STEP maths I, II, III 1992 solutions
Maths and statistics discussion, revision, exam and homework help.
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Re: STEP maths I, II, III 1992 solutionsSTEP 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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Re: STEP maths I, II, III 1992 solutionsSTEP III 1992 Question 6
Using the factor formulae, i.e. sin(A+B)+sin(A-B)=2sinAcosB
as required.
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}](http://www.thestudentroom.co.uk/latexrender/pictures/b3/b3f034da7491b3cad4ce3a358af5adbc.png "\\=\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
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
Last edited by khaixiang; 24-05-2007 at 19:05. -
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Re: STEP maths I, II, III 1992 solutionsgeneralebriety, I think my solution to III/2 is okay (in post 6), have you missed that one
? (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).Last edited by nota bene; 24-05-2007 at 20:59. -
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Re: STEP maths I, II, III 1992 solutionsNice 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.(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
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Re: STEP maths I, II, III 1992 solutionsAh, I didn't see your solution because I assumed new solutions would only be in new posts.(Original post by nota bene)
generalebriety, I think my solution to III/2 is okay (in post 6), have you missed that one ? (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). 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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Re: STEP maths I, II, III 1992 solutionsSTEP 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))Last edited by *bobo*; 24-05-2007 at 21:43. -
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Re: STEP maths I, II, III 1992 solutionsEither that or a mistake, in which case there will have been an erratum notice, which we don't have access to...(Original post by *bobo*)
a trick question?!
Otherwise, m3 = m1 + m2 would suffice for the second part of the question.Last edited by generalebriety; 24-05-2007 at 22:19. -
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Re: STEP maths I, II, III 1992 solutionsI 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.(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. -
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Re: STEP maths I, II, III 1992 solutionsThe 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.(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.
Tension rather than extension is a good thought though.
Think I'll have a go at that question.
Last edited by generalebriety; 24-05-2007 at 23:08. -
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Re: STEP maths I, II, III 1992 solutionsBut 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).(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.
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Re: STEP maths I, II, III 1992 solutionsAs always, you're right, of course.(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). Even so, it was a fairly trivial question. So, here we go:
II/14.
Last edited by generalebriety; 25-05-2007 at 22:16. -
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Re: STEP maths I, II, III 1992 solutionsTrying 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 \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}.](http://www.thestudentroom.co.uk/latexrender/pictures/02/028c28cd6c93237782f132d1eb553adb.png "\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...
Last edited by generalebriety; 25-05-2007 at 02:56. -
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Re: STEP maths I, II, III 1992 solutionsWell, I knew it would be short, but you really did a number on it!(Original post by generalebriety)
As always, you're right, of course. Even so, it was a fairly trivial question.
(There do seem to be some very short mechanics questions on these older papers). -
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Re: STEP maths I, II, III 1992 solutionsII/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 .](http://www.thestudentroom.co.uk/latexrender/pictures/d7/d7606c3a1ceb0eb2f63a21c476bb288f.png "\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 .")
Last edited by generalebriety; 25-05-2007 at 04:05. -
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Re: STEP maths I, II, III 1992 solutionsSTEP 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
Thanks for letting me know.
a trick question?!
![\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} .](http://www.thestudentroom.co.uk/latexrender/pictures/25/25efa3a103f38916814eedf8354f85c0.png "\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} .](http://www.thestudentroom.co.uk/latexrender/pictures/e4/e4c17b05d5cf5946c8f5c7b744b78d51.png "\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} .")
![\overrightarrow{RY} = k_3[\mathbf{a} -\mu\mathbf{b} ]\\
\overrightarrow{AY} = k_3\mathbf{a} + \mu 91-k_3)\mathbf{b} .](http://www.thestudentroom.co.uk/latexrender/pictures/6a/6abe5da13fa5e923115c052b0099a22a.png "\overrightarrow{RY} = k_3[\mathbf{a} -\mu\mathbf{b} ]\\
\overrightarrow{AY} = k_3\mathbf{a} + \mu 91-k_3)\mathbf{b} .")
