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    (Original post by mashmammad)
    Anyone got Jan 2001???? Can't find it anywhere. Its the only one I have not done. Please share if you have it.
    It didn't exist. The A2 modules didn't start until May/June 2001 (just as C3 and C4 didn't start until June 2005 although C1 and C2 started in January 2005).
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    (Original post by Tanmayee)
    Oh! Silly me, I forgot all about BD being vertical.



    If you look at the diagram I posted above, the line BD must go through OB for the toy to stay in equilibrium (same idea as in Ex5C in the textbook). The diagram shows the point before it topples, and for the conditions in the diagram, CD is (r/3). You can work that out using the fact that the angles are the same. For the toy to move, CD needs to be greater than (r/3) so that BD no longer goes through OB. CD is the value that you worked out in part a, so if you substitute that into CD > (r/3), you get [3(M-2m)] / [8(M+m)] > (r/3) and once you simplify this, you should get M > 26m.
    thankyou for explaining, makes sense now
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    Hi, for M3 jan 2012 question 6c, why can i not use energy to find out the maxium high B reaches after the string become slack? I know how to I need to work out the vertical component of the velocity and suvat, but I though i can also use energy mgh=1/2 mv^2, where v is the original velocity?? please can anyone explain it to me!!
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    Also can anyone do M3 Delphis paper2 question 7c and paper3 7c,d?? here is the link to the papers, I do not understand the marks scheme for those questions
    http://papers.xtremepapers.com/Edexc...ed/M3/Delphis/
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    (Original post by *mike)
    Thanks to both of you, I think I didnt fully understand the qesution, so does the particle literally go through point A. which is the only way displacement could be a/2, when I first read the question I thought he went below point A to a point D.
    Yes the particle literally looses contact with the surface but doesn't go through A. It descends and at one instant its at the same hight as A. This is the instant the question is talking about.
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    (Original post by mashmammad)
    Yes the particle literally looses contact with the surface but doesn't go through A. It descends and at one instant its at the same hight as A. This is the instant the question is talking about.
    OHHHHHH!! Thank you!!! I was thinking that it left the circle at point D!!! Thank you, when I read your comment I could hear the glass shattering in my head.
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    (Original post by milliezhao)
    Hi, for M3 jan 2012 question 6c, why can i not use energy to find out the maxium high B reaches after the string become slack? I know how to I need to work out the vertical component of the velocity and suvat, but I though i can also use energy mgh=1/2 mv^2, where v is the original velocity?? please can anyone explain it to me!!
    Firstly I was just about to ask how to do this exact question, I was trying energy and failing. and you're not by any chance the same millie on the imperial fb group?
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    (Original post by JenniS)
    Firstly I was just about to ask how to do this exact question, I was trying energy and failing. and you're not by any chance the same millie on the imperial fb group?
    I am indeed! haha what a coincident!!
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    (Original post by milliezhao)
    Hi, for M3 jan 2012 question 6c, why can i not use energy to find out the maxium high B reaches after the string become slack? I know how to I need to work out the vertical component of the velocity and suvat, but I though i can also use energy mgh=1/2 mv^2, where v is the original velocity?? please can anyone explain it to me!!
    The particle still has kinetic energy at the top, its vertical velocity=0 but it still has a horizontal component. Remember me from the Imperial thread btw?

    EDIT: So you'd have to use
    0.5m(vcosθ)^2=mgh

    Where θ is the angle between the velocity of the particle and the upwards vertical.
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    Soooooooooooo, tensed. I've now done all the papers, for some reason I make a lot of silly mistakes. BTW, o read someone mentioned imperial. My offer is imperial as well. I really want to get the grades for it.
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    haha every one of us further maths freaks seems to be aiming for imperial, I'm worried M2 blew my chances
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    Found this great resource that covers the whole of M3. it actually explains why you do what you do when solving a question, It's really good.

    http://www.mrbartonmaths.com/resourc...evision/M3.pdf
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    I didn't even apply to Imperial :woo:

    Good luck tomorrow everyone! :goodluck: It will be fine
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    Jan 2011 Question 2b, another one of those darned centre of mass questions, could somebody help me out with that please
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    (Original post by Serpentine111)
    The particle still has kinetic energy at the top, its vertical velocity=0 but it still has a horizontal component. Remember me from the Imperial thread btw?

    EDIT: So you'd have to use
    0.5m(vcosθ)^2=mgh

    Where θ is the angle between the velocity of the particle and the upwards vertical.
    oh yeah!! thank you that makes so much sense!! Sorry I can't tell who you are from your TSR account :confused:
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    (Original post by milliezhao)
    oh yeah!! thank you that makes so much sense!! Sorry I can't tell who you are from your TSR account :confused:
    I think you posted about being worried because you'd not heard from them for ages, and got an interview invitation later that day
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    When doing c.o.m of rotated curves does anyone know if we have to write out our full method with substitution of the limits etc or can we just bang it into our calculators and write the answer and get full marks?


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    (Original post by AB07)
    When doing c.o.m of rotated curves does anyone know if we have to write out our full method with substitution of the limits etc or can we just bang it into our calculators and write the answer and get full marks?


    Posted from TSR Mobile
    I tend to write

    "\displaystyle M\bar{x} =\sum_{a}^{b} m_{i}x_{i} = \pi\int_{a}^{b}y^2x dx"

    to start with, but then I usually use my calculator sparingly anyway. If you were really looking to use your calculator as much as possible for some reason, then I would at least also write:

    \displaystyle M= \pi\int_{a}^{b}y^2 dx

    before just coming up with any answers for M and the c.o.m.
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    (Original post by AB07)
    When doing c.o.m of rotated curves does anyone know if we have to write out our full method with substitution of the limits etc or can we just bang it into our calculators and write the answer and get full marks?


    Posted from TSR Mobile
    Write out the integration step, then show that you're using the limits with the square brackets. Then just give the final answer/fraction.
    Personally I write out the separate fractions after the using the limits and then simplify it.

    Can someone help me out with this question? I always mess up statics of rigid bodies when it comes to angles:
    2b:
    http://www.edexcel.com/migrationdocu...e_20110128.pdf

    Thanks.
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    Hi, hopefully this is the right place to post this, I have a question regarding the m3 `prove that` questions:
    does anyone happen to know if you have to have exactly what they have in the mark schemes in the questions that ask you to prove something?

    Sometimes I still get to the right answer, but its different from the method in the mark scheme - for example, I just did a question where you had to prove something by resolving forces, and I used cos x, found the angle, and used this to get to the right answer, but they didn't have any angles in their answer as they eliminated the angle straight away so my working wasn't exactly the same as any of the marking points; would I still get the marks?

    thanks
 
 
 
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