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2016 AS physics new spec MARK SCHEME Watch

  • View Poll Results: what do you think you got out of 70
    Below 10
    3
    1.91%
    11-20
    4
    2.55%
    21-30
    12
    7.64%
    31-40
    26
    16.56%
    41-50
    51
    32.48%
    51-60
    39
    24.84%
    61-70
    22
    14.01%

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    How were you meant to find the resistance across the probe and what was it
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    (Original post by SaltandSugar)
    I got 2.5% lol but some people got 0.79% and 25%, apparently 25% worked if they substituted the numbers in but I'm not sure
    yeah i remember getting something similar to 0.8%, although a lot of other people i know got answers that were >20%.
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    (Original post by somerandomer654)
    Hey,

    For material D, wouldnt the cord just stretch to such an extent that the person hits the ground? So it must be C?
    I think it must be D, because C looks very similar to the graph for Copper. Also, you want the material to extend a lot as soon as it gets any tension in it to minimise the forces on the bungee jumper.

    And we don't know the height from which the person will be falling, so we can't rule out D.
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    What I thought for the stopping potential:

    The stopping potential applied across the metal will make the electrons do extra work reducing their KE to 0. To work this out you just had to take the EKmax you worked out from the previous question, and divide that by the charge of an electron (1.6x10^-19). This is because the stopping voltage is the extra work done per unit charge.
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    I think there will be some flexibility with the 6 marker, and that's why the question told you to justify your choice compared to the other materials. Hopefully they will award marks for any answer that makes sense as you can argue that B or C are better for the lift, and D or C for the bungee.
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    (Original post by somerandomer654)
    Hey,

    For material D, wouldnt the cord just stretch to such an extent that the person hits the ground? So it must be C?
    we don't know because we have no idea what it's max extension is...
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    how hard was the paper compared to the specimen papers?
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    (Original post by MathsMPhys)
    100% a photon has 0 mass thats..
    (Not quite, it has some mass but not exactly) 0; < 1×10−18 eV/c2 (It is a virtual mass to make the equations work)
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    What do you guys think A will be
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    Have compiled a near complete unofficial markscheme, missing 1 or 2 subquestions. Feel free to give improvements.
    Attached Files
  1. File Type: docx AS Physics Paper 1 new spec unofficial mark scheme.docx (52.7 KB, 265 views)
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    (Original post by Mysteryman95)
    Have compiled a near complete unofficial markscheme, missing 1 or 2 subquestions. Feel free to give improvements.
    Please post it in a PDF format since not everyone has microsoft word on their laptops...
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    (Original post by Al_YG)
    Please post it in a PDF format since not everyone has microsoft word on their laptops...
    PDF format
    Attached Images
  2. File Type: pdf AS Physics Paper 1 new spec unofficial mark scheme.pdf (430.3 KB, 449 views)
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    (Original post by Mysteryman95)
    PDF format
    Cheers mate
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    (Original post by MathsMPhys)
    Any predictions for grade boundaries??
    Not 100% sure. From what I thought of the exam and from what I have heard, I would guess at around standard percentages for grades. I.e. 80% for an A, 70% for a B etc. Wouldn't be too surprised if it deviated a bit though, so don't take my word for it.
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    Anyone have the paper? If so can you pm me
    Is the a thread for unit 2?

    Posted from TSR Mobile
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    (Original post by T17MA)
    Anyone have the paper? If so can you pm me
    Is the a thread for unit 2?

    Posted from TSR Mobile
    Slightly updated version of the unofficial markscheme I posted yesterday. As far as I know, accounts for all 70 marks. Though there may be some slight errors/little details how answers were reached. Closest thing I've seen to a complete one though.
    Attached Images
  3. File Type: pdf AS Physics Paper 1 new spec unofficial mark scheme (1) (1).pdf (448.5 KB, 364 views)
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    (Original post by Mysteryman95)
    Slightly updated version of the unofficial markscheme I posted yesterday. As far as I know, accounts for all 70 marks. Though there may be some slight errors/little details how answers were reached. Closest thing I've seen to a complete one though.
    Do you have the paper? Can you pm me it please

    Posted from TSR Mobile
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    Thanks for tht ms man aha pretty sure i got 60/70
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    [QUOTE=Mysteryman95;65249087]Slightly updated version of the unofficial markscheme I posted yesterday. As far as I know, accounts for all 70 marks. Though there may be some slight errors/little details how answers were reached. Closest thing I've seen to a complete one though.[/QUOTEbt

    Btw your answer to the probe resistance is incorrect...
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    Sorry gotta interject I've read through and there are some things I wanna go over:
    5B - now I'm not 100% about this esp. as there isn't the diagram but the current produced is proportional to the number of electrons produced so it would follow that increasing the intensity would increase the current. Perhaps I have misunderstood the question.
    5E - The intensity of the radiation has no effect on the kinetic energy of the photons. "What [Philipp] Lenard found was that the intensity of the incident light had no effect on the maximum kinetic energy of the photoelectrons." - http://physics.info/photoelectric/. The energy of the electrons is dependent on the frequency of the EM radiation only as increasing the intensity only increases the no. photons --> no. electrons emitted per second.
    4D - this one again I'm not entirely sure about and so would like an alternative explanation by all means. But increasing the internal resistance keeping EMF the same would surely decrease the current? EMF = IR + Ir so if you increase the r then for the EMF to remain the same either just I or IR would have to decrease. Now R is constant - just the total resistance of the various loads - which means the current would decrease. If the current decreases surely that means IR decreases so reading on voltmeter goes down?
    Electricity really isn't my strong point but these points are bugging me. Prove me wrong or right I don't care - just end my suffering!
 
 
 
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