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OCR G482 Physics A EWP Unofficial mark scheme watch

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    OCR Physics A G482 Unoffical Mark scheme

    Its become something of a tradition now.
    Usual disclaimer. This is is no sense an offical mark scheme.
    It may and probably will contain tpyos and errors. I'll update any I find.

    First impression. That electricity section was a beast.

    Q1 a) i) P = V^2 / R so R = 230^2/1500 = 35.3 ohms (2)
    ii) R = rho L / A so L = 35.3 x 7.5E-8/1.1E-6 = 2.40m (3)
    b) tricky to draw this - I'll attach a picture.
    components in parallel with switches in right places (3)
    c) i) If P is less, R must be more (same V)
    R is prop to 1/A so A must be less
    A is prop to d squared so d must be less. (2)
    ii) Same Q with numbers
    P is prop to 1/R is prop to A is prop to D squared
    so P1/P2 = d^2/D^2
    so ratio of dia = sqrt (1.0/1.5) = 0.82 (3)
    d) P=VI so I = P/V = 2.6E3/230 = 11.3A so need 13A fuse (2)
    e) i) 1kW.h = energy converted by a device of power 1kW operating for a time of 1 hour (1)
    ii) cost = 18 x 1.6E3(fan + 1 heater) x 4.0 = 115p (2)
    Total : 18
    Phew - b and c not straightforward. a, d and e Ok

    Q2 Potential dividers - everybody's favourite - not.
    a) variable resistor has an arrow through it
    thermistor has a dot and an ice hockey stick (2)
    b) i) read off 180 -> 500ohms (1)
    ii) I = V/R = 7.0/500 = 0.014A (1)
    iii) PD across R1 = 12-7=5v
    same current so R = 5.0/0.014 = 357 ohm (2)
    iv) PD in ratio 5:7 so 5/7 = R2/357
    so R2=255ohms so temp = 200 (2)
    c) on R1/r2 = 7/5 so R1 = 7/3 x 250 = 350 ohm so T = 192
    Off R1/R2 = 5/7 so R1 = 5/7 x 250 = 179 ohm so T = 210 (4)
    Total : 17
    a,bi and bii easy Rest tricky.

    Q3 Multi-cell and internal resistance. Thanks a lot
    What happened to series and parallel resistors?
    a) i) Q=It = 450E-3 x 4.67 x 3600 = 7560C (3)
    ii) 1)positive
    2) current flows backwards through cell to charge it.
    charger has larger emf than cell so current flows clockwise so top of charger must be + (2)
    3) v - 1.5 = Ir (Kirchoff 2) so V = 1.5+ (0.45x0.9) = 1.905v (2)
    4) rate of increase of energy = emf x current = 1.5 x 450E-3 = 0.675 J/s (2)
    b) Standard experiment : cell with variable resistor.
    Need voltmeter across cell to measure terminal PD; ammeter inseries to measure current.
    Measure V and I. Vary resistor. Repeat
    Plot V (y axis) against I (x axis)
    Internal resistance = - gradient
    Emf = y intercept (5)
    c) i) Assuming Rlamp = 18; internal resistance = 4 x 0.9 = 3.6ohm
    PD across bulb = 6.0 x 18/(18+3.6) = 5v so wont be fully lit.(3)
    ii) n = no of cells needed, we want 6v across bulb so
    (n x 1.5) x 18/(18+ (nx0.9)) = 6
    (algebra)
    so n = 5 (2)
    Total: 19
    Another beast. Not much easy there.

    Q4 a) Nice easy defintions to kick off
    i) amplitude = maximum displacment
    frequency = no of oscialltions per unit time
    phase difference = how out of step two oscialltions are expressed as fractions of a cycle or an angle.(aw) (3)
    ii) 1) similar: same frequncy / period / amplitude / all SHM
    2) diff : phase / reach max at different times (2)
    b) i) different amplitudes at different points
    so has nodes (points where displacment is always zero) and antinodes (points where displacment is always max)
    doesnt tranfer energy
    particles between nodes in phase
    adjacent segments have phase diff of 180
    is made from 2 waves superposing (2)
    ii) waves reflect off ends
    incident wave and reflected wave superpose and interfere.
    some points constructively/reinforce giving antinodes
    some points desctructively/ cancel giving nodes (3)
    iii) 1) same frequency
    2) different amplitude / phase (3)
    iv) 1) must be in same segment so 1 loop (fundamnetal) so 15hz
    2) must all be nodes so 8 loops so f=8x15 = 120Hz (4)
    Total 17
    Not too bad. Familiar ground at least.

    Q5 a) i)interference : waves superpose. At points where in phase, reinforce -> max
    and at points where 180 out of phase, cancel -> min (2)
    ii) cohernet = constant phase difference (1)
    b) v = f x lambda so lambda = 3.0E8 / 10E9 = 0.03m (microwaves are EM so travel at speed of light)
    so length = lambda / 2 = 0.015m (microwaves are nearly always 3cm!) (2)
    c) i) 1) if move further away amplitude of waves from sources decrease (inverse square law)
    so amplitude of sum decreases. (2)
    2) lambda = ax/D so x =0.03 x 4.0 / 0.2 = 0.6m
    so max to min = lmbda / 2 = 0.3m (2)
    ii) Path difference between sources changes so phase difference changes so sum changes (2)
    iii) Path diff from A increases.
    when extra lmbda/2 waves arrive at o 180 out of phase and cancel
    so need to move 0.015m (2)
    d) i) If double amp of A and B, amp at O will double
    I is prop to amp squared so it will x4 (2)
    ii) Microwaves are polarised (I guess you're supposed to know this or have seen demos)
    so intensity will drop to zero
    Malus' law I is prop to cos squared andtheta =90 so I -> 0 (3)
    Total :18
    First bit Ok. Last bits wordy whoch always causes problems

    Q6 Looks routine - but why use Einstein without mentioning Plank or Millikan
    a) Photoeletric effect (1)
    b) One photon absorbed by one electron.
    Work function energy of metal (phi) = min energy needed to release electron from surface of metal
    if energy of photon > work function enery, electron is released.
    threshold freq hfo = phi
    hf = phi + (KE)max
    Blue light has higher freq than red light so has more energy
    hf(blue) > wfe and hf (red) < wfe so blue light releases electrons and red light doesnt (4)
    c) i) fo = c/lambda = 3.0E8/480E-9 = 6.25E14 Hz
    ph = hfo = 6.6E-34 x 6.25E14 = 4.125E-19J (3)
    ii) hf = phi + (KE)max so KEmax = 5.2E-19-4.125E-19=1.075E-19
    1/2mv^2 = KE max so v = sqrt(2 x 1.075E-19/9.11E-31) = 4.85E5 ms-1 (3)
    d) i) electron diffraction.
    fire beam of electrons at graphite crystal.
    see interference pattern of bright and dark rings on screen (2)
    ii) lambda = h/mv = 6.6E-34 / (9,11E-31 x 500E3) = 1.45E-9m (3)
    Total = 16
    Not too bad to finish with.


    Overall I think this was another tough paper. not quite as tough as last year.
    Not too many easy bits and lots of parts that test understanding.
    The electricity questions in particular were mean. Pot dividers and internal resistance are the two hardest topics. Both are here.

    I'd go with A=65 B=58 C=51 D=44 and E around 37

    Good Luck
    Col
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    Thanks Teachercol!
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    I think I agree with your conclusions on each question, although I found q4 harder than q3 but that was probably just the wordiness of q4.

    Thanks!

    *Good Luck to everyone else too
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    Thanks. I didn't get that microwaves were polarised, how the hell does OCR expect you to know this when it doesn't come up in their official book(s) or specification?

    I am seriously considering writing to Ofqual after that because it's not in the textbook.
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    (Original post by teachercol)
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    Thanks for your effort man However I think the fan had to be a motor on the circuit above.
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    (Original post by phoenixsilver)
    Thanks. I didn't get that microwaves were polarised, how the hell does OCR expect you to know this when it doesn't come up in their official book(s) or specification?

    I am seriously considering writing to
    Ofqual after that because it's not in the textbook.
    All EM waves are transverse and only transverse waves can be polarised due to the perpendicular oscillations.
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    If the grade boundaries are 20% for an A then i smashed it!
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    Thank you!

    You see the d=0.8D one
    I did P=V^2/R and R=pL/A so P=V^2/(pl/A) so both then found that 0.8D ~ d... Would I still get the marks?

    Also for some reason I decided to forget that microwaves also travel at the speed of light -_- So knowing I got wavelength (I just assumed it was 1x10^-2 as microwaves are around that) and length of aerial wrong, may I get follow through mark for still using lambda=ax/D but just using the wrong value for wavelength so get all the marks on that sub-part or would I only get mark for applying equation?

    Also I didn't talk about any sort of graph for the e.m.f and internal resistance experiment question? I just drew a series circuit, with a voltmeter across cell as well as a voltmeter across a variable resistor and also an ammeter in series. THen I said terminal p.d = p.d across variable resistor, so e.m.f = sum of values on voltmeters. So E=V+Ir so you can find r for this. Then change resistance and test for many values... Have I completely screwed that up?

    Also It would really help if you lightly described question before each answer, hard to remember and discuss when all you have is an answer :P

    And are you 100% sure about the cost one? I'm not too sure but It only said for 1 heater so I did only 1.5kw :/
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    For the similar and difference questions with waves, I made reference to the points on the wave, stating the phase difference between two. Where you supposed to use the points as examples?


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    I don't know why but I remember getting I=10. something so still picked a fuse of 13A... is that 1/2 then?
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    (Original post by MathsAstronomy12)
    All EM waves are transverse and only transverse waves can be polarised due to the perpendicular oscillations.
    Yes, of course, I know that.
    The question in G482 yesterday did not state that the two coherent microwave sources were plane polarised. If it's not plane polarised, rotating the microwave receiver should have no effect.
    Nowhere in the syllabus are we told that we should assume that the two waves are plane polarised. It's a mere guess, which is appalling considering one can lose 3% based on the fact that the question didn't state what it needed to.
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    What was Q 3 a)ii part 3 like what was the question?

    Thanks so much for this col, we all appreciate it so much!
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    Anyone know roughly what 70/100 would be in UMS?
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    Thanks very much!
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    Hey teachercol, if you don't mind please, could you give me the question for Q5a to 5d please. I have forgotten what they were so having a hard time remembering if what I wrote is correct. Will really appreciate it. Thanks a lot.
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    (Original post by caldemby)
    Anyone know roughly what 70/100 would be in UMS?
    about 130 I reckon.
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    thanks for this teachercol.
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    Got anywhere from 67/100 to 77/100 :/ No chance of full UMS for me
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    Typo on Q1 part 2 shouldn't it be 2.5m?

    Thanks for the mark scheme!
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    (Original post by joe12345marc)
    I don't know why but I remember getting I=10. something so still picked a fuse of 13A... is that 1/2 then?
    Yeah man I got 10. Something as well. I'm not sure, cos it was two marks so it may be 1 mark for the correct fuse and one for the reason it needs to be higher so just talking about the fuse? However if you get a stern marker and you've got incorrect physics getting you to the answer they may not give it. So 1/2 is realistic
 
 
 
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