STEP 2013 Solutions Watch

mikelbird
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#41
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#41
Just a quick note about the end (part (iii)) of question 7.

in part (i) you have the solution x=17,y=12

using part (ii) this gives m=8,n=6 which means 6^2=1/2x8x(8+1) which is exactly what you are looking for...giving you the solution a=28,b=8,c=6
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Jkn
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(Original post by cpdavis)
Why yes, I knew I forgot something will update solution later today in addition to the original OP. and will also stick in a sketch too

also, did someone do a complete solution to Q3? I've not had a chance to look at solutions as I'm finishing off my work for supervisor :ahee:
Cheers :ahee:
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Slowbro93
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#43
OP updated! Only missing solution is 3
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mikelbird
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Will this do for Q3??
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mathsymathsy
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(Original post by DJMayes)
Damn. When I did the question I missed the "hence find". However, I had x as 3 and y=2 as my first solution, so would this likely lose marks?
Maybe just 1 mark? since it's very easy to deduce another solution. dw, you will still get the S :rolleyes:
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talstactician
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What time are we allowed to discuss/post solutions for STEP I? Also if we post a solution, does it have be done using LaTeX? I have never used it before, but I'd like to post my solution to STEP I Q1 on here...


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DJMayes
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(Original post by talstactician)
What time are we allowed to discuss/post solutions for STEP I? Also if we post a solution, does it have be done using LaTeX? I have never used it before, but I'd like to post my solution to STEP I Q1 on here...


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No discussion until 1 pm tomorrow.
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talstactician
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(Original post by DJMayes)
No discussion until 1 pm tomorrow.
Ok thanks, and is it ok if my solution isn't done in LaTeX?


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Slowbro93
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(Original post by talstactician)
Ok thanks, and is it ok if my solution isn't done in LaTeX?


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That all depends on how clear your handwriting is
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talstactician
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(Original post by cpdavis)
That all depends on how clear your handwriting is
I was going to write out my solution manually on TSR, rather than upload a link to my working on paper, but I can scan it in if you'd like


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DJMayes
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(Original post by talstactician)
Ok thanks, and is it ok if my solution isn't done in LaTeX?


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I think it's generally preferred in LateX, but you can still post it up raw and hopefully someone will Tex it for you.
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HClO
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STEP I 2013 Q12

(i)
Spoiler:
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Part I
\mathrm{Total\ number\ of\ tablets} = kn = 9


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) =^3P_3 \times \frac{3}{9} \times \frac{3}{8} \times \frac{3}{7}


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) = 3!\times \frac{1}{3} \times \frac{3}{8} \times \frac{3}{7}


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) = \frac{9}{28}
Part II
\mathrm{Total\ number\ of \ possibilities} =^9C_3 = 84
\mathrm{P(one\ tablet\ of\ each\ type\ left}) = \frac{[^3C_1]^3}{84}= \frac{27}{84}= \frac{9}{28}

(ii)
Spoiler:
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\mathrm{Total\ number\ of\ tablets} = kn = 3n


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) =^3\mathrm{P}_3 \times \frac{n}{3n} \times \frac{n}{3n-1} \times \frac{n}{3n-2}


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) = \frac{2n^2}{(3n-1)(3n-2)}

(iii)
Spoiler:
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\mathrm{Total\ number\ of\ tablets} = kn = 2n


P(\mathrm{one\ of\ each \ type\ on\ the\ first\ day}) =^2\mathrm{P}_2 \times \frac{n}{2n} \times \frac{n}{2n-1}


P(\mathrm{one\ of\ each \ type\ on\ the\ second\ day}) =^2\mathrm{P}_2 \times \frac{n-1}{2n-2} \times \frac{n-1}{2n-3}
\cdot \cdot \cdot


P(\mathrm{one\ of\ each \ type\ on\ the\ last\ day}) =^2\mathrm{P}_2 \times \frac{1}{2} \times \frac{1}{1}


P(\mathrm{one\ of\ each \ type\ every\ day}) =2^n \times \frac{n}{2n} \times \frac{n}{2n-1}\times \frac{n-1}{2n-2} \times \frac{n-1}{2n-3} \times \cdots\times \frac{1}{2} \times \frac{1}{1}
=2^n \times \frac{(n!)^2}{(2n)!}
\mathrm{Use\ the\ given\ approximation} \ n! \approx (\frac{n}{e})^n \times \sqrt {2 \pi n}
P(\mathrm{one\ of\ each \ type\ every\ day})\approx 2^n \times \frac{(\frac{n}{e})^{2n} \times 2 \pi n}{(\frac{2n}{e})^{2n} \times \sqrt {4 \pi n}} =2^{-n} \sqrt {\pi n}
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DJMayes
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Mechanics section for STEP 1:

Q9 – STEP I

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Firstly, set up some equations:

 x_1 = ucos \alpha t ,  y_1 = usin\alpha t - \frac{1}{2} gt^2 (Particle A)

 x_2 = vsin \alpha t ,  y_2 = vsin\alpha t - \frac{1}{2} gt^2 (Particle B)

We know that the particles both collide at the same time and at their greatest height, from which we can deduce that  usin\alpha = vsin\beta

Now, let’s apply conservation of linear momentum to the collision. As they both return to their starting points, their velocities after the collision must be equal and opposite to their velocities before the collision:

 mucos\alpha –Mvcos\beta = Mvcos \beta – mucos\alpha

 2mu cos \alpha = 2Mv cos \beta

 mu cos\alpha =M v cos \beta

Now, using our equation linking u, v, and the sines:

 mucos\alpha = \dfrac{M u sin \alpha cos \beta }{sin \beta }

Which gives us  m cot \alpha = M cot \beta , as required.

For the distance, we know that the ratio of b to d is going to be the same as the ratio of the speed of A to the speed of approach of the two particles:

 \dfrac{b}{d} = \dfrac{ucos\alpha }{ucos \alpha + v cos \beta }

 b = \dfrac{u cot\alpha }{ u cot\alpha + \frac{vcos\beta }}{ sin \alpha }

 b = \dfrac{ \frac{ M cot \beta }{m} }{ \frac{ M cot \beta }{m} + \frac{vcos\beta }{ usin \alpha }}

 b = \dfrac{M}{M + m \frac{v sin \beta }{u sin \alpha }}

From the very beginning, we know the vertical velocities are the same, so the nested fraction is just one, giving us our required answer for b.

Finally, let’s look back at our initial equations:

 b = ucos \alpha t ,  y_1 = usin\alpha t - \frac{1}{2} gt^2

Substitute the first into the second by eliminating t:

 y = b tan \alpha - \frac{gb^2}{2u^2cos^2 \alpha }

We need to eliminate u from this. Remember we are considering the maximum height, at which time vertical velocity is 0, giving:

 0 = usin \alpha  - gt so  t = \frac{usin \alpha }{g}

 b = ucos \alpha t

 b = \frac{u^2sin \alpha cos \alpha t}{g} so  u^2 = \frac{bg}{sin \alpha cos \alpha }

Plugging this into the expression for y, we get:

 y = b tan \alpha - \frac{gb^2}{2cos^2 \alpha }\frac{sin \alpha cos \alpha }{bg}

Which simplifies to  y =\frac{1}{2} b tan \alpha



Q10 – STEP 1

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Resolving, the frictional force acting on the puck will be  \mu mg where m is the mass of the puck, and so the work done in getting from barrier to barrier is  \mu mgd . Now, start after the ith collision:

Energy –  \frac{1}{2}mv_i^2

When it reaches the other barrier, it has KE of  \frac{1}{2}mv_i^2 - \mu mgd

In the collision with the barrier, the fraction of KE lost is  (1-e^2) x KE, and so we have:

 frac{1}{2}mv_{i+1}^2 = r^2 (\frac{1}{2}mv_i^2 - \mu mgd )

This then re-arranges to  v_{i+1}^2 – r^2v_i^2 = -2r^2\mu gd , which is the required result.

For the second bit, look at several successive collisions:

 v_{n}^2 = r^2 v_{n-1}^2 -2r^2\mu gd

 v_{n}^2 = r^2(= r^2 v_{n-2}^2 -2r^2\mu gd) -2r^2\mu gd

As you take more terms, it becomes apparent that the final velocity is a combination of a power and a sum of a geometric progression:

 ] v_{n}^2 = r^{2n} v^2 - \dfrac{2r^2\mu gd(1-r^{2n})}{1-r^2}

The puck comes to rest when v = 0:

 r^{2n}(1-r^2)v^2 – 2r^2 \mu gd + 2r^{2(n+1)} \mu gd = 0

Divide through in order to get in terms of k, and get all powers of n on their own:

 r^{2n} ((1-r^2)k + r^2) = r^2

Take logs:

 2n lnr + ln(((1-r^2)k + r^2) = 2lnr

Which re-arranges to give  n = 1 - \frac{ln(((1-r^2)k + r^2)}{2lnr}

Now, plug in  r = e^{-1} :

 n = 1- \frac{ln(((e^2-1)k + 1)-lne^2}{-2lne}

Which then simplifies to give  n = \frac{1}{2} ln(1+k(e^2 -1 )) , the required result.

Finally, if r=1 no kinetic energy is lost in collisions with barriers and n is then simply the kinetic energy divided by the work done against friction moving between two points, which is simply k.



Q11 – STEP I

Spoiler:
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(A good diagram helps here; I have omitted one but it would be nice if someone could add one)

Let R be the reaction force between the block and the ground, and assume friction is limiting. Then, resolve upwards:

 R + Tsin\alpha cos \beta + Tcos \alpha sin \beta = W

 R + Tsin(\alpha + \beta ) = W so  R = W - Tsin(\alpha + \beta )

(First of all it is useful to note that this is why the inequality regarding W and T is given; without it reaction forces would be negative if the block remained in equilibrium which is nonsensical, so the condition is required for equilibrium.)

Next, resolve horizontally:

 Tcos \alpha cos \beta = T sin \alpha sin \beta + \mu (W - Tsin(\alpha + \beta ))

 T  cos ( \alpha + \beta ) + \mu T sin (\alpha + \beta ) = \mu W

This looks somewhat similar to the expected result, so lets try work towards it. We are given that  tan \lambda = \mu , so we can calculate the sines and cosines of lambda (By right angled triangles to be:

 sin \lambda = \frac{\mu }{\sqrt{1+\mu ^2 }} ,  cos \lambda = \frac{1}{\sqrt{1+\mu ^2 } }

What this immediately suggests is to divide through by the rooted term to get the sine function next to the W:

 \dfrac{ ] T  cos ( \alpha + \beta ) + \mu T sin (\alpha + \beta ) }{ \sqrt{1+\mu ^2 }} = W sin \lambda

And now applying the forms for sine and cosine lambda we worked out earlier as well as the addition formulae, we get the result:

 Wsin \lambda  = T cos(\alpha + \beta - \lambda

Now, this is when friction is limiting, and so for the block to remain at rest the given inequality is achieved.

For the second part, resolve for both horizontal and vertical components of acceleration (Call them x and y):

 my = T sin 2 \phi  – T tan \phi

 mx = Tcos 2 \phi

The tangent of the direction of motion of the block is then given by y/x:

 tan \theta  = \dfrac{ sin 2 \phi  - tan \phi }{ cos 2 \phi }

 = \dfrac{ 2tan \phi – sec^2 \phi tan \phi }{1 – tan^2 \phi }

 = \dfrac{tan \phi + tan\phi (1-sec^2 \phi )}{1- tan^2 \phi }

 = \dfrac{ tan \phi (1- tan^2 \phi )}{1 – tan^2 \phi }

 = tan \phi so the direction of motion makes that angle with the horizontal, as required.

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Mastermind2
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STEP I 2013 Question 7:
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7.i) y = ux ---> dy/dx = xdu/dx + u
Substitute into DE to give xdu/dx + u = u + 1/u ---> Int(u)du = Int(1/x)dx ---> (1/2)u^2 = lnx + c. Now substitute u = y/x back in, to give: (1/2)(y^2/x^2) = lnx + c. Substituting the initial conditions gives c = 2, and we get y = xroot(4 + 2lnx), as required.

ii) Using the same substitution, we get: xdu/dx + u = 1/u + 2u ---> xdu/dx = (u^2 + 1)/u ---> 1/2Int(2u/(u^2 + 1))du = Int(1/x)dx ---> (1/2)ln(u^2 + 1) = lnx + c. Substituting u = y/x back in gives: (1/2)ln((y^2/x^2)+1) = lnx + c. Substituting the initial conditions gives c = (1/2)ln5, and we get ln((y^2 + x^2)/x^4) = ln5 ---> y^2 + x^2 = 5x^4 ---> y = xroot(5x^2 -1)

iii) We need a different substitution for this part. Let y = (1/2)ux^2 ---> dy/dx = ux + (1/2)x^2du/dx. Substituting this into the DE gives: (1/2)x^2du/dx = 2/u ---> Int(u)du = Int(4x^-2)dx ---> (1/2)u^2 = -4/x + c. Substituting the initial conditions gives: c = 12. So we have: 2(y^2)/(x^4) = 12 - 4/x ---> y = x^2root(6 - 2/x)
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Vaz_Tê
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I used the substitution y=(x^2)(u) and it worked, though I might have made a mistake. I'll have to see the paper again.
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MathsNerd1
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(Original post by Vaz_Tê)
I used the substitution y=(x^2)(u) and it worked, though I might have made a mistake. I'll have to see the paper again.
I also used this solution so I hope its not wrong
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Mastermind2
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(Original post by MathsNerd1)
I also used this solution so I hope its not wrong
Did you get the same final solution as me?
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MathsNerd1
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(Original post by Mastermind2)
Did you get the same final solution as me?
I got my value of c being 6 so I don't think it was the same at all :-/ I'm really doubting my performance in this paper now
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Mastermind2
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(Original post by MathsNerd1)
I got my value of c being 6 so I don't think it was the same at all :-/ I'm really doubting my performance in this paper now
I don't remember what I did in the exam tbh, my answer might have been different to what I put here.
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GeneralOJB
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What do you think the boundaries will be for STEP 1? I thought it was a really nice paper, managed to get five and a half solutions. Would be really nice to get a grade 1 in lower 6th.
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