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STEP 2017 Solutions

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Reply 40
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Zacken
OP
22
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.
Reply 41
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Zacken
OP
22
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.
Q10 anybdody maybe please?
Reply 43
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S2M
22
Original post by dididid
Q10 anybdody maybe please?


Go on page 1. You should see a link to Q10 or a solution posted there.
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
Reply 45
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S2M
22
Original post by dididid
ooh just seen it thanks


No problem. :smile:
Reply 46
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math42
20
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..
Original post by 1 8 13 20 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 :smile:
[QUOTE="FractalSteinway;72020702"]Here's my solution to q10.

Question 10

Part (i)

For the first collision between PP and P1P_1, by conservation of momentum mu=mv+λmu1mu = m v + \lambda m u_1 where vv is the final velocity of particle PP. Also, the law of restitution gives e=u1−vue = \frac{u_1 - v}{u}. So, eliminating vv,
mu−λmu1m=u1−eu\frac{mu - \lambda m u_1}{m} = u_1 - eu
  ⟹  u+eu=u1+λu1\implies u + eu = u_1 + \lambda u_1
  ⟹  u1=1+e1+λu\implies u_1 = \frac{1+e}{1+ \lambda} u
as desired. Now for every subsequent collision the ratio of masses is still λ\lambda and the coefficient of restitution is still ee, so the same equations hold and
un=1+e1+λun−1u_n = \frac{1+e}{1+\lambda}u_{n-1}
for every nn. Iterating this gives un=(1+e1+λ)nuu_n = \big(\frac{1+e}{1+\lambda}\big)^n u.

Now eliminating u1u_1 instead from the first two equations, we have
mu−mvλm=eu+v\frac{mu - mv}{\lambda m} = eu + v
  ⟹  u−v=λeu+λv\implies u - v = \lambda e u + \lambda v
  ⟹  v=1−λe1+λu\implies v = \frac{1 - \lambda e}{1 + \lambda} u
and so similarly, for every n n we know
vn=1−λe1+λunv_n = \frac{1 - \lambda e}{1 + \lambda} u_n
  ⟹  vn=1−λe1+λ(1+e1+λ)nu.\implies v_n = \frac{1 - \lambda e}{1 + \lambda} \big(\frac{1+e}{1+\lambda}\big)^n u.

Part (ii)

If e>λe > \lambda then 1+e1+λ>1\frac{1+e}{1+\lambda} > 1 and so un>un−1u_n > u_{n-1} for every nn. Since vnv_n and unu_n are proportional, this implies vn>vn−1v_n > v_{n-1} for every nn, meaning that after its first two collisions every particle (with velocity vnv_n) moves forwards faster than the one preceding it, so no further collisions can occur.

Part (iii)

If e=λe = \lambda then 1+e1+λ=1\frac{1+e}{1+\lambda} = 1 and so un=un−1=uu_n = u_{n-1} = u for every nn. Therefore also vn=vv_n = v for all nn, and so the final velocities of all the particles are equal. In this case, the total fractional kinetic energy change η\eta after the nnth collision is
η=ΔEKinitial EK=12mu2−12m(v2+λv12+λ2v22+⋯+λn−1vn−12+λnun2)12mu2\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,
η=1−v2+λv12+λ2v22+⋯+λn−1vn−12+λnun2u2\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=λe = \lambda and vn=vv_n = v and un=uu_n = u for every nn,
Unparseable latex formula:

\eta = 1 - \frac{v^2(1 + e + e^2 + \cdots + e^{n-1}) + \e^n u^2}{u^2}


Unparseable latex formula:

\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}


  ⟹  η=1−en−(1−e21+e)2(1+e+e2+⋯+en−1)\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:
  ⟹  η=1−en−(1−e21+e)2(en−1e−1).\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:
η=1−en−((1−e)(1+e)1+e)2(en−1e−1)\eta = 1 - e^n - \left(\frac{(1-e)(1+e)}{1+e}\right)^2\left(\frac{e^n-1}{e-1}\right)
  ⟹  η=1−en−(1−e)2(en−1e−1)=1−en−(e−1)(en−1).\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<10 < e < 1,
lim⁡n→∞η=1−0−(e−1)(0−1)=1+(e−1)=e\displaystyle \lim_{n \to \infty} \eta = 1 - 0 - (e-1)(0-1) = 1 + (e-1) = e
and we're done.

Part (iv)

If λe=1\lambda e = 1 then 1−λe1+λ=0\frac{1 - \lambda e}{1 + \lambda} = 0 and so vn=0v_n = 0 for every nn. This means in every collision the first particle comes to rest. So, in this case, returning to our previous expression,
η=1−0+λnun2u2=1−λn(1+e1+λ)2nu2u2\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}
  ⟹  η=1−λn(1+e1+λ)2n=1−((λ+λe)(1+e)(1+λ)2)n\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 λe=1\lambda e = 1,
  ⟹  η=1−(1+e1+λ)n.\implies \eta = 1 - \big(\frac{1+e}{1+\lambda}\big)^n.

This is the final fractional KE loss after
Unparseable latex formula:

n [\latex] collisions.


thanks for the solution :smile:, how many marks do you reckon id get if i got all the same anwsers as you apart from the last 2 energy parts.
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?
Anyone done Q9? Just interested in the last part mainly. Thanks in advance :smile:

Also how many marks just for the proof by induction in Q8?
(edited 6 years ago)
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.
(edited 6 years ago)
Reply 52
Avatar for Zacken
Zacken
OP
22
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.
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?
Reply 54
Avatar for Zacken
Zacken
OP
22
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).
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...
Question 9

Part (i)

Applying the constant acceleration formulae horizontally and vertically, we have y=usin⁡αt−12gt2y = u \sin \alpha t - \frac{1}{2}gt^2 and x=ucos⁡αtx = u\cos\alpha t where xx and yy are the horizontal and vertical displacements respectively of the particle from point OO at time tt. The latter gives t=xucos⁡αt = \frac{x}{u\cos\alpha} which when substituted into the first equation yields
y=xtan⁡α−gx22u2cos⁡2α.y = x \tan \alpha - \frac{gx^2}{2u^2\cos^2\alpha}.
We know the particle passes through point PP, so (d,dtan⁡β)(d,d\tan\beta) must be a solution. Hence,
dtan⁡β=dtan⁡α−gd22u2cos⁡2αd\tan\beta = d\tan\alpha - \frac{gd^2}{2u^2\cos^2\alpha}
and so as d≠0d \ne 0 for the situation to be physical,
tanβ=tan⁡α−gd2u2cos⁡2α.tan\beta = \tan\alpha - \frac{gd}{2u^2\cos^2\alpha}.
Differentiating implicitly with respect to α\alpha with d,βd,\beta held constant,
0=sec⁡2α−(−2)gd2(ucos⁡α)−3⋅(dudαcos⁡α−usin⁡α).0 = \sec^2\alpha - (-2)\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(\frac{du}{d\alpha}\cos\alpha - u\sin\alpha).
Now as uu is minimised over α\alpha, we set dudα=0\frac{du}{d\alpha} = 0 (we will not be required to show this is a minimum) and so,
0=sec⁡2α+2gd2(ucos⁡α)−3⋅(−usin⁡α)0 = \sec^2\alpha + 2\frac{gd}{2}(u\cos\alpha)^{-3}\cdot(-u\sin\alpha)
  ⟹  0=1cos⁡2α−gdusin⁡αu3cos⁡3α\implies 0 = \frac{1}{\cos^2\alpha} - \frac{gdu\sin\alpha}{u^3\cos^3\alpha}
  ⟹  gdsin⁡αu2cos⁡α=1\implies \frac{gd\sin\alpha}{u^2\cos\alpha} = 1
  ⟹  u2=gdtan⁡α\implies u^2 = gd\tan\alpha
as was to be shown.

Now substituting this back into our initial path equation,
tanβ=tan⁡α−gd2(gdtan⁡α)cos⁡2αtan\beta = \tan\alpha - \frac{gd}{2(gd\tan\alpha)\cos^2\alpha}
  ⟹  tan⁡β=tan⁡α−12sin⁡αcos⁡α\implies \tan\beta = \tan\alpha - \frac{1}{2\sin\alpha\cos\alpha}
  ⟹  tan⁡β=sin⁡αcos⁡α−12sin⁡αcos⁡α\implies \tan\beta = \frac{\sin\alpha}{\cos\alpha} - \frac{1}{2\sin\alpha\cos\alpha}
  ⟹  tan⁡β=2sin⁡2α−12sin⁡αcos⁡α.\implies \tan\beta = \frac{2\sin^2\alpha - 1}{2\sin\alpha\cos\alpha}.
and since sin⁡2x≡2sin⁡xcos⁡x\sin 2x \equiv 2\sin x \cos x, and cos⁡2x≡cos⁡2x−sin⁡2x≡1−2sin⁡2x\cos 2x \equiv \cos^2 x - \sin^2 x \equiv 1 - 2 \sin^2 x,
tan⁡β=−cos⁡2αsin⁡2α=−cot⁡2α\tan\beta = \frac{- \cos 2\alpha}{\sin 2\alpha} = -\cot 2\alpha
and since the tangent function is odd,
tan⁡(−β)=cot⁡2α.\tan(-\beta) = \cot 2\alpha.
But by the definition of the cotangent function, cot⁡x≡tan⁡(π2−x)\cot x \equiv \tan (\frac{\pi}{2} - x) and so
tan⁡(−β)=tan⁡(π2−2α\tan(-\beta) = \tan(\frac{\pi}{2} - 2 \alpha
and since the angles involved are acute,
−β=π2−2α-\beta = \frac{\pi}{2} - 2\alpha
  ⟹  2α=β+π2\implies 2\alpha = \beta + \frac{\pi}{2}
as desired.

Part (ii)

Let γ\gamma be the angle asked for. So, the gradient of the path's curve at point PP must be tan⁡γ\tan \gamma. Differentiating our initial path equation with respect to xx to find this gradient (holding all other values constant),
dydx=tan⁡α−2gx2u2cos⁡2α\frac{dy}{dx} = \tan\alpha - \frac{2gx}{2u^2\cos^2\alpha}
and so at point PP where x=dx=d,
tan⁡γ=tan⁡α−2gd2u2cos⁡2α\tan\gamma = \tan\alpha - \frac{2gd}{2u^2\cos^2\alpha}
and therefore substituting in u2=gdtan⁡αu^2=gd\tan\alpha,
tan⁡γ=tan⁡α−2gd2gdtan⁡αcos⁡2α\tan\gamma = \tan\alpha - \frac{2gd}{2gd\tan\alpha\cos^2\alpha}
  ⟹  tan⁡γ=sin⁡αcos⁡α−1sin⁡αcos⁡α=sin⁡2α−1sin⁡αcos⁡α.\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⁡2x+sin⁡2x≡1\cos^2 x + \sin^2 x \equiv 1,
tan⁡γ=−cos⁡2αsin⁡αcos⁡α\tan\gamma = \frac{-\cos^2\alpha}{\sin\alpha\cos\alpha}
  ⟹  tan⁡γ=−cos⁡αsin⁡α=−cot⁡α\implies \tan\gamma = -\frac{\cos\alpha}{\sin\alpha} = -\cot\alpha
and so by the same reasoning as before,
tan⁡(−γ)=tan⁡(π2−α)\tan(-\gamma) = \tan(\frac{\pi}{2} - \alpha)
  ⟹  γ=α−π2\implies \gamma = \alpha - \frac{\pi}{2}.
(edited 6 years ago)
Original post by dididid
thanks for the solution :smile:, 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.
Reply 58
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Zacken
OP
22
Original post by LaurenLovesMaths
Also how many marks just for the proof by induction in Q8?


Probably 3-5.
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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