Edexcel A2 Paper 2 Physics Unofficial Markscheme 21/06/17

Hey all. Got a copy of the paper from my school. Gonna post my answers to them below, but they are definitely not all right. If you think i'm wrong somewhere, point it out please!
Key: xEy = x*10^y

MCQ
1) C - Polarisation Same
2) B - 4cm same
3) C - 2.5 Hz Not sure may have put 0.4
4) D - Both fields act on all particles same
5) D - 838000/545 same
6) B - main sequence same
7) D - 8L same
8) D - (656.3-654.9)/654.9 * 3E8 same
9) C - 1.5 = magnification same
10) D - Light with higher frequency I got this one wrong

11) P=E/t so E=Pt = 2600*60 = 156KJ in one minute
q=mcT = 0.89kg*450K/kg/K*(215-18)=78900J
So plate can reach working temp in less than one minute yes i got the same

12) a) pretty sure I got this wrong
b) for A: F/mr = w^2 so Method looks good, i think i did an arithmetic mistake though

13) An astronomical object of known luminosity yep
b) The angle of parallax becomes too small to measure beyond a certain distance because the radius of earths orbit becomes too small in comparison to the distance of the star/object yes and i said uncertainty in distance becomes very large as angle is very small and u is large
c) use z= dWavelength/wavelength * c = v and then v*t=d with t=age of the universe Yep sounds good, think i messed up a little here talked about received intensity on earth and find distance using luminosity formulaes, probably lost 2

14) fission from heavier elements to lighter ones releasing binding energy of releases neutrons, fusion from light to heavier (up to Fe56) energy released is the mass deficit (E=mc^2) yeah
b) Hot, high pressure, abundance of fuel, explain one yes

15) light source emits light in every plane, filters only allow light travelling parallel to plane of polarization to pass. When both filters are in the same plane (whole multiple of pi) all light that is polarized by the first filter is passed through the second. As you rotate the second through an odd multiple of pi/2, less light gets through until the planes are exactly pi/2 radians apart, at which the planes of polarization are orthogonal to one another and hence there is no intensity measured. looks good to me

16) (8E-6*1E3*9.81)-(9.81*0.0059)=0.02N same
b) mg=upthrust so pVg=mg so 9.81pV=0.0059*9.81
so pV=0.0059 same
so 0.0059/1E3 = 5.9E-6 m^3
c) PVi=PVf so P=(1.01E5*8E-6)/6E-6 = 1.35E5 Pa yes

17) super standard photoelectric effect question, loads of past paper markschemes available. yes not too bad there
b) plot a graph of ln(q)/ln(t) and it would be a linear relationship (y=mx+c) if the relationship (and hence curve) is exponential i said plot a graph of ln q against t, gradient is negative and straight line and y intercept ln q0, you dont do lnt

18) Pc = 44.8D
Pl = 18.98D
44.8/(44.8+18.98) *100 = 70.3% of power from cornea same, cheeky 4 marks there
b) distance = 4.8cm yeah, wasnt sure on this one but got same so good
c) angle: 10.8 degrees yep
d) light moves more slowly in water so is refracted more by eye, goggles fix this etc etc yes

19) (40)Ca(20) + (0)B(-1) yeah
b) 300E-3 * 8.1E21 * 0.012/100 = 2.92E17 nuclei same
c) x = decay constant!!
x=ln2/(1.25E9*86400*365)=1.76E-17, A=xN = 1.76E-17*2.92E17 = 5.13Bq yeah
d)176-(15*10)=26
26 counts in 10 minutes so 0.043 per second
0.043/5.13 = 0.84% so not efficient same
e) Not all beta particles emit directly into the tube so some will be lost to different direction and absorption by air molecules etc wrote the same
f) The activity of KCl will not drop notable due to it's long halflife so more suitible
something along those lines

20) f=~40Hz and v=lambda*f
so wavelength = C/40 = 7.5E6m of course im an idiot i messed this up
b) 4000*2.35E-21=9.4E-18, 0.001*proton diameter = 1E-19 so statement doesn't apply as almost factor of 100 out. i think i did this one backwaards where i used the value and then compared with graph but said not supported.
c) at X/2 path diff destructive interference happens so 0 amplitude
at point where path is not (2x+1)/2 off there will be amplitude so points with amplitude are changes in length (from 0 amplitude being equilibrium) lost 2 marks here i reckon, i swapped node and antinode around and said constructive occurs instead of destructive
d) using X/2 is more sensitive because observing a change in amplitude from zero amplitude is easier than from maximum amplitude[messed this up, but yours sounds right

you think 2 wave cycles will be alright?

also did time period have to be in days for the black holes?

Done in bold

I agree with pretty much everything apart from 2 questions.
For the refraction one (18d), I said that the speed of light in water is similar to speed of light in cornea, therefore very little refraction will occur, so the light rays will not be focused onto cornea. Speed of light in goggles is higher than in cornea, so light will refract more in goggles which corrects this. I think question was talking about lack of refraction.
Also, for 20b, did you take the amplitude of the tallest wave to be strain. If so, shouldn't you have got something like 1.2x10^-21 x 4000 = 4.8x10^-18. Where did 2.35x10^-21 come from?

Done in bold

I agree with pretty much everything apart from 2 questions.
For the refraction one (18d), I said that the speed of light in water is similar to speed of light in cornea, therefore very little refraction will occur, so the light rays will not be focused onto retina. Speed of light in goggles is higher than in cornea, so light will refract more in goggles which corrects this. I think question was talking about lack of refraction.
Also, for 20b, did you take the amplitude of the tallest wave to be strain. If so, shouldn't you have got something like 1.2x10^-21 x 4000 = 4.8x10^-18. Where did 2.35x10^-21 come from?

Guys wasnt the question that asked how standard candles are used to find distances this:
http://www.astro.ex.ac.uk/people/hatchell/rinr/candles.pdf
similar to this kind of thing,why is everyone talking about hubbles law?

What was the frequency question again? (the MC one, question 3?)

Guys wasnt the question that asked how standard candles are used to find distances this:
http://www.astro.ex.ac.uk/people/hatchell/rinr/candles.pdf
similar to this kind of thing,why is everyone talking about hubbles law?

Also, did you have to talk about nodes and antinodes on the last question? I don't think it's relevant on that question because the standing isn't what causes the change in detection (because the superposition occurs between the two waves which are perpendicular. I thought it was because the change in length leads to a path difference so the phase difference changes so the 2 beams of light are not coherent anymore (no longer constant phase difference) so complete destructive interference doesn't occur at an odd no. ½ wavelengths.

24 in a minute, so time period is 2.5s and f=0.4 right?
did i mess this one up?

RTQ - standard candles only for short distances mate. Very easy to get caught out on that what, originally I thought they made a mistake in the question.

24 in a minute, so time period is 2.5s and f=0.4 right?
did i mess this one up?

The question was "Describe how distances too large for the use of standard candles can be determined" Hubbles law is right I think, which I think is what I did but I didn't explicitly mention it so probably lost one

Oh well i just wrote what was in the book, parallax was for short but why is standard candles for short? if you know luminisoty and can measure intensity it doesnt have to be short distances? or is it some distances are so great that the intensity on earth is so low we cant use it?

I put 0.4 in the exam, but when I got out all my class told me 2.5, I still feel like it's 0.4 cos:
24 in a minute => T = 2.5 so f=0.4