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    Yup 2/7u is correct, which is smaller than 2/3u, therefore no collision
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    i didnt think it matters whether its a lamina or a wire system???
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    (Original post by chenners)
    For the centre of mass i got 3cm from AB
    So taking moments about midpoint of BC gave
    0.5 M = 3.5 k M
    So 1 = 7 k

    k = 1/7 maybeeee??!!!!
    yeah! k= 1/7
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    Yeah same here, I don't think it matter if it was a wire or not it still considered as a lamina.
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    Did anyone manage to get simply a vertical component to part b) of the rod question? I managed to get the x-component to cancel out completely and equal zero! I know that is so obviously wrong because they always ask for a bit of pythagoras' for the magnitude of reaction forces...

    Can anyone clarify?!
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    (Original post by jmzcherry)
    i didnt think it matters whether its a lamina or a wire system???
    yes it does. If it is a uniform wire, then the mass will act at the midpoint of each wire. So you take that into consideration.. if it was a lamina, then it is much easier (I started by assuming that it was a lamina,....then changed my answer :O)
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    The only difference between a wire and lamina system is that you'd let 'm' equal mass per unit length, instead of mass per unit area for a lamina
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    Oh and was there a loss of PE and a loss of KE?
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    But surely the centres of masses of the individual wire conform together to give the centre of mass that would be the same as if it were a lamina?
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    (Original post by Mathemagician)
    yes it does. If it is a uniform wire, then the mass will act at the midpoint of each wire. So you take that into consideration.
    yup, it was like that.
    stupidly enough i made a arithmetic mistake that i realized just as time ended, n got an answer of 2 2/3
    i feel so bad about that cuz i KNEW i had it written wrong but cudnt change it
    o well..the rest of it was all good..that question was easy as well -- i just dont kno how many marks i'd lose for that!
    do u think it will just b the answer marks n not the method marks? cuz i used the wrong value for part b as well getting a wrong k :rolleyes:
    anyway, im attaching the working!
    what does everyone think the boundary will be? o n how many marks was the COM question in total?
    Attached Images
     
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    (Original post by simiddd)
    Wot did u people get for Question 6 heres my answers
    1)a)35
    b)14.9
    2a)31/11
    b)15/77
    3a)c=4
    b) -36i + 8j
    4a) 1.8m
    b)6.75
    5a) 2/3
    b) proof
    6a) -----------
    b) ---------
    c)--------
    7a)117.6J
    b) 10N
    c) 0.393
    d) 5.4
    what was question 4. was it that one about the bricks falling. I remember getting 1.8 for somethinng but I cannot remember any 6.75
    for some of my answers I rounded and did not write rational numbers. will I lose the accuracy marks. I thought that accuracy and exact number were more important in futher pure maths. i got pretty much the same. any ideas about the grade boundaries?
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    (Original post by Eddie K)
    The only difference between a wire and lamina system is that you'd let 'm' equal mass per unit length, instead of mass per unit area for a lamina
    then why is the centre of mass different for a wire shaped cone, and a lamina(cone) ?

    I am 100% sure that it makes a difference as to whetehr it is a lamina or a wire.
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    If you consider just the 4X4 square, by considering them as wires, 4x2+4x2+4x4= 32
    By considering them as a lamina 16x2=32
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    (Original post by jmzcherry)
    If you consider just the 4X4 square, by considering them as wires, 4x2+4x2+4x4= 32
    By considering them as a lamina 16x2=32
    thats exactly what makes the difference...it wasn't a square
    try the same thing including the triangle n u'll see it wont work
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    it does make a big difference if you take it as a lamina... as then the CoM for each shape is at a different point.
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    jmzcherry yeah, that's what I did
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    was 6b 778 N?

    and no you can't treat it as a lamina...
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    (Original post by jmzcherry)
    If you consider just the 4X4 square, by considering them as wires, 4x2+4x2+4x4= 32
    By considering them as a lamina 16x2=32
    Lamina and wire does matter...
    Lamina mass for the square is 4x4 unit²
    The mass of the wire was 4x3 unit²
    this is beacuse there were only 3sides of the square!
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    What sort of raw mark do you usually need for an A?
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    for the first question, part (b) - the new max speed the car can travel up the slope - did anyone get 14∟7∟12 (thats supposed to be 14 and 7/12)?
 
 
 
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