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EDEXCEL A2 Physics EXAM Unit 4 Physics On The Move 20th June 2016 (NOT I-A-L) Watch

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    (Original post by jtebbbs)
    Resolve vertically: F*sin22 = 80 * 9.81 gives F = 846

    Set horizontal component equal to mv^2/r, rearrange to get

    r = (80 * 9^2) / (846 * cos22) which gives you 20.4m
    you got your sin and cosine mixed up, but everything else is right.
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    It's not gonna be 70+ for A*, right? 66?
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    I got 0.14m for the radius??? Which I thought was right because cycles can't tilt like 20m
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    (Original post by Z1228)
    I got 0.14m for the radius??? Which I thought was right because cycles can't tilt like 20m
    It's the radius of the bend, i.e. the radius of the circle the bike covers.
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    has anyone seen an unofficial mark scheme?
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    I think the paper was pretty good overall, as normal with physics doing further mechanics modules in maths always helps. Any other ideas on why the H+ ion left no track after the photon interaction? I said it was because it wasn't moving, something about momentum
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    (Original post by guidoblee)
    I think the paper was pretty good overall, as normal with physics doing further mechanics modules in maths always helps. Any other ideas on why the H+ ion left no track after the photon interaction? I said it was because it wasn't moving, something about momentum
    Same. What did you put for the bit about the spokes and the speedometer?
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    (Original post by oShahpo)
    Same. What did you put for the bit about the spokes and the speedometer?
    The speedo will overestimate the speed - decrease in radius means the bike travels less far for the same rotation.
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    (Original post by guidoblee)
    The speedo will overestimate the speed - decrease in radius means the bike travels less far for the same rotation.
    Yea I said the same thing, but I also that the value itself wouldn't change as the angular velocity stays the same, does that negate the first bit I said?
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    (Original post by jtebbbs)
    I put the velocity was greater for the high speed electron, and r=mv/BQ so its path would have a greater radius i.e. it would be less deflected

    Struggled with the friction one, I put so that no momentum was transferred to the track and the system remained 'closed'...? Also struggled with that 5 marker with light gates and gliders, what did other people get for that?

    For the flat tire, the radius would decrease, and the speed on the speedo is inversely proportional to radius so the speedo would display a higher speed than the true value
    Surely the speed is directly proportional? v=2pir/T?
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    (Original post by target21859)
    What did people put for why it wasn't deflected as much, effect of flat tire, need for low friction and graph question? Grade boundaries?
    For the low friction one I put that there would be no external forces acting on the gliders (the COM states that momentum is conserved provided there are no external forces).
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    (Original post by guidoblee)
    I think the paper was pretty good overall, as normal with physics doing further mechanics modules in maths always helps. Any other ideas on why the H+ ion left no track after the photon interaction? I said it was because it wasn't moving, something about momentum
    Yeah I said there was no track because the H+ was stationary too!
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    anyone else notice how little maths was in it? Pretty much all qualitative!

    What did people get for q10? The kinetic energy and momentum multiple choice?
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    (Original post by kidlikethat)
    anyone else notice how little maths was in it? Pretty much all qualitative!

    What did people get for q10? The kinetic energy and momentum multiple choice?
    1/root6
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    (Original post by kidlikethat)
    anyone else notice how little maths was in it? Pretty much all qualitative!

    What did people get for q10? The kinetic energy and momentum multiple choice?
    1/sqrt6
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    What are the marking points going to be for the apparatus light-gate-momentum-gliders question?
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    (Original post by TheBigBo)
    I got 20.4
    I got 7.66 metres. You had to get the weight first. Then you do weight/cos(22) for the resultant force. Then you plug that force into r=mv^2/F
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    (Original post by regibayoan)
    I got 7.66 metres. You had to get the weight first. Then you do weight/cos(22) for the resultant force. Then you plug that force into r=mv^2/F
    r = mv^2 / F sin22 .
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    For C=k/d what did you guys put? I said C is inversely proportional to d so C decreases as d increases
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    (Original post by target21859)
    For C=k/d what did you guys put? I said C is inversely proportional to d so C decreases as d increases
    I put that as d increases, C decreases, C= Q/V, V is constant so Q decreases too and there's a current.
 
 
 
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