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    (Original post by Rishabh_01)
    You will get full marks as it stated that prove by contradiction or otherwise
    Yes, but the point of the question is whether that's a valid proof... it seems very handwavy but then STEP has almost no rigour, so I don't know.
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    (Original post by peskid)
    They say students who have finished four questions reasonably well will be given a grade 1.
    This used to be true - it's not so much so anymore, at least not on STEP 1.

    Do you think I can get by with a grade 2?
    20 + 13 + 16 + ? + 10 + 3 -> should be a reasonably high 2 or a low 1.
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    Q10 anybdody maybe please?
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    (Original post by dididid)
    Q10 anybdody maybe please?
    Go on page 1. You should see a link to Q10 or a solution posted there.
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    (Original post by S2M)
    Go on page 1. You should see a link to Q10 or a solution posted there.
    ooh just seen it thanks
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    (Original post by dididid)
    ooh just seen it thanks
    No problem.
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    Man 12 is easy. Might do it at some point if nobody else does soon. I feel like people should be more clued in to how nice the Probability is on average; remember when I sat it in 2015 I did a great deal of one of the prob questions in 10 minutes at the end. At a glance Q13 looks a lot more interesting so if I can get that I'll be more liable to put one up, although I dunno how useful it will be considering barely anybody does them..
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    (Original post by 13 1 20 8 42)
    Man 12 is easy. Might do it at some point if nobody else does soon. I feel like people should be more clued in to how nice the Probability is on average; remember when I sat it in 2015 I did a great deal of one of the prob questions in 10 minutes at the end. At a glance Q13 looks a lot more interesting so if I can get that I'll be more liable to put one up, although I dunno how useful it will be considering barely anybody does them..
    would be interesting to see your solution none the less
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    (Original post by FractalSteinway)
    Here's my solution to q10.

    Question 10

    Part (i)

    For the first collision between P and P_1, by conservation of momentum mu = m v + \lambda m u_1 where v is the final velocity of particle P. Also, the law of restitution gives e = \frac{u_1 - v}{u}. So, eliminating v,
    \frac{mu - \lambda m u_1}{m} = u_1 - eu
    \implies u + eu = u_1 + \lambda u_1
    \implies u_1 = \frac{1+e}{1+ \lambda} u
    as desired. Now for every subsequent collision the ratio of masses is still \lambda and the coefficient of restitution is still e, so the same equations hold and
    u_n = \frac{1+e}{1+\lambda}u_{n-1}
    for every n. Iterating this gives u_n = \big(\frac{1+e}{1+\lambda}\big)^  n u.

    Now eliminating u_1 instead from the first two equations, we have
    \frac{mu - mv}{\lambda m} = eu + v
    \implies u - v = \lambda e u + \lambda v
    \implies v = \frac{1 - \lambda e}{1 + \lambda} u
    and so similarly, for every  n we know
    v_n = \frac{1 - \lambda e}{1 + \lambda} u_n
    \implies v_n = \frac{1 - \lambda e}{1 + \lambda}  \big(\frac{1+e}{1+\lambda}\big)^  n u.

    Part (ii)

    If e > \lambda then \frac{1+e}{1+\lambda} > 1 and so u_n > u_{n-1} for every n. Since v_n and u_n are proportional, this implies v_n > v_{n-1} for every n, meaning that after its first two collisions every particle (with velocity v_n) moves forwards faster than the one preceding it, so no further collisions can occur.

    Part (iii)

    If e = \lambda then \frac{1+e}{1+\lambda} = 1 and so u_n = u_{n-1} = u for every n. Therefore also v_n = v for all n, and so the final velocities of all the particles are equal. In this case, the total fractional kinetic energy change \eta after the nth collision is
    \eta = \frac{\Delta E_K}{\textrm{initial } E_K} = \frac{\frac{1}{2}mu^2 - \frac{1}{2}m \left( v^2 + \lambda v_1^2 + \lambda^2 v_2^2 + \cdots + \lambda^{n-1} v_{n-1}^2 + \lambda^n u_n^2 \right)}{\frac{1}{2}mu^2}
    as the last particle will still not have collided for the second time. So,
    \eta = 1 - \frac{v^2 + \lambda v_1^2 + \lambda^2 v_2^2 + \cdots + \lambda^{n-1} v_{n-1}^2 + \lambda^n u_n^2}{u^2}
    but since e = \lambda and v_n = v and u_n = u for every n,
    \eta = 1 - \frac{v^2(1 + e + e^2 + \cdots + e^{n-1}) + \e^n u^2}{u^2}
    \implies \eta = 1 - \frac{\left(\frac{1 - \lambda e}{1 + \lambda}\right)^2u^2(1 + e + e^2 + \cdots + e^{n-1}) + \e^n u^2}{u^2}
    \implies \eta = 1 - e^n - \left(\frac{1 - e^2}{1 + e}\right)^2 (1 + e + e^2 + \cdots + e^{n-1})
    and this geometric sequence simplifies nicely:
    \implies \eta = 1 - e^n - \left(\frac{1 - e^2}{1 + e}\right)^2\left(\frac{e^n-1}{e-1}\right).
    Taking the difference of two squares:
    \eta = 1 - e^n - \left(\frac{(1-e)(1+e)}{1+e}\right)^2\left(\fra  c{e^n-1}{e-1}\right)
    \implies \eta = 1 - e^n - (1-e)^2 \left(\frac{e^n-1}{e-1}\right) = 1 - e^n - (e-1)(e^n-1).
    So, as 0 < e < 1,
    \displaystyle \lim_{n \to \infty} \eta = 1 - 0  - (e-1)(0-1) = 1 + (e-1) = e
    and we're done.

    Part (iv)

    If \lambda e = 1 then \frac{1 - \lambda e}{1 + \lambda} = 0 and so v_n = 0 for every n. This means in every collision the first particle comes to rest. So, in this case, returning to our previous expression,
    \eta = 1 - \frac{0 + \lambda^n u_n^2}{u^2} = 1 - \frac{\lambda^n \big(\frac{1+e}{1+\lambda}\big)^  {2n} u^2}{u^2}
    \implies \eta = 1 - \lambda^n \big(\frac{1+e}{1+\lambda}\big)^  {2n} = 1 - \left(\frac{(\lambda + \lambda e)(1 + e)}{(1 + \lambda)^2} \right)^n
    so as \lambda e = 1,
    \implies \eta = 1 - \big(\frac{1+e}{1+\lambda}\big)^  n.

    This is the final fractional KE loss after [latex] n [\latex] collisions.
    thanks for the solution , how many marks do you reckon id get if i got all the same anwsers as you apart from the last 2 energy parts.
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    Hi ppl
    I got all of Q3 Q4
    All of Q2 except for that (x/y is less than 1 and it still holds) bit
    First two integrals of Q1 (embarrassing)
    (i) half proved(ii) and (iv) of Q10
    Proved (i) and (ii) and obtained the three equations of the parameters of the line in Q6

    Do I have a shot at 1?
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    Anyone done Q9? Just interested in the last part mainly. Thanks in advance

    Also how many marks just for the proof by induction in Q8?
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    I swear they stole Q6 from a Putnam competition lol.
    Nice looking paper though.
    EDIT: Lol @ Q13, STEP examiners having way too much fun again. Almost as bad as the 'crusty bread' question from long ago.
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    (Original post by IrrationalRoot)
    Lol @ Q13, STEP examiners having way too much fun again. Almost as bad as the 'crusty bread' question from long ago.
    It gets even worse at Tripos! We had something about Davros the Dalek overlord jumping around Skaro in his rocket last year in our dynamics and relativity exam.
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    (Original post by Zacken)
    It gets even worse at Tripos! We had something about Davros the Dalek overlord jumping around Skaro in his rocket last year in our dynamics and relativity exam.
    Oh dear... was that all taking place on the flat planet Zog?
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    (Original post by IrrationalRoot)
    Oh dear... was that all taking place on the flat planet Zog?
    Not quite, but you'll be pleased to find that spaceships do random landing on Zog (12F).
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    (Original post by Zacken)
    Not quite, but you'll be pleased to find that spaceships do random landing on Zog (12F).
    Zog is one unstable planet, in just 5 years it transforms from a ball into a plane...
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    Question 9

    Part (i)

    Applying the constant acceleration formulae horizontally and vertically, we have y = u \sin \alpha t - \frac{1}{2}gt^2 and x = u\cos\alpha t where x and y are the horizontal and vertical displacements respectively of the particle from point O at time t. The latter gives t = \frac{x}{u\cos\alpha} which when substituted into the first equation yields
    y = x \tan \alpha - \frac{gx^2}{2u^2\cos^2\alpha}.
    We know the particle passes through point P, so (d,d\tan\beta) must be a solution. Hence,
    d\tan\beta = d\tan\alpha - \frac{gd^2}{2u^2\cos^2\alpha}
    and so as d \ne 0 for the situation to be physical,
    tan\beta = \tan\alpha - \frac{gd}{2u^2\cos^2\alpha}.
    Differentiating implicitly with respect to \alpha with d,\beta held constant,
    0 = \sec^2\alpha - (-2)\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(\frac{du}{d\alpha}\cos\a  lpha - u\sin\alpha).
    Now as u is minimised over \alpha, we set \frac{du}{d\alpha} = 0 (we will not be required to show this is a minimum) and so,
    0 = \sec^2\alpha + 2\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(-u\sin\alpha)
    \implies 0 = \frac{1}{\cos^2\alpha} - \frac{gdu\sin\alpha}{u^3\cos^3\a  lpha}
    \implies \frac{gd\sin\alpha}{u^2\cos\alph  a} = 1
    \implies u^2 = gd\tan\alpha
    as was to be shown.

    Now substituting this back into our initial path equation,
    tan\beta = \tan\alpha - \frac{gd}{2(gd\tan\alpha)\cos^2\  alpha}
    \implies \tan\beta = \tan\alpha - \frac{1}{2\sin\alpha\cos\alpha}
    \implies \tan\beta = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{2\sin\alpha\cos\alpha}
    \implies \tan\beta = \frac{2\sin^2\alpha - 1}{2\sin\alpha\cos\alpha}.
    and since \sin 2x \equiv 2\sin x \cos x, and \cos 2x \equiv \cos^2 x - \sin^2 x \equiv 1 - 2 \sin^2 x,
    \tan\beta = \frac{- \cos 2\alpha}{\sin 2\alpha} = -\cot 2\alpha
    and since the tangent function is odd,
    \tan(-\beta) = \cot 2\alpha.
    But by the definition of the cotangent function, \cot x \equiv \tan (\frac{\pi}{2} - x) and so
    \tan(-\beta) = \tan(\frac{\pi}{2} - 2 \alpha
    and since the angles involved are acute,
    -\beta = \frac{\pi}{2} - 2\alpha
    \implies 2\alpha = \beta + \frac{\pi}{2}
    as desired.

    Part (ii)

    Let \gamma be the angle asked for. So, the gradient of the path's curve at point P must be \tan \gamma. Differentiating our initial path equation with respect to x to find this gradient (holding all other values constant),
    \frac{dy}{dx} = \tan\alpha - \frac{2gx}{2u^2\cos^2\alpha}
    and so at point P where x=d,
    \tan\gamma = \tan\alpha - \frac{2gd}{2u^2\cos^2\alpha}
    and therefore substituting in u^2=gd\tan\alpha,
    \tan\gamma = \tan\alpha - \frac{2gd}{2gd\tan\alpha\cos^2\a  lpha}
    \implies \tan\gamma = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{\sin\alpha\cos\alpha} = \frac{\sin^2\alpha - 1}{\sin\alpha\cos\alpha}.
    And since \cos^2 x + \sin^2 x \equiv 1,
    \tan\gamma = \frac{-\cos^2\alpha}{\sin\alpha\cos\alp  ha}
    \implies \tan\gamma = -\frac{\cos\alpha}{\sin\alpha} = -\cot\alpha
    and so by the same reasoning as before,
    \tan(-\gamma) = \tan(\frac{\pi}{2} - \alpha)
    \implies \gamma = \alpha - \frac{\pi}{2}.
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    (Original post by dididid)
    thanks for the solution , how many marks do you reckon id get if i got all the same anwsers as you apart from the last 2 energy parts.
    Well I'm no better at predicting STEP mark schemes than anyone else but based on the difficulty of each part I'd guess the marks might be split 7/3/6/4 roughly. That being said the third part was just algebraically most dense so it might not be weighted as much as it seems like it should be.
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    (Original post by LaurenLovesMaths)
    Also how many marks just for the proof by induction in Q8?
    Probably 3-5.
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    (Original post by FractalSteinway)
    Well I'm no better at predicting STEP mark schemes than anyone else but based on the difficulty of each part I'd guess the marks might be split 7/3/6/4 roughly. That being said the third part was just algebraically most dense so it might not be weighted as much as it seems like it should be.
    oh thanks, hopefully its not then haha,
 
 
 
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