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OCR Physics A newtonian World 20/6/16 Unofficial Mark Scheme Watch

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    (Original post by apache324)
    I got 1.02*10^33, anyone else get this?
    I can't remember the specific number but I got E^33
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    (Original post by apache324)
    I got 1.02*10^33, anyone else get this?
    Yep and it's right this teacher guy who uploaded is wrong


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    Thankyou!

    Not too happy considering past paper performance and the amount of revision I put in. I'm guessing maybe 42/60.
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    (Original post by Kevinwong)
    Yep and it's right this teacher guy who uploaded is wrong


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    Teachercol is a real life Physics teacher who gets to see the actual paper, so I would wager he is right. Good luck though?
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    (Original post by NickLCFC)
    I think it will be similar to last years...

    FULL 51
    A* 47
    A 43
    B 39
    C 35

    Then again, that's probably me being a bit too hopeful. A lot of people seemed to find it much easier than I did
    Damn Math students...
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    (Original post by Asiimov)
    Teachercol is a real life Physics teacher who gets to see the actual paper, so I would wager he is right. Good luck though?
    I don't know if you've seen my earlier post but you will find that he messed up on equating the formulae. Not much point repeating what i said (you can read it yourself if you want)

    but essentially, yes he is wrong (surprise!), and 1.0x10^33 is the correct answer.
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    Damn, seems I did a lot better than I thought when leaving the exam, if i remember my answers correctly, then it means i got at least a B with reference to last years grade boundaries which i'm happy with. Thanks for the MS.
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    Sorry can someone tell me why for Q2 it was mg + Fsin55 rather than mg - Fsin55 when both forces are acting in opposite directions of each other?

    Other than that, it looks like I've done better than expected:o:o

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    (Original post by apache324)
    I got 1.02*10^33, anyone else get this?
    Yeah exactly the same as mine
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    (Original post by Leechayy)
    Sorry can someone tell me why for Q2 it was mg + Fsin55 rather than mg - Fsin55 when both forces are acting in opposite directions of each other?

    Other than that, it looks like I've done better than expected:o:o

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    If you use mg with g as (-9.81), then youd be right. The hose is placed at an angle of 55 to the horizontal, meaning that Fsin 55 would give you the horizontal force (acting downwards) from the hose. You also have mg acting downwards, meaning they add up to form R. Newtons 2nd and 3rd law, as the change in momentum/t is equal to the force, and there is an equal and opposite force downwards.


    TL;DR The water is pushed upwards from the hose, meaning it pushes the hose downwards, with an equal and opposite force.
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    (Original post by Brokensteps)
    If you use mg with g as (-9.81), then youd be right. The hose is placed at an angle of 55 to the horizontal, meaning that Fsin 55 would give you the horizontal force (acting downwards) from the hose. You also have mg acting downwards, meaning they add up to form R. Newtons 2nd and 3rd law, as the change in momentum/t is equal to the force, and there is an equal and opposite force downwards.


    TL;DR The water is pushed upwards from the hose, meaning it pushes the hose downwards, with an equal and opposite force.
    Ohhh okay, I think I get it now. So based off the question's context (which I don't really remember) you lead to using the equal and opposite force?

    So I'm guessing I'd lose 2/3 marks then

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    (Original post by Danny.L)
    Ignore the number in the brackets, thats me tallying how many marks i dropped worst case.
    If anyone could help put my mind at ease for some of these that'd be great.
    2)
    c)ii)
    R = mg + Fsin55 = 92 x 9.81 + (8.25 x 25)sin 55
    = 902.5+169 = 1071N (3)
    I omitted the Multiple by 25 by 8.25 and instead did Weight plus 8.25sin55.
    Is this likely to get 1 or 2 marks out of the 3? (-2/3)

    Q3 a)iii)
    Left in Pi :/
    b) i) I just flipped the EPE curve, is that both marks out the window?
    ii)&iii) I then calculated these using 80mJ opposed to 50, would I drop all of these marks, or would I get e.c.f? (-3/5)


    Q5e) Used keplers third law and got 1.73*10^33 :/
    Is that 3 marks out the window? Or could I (hopefully) get 2 for this method? (-2/3)

    Q6)a) Possibly didnt answer in enough detail. (-1/2)
    d) So stupid, was rushing at this point, stated 2x the energy input, but not 2x mass so said temperature would increase by 2. Ridiculously stupid. Would I get 1 mark for stating that energy input is doubled, or is it 1 mark for saying constant and 1 mark for correct explanation? (-2/2)

    Q7)a) Not enough detail (-1/2)

    Worst case 49/60. Very disappointed considering I was getting 58+/60 on my past papers. Absolutely kicking myself.
    Oh God man you're looking at a A/B grade in that paper from what I can infer from your mistakes, sorry I hope I'm wrong. I probably dropped similar amount of marks as you. The only *mathematical* mistakes I made was the mass of S2 and the hosepipe, and goddamn phase difference had to be in number radians not in terms of 'pi'. other than that I think got all the other maths bits right :O So glad I got the shm question right tho xd not bragging
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    (Original post by Spectral)
    I don't know if you've seen my earlier post but you will find that he messed up on equating the formulae. Not much point repeating what i said (you can read it yourself if you want)

    but essentially, yes he is wrong (surprise!), and 1.0x10^33 is the correct answer.
    My answer agreed with the teacher
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    I've redone 5e

    Ok I've had an overnight rethink on this one. I'm pretty confident with this now
    Its what parts A I) and ii) were intended for.
    M1v1^2/r1 = GM1M2/(r1+r2)^2
    so M2= v1^2 (r1+r2)^2/Gr1
    = 6.08E4^2 x 4.8E12 ^2 / ( 6.67E-11 x 1.2E12)
    =1.06E33kg
    M2 = 1.06E33kg
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    (Original post by TheFarmerLad)
    teachercol For the binary star question, could you treat the centre of the two circular motions to be a body of mass (M1 + M2 all divided by two) as the question said the two stars orbit about their centre of masses. By doing this could one then say : (v1)^2/R1 = (G)(M1+M2/all divided by two)/ (R1)^2 or (V2)^2/R2 = G(M1+M2/all divided by two)/ (R2)^2? If unclear I'll attach a diagram
    No - I don't think you can
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    (Original post by Asiimov)
    Teachercol is a real life Physics teacher who gets to see the actual paper, so I would wager he is right. Good luck though?
    Not this time. When I've had time to think about it, I've redone that question.
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    (Original post by Spectral)
    I don't know if you've seen my earlier post but you will find that he messed up on equating the formulae. Not much point repeating what i said (you can read it yourself if you want)

    but essentially, yes he is wrong (surprise!), and 1.0x10^33 is the correct answer.
    Quite right - redone that one. Did too fast and should have thought more.
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    ^33 though. Isn't that a thousand solar masses? Doesn't that seem a bit.... Odd?
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    R136a1
    Largest known star. 265 solar masses.
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    (Original post by speed1✈️✈️)
    Oh God man you're looking at a A/B grade in that paper from what I can infer from your mistakes, sorry I hope I'm wrong. I probably dropped similar amount of marks as you. The only *mathematical* mistakes I made was the mass of S2 and the hosepipe, and goddamn phase difference had to be in number radians not in terms of 'pi'. other than that I think got all the other maths bits right :O So glad I got the shm question right tho xd not bragging
    49/60 isnt an A/B calm down son.
    That's A*/100% ums territory. lmfao
 
 
 
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