You are Here: Home

# G485 June 2012 Paper and unofficial mark scheme Tweet

Physics exam discussion - share revision tips in preparation for GCSE, A Level and other physics exams and discuss how they went afterwards.

Announcements Posted on
1. Re: G485 June 2012 Paper and unofficial mark scheme
I think it'll be 1/3 actually, lose one for incorrect denominator, and lose one for wrong answer
2. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by zabuzar)
i got 1320Mpc
there was a decimal point and a digit after but i rounded to 3s.f.
Same

480nm is a 22% expansion of the origianl, which gives 0.22c as speed of recession, dividing by 50 gives 1320Mpc
3. Re: G485 June 2012 Paper and unofficial mark scheme
i managed to get something ^ -4 for the very last question - knew that couldn't really be right but didn't have the time to go over it again, just hoping the Newtonian World paper is nice!
4. Re: G485 June 2012 Paper and unofficial mark scheme
Am I the only one who's wondering how the bloody hell do you work out the wavelength of gamma rays :/??
5. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by Rmaoan)
Am I the only one who's wondering how the bloody hell do you work out the wavelength of gamma rays :/??
well what i did and i think others seem to agree with is using E=mc^2 to calculate the mass lost from an electron and positron annihilating. so total mass lost is 2x (9.11x10^-31) and from this we can work out energy. once we have energy we can work out from the formulae E=hc/lambda or E=hf then work out lambda seperately. remember that there is 2 waves so essentially ou half the energy again leaving the mass of an electron to produce 1 gamma photon.

i looked at it initially when doing the paper and figured it might be loss in mass but skipped it and came back because i was unsure. i seems like a correct method because the KE that the electron had was to be considered negligable so i cant see another source of energy for the photons.
6. Re: G485 June 2012 Paper and unofficial mark scheme
i think i've got about 80 - 83 , what do you reckon that'll be?
7. Re: G485 June 2012 Paper and unofficial mark scheme
Thanks for uploading ... by looking though the paper I reckon I got about 77 to 82... which pretty much has to be an A right, I really need it !

What do other people think they got ?
8. Thanks for the mark scheme,
Pretty my calculations are right just may drop some marks in the qualitative questions :/

This was posted from The Student Room's iPhone/iPad App
9. Anyone got 0.112kg for the mass of uranium?
I rounded to 3sf

This was posted from The Student Room's iPhone/iPad App
10. Re: G485 June 2012 Paper and unofficial mark scheme
for the x ray question would you get the mark for saying they travel at c
11. Re: G485 June 2012 Paper and unofficial mark scheme
I think I probably got an A, *fingers crossed* for 90%
I checked the june 2011 boundaries and 90% was 71. I think the grade boundaries will be similar as not may people will have bothered to memorise the MRI section, as if has already come up. Plus not many people will have gotten full marks on the universe question. I'm definitely hopeful for low boundaries
12. Nice, i think i can get more than 90% in UMS too,
I think the boundary will be lower than june 11 as that one was fairly easy and some people found the calculations in this paper hard, also use of nuclear reactor and evolution of universe are the first time coming up in new syllabus, so pretty sure the grade boundary will be low!

This was posted from The Student Room's iPhone/iPad App
13. Re: G485 June 2012 Paper and unofficial mark scheme
Shouldn't question 10 a) F=GMm/r^2 be Mm: (10^11 x 10^30)^2and not (10^11 x 10^30)x2?
14. Re: G485 June 2012 Paper and unofficial mark scheme
Also in dispair about people's reply to the red shift question, maybe if they had an extra question before hand which asked 'what happens to the wavelength of em radiation as it undergoes red shift' people may have put 2 and 2 together.

I see no one has answered question 3 yet so this was my attempt:

3ai) Frequency = 1/Period
Frequency = 1/10x^-3
Frequency = 100 Hz

3aii) Magnetic flux linkage = B A N
B = Magnetic flux linkage / A N
B = 2x10^-2/1.6x10^-3 x 400
B = 0.03125 (0.0313 to 3 s.f.)

3aiii) For this one i started to find the equation of the line to differentiate it but then realised it said 'using figure 3.1'. For this question you needed to make a gradient triangle between 2 seconds and 3 seconds. The magnetic flux density changes from 0.6x10^-2 to -0.6x10^-2 in 1x10^-3 seconds:

Emf = Change in magnetic flux density / Change in time
Emf = 1.2x10^-2/1x10^-3
Emf = 12V

3b) P = V^2/R
P = 12^2/150
P = 0.96W

Think there were a few difficult questions, lots of essays which was a pain. Was happy with my answer to most questions and will post more later
15. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by welshy93)
Shouldn't question 10 a) F=GMm/r^2 be Mm: (10^11 x 10^30)^2and not (10^11 x 10^30)x2?
I think you're right, its F= GM^2/r^2 when the masses are the same
16. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by welshy93)
Shouldn't question 10 a) F=GMm/r^2 be Mm: (10^11 x 10^30)^2and not (10^11 x 10^30)x2?
Your right sorry, that was what I meant to write, I'll change it.
17. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by Hamer)
Also in dispair about people's reply to the red shift question, maybe if they had an extra question before hand which asked 'what happens to the wavelength of em radiation as it undergoes red shift' people may have put 2 and 2 together.

I see no one has answered question 3 yet so this was my attempt:

3ai) Frequency = 1/Period
Frequency = 1/10x^-3
Frequency = 100 Hz

3aii) Magnetic flux linkage = B A N
B = Magnetic flux linkage / A N
B = 2x10^-2/1.6x10^-3 x 400
B = 0.03125 (0.0313 to 3 s.f.)

3aiii) For this one i started to find the equation of the line to differentiate it but then realised it said 'using figure 3.1'. For this question you needed to make a gradient triangle between 2 seconds and 3 seconds. The magnetic flux density changes from 0.6x10^-2 to -0.6x10^-2 in 1x10^-3 seconds:

Emf = Change in magnetic flux density / Change in time
Emf = 1.2x10^-2/1x10^-3
Emf = 12V

3b) P = V^2/R
P = 12^2/150
P = 0.96W

Think there were a few difficult questions, lots of essays which was a pain. Was happy with my answer to most questions and will post more later
Thanks
18. Re: G485 June 2012 Paper and unofficial mark scheme
(Original post by chris669)
for the x ray question would you get the mark for saying they travel at c
no you wouldn't sorry!
19. Re: G485 June 2012 Paper and unofficial mark scheme
Might as well do another, here's my go at question 5 (only like the maths ones)

5a) 144 neutrons in the plutionium nucleus - 2 from alpha particle = 142 neutrons

5bi) From the equation sheet:
1 eV = 1.6x10^-19 J
so 1 MeV = 1.6x10^-13J
5.6x1.6x10^-13 = 8.96x10^-13J

5bii) Ke = 1/2 m v^2
v^2 = 2ke/m
v^2 = 2x8.96x10^-13/6.65x10^-27
v^2 = 2.69x10^14
v =1.64x10^7 ms-1

5ci) 62/8.96x10^-13 = 6.92x10^13

cii) (going to call lambda Y)
Y t1/2 = ln2
Y = ln2/(88x3.16x10^7)
Y = 2.49x10^-10

ciii)
1) A=YN so N = A/Y
N = 6.92x10^13/2.49x10^-10
N =2.78x10^23

2) 2.78x10^23 Atoms/6.02x10^23 (avagadro) = 0.46... moles
0.46... x 0.24 = 0.111 kg
(answers between 0.11 and 0.12 will probably be allowed)
20. Re: G485 June 2012 Paper and unofficial mark scheme
anyone get something like this? I think the lines have to be more curvy and going backwards, but i don't think it matters that much.

http://imgur.com/7OZCV
Useful resources

## Articles:

Study Help rules and posting guidelinesStudy Help Member of the Month nominations

## Groups associated with this forum:

View associated groups