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AQA Maths Mechanics 2B June 2012 Unofficial Mark Scheme

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    For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
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    (Original post by Signed Solution)
    For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
    That sounds like what I got.
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    0.0668 Correct. The rebound velocity was v = ((96-11gu)^0.5)/2

    0.5mv^2 - 5,5mgu - 120e^2 /2l =0 (e=0.5)
    this leads to 96-11gu /20= 2.2gu +3
    which then leads to
    u=36/55g

    which is, 0.0668 ( you need to do it to 3sf)
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    (Original post by Signed Solution)
    For the two tensions I got 7.84N (or around there, can't remember) at A and 11.76N at B.
    got same
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    Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
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    (Original post by Iepnauy)
    Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
    If you only dropped five marks, then you'll definitely get above 90% on this paper. (:
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    The highest i have ever seen for an a* is 69 so you should be fine
    (Original post by Iepnauy)
    Dropped 5 marks-ish. Knowing how high the boundaries will be, I doubt it'll be an A*
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    Can anyone tell me what they got for part b of 6.. swinging child tenison in string one...?
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    (Original post by jph12)
    Question one: Ski jumper
    1a)29792
    b)23088.8
    ci)52880.8
    cii)37.3

    Tick

    Question 2:
    2ai)a=12t+8e-4t
    aii)7.08
    b)28.3
    c)2t2+0.5e-4t+8t-0.5

    Tick
    Question 3: Centre of mass
    3ai)4.8
    aii)7.2
    aii)A=7.84N, B=11.76N

    Tick

    Question 4: Trig motion vectors
    4a) Didn't complete. ): Hard question! I mentioned that r would never equal zero, don't know if that will get me marks.
    b)-12sin3ti-12cos3tj
    c)-36cos3ti+36sinstj
    d)k=-9
    e)Toward centre of the circle? North west?

    4a - I showed that the modulus of r is always 4 and hence it must be going in a circle. I then showed that at t=0 r = 4i and at t=60 r = -4i and thus if radius is 4 then centre must be at origin. Anyone to back me up ?
    4e - If something is moving in a circle about the origin then surely the acceleration must be towards the center ? Where are people getting NW from ?

    Question 5: Horizontal circular motion
    5a)7
    b)2.1
    c)0.27

    5c - w = 2pi/t so t = 2pi/7 and so t= 0.898 sec, no ?
    Question 6: Swinging child
    6a)1.51
    b)236.7

    Tick

    Question 7: Differential equation
    7a)-1.96(v-5)
    b)v=2e-1.96t+5
    Tick

    Question 8: Elasticity question
    8a)9.80
    bi)sqroot(96-11gu)
    bii)mew=0.334

    8b - 0.668.
    I will edit according to the general concensus. Some right answers for question eight, please, and the ones I couldn't do. (:

    (Original post by vedderfan94)
    Nice job man, I've got these:

    3b) Tension at A = 18.4N, Tension at B = 1.2N

    4a) I found the magnitude and showed that r always equals 4, therefore it moves in a circle. Not sure if that's right.

    8a) 9.80
    bi) sqroot(96-11gu)
    bii) Think I got this wrong. Correct answer seems to be 0.067 or around that :/
    Yea I got all your answers except for 3b I got TA=7.84 N and TB= 11.76N
    (Original post by SwimGood)
    Also, I think I wrote North West for the direction on the vectors question
    But if it's in circular motion then isn't the acceleration to the center ? Where did you get NW ?
    (Original post by Signed Solution)
    I think in the first question there shouldn't be any newtons.

    For question 8, I remember getting root(96-11gmew) or something like that and ended up with mew=0.0668.
    I got that value for mew

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Updated: June 21, 2012
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