 The planets in our solar system orbit the Sun in almost circular orbits.
a Show that the orbital speed v of a planet at a distance r from the centre of the Sun is
given by:
v =
r
GM
[4]
b The mean distance between the Sun and the Earth is 1.5 × 1011
m and the mass of the Sun
is 2.0 × 1030
kg.
Calculate the orbital speed of the Earth as it travels round the Sun. [2]
12 There is a point between the Earth and the Moon where the net gravitational field strength
is zero. At this point the Earth’s gravitational field strength is equal in magnitude but
opposite in direction to the gravitational field strength of the Moon.
Given that:
Moon
of
mass
Earth
of
mass
= 81
calculate how far this point is from the centre of the Moon in terms of R, where R is
the separation between the centres of the Earth and the Moon. [4]
Total:
57
Score: %

20 Worksheet (A2)
1 For an oscillating mass, define:
a the period [1]
b the frequency. [1]
2 The graph of displacement x against time t for an object executing simple harmonic motion
(s.h.m.) is shown here.
a State a time at which the object has maximum speed. Explain your answer. [2]
b State a time at which the magnitude of the object’s acceleration is a maximum.
Explain your answer. [2]
3 An apple is hung vertically from a length of string to form a simple pendulum.
The apple is pulled to one side and then released. It executes 12 oscillations in a time of 13.2 s.
a Calculate the period of the oscillations. [2]
b Calculate the frequency of the oscillations. [2]
4 This is the graph of displacement x against time t for an oscillating object.
Use the graph to determine the following:
a the amplitude of the oscillation [1]
b the period [1]
c the frequency in hertz (Hz) [2]
d the angular frequency in radians per second (rad s−1
). [2]
e the maximum speed of the oscillating mass. [2]

20 Worksheet (A2)
5 Two objects A and B have the same period of oscillation. In each case a and b below,
determine the phase difference between the motions of the objects A and B.
a [2]
b [2]
6 A mass at the end of a spring oscillates with a period of 2.8 s.
The maximum displacement of the mass from its equilibrium position is 16 cm.
a What is the amplitude of the oscillations? [1]
b Calculate the angular frequency of the oscillations. [2]
c Determine the maximum acceleration of the mass. [3]
d Determine the maximum speed of the mass. [2]
7 A small toy boat is floating on the water’s surface. It is gently pushed down and then
released. The toy executes simple harmonic motion. Its displacement–time graph
is shown here.
For this oscillating toy boat, calculate:
a its angular frequency [2]
b its maximum acceleration [3]
c its displacement after a time of 6.7 s, assuming that the effect of damping on the boat is
negligible. [3]

20 Worksheet (A2)
8 The diagram shows the displacement–time graph for a particle executing simple
harmonic motion.
Sketch the following graphs for the oscillating particle:
a velocity–time graph [2]
b acceleration–time graph [2]
c kinetic energy–time graph [2]
d potential energy–time graph. [2]
9 A piston in a car engine executes simple harmonic motion.
The acceleration a of the piston is related to its displacement x by the equation:
a = −6.4 × 105
x
a Calculate the frequency of the motion. [3]
b The piston has a mass of 700 g and a maximum displacement of 8.0 cm.
Calculate the maximum force on the piston. [2]
10 The diagram shows a trolley of mass m attached
to a spring of force constant k. When the trolley
is displaced to one side and then released, the
trolley executes simple harmonic motion.
a Show that the acceleration a of the trolley is given by the expression:
a = x
m
k
⎟
⎠
⎞
⎜
⎝
⎛
−
where x is the displacement of the trolley from its equilibrium position. [3]
b Use the expression in a to show that the frequency f of the motion is given by:
f =
m
k
π
2
1
[2]
c The springs in a car’s suspension act in a similar way to the springs on the trolley.
For a car of mass 850 kg, the natural frequency of oscillation is 0.40 s.
Determine the force constant k of the car’s suspension. [3]
Total:
59
Score: %

21 Worksheet (A2)
Specific heat capacity of water = 4200 J kg−1
K−1
Specific latent heat of fusion of water = 3.4 × 105
J kg−1
1 Describe the arrangement of atoms, the forces between the atoms and the motion of the
atoms in:
a a solid [3]
b a liquid [3]
c a gas. [3]
2 A small amount of gas is trapped inside a container. Describe the motion of the gas atoms
as the temperature of the gas within the container is increased. [3]
3 a Define the internal energy of a substance. [1]
b The temperature of an aluminium block increases when it is placed in the flame of a
Bunsen burner. Explain why this causes an increase in its internal energy. [3]
c A lump of metal is melting in a hot oven at a temperature of 600 °C.
Explain whether its internal energy is increasing or decreasing as it melts. [4]
4 Write a word equation for the change in the thermal energy of a substance in terms of its
mass, the specific heat capacity of the substance and its change of temperature. [1]
5 The specific heat capacity of a substance is measured in the units J kg−1
K−1
, whereas
its specific latent heat of fusion is measured in J kg−1
. Explain why the units are different. [2]
6 During a hot summer’s day, the temperature of 6.0 × 105
kg of water in a swimming pool
increases from 21 °C to 24 °C. Calculate the change in the internal energy of the water. [3]
7 A 300 g block of iron cools from 300 °C to room temperature at 20 °C. The specific heat
capacity of iron is 490 J kg−1
K−1
. Calculate the heat released by the block of iron. [3]
8 Calculate the energy that must be removed from 200 g of water at 0 °C to convert it all
into ice at 0 °C. [2]

21 Worksheet (A2)
9 a Change the following temperatures from degrees Celsius into kelvin.
i 0 °C
ii 80 °C
iii −120 °C [3]
b Change the following temperatures from kelvin into degrees Celsius.
i 400 K
ii 272 K
iii 3 K [3]
10 An electrical heater is used to heat 100 g of water in a well-insulated container at a steady
rate. The temperature of the water increases by 15 °C when the heater is operated for a
period of 5.0 minutes. Determine the change of temperature of the water when the same
heater and container are individually used to heat:
a 300 g of water for the same period of time [3]
b 100 g of water for a time of 2.5 minutes. [3]
11 The graph below shows the variation of the temperature of 200 g of lead as it is heated
at a steady rate.
a Use the graph to state the melting point of lead. [1]
b Explain why the graph is a straight line at the start. [1]
c Explain what happens to the energy supplied to the lead as it melts at a constant
temperature. [1]
d The initial temperature of the lead is 0 °C. Use the graph to determine the total
energy supplied to the lead before it starts to melt. [3]
(The specific heat capacity of lead is 130 J kg−1
K−1
.)
e Use your answer to d to determine the rate of heating of the lead. [2]
f Assuming that energy continues to be supplied at the same rate, calculate the
specific latent heat of fusion of lead. [3]

21 Worksheet (A2)
12 The diagram shows piped water being heated by an electrical heater.
The water flows through the heater at a rate of 0.015 kg s−1
. The heater warms the water
from 15 °C to 42 °C. Assuming that all the energy from the heater is transferred to
heating the water, calculate the power of the heater. [5]
13 A gas is held in a cylinder by a friction-free piston. When the force holding the piston in
place is removed, the gas expands and pushes the piston outwards.
Explain why the temperature of the gas falls. [2]
14 Hot water of mass 300 g and at a temperature of 90 °C is added to 200 g of cold water
at 10 °C. What is the final temperature of the mixture? You may assume there are no
losses to the environment and all heat transfer takes place between the hot water and
the cold water. [5]
15 A metal cube of mass 75 g is heated in a naked flame until it is red hot. The metal block is
quickly transferred to 200 g of cold water. The water is well stirred. The graph shows the
variation of the temperature of the water recorded by a datalogger during the experiment.
The metal has a specific heat capacity of 500 J kg−1
K−1
. Use the additional information
provided in the graph to determine the initial temperature of the metal cube. You may
assume there are no losses to the environment and all heat transfer takes place between
the metal block and the water. [5]
Total:
71
Score: %

22 Worksheet (A2)
1 Determine the number of atoms or molecules in each of the following.
a 1.0 mole of carbon [1]
b 3.6 moles of water [1]
c 0.26 moles of helium [1]
2 The molar mass of helium is 4.0 g.
Determine the mass of a single atom of helium in kilograms. [2]
3 The molar mass of uranium is 238 g.
a Calculate the mass of one atom of uranium. [2]
b A small rock contains 0.12 g of uranium. For this rock, calculate the number of:
i moles of uranium [2]
ii atoms of uranium. [1]
4 Explain what is meant by the absolute zero of temperature. [3]
5 a Write the ideal gas equation in words. [1]
b One mole of an ideal gas is trapped inside a rigid container of volume 0.020 m3
.
Calculate the pressure exerted by the gas when the temperature within the
container is 293 K. [3]
6 A fixed amount of an ideal gas is trapped in a container of volume V.
The pressure exerted by the gas is P and its absolute temperature is T.
a Using a sketch of PV against T, explain how you can determine the number of moles
of gas within the container. [4]
b Sketch a graph of PV against P when the gas is kept at a constant temperature.
Explain the shape of the graph. [3]
7 A rigid cylinder of volume 0.030 m3
holds 4.0 g of air. The molar mass of air is about 29 g.
a Calculate the pressure exerted by the air when its temperature is 34 °C. [4]
b What is the temperature of the gas in degrees Celsius when the pressure is twice
your value from part a? [4]

22 Worksheet (A2)
8 The diagram shows two insulated containers holding gas.
The containers are connected together by tubes of negligible volume.
The internal volume of each container is 2.0 × 10−2
m3
.
The temperature within each container is −13 °C. The gas in container A exerts a pressure
of 180 kPa and the gas in container B exerts a pressure of 300 kPa.
a Show that the amount of gas within the two containers is about 4.4 moles. [3]
b The valve connecting the containers is slowly opened and the gases are allowed to mix.
The temperature within the containers remains the same.
Calculate the new pressure exerted by the gas within the containers. [3]
9 The diagram shows a cylinder containing air
at a temperature of 5.0 °C.
The piston has a cross-sectional area 1.6 × 10−3
m2
It is held stationary by applying a force
of 400 N applied normally to the piston.
The volume occupied by the compressed air
is 2.4 × 10−4
m3
.
The molar mass of air is about 29 g.
a Calculate the pressure exerted by the compressed air. [2]
b Determine the number of moles of air inside the cylinder. [3]
c Use your answer to b to determine:
i the mass of air inside the cylinder [1]
ii the density of the air inside the cylinder. [2]
10 The mean speed of a helium atom at a temperature of 0 °C is 1.3 km s–1
. Estimate the mean
speed of helium atoms on the surface of a star where the temperature is 10 000 K. [6]
11 The surface temperature of the Sun is about 5400 K. On its surface, particles behave like the
atoms of an ideal gas. The atmosphere of the Sun mainly consists of hydrogen nuclei.
These nuclei move in random motion.
a Explain what is meant by random motion. [1]
b i Calculate the mean translational kinetic energy of a hydrogen nucleus
on the surface of the Sun. [2]
ii Estimate the mean speed of such a hydrogen nucleus.
(The mass of hydrogen nucleus is 1.7 × 10−27
kg.) [3]
12 a Calculate the mean translational kinetic energy of gas atoms at 0 °C. [2]
b Estimate the mean speed of carbon dioxide molecules at 0 °C.
(The molar mass of carbon dioxide is 44 g.) [5]
c Calculate the change in the internal energy of one mole of carbon dioxide gas when its
temperature changes from 0 °C to 100 °C. [3]

22 Worksheet (A2)
13 The diagram below shows three different types of arrangements of gas particles.
A gas whose particles consist of single atoms is referred to as monatomic – for example
helium (He). A gas with two atoms to a molecule is called diatomic – for example oxygen
(O2). A gas with more than two atoms to a molecule is said to be polyatomic – for example
water vapour (H2O).
A single atom can travel independently in the x, y and z directions: it is said to have three
degrees of freedom. From the equation for the mean translational kinetic energy of the atom,
we can generalise that a gas particle has mean energy of kT
2
1
per degree of freedom.
Molecules can also have additional degrees of freedom due to their rotational energy.
a Use the diagram above to explain why:
i the mean energy of a diatomic molecule is kT
2
5
[2]
ii the mean energy of a polyatomic molecule is 3kT. [2]
b Calculate the internal energy of one mole of water vapour (steam) per unit kelvin. [3]
Total:
75
Score: %

23 Worksheet (A2)
1 a Explain what is meant by the electric field strength at a point. [1]
b Explain what is meant by the electric potential at a point. [1]
2 A pair of parallel metal plates has a potential difference of 5000 V across them.
The electric field strength between them is 400 kN C−1
. Calculate:
a the separation between the plates [2]
b the force on a dust particle between the plates which carries a charge of 1.6 × 10−19
C. [2]
3 The electric field strength E at a distance r from a point charge Q may be written as:
E = k 2
r
Q
What is the value for k? [1]
4 The diagram shows a point charge +q placed in the electric field of a charge +Q.
The force experienced by the charge +q at point A is F. Calculate the magnitude of the
force experienced by this charge when it is placed at points B, C, D and E. In each case,
explain your answer. [9]
5 A spherical metal dome of radius 15 cm is electrically charged. It has a positive charge of
+2.5 µC distributed uniformly on its surface.
a Calculate the electric field strength on the surface of the dome. [3]
b Explain how your answer to a would change at a distance of 30 cm from the surface of
the dome. [2]

23 Worksheet (A2)
6 The diagram shows two point charges.
The point X is midway between the charges.
a Calculate the electric field strength at point X due to:
i the +20 µC charge [3]
ii the +40 µC charge. [2]
b Calculate the resultant electric field strength at point X. [2]
7 The dome of a van de Graaff generator has a diameter of 30 cm and is at a potential of
+20 000 V. Calculate:
a the charge on the dome [2]
b the electric field strength at the surface of the dome [2]
c the force on a proton near the surface of the dome. [1]
8 a An isolated charged sphere of diameter 10 cm carries a charge of −2000 nC.
Calculate the potential at its surface. [3]
b Calculate the work that must be done to bring an electron from infinity to the
surface of the dome. [2]
9 Describe some of the similarities and differences between the electrical force due to a point
charge and the gravitational force due to a point mass. [6]
10 The diagram shows two point charges.
Calculate the distance x of point P from
charge +Q where the net electric field
strength is zero. [6]
11 Show that the ratio:
protons
o
between tw
force
nal
gravitatio
protons
o
between tw
force
electrical
is about 1036
and is independent of the actual separation between the protons. [6]
12 A helium nucleus consists of two protons and two neutrons. Its diameter is about 10−15
m.
a Calculate the force of electrostatic repulsion between two protons at this separation. [2]
b Calculate the potential at a distance of 10–15
m from the centre of a proton. [2]
c How much work would need to be done to bring two protons this close to each other? [2]
d If one proton were stationary, at what speed would the second proton need to be fired
at it to get this close? (Ignore any relativistic effects.) [3]
Total:
66
Score: %

24 Worksheet (A2)
1 A 30 µF capacitor is connected to a 9.0 V battery.
a Calculate the charge on the capacitor. [2]
b How many excess electrons are there on the negative plate of the capacitor? [2]
2 The p.d. across a capacitor is 3.0 V and the charge on the capacitor is 150 nC.
a Determine the charge on the capacitor when the p.d. is:
i 6.0 V [2]
ii 9.0 V. [2]
b Calculate the capacitance of the capacitor. [2]
3 A 1000 µF capacitor is charged to a potential difference of 9.0 V.
a Calculate the energy stored by the capacitor. [2]
b Determine the energy stored by the capacitor when the p.d. across it is doubled. [2]
4 For each circuit below, determine the total capacitance of the circuit. [13]
5 The diagram shows an electrical circuit.
a Calculate the total capacitance of the two capacitors in parallel. [2]
b What is the potential difference across each capacitor? [1]
c Calculate the total charge stored by the circuit. [2]
d Calculate the total energy stored by the capacitors. [2]

24 Worksheet (A2)
6 A 10 000 µF capacitor is charged to its maximum operating voltage of 32 V.
The charged capacitor is discharged through a filament lamp. The flash of light from
the lamp lasts for 300 ms.
a Calculate the energy stored by the capacitor. [2]
b Determine the average power dissipated in the filament lamp. [2]
7 The diagram shows a 1000 µF capacitor charged to a p.d. of 12 V.
a Calculate the charge on the 1000 µF capacitor. [2]
b The 1000 µF capacitor is connected across an uncharged 500 µF capacitor by closing
the switch S. The charge initially stored by the 1000 µF capacitor is now shared with
the 500 µF capacitor.
i Calculate the total capacitance of the capacitors in parallel. [2]
ii Show that the p.d. across each capacitor is 8.0 V. [2]
8 The diagram shows a circuit used to measure the capacitance of a capacitor.
The reed switch vibrates between the two contacts with a frequency of 50 Hz. On each oscillation
the capacitor is fully charged and totally discharged. The current through the milliammeter is
225 mA.
a Calculate the charge that flows off the capacitor each time it is discharged. [1]
b Calculate the capacitance of the capacitor. [2]
c Calculate the current through the milliammeter when a second identical capacitor
is connected:
i in parallel with the original capacitor [1]
ii in series with the original capacitor. [1]
reed switch
mA
9.0 V

24 Worksheet (A2)
9 A capacitor of capacitance 200 µF is connected across a 200 V supply.
a Calculate the charge stored on the plates. [1]
b Calculate the energy stored on the capacitor. [1]
The capacitor is now disconnected from the power supply and is connected across a 100 µF
capacitor.
c Calculate the potential difference across the capacitors. [3]
d Calculate the total energy stored on the capacitors. [2]
e Suggest where the energy has been lost. [1]
10 The diagram below shows a charged capacitor of capacitance C. When the switch S is closed,
this capacitor is connected across the uncharged capacitor of capacitance 2C.
Calculate the percentage of energy lost as heat in the resistor and explain why the actual
resistance of the resistor is irrelevant. [7]
Total:
64
Score: %

25 Worksheet (A2)
1 The diagram shows the magnetic field pattern for a
current-carrying straight wire drawn by a student in her notes.
List two errors made by the student. [2]
2 A current-carrying conductor is placed in an external magnetic field. In each case below, use
Fleming’s left-hand rule to predict the direction of the force on the conductor.
a b c
[3]
3 The unit of magnetic flux density is the tesla.
Show that: 1 T = 1 N A−1
m−1
[2]
4 Calculate the force per centimetre length of a straight wire placed at right angles to a uniform
magnetic field of magnetic flux density 0.12 T and carrying a current of 3.5 A. [3]
5 A 4.0 cm long conductor carrying a current of 3.0 A is placed in a uniform magnetic field
of flux density 50 mT. In each of a, b and c below, determine the size of the force acting
on the conductor. [6]
a b c
current into paper

25 Worksheet (A2)
6 The diagram shows the rectangular loop PQRS of a simple electric motor placed in a
uniform magnetic field of flux density B.
The current in the loop is I. The lengths PQ and RS are both L and lengths QR and SP are both x.
Show that the torque of the couple acting on the loop for a given current and magnetic flux
density is directly proportional to the area of the loop. [5]
7 The diagram shows a rigid wire AB pivoted at
the point A so that it is free to move in a vertical
plane. The lower end of the wire dips into
mercury. A uniform magnetic field of
6.0 × 10−3
T acts into the paper throughout the
diagram.
a When the current is switched on, the wire
continuously moves up out of the mercury
and then falls back again. Explain this
motion. [4]
b The force on the wire due to the current may
be taken to act at the midpoint of the wire.
When the current is first switched on, the
moment of this force about A is 3.5 × 10−5
N m.
Calculate:
i the force acting on the wire [2]
ii the current in the wire. [2]
8 The coil in the d.c. motor shown in question 6 has a length L = 7.0 × 10−2
m and width
x = 3.0 × 10−2
m. There are 25 turns on the coil and it is placed in a uniform magnetic field of
0.19 T. The coil carries a current of 2.8 A. The coil is free to rotate about an axis midway between
PQ and RS.
a Calculate the force on the longest side of the coil. [2]
b Calculate the maximum torque (moment) exerted on the coil. [2]
c Explain why the force acting on the long side of the coil does not change as the coil rotates
but the torque exerted on the coil varies. [2]
axis

8 Marking scheme Worksheet (A2.pdf

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