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Edexcel Mathematics: Mechanics M3 6679 01 - 17 May 2017 [Exam Discussion] Watch

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    (Original post by solC)
    There are some notes on Physicsandmathstutor by a guy called Simon Baxter, I found them quite helpful for FP3 so they are probably quite helpful for M3 too.

    http://pmt.physicsandmathstutor.com/...M3%20Notes.pdf
    Thanks this is quite useful ^^^
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    Misread question
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    Part of a hollow spherical shell, centre O and radius r, forms a bowl with a plane circular rim. The bowl is fixed to a horizontal surface at A with the rim uppermost and horizontal. The point A is the lowest point of the bowl. The point B, where !AOB = ! and tan !"#" 3 4 , is on the rim of the bowl, as shown in Figure 2. A small smooth marble M is placed inside the bowl at A!"#$%"&'()$"#$"'$'*'#+",-.'/-$*#+"01))%"23gr). The motion of M takes place in the vertical plane OAB. (a) Show that the speed of M as it reaches B is ! 3 5 gr ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ . (4) After leaving the surface of the bowl at B, M moves freely under gravity and first strikes the horizontal surface at the point C. Given that r = 0.4m, (b) find the distance AC

    Sorry copy and paste didnt want to work but can someone help me with part b) its june 2013 R q5
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    Hi, could someone please tell me what M3 requires in terms of knowledge about volumes, surface areas etc of shapes. Thank you.
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    (Original post by Major-fury)
    Hi, could someone please tell me what M3 requires in terms of knowledge about volumes, surface areas etc of shapes. Thank you.
    Same as C4
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    (Original post by TSRUser12345)
    Same as C4
    what are the ones needed for c4 lol. I've already taken c4 a while ago :/
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    Hi guys,

    I'm a bit confused on a couple calculus examples in Chapter 5 (Statics) in the Edexcel M3 textbook.

    First off, do we have to use the formula for Mx̅ and M then divide Mx̅ by M to get x̅ (and the same thing for ȳ?)
    It just doesn't make sense to do that when the x̅ and ȳ formulae are given to us. Like in one example they'll do it the way I described above and in another example they'll just use the x̅ and ȳ formulae to get the answers straight away. Is there a reason to do it the first method or is it okay to do it just using the x̅ and ȳ formulae without needing to calculate the mass and stuff?

    Many thanks
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    Could someone help me with Question 2 jan 2007 please. I can get to the bit where we find the centre of mass of the cone etc. but i can't seem to visualise where the triangle comes from with the upward vertical.
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    (Original post by Major-fury)
    Could someone help me with Question 2 jan 2007 please. I can get to the bit where we find the centre of mass of the cone etc. but i can't seem to visualise where the triangle comes from with the upward vertical.
    The triangle goes from A to G to the centre of the circular base, with the right angle at the centre of the base.
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    (Original post by tiny hobbit)
    The triangle goes from A to G to the centre of the circular base, with the right angle at the centre of the base.
    Sorry but I don't quite follow. Aren't we trying to calculate the angle made with the upward vertical? If possible could you draw a diagram, I think i'm overlooking something.
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    (Original post by Major-fury)
    Sorry but I don't quite follow. Aren't we trying to calculate the angle made with the upward vertical? If possible could you draw a diagram, I think i'm overlooking something.
    Yes. This is the same size as the angle at G in the triangle I was talking about.

    Diagram from mark scheme attached.
    Attached Files
  1. File Type: docCone question.doc (30.0 KB, 18 views)
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    (Original post by tiny hobbit)
    Yes. This is the same size as the angle at G in the triangle I was talking about.

    Diagram from mark scheme attached.
    Could you explain why theta is equal to that angle please?
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    (Original post by Major-fury)
    Could you explain why theta is equal to that angle please?
    G (the centre of mass) will be vertically below A, the point of suspension.

    So this angle is, like the one marked theta, between a vertical line and the axis of the cone.
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    (Original post by tiny hobbit)
    G (the centre of mass) will be vertically below A, the point of suspension.

    So this angle is, like the one marked theta, between a vertical line and the axis of the cone.
    Thanks. I think I understand it. At least a little better
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    (Original post by physicalgraffiti)
    Hi guys,
    It just doesn't make sense to do that when the x̅ and ȳ formulae are given to us.
    I don't think those formulae are given - you have to memorise them.
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    (Original post by physicalgraffiti)
    Hi guys,

    I'm a bit confused on a couple calculus examples in Chapter 5 (Statics) in the Edexcel M3 textbook.

    First off, do we have to use the formula for Mx̅ and M then divide Mx̅ by M to get x̅ (and the same thing for ȳ?)
    It just doesn't make sense to do that when the x̅ and ȳ formulae are given to us. Like in one example they'll do it the way I described above and in another example they'll just use the x̅ and ȳ formulae to get the answers straight away. Is there a reason to do it the first method or is it okay to do it just using the x̅ and ȳ formulae without needing to calculate the mass and stuff?

    Many thanks
    If you can do it using the formulae then you probably should, unless the question specifies not to, but its not always possible to do it using the formulae, so you'd have to use the first method

    (Original post by kavindajd)
    I don't think those formulae are given - you have to memorise them.
    Pretty sure we're given formulae for centres of mass of hemispheres and cones
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    (Original post by Jamvicious)
    Pretty sure we're given formulae for centres of mass of hemispheres and cones
    Yes you are right, I think we are given 4 formulae in total for the final chapter.
    I thought he was talking about the formulae for the solid of revolution which isn't given.
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    Guys june 2016 IAL m3 q5 is confusing me its SHM can someone help me

    5. A vertical ladder is fixed to a wall in a harbour. On a particular day the minimum depth of water in the harbour occurs at 0900 hours. The next time the water is at its minimum depth is 2115 hours on the same day. The bottom step of the ladder is 1m above the lowest level of the water and 9m below the highest level of the water. The rise and fall of the water level can be modelled as simple harmonic motion and the thickness of the step can be assumed to be negligible.

    Find (a) the speed, in metres per hour, at which the water level is moving when it reaches the bottom step of the ladder, (7)

    (b) the length of time, on this day, between the water reaching the bottom step of the ladder and the ladder being totally out of the water once more. (4)
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    Mad how close the exam is now! Ive done
    - ALL M3 edexcel papers done and redoing those which I found tough.
    - Every question in the textbook done.
    - All solomon papers done.
    - Loads of madasmaths too
    My weakest area is definitely SHM. All I want is a standard paper, like not easy so the boundaries become absurd but not that hard that it is demoralising - guess we will have to just show up and do the best on the day.
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    Has anyone got any predictions for this year's paper?
    http://pmt.physicsandmathstutor.com/...%20Edexcel.pdf
    Also, part B of Q6 terrifies me! Can someone please show me a solution with a diagram?
 
 
 
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