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STEP I, II, III 2002 Solutions

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    (Original post by ghostwalker)
    Wonder why Simon didn't include it in the solved list in the OP, particularly as it's on the first page. Wish I'd seen it earlier, thanks.
    I'm being particularly lazy today, could you link me please?
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    (Original post by SimonM)
    I'm being particularly lazy today, could you link me please?
    If I knew how. The format for the links in your OP doesn't match what displays as a URL, so, no can do; sorry.
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    (Original post by SimonM)
    ..
    (Original post by ghostwalker)
    ..
    I've added the link to the index post.
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    Using NLR and a couple of momentum equations as detailed below:

    Perpendicular to line of "centres":

    u\sin\phi=v_1\sin(\theta+\phi)

    Parallel to line of centres:

    u\cos\phi=v_1\cos(\theta+\phi)+k  v_2

    NLR:

    v_2-v_1\cos(\theta+\phi) = u \cos\phi

    and after some nasty algebra (hence I'm not inclined to put the latex in at the moment), I end up with:

    \sin\theta=k\sin (\theta+2\phi)

    Which, whilst reasonably elegant, bears little resemblence to the desired equation.

    However, it does still satisfy the requirement that k = 1, when the two paths are perpendicular, which give me some small faith in it.
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    (Original post by Unbounded)
    ...
    (Original post by DFranklin)
    ...
    Thanks
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    (Original post by ghostwalker)
    \sin\theta=k\sin (\theta+2\phi)
    This line appears in Unbounded's solution, so it does end up in the right place eventually.
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    (Original post by Unbounded)
    The page for a single post can be found by clicking on the post number at the top of the post.
    Thanks; SimonM has introduced an entire new level of laziness, a quantum leap in inactivity - an oxymoron, if ever there was one.
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    (Original post by DFranklin)
    This line appears in Unbounded's solution, so it does end up in the right place eventually.
    Ta, Perhaps I ought to actually read it then. Lol.

    Edit: Not too far from the end either. My faith in mechanics is restored and all is right with the world once more. Deep sigh!

    Nasty bit of STEPish level work at the end, IMHO. Interestingly, I didn't use the conservation of KE directly in this second one (although implicit in using e=1).

    I have to say, of the three solutions, as an impartial observer (not), I prefer my first one.
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    What was the difference between yours and Unbounded? As described it looked similar, but from what you're saying, his turned out a bit nastier.
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    (Original post by DFranklin)
    What was the difference between yours and Unbounded? As described it looked similar, but from what you're saying, his turned out a bit nastier.
    You disagree then?

    Well it avoided the (\theta+2\phi), avoided the product-sum, seemed, to me, a hell of an lot simpler with the trig manipulations, and really reduced the problem to more standard A-level, rather than STEP, IMHO.

    P.S. I did say I was impartial (not)!
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    Wasn't disagreeing, just trying to ascertain where the big "don't do this, do that" point of decision was. It wasn't obvious to me on a cursory glance why your algebra was so much nicer - largely because I wasn't sure what was involved in your "sub (4) into (3) and fiddle" comment. Looking a bit more carefully, I see that there is very little fiddling required, but I didn't notice that at the time.
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    (Original post by DFranklin)
    ...
    :cool: Yep, "fiddle" with me is a small thing (though someone else may use the term compeletely differently), as opposed to the "after some nasty algebra" in my second solution which involved 8 terms each comprising a combination of three, sines and cosines; yuck!

    I guess the main difference was that I didn't square the "momentum in the direction of the initial velocity" equation early on; and later didn't need to.

    After stating the initial equations, putting v2 in terms of v1 seemed a natural first step, which would almost certainly come in handy later.

    Then "putting (4) into (3) and fiddle" so u is just in terms of v1 produced something akin to the one side of the desired equation (fortuitously perhaps), and the next step was obvious.
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    (Original post by SimonM)
    STEP I, Question 4

    \displaystyle y' = - \frac{2x}{(1+x^2)^2}

    Therefore the equation of the tangent is \displaystyle y = - \frac{2a}{(1+a^2)^2} x + \frac{1}{1+a^2} +\frac{2a^2}{(1+a^2)^2}
    Where did that part on the end (not the coefficient of x, everything else) come from?
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    (Original post by ziedj)
    Where did that part on the end (not the coefficient of x, everything else) come from?
    Suppose it is tangent at x=a. Then this tangent has equation  y = -\dfrac{2a}{(1+a^2)^2} x + c.

    But it passes through  \left( a, \frac{1}{1+a^2} \right) , and so c equals...
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    (Original post by Unbounded)
    Suppose it is tangent at x=a. Then this tangent has equation  y = -\dfrac{2a}{(1+a^2)^2} x + c.

    But it passes through  \left( a, \frac{1}{1+a^2} \right) , and so c equals...
    Oh.. isn't that unnecessarily complicated when you have already found via sketching that the curve passes through (0,1), so c = 1? Is there a particular reason?
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    (Original post by ziedj)
    Where did that part on the end (not the coefficient of x, everything else) come from?
    For what it's worth I set it out a slightly different way
    Tangent is at x=a (>0) and passes through (0,1)
     y - 1 = \dfrac{-2a}{(1+a^2)^2}x

    and  \dfrac{y-1}{x} = \dfrac{\dfrac{1}{1+a^2} - 1}{a} = \dfrac{-a}{1+a^2}

    So we must have

     \dfrac{-2a}{(1+a^2)^2} = \dfrac{-a}{1+a^2}

    Which quickly cancels down to give  a^2 = 1
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    My solution to question 14 STEP II
    Attached Files
  1. File Type: pdf STEP II.pdf (42.7 KB, 35 views)
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    Completing Dadeyemi's work on STEP III Q2, last part
    Attached Files
  2. File Type: pdf STEP III Q2.pdf (32.5 KB, 69 views)
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    (Original post by SimonM)
    STEP III, Question 1

    Spoiler:
    Show


    As  a \to \infty, the volume tends to 2\pi
    Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

    Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
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    (Original post by Diego Granziol)
    Sorry could you explain how you get to 2 divided by pi? I got the -(ln^2 (a) +2 ln(a) +2)/a but could not justify whether that expression would tend to 0 or 1, im thinking now probably 0 as if you sub in x = e^2 then e^5 that gets progressively smaller and smaller, so did you mean 2pi?

    Also a little suggestion maybe it would have been easier to use the substitution u = ln x.
    It's 2 pi, in latex that is 2\pi. Perhaps I forgot the tags?

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