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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 Shadez)
    I'm stuck on question 5b from this exercise sheet:http://www.madasmaths.com/archive/ma...gs_springs.pdf

    The answer is 1.65m
    Here's what I've done could someone point out where I went wrong
    Attachment 432841

    Attachment 432841432843
    Doesn't matter I've figured it out
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    Can anyone try to explain question 5 from january 2012 please?
    http://filestore.aqa.org.uk/subjects...B-QP-JAN12.PDF

    Edit: I read the question wrong, I thought it was going down a hill instead of around a bend :/
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    (Original post by samlyon)
    Can anyone try to explain question 5 from january 2012 please?
    http://filestore.aqa.org.uk/subjects...B-QP-JAN12.PDF
    Centripetal force in the case of the parcel is provided by friction cause surface is rough

    f<=uR but for max speed, centripetal force is max so friction is max so

    uR=mv^2/r

    R=mg so

    umg=mv^2/r

    you can see the mass cancels so 'm' is not required

    ug*r=v^2
    v = root(ugr)

    sub in the values and find v
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    (Original post by sum1random)
    Centripetal force in the case of the parcel is provided by friction cause surface is rough

    f<=uR but for max speed, centripetal force is max so friction is max so

    uR=mv^2/r

    R=mg so

    umg=mv^2/r

    you can see the mass cancels so 'm' is not required

    ug*r=v^2
    v = root(ugr)

    sub in the values and find v
    Thank you
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    We wont lose marks for using column vectors as opposed to using the i and j format aren't we? Because i and j require so much more effort to work with than column vectors.
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    (Original post by Jetraft87)
    We wont lose marks for using column vectors as opposed to using the i and j format aren't we? Because i and j require so much more effort to work with than column vectors.
    You shouldn't do, it means the same thing! My teacher recommends writing them out in the same format as the question for your final answer but I usually forget
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    Why are the grade boundaries for M2 so high 😩 I don't understand, it's quite a big step up from M1 in my opinions and the grade boundaries are so much higher
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    (Original post by JackNorman)
    Why are the grade boundaries for M2 so high 😩 I don't understand, it's quite a big step up from M1 in my opinions and the grade boundaries are so much higher
    It's a unit that a lot of people do as a further maths module. I'm doing it as part of normal maths A level as I did M1 last year, but the boundaries are always so high because generally the people who do it are insanely clever 👀


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    Can u explain 9c to me, please? I do not get where the cos 2theta come from.

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    (Original post by ike96)
    Can u explain 9c to me, please? I do not get where the cos 2theta come from.

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    https://m.youtube.com/watch?v=kA4F-p5Hj_4


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    Hey can someone do question 9 part c june 2014 i dont understand how they got 2.04m
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    june 2013*
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    (Original post by olatee)
    Hey can someone do question 9 part c june 2014 i dont understand how they got 2.04m
    The extension was 2.5m initially but A has moved 0.46m, so the extension now is 2.5-0.46 = 2.04m. Using a quadratic where the distance travelled is defined to be x, the extension at x metres can be expressed as (2.04 - x).


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    For Jan 09 question 7b, how do they expect me to know the angle.....
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    Thanks for the video.

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    (Original post by Tiwa)
    For Jan 09 question 7b, how do they expect me to know the angle.....
    Its a standard rule in vertical circles that when the object is in the below half of the circle the height it has gained is
    =r-rcostheta

    this is cause the angle the vertical line makes with the radius line at point B is theta as well (alternate angles)
    where theta is the angle between radius at A and radius at B

    so r-rostheta = 2m (height)

    rearrange to get theta=60deg
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    Can anyone quickly go over the rules needed for a complete circle in each situation?

    Bead on a wire
    Ball in a cylinder
    Ball on a hill
    Ball on a string


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    Im so confused as to how to work this out, supposedly the answers are Ta = 24.2N and Tb = 4.6N.Name:  Photo on 21-06-2015 at 12.30.jpg
Views: 134
Size:  127.4 KB

    If anyone can help out I would be massively grateful - I've wasted an hour trying to figure it out haha.
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    Has anyone done Question 7 on page 136 Exam style practice papers in the M2 textbook?
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    (Original post by undead345)
    Has anyone done Question 7 on page 136 Exam style practice papers in the M2 textbook?
    Do you mean this book?

    https://wordery.com/mechanics-2-for-...FQsCwwodKhEARQ


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