Question Answer Mark

Number
1 1
The only correct answer is B
A is not correct because if the half-life were 2.4 hours the activity would be 150 Bq
C is not correct because if the half-life were 4.0 hours the activity would be 600 Bq
D is not correct because if the half-life were 12 hours the activity would be 2400 Bq

Question Answer Mark

Number
2 1
The only correct answer is B
A is not correct because standard candles enable distances to be determined
C is not correct because the age of the universe is not related to the Planck constant
D is not correct because the age of the universe is not simply related to the
temperature of deep space

Question Answer Mark

Number
3 1
The only correct answer is C
A is not correct because the smaller the number of nuclei the lower the activity
B is not correct because the larger the half-life of the source the lower the activity
D is not correct because the larger the half-life of the source the lower the activity

Question Answer Mark

Number
4 1
The only correct answer is A
B is not correct because P is further away with a greater flux, so luminosity is greater
C is not correct because a smaller parallax angle indicates a greater distance
D is not correct because a smaller parallax angle indicates a greater distance

Question Answer Mark

Number
5 1
The only correct answer is C
A is not correct because nuclear decay releases only a small fraction of the total energy
B is not correct because nuclear emission is another way to describe nuclear decay
D is not correct because uranium fuel does not undergo fusion
 The only correct answer is C

A is incorrect, as radiation flux decreases with distance from the source and
Q is further away than P.
6 (1)
B is incorrect as radiation flux decreases with distance from the source and
Q is further away than P
D is incorrect as, the parallax angle for P is greater than that for Q and so P
is closer than Q

The only correct answer is B

7 A is incorrect, as X has a smaller amplitude than Y. (1)

C is incorrect, as Z has a smaller amplitude than Y.
D is incorrect, as X and Z have a smaller amplitude than Y.

The only correct answer is D

A is incorrect, as the cosmic background radiation suggests the Big Bang.

8 B is incorrect, as the gravitational constant determines the strength of the (1)
gravitational interaction between two masses.
C is incorrect, as a Hertzsprung-Russell diagram shows the way in which
luminosity depends upon temperature for a range of stars.

The only correct answer is D

A is incorrect, as the temperature scale is increasing from X to Y.

9 (1)
B is incorrect, as the temperature scale is increasing from X to Y.
C is incorrect, as although the temperature scale is linear and not
logarithmic.

The only correct answer is B

10 A is incorrect, as elastic deformation returns energy to the building (1)

C is incorrect, as stiffness is unrelated to energy dissipation.
D is incorrect, as strength is unrelated to energy dissipation.

Total for multiple choice questions 10

Q.11
Question
Answer Mark
Number
17(a) Top line correct (1)
Bottom line correct (1) 2

0 
239
93 Np239
94 Pu  -1 β
17(b)(i) The rate of decay of the neptunium nuclei is proportional to the number of
neptunium nuclei
∆
Or ∝ with symbols defined (1) 2
∆

Hence the number of neptunium nuclei decaying per second increases

(1)
over time
17(b)(ii) When the rate of decay of neptunium nuclei is equal to the rate of
(1) 1
production of neptunium nuclei
17(b)(iii)
Use of t1/2  loge 2 (1)

Use of A  ()N (1)

N = 5.44 × 1012 [may see 5.45 × 1012 depending upon rounding] (1)

Example of calculation
3
loge 2 0.693
   3.39  106 s 1
t1/2 2.04  105 s
1.85  10 7 s 1  ( -)3.39  10 -6 N
1.85  10 7 s 1
N   5.44  10 12
3.39  10 -6 s 1
17(c) The decay products of the fission are radioactive
Or the decay products emit (very) penetrating radiation (1)

Neutrons can only be stopped by large thicknesses of concrete

[accept -radiation for neutrons] (1) 2

Total for question 17 10

 Q.12.

38. (a) Expression for gravitational force:

F = GMm/r2 (1) 1

(b) Expression for gravitational field strength:

g = force on 1 kg, so g = GM/r2, or g = F/m so g = GM/r2 1

(c) Radius of geostationary orbit:

Idea that a = g, and suitable expression for a quoted [can be
in terms of forces] (1)
substitution for velocity in terms of T (1)
algebra to obtain required result (1) 3
Example of derivation:
g = v2/r or g = ω2r
and v = 2πr/T or ω = 2π/T
so (2πr/T)2/r = GM/r2 or (2π/T)2r = GM/r2, leading to expression given

(d) Calculation of radius:

Substitution into expression given (1)
Correct answer [4.2 × 107 m] (1) 2
Example of calculation:
r3 = 6.67 × 10–11 N m2 kg–2 × 6.0 × 1024 kg × (24 × 60 × 60 s)2 /4π2
= 7.6 × 1022 m3
So r = 4.2 × 107 m

(e) (i) Satellite with greater mass:

Yes – because, in geostationary orbit, r constant so
acceleration remains the same, regardless of mass (1)

(ii) Satellite with greater speed:

No + suitable argument (1) 2
[e.g. for geostationary orbit, T and r are fixed, so v cannot
increase (v = 2πr/T)]

(f) Why satellite must be over equator:

Idea that centre of satellite’s orbit must be the centre of the
Earth (can be shown on diagram) (1)
there must be a common axis of rotation for the satellite and
the Earth / the satellite’s orbit must be at right angles to the
spin axis of the Earth (1) 2
[11]
 Q.13.

8. (a) Change in nuclear composition

• Nucleus has one less neutron OR nucleus has one more proton) 1

(b) (i) Calculation of age of skull

• Use of λ = ln2/t½ to obtain value for λ
• Use of N = Noe–λt
• Correct answer for age of skull [1.2 × 104 y; 3.83 × 1011 s]
Example of calculation:
λ = ln 2/t½ = ln 2/5730 y = 1.2 × 10–4 y–1 [3.84 × 10–12 s–1]
ln(N/No) = –λt
ln(2.3 × 10–11/1.0 × 10–10) = –(1.2 × 10–4 y–1)t
t = 1.2 × 104 y
Alternative mark scheme
• Use of half life rule
• Correct answer for number of half lives [2.12]
• Correct answer for age of skull [1.2 × 104 y]
Example of calculation:
N/ No = (0.5)n
(2.3 × 10–11)/(1 × 10–10) = (0.5)n
log(0.23) = n log(0.5)
n = log(0.23)/log(0.5) = 2.12
t = 2.12 × 5730 = 1.2 × 104 y 3

(ii) Reason for inaccuracy

• Idea that it is impossible to know the exact proportion of
14
C in the atmosphere when the bones were formed OR
reference to the difficulty of measuring such small
percentages of 14C. 1

(iii) Why 210Pb is more suitable:

• Idea that the half life of 210Pb is closer to the age of
recent bones [e.g. a greater proportion of 210Pb will have
decayed as the time elapsed is one or more half lives] 1
[6]
 Q.14.

33. (a) Two deductions [not simplv word descriptions of features of the diagram]

The gravitational potential is increasing with height / when moving

away from the Earth / Work must be done to move away from the Earth (1)
1
[Ignore idea that V ∝ ; in words or symbols]
r
The field is non-uniform / radial / Field strength decreases with
height / when moving away from the Earth (1) 2

(b) Entry speed at Earth’s atmosphere

| –1MJkg–1 – (– 61MJkg–1) = 60MJkg–1 [accept ± 60MJkg–1] (1)
Loss of GPE/ Gain in KE of spacecraft = m (1)
Statement / use of ½ mv2 = m∆V OR ½v2 = ∆V/(= 60 MJ kg–1) (1)
[either of these statements also earns the second mark, if not already awarded]
[See v2 = 1.2 × 108
Answer 1.095 × 103 m s–1 /10950 m s–1/11.0 km s–1 [more than 2 sf] (1) 4

(c) Showing relative distance

GM E m GM M m
Use of Newton’s Law; FE = or FM = (1)
rE rM
2 2

rE ME r M
2
= (or) E = E
(1)
rM
2
MM rM MM
[or equivalent re-arrangement]
rE 6.0 × 10 24 kg
2
= = 81(.1) [or equivalent] (1)
rM
2
7.4 × 10 22 kg
[correct relationship, expressed in terms of numerical values]
[If inverted, then MM:ME = 0.0123]
r r
So E = 81 = 9 [or M = 0.11] (independent mark) (1)
rM rE
[Stating 81 =100 at the third mark stage does not, alone, earn the 4th mark]
[Beware ambiguity or transposition of r values at steps 2 or 3] 4
[10]
 Q.15.

41. H–R Diagram

(i) L and T (1)
L and K (1) 2
or L and L (1), T and K (1), not W]

(ii) Any 2 correct [of 102, 1 or 10°, 10–2] (1)

All 3 correct (1) 2

(iii) 20 000 and 5 000 (1) 1

Identify stars
(iv) Red giant = (B and) C (1)
Low mass ms star = E [ignore X] (1) 2

Zeta Tauri Luminosity

(v) Use of L = 4 π D2 I (1)
Correct substitution (1)
3.8(2) × 1030(W) (1) 3

Zeta Tauri identification (ecf)

(vi) 3.8(2) × 1030 W ÷ 3.9 × 1026 W [or 4 × 1030 W used] (1)
Correct ratio [e.g. 9700, 9800, 10300 or 104, etc.] (1)
Hence A [from answer in range 9700 to 10300] (1) 3
[13]
 Q.16.

28. Displacement-time graph:

Cosine curve (1)

Constant period and amplitude (1) 2

Displacement

0 Time
Velocity-time and acceleration-time graphs:
Velocity: sine curve; 90° out of phase with displacement-time (1)
Velocity

0 Time

Acceleration: cosine curve; 180° out of phase with displacement-time (1) 2

Acceleration

0 Time

Two requirements:
Force towards centre (1)
a (or F) ∝ x (1)
opposite direction [acceleration and displacement acceptable] (1) Max 2

Displacement-time and acceleration-time:

Starts positive, curved, always > 0 (1)
Period same as velocity (1) 2

Displacement

0 Time
Acceleration constant not equal to 0 (1)
Sharp peak when ball in contact with floor (1) 2
Acceleration

0 Time

Explanation:
Not simple harmonic motion (1)
Reason e.g. acceleration constant when ball in free fall / period not constant /
acceleration not ∝ displacement. (1) 2
[12]
Q.17.

23. Definition of specific heat capacity

Energy (needed) 1
(per) unit mass/kg ) 1
(per) unit temperature change/ °C / K )
OR
Correct formula [does not need to be rearranged] 1
with correctly defined symbols 1

Circuit diagrams

A OR
A

V V

Accept voltmeter across heater and ammeter as well as voltmeter across heater only
Means of varying p.d./current 1
Voltmeter in parallel with a resistor symbol 1
Ammeter in series with any representation of heater 1

Other apparatus
• (Top pan) balance / scales 1
• Stopwatch / timer / clock 1
Explanation
Energy/heat loss to surroundings/air/bench
OR
mc∆θ +∆Q = Vlt or equivalent in words (e.g. student ignores
energy loss in calculations)
OR
mc∆T +∆Q = Vlt or equivalent words 1

Modifications
Any two from
• Use of insulation around block
• Ensure all of heater is within block
• Grease heater/thermometer 2
[10]
Q.18.
11. (a) Meanings

Spontaneous: Happens independently of/cannot be controlled by/

unaffected by chemical conditions/physical conditions/temperature/
pressure or without stimulation/without trigger. (1)
[Do not accept random/cannot be predicted]
Radiation: alpha, beta and gamma and positron
[give the mark if they name one of these] (1)
Unstable: (Nuclei) [not atoms] are (liable) to break up / decay /
disintegrate or nucleus has too much energy or too many nucleons
[not particles]/may release radiation/[Accept] binding force is
not sufficient/[Accept] binding energy is not sufficient/
[Accept] too many/too few protons/neutrons (1)
[For this mark do not accept ‘nucleus has high energy’ or ‘..has
many particles’] 3

(b) (i) Half life

Evidence of an average calculated ie have used more than just
one value (1)
[Make sure to look at graph, if 2 sets of lines are seen,
award this mark, even if there is no evidence in written answer]
Answer [(5.6 – 6) hours (20160 s – 21600 s)] (1) 2

(ii) Decay constant

Answer [Accept answers in the range 3.1 – 3.5 × 10–5 s–1 /
0.11(5) – 0.12(3) h–1] (1) 1
[ecf their value of half life]
[Do not accept Bq for the unit]
0.69 0.69
Eg λ = / = 3.19 × 10–5 s–1 / 0.12 h–1
6 × 3600 s 6 h

(iii) Number of atoms

Use of A = λN
Answer [in range (1.50 – 1.65) × 1011]
0.5 × 10 7 Bq
Eg N =
3.2 × 10 −5 s −1
= 1.56 × 101 2
[8]