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STEP Maths I,II,III 1987 Solutions watch

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    Go for it, I maybe got about a third of the way through before i got stumped.
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    (Original post by Square)
    Go for it, I maybe got about a third of the way through before i got stumped.
    So where did you get to? (Slightly curious, as it's a two part question and you say you got a third of the way...)
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    I made the subsitution for dx and then substituted for x on the denominator whereas coffeym has substituted for x-a and x-b - which makes more sense.

    I think sort of fiddled around abit and got a pretty nasty expression - it would've worked if I'd changed my cos^2 into sin^2 and such would've had all the individual alphas and betas being subtracted from each other.
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    (Original post by DFranklin)
    No.

    \int_0^2 t \, dt = \int_0^2 1\, dt but this doesn't mean t = 1.

    On the other hand, you can correctly argue that the -t^2 integral is negative and the other integral is positive, so they can't be equal. But even then, all you are really saying is "these two integrals aren't the same, so something must have gone wrong". To get the marks, I think you need to identify that something.

    ok, fair enough.

    but your evidence that when the original expression is integrated by parts, two distinct equations are produced (and added), one of which is the second integrand and hence the two cannot be equivalent would be fine for all the marks, right?

    thanks for the help with this btw, it's helped clear up some confusion i had with this kind of question
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    (Original post by kabbers)
    but your evidence that when the original expression is integrated by parts, two distinct equations are produced (and added), one of which is the second integrand and hence the two cannot be equivalent would be fine for all the marks, right?
    No, not really. The question isn't asking you "prove this method gives the wrong result", it's asking "what is wrong with this method?".

    Suppose I ask you "what is wrong with the following argument?":

    Suppose x = y. Then:

    x^2 = xy

    x^2-y^2= xy-y^2

    (x+y)(x-y) = y(x-y)

    (x+y) = y

    2x = y

    Taking the particular choice x=y=1, we find 2 = 1.

    If you say "the argument is wrong because 2 \neq 1", you won't get any marks. Instead you would need to say something like "You can't go from (x+y)(x-y) = y(x-y) to x+y = y, because you're dividing by (x-y) which is zero".

    To look at it another way, the question asks "explain why the substitution x=1/t does not show the integrals are equal" (emphasis mine). So if your explanation doesn't even mention the substitution x=1/t, then you're in trouble!

    In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to \int_0^1 \frac{1}{(1+x^2)^2}\, dx and \int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/t.

    This is somewhat belabouring the point for this particular question - I can't think the explanation would be worth many marks, and I'd expect you do get most of them for any explanation mentioning the behaviour of 1/x at x=0.
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    (Original post by DFranklin)
    In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to \int_0^1 \frac{1}{(1+x^2)^2}\, dx and \int_{-1}^1 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/t.
    You didn't happen to mean \int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx the second time?

    At least that was what I did in my working...
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    (Original post by nota bene)
    You didn't happen to mean \int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx the second time?
    Yes, thanks. Darned cut'n'paste errors...
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    (Original post by DFranklin)
    No, not really. The question isn't asking you "prove this method gives the wrong result", it's asking "what is wrong with this method?".

    Suppose I ask you "what is wrong with the following argument?":

    Suppose x = y. Then:

    x^2 = xy

    x^2-y^2= xy-y^2

    (x+y)(x-y) = y(x-y)

    (x+y) = y

    2x = y

    Taking the particular choice x=y=1, we find 2 = 1.

    If you say "the argument is wrong because 2 \neq 1", you won't get any marks. Instead you would need to say something like "You can't go from (x+y)(x-y) = y(x-y) to x+y = y, because you're dividing by (x-y) which is zero".

    To look at it another way, the question asks "explain why the substitution x=1/t does not show the integrals are equal" (emphasis mine). So if your explanation doesn't even mention the substitution x=1/t, then you're in trouble!

    In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to \int_0^1 \frac{1}{(1+x^2)^2}\, dx and \int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/t.

    This is somewhat belabouring the point for this particular question - I can't think the explanation would be worth many marks, and I'd expect you do get most of them for any explanation mentioning the behaviour of 1/x at x=0.
    right, ok.. i understand
    thanks!
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    Step II, Q1

     (\mathrm{i}) (x-\mathrm{a})^2 + (2x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = x^2 - 2\mathrm{a}x + \mathrm{a}^2 + 4x^2 - 4\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

                           = 14x^2 - x(2\mathrm{a} + 4\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

     = 14x^2 - x(14x) + \mathrm{a}^2 + \mathrm{b}^2  + \mathrm{c}^2 \,\, \mathrm{since} \,\,7x = \mathrm{a} + 2\mathrm{b} + 3\mathrm{c}

     = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2



     (\mathrm{ii}) (2x-\mathrm{a})^2 + (3x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = 4x^2 - 4\mathrm{a}x + \mathrm{a}^2 + 9x^2 - 6\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

                           = 22x^2 - x(4\mathrm{a} + 6\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

     = 22x^2 - x(22x) + \mathrm{a}^2 + \mathrm{b}^2  + \mathrm{c}^2 \,\, \mathrm{since} \,\,11x = 2\mathrm{a} + 3\mathrm{b} + 3\mathrm{c}

     = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

     \mathrm{if}\,\, \alpha\mathrm{a} + \beta\mathrm{b} + \gamma\mathrm{c} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2}x

    then  \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 = (\alpha x - \mathrm{a})^2 + (\beta x  - \mathrm{b})^2 + (\gamma x - \mathrm{c})^2

    Apologies if there are errors, rather rushed it.
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    Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

    And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?
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    (Original post by DFranklin)
    Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

    And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?
    OK, I think this is a good idea
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    So do I :P

    I have to pop out for abit then I'm going to try some more.
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    (Original post by DFranklin)
    Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?
    I don't think any of the new comers could do STEP III yet since most of us haven't done any of the further maths modules, especially not FP3.
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    Hey.

    I'm new to STEP. I'm trying to do STEP I/12. Any help would be great!

    So far, I have got some equations for the vertical collisions of the ball with the steps. But in order to find the coefficient of restitution, I'm thinking that I'd need some equations for the horizontal motion of the ball. This is what I'm struggling with. My current thinking is that surely the horizontal velocity of the ball will just remain constant.

    Here is my working so far.

    Let u_{n} be the initial velocity of collision n, v_{n} be the final velocity of collision n and e be the coefficient of restitution.

    Then for the first collision e u_{1} = v_{1}. However from conservation of energy u_{1} = \sqrt{2 g h} \Rightarrow e = \frac{v_{1}}{\sqrt{2 g h}}.

    For the second collision u_{2} = v_{1} + \sqrt{2 g h} = \sqrt{2 g h} (e + 1) and so we get v_{2} = e u_{2} = \sqrt{2 g h} (e) (e + 1).

    Unfortunately that's all I've got up to.
    I need some help in getting equations involving d which will give me more ammo for finding e.
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    (Original post by Dharma)
    Hey.

    I'm new to STEP. I'm trying to do STEP I/12. Any help would be great!

    So far, I have got some equations for the vertical collisions of the ball with the steps. But in order to find the coefficient of restitution, I'm thinking that I'd need some equations for the horizontal motion of the ball. This is what I'm struggling with. My current thinking is that surely the horizontal velocity of the ball will just remain constant.
    Yes, horizontal velocity is constant.

    I think the obvious thing to do next is to consider how long it is between the initial 'kick' and the first bounce, and how long between the first bounce and the 2nd bounce.
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    (Original post by Dharma)
    However from conservation of energy u_{1} = \sqrt{2 g h}
    Not really, there's an initial kinetic energy as well. (I assume your u, v are not just the vertical components of the velocity)

    For the second collision u_{2} = v_{1} + \sqrt{2 g h}
    Again, not really: what conservation of energy gives you is u_2^2 = v_1^2 + 2gh, which is not the same thing.

    (disclaimer: I have not yet solved the question)

    Question to anybody: Is the coefficient of restitution, in an oblique impact, equal to the ratio of the speeds after and before impact, or is it the ratio of the vertical (or in general perpendicular to the plane tangent to both the impacting objects at the point of impact at the moment of impact) components of the velocity?

    I don't know this and it seems kinda essential to solving the question *shrugs*
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    (Original post by ukgea)
    Not really, there's an initial kinetic energy as well. (I assume your u, v are not just the vertical components of the velocity)
    I was assuming they were, in which case
    what conservation of energy gives you is u_2^2 = v_1^2 + 2gh, which is not the same thing.
    is effectively the same thing - the K.E. due to the horizontal velocity is just a constant, so as you only care about the change in K.E. v.s. G.P.E. to work out the vertical velocity the end result is the same. (It's perhaps not totally obvious, but do these questions enough and you tend to get a bit lazy about showing this to be the case).

    Edit: ukgea is right. Ignore me about this bit...

    But yes, you are right, if u,v, aren't the vertical components, Dharma's equations are incorrect. Although I'd say the only sensible way to approach this question is to consider the vertical and horizontal motion separately.

    Question to anybody: Is the coefficient of restitution, in an oblique impact, equal to the ratio of the speeds after and before impact, or is it the ratio of the vertical (or in general perpendicular to the plane tangent to both the impacting objects at the point of impact at the moment of impact) components of the velocity?
    The latter. The normal (pun intended) approach is to resolve the velocity into components parallel and perpendicular to the surface normal. Only the component parallel to the normal is affected by the impact (neglecting friction, which of course you do...)
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    (Original post by DeathAwaitsU)
    I don't think any of the new comers could do STEP III yet since most of us haven't done any of the further maths modules, especially not FP3.
    Although true for the majority of the questions, I don't see STEP III, Q1 requires anything beyond GCSE maths, let alone FM. There are a few others that should also be reasonably accessible I think.
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    (Original post by DFranklin)
    in which caseis effectively the same thing - the K.E. due to the horizontal velocity is just a constant, so as you only care about the change in K.E. v.s. G.P.E. to work out the vertical velocity the end result is the same. (It's perhaps not totally obvious, but do these questions enough and you tend to get a bit lazy about showing this to be the case).
    This I still don't get. Perhaps I just haven't done these questions enough. But I get Conservation of energy:

    u_{2 x}^2 + u_{2y}^2 = v_{1 x}^2 + v_{1 y}^2 + 2gh

    and since horizontal velocity is constant they cancel:

    u_{2y}^2 = v_{1y}^2 + 2gh (1)

    . Now if

    u_{2y} = v_{1y} + \sqrt{2gh} (2)

    also were to hold, wouldn't we by squaring (2) and subtracting (1) get

    0 = 2v_{1y}\sqrt{2gh}

    which I can't see hold true in any case (obviously neither of the factors is zero). So what am I missing?

    The latter. The normal (pun intended) approach is to resolve the velocity into components parallel and perpendicular to the surface normal. Only the component parallel to the normal is affected by the impact (neglecting friction, which of course you do...)
    Okay, thanks *heads off to solve the question*
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    (Original post by DFranklin)
    Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

    And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?
    Could I just ask that, on a point of pragmatism, someone write up the solutions in LaTeX once they're done? We're currently putting the solutions onto the wiki (link will be provided as soon as the damn thing is unfrozen) and it's much more difficult if we have to essentially re-write them ourselves.
 
 
 
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