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    (Original post by kcfch)
    i really dont remember which question is was but one of them we had to find the tension, does anyone recall getting a T = 3/2 mg? im convinced i did it wrong oh and i think the second part of that question was a find w (omega) in terms of r and h or something if that helps jog memories
    T=3/2mg. Why do you think thats wrong.

    .A.
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    agreed.

    cos15/16
    .A.
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    (Original post by luckyu)
    i got for the circular motion cos = 15/16 does any one agree?
    yup i got that too
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    (Original post by .A.)
    1.6 cm :confused:

    The time period was 3s it was initially moving towards A away from O.

    The question said find the distance away from B in 2 s. So in 1.5 sec it goes from O to A and then comes back to O.

    Then you use X=asinwt. t=5, a=1.2 (i think) This gives you distance away from O. Then you subtract this from amplitude to get distance away from B.

    .A.
    I done the same methood..
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    (Original post by .A.)
    1.6 cm :confused:

    The time period was 3s it was initially moving towards A away from O.

    The question said find the distance away from B in 2 s. So in 1.5 sec it goes from O to A and then comes back to O.

    Then you use X=asinwt. t=5, a=1.2 (i think) This gives you distance away from O. Then you subtract this from amplitude to get distance away from B.

    .A.
    i used that way and got 1.6cm
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    (Original post by .A.)
    T=3/2mg. Why do you think thats wrong.

    .A.
    is that right? i dont know, i just remember thinking hmmm is this the right method...
    i think i was resolving vertically and horizontally and somehow arrived at that answer
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    (Original post by MaLvI)
    i found that it wouldnt topple
    In the limiting case angle comes out to be 12.9 which is greater than 12 so toppling does not occur.
    Same here, I got the limiting angle as 12.94 using:
    \displaystyle

tan \alpha = \frac{2a}{12a - y}
    where y was the distance given in part (b).
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    (Original post by luckyu)
    I done the same methood..
    But dude the amplitude is 1.2 how could you get that answer.

    .A.
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    (Original post by alex_hk90)
    Was I the only one who found that paper very tricky?

    As for question 6(c):
    Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
    \displaystyle

    v^2 = 6(1 - \frac{1}{(x + 1)^2}) = 6(\frac{x^2 + 2x}{(x + 1)^2}) \newline

    v = \frac{dx}{dt} = \sqrt{6}\frac{\sqrt{x^2 + 2x}}{x + 1} \newline

    \int{\sqrt{6}}dt = \int{\frac{x + 1}{\sqrt{x^2 + 2x}}dx \newline

    \displaystyle

\sqrt{6}t = \sqrt{x^2 + 2x} + c \newline

t = 0, x = 0, \Rightarrow c = 0 \newline

\to \sqrt{6}t = \sqrt{x^2 + 2x} \newline

t = 2, \Rightarrow 2\sqrt{6}t = \sqrt{x^2 + 2x} \newline

24 = x^2 + 2x \newline

x^2 + 2x - 24 = 0 \newline

(x + 6)(x - 4) = 0 \newline

x > 0 \to x = 4 \newline

    Ahhhhhh.... i started off like that, and then didn't have time to finish it!!

    Yes it was a tricky paper
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    (Original post by kcfch)
    is that right? i dont know, i just remember thinking hmmm is this the right method...
    i think i was resolving vertically and horizontally and somehow arrived at that answer
    I am sure its right. Although it would be nice to have others agree with us.

    .A.
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    (Original post by ahezhara)
    For the vertical circular motion question, I got something like 20.4 degrees.
    I had 1.6 cm away from B...
    My w was 2PI/3, a was 0.12m, maximum v was 2PI/25... 0.016...m away from B...
    I got that as well - 20.4 degrees cos 20.4 ... = 15/16

    Got the same dist from B as well.
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    I got both cos x = 15/16 and x = 0.016 m (to 2 s.f.) as well.
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    what about Question 3 horizontal circular motion how do you do part a and what did u get for part b??????????
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    (Original post by luckyu)
    what about Question 3 horizontal circular motion how do you do part a and what did u get for part b??????????
    What was the question again, was it the one with a particle on a table?
    And what were part (a) and part (b)?
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    Considering 65/75 (Jun 07) for A. What do you think it would be for this.

    .A.
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    What was question 3? The one on the table?

    Basically, resolve vertically and horizontally. I can't remember the exact values, so I won't do it here.

    Resolving vertically, consider the limiting position where the thing is spinning JUST fast enough so that it's about to leave the surface of the table. Hence, take R=0 at this point and carry and resolve to find w^2 . But then realise that w^2 has to be less than or equal to this value so that the particle remains on the table, as stated in the question - that's how the inequality came into it.

    Was part 'b' finding the tension in the string? I cannae remember.

    EDIT: .A. I'd say lower, those last 7 marks of question 7 are going to have eluded far too many people for it to be ignored. None of my further maths group were able to do it, and by the looks of it on here, hardly anyone else was able to either.
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    I think it had fairly tricky bits, so probably about 58/75 for an A? Might even be 60 for an A. So hopefully a B will be 50-53. I hope I can get a B on that paper
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    (Original post by .A.)
    Considering 65/75 (Jun 07) for A. What do you think it would be for this.
    I'm hoping it will be something like 54/75 for an A.
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    (Original post by .A.)
    Considering 65/75 (Jun 07) for A. What do you think it would be for this.

    .A.
    o ****, are you serious?! well i hope these boundaries will be lower, otherwise my A is in jeopardy
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    [QUOTE=alex_hk90]What was the question again, was it the one with a particle on a table?
    And what were part (a) and part (b)?[/QUOTE
    it was the one with the particle on the table. i know im wrong about not including the reaction of the table but how many marks do you think i can get because i gt the answer since i ingnored the reaction of table since it was greater than zero but then realised i was to l8 to change it since the inequality came into it.. And part b was make w the subject
 
 
 
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