OCR AS Physics A Legacy G482 EWP Thursday 9th June 2016 Unofficial mark scheme

OCR AS Physics A Legacy G482 EWP Thursday 9th June 2016

Usual disclaimers. These are just my opinions - did this paper in 25 mins late last night so there may be errors
omissions and typos.

My students seemed very positive when they came out of the exam room and sure they had improved their mark.
I thought it was much more accessible than last years EWP. Well done examiners.

Q1 a) resistivity = resistance x cross sectional area / length (1)
b) i) rho = RA/L = 8.0 x pi x (2.0E-3/2)^2 / 150E-3 = 1.68E-4 ohm. metre (4)
ii) 1) Parallel so same PD so if current splits equally then resistances must be same
so resistance below X = 4.0 ohm so must be in middle (3)
2) Total resistance = (4?4 in parallel + 4 ) = 6.0 ohms
so I= V/R = 3.0/6.0 = 0.50A (2)
c) I = nAvq so v = 0.40 / (3.6E26 x pi x (2.0E-3/2)^2 x 1.6E-19) = 2.2E-3 ms-1 (3)
d) i) Copper has many more free charge carriers per unit volume so is better conductor so has lower resistivity (2)
ii) Copper : rho increase with temp due to increased vibration of ions and more collisions
Graphite: R decreases with temp due to bonds breaking and releasing more charge carriers (2)
Total: 17

Q2 a) i) calculate a point and draw a straight line through the origin (2)
ii) R: graph is straight line through origin so I is prop to V
LED: not straight line through origin so I is not prop to V (2)
b) i) When I = 30mA; PD across 67R = 2.0V so PD across LED = 3.0V so lights normally (3)
ii) 1) Q = 30E-3 C = 3.0E-2C (1)
2) W=VQ = 3.0 x 30E-3 = 9.0E-2J (2)
3) P=I^2R = (30E-3)^2 x 67 = 6.0E-2J (2)
iii) Total current = 30mA x 7 x 3 = 0.63A
so choose 1.0A fuse. (2)
c) Car headlights / torches / street lights: more efficient / produces less heat / cheaper to run (2)
Total: 12

Q3 a) energy is conserved (1)
b) i) terminal PD : energy converted per unit charge in circuit outside the cell. (or WTTE) (2)
ii)internal resistance = resistance inside supply=y which converts some electrical energy to heat (1)
c) i) Measure terminal PD and current. Adjust variable resistor to get several sets of measurements.
Plot graph of I (yaxis) x V (xaxis); gradient =-r
Straight line if r is constant. (5)
ii) 3.0R limts maximum current when variable resistor is zero so wires / cell dont get too hot. (1)
d) i) clockwise (1)
ii) 4.5 - 2.4 = I x (0.6+2.0+0.4) so I = 0.70A (2)
iii) 1) V = IR = 0.70 x 2.0 = 1.4v (1)
2) Vx = E - I r = 4.5 - 0.70 x 0.60 = 4.08v (2)
Total: 16

Q4 a) At certain angles, path difference is a whole number of wavelengths so all waves arrive in phase
so get constructive interference / reinforcement and get bright line
d sin theta = n lambda (4)
b) i) 1) C is a bit more than half way between o and E (1) (12.5 degrees if work it out)
2) 579E-9 = d sin 20 so d = 1.69E-6m (3)
ii) lowest energy = longest wavelength E (1)
iii) CDE are visible (1)
iv) Energy change = 3.64E-19J
so f= 3.64E-19/6.63E-34 = 5.49E14Hz
so lambda = c/f = 5.46E-7 = 546nm D (3)
v) sim: same frequencies
diff : emission bright lines / abs dark lines 2)
vi) f = c/lambda = 1.18E15 Hz
E = hf = 7.83E-19J
This is greater than wfe so Max KE = 7.83E-19 - 4.7E-19 = 3.13E-19J (3)
Total: 18

Q5 a) i) wavelength: distance between 2 successive points that are in phase
frequency: no of cycles per unit time
period: time for 1 complete cycle (3)
ii) Wave moves a distance lambda in a time equal to the period
so v = lambda / T = f x lambda (3)
b) i) 1) 2 nodes and 2 AN , Node at closed end and AN at open end (1)
2) lambda / 4 = 16.5 cm so lambda = 66cm = 0.66m
v = f x lambda = 512 x 0.66 = 338 ms-1 (3)
c) i) AN at each end and in middle (1)
ii) 1) 512 Hz (1)
2) AN at each end ; N in middle (1)
iii) Length of pipe = lambda so must have AN - N - AN - N - AN (2)
Total : 11

Q6 a) i) lambda = h/mv
lambda: wavelength associated with particle
h: Planks constant
m: mass of particle
v: velocity of particle (3)
ii) Electrons show interference pattern when fired through graphite crystal. (2)
iii) 1) eV = 1/2 mv^2 so v = 1.33E8 ms-1 (3)
2) lambda = h/mv = 5.47E-12m (2)
3) this is 10^5 times shorter than visible (5E-7m) (2)
b) Photoelectric effect
If shine light on surface of a metal electrons may be emitted
Max energy of electrons depends on freq of light
Threshold frequency
No of electrons depends on brightness of light
etc (4)

No - I dont know what the grade boundaries will be.
It seems a straightforward paper to me so they wont be as low as last year.

Get revising for Newtonian World
Good Luck
Col

Good Luck

Reply 1

seanthubb

Thanks for this!

Think I might have managed 70/100 or so, so not as much as I wanted but still ok I hope?

(edited 7 years ago)

Reply 2

Revision King98

Thanks for this :smile:

will 75 definitely be an A ?

(edited 7 years ago)

Thanks a lot :smile:

Original post by Revision King98

Thanks for this :smile:

will 75 definitely be an A ?

Last year I think 62 was an A and because this year it was just the year 13 that didnt do well in that paper re-sat and majoirty of the people I spoke to found it weird so I think it may either slightly lower by one or two marks or higher by one or two

I got 65 :frown:

not what I hoped I would of got..

Reply 5

seanthubb

Original post by FemaleBo55

not what I hoped I would of got..

It's still a good score so don't feel down :smile:

You're getting at least a B, and likely an A!

Reply 6

Revision King98

Original post by FemaleBo55

not what I hoped I would of got..

Hopefully still an A, just make sure u get the marks u lost back in 484 and 485

Reply 7

Oalasow

Thank you so much. I'm very happy with my mark considering how unsure I was of my answers after the exam. I've at least got 75 marks and at most 82. Hopefully I can perform well in G484 and G485

Reply 8

FemaleBo55

Original post by Revision King98

Hopefully still an A, just make sure u get the marks u lost back in 484 and 485

I'm not doing A2 physics LOL, it was just an AS resit 🙈
Hopefully, the thing was that in past papers I kept getting 73-78 :// and now it's like really?!!

Posted from TSR Mobile

Reply 9

username1969391

Original post by FemaleBo55

I'm not doing A2 physics LOL, it was just an AS resit 🙈
Hopefully, the thing was that in past papers I kept getting 73-78 :// and now it's like really?!!

Posted from TSR Mobile

Just out of curiosity; why would you want to retake it if you're not doing it at A2? AS levels don't really count for anything do they?

Reply 10

Revision King98

Original post by FemaleBo55

I'm not doing A2 physics LOL, it was just an AS resit 🙈
Hopefully, the thing was that in past papers I kept getting 73-78 :// and now it's like really?!!

Posted from TSR Mobile

Why are u retaking it then , is there any point?

Reply 11

shahryair

can we get the question paper uploaded as well?

Reply 12

Brokensteps

Original post by teachercol

iii) Total current = 30mA x 7 x 3 = 0.63A
so choose 1.0A fuse. (2)

Hey, could you please Check this question? I remember it saying that there was a fuse per Number, so, you timing it by 3 at the end there seems wrong.

I could be wrong, but I remember seeing that and changing my answer in the exam