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AQA Physics PHYA4 - Thursday 11th June 2015 [Exam Discussion Thread] Watch

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    (Original post by CD223)
    Here is PotterPhysics' answer to the first question: The answer is 800 (C)

    http://www.thestudentroom.co.uk/show...9#post56268219

    And the second one comes out to be -13 (A) as above


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    My bad; I'll remove the error to avoid confusion. :yy:
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    (Original post by Fvthoms)
    My bad; I'll remove the error to avoid confusion. :yy:
    No problem! If they generally state an effect on the radius/diameter without stating the effect on the mass, then both the numerator AND denominator change when working out g or V


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    Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

    The questions are 2, 6, 10, 16, 24.

    http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
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    (Original post by aprocrastinator)
    Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

    The questions are 2, 6, 10, 16, 24.

    http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
    Question 2:
    Name:  ImageUploadedByStudent Room1432810118.018798.jpg
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    (Original post by aprocrastinator)
    Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

    The questions are 2, 6, 10, 16, 24.

    http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
    Question 6:
    Name:  ImageUploadedByStudent Room1432810222.548919.jpg
Views: 38
Size:  79.0 KB


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    (Original post by aprocrastinator)
    Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

    The questions are 2, 6, 10, 16, 24.

    http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
    Question 10:
    The answer is B.

    This is because resonance curves show that a system reaches maximum amplitude (a sharp resonance peak) when the driving frequency equals the natural frequency if the damping is light.

    Name:  ImageUploadedByStudent Room1432810373.619885.jpg
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    When a system is damped more and more, the amplitude at resonance is reduced and moreover, the resonant frequency is slightly reduced.


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    (Original post by aprocrastinator)
    Hi! I just did the June 2012 paper for Unit 4 and I don't understand how to do several of the questions in section A so I was hoping someone could explain them to me?

    The questions are 2, 6, 10, 16, 24.

    http://filestore.aqa.org.uk/subjects...1-QP-JUN12.PDF
    Question 16:
    The answer is C.

    At the point of closest approach, the velocity is zero, not the acceleration.

    If the velocity is zero, the kinetic energy is zero.

    The closer the particle gets to the nucleus, the greater its potential energy becomes, ruling out A, B and D.

    Question 24:
    The answer is C.

    When the current is at a maximum and suddenly switched off, there is the greatest change in flux linkage through the coil out of all scenarios given.

    A greater change of magnetic flux linkage produces a greater emf, and as the circuit is complete, the greatest induced current reading is observed.

    Hope that helps


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    (Original post by CD223)
    Hahaha, I'm flattered, but no. I got rejected by Oxford after my interview in December. Being totally honest, I'm happy about it because I don't think the atmosphere would have been healthy for me - I need a place I feel I can chill out!

    I'm (hopefully) going to Bath to do Computer Science in September. How about you?


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    Well Oxford have missed out! Sheffield hopefully - Aerospace Engineering
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    Anyone care to explain why sharp points produce intense fields?

    Is it something to do with:
    

E = \dfrac{\Delta V}{\Delta d}

    Where there is a large potential across a small distance?


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    (Original post by EmiratesCaptain)
    Well Oxford have missed out! Sheffield hopefully - Aerospace Engineering
    That's really nice of you, thank you
    Good luck! What grades do you need? 😁


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    (Original post by CD223)
    That's really nice of you, thank you
    Good luck! What grades do you need? 😁


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    No problem AAB - The B will probably be in physics.. I find it quite hard :/
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    (Original post by EmiratesCaptain)
    No problem AAB - The B will probably be in physics.. I find it quite hard :/
    Physics is my hardest subject too - which is disheartening because it's one of my more interesting subjects!


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    (Original post by CD223)
    Physics is my hardest subject too - which is disheartening because it's one of my more interesting subjects!


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    Haha agreed! I think unit 5 is harder than 4 though
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    Hey guys. Can anyone help me with part (c)ii of this question? I just don't understand the mark scheme. They seem to be using eV for a potential difference? I thought eV was a measurement of energy...

    Thanks. Question is attached. This is from June 2012.
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    Yeah I was stuck on that one too, You can use work done= force x distance and use the energy they gave to find the force

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    Does anyone know why the particle would be positive?

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    (Original post by huniibehi)
    Does anyone know why the particle would be positive?

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    Electric field line always act from positive to negative, so if the electric field in upwards the particle must have a negative charge. I think this is why, but not 100% sure.


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    (Original post by bwr19)
    Hey guys. Can anyone help me with part (c)ii of this question? I just don't understand the mark scheme. They seem to be using eV for a potential difference? I thought eV was a measurement of energy...

    Thanks. Question is attached. This is from June 2012.
    Can't you do:

    

F = \dfrac{W}{s} = \dfrac{QV}{d}

    Where:
    

Q = 1.6 \times 10^{-19} C

V = 4500 V

d = 180 \times 10^{-3}m

    Then the minimum distance equals

    

\dfrac{\text{Electron volts} \times 1.6 \times 10^{-19}}{F}

    ?


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    (Original post by huniibehi)
    Does anyone know why the particle would be positive?

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    Is there a diagram with that question? It's just I'm unsure if you could answer it without one as all it says is "vertical" field?


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    I know that's what I thought! But no diagram so I don't get why it would be positive

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