Specimen HSC Physics Exam

Download Syllabus (pdf format) 

Download Sample Examination Paper  (pdf format) 

Board of Studies, New South Wales
Physics Stage 6 Syllabus, 1999 © Board of Studies NSW, 1999.

Comments and Solutions on the Specimen Exam Paper for the HSC Physics Syllabus 

Part A Multiple Choice
 

  • 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 navigation.
    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 be A.
  • 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!
  • Q6 A (the photons each have hf, so an ejected electron has an energy hf less the energy required to remove it from the material)
  • Q7 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)
  • Q8 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.
  • Q9: C.
  • Q10: C.
  • Q11 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.
  • Q12 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.
  • Q13 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.
  • Q15: C.

Part B Calculations / Descriptions

  • Q16.  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 kg.
    • (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 experiment?
  • This is obviously an open-ended question with many possibilities, and will probably be a nightmare to mark!  Here are a couple of examples: 

    • (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 times.
    The conclusion is that events which are simultaneous in one reference frame need not be simultaneous to an observer in another reference frame.
  • Q19.  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 K3L.
    • (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. 

     
  • Q20.
    • (a) See the notes in 9.2 Space - this is obviously rather open-ended! 
    • (b) Some factors for successful re-entry:
      • 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 is dissipated.
      • Heat-shielding around spacecraft, to dissipate away heat and prevent craft burning up.
      • Pressurised compartment, with life-support systems (e.g. a breathable atmosphere).  Possibly wearing a pressurised suit as an additional safety measure.
      • Contoured seating to provide support and protection for the high deceleration during re-entry.

       

  • Q21 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 x axis.
  • 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 unexpected feedback.
  • 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.)

  • Q22 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, to
    Michael Faraday's great surprise).
    We rotated a very large coil in the Earth's field.
    We sang "eeee" or whistled into a dynamic microphone.
    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 same direction).

  • Q23 I'm guessing here, as I have never seen a school power pack.
    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
  • The purpose 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.
  • (The diagram 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.)
  • b) This 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)

  • Q25
    • (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.
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