Since most of the discussion has died down now and the most common thing being posted in those threads are now solutions, I thought I may as well make a thread for it. If you spot any mistakes, feel free to post in this thread and we'll make changes. Cheers!
STEP I Paper here. STEP II Paper here. STEP III paper here.
OK, let's finish STEP II. Here's Q11 (I hope. This was very fiddly even before I started Latexing it...)
To start, lets ignore the vertical dimension, and try to get a mental picture of where i and j go. It's not that hard. If we do the normal 2D thing of j being the y-axis and i being the x-axis, then OA lies on the Y axis, OC is at roughly 4 o'clock (i.e. x > 0, y < 0) and OB is at roughly 8 o'clock (i.e. x < 0, y < 0). As we progress, we're going to have quite a few possible sign ambiguities (basically we're going to want to work out a unit vector (sin t, cos t) knowing only what tan t is); this should be obvious given the diagram and a little thought.
(i) Let the unit vector in direction PB = ai + bj + ck. If we write h for the size of the horizontal component (so that h^2 = a^2 + b^2), then we have h^2+c^2 = 1 (since we have a unit vector), and also ch=tan(90−θ)=tanθ1=21. So h=2c and so h2+c2=21c2+c2=23c2.
So in fact we have c2=2/3 (so that c=32, since c > 0) and h2=1/3 so that a2+b2=1/3. Now consider a/b (keeping in mind OB is at roughly 8 o'clock). We have ab=tanθ=2, so b=a2, so a2+b2=3a2. So 3a2=1/3, so a=±31 and since "x < 0" we choose a = -1/3. Then b=a2=3−2.
Thus our unit vector is 3−1i−32j+32k as desired.
(ii) Having done (i), it's fairly clear the plan should be to find unit vectors for PA and PC also. The force corresponding to each string will just be U, V, W times the appropriate unit vector.
So again, write the unit vector in direction PA in the form ai + bj + ck. In this case we know a = 0 (since OA is in the direction of j), and we also know b=ctan30o=c/3. So 1=b2+c2=4c2/3, from which b=21,c=23, giving a final unit vector 21j+23k. Finally write the unit vector in direction PC in the form ai + bj + ck. Again, let h^2 = a^2 + b^2, then h=ctan60o=c3 and so c=21,h=23. Now we want to find a and b. We have a^2+b^2 = 3/4. Considering the position of C we have a > 0, b < 0 and ab=−tanϕ. So b=−atanϕ=4−a2=8−a. So b2+a2=8a2+a2=89a2. So a2=4398=32. So a=32,b=−382=23−1. Thus our final unit vector is 32i+23−1j+21k And of course the unit vector in the direction of W is just -k.
Dotting with i, we get: 3−U+3V2=0, and so U=3V32=V6. Dotting next with j we get: 2T−32U−231V=0. Replacing U with V6 we end up with T=V3. Finally dot with k to get: 2T3+3U2+2V=W. Replace U and T with the appropriate multiples of V and we find 23V+3V26+2V=W and so (23+2+21)V=W⟹V=4W. So finally T=43W,U=4W6=22W3.
(i) Can always give change unless the first person has a £2 coin. There are m+1 people, only one of which has a £2 coin. Therefore p = m/(m+1). (ii) As in (i), we fail if the first person has a £2 coin. This has probability 2/(m+2) We also fail if the first person has a £1 coin but the next 2 people have 2 pound coins. This has probability m+2mm+12m1=(m+1)(m+2)2. In all other cases we succeed. So p(fail) = m+22(1+m+11)=m+22m+1m+2=m+12. So p(succeed) = 1-p(fail) = m+1m−1 as desired. (iii) So now, the failure cases are: 2, any, ... with p = 3/(m+3) 1,2,2, any, with p = m+3mm+23m+12=m+36(m+2)(m+1)m. 1,1,2,2,2, any with p = m+3mm+2m−1m+13m2m−11=m+36(m+2)(m+1)1 1,2,1,2,2, any, with p = m+3mm+23m+1m−1m2m−11=m+36(m+2)(m+1)1 Adding these, p(fail) = m+33(1+(m+2)(m+1)2m+(m+2)(m+1)2+(m+2)(m+1)2) =m+33(1+(m+2)(m+1)2(m+2))=m+33(1+m+12)=m+33m+1m+3=m+13. Then p(succeed) = 1-p(fail) = m+1m−2 as desired.
STEP I question 10 Let collision occur at a height h above the ground Taking upwards as positive Let velocities of A and B at collision be uA and uB then, since both have same displacement from starting position uA2=2g(h−x) and uB2=2g(h−x)⟹uA=uB=u say If velocities after collision are vA and vB then by conservation of momentum Mu−mu=MvA+mvB and since collision is perfectly elastic vB−vA=2u⟹vB=2u+vA
vB=2u+M+m(M−3m)u which is clearly also positive so both particles move upwards after the collision considering motion of B before collision h−x=2gu2 considering motion of B after collision, if maximum height attained is y above point of collision y=2gvB2=2(M+m)2g(3M−m)2u2 so maximum height above the plane is2(M+m)2g(3M−m)2u2+h−gu2 =h+2g(M+m)2[(3M+m)2−(M+m)2]u2=h+2g(M+m)2(8M2−8Mn)u2=h+g(M+m)24M(M−m)u2 as required]
To do this equation, we need only the equation for the trajectory of the particle, in terms of t=tan(θ), x and the inital velocity, which I choose to be ums−2. We let t=tan(θ) as time.
It is quite easy to derive the equation of motion, which is:
y=xt−2u2gx2(1+t2). Now (d1,d2) and (d2,d1) both lie on the trajectory of the particle. Hence, on rearranging:
STEP III Q12 let D(k)[f(t)] denote the kth derivative of f(t) w.r.t. t. To find the expected value of Y we have to use the fact that E(Y)=D′[G(H(t))]∣t=1 so, using the chain rule: E(Y)=D′[G(H(t))]∣t=1=G′(H(t))H′(t)∣t=1=G′(H(1))H′(1)=G′(1)H′(1)=E(N)E(Xi) as required. Now, to find the variance: Var(Y)=E(Y2)−E(Y)2=E(Y2)−E(Y)+E(Y)−E(Y)2=E(Y(Y−1))+E(Y)−E(Y)2[br]=D′′[G(H(t))]∣t=1+D′[G(H(t))]∣t=1−(D′[G(H(t))]∣t=1)2][br]=[G′′(H(t))H′(t)2+H′′(t)G′(H(t))]∣t=1+E(N)E(Xi)−(E(n)E(Xi))2[br]=E(N(N−1))E(Xi)2+E(Xi(Xi−1))E(N)+E(N)E(Xi)−(E(n)E(Xi))2[br]=E(N2)E(Xi)2−E(Xi)2E(N)+E(Xi2)E(N)−(E(Xi)E(N))2[br]=Var(N)E(Xi)2+E(N)Var(Xi) as required. For the next part, notice the perfect fit between what the examiners have asked us to prove in the first part and the scenario of the second, the only thing left to see that we haven't been explicitly told is that Xi is a random variable denoting the outcome of the ith toss (1=heads, 0=tails). Using the first part, since G(t) is the pgf of N and H(t) is the pgf of Xi then the pgf of Y is G(H(f)) so we need to find H(t) and G(t): G=E(tN)=n=1∑∞P(N=n)tn=n=1∑∞(21)ntn=1−t/2t/2=2−tt[br]H(t)=E(tXi)=x=0∑1P(Xi=x)tx=21×1+21×t=21(1+t)[br]∴pgfY=G(H(t))=2(2−0.5(1+t))1+t=3−t1+t[br] To find the expected value: E(Y)=D′[3−t1+t]∣t=1=(3−t)24∣t=1=1 (You can alternatively do this by using E(Y)=E(N)E(Xi). I have done both methods and I got the same answer for both). To find the variance, use the first few steps in your derivation of Var(Y) in the first bit of the question to give yourself a shortcut where you can do it in terms of Y's pgf alone: Var(Y)=D′′[3−t1+t]∣t=1+D′[3−t1+t]∣t=1−(D′[3−t1+t]∣t=1)2=[4(−2)(3−t)−3(−1)]∣t=1+1−1=1. Now to to find P(Y=r). First, consider G(H(t)): G(H(t))=P(Y=1)t+P(Y=2)t2+...+P(Y=r)tr Our most immediate problem is the 't' term which makes it difficult to get the term we want on it's own so let's differentiate r times and then set t=0 to isolate the P(Y=r) term: D(r)[G(H(0))]=r!P(Y=r) which means we only have to divide both sides by r! to get the required result: P(Y=r)=r!F(r)[G(H(0))]=r!4r!(3)−(r+1)=4(3)−(r+1)
What do you guys reckon will be the question that the fewest number of people will attempt? After doing STEP III Q12 I have a feeling no-one will do that one. What modules are probability generating functions even on?
What do you guys reckon will be the question that the fewest number of people will attempt? After doing STEP III Q12 I have a feeling no-one will do that one. What modules are probability generating functions even on?
I didn't think STEP III Q12 was that bad, but for sure, not many will probably have covered PGFs. Generally no-one touches the "post-S2" probability questions, so I suspect you might be right.
STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).
STEP II, Q11 not only looks horrendous, but I found it to be pretty long and tricky too. I suspect it will be one of the least answered questions on STEP II.
STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).
This question is pretty tough. The fact that they've given so many 'show that's gives a good reflection of it's trickiness. I've given it a little thought so far and I'm already stuck on the magnitude of the couple, even though it feels like I shouldn't be. Probably has something to do with the fact that I don't know much of the stuff on M4 and thus know very little about couples. I should really be learning rotational dynamics properly first...
I didn't think STEP III Q12 was that bad, but for sure, not many will probably have covered PGFs. Generally no-one touches the "post-S2" probability questions, so I suspect you might be right.
STEP III, Q11 looks horrendous, but I suspect is not that bad (I am assuming that you get the right answer by assuming the vertical velocity = 0 at the point where the strings go taut - if that doesn't give the right answer then the difficulty has suddenly taken a quantum leap).
STEP II, Q11 not only looks horrendous, but I found it to be pretty long and tricky too. I suspect it will be one of the least answered questions on STEP II.
STEP III Q12 is definitely not necessarily 'hard' by STEP standards, I mean, I have only done S1 and had never heard of a pgf before today and I managed to get it out so it can't be that bad. On the other hand, I've been looking and I can't find generator functions anywhere on the whole edexcel syllabus which is a bit tragic for those on edexcel who made the effort to do up to S4. Did you do generator functions at a-level? OMG, I hadn't noticed STEP III Q11. That is a stomach wrenchingly horrible question. The amount of reading you have to do before you could even get into the question was already a bad sign.
Consider the point P. It's hanging from the point (a, 0, 0). After rotation, P has position (acosθ,asinθ,h), where h is the height of the disk. So, what's the horizontal displacement? It's (a,0)−(acosθ,asinθ)=a(1−cosθ,sinθ). So its size is a(1−cosθ)2+sin2θ=a1−2cosθ+cos2θ+sin2θ=a2(1−cosθ)=a4sin22θ=2asin2θ. But of course the horizontal displacement is also bsinϕ, hence the first result.
Now let bT be the tension in the string. Then the horizontal component of the force from the string is just Ta(1−cosθ,sinθ). We want the size of the tangential component of this, which is going to be Ta(1−cosθ,sinθ)⋅(asinθ,acosθ)=Ta2sinθ.
So each string provides a turning moment Ta2sinθ. Suppose we have n strings. Resolving vertically, nbTcosϕ=mg. But bcosϕ=b2−b2sin2ϕ=b2−4a2sin22θ.
So T=nb2−4a2sin22θmg.
So the n strings provide a total couple of nTa2sinθ=b2−4a2sin22θmga2sinθ as desired.
At this point, the disc is bcosϕ=b2−4a2sin22θ below the ceiling. When the strings go taut, the disc is b below the ceiling.
So the loss in GPE is mg(b−b2−4a2sin22θ)
This must equal the rotational KE.
So 21ω2I=mg(b−b2−4a2sin22θ), where I is the moment of inertia of the disk.
That is, 21ω22ma2=mg(b−b2−4a2sin22θ)
So 4ga2ω2=b−b2−4a2sin22θ as desired.
This took about 26 minutes, including LaTeX. I'd say that puts it in the "not too bad" category for STEP III
Edit: on a little thought, I'm not 100% convinced about the method for calculating the "tangential component" I've used; in terms of the picture I had in my mind, I think there are 2 compensatory sign errors (one for each x-component). I'd be very surprised to lost more than 1 mark for it though - it would be fine for a different mental picture. On the other hand, if you draw an actual diagram, it's obvious - it's just that it's hard to draw diagrams on here.
STEP III Q12 is definitely not necessarily 'hard' by STEP standards, I mean, I have only done S1 and had never heard of a pgf before today and I managed to get it out so it can't be that bad. On the other hand, I've been looking and I can't find generator functions anywhere on the whole edexcel syllabus which is a bit tragic for those on edexcel who made the effort to do up to S4. Did you do generator functions at a-level? I believe pgfs and mgfs used to be in the S5 module that no longer exists.
When I did Further Maths, the applied was very mechanics heavy. (From memory, only 2 of 12 questions on the applied paper would be probability based, although it might have been 3). I also did "Maths with Stats", and I believe they may have been mentioned there, although I'm not sure if that was our teacher going beyond the syllabus. It must have been either taught or in a textbook I had, however, as I recall doing CCE questions involving pgfs. (And we didn't have t'internet then).