PEM exam
HSC Trial Examination 2014
Physics
Solutions and marking guidelines
Physics HSC Trial 2014 PEM
Page 1
Section I
Part A
Answer and explanation
Question 1
Answer D
Kepler’s law of periods used to derive answer.
Question 2
Answer B
North pole is induced in the side closest to the magnet to oppose
the motion of the magnet (Lenz’s law). This requires current to
flow towards 2 hence answer B.
Question 3
Answer A
AC induction motors have a rotor.
Question 4
Answer B
Length is contracted in the direction of motion.
Question 5
Answer C
2
Application of E = mc
Question 6
Answer D
Current is induced to oppose motion on the coil entering the
field. Then no current flows while it is in the magnetic field. As
the coil leaves then current is induced in the opposite direction
to oppose the motion.
Question 7
Answer A
Equations of motion used to find the answer.
Question 8
Answer B
The stone thrown by Ben will return at the same speed it was
thrown at hence Bill and Ben’s stone will have the same speed at
the edge of the cliff and hence answer B.
Question 9
Answer D
Situation D has the greatest rate of change of flux and hence the
greatest current.
Question 10
Answer D
UV light has a greater energy hence it will also release
photoelectrons.
Question 11
Answer A
Application of the universal law of gravitation and weight force.
Question 12
Answer C
Speed of light is independent of velocity of the source.
Question 13
Answer B
Red light has a lower frequency f and hence less energy. This
means more red photons are needed if they are to have the
same power output.
Question 14
Answer D
UV catastrophe.
Question 15
Answer B
Application of the transformer equation.
Question 16
Answer A
Right hand solenoid rule used to find magnetic field inside the
coil.
Physics HSC Trial 2014 PEM
Syllabus content and course outcomes
9.2, 2, C3, D5
H6
9.3, 2, C2, D7
H9
9.3, 5, C3, D1
H9
9.2, 4, C2, D9
H6
9.2, 4, C3, D5
H6
9.3, 2, C2, D4
H6
9.2, 2, C3, D1
H6
9.2, 2, C3, D1
H6
9.3, 2, C3, D2
H9
9.4, 2, C3, D4
H10
9.2, 3, C3, D2
H6
9.2, 4, C2, D6
H6
9.4, 2, C2, D4
H10
9.4, 2, C3, D2
H10
9.3, 4, C3, D2
H7
9.3, 1, C2, D1
H14
Page 2
Question 17
Answer A
Increase in diameter allows more current to follow hence
reducing the resistance.
Question 18
Answer C
Semiconductor is doped with a group three element.
Question 19
Answer C
UV light which has a high energy will knock out electrons from
the metal in the receiving coil so allowing a stronger spark.
Question 20
Answer B
Planck’s hypothesis.
Physics HSC Trial 2014 PEM
9.4, 3, C2, D4
H14
9.4, 3, C2, D7
H9
9.4, 2, C2, D1
H10
9.4, 3, C2, D3 and 6
H6
Page 3
Part B
Sample answer
Syllabus content, course outcomes and marking
guide
Question 21
9.3, 2, C3, D1
H7 and 9
The sound waves make the diaphragm vibrate. This Coherent description
3 marks
vibration makes the solenoid vibrate. The moving
solenoid is in a magnetic field and hence a current is Limited coherence
2 marks
induced due to Lenz’s law. This is detected by the
amplifier.
A statement of Faradays/Lenz’s law
1 mark
Question 22
9.2, 1, C2, D3
H6
7
(a) (i) EP = - 6.0 x 10 J (variations allowed e.g. - 6.2 x (i)
107 J)
Correct answer
1 mark
(a) (ii) R = (25 + 6.4) x 106 m = 31.4 x 106 m
EP = - 1.4 x 107 J
(b) Change in EP = EK required
= - 1.4 x 107 - - 6.0 x 107 = 4.6 x 107 J
(c) Air friction acts on the satellite hence increasing
the energy required.
The satellite needs to be launched by a rocket to
achieve orbit which increases the mass and thus
energy required.
Question 23
(a) Metals – most number of free electrons
n type semiconductor has less free electrons
p type semiconductor has the least free electrons.
(b)
Metals
Conduct through the outer free valence electrons.
An electric field connected across the metal allows
movement of the valence electrons which makes up
current flow.
9.2, 1, C2, D3
(ii)
Correct answer
9.2, 1, C2, D3
(b)
Correct answer
9.2, 2, C2, D8
(c)
two points covered
one point covered
9.4, 3, C2, D4
(a)
Correct answer
9.4, 3, C2, D1-3
(b)
Complete coherent answer
Limited answer
Attempt at an answer
H6
1 mark
H6
1 mark
H6 and 7
2 marks
1 mark
H14
1 mark
H3 and 13
3 marks
2 marks
1 mark
n- type semiconductor
Electrons move from the donor level into the
conduction band allowing conduction to take place.
p- type semiconductor
Electrons move from a filled band into the acceptor
band. This leaves behind a positive hole which
allows conduction to take place.
Physics HSC Trial 2014 PEM
Page 4
Question 24
(a) move to a position so that the gravity of the
planet is zero (or effectively zero).
9.2, 2, C2, D4
(a)
Coherent answer
(b)
EP + EK = 0
(b)
9.2, 2, C2, D3 and 4
correct answer
incorrect substitution
H6
2 marks
1 mark
(c) (i)
9.2, 2, C2, D8
correct answer
incorrect substitution
H6
2 marks
1 mark
m1m2
+ ½ mv2 = 0
r
Ep   G
2𝐺𝑀
𝑟
𝑣=√
2 𝑥 6.67 𝑥 10−11 𝑥 3.5 𝑥 1016
40 𝑥 103
= √
𝐹𝑐 = 𝐹𝑔
𝑚𝑣 2
𝐺𝑀𝑚
=
𝑟
𝑟2
𝐺𝑀
𝑟
= √
6.67 𝑥 10−11 𝑥 3.5 𝑥 1016
40 𝑥 103
(c) (ii)
1 mark
= 10.8 m s-1
(c) (i)
𝑣=√
H6 and 13
= 7.6 m s-1
(c) (ii)
9.2, 2, C2, D4
H6
Any suitable shape that indicates that the
astronaut does not reach orbital velocity.
1 mark
Physics HSC Trial 2014 PEM
Page 5
Question 25
(a)
(i) s = ut
(ii) t v 
4.2 = 0.84t
to
1
v2
c2
hence t = 5 years
t0 = 2.7 years
(b)
(i) 5 x 2 = 10 years
(ii) 2.7 x 2 = 5.4 years
(c) (i)
The term paradox is used because both answers are
correct which is a contradiction.
(ii)
The paradox is resolved as two reference frames are
involved and time passes more slowly for the
moving reference frame compared to the Earth’s
reference frame however both are correct for their
own reference frame.
Question 26
(a)
Top coil is labelled secondary; bottom coil is labelled
primary.
(b)
The transformer is step down. This means that the
secondary coils should have less turns that the
primary coil. The photograph shows that the top coil
has less turns than the bottom hence the top coil is
the secondary.
(c)
The function of the transformer core is to provide
magnetic linkage between the primary and
secondary coils so that the changing magnetic field
in the primary is passed onto the secondary coil.
The core is made of soft iron which passes the
magnetic flux from one coil to another. Soft iron is
used as it does not keep its magnetism so switches
polarity easily hence reducing energy losses. The coil
is made of laminated iron to reduce eddy current
induced in the core so reducing heat losses.
Physics HSC Trial 2014 PEM
9.2, 4, C2, D9
(a) (i)
Correct answer
9.2, 4, C2, D9
(a) (ii)
Correct answer
9.2, 4, C2, D9
(b) (i)
Correct answer
9.2, 4, C2, D9
(b) (ii)
Correct answer
H6
1 mark
H6
1 mark
H6
1 mark
H6
1 mark
9.2, 4, C3, D3
(c) (i)
Coherent answer
H4 and 13
9.2, 4, C2, D10
(c) (ii)
coherent answer
H13
(a)
9.3, 4, C2, D3
Correct labels
1 mark
1 mark
H7, 9 and 13
1 mark
9.3, 4, C2, D2 and 3
(b)
Correct answer
H7 and 9
9.3, 4, C3, D 1 and 3
(c)
coherent answer
limited coherence
attempt made to answer question
H7 and 9
1 mark
3 marks
2 marks
1 mark
Page 6
Question 27
(a) Current flowing in the coil produces a magnetic
field. This magnetic field depending on its direction
either adds or subtracts with the permanent
magnetic field so producing a force on the sides of
the coil.
(b)
(i) F = n B I L = 60 x 0.015 x 4 x 0.24 = 0.864 N
(ii) Torque = Fd + Fd = 0.864 x 0.04 + 0.864 x 0.04
= 6.9 x 10-2 Nm
(c)
Torque due to the magnetic field = torque due to
the spring.
Question 28
(a) the output is AC. The output is from slip rings
which does not reverse the output voltage.
(b)
the frequency of the ac is increased
the output voltage is greater
(c) the position shown in the diagram gives
maximum output voltage.
Physics HSC Trial 2014 PEM
9.3, 1, C3, D5
(a)
Coherent answer
Limited coherence
H 7and 9
9.3, 1, C3, D5
(b) (i)
Correct answer
H7 and 9
9.3, 1, C2, D3
(b) (ii)
Correct answer
H7 and 9
9.3, 1, C3, D5
(c)
Correct answer
H7 and 9
9.3, 3, C2, D3
(a)
Correct answer AND reason
2 marks
1 mark
1 mark
1 mark
1 mark
H 14
1 mark
9.3, 3, C3, D1
(b)
two effects correct
one effects correct
H7 and 14
9.3, 2, C2, D4
(c)
Correct response
H7 and 14
2 marks
1 mark
1 mark
Page 7
Question 29
(a)
(a) Line of best fit drawn with the same number of Correct line of best fit
points below as above the line.
If poor line of best fit
(b) (i) Gradient = h = 1.77 x 10-19/(7.5 – 4.8) x 1014
h = 6.6 x 10-34 J s
(ii) the threshold frequency f0 = 4.8 x 1014 Hz
Work function = hf0
= 6.626 x 10-34 x 4.8 x 1014 = 3.2 x 10-19 J
(c)
Electrons are held to the metal ion by electrostatic
attraction. This is the reason for the work function
of a metal and differs for different metals. A photon
of light needs to possess at least this energy if it is to
eject an electron. This is described as the threshold
frequency.
Question 30
Before the invention of the transistor,
communication devices (e.g., radios) used
thermionic devices (vacuum tubes). These use a lot
of energy in heating filaments, were slow to
operate, bulky, expensive and fragile. The invention
of the transistor helped overcome these problems in
that transistors are small, robust, energy efficient
and because they do not require a cathode to be
heated are much faster. Hence transistors have led
to a much improved communication technology
over the old style vacuum tubes.
Question 31
Advantages
AC generators can have their output voltage
increased by transformers for transmission over
long distances.
9.4, 2, C2, D6
(b) (i)
Gradient measured correctly
Incorrect substitution
1 mark
0 marks
H13 and 14
2 marks
1 mark
9.4, 2, C2, D6
H13
(b) (ii)
Frequency measured from intercept correctly
1 mark
9.4, 2, C3, D2
(c)
Coherent answer
Limited coherence
H10 and 13
9.4, 3, C3, D2
Coherent answer covering all points
limited coherence
attempt at answer
H1, 3-5
5 marks
3 - 4 marks
1 – 2 marks
2 marks
1 mark
9.3, 1, C, D
H4 and 13
One correct advantage and one correct
disadvantage clearly provided
2 marks
Only one advantage or disadvantage provided.
1 mark
Disadvantages
AC generators require their output voltage changed
by diodes (rectifiers) if DC is required.
Physics HSC Trial 2014 PEM
Page 8
Question 32
9.4, 4, C2, D4, 5 and 6
H9, 13 and 14
Above the critical temperature
The metals resistance decreases as temperature Coherent answer covering all points
4 marks
decreases due to less lattice vibration and less
interaction with the conduction electrons and Limited answer
1 to 3 marks
lattice. It can be seen that resistance decreases as
temperature decreases.
At and below the critical temperature
From the graph it can be seen that resistance drops
to zero as the electrons form Cooper pairs. These
electron pairs interact with the lattice causing the
electrons to pass unimpeded through the lattice
with no resistance.
Physics HSC Trial 2014 PEM
Page 9
Section II
Question 33
Geophysics
Sample answer
Part (a)
S waves are shear waves and are transverse,
whereas P waves are compression waves and
therefore longitudinal. The velocity of both types of
waves increases with depth. P waves are faster than
S waves but only P waves can travel through a
liquid.
Part (b)
Geophysicists have developed remote sensing
technologies which have led to increased knowledge
of deforestation, crop disease, salinity, climate
change, air pollution, etc. The geophysical methods
developed in this area have an important role in
forecasting and monitoring society’s efforts to
reduce these hazards.
Syllabus content, course outcomes and marking
guide
9.5, 3, C2, D1 and 2
H8
(a)
coherent answer
4 marks
limited coherence
2 - 3 marks
attempt at answer
1 mark
9.5, 5, C2, D2
(b)
coherent answer
limited coherence
attempt at answer
H4
6 marks
2 - 5 marks
1 mark
Developments in the area of seismology have
allowing early warnings to be given and this has
saved the lives of many people living in tectonically
active areas.
Nuclear test ban treaties are based on seismic
monitoring networks. These have led to the world
being a safer place in terms of nuclear weapons
development. No nations can now test nuclear
weapons in secret. This has led to more open
knowledge and discussion of developments in
nuclear weapon technology.
Part (c) (i)
3
r
GM

2
T
4 2
9.5, 2, C3, D4
(c) (i)
correct calculation
H6
1 mark
Substitution changing 7.1 hours into seconds, gives
1.88 x 107 m
(c) (ii)
The height of the orbit of the satellite changes as it
passes over varying gravitational fields caused by
different densities of the Earth’s crust. These
changes in height can be used to work out the
density of the crust.
Physics HSC Trial 2014 PEM
9.5, 2, C2, D4 and 5
(ii)
coherent answer
limited coherence
H4
3 marks
1 - 2 marks
Page 10
Part (d)
The original theory of continental drift was that the
continents moved relative to the sea floor. This idea
was opposed as there was no mechanism known to
provide the energy needed to move the continents
across the sea floor. However the theory of plate
tectonics has the continents sitting on crustal plates
which include the sea floor and the continents move
with those plates, not relative to them.
Sea-floor spreading, the rising of molten mantle
rock through crustal plate boundaries, pushing the
plates apart provides the mechanism for motion of
the crustal plates. As the molten rock cools the
magnetite and haematite parts of the rock indicate
the direction of the magnetic field at that time and
place. Due to random reversals of the magnetic
field, rocks solidifying at different times can have
different directions of magnetisation.
Mapping of the sea floor with magnetometers
shows alternating regions of magnetisation around
plate boundaries. Dating of isotopes gives the time
of magnetic field reversals and shows that younger
rock are nearest to the plate boundaries, and older
rocks are further away, showing that crustal plates
have moved apart due to the up-welling molten
mantle rock pushing the plates apart. This shows
that the crustal plates move apart, the continents
moving with the sea floor, and no other method is
needed to move the continents relative to the sea
floor.
Part (e)
Description of the investigation carried out by the
students with clear labelled diagrams.
Physics HSC Trial 2014 PEM
9.5, 4, C2, D3 and 4
H1 and 2
(d)
coherent answer with most points covered
6 marks
coherent answer with a few points missing
3 - 5 marks
attempt at answer
1 - 2 marks
9.5, 4, C3, D1
H2
(e)
Provides a complete and logical description of the
investigation
4-5 marks
Provides a partial description of the investigation
2 - 3 marks
Provides a statement that is relevant to how the
Earth’s magnetic field varies with latitude.
1 mark
Page 11
Question 34
Medical Physics
Sample answer
Part (a)
The two gamma rays produced move away in
opposite directions due to conservation of
momentum. Gamma ray detectors arranged around
the patient can determine the position of the
emission within the patient from the time delay
between each gamma ray. A composite image is
formed from all pairs of gamma rays allowing the
position of emission to be found.
Part (b)
Many possible answers are possible, for example:
The radioactive tracer, technetium-99m, is put into a
pharmaceutical product, which is ingested/injected
into body.
The technetium emits gamma rays so its path through
the body can be monitored.
The properties of the radioisotope required are:
Must decay mainly by gamma radiation of a suitable
energy value, alpha and beta would damage and be
absorbed by the body.
Half-life must be long enough for diagnosis to take
place but short enough for it not to remain in the
body for a long time.
Must be possible to monitor the gamma rays emitted.
Must not be chemically poisonous.
Must be able to reach part of body being diagnosed.
From the gamma ray images produced it is possible to
diagnose if an organ, e.g. the lungs are working
correctly.
Part (c)
Coherent description with labelled diagrams of
experiment performed by students.
Part (d)
X-ray are passed onto the leg. X-rays pass through the
leg however bones absorb more X-rays than soft
tissue because bones have a greater density. The Xrays passing through the leg expose a photographic
film, showing the image of the bones which can be
checked for fractures, etc.
Physics HSC Trial 2014 PEM
Syllabus content, course outcomes and marking
guide
9.6, 3, C2, D5
H7 and 13
(a)
Coherent answer
3 marks
Limited coherence
1-2 marks
9.6, 3, C2, D1 and 2
(b)
Coherent answer
Limited coherence
H4
4 marks
2 – 3 marks
9.6, 2, C3, D3
H8
(c)
Provides a thorough description of an
investigation into optical fibre technology.
4-5 marks
Provides a partial description of an investigation
into optical fibre technology.
2-3 marks
Provides a correct statement connected to optical
fibre technology.
1 mark
9.6, 2, C3, D1
H8
(d)
Coherent answer
3 marks
Limited coherence
1-2 marks
Page 12
Part (e)
Ultrasound is a wave motion and if there is any
motion of the object that reflects the ultrasound, then
the frequency detected at the transducer will be
different from that emitted.
The ultrasound source is stationary and blood in a
blood vessel is moving towards the transducer. The
apparent frequency of the ultrasound increases
because the reflector, in this case blood, is moving
towards the transducer. By monitoring the change in
frequency of the ultrasound it is possible to obtain the
average speed of the blood. This technique can be
used to obtain information on blood flow
characteristics in the heart.
Part (f)
(i)
Z = ρv = density x velocity
Muscle: Zmuscle = 1070 x 1585 = 1.70 x 106 kg m-2 s-1
9.6, 1, C2, D8
(e)
Coherent answer
Limited coherence
9.6, 1, C3, D5
(f)(i)
Both values correct
One value correct
H8 and 13
3 marks
1-2 marks
H8
2 marks
1 mark
Bone: Zbone = 1910 x 4080 = 7.79 x 106 kg m-2 s-1
(ii)
Substituting into the equation


I r Z 2  Z1 
1.70 x10 6  7.79 x10 6

=
I o Z 2  Z1 2
1.70 x106  7.79 x106
2


2
2
gives a value of Ir/Io of 0.412
9.6, 1, C3, D1 and 5
H8
(ii)
Correct substitution into equation using values
from part (i). Note carry through errors allowed.
2 marks
Incorrect attempt made to use the equation.
1 mark
Hence 41.2% would be reflected by the boundary.
Part (g)
Precession is when the axis of rotation of the nuclei
describes a circle. The spin of the nuclei – in the case
of Hydrogen the spin of a proton, generates a
magnetic field. Different nuclei produce different
magnetic fields. When this magnetic field is subjected
to an external magnetic field the forces interact and
precession results. The frequency of precession is
increased with the strength of the external field and is
also determined by the composition of the nuclei.
Physics HSC Trial 2014 PEM
9.6, 4, C2, D5
coherent answer.
limited coherence
attempt at answer
H7, 9 and 13
3 marks
2 marks
1 mark
Page 13
Question 35
9.7 - Astrophysics
Sample answer
Part (a)
Advantages:
No atmospheric distortion.
No problems with weather e.g. clouds, as the
telescope is well above the atmosphere.
Wavelengths that are absorbed by the Earth’s
atmosphere can be observed in space giving
information that is otherwise unobtainable from the
Earth’s surface.
No seeing problems.
Disadvantages
Very expensive and in general if the telescope has a
problem is it not possible to repair.
The size of the telescope is limited, i.e. the primary
mirror will be a lot smaller than some of the larger
telescopes on Earth.
Part (b)
tan θ = r/d
r = 6370 km
θ = 0.00250
d = 1.46 x 108 km
Physics HSC Trial 2014 PEM
Syllabus content, course outcomes and marking
guide
9.7, 1, C2, D2 and 4
H3 and 13
9.7, 1, C3, D1
(a)
All points substantially covered
5 marks
One or two errors or missing detail
4 marks
A reasonable attempt made
2-3 marks
Limited attempt
1 mark
9.7, 2, C3, D1
H8
(b)
correct formula and substitution
2 marks
Incorrect substitution or incorrect formula 1 mark
Page 14
Part (c)
Visual binaries
These are observed by observing the star as two
separate objects. The limitation is the resolution of
the telescope. Due to the distortion of the
atmospheric space telescopes are better at
observing visual binaries.
9.7, 5, C2, D1
H1 and 13
(c)
Complete coherent answer including limitations
6 marks
most points covered
4-5 marks
limited answer
1-3 marks
Astrometric binaries
Telescopic observation of the binary shows that the
observable star wobbles showing the gravitational
presence of a second unseen star. Limitations of this
method are that the unseen star needs to be large
enough to cause the observable star to wobble.
Eclipsing binaries
The light curve of the star shows the presence of the
component stars. This is illustrated below.
The limitations are that changes in the intensity of
light must be large enough to be measured.
Spectroscopic binaries
The spectra of the star shows a Doppler shift. As one
of the stars moves away from the Earth the light is
red shifted and when the star moves towards the
Earth the light is blue shifted. The limitation of this
method is that measurements of small wavelength
changes are necessary.
Part (d)
M = - 1.0
m = 0.5
M  m  5 log (
d
)
10
 1.0  0.5  5 log (
d
)
10
d  19.95 pc
Physics HSC Trial 2014 PEM
9.7, 4, C3, D1
H14
(d)
correct formula and correct substitution shown
2 marks
or
correct formula and an incorrect attempt at
substitution
1 mark
Page 15
Part (e)
(i) globular cluster
(ii) young open cluster
9.7, 6, C3, D1
H7
(e) (i)
Correct HR diagram with labels
1 mark
Globular cluster should have no blue main sequence
stars but should have many giants.
(e) (ii)
Correct HR diagram with labels
1 mark
Young open cluster should have blue main sequence
stars and few giants.
(iii)
Globular clusters are mainly made up of old stars
and so are missing the massive main sequence stars
which have evolved off from the main sequence.
9.7, 6, C3, D2
(e) (iii)
coherent answer
limited coherence
H13 and 14
2 marks
1 mark
Young open clusters have not had enough time for
the larger stars to evolve and move off the main
sequence hence they contain a full range of main
sequence stars but few giants.
Physics HSC Trial 2014 PEM
Page 16
Part (f) (i)
Star X has the largest radius as it has a high
luminosity but lower temperature and as luminosity
is dependent on both radius and temperature then
it must be very large.
Part (f) (ii)
Physical properties
star V – very large, high temperature and luminosity
star Z – very small, low temperature and luminosity
9.7, 6, C3, D2
(f) (i)
coherent answer
limited coherence
9.7, 6, C2, D3 and 4
(f) (ii)
All points substantially correct
One or two errors or missing detail
A reasonable attempt made
Limited attempt
H7 and 14
2 marks
1 mark
H7, 9 and 13
4 marks
3 marks
2 marks
1 mark
Nucleosynthesis reactions
star V – produces Helium at the core through the
C N O cycle
Star Z – produces Helium at the core through the
P-P cycle
Eventual fate
Star V has a very short lifetime – supernovas to
produce either a neutron star or a black hole.
Star Z– has a very long lifetime – no stars of this
type are old enough to have evolved off of the main
sequence as this takes longer that the current age of
the universe.
Physics HSC Trial 2014 PEM
Page 17
Question 36
9.8 - From Quanta to Quarks
Sample answer
Part (a)
The neutron was identified by James Chadwick.
Alpha particles bombarded either Boron or
Beryllium and a proton was ejected. These neutrons
were not easy to detect as they are uncharged. The
neutrons were made to interact with paraffin wax
and it was noticed that protons were ejected. Using
conservation of momentum and energy Chadwick
worked out that the uncharged radiation emitted
was similar in mass to that of a proton and was the
particle we now call the neutron.
Part (b)
Neutrons because they are uncharged can easily
penetrate materials. Because they have a wave
nature (according to de Broglie) which can be
altered they are used to work out the properties of
materials by scattering experiments.
Part (c)
The LHC has played an important part in verifying
the standard model of matter. The LHC allows
charged particles (protons) to be accelerated up to
speeds approaching that of light. When two proton
beams collide they have enough energy to produce
new particles from which can be inferred the
various range of fundamental particles, quarks,
leptons, etc. The LHC discovered the Higgs Boson
which is an essential requirement of the standard
model of matter. The expenditure on the LHC has
therefore been worthwhile as it has verified the
standard model which explains matter and its
constituents. It also may lead to other discoveries
which may have commercial/financial benefits for
mankind.
Physics HSC Trial 2014 PEM
Syllabus content, course outcomes and marking
guide
9.8, 3, C2, D2
H1
(a)
coherent answer
3 marks
limited coherence
2 marks
attempt made to answer question
1 mark
9.8, 4, C2, D3
H6 and 8
(b)
Neutrons easily penetrate matter as they are
uncharged.
1 mark
The wavelength of the neutrons is suitable for
scattering experiments.
1 mark
9.8, 4, C2, D4
H1, 3, 5 and 13
(c)
Coherent answer with most points
5 marks
Limited coherence
2 - 4 marks
Attempt at an answer
1 mark
Page 18
Part (d)
(i) Nuclear fission is the process where a nucleus is
bombarded with neutrons causing it to split into
two or more parts with the release of more
neutrons.
9.8, 4, C2, D1
(d) (i)
Coherent answer
Limited coherence
H7 and 13
(ii) The moderator slows down the fast fission
neutrons which are emitted from the nucleus. This is
necessary as fission with Uranium requires slow
speed neutrons.
9.8, 4, C2, D1
(d) (ii)
Coherent answer
Limited coherence
H7
(iii) To control the nuclear reaction the number of
neutrons causing fission must equal the number of
neutrons released from fission. The cadmium rods,
which absorb neutrons, are lowered between the
fuel rods to reduce the number of neutrons and
hence have a 1:1 relationship which provides a
stable reaction.
Part (e) (i)
Bohr suggested that the electrons orbited at
discrete or fixed energy levels and that an electron
could only be in one of these fixed energy levels.
This overcame the problem with the Rutherford
model which should have been unstable as by
classical Physics the electron should have emitted
energy and spiralled into the nucleus.
9.8, 4, C2, D1
(d) (iii)
Coherent answer
Limited coherence
H6 and 7
9.8, 1, C3, D4
(e) (i)
Coherent answer
Limited coherence
H1 and 2
(e) (ii)
The problems with the Bohr model were:
It only worked with the hydrogen atom and did not
work with larger atoms.
It did not explain the relative brightness of the
spectral lines.
It did not explained Hyperfine lines.
It did not explain the splitting of spectral lines in a
magnetic field.
Part (f)
Diagrams showing the constituents of the atom and
how protons and neutrons are made of both up and
down quarks.
Strong nuclear force with gluons shown holding the
quarks together and also holding the protons and
neutrons together. Electromagnetic force between
the proton and electron shown (note not shown on
the following diagram).
Students might also show the gravitational force
between particles.
Physics HSC Trial 2014 PEM
9.8, 1, C2, D6
(e) (ii)
Any two problems outlined correctly
Any one problem outlined correctly
2 marks
1 mark
2 marks
1 mark
2 marks
1 mark
2 marks
1 mark
H13
2 marks
1 mark
9.8, 1, C3, D4
H7, 9 and 13
(f)
Comprehensive answer with most points
5 marks
Some points missing
2 - 4 marks
Attempt at an answer
1 mark
Page 19
Strong nuclear force
Physics HSC Trial 2014 PEM
Page 20
Question 37
The Age of Silicon
Sample answer
Part (a)
Similarities
The transistor and integrated circuit are both built
with the same material, that is, silicon. Both devices
operate at low voltages.
Differences
The differences are that transistors do not have
other components in them unlike the integrated
circuit which has resistors, capacitors and many
transistors.
Part (b)
Light emitting diodes (LEDs) are made of P and Ntype silicon joined together. The LED is contained in
a plastic case which is transparent to the light
frequency the LED emits.
Forward-biasing the diode makes charge carriers
move across the P-N junction. The P-N junction then
gives off light. The colour of the LED depends on the
semiconductor material and the doping chemicals.
Part (c)
Transducers are used to measure environmental
processes for example: temperature, brightness, air
pressure, humidity, etc. Transducers respond to the
environment and change their response into a
continuously varying electrical signal, usually a
voltage, which corresponds to the environmental
processes that produced it. In this way
environmental properties can be measured.
Part (d)
Silica which is silicon dioxide has a refractive index
that is suitable for total internal reflection. The
refractive index can be changed by doping allowing
a gradual change in the critical angle which is
required for optical fibres. This is called optical
non-linearity. Silica can also be easily formed into
thin fibres.
Part (e)
(i)
Correct calculation (800/800 + 450) x 12 volts
= 7.7 volts
(ii)
Electronic circuits require a range of voltages to
operate correctly. The supply voltage provided
(from a battery or power supply) needs to be
reduced to the correct voltage to operate many
components e.g. to bias a transistor and this can be
Physics HSC Trial 2014 PEM
Syllabus content, course outcomes and marking
guide
9.9, 1, C2, D4
H13
(a)
Coherent answer with both similarities and
differences
3 marks
Limited coherence
2 marks
Attempt at an answer
1 mark
9.9, 4, C2, D3
(b)
Coherent answer
Limited coherence
9.9, 3, C2, D1
9.9, 3, C3, D3
(c)
Coherent answer
Limited coherence
Attempt at an answer
9.9, 1, C3, D2
(d)
Coherent answer
Limited coherence
Attempt at an answer
9.9, 2, C3, D3
(e)(i)
Correct answer
9.9, 2, C2, D5
(e) (ii)
Coherent answer
Limited coherence
H7
2 marks
1 mark
H7 and 13
4 marks
2 -3 marks
1 mark
H3 and 13
3 marks
2 marks
1 mark
H9
1 mark
H3
2 marks
1 mark
Page 21
provided by a potential divider circuit.
Part (f)
A non-inverting operational amplifier produces an
output signal that is in phase with the input signal.
The inverting operational amplifier produces an
output signal that is 1800 out of phase with the
input. The inverting input is marked with a negative
on the diagram below and the non-inverting output
is marked with a positive.
Physics HSC Trial 2014 PEM
9.9, 6, C2, D7 and 9
Coherent answer
Limited coherence
Attempt at an answer
H7
4 marks
2 - 3 marks
1 mark
Page 22
Part (g)
(i)
AND gate
9.9, 5, C2, D1
H9 and 13
9.9, 5, C3, D1
Three correct diagrams and truth tables
3 marks
Two correct diagrams and truth tables
2 marks
One correct diagram and truth tables
1 mark
OR gate
Invertor gate
(ii)
Half adder
Physics HSC Trial 2014 PEM
(ii)
9.9, 5, C2, D2
Correct diagram and truth table
Some errors
H9 and 13
3 marks
1 – 2 marks
Page 23