OCR G482 Physics A EWP Unofficial mark scheme

To work out the ratio d=0.8D, I worked out their cross sectional areas with the resistivity equation, found their diameter and divided one by the other to get 0.81, do you think I'll get the marks?
Thanks for making this mark scheme!

Crazy how you can get a method technically like entirely wrong but get the actual answer right xD

What's your opinion on grade boundaries? I don't see an A being any higher than 63 but a lot of others do :/

Haha yeah I know.

I think 63 is about right. I think it will be somewhere between 60 - 65. If this markscheme is correct I've gotten somewhere between 85 and 95, can't believe it I thought i did mediocre at best. Means I've hopefully got full ums in both now. Hard work paid off😎

May I ask why it is important to get full UMS? (Other than personal achievement/ satisfaction). The A* depends on the A2 papers, so an A in AS should suffice?

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.

Haha yeah I know.

I think 63 is about right. I think it will be somewhere between 60 - 65. If this markscheme is correct I've gotten somewhere between 85 and 95, can't believe it I thought i did mediocre at best. Means I've hopefully got full ums in both now. Hard work paid off😎

May I ask why it is important to get full UMS? (Other than personal achievement/ satisfaction). The A* depends on the A2 papers, so an A in AS should suffice?

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 reistance = - gradient (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

Teacher col, did you make a mistake on 3aii) 3) you seem to have used a rearrangment of V = E + Ir as opposed to E = V + Ir. (and you got a terminal p.d. of greater than the E.M.F. of 1.5V?

Couple quick things, what are A* conditions for A2 OCR Physics A?

Also what did you think about the coursework this year - same grade boundaries as last year?

Couple quick things, what are A* conditions for A2 OCR Physics A?

Also what did you think about the coursework this year - same grade boundaries as last year?

Don't want to be that guy but 3aii 3 is wrong, as V= E-Ir, not E +Ir.

Thank you very much for the rest of it though!

PS I might be wrong myself

For an A* you have to score 270/300+ at A2 and 480/600+ overall - which basically means 90% + at A2.

Coursework marks look pretty similar to last year - I'm moderating G483 right now. - so I'd expect grade boundaries to be pretty much the same.

See my previous reply

See my previous reply