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AQA A2 Mathematics MM2B Mechanics 2 - Monday 22nd June 2015 [Exam Discussion Thread] Watch

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    (Original post by CD223)
    I heard that was a tough paper!


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    Yep toughest chemistry paper I've ever done hopefully I get the C grade needed to boost my overall grade to an A
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    (Original post by Shadez)
    Yep toughest chemistry paper I've ever done hopefully I get the C grade needed to boost my overall grade to an A
    Ah well good luck!


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    Does anyone else find elasticity questions hard? I just can't seem to get hang of what the question is actually requiring me to do.
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    (Original post by Artur96)
    Does anyone else find elasticity questions hard? I just can't seem to get hang of what the question is actually requiring me to do.
    They're typically the hardest question on the paper. The ones where you form a quadratic are never the same method which is why it's difficult. I'm sure if you practice it enough before the day you'll be okay though.


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    (Original post by Artur96)
    Does anyone else find elasticity questions hard? I just can't seem to get hang of what the question is actually requiring me to do.
    I see what you did there :laugh:

    Anyways try all the questions from here should help

    http://www.madasmaths.com/archive/ma...gs_springs.pdf
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    (Original post by Shadez)
    I see what you did there :laugh:

    Anyways try all the questions from here should help

    http://www.madasmaths.com/archive/ma...gs_springs.pdf
    Thanks for sharing!


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    Also, how do you get around determining whether you have to use sin or cos when taking moments on an incline?
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    I reckon after last year's odd variation on ladders (with it leaning over a short wall rather than against it) we might have another odd take on moments to see if we've paid attention to the examiners reports, like maybe leaning against a rough wall
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    Can anyone suggest the most pain-free way of learning M2 before monday? Haven't really done anything yet
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    Well now PHYA5 is over it's all aboard the M2 hype :/ got computing Tuesday though which takes priority.


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    Doing june 2012, on the very last question , I've found that WD by friction is equal to the kinetic energy of the block - 3, and that that is also equal to 2.2μg. But when I try to work out the initial kinetic energy, I can't seem t get the same as the mark scheme, there must be something really stupid I'm missing out. I get EK=1/2*0.4*1/2(96-11μg), but the mark scheme says 1/2*0.4*1/4(96- 11μg). Not sure where the quarter comes from
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    (Original post by JackNorman)
    Doing june 2012, on the very last question , I've found that WD by friction is equal to the kinetic energy of the block - 3, and that that is also equal to 2.2μg. But when I try to work out the initial kinetic energy, I can't seem t get the same as the mark scheme, there must be something really stupid I'm missing out. I get EK=1/2*0.4*1/2(96-11μg), but the mark scheme says 1/2*0.4*1/4(96- 11μg). Not sure where the quarter comes from
    Doesnt matter, I figured it out 😓
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    Can somebody please help me with part b? I really don't understand Name:  Screen Shot 2015-06-18 at 19.36.42.png
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    (Original post by smilejensmile)
    Can somebody please help me with part b? I really don't understand Name:  Screen Shot 2015-06-18 at 19.36.42.png
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    You have to use conservation of energy does that help?
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    (Original post by smilejensmile)
    Can somebody please help me with part b? I really don't understand Name:  Screen Shot 2015-06-18 at 19.36.42.png
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    You need to resolve the forces at Q.

    T + mg = mv^2/a because both the weight and the tension provide the centripetal force.

    So for the tension: T = mv^2/a - mg and v will be the speed you found in part (a), so just substitute in v and simplify.
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    Can someone please explain to me why 8b has been resolved as R= -mgcos60 + (mv^2)/a

    i thought finding the reaction would involve resolving as mgcos60-R = (mv^2)/a
    ??
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    (Original post by 221loki)
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    Can someone please explain to me why 8b has been resolved as R= -mgcos60 + (mv^2)/a

    i thought finding the reaction would involve resolving as mgcos60-R = (mv^2)/a
    ??
    Because both R and the component of the weight towards the centre of the circle provide the centripetal force. So R + mgcos60 = mv^2/a
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    (Original post by FeelsToWaltz)
    Because both R and the component of the weight towards the centre of the circle provide the centripetal force. So R + mgcos60 = mv^2/a
    Oh Okay but how do I know if it's acting towards or away from the centre?
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    (Original post by 221loki)
    Oh Okay but how do I know if it's acting towards or away from the centre?
    Basically if its in the top half of the circle then the weight provides some of the centripetal force, if it's in the bottom half then the weight opposes the centripetal force.

    So at the top of the circle, R + mg = mv^2/a and at the bottom R - mg = mv^2/a, then if its at an angle you change mg to mgcos(theta)
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    (Original post by FeelsToWaltz)
    Basically if its in the top half of the circle then the weight provides some of the centripetal force, if it's in the bottom half then the weight opposes the centripetal force.

    So at the top of the circle, R + mg = mv^2/a and at the bottom R - mg = mv^2/a, then if its at an angle you change mg to mgcos(theta)
    Oh that makes so much sense thank you!!

    So what would happen if it's right in the middle of the circle?
 
 
 
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