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# AQA Physics PHYA5 - 28th June 2016 [Exam Discussion Thread] watch

1. Does anyone remember the questions and/or have the paper?

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2. (Original post by Manam786)
Depends on what option you did. I found last year's astro paper difficult

what was last year's astro grade boundary
25 for an A on the astro section. I've noticed that the astrophysics boundary is typically the highest. A* was 62-60 depending on option, whereas usually it's about 57-59 for an A* aside from astrophysics which always seems to be about 60
3. (Original post by particlestudent)
For Medical I'd say 53 for an A*
was medical really that hard
4. 55/75 would be an A, hopefully? i did astro btw
5. (Original post by Manam786)
was medical really that hard
I didn't find it hard at all but that's how low they've been in one period. June 2010 was 56 for an A* and that one was much easier than this one. Again June 2011 was 57 for an A* and it was very easy.

5 sub questions from June 2012 was repeated (closely) on this paper, and 4 from June 2013. The parts that weren't repeated were very weird unusual calculations

June 2012 boundaries were 53 for an A*
June 2013 boundaries were 54 for an A*

That's why I think it will be 53. June 2014 was 52 for an A* and it wasn't even that hard
6. For question 5 on section A, to find the pressure was the correct way using boyle's law? P2 = P1V2/V1?
7. (Original post by slb971)
For question 5 on section A, to find the pressure was the correct way using boyle's law? P2 = P1V2/V1?
No it was more complicated because that only works at constant temperature. They told you that the pressure, volume and temperature all changed.

So it was P1V1/T1 = P2V2/T2 - From pV = nRT your constant becomes nR so that's what you need to equate.
8. what did people put for where fusion and fission nuclei are placed on the graph?

i just said that fusion nuclei are found when A<50 and fission when A>50, because fusion nuclei are smaller and more stable and fission nuclei are large and unstable?
9. (Original post by -jordan-)
No it was more complicated because that only works at constant temperature. They told you that the pressure, volume and temperature all changed.

So it was P1V1/T1 = P2V2/T2 - From pV = nRT your constant becomes nR so that's what you need to equate.
Did you get a volume of 4140 or something or 5140. How did you find the exam and what do you expect the GB to be ?
10. (Original post by koolgurl14)
Did you get a volume of 4140 or something or 5140. How did you find the exam and what do you expect the GB to be ?
No, my volume was 0.00044, very similar order to the other volumes it gave.

Found it reasonable, think 58 for an A* on 2D (turning points)
11. Was the angle you needed to use in the electron diffraction question 42.5 degrees?
12. ohh **** when you found the distance of the object with 1/f=1/u+1/v it was a virtual image wasnt it>?
13. (Original post by PhyM23)
Was the angle you needed to use in the electron diffraction question 42.5 degrees?
I used 42 degrees, anything around that I assume they'll accept.
14. (Original post by lucabrasi98)
For the distance question in turning points I just did wavelength/2. What were you supposed to do?
that's right because the distance between two nodes is 1/2 wavelength which is the same as the distance between a maxima and minima
15. (Original post by Aprkrheiqkk)
that's right because the distance between two nodes is 1/2 wavelength which is the same as the distance between a maxima and minima
I think it's 1/4 since that's the distance from an antinode to a node
16. (Original post by a4567)
3)c) I got 0.0375m, as I got the wavelength as 0.15m and it asked for the distance between a maximum and a minimum recording, i.e. the distance between an antinode and a node, which is a quarter of a wavelength (0.15/4) = 0.0375m

Anyone else do it this way?
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
17. (Original post by Jay1421)
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
the distance between a maxima and minima is not the same as the distance between a node and antinode.
18. (Original post by PhyM23)
I think it's 1/4 since that's the distance from an antinode to a node
the node is not a minima, its the point with zero displacement
19. (Original post by Jay1421)
The wavelength wasn't 0.15m. They gave you 2.2GHz as the frequency (2.2*10^9), so 3*10^8 / 2.2*10^9 = wavelength - this gives 0.136m, which you then divide by 4 to get 0.341m between a node and antinode - your method was right, I think you must have miscalculated wavelength though.
I think it's half a wavelength because you were moving the reflector not the detector. So the position of the nodes changes due to more nodes being made. This caused the difference to be half a wavelength from the original node to a newly formed antinode. Not entirely sure though, it was a weird question.
20. (Original post by SirRaza97)
I think it's half a wavelength because you were moving the reflector not the detector. So the position of the nodes changes due to more nodes being made. This caused the difference to be half a wavelength from the original node to a newly formed antinode. Not entirely sure though, it was a weird question.
This is what I thought too.

This is 1/4 of a wavelength:

This is 1/2 a wavelength:
Attachment 557587557589
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