Physics A G482 EWP Unofficial mark Scheme

But that isn't using the diagram

But that isn't using the diagram


85 is average so 82/83 maybe?

I don't know how full UMS is calculated.
For physics its always like 15 marks above A boundary. So if A is 65 or 64 then it will be like 80/81 but if an A is 69 then it will be around 85.

OCR Physics A G482 EWP 9th June 2014

Usual disclaimers.
These are just my answers and are in no sense 'official'.
They may contain errors / typos.

I think this paper was one the hardest I've seen for a long time. Lots of tricky bits.
Expecting low grade boundaries.

Q1 a) i) Not a straight line through origin
so I is not prop to V
so doesn't obey Ohm's law [1]
ii) Line is approx straight so R is approx constant
R = V / I = 0.5v / 50mA = 10ohms [2]
iii) At 6v filament gets hot ( more power dissipated)
Atoms oscillate with bigger amplitude so more collisions with electrons so more energy converted to heat
so resistance increases [2]
b) In the daytime Rldr = 1ohm
so Rcombination is approx 1 ohm
so Pd divides 25:1 with resistor getting 25/26 of 12v and LDR getting 1/26 of 12v
PD across light bulb = PD across LDR is small and doesnt light
At night Rldr = 1000ohm
so R combination = R light bulb approx = 25ohm
so light bulb gets 1/2 of 12v = 6v
so lights up. [5]
Tricky potential divider with that parallel combination.
TOTAL 10

Q2 a) Emf = J/C; resistance = V/A; energy = VC; charge = As [2]
b) i) PD is the energy converted per unit charge from electrical to other forms of energy in a load. [2]
ii) Internal resistance = resistance between the terminals of the cell. [1]
iii) Same PD (in parallel) I = V/R so if half the resistance in branch (6ohm and 12 ohm)) then it
must have twice the current [2]
iv) PD across top branch = I x total R = 0.16x6 = 0.96v
so 0.96v across bottom branch
Potential divider so PD across 6ohm = 6/12 x 0.96 = 0.48v
[A lot of work for 2 marks]
v) PD splits 3:3 in top branch and 6:6 in bottom branch
same potential at each point so no difference [2]
vi) Current in bottom branch = 0.08A (half the other one)
so current through cell = 0.16 + 0.08 = 0.24A
V = E - Ir so 0.96 = 1.2 - 0.24 r
so r= 1.0ohm (possible -1 for sf of 1ohm answer) [4]
A tough question after the easy definitions
TOTAL 15

Q3 a) resistance = PD / Current [1]
b) i) strange setup which will throw many
R = rho x L / A = 1.7E-8 x 20 x 3.8E-10 / (4 x 3.8E-10) ^2
= 224 ohm [3]
ii) no density of electrons = no of atoms per unit vol
= 4 x 20 x 1 / ( 4d x d x 20 d) = 1/ 3.8E-10^3 = 1.82E28 /m3 [1]
Hard work for 1 mark
iii) I = nAvq = 1.82E28 x (4 x 3.8E-10 x 3.8E-10) x 1.9E-5 x 1.6E-19
= 3.2E-14A (amazingly small) [2]
iv) Power =I^2 R for each conductor
so total power = 1E9 x (3.2E-14)^2 x 224 = 2.29E-16W [3]
c) Electrons undergo many collisions with atoms so mean drift velocity is very small.
EM waves travel at speed of light (ish) so v fast and very short time [2]
Not a straightforward resistivity question.
TOTAL 12

4 a) i) Ammeter in series + voltmeter in parallel with LED [1]
ii) PD across LED = 4.0v (from graph)
PD across R = 100 x 20mA = 2.0v
so Terminal PD = 6.0v [3]
b) i) E = 4.1E-19 / 1.6E-19 = 2.56eV [1]
ii) Turn on PD = 2.56v from graph
Photon is emitted when electrons have enough energy to excite atoms in LED
energy lost by electron = Energy of photon [2]
c) i) n = I / Q = 20E-3/1.6E-19 = 1.25E17 [2]
ii) Total = energy of 1 photon x no of photons /s (possible ecf)
= 1.25E17 x 4.1E-19 = 5.125E-2 [2]
iii) Efficiency = useful power out / power in = power out / VI
= 5.125E-2 / (4.0 x 20E-3) = 0.64 [2]
d) graph starts at 2.0ev and goes through 3.4v/20mA parallel to first time [2]
Not too bad if you know what you're doing but tricky all the same
TOTAL 15

5 a) yes coherent -constant phase difference
b) Superposition principle : At min add displacements together.
Min displacement = 4.0 - 2.0 = 2.0 so not zero
Waves have different amplitudes so cant cancel perfectly [2]
c) i) 90 ii) 135 [2]
d) i ) T = 0.80ms so f = 1250Hz
v = f x lambda so lambda = 340/1250 = 0.272m [3]
ii) lambda = ax/D so D = 0.40 x 2.4 x 2 / 0.272 (need max to max distance)
= 7.1m [3]
e) i) Intensity = power per unit area [1]
ii) Min amp = 2 units
Max amp = 6 units
Intensity is prop to amp squared
so max intensity = 9 x min intensity = 9 x 4.0E-6
= 3.6E-5 [3
Finishes with a real tricky bit.
TOTAL 15

Q6 nearest thing to straightforward!
a) Travel at speed of light
through a vacuum [2]
b) i) plane polarised means oscillations are confined to a single plane [2]
ii) Polariser 1 only allows vert components through and absorbs horiz
Polariser 2 only allows horizontal components through and absorbs vert
so all components absorbed and dont see anything [2]
iii) Yes - will see something.
Easiest to draw diagrams showing components - I demo this with microwaves each year.
Hopefully my students remembered it! Difficult if youve never seen it before [3]
TOTAL 9

Q7 a) Standard stuff . Waves reflect from walls.
Interference / superposition.
Cancellation - . nodes. Reinforcement -> antinodes etc [3]
b) Antinodes ; particles get hot because oscillating with max amplitudes at these points [2]
c) Some confusion about scales
Antinode to antinode = 6cm = lambda /2
so lambda = 12cm
so v = f x lambda = 2.5E9 x12E-2 = 3.0E8 as expected [4]
TOTAL 9
Not too bad!

Q8 a) i) wfe = minimum energy needed to release an electron from surface of metal [2]
ii) wfe = energy of photon that just releases electron = h x threshold frequency [2]
iii) wfe = hc/ lambda = 6.63E-34 x 3.0E8 / 55E-9 = 3.6E-19 [2]
b) i) Energy of photon = hc/lambda = 4.5E-19
so max KE = 4.5E-19 - 3.6E-19 = 9.0E-20J
so KE = 1/2 mv^2 so v = sqrt( 2 x 9.0E-20/9.11E-31 = 4.45E5 [3]
ii) lambda = h /mv = 1.63E-9m [2]
c) i) Longer wavelength so smaller energy gap so 3 -> 2 [1]
ii) 3rd transition means 3 -> 1
Calculate energies of2->1 and 3->2 , add then use E=hf
or add frequencies etc
Ans = 252nm [3]
Tough ending to a tough paper

Grade boundaries??

Last year it was 69 for an A and 39 for an E
They are going to come down from that
Could be 65 and 35.
Might be even lower

Best guess?

A 65
B 58
C 50
D 43
E 35

Good Luck.
Everyone is going to find this paper hard.


(Edited for various typos)

I think I got a different answer for the potential divider question but I don't understand how because I just used the formula. I got like 11.7 V or something across the lamp for the night.

I had that too

Also where you said it's a lot of work for two marks.

Could just do

Current = 0.08A (half that of X)

V = IR = (0.08)(6) = 0.48 V

Also got a different answer to 3b)i)

Instead of dividing by (4 x 3E-10), I worked out the area of the circle and multiplied by 4, which I don't think is right.

But I don't really get the question because you have to divide it by the cross sectional area but since it's made out of spherical atoms lined up, the cross sectional area is constantly changing...

That's what I did as well.

Would I lose a mark for writing the velocity of the electron as 4.4x10^5 instead of 4.45x10^5? I don't remember my answer correctly but I'm sure there was a rounding error.


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Think I got around 83

Well done, hope I've done enough for dat 150

Same
It was tricky and I made some silly mistakes, and gained marks I didn't expect to!

No that was the actual answer, same as mine.
It said show its around 4.5 and 4.4 is around so you get marks.
And the next question if you use 4.4 you also get marks.

Thanks, how did you find the exam?


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