Hey guys, I'm at school now looking at a spare copy of the exam (which I think my teacher wants back). I'll scan it when I get home and whatnot
Here are my solns so far:
(will make some nice LaTeXed Worked solns later
)
All done now, here: http://www.mediafire.com/download/d5a3w6a3z6nj85o/VCAA_Physics_2013_15.11.2013.pdf#1 Attached worked solns for motion, sorry for all the preamble and stuff. Only working LaTeX doc I had 
#2 Rest still coming, then I'll look at all the detail studiesMotion1. a) a = 1.73 m/s/s if you find the net force and divide by the mass mgsin(10)/m
a = 1.75 m/s/s if you use a constant acceleration formula x = ut + 1/2 a t^2
b) Using constant acceleration formulas, a= 0.194 so then Ffriction = 0.50(1.75-0.194)=0.78N
2. a) W = 20 N
b) T = a x m2 = 2.5 x 6 = 15 N
3. a) 6 x 2 = 6 v so speed is 2m/s
b) KE before = 36 J and KE after = 12 J so inelastic
c) I = Change momentum of p of m1 = 2 x 2 - 6 x 2 = -8 Ns. Same as 8Ns to the left
4. a) One arrow “N” or “R” perpendicular to the slope
One arrow straight down “W=mg”
One arrow parallel to horizontal pointing to the left
b) tan(x) = v^2/gr so tan(x) = 0.125 hence the angle is 7.13 degrees
You could do a “from scratch” approach by only 2 mark q. One for formula, one for ans
5. a) Net force always towards centre. At P, direction is D
b) Fc = T – W so mv^2/r = T - mg --> T = 2x49/1 + 2x10 = 118 N
Didn't read the question properly first time sorry if I scared anyone!
6. THE TRICK WITH THIS Q IS THAT KE IS NOT ON THE GRAPH
a) At Z KE = 0 and GPE = 0. So total energy = EPE = 20J
b) At any point KE + GPE + EPE = total energy = 20J
So at Y, 5 + 10 + KE = 20 so KE = 5 = 1/2 m v^2 so then v = 3.16 m/s
c) They At Q there already is spring potential energy since it is extended.
Use EPE = 1/2 k x^2 to prove this
7. a) Period is the length of one day since it stays in the same spot, so 24 hours
b) Use Kepler’s law or similar, converting units THEN minus the radius of the earth
R = 4.24 e7 so then Radius
is 4.27 E7 – 6.4E6 = 3.60 E7 oops I read altitude, not radius my bad
Supposed to be just R = 4.24 E7 thanks again for picking that up!
c) Only apparent weightlessness applies
8. a) vertically and up is positive -15 = 10 t + 1/2 (-10) t^2 solving a quadratic of formula
Or find the time separately in sections. t = 3 sec
b) vertically: v^2 = u^2 + 2 a x --> v^2 = 100 + 20x15
so v=-11.61 m/s thanks guys! the working was right... I put it in the calc wrong lolol... v = -20m/s
horizontally: v = 20 cos30 = 17.32
so then |v| = 26.46 m/s and angle = inverse tan = (-20/17.32) = -49.1 degrees
… although this seems like a spesh question
Electronics9 Total resistance = 5, Vout = 15V. Total resistance = 15, Vout = 5V
10. Two 50 ohm in parallel then followed by two more resistors in series
11.a) R = 2.5 / 0.005 = 500 ohm
b) Rthermistor = 1.0/0.005= 200 ohm. From graph, 20 degrees
c) Rldr = 10/0.005 = 2000 ohm. From graph, 15 lux
d) Change 1: increase the temp, smaller Vout
Change 2: decrease the lighting, smaller Vout
12.a) Vled = 2.0V Vresistor = 0.010 x 450 = 4.5. EMF = 6.5 V
b) Resistor: electrical -> heat. LED: electrical -> light
13.a) 5/0.012 = 416.6
b) The signal is changed
Electric Power14. Should look like two north poles facing each other
15.a) 1:6 ratio so 18.0 V
b) Peak = root(2) x 18 = 25.5 V
c) P=V^2/R = 18^2/1200 = 0.27 W
d) Explain general principles of how transformers work (one mark)
Relate it to this situation here with the DC input and the on and off position (one mark)
Explain what happens when it’s turned on using Lenz’s Law (one mark)
16.a) Current going from W to X. Use slap rule, Anticlockwise
b) F=nBIL so F = 20x0.500x0.50x0.050 = 0.25 N
c) Slips rings still make contact so there still is friction present (one mark maybe?)
Explain this is a DC motor and the direction of the current in the coil needs to be altered
every half turn which is the function of the slip ring (one mark)
Explain if a slip ring commutator was used in this motor the armature would oscillate
perpendicular to the field then stop turning. So the motor does not work (one mark)
17.a) EMF=change in flux/time = (0.6-0.2)/0.5 = 0.8V
I = 0.8/0.1=8A …note that I don’t think you’d get this formula if you did the sin(2pif) rule…
b) EMF zero when flux is max/min, think of derivatives if it helps. So t=0, 0,5, 1, 1.0, 1.5, 2.0, 2.5
c) Moving towards a North Pole.
To oppose this motion, a North Pole is induced on the side of the ring closest to the N Pole
So, induced field direction is down towards the magnet.
Using RH Grip rule, the current flows clockwise
Or, the magnetic field strength is increasing so magnetic flux increases as the ring moves down
To oppose this, the induced current will create a field in the opposite direction, etc etc
As it doesn’t specify if the speed is constant and VCE doesn’t require you to know about
strength of magnetic fields, I don’t think you would be penalised if you didn’t mention how
the current changes over time (decreases for those interested)
d) Flux is relative to the magnetic field strength/density. Technically strongest part is around the
poles, 1/6th the total length I think I read somewhere last year, but we assume it’s at the
middle of the magnet. When it’s far away, weaker field so less current induced (think of Lenz’s
Law if it helps).
So, first min of graph is A (given),the first max is
B C (middle of magnet), next min is B, then C etc
Aka A: t=0, 2, B: t=1.0 C: t=0.5, 1.5, 2.5
Sorry guys, didn't look at the diagram right... thought it was B in the middle gosh
This is main reason I do crap on exams =P18.a) R = Vloss/I = 24/6=4.0 ohm
b) Voutput = 1200/6.0 = 200V
c) Ploss = 4 x 6^2 =144. So 96/1200 = 12% -- cheers Jtc + nliu, my mistake sorry!
d) I = 10/2.0 = 5A. Voutput = 1200/5 = 240V
Light and Matter19.a) E = 6.63 E-34 x 6.7E14 = 4.44 e-19
b) wavelength = 3e8/6.67e14 = 4.48e-7 m
20.a) An atom is at 3.19eV state? That’s interesting wording…
f = 3.19/4.14e-15 = 7.71e+14 Hz. Wavelength = 3e8 / 7.71e+14 = 3.89e-7 m
b) 588.63nm, so f = 3e8/588.63e-9 = 5.10e+14 Hz. So E = 5.10e+14 x 4.14e-15 = 2.11 eV
This exactly corresponds to electron falling (emission spectra) from 2.11eV level to ground state
So, there is a spectral line at 588.63 nm
21.a) Max KE = change of e x Stopping voltage = 1.6e-19 x 1.85 = 2.96e-19 J
b) Work function =
KE max / hf = 2.96e-19/(6.63e-34 x 1e15) = 0.446 **************** double-checking
c) Less intensity, smaller max photocurrent but same stopping voltage
d) The frequency is less than the threshold frequency. Thus, no current produced
22.a) The intensity at the centre will be the greatest. This is because it is equidistant from both slits
Constructive interference occurs
b) wavelength (lambda) = fringe width x slit separation / distance from slits to screen
so, using a lower frequency means bigger wavelength so bigger fringe width. So D. further away
c) centre-dark-bright-dark-bright = 1.4e3 nm. centre-dark = 1.4e3 / 4 = 3.5e2 nm
d) Particle model only predicts two central light band (one mark)
Wave model predicts bands formed by interference patterns (one mark)
Young’s double slit experiment showed constructive and destructive interference so supports
the wave model (one mark)
Moto: never trust uncyclopedia (bonus ten marks)
23.a) E = 80e3 eV = hf so f= 80e3/4.14e-15 = 1.93e+19 and wavelength = 3e8/1.93e+19=1.55e-11m
p = 6.63e-34 / 1.55e-11 = 4.27e-23 kgm/s
b)
*****indicates I'm double checking my solns for these questions
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