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STEP maths I, II, III 1992 solutions

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    Doing II/10.

    (i) arg(z-1) - arg z = alpha
    (ii) arg(z-1) - arg z = alpha - arg(-1)
    arg(1-z) - arg z = alpha
    (iii) |z-1| = |z|

    Stuck on (iv)... anyone?
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    STEP I
    9)dV/dt=-Ax
    V=(1/3)pi(ax/H)^2x= (1/3)(a/H)^2 pi x^3
    dV/dx=pi(ax/H)^2

    dx/dt= -(AH^2)/(pi xa^2)
    => 0.5x^2= -t(AH^2)/(pi a^2) + 0.5h^2 as x=h initially
    so when x=0:
    t=(pi a^2h^2)/(2AH^2)

    dV/dt=B- Ax
    if B>Ah, then the water level will rise until x=B/A
    if B<Ah then the water level will initially drop, and then remain at a level where x=B/A
    if B=Ah then the water level will remain at h
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    STEP III 1992 Question 6

    \\I_{n}=\int_{0}^{\pi} \frac{x\sin^{2}{nx}}{\sin^2{x}}d  x\\ I_{n}-I_{n-1}=\int_{0}^{\pi} \frac{x}{\sin^{2}{x}}\{\sin^{2}{  nx}-\sin^{2}{(nx-x)}\}dx\\=\int_{0}^{\pi} \frac{x}{\sin^{2}{x}}\{(\sin{nx}  +\sin{(n-1)x})(\sin{nx}-\sin{(n-1)x})\}dx
    Using the factor formulae, i.e. sin(A+B)+sin(A-B)=2sinAcosB
    \\=\int_{0}^{\pi} \frac{x}{\sin^{2}{x}}\{(2\sin{(n  x-x/2)}\cos{x})(2\sin{(x/2)}\cos{(nx-x/2)})\}dx \\=\int_{0}^{\pi} \frac{x}{\sin^{2}{x}}\{(2\sin{(x/2)}\cos{(x/2)})(2\sin{(nx-x/2)}\cos{(nx-x/2)})\}dx\\ =\int_{0}^{\pi} \frac{x}{\sin^{2}{x}}\{\sin{x}\s  in{(2n-1)x}\}dx\\ =\int_{0}^{\pi} \frac{x}{\sin{x}}\{\sin{(2n-1)x}\}dx\\=J_{n}
    as required.

    \\J_{n}=\int_{0}^{\pi} \frac{x\sin{(2nx-x)}}{\sin{x}}dx\\ =\int_{0}^{\pi} \frac{x(\sin{2nx}\cos{x}-\sin{x}\cos{2nx})}{\sin{x}}dx\\ =\int_{0}^{\pi} \frac{x\sin{2nx}\cos{x}}{\sin{x}  }dx-\int_{0}^{\pi} x\cos{2nx}dx
    Applying Integration by Parts for the second integral and rewriting the first integral using the factor formula,
    \\=\int_{0}^{\pi} \frac{x\sin{(2n+1)x}}{2\sin{x}}+  \frac{x\sin{(2n-1)x}}{2\sin{x}}dx-[\frac{x\sin{2nx}}{2n}]^{\pi}_{0}+\int^{\pi}_{0}\frac{\  sin{2nx}}{2n}dx\\ J_{n}=\frac{1}{2}J_{n+1}+\frac{1  }{2}J_{n}+\frac{1}{2n}[\frac{-\cos{2nx}}{2n}]^{\pi}_{0}\\ J_{n}=J_{n+1}
    This is beautifully surprising!

    Area under curve C is I_n
    \\I_{n}-I_{n-1}=J_{n}=J_{1}=\frac{\pi^2}{2}\\ I_{1}-I_{0}=\frac{\pi^2}{2}\\ I_{2}-I_{1}=\frac{\pi^2}{2}\\ I_{3}-I_{2}=\frac{\pi^2}{2}\\. \\. \\. \\I_{n}-I_{n-1}=\frac{\pi^2}{2}\\ \text{sum them together}\\ I_{n}-I_{0}=\frac{n\pi^2}{2}\\ I_{0}=0\\ I_{n}=\frac{n\pi^2}{2}\\ \text{as required}

    This is a very nice question. But the latexing is as time-consuming as ever. And I must get started on my chemistry revision now :p:
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    generalebriety, I think my solution to III/2 is okay (in post 6), have you missed that one:p: ? (also I/2 (post 36) is okay although Davids is neater).

    I'll attempt some statistics ones now but I'm not sure I'll be able to do them, doesn't seem to be any nice ones

    edit: coincidentally when looking at the statistics ones I saw that TheDuck's solution is of STEP I Q 15 (as the papers are mislabelled).
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    (Original post by khaixiang)
    This is a very nice question. But the latexing is as time-consuming as ever. And I must get started on my chemistry revision now :p:
    Nice solution! I started up on this question earlier and ended up with the desired reduction formulae for the first parts via. some short cuts, but I was attempting J_n reduction by a bit of a complicated method which would never have got me there. Well done. Also best of luck for chem exam.
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    (Original post by nota bene)
    generalebriety, I think my solution to III/2 is okay (in post 6), have you missed that one:p: ? (also I/2 (post 36) is okay although Davids is neater).

    I'll attempt some statistics ones now but I'm not sure I'll be able to do them, doesn't seem to be any nice ones

    edit: coincidentally when looking at the statistics ones I saw that TheDuck's solution is of STEP I Q 15 (as the papers are mislabelled).
    Ah, I didn't see your solution because I assumed new solutions would only be in new posts. :p: Sorry, it wasn't clear whether I/2 was ok. I'll keep the link to David's solution in there but will change the wording. Thanks for letting me know.
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    STEP III
    12)i)Conservation of momentum: (horizontal)
    w= angular speed, ( velocity=radiusxangular velocity)

    total mass of bowl=3m
    3mv= mwasin(theta)
    3v=awsin(theta)

    ii)v(p)=velocity of particle

    conservation of energy:
    -1.5mv^2 + 0.5mv(p)^2= magsin(theta) (v(p)=aw)
    -3v^2 + a^2w^2=2agsin(theta) substituing from part (i)
    -3v^2 + 9v^2/sin^2(theta)= 2agsin(theta)
    3v^2(3-sin^2(theta))=2agsin^3(theta)
    v^2= (2agsin^3(theta)/(3(3-sin^2(theta)))

    iii)R-mgsin(theta)= maw^2
    w^2=9v^2/(a^2sin^2(theta))= (6gsin(theta))/(a(3-sin^2(theta))
    =>R= mgsin(theta)(9-sin^2(theta))/(3-sin^2(theta))
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    Remark on II/14:
    "A smooth light inextensible string runs over P1, under P3 and over P2, as shown.
    [...]
    Find the extension in the string."

    Erm? :p:
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    a trick question?!
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    (Original post by *bobo*)
    a trick question?!
    Either that or a mistake, in which case there will have been an erratum notice, which we don't have access to...

    Otherwise, m3 = m1 + m2 would suffice for the second part of the question.
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    (Original post by generalebriety)
    Either that or a mistake, in which case there will have been an erratum notice, which we don't have access to...

    Otherwise, m3 = m1 + m2 would suffice for the second part of the question.
    I think it might be supposed to be "tension" rather than "extension". I'm not sure it's as trivial a question as it looks. I'm thinking it might be possible you can have a situation where m1=/=m2 (so the particles on the string are moving), but the middle pulley is still stationary.
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    (Original post by DFranklin)
    I think it might be supposed to be "tension" rather than "extension". I'm not sure it's as trivial a question as it looks. I'm thinking it might be possible you can have a situation where m1=/=m2 (so the particles on the string are moving), but the middle pulley is still stationary.
    The middle pulley would be stationary though, as you said. Although obviously this causes a slight problem when one of the particles falls over the top of a pulley - I did consider that in writing my last post but couldn't be bothered. The pulley still stays stationary. :p:

    Tension rather than extension is a good thought though.

    Think I'll have a go at that question. :p:
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    (Original post by generalebriety)
    The middle pulley would be stationary though, as you said. Although obviously this causes a slight problem when one of the particles falls over the top of a pulley - I did consider that in writing my last post but couldn't be bothered. The pulley still stays stationary. :p:
    But I think the m3=m1+m2 solution only works if you assume the m1,m2 are in equilibrium. (Which is obviously the case when m1=m2. Interestingly, the formula given in the question reduces to m3=m1+m2 in this case).
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    (Original post by DFranklin)
    But I think the m3=m1+m2 solution only works if you assume the m1,m2 are in equilibrium. (Which is obviously the case when m1=m2. Interestingly, the formula given in the question reduces to m3=m1+m2 in this case).
    As always, you're right, of course. Even so, it was a fairly trivial question. So, here we go:

    II/14.

    \displaystyle \text{Newton's second law on }P_2:\; m_2g-T=m_2a \\

\text{Newton's second law on }P_1:\; T-m_1g=m_1a\\

\text{Dividing, }\frac{m_2g-T}{T-m_1g} = \frac{m_2}{m_1}\\

m_1m_2g - m_1T = m_2T - m_1m_2g\\

T(m_1+m_2) = 2m_1m_2g\\

T = \frac{2m_1m_2g}{m_1+m_2}.\\

P_3 \text{ remains at rest if there is no acceleration} .\\

\Leftrightarrow m_3g = 2T.\\

m_3g = \frac{4m_1m_2g}{m_1+m_2}

\text{and result follows} .
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    Trying III/9. Absolute bastard of a question so far.

    Change axes. x- and y-axes are chosen to meet at S. In polar coordinates with the positive x-axis as initial line, P is the point (r, φ), where \displaystyle r = \frac{\l}{1+e\cos\theta},\; \phi = 45^\text{o} + \theta.
    \displaystyle \therefore r = \frac{\l}{1+e\cos (\phi -45^\text{o} )} = \frac{l}{\sqrt{2}} \cdot \frac{1}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi}.\\

\tfrac{l}{\sqrt{2}} \cdot x = \frac{\cos\phi}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi} \\

\tfrac{l}{\sqrt{2}} \dot{x} = \frac{-[(1/\sqrt{2}) + e\cos\phi + e\sin\phi ]\sin\phi - \cos\phi (e\cos\phi - e\sin\phi)}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} \\

=\frac{-\tfrac{1}{\sqrt{2}} \sin\phi - e}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} . \\

\tfrac{l}{\sqrt{2}} \cdot y = \frac{\sin\phi}{(1/\sqrt{2}) + e\cos\phi + e\sin\phi} \\

\tfrac{l}{\sqrt{2}} \dot{y} = \frac{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi ]\cos\phi - \sin\phi (e\cos\phi - e\sin\phi)}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} \\

=\frac{\tfrac{1}{\sqrt{2}} \cos\phi + e}{[(1/\sqrt{2}) + e\cos\phi + e\sin\phi]^2} . \\

\therefore \frac{\text{d}y}{\text{d}x} = \dot{y} / \dot{x} = \frac{\cos\phi + \sqrt{2} e}{-(\sin\phi + \sqrt{2} e)} .\\

\theta = \alpha \Rightarrow \phi = 45^\text{o} + \alpha \\

\therefore \bigg( \frac{\text{d}y}{\text{d}x} \bigg)_{\theta =\alpha} = \frac{\cos(45^\text{o} + \alpha) + \sqrt{2} e}{-(\sin(45^\text{o} + \alpha) + \sqrt{2} e)} \\

= \frac{\sqrt{2} \cos\alpha - \sqrt{2} \sin\alpha + \sqrt{2} e}{-(\sqrt{2} \sin\alpha + \sqrt{2} \cos\alpha + \sqrt{2} e)} = \frac{\sin\alpha - \cos\alpha - e}{\sin\alpha + \cos\alpha + e}.

    Hmm... what next...
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    (Original post by generalebriety)
    As always, you're right, of course. Even so, it was a fairly trivial question.
    Well, I knew it would be short, but you really did a number on it!

    (There do seem to be some very short mechanics questions on these older papers).
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    II/3.

    The circumference of the larger circle is four times the circumference of the smaller circle. Hence, the smaller circle rotates four times as quickly as OP. Let the (variable) point at which the two circles touch be Q, so that OQ and OP are collinear, with OQ = 4. Let the angle between QP and PB be θ; then θ = -4φ.

    \displaystyle x = OP\cos\phi + PB\cos (\phi +\theta ) = 3\cos\phi + \cos (-3\phi ) = 3\cos\phi + \cos 3\phi \\

y = OP\sin\phi + PB\sin (\phi +\theta ) = 3\sin\phi + \sin (-3\phi ) = 3\sin\phi - \sin 3\phi \\

\therefore (x, y) = (3\cos\phi + \cos 3\phi , 3\sin\phi - \sin 3\phi ).\\

A = \int_0^{2\pi} x \frac{\text{d}y}{\text{d}\phi} \text{d}\phi = \int_0^{2\pi} (3\cos\phi + \cos 3\phi )(3\cos\phi + 3\cos 3\phi ) \text{d}\phi \\

= \int_0^{2\pi} (9\cos^2 \phi - 6\cos\phi\cos 3\phi - 3\cos^2 3\phi) \text{d}\phi \\

= \frac{1}{2} \int_0^{2\pi} (9\cos 2\phi + 9 - 6\cos 2\phi - 6\cos 4\phi - 3\cos 6\phi - 3) \text{d}\phi \\

= \frac{1}{2} \left[6\phi + \tfrac{9}{2} \sin 2\phi - 3\sin 2\phi - \tfrac{3}{4} \sin 4\phi - \tfrac{1}{2} \sin 6\phi \right]_0^{2\pi} \\

= 3\left[\phi\right]_0^{2\pi} = 6\pi .
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    STEP III
    13)
    I= 0.5.4b^2m + m4b^2= 6mb^2
    conservation of energy:
    EPE= GPE+KE
    kgmb(pi)/2 = 2bmg + 0.5(6mb^2)w^2
    => w^2= g(k(pi)- 4)/6b
    so P does reach C as k>4/(pi) so w does exist

    force F acting at Q= mg(kpi- 4)/3 (F=mrw^2)= m2bw^2)
    reaction produced by axis on Q (R(1)):
    R(1)-mg= mg( 4-k(pi))/3
    R(1)= mg(7-kpi)/3

    reaction R(2) produced by axis to counter the weight of the circle= mg
    total upward vertical reaction produced by axis= R(1)+R(2)= mg(10- kpi)/3
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    III/7.

    The lines RB and CQ produced meet at X; the lines RB and DP meet at Y; the lines DP and CQ produced meet at Z.

    \displaystyle

\text{Let } \overrightarrow{AB} = \mathbf{a}, \overrightarrow{AD} = \mathbf{b} .\\

\overrightarrow{AP} = \lambda\mathbf{a} \\

\overrightarrow{AR} = \mu\mathbf{b}\\

\overrightarrow{AQ} = \lambda\mathbf{a} + \mu\mathbf{b}\\

\overrightarrow{AC} = \mathbf{a} + \mathbf{b}\\

\overrightarrow{CQ} = (\lambda -1)\mathbf{a} + (\mu -1)\mathbf{b}\

\overrightarrow{CX} = k_0 [(\lambda -1)\mathbf{a} + (\mu -1)\mathbf{b} ]\\

\therefore \overrightarrow{AX} = (1 + k_0 \lambda - k_0)\mathbf{a} + (1 + k_0\mu - k_0)\mathbf{b} .

    \overrightarrow{AB} = \mathbf{a}\\

\overrightarrow{RB} = \mathbf{a} - \mu\mathbf{b}\\

\overrightarrow{RX} = k_1 [\mathbf{a} - \mu\mathbf{b} ]\\

\overrightarrow{AX} = k_1\mathbf{a} + \mu (1-k_1)\mathbf{b} .

    \therefore k_1\mathbf{a} + \mu (1-k_1)\mathbf{b} = (1 + k_0 \lambda - k_0)\mathbf{a} + (1 + k_0\mu - k_0)\mathbf{b} \\

\text{Equating coefficients, }\\

1 + k_0 \lambda - k_0 = k_1 \Rightarrow k_0 = \frac{k_1 -1}{\lambda -1} \\

1 + k_0\mu - k_0 = \mu (1-k_1) \Rightarrow k_0 = \frac{\mu (1-k_1)-1}{\mu -1} \\

\therefore \frac{k_1 -1}{\lambda -1} = \frac{\mu (1-k_1)-1}{\mu -1} \\

(k_1 -1)(\mu -1) = (\mu (1-k_1)-1)(\lambda -1) \Rightarrow k_1 = \frac{\lambda (1-\mu )}{1-\lambda\mu} .

    \overrightarrow{DP} = \lambda\mathbf{a} - \mathbf{b} \\

\overrightarrow{DY} = k_2(\lambda\mathbf{a} - \mathbf{b} )\\

\overrightarrow{AY} = k_2\lambda\mathbf{a} + (1-k_2)\mathbf{b} .

    \overrightarrow{RY} = k_3[\mathbf{a} -\mu\mathbf{b} ]\\

\overrightarrow{AY} = k_3\mathbf{a} + \mu 91-k_3)\mathbf{b} .

    \text{Equating coefficients and solving simultaneously gives} \\

k_3 = \frac{\lambda (1-\mu )}{1-\lambda\mu} = k_1. \\

\therefore \overrightarrow{RX} = \overrightarrow{RY} \Rightarrow X \text{ and } Y \text{ are the same point} .\\

\Rightarrow Z \text{ must also be this point} .
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    for some reason my computer won't let me access paper III anymore, says its damaged? :confused:

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