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AQA Physics PHYA5 - Thursday 18th June 2015 [Exam Discussion Thread] Watch

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    (Original post by Puzzled_Logician)
    For anyone else who did the applied module; how do you think it compared difficulty wise to previous years?

    Personally I found it harder than some of the previous years; but mainly due to mostly being word based questions (there didn't seem to be much maths this year).

    I am better at maths than explaining stuff; but it could've been much worse I guess.
    more difficult indeed, slightly odd question. Rest of it was fine thankfully
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    you know the one where you had to work out temperature and the radius and then tick the box for what type of star it was. What is the right answer for that radius?
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    Possible grade boundaries for astro?
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    (Original post by simonli2575)
    Do you think I might get marks for 0.05m and 1.0m for their ranges?

    And what's that about the shield becoming radioactive?
    It's 0.06m and 1m in the text book, so yeah.
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    (Original post by Lau14)
    Probably for the best, nothing to be done about it until results day now anyway!
    Indeed. Good luck
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    For nuclei that large using the atomic mass unit is more accurate.
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    (Original post by StarvingAutist)
    Indeed. Good luck
    And you!
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    Anyone who did Medical Physics, quote me, how did you find the paper?
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    Any thoughts on grade boundaries then, folks? I don't think either the Thermal or Astro were much harder or easier than last years (astro was probably a bit harder?). Last years boundaries were really low - something like 58/75 for 120 UMS.

    I really hope I get full UMS in this one to make up for my terrible unit 4. Not sure if I can still rescue my A grade even if I got full UMS.
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    (Original post by Doomlar)
    My answers for the questions that didn't have a "show that the answer is this" were:

    Thermal and Nuclear
    Range of \alpha = 0.1m
    Range of \beta = 2.0m
    Density = 1.63 \times 10^{17}  kg  m^{-3}
    Number of nuclei = 2.35 \times 10^{17}
    Energy used = 47000J
    \left(C_{rms}\right)^{2} = 2.07 \times 10^{7} m^{2} s^{-2}
    Temperature = 319K

    Astrophysics
    Maximum distance = 5.36 \times 10^{11} m
    Diameter = 15m (unsure about this)
    In the 6 mark essay my period was 4 days, and my velocity was 109800 ms-1
    Magnitude = -0.01 (definitely derped on this; 1sf and I missed the minus lmao)
    Temperature = 16100K
    White dwarf radius = 9.37 \times 10^{6} m
    Black hole radius = 2.95 \times 10^{13} m
    Age of the universe = 4.96 \times 10^{17} s
    I got 48000J for energy used was it Q=mcdeltaT+ml?
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    (Original post by Disney0702)
    But the neutron number changes so wouldn't the nucleon number be bigger or smaller so the size of the nucleus should be different.

    So shouldn't the dispersion be different?
    I think you are confusing scattering with diffraction. It is true that in diffraction mass would make a difference as the radius would change but with scattering of two charged particles only the charge will make a difference.
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    I think anywhere between last years and 2013 grade boundaries I belive, ie 52-56 for an A*
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    But for alphas hitting the nucleus, and being deflected back on themselves surely the size of the nucleus matters?

    (Original post by Jamesbe1993)
    I think you are confusing scattering with diffraction. It is true that in diffraction mass would make a difference as the radius would change but with scattering of two charged particles only the charge will make a difference.
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    What was the density question? I did (51*1.67*10^-27)/(4/3 *pi*r^3) but I can't remember what r was.

    Also, going off of the past grade boundary trends, A* will be 57 and full UMS will be 62
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    (Original post by JacquesCustard)
    What was the density question? I did (51*1.67*10^-27)/(4/3 *pi*r^3) but I can't remember what r was.

    Also, going off of the past grade boundary trends, A* will be 57 and full UMS will be 62
    Was it 1.4x1017? I think r was about 5fm...

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    (Original post by _Caz_)
    Was it 1.4x1017? I think r was about 5fm...

    Posted from TSR Mobile
    Yeah, 5fm or so - think I got 1.6 or 1.4 *10^-17

    How did you guys do the number of Neptunium nuclei? N=N_0e^(-lambda*t)?
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    (Original post by sophiebeth100)
    Anyone who did Medical Physics, quote me, how did you find the paper?
    Hey i did the medical paper! I found it alright, except the ear question! How did you find it?
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    (Original post by PoojaRathod)
    But for alphas hitting the nucleus, and being deflected back on themselves surely the size of the nucleus matters?
    Under the assumption of equally distributed charge the electrostatic force on the alpha particle will not change even for an increase in radius of the gold nuclei due to an increase in mass. The charge density of the nucleus will change but the strength of the field produced by it will not. Ergo the scattering effect will remain the same.
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    (Original post by BolognaBoy)
    Yeah, 5fm or so - think I got 1.6 or 1.4 *10^-17

    How did you guys do the number of Neptunium nuclei? N=N_0e^(-lambda*t)?
    used A=lambda x N ?
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    (Original post by chizz1889)
    Yeah there was, the exam code was PHYA5R

    I was expecting it to be like that (I heard about the van getting stolen etc.) so I checked and there was no R, I was looking for it !
 
 
 
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