Doing II/10.
(i) arg(z1)  arg z = alpha
(ii) arg(z1)  arg z = alpha  arg(1)
arg(1z)  arg z = alpha
(iii) z1 = z
Stuck on (iv)... anyone?
STEP maths I, II, III 1992 solutions
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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 
STEP III 1992 Question 6
Using the factor formulae, i.e. sin(A+B)+sin(AB)=2sinAcosB
as required.
Applying Integration by Parts for the second integral and rewriting the first integral using the factor formula,
This is beautifully surprising!
Area under curve C is I_n
This is a very nice question. But the latexing is as timeconsuming as ever. And I must get started on my chemistry revision now 
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). 
(Original post by khaixiang)
This is a very nice question. But the latexing is as timeconsuming as ever. And I must get started on my chemistry revision now 
(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). 
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(3sin^2(theta))=2agsin^3(theta)
v^2= (2agsin^3(theta)/(3(3sin^2(theta)))
iii)Rmgsin(theta)= maw^2
w^2=9v^2/(a^2sin^2(theta))= (6gsin(theta))/(a(3sin^2(theta))
=>R= mgsin(theta)(9sin^2(theta))/(3sin^2(theta)) 
Remark on II/14:
"A smooth light inextensible string runs over P_{1}, under P_{3} and over P_{2}, as shown.
[...]
Find the extension in the string."
Erm? 
a trick question?!

(Original post by *bobo*)
a trick question?!
Otherwise, m3 = m1 + m2 would suffice for the second part of the question. 
(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. 
(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. 
(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. 
(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).
II/14.

(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). 
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.

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( 4k(pi))/3
R(1)= mg(7kpi)/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 
for some reason my computer won't let me access paper III anymore, says its damaged?
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