Specimen HSC Physics
Syllabus (pdf format)
Sample Examination Paper (pdf format)
Studies, New South Wales
6 Syllabus, 1999 © Board of Studies NSW, 1999.
and Solutions on the Specimen Exam Paper for the HSC Physics
Part A Multiple
- Q1: B. Activity
in the Van Allen belts, e.g. interaction with the protons
in the wind from a solar storm can certainly affect
communications - producing auroral displays, for instance.
The spiralling electrons around the magnetic field lines
will emit synchrotron radiation at radio wavelengths.
The speed of a satellite has no affect, though of course
if it is in inter-planetary space there will be a light-travel
delay time. Relativity does not affect communication,
though in fact relativity needs to be taken into account
when using the GPS to determine the precise position
of an object. The Transverse Doppler Effect needs
to be included when calculating the shift in frequency
due to the relative motion of observer and satellite
in order to calculate positions accurate enough for
However, the true answer to
this question probably has to do with porous tiles being
able to reduce the weight, even though this is not offered
as an option! The combination of low thermal conductivity
plus low mass is what makes porous tiles suitable material
for the Space Shuttle.
- Q2: D. Frictional
heating with the atmosphere generates heat as
the Shuttle decelerates from Mach 25. This must
be radiated away, otherwise the Shuttle would burn up.
The blunt face the Shuttle presents on re-entry helps
act as a shock absorber - by transferring much of the
energy into a bow-shock surrounding the Shuttle.
There is no attempt to minimise drag on re-entry - the
speed is being reduced rapidly, quite unlike take-off
when an aerodynamic shape helps. Heat is being
conducted, not reflected.
- Q3: B. This
is simply comparing the value of M/R2 for
the Earth and Mars.
- Q4: D. Remembering
N2L and a=F/m, then acceleration, a, is
increasing while thrust is applied to a rocket as the
fuel mass is decreasing. Thus (A) applies when
a=constant, (B) when a=0, (C) when a
decreases (e.g. as for a car ) and (D) when a
is increasing. Acceleration peaks at 3g
for human-flight launch vehicles as we cannot endure
much higher values. Note: I have assumed
that the mass of the fuel is a significant portion of
the total mass of the rocket on launch. If this
were not the case then the answer to this question would
- Q5: A. The
Celestial Castle is at geosynchronous orbit, so an unattached
object will orbit the Earth at a rate of 1 revolution
per period of rotation of the Earth; ie 1 day.
Thus the hammer will appear to remain fixed in place.
In practice the hammer is likely to follow a trajectory
that keeps it near, but not actually at, the Castle
(assuming the Castle is in fact attached to the Earth),
as the gravitational acceleration is not completely
uniform during orbit. The concept of the Celestial
Castle has evolved into what is now termed a Space Elevator
- popular in Science Fiction novels!
A (the photons each have hf, so an ejected electron
has an energy hf less the energy required to remove
it from the material)
A (although magnetic fields obey an inverse square
law---that of Biot-Savart---the field of a long
wire has cylindrical symmetry and so goes as 1/r)
A, B or none, depending upon circumstances. If the
potential difference were held constant (source
with low output impedance), it would be A. If the
current were held constant (source with high output
impedance), it would be B. If the field were held
constant, it would be none of the above. The foregoing
neglects skin effects. If these are substantial,
then the answer is B.
- Q10: C.
A. If the field is approximately uniform, then the
alternate loops in a twisted pair have opposite
flux, and so induced emfs tend to cancel. Conversely,
if the wires are carrying large current, then alternate
loops are magnetic dipoles in opposite directions,
and so tend to cancel in the far field.
C. In first sentence, flux out of page is being
reduced, so current flows as indicated to oppose
this change. In second sentence, flux into
page is being increased, so current flows as indicated
to oppose this change. In A, B and D, the two options
produce opposite effects.
B is most likely. This effect is due to Faraday's
Law. A doesn't (directly) involve magnetism, C is
possibly true, and D is false.
- Q14: D.
It might be noted that the effect is small. The
refractive index change for radio waves entering the
atmosphere from space is very small. This effect
is, however, much more noticeable for visible light.
Part B Calculations
Remember M = Mcraft + Mfuel and
that Mfuel is initially large but then decreases
to near zero. Since Mfuel = 1.9 x 106
kg initially, then Mcraft = 0.3 x 106
- (a) From graph, loss
rate = (1.0-0.5) x 106 / (2-0) =
0.7 x 106 kg/min.
- (b) a=F/M with M =
(0.3 + 0.5) x 106 kg at t = 2 mins.
So a=3.0 x 107 / 0.8 x 106
= 37 m/s2 to 2SF.
- (c) Momentum of entire
system (ie rocket + fuel ejected through nozzle) must
remain constant, so graph is a straight line with
y-axis (momentum) constant against x-axis (time).
- Q17. Thrust to
a ballistically launched satellite, e.g. by a cannon,
is only applied at launch. The orbital trajectory
must therefore pass through this point. An elliptical
or parabolic trajectory could ensue, but not a circular
trajectory. There has to be additional thrust
applied through a course correction once out of the
Earth's atmosphere to change the orbit into a circular
one - hence multi-stage rockets.
Furthermore, though probably
not what is required for the solution, the acceleration
of the satellite during such a ballistic launch depends
inversely on the length of the cannon. In the
best case, where acceleration remains constant down
the barrel, then a = v2/2L. Given that
we need v to be several km/s to achieve orbit, then
in order to keep the acceleration to below 10g we would
need a cannon with barrel length around 100 km long!
- Q18. Describe
a thought experiment that demonstrates the relativity
of simultaneity, and what can be concluded from the
This is obviously an open-ended
question with many possibilities, and will probably
be a nightmare to mark! Here are a couple of examples:
The conclusion is that events
which are simultaneous in one reference frame need not
be simultaneous to an observer in another reference frame.
- (i) Consider astronauts
in two space-craft, one of which is overtaking the
other at a speed v close to c. As this happens,
the first astronaut sets off two flares, at either
end of their spaceship. He sees the flares simultaneously.
The astronaut in the other spaceship, however, sees
the flare from the front-end of the first spacecraft
first, followed by the flare from the rear-end - a
consequence of the different light-travel times for
information on the two events to reach him, resulting
from the relative motion. The lighting
of the flares are not seen as simultaneous events.
- (ii) A second, slightly
more exotic experiment, but one that in fact could
be observed, is to consider the decay of an elementary
particle, such as a pion, into two photons, travelling
in opposite directions at c. The pion is travelling
at speed v relative to a laboratory frame and the
decay occurs mid-way between two detectors in the
lab. In the lab frame the arrival of the two
photons at the two detectors is seen simultaneously.
However in the rest frame of the pion (ie before decay)
they are not, the detectors being struck at different
We need to remember Kepler's Third Law, that T2/R3
is a constant, though it's not necessary to remember
what the constant actually is.
- (a) Show that T2/R3
= 4 for both moons Alpha and Beta, hence verifying
- (b) Since K3L must
apply for this, this implies that T2/R3
= 4 so that T=7.01 reps. So, from V=2pR/T
we get V=2.0 orbs/rep, to 2SF. Or do the algebra
to show that V=pR-0.5
and then substitute for R to get the same result.
- (a) See the notes in
Space - this is obviously rather open-ended!
- (b) Some factors for
- Correct angle of
attack. Too steep, then too much frictional
heating and burn-up. Too shallow, then skips
off atmosphere into extended elliptical orbit.
- Blunt profile presented
to atmosphere, ensuring a bow-shock set up around
the space-craft, in which most of the heat generated
- Heat-shielding around
spacecraft, to dissipate away heat and prevent craft
- Pressurised compartment,
with life-support systems (e.g. a breathable atmosphere).
Possibly wearing a pressurised suit as an additional
- Contoured seating
to provide support and protection for the high deceleration
a) The first sketch is a retangular hyperbola in the
fourth quadrant, taking length as the x axis. Also the
first quadrant, if currents may be negative. The second
sketch is a straight line through the origin, with negative
(or possibly positive) slope, taking separation as the
b) If currents
are positive, field due to upper wire is into page at
position of lower wire. l X B on lower wire is therefore
up. Field due to lower wire is out of page at position
of upper wire. l X B on upper wire is therefore down.
a and b are rather odd experiments, by the way, requiring
c) A subtle
question because at first it appears that they may not
be neglected, unless the currents are equal, whereas
the separate subscripts suggests that they were varied
independently during the experiment. AB and CD produce
fields out of the page at the position of BC and FG.
EF and GH produce fields into the page for both these
segments. But unless they cancel due to equal currents,
one would expect their effects to be non-negligible.
Note, however, that the experimentalists find a 1/d
law. This suggests that, to the accuracy of the experiment,
d/l = 0. One could therefore argue that, to an approximation
that is not much worse AB/BC = 0 etc and so these lengths
have little effect because they are short. But if they
are short, then one would have to worry about the wires
the left of A and E and to the right of D and H....
Well let's have d<<AB and also AB<<l...
Perhaps it's easier to say that they may not be neglected
d) The fields
acting on the wires still go approximately as 1/separation,m
although the approximation is slightly poorer. However,
when you double d, you change the separation at other
points by a factor less than 2. (Doing the calculation
explicitly using the Biot-Savart law is messy in html.)
lots of possibilities here:
a) we spun a wire coil near a magnet.
We rotated a magnet near a coil.
We wiggled a magnet backwards and forwards near a coil
We wiggled a coil backwards and forwards near a magnet
We rotated a homopolar disc near a magnet
We rotated a magnet near homopolar disc and this didn't
work, much to our surprise and also, we discovered,
Michael Faraday's great surprise).
We rotated a very large coil in the Earth's field.
We sang "eeee" or whistled into a dynamic
We sang "eeee" very loudly into a loudspeaker.
We plucked the string of an electric guitar.
b) the ends
of the coil are connected to either half of a split
ring, and this ring is contacted by brushes connected
to the external circuit. Thus the side of the coil passing
a particular pole of the magnet is always connected
to the same connection to the external circuit and so
always has the same sign potential (for motion in the
I'm guessing here, as I have never seen a school power
a) it has a core which is made out of many thin layers
of a high permeability metal, probably forming a loop.
Around one part of this loop is wound a large number
N turns of the primary coil, and around another part
is wound a small number n turns of the secondary. The
primary coil is (probably) made of thin wire and the
secondary of thicker wire. The primary coil is connected
to the two symmetric pins on the plug and thus to the
active and neutral wires of the 240 V circuit. This
produces a varying flux phi(t) in the metal core. Because
the core is shaped in a loop, and because its permeability
is much larger than that of air (several thousand times
greater), the flux remains almost entirely within the
core: nearly all of the flux lines go out of one end
of the primary coil, stay within the core, and go back
in the other end. Because of this, the flux phi in any
loop of the secondary coil is almost exactly the same
as that through any loop in the primary coil. Thus we
may relate the secondary voltage Vs to the primary voltage
Vp using Faraday's law:
Vs = - n
dphi/dt = - (n/N) N dphi/dt = (n/N) Vp
of the laminations is to reduce the size of eddy currents.
Another important component is the insulation of the
wire of the coils (made with a thin coat of transformer
laquer) and other insulation to prevent currents travelling
between coils and the core or between coils.
I put on the web page at http://www.phys.unsw.edu.au/~jw/HSCmotors.html
shows a transformer connected to step-up the voltage.
In the power pack, it would be connected to step-down,
so exchange primary and secondary in the diagram.)
is called a step down transformer. Not a very good name,
because it is the way it is connected rather than the
transformer that steps-down. The transformer itself
could easily be used as a step-up, just by connecting
an input to the wires normally used as output.
c) If no
energy is lost, and if the phase relation between voltage
and current is the same in the primary and secondary,
then VpIp = VsIs so Is/Ip = Vp/Vs = N/n (More formally:
power in = VpIp cos alpha power out = VsIs cos beta
If alpha = beta and if no power is lost, then the first
equation above holds)
- (a) Electrons are emitted
by a hot filament at the cathode by thermionic emission,
focussed by a magnetic condensing lens (focussing
coil) and accelerated towards the positive anode.
- (b) Electric field
E is applied at the deflecting plates (which should
be labelled in the diagram!). If E is downward
then there will be an upward force on the electrons,
F=eE, producing an acceleration a=eE/m. Suppose
it takes time t for the electron to pass through the
region of the plates, of length L. The the deflection
y = 0.5 a t2, with L=vxt.
This leads to y=eEL2/2mvx2.
After leaving the plates the electron is no longer
accelerated, and continues in a straight line (ie
uniform motion). The deflection at position
A on the screen can then be calculated knowing the
dimensions of the tube.
- (c) In a television
set the electron beam is passed over the screen in
a set pattern and its intensity varied to produce
a picture. The magnetic force F=eVxB
is used to steer the beam along its path.