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AQA A2 MM2B Mechanics 2 – 27th June 2016 [Exam Discussion Thread] Watch

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    (Original post by C0balt)
    7.58
    Yeah if you use mgdsin(30) you get 7.58, how many marks do you think using h instead of mgh for loss of gpe will be
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    (Original post by IrrationalRoot)
    Yep this is correct reasoning. Bead on a wire can have a reaction force upwards but tension will not act upwards.
    I have seen questions where the reaction force on a bead in a wire is resolved to the centre of the circle but you're probably right.
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    (Original post by Andrew Brockbank)
    Yeah if you use mgdsin(30) you get 7.58, how many marks do you think using h instead of mgh for loss of gpe will be
    You'd lose 3 or so I guess
    Honestly I'm not good at guessing maths marks lol
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    Did anyone else at all get (1 - mu)/4mu
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    Is 7.58 correct for q8?
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    (Original post by IrrationalRoot)
    Is this the method that anyone else used for Q8? Not sure where I went wrong.

    No KEs to consider first of all. Now by conservation of energy,

    GPE+EPE before = GPE+EPE+Work after

    Taking x to be the distance from A when next at rest, height for GPE is x/2. (xsin30.)
    This is taking the 0 GPE level to be at the point when next at rest.

    EPE is for one string, since extension 0 for one string, extension 5 for the other so EPE from that string.

    Because of our choice of GPE level, GPE=0 after.

    EPE after is total of the two EPEs for both strings, so with extension of x in one of them and extension of x-6 in another.

    Work found by x multiplied by coefficient of friction multiplied by reaction force which is mgcos30.

    Then I got two answers, chose the larger one, added 4 (got 7.05m).
    I did the same as you, but chose the smaller value of x, so got 5.20 m instead of 7.05m.
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    How many UMS would 68/69 marks be?
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    (Original post by ThusSpakeAlex)
    How many UMS would 68/69 marks be?
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    I'm thinking similar to 2014 so like that
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    (Original post by rainbowtwist)
    Did anyone else at all get (1 - mu)/4mu
    Got that first time. Then corrected myself and got the right answer.


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    (Original post by jjsnyder)
    Got that first time. Then corrected myself and got the right answer.i


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    What is the correct answer
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    (Original post by Hjyu1)
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    I'm thinking similar to 2014 so like that
    Seemed harder than 2014 imo.

    Especially the last two questions.
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    Name:  Screen Shot 2016-06-27 at 15.51.20.png
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Size:  86.2 KBsee like that's a string question in a vertical circle in jan 12
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    This is my 6)b) Name:  13555516_10205161123587299_392235323_o.jpg
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    (Original post by maisym00)
    What is the correct answer
    correct answer is \frac{1-2\mu^2}{4\mu}


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    Anyone else get this for the separation of variables question? (I think it was Q6b)Name:  ImageUploadedByStudent Room1467041052.503698.jpg
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    (Original post by jjsnyder)
    correct answer is \frac{1-2\mu^2}{4\mu}


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    Where does the mu^2 come from? I didn't see any reason to square mu :/
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    How many marks would i lose :s (7 marker)
    I've got my friction on the wall the wrong way up hence
    2u^2 +1 /4u not 1-2u^2/4
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    Should the two frictions act upwards and to the left on the moments question?
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    (Original post by rainbowtwist)
    Should the two frictions act upwards and to the left on the moments question?
    I've got my friction from the wall the wrong way up lol
    The friction from the ground acts left as the ladder wants to slide down so the friction opposes that.
    The friction from the wall, since the ladder wants to slide down the friction should be pointing upwards hence why my answer + not -. ffs

    I'm dropping marks everywhere
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    I DIDN'T DO THE ENERGY LOST BY FRICTION CORRECTLY!!! FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU UUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU UU.....................

    Couldn't get an answer for the ladder one either but showed some working out on moments and resolving forces... just gonna casually slip away from this thread, not even 24 hours passed and I'm already dreading it by looking here xD
 
 
 
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