Download QAT 2014 Answers part B

2014 HSC TRIAL EXAMINATION
PHYSICS – MAPPING GRID
Exam
Section
Section I:
Part A:
Multiple
Choice
Section I:
Part B:
Extended
Response
Question
Marks
Syllabus/Course
Outcomes
Content
1
2
1
1
H6
H9
3
1
H1, H2, H3, H6
4
1
H6, H9
5
6
7
8
9
10
11
1
1
1
1
1
1
1
H1, H5, H9
H1, H2
H6
H9, H13
H3, H9
H9
H7, H9
12
1
H2, H3, H9
13
14
15
1
1
1
H7, H9
H7, H9, H10
H1, H3, H8
16
17
18
19
20
21
1
1
1
1
1
5
H1, H2, H10, H14
H3, H7, H10
H2, H3
H1, H10, H14
H2, H3
H5, H6, H7, H9, H16
22
23
24
25
5
4
4
8
H6, H7
H1, H3, H7, H9
H1, H6, H12, H10, H14
H2, H9, H12, H13
26
27
6
4
H3, H4, H7, H13, H16
H3, H7, H9, H14
28
5
H1, H4, H13, H16
9.2.1
9.2.1,
9.2.2
9.2.2,
9.4.4
9.2.2,
9.2.4
9.2.2
9.2.4
9.2.4
9.3.1
9.3.1
9.3.1
9.3.2,
9.3.3
9.3.2,
9.4.1
9.3.2
9.3.2
9.4.2,
9.4.3
9.4.1
9.4.2
9.4.4
9.4.3
9.4.4
9.2.1,
9.2.3
9.2.2
9.2.2
9.2.4
9.2.1,
9.3.1
9.3.4
9.3.3,
9.3.4
9.4.2
Targeted
Performance
Bands
1–2
3–4
Answer
3–4
A
4–5
B
2–3
5–6
4–5
2–3
1–2
3–4
5–6
D
A
A
C
A
B
B
3–4
D
2–3
5–6
1–2
B
C
C
3–4
5–6
1–2
5–6
2–3
1–4
A
D
D
C
A
D
B
1–4
1–5
2–6
1–6
1–6
1–5
1–6
Disclaimer
Every effort has been made to prepare this Examination in accordance with the Board of Studies documents. No guarantee or warranty is
made or implied that the Examination paper mirrors in every respect the actual HSC Examination question paper in this course. This paper
does not constitute ‘advice’ nor can it be construed as an authoritative interpretation of Board of Studies intentions. No liability for any
reliance, use or purpose related to this paper is taken. Advice on HSC examination issues is only to be obtained from the NSW Board of
Studies. The publisher does not accept any responsibility for accuracy of papers which have been modified.
PHYTR14B_GUIDELINES
Page 1
Section II
29
30
31
32
6
4
4
25
33
25
34
25
PHYTR14B_GUIDELINES
H1, H9, H10
H1, H2, H3, H10
H1, H3, H5, H10, H13
H1, H2, H3, H4, H7,
H8, H10, H12, H13,
H14
H1, H2, H3, H5, H7,
H8, H9, H11, H13, H14,
H1, H2, H3, H5, H10,
H11, H13, H14, H15,
H16
9.4.1
9.4.2
9.4.3
1–5
1–6
1-5
1-6
1-6
1-6
Page 2
2014 HSC TRIAL EXAMINATION
PHYSICS – MARKING GUIDELINES
Section I
Part A – 20 marks
Questions 1-20 (1 mark each)
Question
Correct Response
Outcomes Assessed
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
D
B
A
B
D
A
A
C
A
B
B
D
B
C
C
A
D
D
C
A
H6
H9
H1, H2, H3, H6
H6, H9
H1, H5, H9
H1, H2
H6
H9, H13
H3, H9
H9
H7, H9
H2, H3, H9
H7, H9
H7, H9, H10
H1, H3, H8
H1, H2, H10, H14
H3, H7, H10
H2, H3
H1, H10, H14
H2, H3
PHYTR14B_GUIDELINES
Targeted
Performance Bands
1–2
3–4
3–4
4–5
2–3
5–6
4–5
2–3
1–2
3–4
4–5
3–4
2–3
5–6
1–2
3–4
5–6
1–2
5–6
2–3
Page 3
Section I
Part B – 55 marks
Question 21 (5 marks)
21 (a) (1 mark)
Outcomes Assessed: H6, H9
Targeted Performance Bands: 1-2
Criteria
 Correct substitution into a correct formula, yielding an answer
Sample answer
m m
EP   G 1 2
r
 EP  
Mark
1
6.67  10 11  6.0  10 24  12
  1.2  10 8 J
40 000  10 3
21 (b) (2 marks)
Outcomes Assessed: H6, H7, H9
Targeted Performance Bands: 2-4



Criteria
Clear statement that an inertial frame of reference cannot be accelerating
Clear statement that the rock’s frame of reference is not inertial
Only one of the above
Marks
2
1
Sample answer
In order for a frame of reference to be inertial it is necessary that it not be accelerating. The
rock is attracted by Earth’s gravity (and the Sun, Moon etc.). While it is true that every body
in the Universe is similarly affected by gravity, so strictly none can be perfectly inertial, the
frame of reference of this rock certainly cannot be considered to be inertial.
21 (c) (2 marks)
Outcomes Assessed: H5, H9, H16
Targeted Performance Bands: 1-3





Criteria
Not all the laws of physics hold in a non-inertial frame of reference
The moving Earth or the air has kinetic energy and
Interaction (or friction) changes the KE to heat
One of the above or
The rock actually did possess KE because its frame is not inertial
Marks
or
2
1
Sample answer
Although the rock had no kinetic energy in its own reference frame, the Earth certainly did
have kinetic energy as it approached the rock. When there was an interaction between the air
(moving with the Earth) and the rock, the air provided the energy to rapidly heat the rock.
PHYTR14B_GUIDELINES
Page 4
Question 22 (5 marks)
22 (a) (2 marks)
Outcomes Assessed:
Targeted Performance Bands: 2-4


Criteria
Correct substitution into the correct formula to yield an answer
Incorrect substitution into a correct formula
Marks
2
1
Sample answer
Vertical motion: Δ y = ? uy = 16 sin 60 = 13.856 m s-1 ay = –9.8 m s-2 t = 1.5 s
Δ y = uy t + ½ ay t2  Δ y = 13.856  1.5 – 0.5  9.8  1.52 = 9.759 [9.8 m ↑]
22 (b) (2 marks)
Outcomes Assessed:
Targeted Performance Bands: 1-3



Criteria
Correct substitution into a correct formula to obtain the final velocity
Correct method to find time for parachute to descend
Only one of the above attempted correctly
Marks
2
1
Sample answer
Vertical motion: uy = 13.856 m s-1 vy = ? ay = –9.8 m s-2 t = 1.5 s
vy = uy + ay t
 vy = 13.856 – 9.8  1.5 = – 0.844 m s-1↓
Time required to drop 9.759 m at 0.844 m s-1 = 11.56 s.
22 (c) (1 mark)
Outcomes Assessed:
Targeted Performance Bands: 1-2

Criteria
Correct substitution into a correct formula to yield a displacement
Mark
1
Sample answer
Horizontal motion: ux = vx = 16 sin 60 = 8.0 m s-1
Δ x = ux  t = 8.0  1.5 = 12 m →
PHYTR14B_GUIDELINES
Page 5
Question 23 (4 marks)
23 (a) (3 marks)
Outcomes Assessed:
Targeted Performance Bands: 2-5
Criteria
Correct substitution into the correct equation to obtain an answer
Correctly determining r3, but not r or
Incorrectly substituting T = 96 into the correct equation, producing an answer
Other incorrect substitution into the correct equation or
Correctly identifying that T = 96  60 seconds





Marks
3
2
1
Sample answer
GM
r3

2
T
4 2

r3
96  60
2
6.67  10 11  6.0  10 24

4 2
r 3  3.363  10 20
 r = 6.954  10 m [= 7 000 km]
6
23 (b) (1 mark)
Outcomes Assessed: H5, H14, H15
Targeted Performance Bands: 1-2
Criteria
 Correct substitution into an appropriate formula yielding an answer
Sample answer
2r
2   6.954  10 6
v 
v 
96  60
T

 v  7.586  10 3 m s 1 7.6 km s 1
Mark
1

Question 24 (4 marks)
24 (a) (2 marks)
Outcomes Assessed: H1, H3, H7, H9
Targeted Performance Bands: 3-5
Criteria
 Both principles correctly stated
 One principle correctly stated
Marks
2
1
Sample answer
(The relativity principle): The laws of physics are the same in all inertial frames of reference.
(Constant speed of light): Light travels through a vacuum at a fixed speed c, independent of
the source or the observer.
24 (b) (1 mark)
Outcomes Assessed: H7, H10
Targeted Performance Bands: 2-3

Criteria
A correct statement relating to either of Einstein’s postulates
Mark
1
Sample answer
For the speed of light in a vacuum to be unchangeable, length and time cannot also be fixed.
As objects travel at relativistic speeds their length is observed shorter and their time dilated.
PHYTR14B_GUIDELINES
Page 6
24 (c) (1 mark)
Outcomes Assessed: H2, H3, H8, H11
Targeted Performance Bands: 5-6
Criteria
 A correct definition involving the speed of light in a vacuum
Mark
1
Sample answer
One metre is now defined to be the distance light will travel through a vacuum in 1/c seconds,
where c is the velocity of light in a vacuum, i.e. 2.99792  108 m s-1.
Question 25 (8 marks)
25 (a) (4 marks)
Outcomes Assessed: H2, H9
Targeted Performance Bands: 1-5
Criteria
 A correct label for both axes, and
 Correct and uniform scaling for both axes, and
 Marking in all five points correctly (within 0.5 of a unit either direction),
 Drawing a neat, straight, appropriate line of best fit
Successfully completing three of the required outcomes
Successfully completing two of the required outcomes
Successfully completing one of the required outcomes
Marks
4
and
3
2
1
+
16
+
+
12
+
+
8
4
0
Reading on the balance [ 10-3 kg]
Sample answer
0
0.50
1.00
1.50
2.00
Current through the wire
[A]
25 (b) (1 mark)
Outcomes Assessed: H9, H12
Targeted Performance Bands: 2-3

Criteria
Correctly identifying the value of the drawn line of best fit when I = 0
Mark
1
Sample answer
The wire’s.mass is 2.5  10-3 kg; I = 0 (no magnetic force) when the graph line is at that point.
PHYTR14B_GUIDELINES
Page 7
25 (c) (3 marks)
Outcomes Assessed: H H9, H13
Targeted Performance Bands: 3-6





Criteria
Determination of the slope of the graph as F/I, and attempting to find it
Conversion of the weight force into newtons
Identification of the direction of the magnetic field
One of the above required outcomes is missing
One of the above outcomes is present
Marks
3
2
1
Sample answer
The slope of the line of best fit is found as:
 m 15.3  10 3  2.5  10 3

 6.4  10 3 kg A 1
I
2.00  0
This represents the rate of change of the apparent mass of the wire as the current varies, but
since the scale is measured in kg it must be converted to newtons, i.e.
 FB
 6.4  9.8  10 3  6.272  10 2 N A 1
I
 FB
 B   6.272  10 2  0.40  B  B  0.1568 [ 0.16] T
Since FB  B I  
I
At any time the scale must apply an upward force that balances the weight of the length of
wire plus the magnetic force, which in this case is clearly downwards. The direction of the
current is due north, so the direction of the magnetic field must be due east.
Question 26 (6 marks)
Outcomes Assessed: H3, H4, H7, H13, H16
Targeted Performance Bands: 1-6
Criteria
 Response includes correct comparisons of resistivity, density and
stress/strain
 Response explains preferred use of copper or aluminium in distinct
situations
 Response lacking one or two of the facets of the above outcomes
 Identifies two differences in the usage of copper and aluminium wires or
cables
 Makes one correct statement about copper or aluminium based on the given
data
Marks
5–6
3–4
2
1
Sample answer
The significantly lower resistivity of copper means that for any particular length and diameter
of wire there is lower energy loss if it is used to carry electric current rather than aluminium.
In homes, workplaces and commercial locations this is favourable because no other factors
need consideration. However, electricity generators are not popular in residential areas as they
are noisy, always lit up, polluting (due to fuel wastes and wasted heat), and work continuously
24 hours each day, 365 days per year. With vast amounts of electric power to be transferred
over long distances from generators to cities for use, other factors do need to be considered.
Since the power loss in long-distance cables is high, and PLOSS = I2 R it is essential to minimise
the current carried by each cable, whilst maintaining the total power supplied. To permit this
the cables’ potential is high, so the wires are strung between tall pylons, and heavily insulated
against short-circuiting with the ground, the pylons, or other objects such as trees. Pylons are
expensive to construct and maintain, so minimising their number is economically important.
PHYTR14B_GUIDELINES
Page 8
Copper’s density is 3.3 times that of aluminium, so it is much harder to lift, carry and work
with, and more pylons are needed to sustain copper cables than otherwise identical aluminium
ones. Furthermore, the stress:strain ratio of aluminium is about half that of copper, meaning it
sags far less, also reducing the number of pylons required to support those cables.
Question 27 (4 marks)
27 (a) (2 marks)
Outcomes Assessed: H3, H7
Targeted Performance Bands: 2-4




Criteria
An appropriate labelled diagram
A correct description of the principle of mutual induction
An unlabelled, but otherwise correct diagram of a two-coil transformer
An outline of the principle of mutual induction
Marks
2
or
1
Sample answer
The principle of mutual induction for transformers is that for two solenoids that are linked, a
change in the magnetic flux threading one coil induces an emf within the other coil (in
accordance with Faraday’s law), and the induced emf is generated such that it opposes the
initial flux change in the first coil (in accordance with Lenz’s law).
Coil 1
Coil 2
Switch
~
AC source
Galvanometer
27 (b) (1 mark)
Outcomes Assessed: H7
Targeted Performance Bands: 1-2

Criteria
Correct substitution into the correct formula, giving an answer
Mark
1
Sample answer
VP
n
n
240
 P  P 
 12.24  12
VS
nS
nS
19.6
27 (c) (1 mark)
Outcomes Assessed: H9, H14
Targeted Performance Bands: 4-5

Criteria
Identification of the name rectification, or a description of the output
Mark
1
Sample answer
In order for the AC mains supply to become DC it is necessary to rectify the AC output of the
transformer. A diode could be used for the purpose.
PHYTR14B_GUIDELINES
Page 9
Question 28 (5 marks)
Outcomes Assessed: H1, H4, H13, H16
Targeted Performance Bands: 1-6
Criteria
 At least one example supporting the opinion of the article
 At least one example opposing the opinion of the article
 At least one clear reference to the Einstein-Planck ‘disagreement’
 A stated assessment of the correctness of the article provided
 A substantial response, but lacking in one or two of the above
 One clear example either supporting or opposing the article’s comments or
 One statement in support of the article and one opposing it
 A statement relating the article’s contents to the Einstein-Planck
‘disagreement’
Marks
5
3–4
2
1
Sample answer
Einstein was a renowned believer that human knowledge belongs equally to all people, and
that all research discoveries should be available to anyone interested, especially to those who
are intending research along similar lines. In this way the trend towards advancing knowledge
is most likely to be maximised. There is a common saying about not needing to “reinvent the
wheel”, but whenever the results of research are concealed others must independently follow
the same path to rediscover similar outcomes, instead of being able to progress from there.
The great inventor Nikola Tesla was determined that his genius should be useful for all, so he
did not patent a number of his greatest inventions which could have made him an extremely
wealthy man. Unfortunately, as we know from coursework, those whom he worked for were
not of the same inclination. They became very rich, while Tesla died in poverty.
The First World War began a century ago. At that time there was an intense patriotic feeling
in Germany, that discoveries of all kinds belonged to the Fatherland, and Max Planck
supported this, at least at the time.
Nowadays, however, there is a considerable number of research journals available to readers,
allowing those interested to find a treasure of articles associated with what they may intend to
investigate. Intense efforts are made by researchers to have articles included in the prestigious
journals, and it is similarly a requirement to include supporting material from a large array of
articles found within the journals to support or challenge the findings of investigation in order
to have an article accepted for publication. This would certainly appear to dispute the claims
made in Mr Trounson’s article.
However, in today’s Australia, there is a struggle for research funding. Government money is
available for research of all kinds, but it is limited. Grants are supervised by independent
Boards, but they are competitive. In 2013 applications were invited for grants to improve the
interest in science education in schools right across Australia, due to the National Syllabus,
and because the standard of science and maths education here was seen to be slipping – but it
was competitive! Instead of cooperation between universities, sharing ideas and selecting the
best of them so all researchers could work together, the secrecy certainly limited the potential.
Research money is also available from industry as well as military budgets, but it is rarely in
the interests of industry to share research information before exhausting every avenue to gain
financial benefits from what they have paid for, while discoveries made via military research
are almost inevitably top secret, to be revealed only during conflict. Just as in those days of
World War 1, discoveries such as the Haber process to make nitrates, and hence gunpowder,
from air, are very closely-guarded secrets.
PHYTR14B_GUIDELINES
Page 10
Mr Stone, and Mr Trounson as the author of the article, seem to be correct insofar as Australia
is concerned, though whether we differ from other countries in this is questionable.
Question 29 (6 marks)
29 (a) (1 mark)
Outcomes Assessed: H9
Targeted Performance Bands: 2-3

Criteria
Eight straight lines evenly separated, with arrows directed towards the 0 V
plate
Mark
1
Sample answer
+30 V
0V
29 (b) (2 marks)
Outcomes Assessed: H9, H10
Targeted Performance Bands: 1-3



Criteria
Correct substitution into the correct formula yielding an answer
Correctly identifying the direction of the force
One of the above
Sample answer
qV
FE  q E 
d
and
Marks
2
1
1.602  10 19  30
 FE 
 2.0  10 16 N 
2
2.4  10
29 (c) (1 mark)
Outcomes Assessed: H10
Targeted Performance Bands: 4-5

Criteria
Correct substitution into the correct formula giving an answer, direction not
needed
Mark
1
Sample answer
The impulse of the force is given by:
Fd
2.0  10 16  4.5  10 2
Impulse  F t  Impulse 

 1.67  10 24 N s 
6
v
5.4  10
PHYTR14B_GUIDELINES
Page 11
29 (d) (2 marks)
Outcomes Assessed: H1
Targeted Performance Bands: 2-4



Criteria
Correct substitution into the correct formula yielding an answer
Correctly identifying the direction of the magnetic field
One of the above
and
Marks
2
1
Sample answer
FB = B q v For the electron to not be deviated, FE ↑ = FB ↓  FB = 2.0  10-16 N ↓
2.0  10-16 = B  1.602  10-19  5.4  106 = 8.65  10-13 B
 B = 2.312  10-4 T. It must be directed due south to balance the electric force
Question 30 (4 marks)
30 (a) (1 mark)
Outcomes Assessed: H1, H3, H10
Targeted Performance Bands: 1-2

Criteria
Response that correctly identifies a blackbody radiation curve
Mark
1
Sample answer
The controversial graph is a blackbody radiation curve, showing the energy emitted by an
ideal blackbody of a fixed temperature across all wavelengths of electromagnetic radiation.
30 (b) (1 mark)
Outcomes Assessed: H1, H3, H10
Targeted Performance Bands: 3-4

Criteria
Suitable correct answers for both axes required to gain this mark
Mark
1
Sample answer
The x-axis represents the wavelength of the e-m radiation being emitted.
The y-axis represents the amount of energy emitted by the blackbody at the particular
wavelength. (“Intensity” would also be acceptable.)
PHYTR14B_GUIDELINES
Page 12
30 (c) (2 marks)
Outcomes Assessed: H1, H2, H3, H10
Targeted Performance Bands: 4-6
Criteria
 Response describes the major difference between experimental and
theoretical curves and
 The significance of the theoretical curve
 A correct statement about a difference between the two graphs
Marks
2
1
Sample answer
The graph indicates the experimental relationship between the identified variables. It shows
that for any ideal blackbody at a fixed temperature, the amount of radiation emitted increases
steadily for shorter wavelengths up to a specific value, called the dominant wavelength,
beyond which the energy released becomes less with decreasing wavelength, approaching
zero emission as the wavelength approaches zero. The dominant wavelength depends only on
the surface temperature of the radiating blackbody.
It was controversial because no theory based on the Classical physics at that time was able to
generate a curve matching the actual blackbody curve. There was no maximum, so the curve
continued to rise exponentially as the wavelength became shorter, even in the case of very
cold blackbodies, which apart from not matching experiment was also not logical, and did not
obey the fundamental Law of Conservation of Energy.
Question 31 (4 marks)
31 (a) (1 mark)
Outcomes Assessed: H3
Targeted Performance Bands: 1-2

Criteria
Response identifies germanium
Mark
1
Sample answer
The first semiconductor in common use was germanium.
31 (b) (2 marks)
Outcomes Assessed: H1, H3, H5, H13
Targeted Performance Bands: 1-3
Criteria
 Outlines two correct reasons why silicon replaced germanium for common
use
 Identifies one reason why silicon replaced germanium
Marks
2
1
Sample answer
Germanium is a very rare element, so much so that although when developing his periodic
table of the elements Mendeleev predicted its existence and several of its physical and
chemical properties, it had not been identified at that time. By contrast, silicon is abundant,
the second most common element in the Earth’s crust. For this reason, silicon is far cheaper
than germanium.
Furthermore, germanium loses its semiconducting properties at fairly low temperatures,
becoming a regular electrical conductor, whereas silicon retains its semiconducting identity to
far higher temperatures.
PHYTR14B_GUIDELINES
Page 13
31 (c) (1 mark)
Outcomes Assessed: H1, H3, H10, H13
Targeted Performance Bands: 4-5
Criteria
 Response describes the actual series of electron jumps making the hole
‘migrate’
Mark
1
Sample answer
An electron hole is a location in a crystal where an electron is “missing”. Frequently they
occur at places where an electron absorbs a photon, and gains sufficient energy to escape into
the conduction band of the crystal. They also occur in crystals doped with a p-type impurity
when an atom of the dopant shares electrons with surrounding atoms, but has insufficient to
covalently bond with all its neighbours.
An electron from a surrounding atom can jump into the vacant hole, in which case it leaves a
new hole where it came from, which may be filled in its turn by another electron – which now
creates another hole, and so on. Electrons are the actual moving charges, but it is the hole that
“migrates” through the crystal in this manner.
PHYTR14B_GUIDELINES
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Section II
Question 32 – Medical Physics (25 marks)
32 (a) (i) (1 mark)
Outcomes Assessed: H2, H10
Targeted Performance Bands: 1-3
Criteria
 Four days
Mark
1
Sample answer
The graph shows a repeated decrease in quantity by half every four days.
32 (a) (ii) (2 marks)
Outcomes Assessed: H1, H3, H7
Targeted Performance Bands: 2-3



Criteria
Response concludes that a proton is likely to change into a neutron and
This conclusion is based on correct and stated evidence
A response stating that a positron is likely to be emitted
Marks
2
1
Sample answer
Since the only stable isotope of iodine is I-127, which has 53 protons and 74 neutrons, whilst
unstable I-124 has 53 protons and only 71 neutrons, it clearly has insufficient neutrons to
balance the number of protons in its nucleus.
In such cases, one proton may change into a neutron by releasing a positron and a neutrino.
32 (a) (iii) (1 mark)
Outcomes Assessed: H10
Targeted Performance Bands: 1-2

Criteria
Positron Emission Tomography
Mark
1
Sample answer
Positron Emission Tomography uses positrons emitted as protons are converted to neutrons.
32 (b) (i) (2 marks)
Outcomes Assessed: H1, H2, H3, H10
Targeted Performance Bands: 1-2
Criteria
 Two correct substitutions into the correct formula yielding an answer
 One correct substitution into the correct formula yielding an answer
Marks
2
1
Sample answer
Z = ρ v  ZM = 1075  1590 = 1.71  106 kg m-2 s-1 (Rayls)
 ZB = 1908  4080 = 7.78  106 kg m-2 s-1
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32 (b) (ii) (2 marks)
Outcomes Assessed: H3, H10, H12, H14
Targeted Performance Bands: 3-6
Criteria
 Correct substitution into the correct formula yielding an answer and
 The correct analysis of the result leading to the correct conclusion
 One of the required outcomes correct
Sample answer
Z 2  Z1 
IR

I0
Z 2  Z1 2


Marks
2
1


2
IR
7.78  10 6  1.71  10 6


 0.409
2
I0
7.78  10 6  1.71  10 6
0.387
Percentage variation from normal 
 100  94.6% Bone loss is not significant.
0.409
2
32 (b) (iii) (2 marks)
Outcomes Assessed: H3, H11, H12, H13
Targeted Performance Bands: 1-2
Criteria
 The response describes the difference between A-scans and B-scans
 Identifies why there would have been an advantage using a B-scan
 Identifies a difference between ultrasound A-scans and B-scans
and
Marks
2
1
Sample answer
A-mode ultrasound scans cannot produce images. Instead, the oscilloscope detector registers a
“spike” as each echo returns after reflection from tissue boundaries. The intensity of the echo
is displayed as the amplitude of the spike. The time delay is measured very accurately.
In B-scans signals are sent in many different directions. The intensity and time information
from the echoes are recorded as spots of different brightness on a TV screen. The location of
the spot depends upon the direction the signal was sent and the time delay, so images are
made as the information provided by each point combines. Using a B-scan images proceeding
from within the bones of the cosmonauts would have provided more detailed information.
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32 (b) (iv) (3 marks)
Outcomes Assessed: H1, H11, H13, H14
Targeted Performance Bands: 4-5
Criteria
 The acoustic impedances of both air and skin are calculated and
 The ratio of reflected U/S to the original signal is found and
 A correct conclusion is reached regarding the use of a coupling gel
 Two of the above outcomes are achieved
 Correct substitution into one correct formula to yield an answer
Sample answer
Z = ρ v ZAIR = 1.3  330 = 429 kg m-2 s-1
ZSK = 1040  1540 = 1.60164  106 kg m-2 s-1
Z 2  Z1 
IR

I0
Z 2  Z1 2


Marks
3
2
1


2
I
1.60164  10 6  429
 R 
 99.9%
2
I0
1.60164  10 6  429
When ultrasound passing through air reaches the skin, almost all the signal is reflected, only
0.1% enters the skin. Also, when the reflected signal carrying data returns to the skin to cross
to the transducer, for the same reason only 0.1% of the signal reaches the transducer.
For this reason a coupling gel must always be placed between the transducer and the skin.
2
32 (c) (i) (1 mark)
Outcomes Assessed: H10
Targeted Performance Bands: 1-2
Criteria
 Answer gives a clear description of the proton’s ability to act as a magnet
Mark
1
Sample answer
Since hydrogen has a single proton it has a net magnetic moment because that proton spins as
the nucleus is spinning. This allows the nucleus to act as a tiny magnet.
32 (c) (ii) (4 marks)
Outcomes Assessed: H1, H3. H8
Targeted Performance Bands: 3-6






Criteria
Response describes the precession of H-nuclei in a strong magnetic field
A radio signal at the resonant frequency transfers energy to bodily tissues
This energy is released when the signal stops, and analysed by computer
Movement of the external signal source creates a sequence of images
One or two of the processes unclear or omitted
One of the above processes outlined
Marks
4
2–3
1
Sample answer
MRI scans produce detailed, high-quality images, especially of soft body tissues. Patients are
located on a horizontal platform which slides into a large cylinder within which a very strong
magnetic field is present. When subjected to a very powerful external magnetic field many of
the hydrogen nuclei in the body of the patient undergoing the MRI scan start to precess
(wobble) around the field at their special resonant frequency, known as the Larmor frequency.
An external radio signal, the gradient field, with precisely the Larmor frequency, are directed
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at the body of the patient. Though these signals are weak, because they are at exactly the same
frequency they produce resonance, so the hydrogen nuclei gain a great deal of energy, causing
their axis of rotation to change. Whenever the radio signal stops the nuclei return to their
original axis of rotation. During this relaxation process they release energy as radio waves that
are detected and converted into signals analysed by a computer. The hydrogen in different
tissues relaxes at distinct rates, so different signals are transmitted, producing images of high
clarity. Since hydrogen is so common throughout the body the scanning technique can be
used everywhere. As the radio transmitter moves along the body it causes responses from the
hydrogen in the body tissues in sequence, producing a series of 2-dimensional “slices”.
32 (c) (iii) (1 mark)
Outcomes Assessed: H1, H4, H12, H14
Targeted Performance Bands: 2-3
Criteria
 A correct advantage of the MRI technique is identified
Mark
1
Sample answer
All the other investigative techniques mentioned require the use of very high frequency
electromagnetic radiation; CAT scans use X-rays and PET uses γ-radiation. MRI does not use
this, so it is intrinsically less hazardous, and will not damage patients unless they have certain
types of metallic implants in their bodies, such as pace-makers.
32 (d) (6 marks)
Outcomes Assessed: H1, H3, H4, H13
Targeted Performance Bands: 1-6
Criteria
 Describes advantages of computer processing of the many CAT images
 CAT scans work well for both hard and soft tissue
 Describes higher risks in CAT images, and other drawbacks
 Gives an assessment based on the identified impacts
 Good explanation of the MRI process, but misses identified feature(s)
 Correctly identifies two features of CAT scans as above
 Correctly identifies one feature of CAT scans
Marks
5–6
3–4
2
1
Sample answer
Following the discovery of X-rays in 1895 and the publication of their capacity to “see inside”
bodies, they started being used throughout the world, even before their exact nature had been
established. Initially they were used mainly to check for broken bones, but a large number of
other uses have developed since then.
Now we have CAT scans. Images produced by this modern branch of X-ray technology leave
the old system in the dark, as seen on the X-ray image introducing this question. The normal
X-ray shown in Image 1 was captured by a single exposure, but the CAT scan (Image 2)
consists of multi-images captured by a rotating gantry camera. Although both techniques are
used successfully in brain scans when searching for signs of tumours, a CAT scan is able to
identify possible tiny new tumours by scanning the patient a second time a half-hour after
being injected with an X-ray fluorescent dye that moves with the blood to the site of a
developing tumour. CAT images can be computer-processed to form 2-D and 3-D, whereas
the single image made by an X-ray cannot be further processed.
There is no doubt that human society has benefited from the appearance of this technology.
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However, it is not all good news. The hazardous nature of X-rays has often been ignored, but
it does exist, and CAT scan images are the result of numerous X-rays, the data from which is
compiled by a computer, as there is a vast amount of data to store and many calculations to be
performed to prepare the images.
CAT scans are far more expensive than regular X-ray images, so some patients may be unable
to afford them. Meanwhile, the government health care system is placed under increased
pressure by subsidising many CAT scans when X-rays would have been sufficient.
However, in spite of the increased potential hazards of CAT scans, in many situations they
can provide comforting certainty to those fearing cancer, or at least earlier detection with an
improved chance of survival, so in all the benefits far outweigh the drawbacks.
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Question 33 – Astrophysics (25 marks)
33 (a) (i) (1 mark)
Outcomes Assessed: H1, H2, H3
Targeted Performance Bands: 1-3

Criteria
Correct substitution into the correct formula yielding an answer
Sample answer
mE  mH
LH
 100 5
LE
L
 H  100
LE
6  20
5
 100  2.8

Mark
1
.
LH
 2.512  10 6
LE
The brightest objects that Hubble camera can observe without being “blinded by the light”
have a luminosity just 2.5 millionths of that of the dimmest objects visible by human eyes.
33 (a) (ii) (2 marks)
Outcomes Assessed: H7, H8, H11
Targeted Performance Bands: 3-5
Criteria
 Clear explanation why direct sunlight does not enter the camera
 Clear explanation why scattered light does not enter the camera
 One of the above outcomes
and
Marks
2
1
Sample answer
Unlike ground-based telescopes, those above the atmosphere do not have problems with light
being scattered off air molecules (unless they are observing Earth). For that reason, provided
their target is in a direction sufficiently apart from the Sun, the Moon and the planets, once set
in a particular direction the camera will continue to receive light from the sources it observes.
Even though there is no “night” for an orbiting telescope, with the Sun always visible unless it
is temporarily eclipsed by the Earth, while the camera is aimed in a different direction light
from the sun will not enter it.
33 (b) (3 marks)
Outcomes Assessed: H1, H3, H11, H14
Targeted Performance Bands: 2-5
Criteria
 Two correct reasons for telescopes on high mountains outlined
 One correct reason outlined or
 Two reasons identified
 One reason stated
Marks
3
2
1
Sample answer
There are several reasons why telescopes are frequently sited at the top of high mountains.
One of these is light pollution; so much scattered light from buildings in large cities produces
a “noise” reducing the sensitivity and resolution of telescopes. Air molecules do cause some
scattering, but more is caused by exhaust gases, dust and other particles polluting the air, but
if the telescope is located where there is less air and little pollution the problem is reduced.
A second factor that improves “seeing” from the top of high mountains is that there is far less
water vapour there. Water vapour also causes scattering, but more significantly with regard to
telescopes, it causes cloud and rain. The air is colder at greater heights, and as shown in the
photograph, the clouds (and rain) are usually at levels below, so viewing is not impeded.
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33 (c) (i) (2 marks)
Outcomes Assessed: H1, H2, H3, H7
Targeted Performance Bands: 2-3
Criteria
 Determining the colour index of the star to be +0.29
 Using the table to estimate the spectral class of Alderamin to be A6 or
A7
 Determining the colour index of Alderamin to be +0.29 or
 Incorrectly finding a CI, but correctly using that CI to find a spectral
class
Marks
2
1
Sample answer
CI = mB – mV
 CI = 2.73 – 2.44 = 0.29.
The CI of an A0 star is exactly 0; the CI of an F0 star is +0.4.
Hence Alderamin is an A6 type.
33 (c) (ii) (2 marks)
Outcomes Assessed: H1, H2, H3, H14
Targeted Performance Bands: 2-4
Criteria
 If the star is determined to be an A6 or A7, estimating from +1 to +1.5
or
 If the star was found to be anything else, estimating a range of ± 0.3
 If the star is determined to be an A6 or A7, estimating from 0 to +2
 If the star was found to be anything else, estimating a range of ± 1
Marks
2
1
Sample answer
Using the centre line of the given H–R diagram, and estimating the location of an A6 star, the
corresponding value of its absolute magnitude would be +1.
33 (c) (iii) (1 mark)
Outcomes Assessed: H1, H2, H3
Targeted Performance Bands: 2-6

Criteria
Substituting the appropriate values into the correct formula to get an
answer
Mark
1
Sample answer
d
d
 1.44
d
 1  2.44  5 log 10

 log 10
10
10
5
10
The distance of Alderamin from Earth is therefore approximately 19.4 parsecs.
M  m  5 log
10
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33 (c) (iv) (1 mark)
Outcomes Assessed: H1, H2, H3, H14
Targeted Performance Bands: 2-4
Criteria
 Correctly naming the process spectroscopic parallax
Mark
1
Sample answer
The name of this process is spectroscopic parallax.
33 (d) (3 marks)
Outcomes Assessed: H2, H8, H11
Targeted Performance Bands: 3-6






Criteria
States that the F-star will be more luminous than the K-star and
Its dominant wavelength will be bluer than that of the K-type and
There will be differences in the absorption spectral lines of the stars
Identifies two differences as above or
Identifies all three, but without stating which star has which properties
Identifies one difference between the star types
Marks
3
2
1
Sample answer
The spectrograms of stars are composed of their blackbody radiation curve superimposed by a
series of dark absorption lines that appear like troughs of lower intensity.
The blackbody curve depends purely on the surface temperature of the star. An F-type star is
yellow-white, with a surface temperature of around 7500K, whereas a K-type is orange, with
a surface temperature around 4000K. Consequently, the F-type has far greater luminosity
(around 16 times greater) than that of the K-type.
Furthermore, the dominant wavelength of the F-type will be located towards the centre of the
visible spectrum, around green or blue, while the dominant wavelength of the K-type would
be located at the red end of the spectrum.
The other major difference will be in the dark absorption lines distinguishing the spectra. An
F-type star will display clear hydrogen lines amongst others. The hydrogen lines in the K-type
star’s spectrum will be far less prominent, whereas there will be clear lines of various metals
such as calcium.
33 (e) (i) (2 marks)
Outcomes Assessed: H2, H13, H14
Targeted Performance Bands: 3-5



Criteria
Correctly converts AU into metres and days into seconds and
Correctly substitutes into the correct formula leading to an answer
One of the above outcomes
Sample answer
The combined mass of the two stars is found using:
M1  M 2
4 2 r 3

GT 2
 M1  M 2 

4  2 0.12  1.5  1011
Marks
2
1

3
6.67  10  4.015  24  3600
11
2
= 2.868  1031 kg
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33 (e) (ii) (2 marks)
Outcomes Assessed: H1, H3, H14
Targeted Performance Bands: 1-4




Criteria
A correct light-curve, having 2 complete cycles
Appropriately labelled axes and eclipses, and correct time scale
An appropriate light-curve or
Appropriately labelled axes and correct time scale
Marks
2
1
Luminosity
Sample answer
Seen from the edge the light-curve of the largest stars of the Spica binary would appear as:
Larger star
eclipses smaller
0
2
Smaller star
eclipses larger
4
6
8
Time (days)
33 (f) (6 marks)
Outcomes Assessed: H1, H3, H5, H9, H11, H12, H13, H14
Targeted Performance Bands: 1-4
Criteria
 A response that describes at least two assumptions about stars in clusters
 The logical basis for the identified assumptions
 How at least one general property of stars was developed from these
 How star evolution or evolutionary paths were inferred from these
 A response that follows the above, but lacks details linking assumptions to
observations to inferences in two or three places
 Two of the required outcomes clearly stated
 One property of star clusters identified
Marks
5–6
3–4
2
1
Sample answer
All the stars in a cluster are assumed to have formed from the same nebula. Consequently,
they are considered to be composed of the usual ratio of hydrogen and helium, as well as the
same mixture of “metals” in the same proportions, and to have been “born” at approximately
the same time, as well as being effectively the same distance from Earth.
These assumptions provide numerous advantages for astronomers when determining the
properties of stars, as demonstrated below.
If all the stars in a cluster are the same distance away the distance factor plays no part in their
observed brightness, so astronomers were able to relate the brightness of a star to its spectral
class, and therefore to its colour, allowing the arrangement of stars into O, B, A, F, G, K and
M classes, not only because of the intensity of the hydrogen absorption lines as was originally
applied, but also in terms of their luminosity.
In this way it was also found that a few K- and M-type stars appeared to be far too luminous
relative to others having the same colour, so the concept of red giant stars was developed. In a
similar manner, white dwarfs were recognised because a few B-, A- and F-types located in
clusters were found to have much lower luminosity than their normal counterparts.
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Assuming all the stars in a cluster have effectively the same material in the same proportions
means that the differing absorption spectral lines observed in their individual spectra must be
caused by factors such as surface temperature, adding a further factor into the arrangement of
stars into their spectral classes.
Assuming that all the stars in a cluster were “born” at roughly the same time allowed the very
significant interpretation to be made that the brightest stars also exhaust their fuel much faster
than their dimmer siblings, because there are far fewer O- and B-type stars present in older
clusters relative to those present in newer clusters, while there are also far more red giants and
white dwarfs present in older clusters. From the data gathered the evolutionary path of stars of
the different spectral classes was gradually understood.
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Question 34 – From Quanta to Quarks (25 marks)
34 (a) (i) (2 marks)
Outcomes Assessed: H1, H2
Targeted Performance Bands: 2-4
Criteria
 Correct substitution into Rydberg’s formula to find an answer for 1/λ and
 Subsequent processing to find the photon energy in either J or eV
 One correct process leading to an answer
Marks
2
1
Sample answer
 1
1
1 
 R 2  2 

ni 
 n f



1
1
1
 1.097  10 7   2  2 

1 
5
  1.053  10 7 m 1
c
 f  3.0  108  1.053  107  3.16  1015 Hz

E  h f  6.63  1034  3.16  1015  2.095  1018 J [ 13.08 eV]
f 
34 (a) (ii) (2 marks)
Outcomes Assessed: H1, H2, H14
Targeted Performance Bands: 2-5
Criteria
 Identifying that nf must be 2
 Correctly substituting into the correct formula to yield an answer
 Substituting incorrect data into the correct formula
Marks
2
1
Sample answer
For the photon to be visible it must belong to the Balmer series, i.e. nf = 2.
1
1
1
1
1
1
 R 2  2 
 1.097  107    2  2   2.303  106 m 1

ni 

5 
2
 n f
 λ = 434 nm
34 (b) (3 marks)
Outcomes Assessed: H1, H3, H13
Targeted Performance Bands: 3-6





Criteria
Original concepts of electron orbits and orbitals are identified
Heisenberg’s uncertainty principle is stated and/or explained in this context
A diagram representing some feature of the probability concept included
One of the above steps either not present, or inappropriately considered
An attempt made to relate the principal to orbits by diagram or writing
Marks
3
2
1
Sample answer
Heisenberg’s uncertainty principle had the effect of changing the
concept of particle-electrons moving in circular (Bohr) or elliptical orbits
(Pauli) with integral numbers of wavelengths (de Broglie), becoming
almost like cloud-electrons. The electrons’ mass is so small and their
speed so great their location becomes impossible to ascertain, becoming
instead like a shadow picture representing the probability of finding the
electron at a point at any instant, as represented in 2-D in this diagram.
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34 (c) (i) (1 mark)
Outcomes Assessed: H3, H10, H13, H14
Targeted Performance Bands: 1-5
Criteria
 The difficulty in determining the chemical identity of product nuclei
explained
Mark
1
Sample answer
Bombardment of nuclei with α-particles, or even neutrons, does not always lead to a change
in the target nucleus. Even if it were 100% successful every time, there is such an enormous
number of nuclei of any substance present in even a microscopic sample that it would take
years for the number of successfully transmutated nuclei to be detectable by chemical means.
34 (c) (ii) (2 marks)
Outcomes Assessed: H1, H3, H10, H11
Targeted Performance Bands: 2-3
Criteria
 Diagram showing a collision of two particles, with two different particles
afterwards and
 At least 2 particles correctly identified
 One of the outcomes identified above
Marks
2
1
Sample answer
Nitrogen
nucleus
Oxygen-17
nucleus
Combined
nucleus
α-particle
proton
34 (c) (iii) (2 marks)
Outcomes Assessed: H10, H13
Targeted Performance Bands: 2-4


Criteria
Response gives a completely balanced nuclear equation
Response gives a nuclear equation that is not balanced
Marks
2
1
Sample answer
4
2
 
14
7
N 
18
9
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F  11 p 
17
8
O
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34 (d) (i) (1 mark)
Outcomes Assessed: H2, H10, H14
Targeted Performance Bands: 2-3
Criteria
Mark
1
 Identifies the antineutrino
Sample answer
The missing reaction product is an antineutrino.
34 (d) (ii) (2 marks)
Outcomes Assessed: H1, H7, H19
Targeted Performance Bands: 3-5
Criteria
 Correctly determines the mass defect and
 Uses a correct method to evaluate the energy released (in joules)
 Correctly applies one of the above outcomes
Marks
2
1
Sample answer
The mass of the reactant (one neutron) is 1.008665 amu.
The mass of the products is 1.007276 + 0.000549 = 1.007825 (ignoring the antineutrino).
Hence the mass defect is 8.4  10-4 amu.
The energy released in this reaction is therefore 8.4  10-4  931.5 = 0.7825 MeV
1MeV = 1.602  10-13 J, so the energy release is 0.7825  1.602  10-13 = 1.25  10-13 J.
34 (d) (iii) (2 marks)
Outcomes Assessed: H1, H2, H10
Targeted Performance Bands: 2-4
Criteria
 Response correctly identifies two properties of an anti-proton
 Correctly identifies one property of an anti-proton
Marks
2
1
Sample answer
An anti-proton will have precisely the same mass as a proton.
Its charge will be negative, identical to that of an electron and opposite to that of a proton.
d
34 (d) (iv) (1 mark)
Outcomes Assessed: H1, H2, H5, H10
Targeted Performance Bands: 3-4
Criteria
d u
d
 Diagram
shows that the anti-proton has 2 anti-up and 1 anti-down quarks
u
Sample answer
An anti-proton:
d
u
represents an anti-up quark
d
represents an anti-down quark
Mark
1
u
u
u
d
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u
u
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34 (e) (7 marks)
Outcomes Assessed: H1, H2, H3, H10, H11, H13, H15, H16
Targeted Performance Bands: 1-6
Criteria
 Historical development of the atomic model
 Problems with the Rutherford atomic model
 Discussion of Balmer’s (and others’) emission spectral lines
 Development of the quantum theory
 Description of the Bohr atomic model
 What the Bohr model explained successfully
 What the model was unable to successfully explain
 A sound consideration of the issue, lacking one or two of the listed issues
 A fair consideration of the issue, using statements rather than descriptions
 Identifying any two of the issues listed above
 A correct statement referring to the issue
Marks
7
5–6
3–4
2
1
Sample answer
Following Thomson’s experimental proof of the existence of electrons as part of atoms, it
took little time before Rutherford’s gold leaf experiment led him to propose a new model of
an atom. It defied Maxwell’s laws of electromagnetism, because electrons were postulated to
orbit around a central positively charged nucleus, kept in orbit by their electrostatic attraction,
but since they were accelerating charges they had to radiate e-m energy, causing the atom to
collapse.
Balmer, a Swedish maths teacher, noticed a mathematical relationship for the wavelengths of
the visible spectral lines of excited hydrogen. Lyman and Paschen discovered similar series
relationships for hydrogen’s U-V and infra-red lines too, and Rydberg united all three with a
single formula to predict them, but it seemed an interesting coincidence that worked only for
hydrogen atoms.
Planck announced his quantum theory in 1900. Although received with great scepticism, with
Einstein’s support it had become accepted in 1905.
Bohr explained away the problem with Rutherford’s atomic model by applying the quantum
theory, postulating that electrons could not gradually lose energy – rather, they could lose (or
gain) only fixed quanta. They were normally located within stable energy shells, and could
only jump to alternate shells either outwards by absorbing a photon of energy, or inwards by
emitting the appropriate photon. Not only did this supply a reason allowing Rutherford’s atom
to exist, it explained the spectral lines of Balmer, etc.
Bohr used Classical theory wherever possible to explain phenomena, with quantum theory to
explain away the difficult points. Being a hybrid theory, it is surprising how long it survived;
in fact it is still taught in school chemistry! However, it was unable to explain 4 phenomena.
In spite of much time and effort, Bohr’s atom could explain only hydrogen and other singleelectron species, He+, Li2+, Be3+ and so on. It could not explain the differences in intensity and
width of spectral lines (called “sharp” – s, “primary” – p and “diffuse” – d). It was unable to
explain “hyperfine splitting” of some spectral lines, and although it could explain the Zeeman
effect (the splitting of single spectral lines into three separate polarised lines when light passes
through a strong magnetic field, it could not explain the ‘anomalous’ Zeeman effect, where
the lines could be split into up to 15 separate lines. Eventually, therefore, it was superseded.
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