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

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    (Original post by misslili118)
    NEED HELP WITh QUESTION 4 JAN 11 M2 PART C AND D
    With part c just set up a diagram using the centre of mass you have worked out and then use trig. Part d is just a moments question where you have to take moments about P in order to work out the mass.
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    (Original post by JC25)
    With part c just set up a diagram using the centre of mass you have worked out and then use trig. Part d is just a moments question where you have to take moments about P in order to work out the mass.
    I used a diagram but i keep getting arc tan 19.4/21.1 , i guess i have the wrong diagram?
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    (Original post by misslili118)
    I used a diagram but i keep getting arc tan 19.4/21.1 , i guess i have the wrong diagram?
    Maybe try the diagram again?
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    (Original post by JC25)
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    Maybe try the diagram again?
    Thank you so much!
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    (Original post by misslili118)
    Thank you so much!
    i am stuck at this question too, markscheme says 24.6 for the angle i keep getting 8.45 because i used tan theta = 19.4/8.45
    d
    did u manage it?
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    (Original post by Husssein)
    i am stuck at this question too, markscheme says 24.6 for the angle i keep getting 8.45 because i used tan theta = 19.4/8.45
    d
    did u manage it?
    its arctan 8.8/19.4 , youve got your sides the other way round
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    (Original post by misslili118)
    its arctan 8.8/19.4 , youve got your sides the other way round
    OMD!, it was a school boy error silly mistake, my method was tan theta = 19.4/8.8 and take that theta away from 90 which gives correct answer but i mustve put it wrong in calc or summink thanks anyways
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    (Original post by Husssein)
    OMD!, it was a school boy error silly mistake, my method was tan theta = 19.4/8.8 and take that theta away from 90 which gives correct answer but i mustve put it wrong in calc or summink thanks anyways
    no problem , i still cant get d around my head
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    HELPPPPPPPPP : jun 11 Q8 b .. why is it 0.5mu^2= 0.5mv^2 + mga(1+sintheta) and not just 0.5mv^2 + mgasintheta?? why is there an extra mga on the markscheme??
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    Because the height of b is a+asintheta hence a(1+sintheta)
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    June 13 question 9)a)i) I used T=lambda*x/original length where x is 0.5, but apparently it's 2.5 and I can't understand why
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    Can someone please explain how to do 8b , I've had a nightmare with it


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    (Original post by RockyG)
    Because the height of b is a+asintheta hence a(1+sintheta)
    Why is it not asintheta since the radius is A


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    (Original post by TheKian)
    June 13 question 9)a)i) I used T=lambda*x/original length where x is 0.5, but apparently it's 2.5 and I can't understand why
    The new length of the string is 5.5 m because from A to O it's 3.5 m and from O to B its 2 m, so the extension is 5.5 - 3 = 2.5
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    (Original post by mamounaltayeb)
    Why is it not asintheta since the radius is A


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    Because the distance from b to the bottom of the circle is more than a , asintheta more so its total distance from the bottom is a + a sintheta
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    For part b in this question in the mark scheme when resolving radially it uses R + mgcos60 = mv^2/r which means that the reaction force acts towards the centre of the circle. My reason of thinking was that the reaction force would act in the opposite direction so that it would act outwards (mgcos60 - R = mv^2/r), directly away from the centre of the circle. Can someone explain why R acts towards the centre of the circle?
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    (Original post by TheLifelessRobot)
    For part b in this question in the mark scheme when resolving radially it uses R + mgcos60 = mv^2/r which means that the reaction force acts towards the centre of the circle. My reason of thinking was that the reaction force would act in the opposite direction so that it would act outwards (mgcos60 - R = mv^2/r), directly away from the centre of the circle. Can someone explain why R acts towards the centre of the circle?
    Centripetal force causes the reaction to act inwards.
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    (Original post by Mm68)
    Can someone confident with moments tell me their technique, my teacher has taught me a variety of methods and I'm getting them confused, mainly for your generic ladder question. I mainly don't understand which way to resolve the forces, been taught in the direction of the line, perpendicular to line, vertical, horizontal, and now I'm getting muddled.
    I always resolve vertically, then horizontally and then take moments about a point
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    (Original post by mamounaltayeb)
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    Can someone please explain how to do 8b , I've had a nightmare with it


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    For the next part you just set the answer to 0 to get sin(theta) = 5/6

    EDIT: Ignore the one that is under 'attached thumbnail', the one above is the correct one
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    Does anyone have a set formula that works for EPE questions.

    Like Final KE - Initial KE = Initial EPE - Final EPE - Work done by friction
 
 
 
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