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

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    Can some people help me with Q8, Q12 and Q24 on June 2013?

    I got answers of C, B, B which is wrong.


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    (Original post by CD223)
    Can some people help me with Q8, Q12 and Q24 on June 2013?

    I got answers of C, B, B which is wrong.


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    Question 8 just uses one ratio of 1:2 (for m and f) so I would use the third type of informal approach in my last answer which is proportions. The other approaches would work but are unnecessary in this 1:something format; I'll show both anyway though. It's hard to distinguish between when it is best to use them but it comes down to how intuitive the question is.
    .Name:  SHM Question.png
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    LHS:
    First two lines are the given ratios rearranged.
    Second two lines show the velocity ratio is equal to the frequency ratio (constant cancels since ratios used) based on the SHM maximum velocity equation.
    Last lines substitute last rearranged equations into the kinetic energy equation.

    The last stage could have just substituted the direct m2/m1 and v2/v1 ratios E=mv^2 (not constant since ratios) to give 8 but wanted to show it in full.

    RHS is intuitive.
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    (Original post by CD223)
    Can some people help me with Q8, Q12 and Q24 on June 2013?

    I got answers of C, B, B which is wrong.


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    This uses the gravitational field equation with two ratios.

    Name:  Gravitation Fields Question 2.png
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    First two equations are the given ratios.
    Second two equations rearrange the gravitational field equation for r.
    Last two equations substitute the given ratios into the rearranged equation, leaving out the constant since ratios are substituted in (cancel if you like).
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    (Original post by JayCS)
    Question 8 just uses one ratio of 1:2 (for m and f) so I would use the third type of informal approach in my last answer which is proportions. The other approaches would work but are unnecessary in this 1:something format; I'll show both anyway though. It's hard to distinguish between when it is best to use them but it comes down to how intuitive the question is.
    .Name:  SHM Question.png
Views: 111
Size:  3.4 KB
    LHS:
    First two lines are the given ratios rearranged.
    Second two lines show the velocity ratio is equal to the frequency ratio (constant cancels since ratios used) based on the SHM maximum velocity equation.
    Last lines substitute last rearranged equations into the kinetic energy equation.

    The last stage could have just substituted the direct m2/m1 and v2/v1 ratios E=mv^2 (not constant since ratios) to give 8 but wanted to show it in full.

    RHS is intuitive.
    Thank you


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    (Original post by JayCS)
    This uses the gravitational field equation with two ratios.

    Name:  Gravitation Fields Question 2.png
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    First two equations are the given ratios.
    Second two equations rearrange the gravitational field equation for r.
    Last two equations substitute the given ratios into the rearranged equation, leaving out the constant since ratios are substituted in (cancel if you like).
    Again - thanks - do you know how to do Q24?


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    (Original post by CD223)
    Again - thanks - do you know how to do Q24?


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    No problem .

    Do you mean a different question to Q24? You said you got B for it and that's correct.
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    (Original post by JayCS)
    No problem .

    Do you mean a different question to Q24? You said you got B for it and that's correct.
    Lol I'm so stupid... I meant I got C and the answer was B

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    Does the equation:

    

(n+1) \times {T_{short}} = (n) \times {T_{long}}

    Always apply for two pendulums where n= the number of oscillations before they are next in phase?

    If so, could I have an explanation of why it works please?

    Brain not functioning today!


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    (Original post by CD223)
    Lol I'm so stupid... I meant I got C and the answer was B

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    I got B because I knew the formulae was Blv and remembered the derivation of it used the same direction (where the rod is moving in the length's direction).
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    (Original post by CD223)
    Lol I'm so stupid... I meant I got C and the answer was B

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    For a start it must be one of the rows with horizontal and perpendicular in, because otherwise it won't cut any field lines and no emf would be induced.
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    (Original post by Lau14)
    For a start it must be one of the rows with horizontal and perpendicular in, because otherwise it won't cut any field lines and no emf would be induced.
    (Original post by JayCS)
    I got B because I knew the formulae was Blv and remembered the derivation of it used the same direction (where the rod is moving in the length's direction).
    That's true. Thanks guys!

    I think I put vertical because I tried to use the right hand rule or something. Stupid mistake.


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    (Original post by CD223)
    That's true. Thanks guys!

    I think I put vertical because I tried to use the right hand rule or something. Stupid mistake.


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    Some form of Fleming's rule would probably have been my first response too to be honest!
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    (Original post by Lau14)
    Some form of Fleming's rule would probably have been my first response too to be honest!
    I used the right hand rule which seemed to not work out?

    Unless I'm mistaken.

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    (Original post by CD223)
    I used the right hand rule which seemed to not work out?

    Unless I'm mistaken.

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    Hi CD223,

    The problem with this question is that it can be quite difficult to visualise at first.

    You don't need to use any of Fleming's hand rules actually, if you imagine the wire moving upwards (bare in mind the diagram is a side view) it doesn't cut through any lines of flux therefore no emf is induced. The wire can only cut through the flux lines if it moves perpendicular to them.

    Hope that helps
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    (Original post by sam_97)
    Hi CD223,

    The problem with this question is that it can be quite difficult to visualise at first.

    You don't need to use any of Fleming's hand rules actually, if you imagine the wire moving upwards (bare in mind the diagram is a side view) it doesn't cut through any lines of flux therefore no emf is induced. The wire can only cut through the flux lines if it moves perpendicular to them.

    Hope that helps
    Ah I see. Thank you. The diagram threw me. I knew the theory of it being perpendicular, but couldn't visualise it.


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    (Original post by CD223)
    Ah I see. Thank you. The diagram threw me. I knew the theory of it being perpendicular, but couldn't visualise it.


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    No worries! It is pretty confusing, I can remember doing that one a while ago and even my teacher couldn't visualise it at first!
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    (Original post by sam_97)
    No worries! It is pretty confusing, I can remember doing that one a while ago and even my teacher couldn't visualise it at first!
    I think if the diagram was clearer then most people would get the right answer!


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    Does anyone has the specimen paper?
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    (Original post by you-only-live-once)
    Does anyone has the specimen paper?
    I have seen some specimen paper on OCR website but i don't know if they are the ones you are looking for


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    (Original post by you-only-live-once)
    Does anyone has the specimen paper?
    My teacher has a paper copy - I'll try and email him asking for a copy! If I'm successful I'll post it here


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