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STEP maths I, II, III 1991 solutions

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    also the first part of paper III number 14. Again can anyone finish it?
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  1. File Type: pdf 1991PAPERIII.14(part).pdf (38.3 KB, 31 views)
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    (Original post by brianeverit)
    Paper III number 13. (first part)
    Would someone like to finish it please?
    I'm only going to do a sketch, because I don't think this is very difficult and you should be able to manage it.

    If the particle hits completely inelastically, then the vertical velocity immediately after the collision is 0. So the time taken to reach the ground is \displaystyle \sqrt{\frac{2h}{g}}. The horizontal velocity is unchanged at \sqrt{8gh} \cos \alpha and so the distance travelled after the collision with the ceiling is 4h \cos \alpha. Obviously the distance travelled before the collision is the same as in the elastic collision case.

    After a bit of algebra grinding, you end up with \displaystyle D = 4h \cos \alpha \left[ 2 \sin \alpha - \sqrt{4 \sin^2\alpha-1} - 1\right]

    From here, just differentiate in the obvious way and group terms and you should be able to get something manageable. You end up with a term involving \displaystyle\sqrt{4 \sin^2 \alpha - 1} and one involving \displaystyle \frac{1}{\sqrt{4 \sin^2 \alpha - 1}} and I found it helpful to rewrite the 1st term as  \displaystyle \frac{4 \sin^2 \alpha - 1}{\sqrt{4 \sin^2 \alpha - 1}} so those terms could be grouped.

    In my working, I did actually get to where you can see that \pi/4 would be a root for the derivative, but you can always substitute \sin \alpha = \cos \alpha = \pi /4 to show you get 0.

    I also got to the point where you could demonstrate algebraically that D'(\pi/4 - \epsilon) > 0, D'(\pi/4+\epsilon) < 0 and so we definitely have a maximum. If you had to, you could always demonstrate this using a calculator, as they were allowed back then.
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    (Original post by brianeverit)
    also the first part of paper III number 14. Again can anyone finish it?
    I think there's a mistake inbetween:

    \displaystyle T=\frac{4}{5}mg +\frac{9}{25}ml\frac{5g}{2a}
    and
    \displaystyle T=\frac{4}{5}mg +\frac{9}{10}mgl

    Shouldn't it be \displaystyle  T=\frac{4}{5}mg +\frac{9}{10}mg \frac{l}{a}?
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    (Original post by DFranklin)
    I think there's a mistake inbetween:

    \displaystyle T=\frac{4}{5}mg +\frac{9}{25}ml\frac{5g}{2a}
    and
    \displaystyle T=\frac{4}{5}mg +\frac{9}{10}mgl

    Shouldn't it be \displaystyle  T=\frac{4}{5}mg +\frac{9}{10}mg \frac{l}{a}?
    Yes, a silly little slip but the later working is correct.
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    (Original post by DFranklin)
    I'm only going to do a sketch, because I don't think this is very difficult and you should be able to manage it.

    If the particle hits completely inelastically, then the vertical velocity immediately after the collision is 0. So the time taken to reach the ground is \displaystyle \sqrt{\frac{2h}{g}}. The horizontal velocity is unchanged at \sqrt{8gh} \cos \alpha and so the distance travelled after the collision with the ceiling is 4h \cos \alpha. Obviously the distance travelled before the collision is the same as in the elastic collision case.

    After a bit of algebra grinding, you end up with \displaystyle D = 4h \cos \alpha \left[ 2 \sin \alpha - \sqrt{4 \sin^2\alpha-1} - 1\right]

    From here, just differentiate in the obvious way and group terms and you should be able to get something manageable. You end up with a term involving \displaystyle\sqrt{4 \sin^2 \alpha - 1} and one involving \displaystyle \frac{1}{\sqrt{4 \sin^2 \alpha - 1}} and I found it helpful to rewrite the 1st term as  \displaystyle \frac{4 \sin^2 \alpha - 1}{\sqrt{4 \sin^2 \alpha - 1}} so those terms could be grouped.

    In my working, I did actually get to where you can see that \pi/4 would be a root for the derivative, but you can always substitute \sin \alpha = \cos \alpha = \pi /4 to show you get 0.

    I also got to the point where you could demonstrate algebraically that D'(\pi/4 - \epsilon) > 0, D'(\pi/4+\epsilon) < 0 and so we definitely have a maximum. If you had to, you could always demonstrate this using a calculator, as they were allowed back then.
    Would it be sufficient to argue that the difference in ranges is zero when particle just touches ceiling and also when \alpha=\pi/2 so
    \alpha=\pi/4 must be a maximum.
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    OK, I wasn't sure if that was what was stopping you.

    From there, what I'd do is resolve horizontally and vertically for the forces at A, splitting the rod reaction into R_x and R_y. Then you know 13R_x=21R_y, which is enough to let you solve for T, and from that, solve for k.

    Edit: crossing of posts doesn't make it terribly obvious, but this is in response to Q14,
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    Paper III number 11. Could someone check it please?
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  2. File Type: pdf 1991PAPERIII.11pdf.pdf (54.3 KB, 29 views)
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    (Original post by brianeverit)
    Would it be sufficient to argue that the difference in ranges is zero when particle just touches ceiling and also when \alpha=\pi/2 so
    \alpha=\pi/4 must be a maximum.
    Not to my eyes, no - there's no reason to expect the different to behave 'symmetrically'.
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    (Original post by DFranklin)
    Not to my eyes, no - there's no reason to expect the different to behave 'symmetrically'.
    But there must be a maximum in between the two zeros and as this is the only time the derivative is zero surely we can conclude that it is a maximum.
    It does not require any assumption of symmetry.
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    (Original post by brianeverit)
    But there must be a maximum in between the two zeros and as this is the only time the derivative is zero surely we can conclude that it is a maximum.
    Sorry, I misunderstood you. I thought you were claiming the derivative must be zero at pi/4 simply because the difference was zero at the point where it touches the ceiling and when theta = pi/2.

    What you are saying is fine if you can show pi/4 is the only place where the derivative is zero. At the point I got to, it was equally easy just to show the derivative had the "/-\" behaviour near pi/4 (which I tend to prefer as being a more "direct", though it doesn't really matter).
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    (Original post by DFranklin)
    OK, I wasn't sure if that was what was stopping you.

    From there, what I'd do is resolve horizontally and vertically for the forces at A, splitting the rod reaction into R_x and R_y. Then you know 13R_x=21R_y, which is enough to let you solve for T, and from that, solve for k.

    Edit: crossing of posts doesn't make it terribly obvious, but this is in response to Q14,
    o.k. so here is the complete solution for III/14
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  3. File Type: pdf 1991PAPERIII.14.pdf (41.9 KB, 38 views)
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    I don't think anyone has posted a solution to this question yet, I had a quick look and couldn't find anything, but sorry I someone has already done it .

    STEP III - Question 8

    I_{k} = \displaystyle\int^{\theta}_0 \cos^{k}(x)\cos(kx) \, dx

     = \displaystyle\int^{\theta}_0 \cos^{k-1}(x)\cos(kx)\cos(x) \, dx

    = \displaystyle\int^{\theta}_0 \cos^{k-1}(x)[ \frac{1}{2}\cos(k+1)x + \frac{1}{2}\cos(k-1)x ] \, dx

     = \frac{1}{2}\displaystyle\int^{\t  heta}_0 \cos^{k-1}\cos(k+1)x \, dx + \frac{1}{2}\displaystyle\int^{\t  heta}_0 \cos^{k-1}(x)\cos(k-1)x ] \, dx

     = \frac{1}{2}I_{k-1} + \frac{1}{2}\displaystyle\int^{\t  heta}_0 cos^{k-1}(x)[ \cos(kx)\cos(x) - \sin(kx)\sin(x) ] \, dx

    = \frac{1}{2}I_{k-1} + \frac{1}{2}I_{k} - \frac{1}{2}\displaystyle\int^{\t  heta}_0 cos^{k-1}(x)\sin(kx)\sin(x) ] \, dx

    = \frac{1}{2}I_{k-1} + \frac{1}{2}I_{k} + \frac{1}{2}[\frac{\cos^{k}(x)}{k}\sin(kx)]^{\theta}_{0} - \frac{1}{2}\displaystyle\int^{\t  heta}_0 cos^{k}(x)\cos(x) ] \, dx

    = \frac{1}{2}I_{k-1} + \frac{1}{2k}\cos^{k}(\theta)\sin  (k\theta).

    \Rightarrow 2kI_{k} = kI_{k-1} + cos^{k}(\theta)\sin(k\theta) (*) (By rearranging etc.)

    b)  1 + m\cos(2\theta) + \displaystyle \binom{m}{2}\cos(4\theta) + ....+ \cos(2m\theta)

    = \Re[ 1 + me^{2\theta i} + \displaystyle \binom{m}{2}e^{4\theta i} + ....+ e^{2m\theta i} ]

    = \Re(1 + e^{2\theta i})^{m} = \Re[\frac{e^{-\theta i} + e^{\theta i}}{e^{-\theta i}}]^{m}

    = \Re(e^{m\theta i}[2\cos(\theta)}^{m}])

    = \Re[ (\cos(m\theta) + i\sin(m\theta))2^{m}\cos^{m}(\th  eta)]

    = \cos(m\theta)2^{m}\cos^{m}(\thet  a). Q.E.D.

    c)

     \frac{m\sin(2\theta)}{2} + \displaystyle \binom{m}{2}\frac{\sin(4\theta)}  {4} + .......+ \frac{\sin(2m\theta)}{2m}

    = \displaystyle\int^{\theta}_{0} 1 + m\cos(2\theta) + \displaystyle \binom{m}{2}\cos(4\theta) + ....+ \cos(2m\theta) \, d\theta - \theta

     = \displaystyle\int^{\theta}_{0} 2^{m}\cos(m\theta)\cos^{m} \, d\theta  -\theta

     = 2^{m}\displaystyle\int^{\theta}_  {0} I_{m} \, d\theta  -\theta

     = 2^{m-1}I_{m-1} + \frac{2^{m-1}}{m}\cos^{m}(\theta)\sin(m\the  ta) - \theta . (using (*)).

     = 2^{0}I_{0} + \cos(\theta)\sin(\theta) + \cos^{2}(\theta)\sin(\theta) + .......+ \frac{2^{m-1}}{m}\cos^{m}(\theta)\sin(m\the  ta) - \theta . (Using (*) repeatedly)

     =  \cos(\theta)\sin(\theta) + \cos^{2}(\theta)\sin(\theta) + .......+ \frac{2^{m-1}}{m}\cos^{m}(\theta)\sin(m\the  ta)
    (Evaluating I_{0} gives \theta, which cancels with the other one.)
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    squeezebox, i think you've done II/8 instead of III/8

    anyway, i've copied your idea and here's III/8 :p:

    there might be some mistakes, so please point them out if you spot them !



    a) Define I = \int^{\pi}_0 x f(\sin x) {\mathrm{d}}x

    substitute y = k-x (so that x = k-y) and dy/dx = -1

    = -\int^{k-\pi}_k (k-y) f(\sin (k-y)) {\mathrm{d}}y

    switching the limits,

    = \int^{k}_{k-\pi} (k-y) f(\sin (k-y)) {\mathrm{d}}y

    the integral turns out into the form f(sin (x)), so guess that sin(k-y) = sin(x) and hence that k = pi


    = \int^{\pi}_{0} (\pi-y) f(\sin (\pi-y)) {\mathrm{d}}y

    \sin(\theta) = \sin(\pi - \theta), so


    = \int^{\pi}_{0} (\pi-y) f(\sin (y)) {\mathrm{d}}y

    = \int^{\pi}_{0} \pi f(\sin (y)) {\mathrm{d}}y - \int^{\pi}_{0} y f(\sin (y)) {\mathrm{d}}y


    I = \int^{\pi}_0 x f(\sin x) {\mathrm{d}}x = \int^{\pi}_{0} y f(\sin (y)) {\mathrm{d}}y, so

    2I = \int^{\pi}_{0} \pi f(\sin (y)) {\mathrm{d}}y


    sin(y) is symmetrical about pi/2 for 0 <= y <= pi, so

    I = \pi \int^{\pi/2}_{0} f(\sin (y)) {\mathrm{d}}y as required.



    Evaluate \int^{\pi}_0 \frac{x}{2 + \sin x} {\mathrm{d}}x

    by the above argument, letting f(x) = 1/(2+x)

    = \pi \int^{\pi/2}_0 \frac{1}{2 + \sin x} {\mathrm{d}}x

    now substitute t = tan(x/2), so

    \sin x = 2\sin(x/2)\sin(x/2) = 2\tan(x/2)\sin^2(x/2) = 2t/\sec^2 x = \frac{2t}{1+t^2}

    dt/dx = \frac{\sec^2(x/2)}{2} = \frac{1+t^2}{2}, so

    dx/dt = \frac{2}{1+t^2}



    \pi \int^{\pi/2}_0 \frac{1}{2 + \sin x} {\mathrm{d}}x

    = \pi \int^{1}_0 \frac{\frac{2}{1+t^2}}{2 + \frac{2t}{1+t^2}} {\mathrm{d}}t

    = \pi \int^{1}_0 \frac{2}{2(1+t^2) + 2t} {\mathrm{d}}t

    = \pi \int^{1}_0 \frac{1}{t^2 + t + 1} {\mathrm{d}}t

    = \pi \int^{1}_0 \frac{1}{(t+\frac{1}{2})^2 + \frac{3}{4}} {\mathrm{d}}t

    let t + 1/2 = \frac{\sqrt 3}{2} \tan \theta

    dt/d{\theta} = \frac{\sqrt 3}{2} \sec^2 \theta


    so \pi \int^{1}_0 \frac{1}{(t+\frac{1}{2})^2 + \frac{3}{4}} {\mathrm{d}}t

    =\pi \int^{\pi/3}_{\pi/6} \frac{\frac{\sqrt 3}{2} \sec^2}{\frac{3}{4}\tan^2 \theta + \frac{3}{4}} {\mathrm{d}}\theta

    =\pi \frac{\frac{\sqrt 3}{2}}{\frac{3}{4}} \int^{\pi/3}_{\pi/6} \frac{\sec^2}{\tan^2 \theta + 1} {\mathrm{d}}\theta

    =\pi \frac{2}{\sqrt{3}} \int^{\pi/3}_{\pi/6} 1 {\mathrm{d}}\theta

    =\pi \frac{2}{\sqrt{3}} [\theta]^{\pi/3}_{\pi/6}

    =\pi \frac{2}{\sqrt{3}}\pi/6

    =\pi^2 \frac{1}{3\sqrt{3}}



    b) evaluate \int^1_0 \frac{(\sin^{-1}t)\cos[(\sin^{-1}t)^2]}{\sqrt{1-t^2}} {\mathrm{d}}t

    let t = sin x

    dt/dx = cos x

    = \int^{\pi/2}_0 \frac{x \cos[x^2]}{\sqrt{1-\sin^2}} \cos x {\mathrm{d}}x

    = \int^{\pi/2}_0 \frac{x \cos[x^2]}{\cos x} \cos x {\mathrm{d}}x

    = \int^{\pi/2}_0 x \cos[x^2]{\mathrm{d}}x

    = \frac{1}{2}\int^{\pi/2}_0 2x \cos[x^2]{\mathrm{d}}x

    = \frac{1}{2}[\sin(x^2)]^{\pi/2}_0

    = \frac{1}{2} \sin(\frac{\pi^2}{4})
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    Really? , I thought Further Maths B was STEP III?

    Oh yeah, I'm a good trend setter :p:
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    Ahh, I see now - on some paper sites the older STEP papers are mislabeled. I think I'm correct in thinking that you have just done STEP I question 8 :p:.
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    (Original post by squeezebox)
    Ahh, I see now - on some paper sites the older STEP papers are mislabeled. I think I'm correct in thinking that you have just done STEP I question 8 :p:.
    aahh ok sorry about that :p: i thought it seemed like an easier question, just imagined it would be III because of the t substitution.. oh well, now i know at the very least i got the question right !
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    Just had a look at STEP I Q2 - what am I missing?

    S_U= 4\pi(2R)^2=16\piR^2 \newline S_L=4\pi(3R)^2=36\pi R^2 \newline V_U=\frac{4\pi(2R)^3}{3}=\frac{3  2\pi R^3}{3} \newline V_L=\frac{4\pi (3R)^3}{3}=36\pi R \newline \frac{dV_T}{dt}=kR^252\pi
    _T means total, _U means upper _L means lower.

    Initial volume is \frac{140\pi R^3}{3}
    Volume when height is halved is (denoted by V') V'_T=\frac{35\pi R^3}{6} So \frac{V_T}{V'_T}=\frac{\pi R^3 140}{3}\frac{ 6}{35\pi R^3}=\frac{56}{7}
    The ratio is 56:7

    \frac{dV}{dS}=\frac{dV}{dR}\frac  {dR}{dS}=140\pi R^2\frac{1}{104\pi R}=\frac{140}{104}R=\frac{140}{1  04}\sqrt{\frac{S}{4\pi}}

    \frac{d^2V}{dS^2}=\frac{70}{208}  \sqrt{\pi}\frac{1}{\sqrt{S}}
    (this is >0 so a min? this is not what is wanted)

    I've got a bit more of working on a few side-tracks but none of which seemed to lead me anywhere good. Any ideas what I'm missing?
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    I've just looked at this really quickly (I'm off to college soon), but I think there is something wrong with your first bit: I don't think that the volume halves when the snowman is halve its initial height (which from your working it seems to). I think you need to find differential equations for the radii of the head and the body, to find the time at which the snowman is half its height. Ive got to go now, but my answer for the ratio is:

    Spoiler:
    Show
    \frac{37}{224}
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    I don't think it's asking for the volume to halve (at least not on the version of the paper that I have). It says "When Frosty is half his initial height, find the ratio of his volume to his initial volume". And half height should mean half radius if the snowballs are right on top of each other. Or am I being stupid? (pretty likely!)

    I don't know if it's relevant but somewhere in my working I got t=\frac{1}{8}(\frac{3V_U}{4\pi})  ^{\frac{2}{3}}+c (after integrating the DE for the volume of the upper sphere). This could possibly help to determine at which time the upper snowball melts completely - but the problem is I don't see how to find the constant, hence this bit is completely useless atm.
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    I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.

    When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.

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