STEP maths I, II, III 2007 solutions

STEP III, Q9.

The tricky bit of this question is hidden in one little word:

\theta

is defined as the acute angle determined by

2\theta =\angle AOB

. So we're going to have to be careful about what happens when

\theta = \frac{\pi}{2}

.

Now P.E. at start =

mk^2a^2(\beta - \alpha)^2

P.E. at

\theta = \frac{\pi}{2}

=

mk^2a^2(\pi/2 - \alpha)^2

.

So we only reach

\theta = \frac{\pi}{2}

if

\alpha - \beta > \pi/2 - \alpha

(i.e.

\beta < 2\alpha - \pi/2

).

So first consider the easy case where

\beta > 2\alpha - \pi/2

).

We have K.E. =

ma^2\dot{\theta}^2

, so conservation of energy gives

\dot{\theta}^2+k^2(\theta-\alpha)^2=const

. Set

\phi = \theta - \alpha

to get

\dot{\phi}^2 + k^2\phi^2 = const

. Then

\dot{\phi}\ddot{\phi} =-k^2\phi\dot{\phi}

so

\ddot{\phi}=-k^2\phi

.

Thus we have simple harmonic motion with period

\frac{2\pi}{k}

.

Now we come to the interesting case. Suppose

\beta < 2\alpha - \pi/2

. We consider the motion up until

\theta = \pi/2

. As before, we have

\ddot{\phi}=-k^2\phi

where

\phi = \theta - \alpha

.
So

\phi = A \cos(kt+\epsilon)

. When t=0,

\dot{\phi} = 0 \implies \epsilon = 0

. When t=0,

\phi =\beta - \alpha \implies A = \beta - \alpha

. So we have

\phi = (\beta-\alpha)\cos kt

.
So, how long does it take to get to

\theta = \pi/2

? Well at this point we have

\phi = \pi/2 - \alpha \implies \cos kt = \frac{\pi/2-\alpha}{\beta-\alpha}

, and so

t = \cos^{-1}\left(\frac{\pi/2-\alpha}{\beta-\alpha}\right)

.

Once we get to

\theta=pi/2

, we have another 3 pieces of motion, each of which will be as we've just analysed. This would be easier with diagrams, but imagine A,B as positions on a clock. Suppose we start with A = 7'o'clock, B = 5'o'clock. We've just analysed the motion up 'til A=9'o'clock, B=3'o'clock. The motion from there up 'til A=11'o'clock, B=1'o'clock looks exactly the same, only with a time reversal. And then we go back to A=9'o'clock, B=3'o'clock and then A = 7'o'clock, B = 5'o'clock.

So the final period is going to be

4 \cos^{-1}\left(\frac{\pi/2-\alpha}{\beta-\alpha}\right)

.

Oscillations will not occur when

\beta = 2\alpha - \pi/2

: instead the two particles will come to rest (in unstable equilibirum) at the point where

\theta = \pi/2

.

Reply 102

STEP II,Q12 (nasty question!):

(i) First observe that p(first six after r throws of a single die) =

pq^{r-1}

.

P(stop after r throws) = p(no sixes in first r-1 throws and 2 sixes on rth throw) +

\sum_{k=1}^{r-1}

p(no sixes in k-1 throws, single six after k throws and 2nd six on rth throw (i.e. r-k throws of single die))

=(q^2)^{r-1}p^2 + \sum_1^{r-1} (q^2)^{k-1} 2pq pq^{r-k-1}

=q^{2r-2}p^2 + 2p^2\sum_1^{r-1} q^{2k-2+1+r-k-1}

=q^{2r-2}p^2 + 2p^2\sum_1^{r-1} q^{k+r-2}

=q^{2r-2}p^2 + 2p^2q^{r-1}\frac{1-q^{r-1}}{1-q}

=q^{2r-2}p^2 + 2pq^{r-1}(1-q^{r-1})

=pq^{r-1}[pq^{r-1} + 2(1-q^{r-1})

=pq^{r-1}[2+(1-q)q^{r-1} -2q^{r-1}

=pq^{r-1}[2 - q^{r-1} -q^r]

Then E(T) =

\sum_0^\infty r pq^{r-1}[2 - q^{r-1} -q^r]

\displaystyle=\frac{2p}{q}\sum_0^\infty rq^r -p\frac{1+q}{q^2} \sum_0^\infty rq^{2r}

\displaystyle=\frac{2pq}{q(1-q)^2} -p\frac{1+q}{q^2} \frac{q^2}{(1-q^2)^2}

\displaystyle=\frac{2p}{(1-q)^2} - \frac{p(1+q)}{(1-q^2)^2}

\displaystyle=\frac{2p}{p^2} - \frac{p(1+q)}{(1-q)^2(1+q)^2}

\displaystyle=\frac{2p}{p^2} - \frac{p}{p^2(1+q)} = \frac{2}{p} - \frac{1}{p(2-p)}

\displaystyle=\frac{3-2p}{p(2-p}

.

So if E(T)=m, we have

p(2-p)m = 3-2p

mp^2-2p(m+1)+3 = 0

\displaystyle p = \frac{m+1 \pm \sqrt{(m+1)^2-3m}}{m}

.

Since

p \leq 1

we need the -ve root, and so we end up with

\displaystyle p = \frac{m+1 - \sqrt{m^2-m+1}}{m}

.

Reply 103

STEP II, Q9:

(i) The only force on the particle at moment of impact is in direction of the normal, so if the path is parallel to the normal before the collision, it is parallel to the normal after the collision.

Let speed of particle after collision be v, speed of cone be w.

Cons of horiz momentum gives:

Mw - mv \cos \alpha = mu \cos \alpha

Newton's law of restitution gives:

w \cos \alpha + v = eu

So

mw \cos^2\alpha + mv \cos \alpha = meu \cos \alpha

So

(M+m\cos^2\alpha) w = mu(1+e) \cos \alpha

, so

w =\frac{mu(1+e) \cos \alpha}{M+m\cos^2\alpha}

.

(ii) Now split the velocity components (post collision) into components

v_n, v_t

in directions normal to and tangential to the cone surface. Then

v_t = u \cos \alpha

(particle momentum in this direction being conserved).

Conservation of horiz momentum gives:

Mw - mv_n \cos \alpha = mv_t \sin \alpha

N's law of restitution gives:

w \cos \alpha +v_n = eu \sin \alpha

(N.B. resolving initial velocity normal to cone gives approach speed =

u \sin \alpha

).

So

mw \cos^2 \alpha + mv_n \cos \alpha = me u \cos \alpha \sin \alpha

.

So

(M+m\cos^2\alpha)w = m\sin\alpha(eu \cos \alpha + v_t) = m\sin\alpha(1+e)u\cos \alpha

.

So

\displaystyle w = \frac{mu(1+e)\sin\alpha \cos\alpha}{M+m\cos^2\alpha}

as required.

For the last part,

\displaystyle \frac{dw}{d\alpha} = mu(1+e)\frac{(M+mc^2)(c^2-s^2)+2ms^2c^2}{(M+mc^2)^4}

(where

c=\cos \alpha, s = \sin \alpha

).

So

\displaystyle \frac{dw}{d\alpha} = 0 \iff (M+mc^2)(c^2-s^2)+2ms^2c^2 = 0

\iff M(c^2-s^2)+mc^2(c^2-s^2+2s^2) = 0

\iff M(2c^2-1)+mc^2 = 0

\iff (2M+m)c^2 = M

Since

M>0, \cos \alpha>0

, deduce

\displaystyle \frac{dw}{d\alpha} = 0 \iff \cos \alpha = \sqrt{\frac{M}{2M+m}}

.

So we've shown w has exactly one turning point in

[0,\pi/2]

. But

w(0)=w(\pi/2) = 0

, while

w(\pi/4) > 0

. Thus our turning point must in fact be the unique maximum point for w and so gives the maximum value for w as required.

Reply 104

STEP I, Q14:

(i)

y=(x+1)e^{-x} \implies y' = e^{-x}-(x+1)e^{-x} = xe^{-x}

. So y has a single turning point at x=0 when y=1, y is increasing for x<0, decreasing for x >0, y=0 when x=-1,

y \to 0

as

x\to\infty

.

P(X\ge 2) = 1-p \implies p(X < 2) = p \implies (1+\lambda)e^{-\lambda} = p

. Consider

y=(x+1)e^{-x}

. When x=0 y=1, and

y\to 0 as x\infty

so we can find x with y(x) < p/2. Then the intermediate value theorem says y has a positive root. Suppose y has two positive roots a,b. Then Rolle's theorem implies y' = 0 for some x between a and b. But we know this isn't the case, so there are no such distinct roots. So y has a unique root, and so there's a unique value for

\lambda

.

(ii)

P(X=1) = q \implies \lambda e^{-\lambda} = q

. Consider

y=xe^{-x}

.

y'=e^-x - xe^{-x} = (1-x)e^{-x}

. So y is increasing for 0<x<1, decreasing for x>1. y(0) = 0,

y\to 0

as

x\to \infty

, while y(1) = 1/e is the maximum value. So we can deduce that if q > 1/e there are no solutions, while if q < 1/e there is one solution in (0,1) and one solution in

(1,\infty)

. The unique solution is when

q=1/e

and

\lambda=1

.

(iii)

P(X=1 | X \leq 2) = \frac{\lambda}{1+\lambda+\lambda^2/2}

. So

P(X=1 | X \leq 2) = r \iff (1+\lambda+\lambda^2/2)r = \lambda

\iff r\lambda^2+2(r-1)\lambda+2r = 0

(*). This has a single unique root when

(r-1)^2=2r^2

, or

r^2+2r-1 = 0

, so

r=(-2\pm\sqrt{4+4})/2 = -1\pm\sqrt{2}

. Since

0<r<1

the only possible solution is

r=\sqrt{2}-1

.

For this value of r, rewrite (*) as

\lambda^2+2(1-\frac{1}{r})+2 = 0

. We know this has repeated roots, so can be written as

(\lambda-a)^2

. Comparing coefficients of x gives

a = \frac{1}{r}-1 = \frac{1}{\sqrt{2}-1} - 1 = \frac{\sqrt{2}+1}{2-1} - 1 = \sqrt{2}

. So the unique value for

\lambda

is

\sqrt{2}

.

Check:

(\sqrt{2}-1)\sqrt{2}^2+2(\sqrt{2}-2)\sqrt{2}+2(\sqrt{2}-1) = 2\sqrt{2}-2+4-4\sqrt{2}+2\sqrt{2}-2 = 0

as expected.

Reply 105

STEP I, Q13:

(i) P(4 numbered) =

5C4/11C4 = 5 / \frac{11.10.9.8}{4.3.2.1} = \frac{5.4.3.2}{11.5.2.3.3.4.2} = \frac{1}{11.3.2} = \frac{1}{66}

.

(ii) P(4 numbered | "3" chosen first) = P(choose 3 of {1,2,4,5} out of 10 discs) =

4C3 / 10C3 = \frac{4.3.2.1}{10.9.8} = \frac{1}{30}

.

(iii) P (2 numbered | "3" taken first) = P(1 numbered when we choose 3 discs from remainder) =

3 \cdot \frac{4}{10}\frac{6}{9}\frac{5}{8} = \frac{3.4.3.2.5}{10.3.3.8} = \frac{1}{2}

.

(iv) P(2 numbered | "3" taken): number of ways of choosing 1 number given "3" taken = 4C1 x 6C2 = 4 x 15 = 60. Number of ways of choosing 3 discs from 10 = 10C3 = 10x9x8/6 = 5x3x8 = 120. So P(2 numbered | "3" taken) = 60/120=1/2.

(v) P(2 numbered | number taken first) = P(1 numbered when we choose 3 discs from remainder) = 1/2 (exactly as (iii)).

(vi) P(2 numbered) = 5C2 x 6C2 /11C4 = 10 x 15 / (11x10x9x8/24)=10x15/(11x10x3) = 5/11.
P(at least 1 number) = 1-P(no numbers). P(no numbers) = 1-6C4/11C4 = 1 - (6x5x4x3)/(11x10x9x8) = (6x3)/(11x9x4)=(2x9)/(11x2x2x9) = 1/22.
So P(at least 1 number) = 21/22.
So P(2 numbered | at least 1 number) = 5/11 x (22/21) = 10/21.

(I hate this kind of question. But at least we're done now...)

Reply 106

coffeym

10

The largest output award goes to...

DFranklin :king1:

Thanks for putting so much into this thread :biggrin:

Reply 107

nota bene

15

DFranklin

So P(at least 1 number) = 21/22.

That was what was missing in my working - had all other bits but couldn't put it together :s-smilie:

I kind of like these not too complicated probability questions, but they're easy to make slips on when there are no answers to work towards.

Reply 108

ukgea

6

So, somebody wanna have a stab at I/10 (ii) and kill this thread? :biggrin:

Reply 109

nota bene

I kind of like these not too complicated probability questions, but they're easy to make slips on when there are no answers to work towards.

I freely admit I often make mistakes on these - when posting on here I generally write a program to do a brute force calculation to check the results.

Reply 110

coffeym

The largest output award goes to...

DFranklin :king1:

Thanks for putting so much into this thread :biggrin:

Agreed.

He went on my rep list half an hour ago when I came to this thread and saw he'd done like 50 questions.

Hmm. Anyone reckon it's worth doing 2003-6 threads? I know solutions are available, but it might be fun to do anyway...? :wink:

Reply 111

ukgea

6

I reckon there's gonna be a new generation of STEPpers coming here in a couple of months or so, maybe we should leave some for them?

(Certainly doing these threads was part of the learning experience for me, but I'd think I wouldn't have done it if there were already solutions threads done, you know...)

Though something we could do is start putting everything up on the wiki or something, or at least get everything in one place.

Reply 112

ukgea

So, somebody wanna have a stab at I/10 (ii) and kill this thread? :biggrin:

Arghh! I just spotted that!

All distances are in terms of Saxon army.

Stage 1: Norman horse is at position d-yt, Saxon horse at position xt.
So they meet when d-yt = xt, so d=(x+y)t, so t = d/(x+y). At this time, the Saxon horse is at position xd/(x+y) as required.

Stage 2: Norman horse reaches Saxon army when d-yt = 0, or t = d/y.
After that, position of Norman horse is y(t-d/y) = yt - d.

Norman army is at position d-ut, Saxon horse at xt. So Saxon horse reaches Norman army when d-ut=xt, so t = d/(u+x). Position at this point is dx/(u+x). Position after this time is dx/(u+x)-x(t-d/(u+x)) =

\frac{2dx-xt(u+x)}{u+x}

So horses meet when

yt-d = \frac{2dx-xt(u+x)}{u+x}

(yt-d)(u+x) = 2dx-xt(u+x)

y(u+x)t +x(u+x)t = d(u+x)+2dx

t = \frac{d(u+3x)}{(u+x)(x+y)}

At this point, the distance from the Saxon Army is

yt - d = \frac{dy(u+3x)-d(u+x)(x+y)}{(u+x)(x+y)}

= \frac{udy + 3xdy - dux -duy-dx^2-dxy}{(u+x)(x+y)}

= \frac{2xdy - dux-dx^2}{(u+x)(x+y)}

= \frac{xd(2y-x-u)}{(u+x)(x+y)}

as required.

The end bit wasn't totally obvious to me, but if you sketch the distance-time graphs, I think what you find is that:

If u > 2y-x, then Saxon rider will overtake the Norman rider before the Norman rider reaches the Saxon army.

If u < y-2x, then the Norman rider will overtake the Saxon rider before the Saxon rider reaches the Norman army.

Reply 113

ukgea

I reckon there's gonna be a new generation of STEPpers coming here in a couple of months or so, maybe we should leave some for them?

(Certainly doing these threads was part of the learning experience for me, but I'd think I wouldn't have done it if there were already solutions threads done, you know...)

Though something we could do is start putting everything up on the wiki or something, or at least get everything in one place.

Hey - just because we've done tons of STEP questions, doesn't mean they can't redo them. :wink:

I reckon getting a complete set of solutions out would be great, you know.

I shall ask around about putting them on the wiki. If it's possible I may start some time soon.

Reply 114

You know what, sod asking around. I'm gonna put 'em on the wiki, it's something I'll end up doing sooner or later anyway and I might as well do it now while I have nothing else to do on it. :biggrin:

Reply 115

Dystopia

10

generalebriety

Hmm. Anyone reckon it's worth doing 2003-6 threads? I know solutions are available, but it might be fun to do anyway...? :wink:

Perhaps it would be better to encourage people to finish off the questions we have yet to do from the years 1990 - 1998, and possibly start a 1987 thread?

Admittedly I may have a somewhat biased opinion, because I want to leave a fair few STEP papers for myself for next year, and just know that I would be unable to resist doing questions early. :redface:

Reply 116

Dystopia

Perhaps it would be better to encourage people to finish off the questions we have yet to do from the years 1990 - 1998, and possibly start a 1987 thread?

Admittedly I may have a somewhat biased opinion, because I want to leave a fair few STEP papers for myself for next year, and just know that I would be unable to resist doing questions early. :redface:

Hmm. Well, I want to do some STEP papers while I still have time! :p:

I think the rest of the threads have been left alone because those earlier questions are, quite frankly, very difficult compared to these ones because of syllabus and style changes.

Reply 117

generalebriety

Hmm. Well, I want to do some STEP papers while I still have time! :p:

I think the rest of the threads have been left alone because those earlier questions are, quite frankly, very difficult compared to these ones because of syllabus and style changes.

That's probably true for the pure, but I think there are a few stats and mechanics questions from 1992-1999 that still need answering.

Reply 118