STEP 2006 Solutions Thread

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Hey guys, I'm having a brain freeze here.

In STEP 1 Q11 Part 3, in the markscheme they say p < q < r at the start, where p, q and r are the velocities of A_(n-2), A_(n-1) and A_n.

The way it's stated so casually at the start, assumes it's something obvious. I don't get why though, and it's frustrating, as this assumption at the start will prove the contradiction needed.

DFranklin in particular, any help?

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DFranklin

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I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

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Physics Enemy

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(Original post by DFranklin)
I can easily justify p <= q <= r (if p > q, then obviously there will be a subsequent collision between A_n-2 and A_n-1, contradicting the assumption that all collisions have taken place, etc.)

Perfect! If all collisions took place, yet p > q, then you'd get another collision as A_(n-2) would catch A_(n-1), which is a contradiction.

(Original post by The God)
It's not obvious to me that you can't have p = q or q = r, but I think the assumption p<=q<=r is sufficient to get the same contradiction and the end of (iii).

Yes they could keep trotting along at the same speed without colliding, but don't worry, all I need at the end is to justify p <= q + (lambda)r which is a formality now.

Thanks so much mate,

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Farhan.Hanif93

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(Original post by SimonM)
...

STEP II Q10

(i)

Let the speed of A after the initial collision with B be

, let the speed of B after the initial collision with A be

, let the speed of B after colliding with C be

and let the speed of C after colliding with B be

.

In order for a second AB collision to occur, clearly

must be larger than

.

Considering first A-B collision:
Newton's law of restitution (NLoR): $\frac{1}{2}u=v-x \implies v=x+\frac{1}{2}u$

Conservation of linear momentum (CoLM): $um=xm+vkm \implies u=x+xk+\frac{1}{2}uk \implies x=\frac{u(2-k)}{2(k+1)}$ $(\alpha)$

Also note that this implies that $v=\frac{2u-uk+uk+u}{2(k+1)} = \frac{3u}{2(k+1)}$ $(\beta)$

Considering first B-C collision:
NLoR: $\frac{1}{4}v=w-y \implies w=\frac{3u}{8(k+1)} + y$

CoLM: $vkm=ykm+3mw \implies \frac{3uk}{2(k+1)} = yk +\frac{9u}{8(k+1)} + 3y \implies y=\frac{3u(4k-3)}{8(k+1)(k+3)}$ $(\gamma)$

if

:
$\Leftrightarrow \frac{u(2-k)}{2(k+1)}>\frac{3u(4k-3)}{8(k+1)(k+3)}$
$\Leftrightarrow 4(2-k)(k+3)>3(4k-3)$
$\Leftrightarrow 4k^2+16k-33 < 0$
$\Leftrightarrow (2k-3)(2k+11) < 0$

Since

, because we can't have negative masses, for two collision to occur between A and B, $\boxed{0<k<\frac{3}{2}}$

(ii)

By subbing

into $(\alpha) , (\beta)$ and $(\gamma)$ :
$x=\frac{u}{4}$ , $y=\frac{3u}{64}$ and $v=\frac{3u}{4}$ .

Using $speed = \frac{distance}{time}$ , the time taken by B to reach C's initial position following the first A-B collision = $\frac{d}{v} = \frac{4d}{3u}$

In that time, A travels a distance of $\frac{dx}{v} = \frac{d}{3}$
Therefore, the distance between A and B after B has collided with C = $\frac{2d}{3}$

Note that the speed of approach of A w.r.t. B after B has first collided with C is given by: $x-y = \frac{13u}{64}$ .
Therefore, the time taken for A to reach B after B has collided with C is given by:
$\dfrac{\frac{2d}{3}}{\frac{13u}{64}} = \frac{128d}{39u}$

So total time between the two collisions of A and B is given by $\frac{4d}{3u} + \frac{128d}{39u} = \boxed{\frac{60d}{13u}}$ as required.

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Farhan.Hanif93

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(Original post by SimonM)
...

STEP I Q12

First part

First notice that the road with the single block-able section (the third road) MUST be blocked.

Let $p_1 = P(\mathrm{Exactly\ 1\ section\ blocked\ on\ 1st\ and\ 2nd\ roads}) = 4p^2(1-p)^2$

$p_2 = P(\mathrm{Exactly\ 2\ sections\ blocked\ on\ 1\ of\ first\ 2\ roads\ AND\ exactly\ 1\ blocked\ on\ other}) = 4p^3(1-p)$

$p_3 = P(\mathrm{All\ 4\ sections\ on\ 1st\ and\ 2nd\ roads\ blocked}) = p^4$

Therefore, $P(\mathrm{Oxtown\ cutoff\ from\ Camville})= p \times (p_1 + p_2 + p_3)$

$=\boxed{p^3(2-p)^2}$ as required.

Second part

Note that:

$P(\mathrm{Remaining\ two\ roads\ both\ blocked\ } \mathbf{|} \mathrm{\ } \mathrm{First\ or\ second\ road\ chosen\ and\ open})$

$= \frac{1}{3}(1-p)^2(2p^2(1-p) + p^3)$

$=\frac{1}{3}p^2(1-p)^2 (2-p)$

And $P(\mathrm{Remaining\ two\ roads\ both\ blocked\ } \mathbf{|} \mathrm{\ } \mathrm{Third\ road\ chosen\ and\ open})$

$= \frac{1}{3}(1-p)(4p^2(1-p)^2 + 4p^3(1-p) + p^4)$

$= \frac{1}{3}p^2(1-p)(2-p)^2$

Therefore:
$P(\mathrm{Remaining\ two\ unchosen\ roads\ both\ blocked\ } \mathbf{|} \mathrm{\ } \mathrm{Chosen\ road\ is\ open})$

$= \dfrac{P(\mathrm{Choosing\ random\ road}) \times P(\mathrm{2\ Unchosen\ roads\ blocked\ and\ chosen\ road\ open})}{P(\mathrm{Choosing\ random\ road }) \times P(\mathrm{Chosen\ road\ open})}$

$= \dfrac{\frac{1}{3}p^2(1-p)(2-p)^2 + 2\times \left( \frac{1}{3}p^2(1-p)^2 (2-p) \right)}{\frac{1}{3}\left( (1-p) + 2(1-p)^2 \right)}$

$= \boxed{\dfrac{p^2(2-p)(4-3p)}{3-2p}}$

As $p\to 1$ , our probability tends to 1, as you would expect. Since it's extremely likely that each of the sections of the three roads are blocked, the chances of having more than one open section is minute. This would need to be the case in our scenario where, in the least case (i.e. third road chosen), one section would definitely have to be open.

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kiddey

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I 2006 q1

hence

satisfies

if

, then

so

,

satisfies

where

$\Rightarrow 33127=(184-27)(184+27)$

$\Rightarrow 33127=211 \times 157$

as

and

are prime
$33127=211 \times 157$ or $33127 \times 1$

so

$\Rightarrow 33127 = 16564^2 - 16563^2$

$\Rightarrow 16564^2 - 33127 = 16563^2$

hence other value of n+m is

, and $n=182 \Rightarrow m=16382$

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Dirac Spinor

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STEP I Q14
Whilst this has been discussed earlier I haven't seen anyone post the actual solution so I'll post mine.

i) P(choosing red on rth go)= $\frac{n^{r-1}}{(n+1)^r}$
(usually I would draw a tree diagram but here it just isn't necessary)
To find the maximum we should differentiate and equate the derivative to zero.
let p=P(choosing red on rth go):
$p=\frac{n^{r-1}}{(n+1)^r}$
take logs of both sides:
$ln[f(n)]=ln(n^{r-1})-ln[(n+1)^{r}]$

Now differentiate both sides:
$\frac{1}{f(n)}f'(n)=(r-1)\frac{1}{n}-r\frac{1}{n+1}$
$f'(n)=[(r-1)\frac{1}{n}-r\frac{1}{n+1}]f(n)$
To find stationary points, let

$[(r-1)\frac{1}{n}-r\frac{1}{n+1}]f(n)=0$ . Since f(n) can not be 0:
$[(r-1)\frac{1}{n}-r\frac{1}{n+1}]=0$
Rearranging we see that:

and so:

ii) A trick in these type of questions where the denominator and the numerator of the probabilities go down by one every time is to notice that the k+1th denominator is canceled by the kth numerator. Therefore all we care about is the first denominator and the last numerator as they are the only ones which don't cancel.
P(choosing red on rth go)= $\frac{1}{n+1}$
so n should be the smallest it can be.
n+1 is greater than or equal to r so the smallest value of n is:
n=r-1

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Farhan.Hanif93

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(Original post by SimonM)
...

STEP II Q14

Solution

Find the graph by noting that $y\to -\infty$ as $x\to 0$ , and considering the behaviour as $x\ \to +\infty$ . Also notice that there are vertical asymptotes at

and

; and you should consider the behaviour of the curve on either side of these. You could also consider the maximum, which occurs at $(\frac{1}{e}, -e)$ .
It should look something like this:

Graph

(i)

Considering the case $a=\frac{1}{4}, b=\frac{1}{2}$ :
Note that $\lambda \displaystyle\int ^{\frac{1}{2}}_{\frac{1}{4}} \dfrac{1}{x\ln x} dx = 1$

$\Leftrightarrow \lambda \displaystyle\int ^{\frac{1}{2}}_{\frac{1}{4}} \dfrac{\frac{1}{x}}{\ln x} dx = 1$

$\Leftrightarrow \lambda \left[\ln |\ln |x||\right]^{\frac{1}{2}}_{\frac{1}{4}} = 1$

$\Leftrightarrow \boxed{\lambda = -\dfrac{1}{\ln 2}}$

(ii)

Since

, we don't have to worry about integrating across the discontinuity at

. It follows that:

$\displaystyle\int ^b_a \dfrac{1}{x\ln x} dx = 1$

$\Leftrightarrow \left[ \ln (\ln x)]^b_a = 1$

$\Leftrightarrow \dfrac{\ln b}{\ln a} = e$

$\Leftrightarrow \ln b = \ln a^e$

$\Leftrightarrow \boxed{b=a^e}$ , as required.

(iii)

Considering the case $\lambda = 1, a=e$ :
Note that

, therefore we still do not have to worry about the discontinuity. It follows that:

$P(e^{\frac{3}{2}} \leq X \leq e^2) = \displaystyle\int ^{e^2}_{e^{\frac{3}{2}}} \dfrac{1}{x\ln x} dx$

$=\left[\ln (\ln x)\right]^{e^2}_{e^{\frac{3}{2}}}$

$=\ln \frac{4}{3}$

Setting $x=\frac{1}{3}$ in the Maclaurin expansion of $\ln (1+x)$ gives:

$\ln \frac{4}{3} \approx \frac{1}{3} - \frac{1}{18} + \frac{1}{81} - \frac{1}{324} = \frac{93}{324} = \frac{31}{108}$

Therefore $\boxed{P(e^{\frac{3}{2}} \leq X \leq e^2) \approx \dfrac{31}{108}}$ , as required.

(iv)

Note that, in the case $a=\exp {\frac{1}{2}}, \lambda =1$ , $b=\exp {\frac{e}{2}}$ from our answer to part (ii).

Also note that $e<3 \implies \exp (\frac{e}{2}) < \exp (\frac{3}{2})$ and that $1<3 \implies \exp (\frac{1}{2}) < \exp (\frac{3}{2})$ . This means that, since our PDF is only non-zero for $\sqrt{e} \leq x \leq \sqrt {e^e}$ , the range $e^{\frac{3}{2}} \leq X \leq e^2$ lies outside the interval for which the probability is non-zero. Therefore, it follows that $\boxed{P(e^{\frac{3}{2}} \leq X \leq e^2) = 0}$

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brianeverit

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$\text {2005 STEP III question 12}$

$\implies \text {E}[X]=at+b(t_1+t_2+t_3+t_4)=at+bt=(a+b)t=t bif a+b=1$
$\text {Var}[X]=a^2 \sigma^2+4b^2 \sigma^2=(a^2+4b^2) \sigma^2 \text { which is as small as possible if }a^2+b^2 \text { is as small as possible}$
$\text {Now }a^2+4b^2=a^2+4(1-a)^2=5a^2-8a+4=5 \left(a^2- \dfrac{8}{5}a+ \dfrac{4}{5} \right)=5 \left[ \left(a-\dfrac{4}{5} \right)^2+\dfrac{4}{25} \right]$
$\text {which has a smallest value of }\dfrac{4}{5} \text { when }a=\dfrac{4}{5} \text { so } b= \dfrac{1}{5} \text { and } X= \dfrac{}1}{5} \left[4T+T_1+T_2+T_3+TR_4 right]$
$\text {Now let }Y=cT+d(T_1+T_2+T_3+T_4), \text{E}[Y]=(c+d)t \text { as above and Var}[Y]=(c^2+4d^2) \sigma^2$
$\text {So E}[Y^2]= \text{Var}[Y]+ (\text {E}[Y])^2=(c^2+4d^2) \sigma^2+(c+d)^2t^2= \sigma^2 \text { independent of lap times if }c+d=0$
$\text {and }c^2+4d^2=1 \implies 5c^2=1 \implies c= \dfrac{1}{\sqrt5}, d=- \dfrac{1}{ \sqrt5} \text { so } Y= \dfrac{1}{ \sqrt5}[T-(T_1+T_2+T_3+T_4)]$
$T=220 \text { and }(T_1+T_2+T_3+T_4)=220.5 \implies \text {E}[X]= \dfrac{1}{5}(880+220.5)=220.1$
$\text {and Var}[X]=\dfrac{4}{5} \sigma^2=\dfrac{4}{5} \times \dfrac{1}{5}[T-(T_1+T_2+T_3+T_4)]^2=\dfrac{4}{25}(220-220.5)^2=0.04$
$\text{we can reasonably expect the true value to be within 2 standard errors of the expected value}$
$\text {i.e. between }220.1+0.4 \text { and } 2-0.4=219.7 \text { to }220.5$

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brianeverit

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$\text {2006 STEP I question 11}$

$\text {(i) If only one particle is moving after all collisions it would have to be }A_n \text { with velocity }v$
$\text {overall conservation of momentum then gives }mu= \lambda mv \text { and conservation of energy gives }$
$mu^2= \lambda mv^2$
$\text {This would therefore require that }u= \lambda v \implies u^2= \lambda^2 v^2, \text { but } u^2= \lambda v^2 \text { so } \lambda = 0 \text { or }1, \text { neither of which is}$
$\text {possible so we cannot have exactly one particle moving after all collisions}$
$\text {(ii) Let final speeds of }A_{n-1} \text { and }A_n \text { be }v \text { and } w \text { respectively}$
$\text {conservation of momentum gives }mu=mu+ \lambda mw \text { \nd of energy } mu^2=mv^2+ \lambda mw^2$
$\text {hence, }v+ \lambda w)^2=v^2+ \lambda w^2 \implies 2v \lambda w+ \lambda^2 w^2= \lambda w^2 \impliea 2v=w(1-\lambda) \lambda \implies v<0 \text { since } \lambda>1$
$\text {this is clearly not possible so }A_{n-1} \text { and }A_n \text { cannot be the only two particles moving}$
$\text {(iii) Let final velocities of }A_{n-2},A_{n-1} \text { and }A_n \text { be } p,q \text { and }r \text { respectively, then}$
$\text {conservation of momentum and energy gives }mu= kmp+mq+ \lambda mr \text { and }mu^2=kmp^2+mq^2+ \lambdamr^2$
$\text {where }km \text { is the unknown mass of }A_{n-2}$
$\implies (kp+q+ \lambda r)^2=kp^2+q^2+ \lambda r^2 \implies 2kpq+2q \lambda r+2k \lambda pr+k^2p^2+ \lambda^2 r^2=kp^2+ \lambda r^2$
$\implies kp(2q+2 \lambda r+kp-p)= \lambda r^2-2 \lambda qr- \lambda^2 r^2=r^2( \lambdea- \lambda^2)-2 \lambda qr<0 \text { since }k>1$
$\text {i.e. }kp(2q+2 \lambda r+kp-p)<0 \text { but } q>p \implies 2q+kp-p>2p+kp-p=p(1+k)>0$
$\text {so }kp(2q+2 \lambda r+kp-p)>0 \text { a contradiction}$
$\text {(iv) If there are only two particles moving they must be }A_0 \text { and }A_n \text { with velocities } x \text { and }y \text { say}$
$\text {so }mu=mx+ \lambda my \text { and } mu^2=mx^2+ \lambda my^2 \implies (u- \lambda y)^2=u^2- \lambda y^2$
$\implies \lambda ^y^2-2u \lambda y=- \lambda y^2 \implies \lambda y-2u=-y \implies y= \dfrac{2u}{1+ \lambda} \text { and } x=u- \lambda y=u-\dfrac{2 \lambda u}{1+ \lambda}= \dfrac{u(1- \lambda)}{1+ \lambda}$

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brianeverit

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$\text {2006 STEP II question 9}$

$\text {The first diagram shows the forces on the ladder when the man is standing a distance }x \text { up from the foot}$
$\text {Resolving horizontally and vertically } F=X \text { and } R=(1+k)W$
$\text {Taking moments about foot of ladder } 3aW \cos \theta +xak W \cos \theta =6aX \sin \theta$
$\text { so } (3+xk)W=6X \tan \theta=12X \implies X= \dfrac{3+xk}{12(1+k)}$
$\text {ladder does not slip if }F< \dfrac{1}{2}R = \dfrac{(1+k)W}{2}$
$\text {maximum value of }F \text { is when }x=6 \text { and }F= \dfrac{3+6k}{12}W$
$\text {i.e. Max }F=\dfrac{1+2k}{4}W \text { which is less than }\dfrac{(1+k)W}{2}$
$\text {so ladder will not slip}$
$\text {(ii) consider now the forces on the table (see second diagram) when it is free to move}$
$\text {Forces on ladder will be as before with }k=9 \text { so }R=10W \text { and }F= \dfrac{3+9x}{12}W$
$\text {Resolving forces }P=F \text { and }N=2W+R=12W$
$\text {if table tilts it will do so about point A so taking moments about A we have }$
$\dfrac{12aW}{4}=aF \implies x= \dfrac{33}{9}= \dfrac{11}{3}$
$\text {table slips before man reaches top of ladder if }P> \dfrac{1}{3}N \text { i.e. If } \dfrac{3+9x}{12}W>4W$
$\text {hence, if }3+9x>48 \text { or } x>5$
$\text {table tilts if }x=\dfrac{11}{3} \text { and slips if }z>5 \text { hence, tilting occurs first}$

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brianeverit

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$\text {2006 STEP II question 12}$

$\text {(i) Number of wickets taken by Arthur will have a B} \left( 30, \dfrac{1}{36} \right) \text { distribution}$
$\text {by Betty a B} \left(30, \dfrac{1}{25} \right) \text { and by Cuba a B} \left(30, \dfrac{1}{41} \right)$
$\text {Pr(1 wicket }= \text {P}(A1 \cap B0 \cap C0)+ \terxt{P}( A0 \cap B1 \cap C))+ \text {P}(A0 \cap B0 \cap C1)$
$= \dfrac{30}{36} \times \left( \dfrac{35}{36} \right)^{29} \times \left( \dfrac{24}{25} \right)^{30} \times \left( \dfrac{40}{41} \right)^{30}+ \dfrac {30}{25} \times \left( \dfrac{24}{25} \right)^{29} \times \left( \dfrac{35}{36} \right)^{30} \times \left( \dfrac{40}{41} \right)^{30}$
$+ \dfrac{30}{41} \times \left( \dfrac{40}{41} \right)^{29} \times \left( \dfrac{24}{25} \right)^{30} \times \left( \dfrac{35}{36} \right)^{30}$
$= \dfrac{(35 \times 24 \times 40)^{30}}{(36 \times 25 \times 41)^{30} (30 \times 35 +25 \times 24+ 41 \times 40) }$
$\text {Pr(Arthur took the one wicket ) is then }$
$\dfrac { \text{P}(A1 \cap B0 \cap C0)}{ \text {P}(A1 \cap B0 \cap C0)+ \text {P} (A0 \cap B1 \cap C0)+ \text {P} ( A0 \cap B0 \cap C1))}$
$= \dfrac{\frac {30}{36} \times \left( \frac{35}{36} \right)^{29} \times \left( \frac{24}{25} \right)^{30} \times \left( \frac {40}{41}\right)^{30}}{\frac{30}{36} \times \left( \frac{35}{36} \right)^{29} \times \left( \frac {24}{25} \right)^{30} \times \left( \frac {40}{41} \right)^{30}+\frac{30}{25} \times \left( \frac{24}{25} \right)^{29} \times \left( \frac {35}{36} \right)^{30} \times \left( \frac {40}{41} \right)^{30}+ \frac{30}{41} \times \left( \frac{40}{41} \right)^{29} \times \left( \frac {24}{25} \right)^{30} \times \left( \frac {35}{36} \right)^{30}}$
$= \dfrac{ \frac{30}{35}}{ \frac{30}{35}+ \frac{30}{24}+ \frac {30}{40}} + \dfrac{ \frac {1}{35}}{ \frac {1}{35}+ \frac {1}{24} + \frac {1}{40}}= \dfrac {1}{1+ \frac {35}{24}+ \frac {7}{8}}= \dfrac{1}{ 1+ \frac{35}{24} + \frac {7}{8}}= \dfrac{1}{ \frac {24+35+21}{24}}= \dfrac{24}{80}= \dfrac{3}{10}$
$\text {(ii) Average number of wickets taken is }30 \times \left( \dfrac {1}{36}+ \dfrac{1}{25}+ \dfrac{1}{41} \right)= \dfrac {5}{6}+ \dfrac{6}{5}+ \dfrac{30}{41}= \dfrac{61}{30}+ \dfrac{30}{41}$
$= \dfrac{61 \times 41+30^2}{30 \times 41}= \dfrac{3401}{1230} = 3 \text { to nearet whole number}$
$\text {(iii) Assuming the distribution B} \left(90, \dfrac {3}{90} \right) \text { for the number of wickets taken which, since }$
$n \text { is large and }p \text { is small may be approximated by the Poisson distribution with mean 3}$
$\text { P(at least 5 wickets is given by}$
$1-[p_0+p_1+p_2+p_3+p_4]=1- \text{e}^{-3} \left( 1+3+ \dfrac {9}{2} + \dfrac {27}{6} + \dfrac {81}{24} \right)$
$=1- \dfrac{1}{20}(1+3+4.5+4.5+3.375)=1-\dfrac{16.375}{P20}=1- \dfrac{131}{160}= \dfrac{29}{160} \approx \dfrac{1}{5}$
$\text {alkternatively use tables of cumulative probability}$

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brianeverit

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$\text {2006 STEP II question 13}$

$\text {There are 24 possible sequences of 4 ice creams as follows:-}$
$\begin{array}{c,c,c,c} 1234 & 2134 & 3124 & 4123 \\ 1243 & 2143 & 3142 & 4132 \\ 1324 & 2314 & 3214 & 4213 \\ 1342 & 2314 & 3241 & 4231 \\ 1423 & 2413 & 3412 & 4312 \\ 1432 & 2431 & 3421 & 4321 \end{array}$

$\text {Of these the 1 will be chosen if the first one is 2,3 or 4 and the 1 is the first number smaller than this}$
$\text {from then above list we see that this occurs 11 times so P}_4(1)= \dfrac{11}{24}$
$\text {number 2 is chosen if the first number is 3 and the 2 comes before 1 or if the first number is 4 and}$
$\text {the 2 comes before either 1 or 3 or if the fiurst number is 1 and the last number is 2}$
$\text {again from the above list this occurs 7 times so P}_4(2)= \dfrac {7}{24}$
$\text {3 is chosen if the first number is 4 and the second one is 3 or if the sequence is 1243}$
$\text {so P}_4(3)= \dfrac {4}{24}= \dfrac {1}{6}$
$\text {whilst 4 is chosen only if the sequence is 1234 or 1324 so P}_4(4)= \dfrac{2}{24} = \dfrac {1}{12}$

$\text {With }n \text { cones to choose from, if the first is 1 then number 1 cannot be chosen so P}_1(1)=0$
$\text {If I am offered the }k^{th} \text { biggest first then P(I choose the biggest) is }\dfrac {1}{n-1}$
$\text {hence, P}_n(1)= \dfrac{1}{n} \displaystyle \sum _{r=1}^{n-1} \dfrac {1}{r}$

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brianeverit

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$\text {2006 STEP III question 9}$

$\text {System will be in stable equilibrium for all positions of the bead if there is no change in P.E. when}$
$\text {moved from one position to another}$
$\text {Taking horizontal plane through O as zero level for P.E. we have }$
$\text {P.E. of system when bead is at H is }mgh-mg(x-h) \text { where }x \text { is the length of the string}$
$\text {P.E. when bead is at B is }mgy-mg(x-r) \text { so change in P.E. is }$

$\text {So there is no change in P.E. if }2h-y-r=0 \text { i.e. }y+r=2h \text { as required}$
$\text {If speed at B is }v \text { then }v=r \dot \theta \text { and speed of P will be } \dot r$
$\theta= \sin^{-1} \left( \dfrac{v}{r} \right)= \sin^{-1} \left( \dfrac {2h-r}{r}\right) \implies \dot \theta \dfrac {-r \dot r-(2h-r) \dot r}{r^2} \times \dfrac {1}{ \sqrt{1- \left( \frac{2h-r}{r} \right)^2}}= \dfrac {-2h \dot r}{r \sqrt {r^2-(2h-r)^2}}$
$\text {i.e. } \dot \theta= \dfrac {-2h \dot r}{r \sqrt{-4h^2+4hr}}= -\dfrac {h \dot r}{r \sqrt {hr-h^2}} \text { as required}$
$\text {P.E. constant } \implies \text { K.E. constant, so } v^2+ (r \dot \theta)^2+v^2=V^2 \text { where }v= \dot r \text { is velocity of P}$
$\text {i.e. } 2v^2+ \dfrac {r^2h^2v^2}{r^2(hr-h^2)}=V^2 \implies 2v^2 \left(1+ \dfrac{h}{2(r-h)} \right)=V^2 \impolies v^2 \left( \dfrac {2r-h}{r-h} \right)=V^2 n\implies v= V \sqrt {\dfrac{r-h}{2r-h}} \text { as required}$

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$\text {2006 STEP III question 10}$

$\text {By conservation of angular momentum } mk^2 \Omega +ma^2 \Omega = mk^2 \omega + mr^2 \omega$
$\text {hence, } \omega= \dfrac { \Omega(k^2+a^2)}{k^2+r^2} \text { as required}$
$\text {K.E. in initial position is } \dfrac {1}{2}mk^2 \Omega^2 + \dfrac {1}{2} ma^2 \Omega^2 + \dfrac {1}{2}mV^2$
$\text {and after time }t, \text { K.E. is } \dfrac {1}{2}mk^2 \omega^2+ \dfrac{1}{2}mr^2 \omega^2 + \dfrac {1}{2} m \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2$
$\text {so, by conservation of energy, } \dfrac {1}{2}mk^2 \omega^2+ \dfrac{1}{2}mr^2 \omega^2 + \dfrac {1}{2} m \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2= \dfrac {1}{2}mk^2 \Omega^2 + \dfrac {1}{2} ma^2 \Omega^2 + \dfrac {1}{2}mV^2$
$\implies \left( \dfrac { \text{d}r}{ \text{d}t} \right)^2= \Omega^2(k^2+a^2)+ \dfrac { \Omega^2 a^2(k^2+a^2)}{k^2}- \omega^2(k^2+r^2) [latex]= \dfra { \Omega^2k^2(k^2+a^2)(k^2+r^2)+ \Omega^2a^2(k^2+a^2)(k^2+r^2)- \Omega^2k^2(k^2+a^2)^2}{k^2(k^2+r^2)}$
$= \dfrac {\Omega^2(k^2+a^2)(k^4+k^2r^2+a^2k^2+a^2r^2-k^4-a^2k^2)}{k^2(k^2+r^2)}= \dfrac { \Omega^2k^2r^2(k^2+a^2)^2}{k^2(k^2+r^2)} \text { as required}$
$\dfrac {\text{d}r}{ \text{d}t}<0 \text { since particle is moving towards the axis and } \dfrac {\text{d}r}{ \text{d} \theta}= \dfrac {\text{d}r}{\text{d}t} \times \dfrac {\text{d}t}{\text{d} \theta}= \dfrac {1}{\Omega} \dfrac {\text{d}r}{\text{d}t}$
$\text {so } \dfrac {\text{d}r}{\text{d} \theta}= \dfrac {\Omega r(k^2+a^2)}{k \sqrt {k^2+ar^2}} \times \dfrac {k^2+r^2}{ \Omega(k^2+a^2)}= -\dfrac {r \sqrt {k^2+r^2}}{k} \implies k \dfrac {\text{d}r}{\text{d} \theta}=-r \sqrt{k^2+r^2}$
$\text {putting }u= \dfrac{k}{r} \text { we have } r= \dfrac {k}{u} \implies \dfrac {\text{d}r}{\text{d} \theta}=- \dfrac{k}{u^2} \dfrac {\text{d}u}{\text{d} \theta} \text { so } \dfrac {k^2}{u^2} \dfrac {\text{d}u}{ \text{d} \theta}= \dfrac {k}{u} \sqrtr k^2+ \frac{k^2}{u^2}}= \dfrac {k^2}{u^2} \sqrt{1+u^2}$
$\text {i.e. } \dfrac {\text{d}u}{\text{d} \theta}= \sqrt {1+u^2} \implies \int \dfrac {\text{d}u}{ \sqrt{1+u^2}}= \int \text{d} \theta \implies \sinh^{-1} u= \theta+\alpha \text { or } u= \sinh( \theta+\alpha)$
$\text {hence, }r= \dfrac {k}{\sinh (\theta+\alpha)} \text { or } r \sinh(\theta+\alpha)=k \text { and }r=a \text { when } \theta=0 \implies \sinh \alpha=\dfrac{k}{a}$
$\text {since we are given that }k>0 \text { it follows that }r \text { can never be zero, i.e. Particle does not reach the axis}$

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$\text {2006 STEP III question 11}$

$\text {After tile falls, by Newton's second law, if tension in cable is }T \text { then for counterweight}$
$Mg-T=Ma \text { and for lift minus tile }T-(M-m)g=(M-m)a \text { where } a \text { is the acceleration}$
$\text {so, acceleration of lift is given by }mg=(2M-m)a \implies a= \dfrac{m}{2M-m}g \text { and acceleration of tile is }g$
$\text {if lift rises a distance }d \text { while tile falls then tile falls a distance }h-d$
$\text {so if velocities of tile and lift just before impact are }u \text {(downwards) and }v \text {(upwards) respectively then}$
$u^2=2g(h-d) \text { and }v^2= \dfrac{2m}{2M-m}gd \implies u^2=2gh- \dfrac{2M-m}{m} v^2 \text { or } u^2+ \dfrac {2M-m}{m}v^2=2gh$
$\text {now, momentum is constant {=0} throughout so }(2M-m)v=mu \implies \dfrac {m}{2M-m}= \theta u \text { say}$
$\text {now let velocities after impact be } w \text { and } x \text {(both upwards) respectively then}$
$(2M-m)x+mw=0 \implies x=- \dfrac {m}{2M-m}w \text { or } x=- \tgheta w$
$\text {by law of restitution at impact }w-x=e(u+v) \implies w(1+ \theta)=eu(1+\theta) \text { so } w=eu$
$\text {change in K.E.}= \dfrac {1}{2}(2M-m)(v^2-x^2)+ \dfrac {1}{2}m(u^2-w^2)$
$= \dfrac {1}{2}(2M-m) \theta^2(u^2-w^2)+ \dfrac {1}{2}m(u^2-w^2)= \dfrac {1}{2}(u^2-w^2)[(2M-m) \theta^2+m]$
$= \dfrac {1}{2} (u^2-w^2) \lefty[\dfrac{m^2}{2M-m}+m \right] \text { since } \theta^2= \left( \dfrac {m}{2M-m} \right)^2$
$= \dfrac{1}{2}u^2(1-e^2) \dfrac {2Mm}{2M-m} \text { since } w=eu$
$= \dfrac {1}{2}(1-e^2) \dfrac {2Mm}{2M-m} \times \dfrac {2M-m}{M} gh \text { since }u= \sqrt {\dfrac{2M-m}{M}gh} \text { from first part of question}$
$= (1-e^2)mgh \text { as required}$

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brianeverit

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$\text {2006 STEP III question 12}$

$\text {If }X \text { is the number of potential passengers then }X \sim \text {B}(1024,0.5) \approx \text {N}(512,16^2)$
$\text {Expected lost profit }(X>480) \text { without another bus is } \displaystyle \sum_{k=1}^{32} k\text{P}(X=480+k)+32 \text{P}(X>512)$
$\text { i.e. } \displaystyle \sum_{k=1}^{32}k \dfrac {1}{16} \phi \left(-2+\dfrac{k}{16} \right)+16 \approx \displaystyle \int_1^{32} \displaystyle \sum_{k=1}^{32} k \dfrac {1}{16} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16$
$= \displaystyle \int _0^{32}\dfrac {x}{16} \dfrac {1}{ \sqrt{2 \pi}} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16= \displaystyle \int _0^{32}\dfrac {1}{16} \dfrac {x}{ \sqrt{2 \pi}}\text {exp} \left(- \dfrac{1}{2} t^2 \right) \text{d}t+16 \text { where }z=-2+ \dfrac {x}{16}$
$=\displaystyle \int _0^{32}\dfrac {1}{16} \dfrac {x}{ \sqrt{2 \pi}}\text {exp} \left(- \dfrac{(x-32)^2}{512} \right) \text{d}x+16$
$\text {now let }t=\dfrac {x-32}{16} \implies \text {d}t= \dfarc {1}{16} \text {d}x, x=0 \implies t=-2, x=32 \implies t=0 \text { so integral becomes}$
$\displaystyle \int_{-2}^0 \dfrac {16t+32}{16 \sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t+16$
$=16 \displaystyle \int_{-2}^0 \dfrac {t}{ \sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t+32\displaystyle \int_{-2}^0 \dfrac {1}{\sqrt{2 \pi}} \text{exp} \left( -\dfrac{1}{2} t^2 \right) \text {d}t +16$
$=16 \left[- \dfrac {1}{ \sqrt{2 \pi}} \text{exp} \left(- \dfrac{1}{2} t^2 \right) \right]_{-2}^0 +32 \left[ \dfrac {1}{2} - \Phi(-2) \right]+16= \dfrac{16}{ \sqrt {2 \pi}} ( \text{e}^{-2}-1)+32 \Phi(2)-16+16$
$\text {In the course of a year the expected loss is 50 times this i.e. }1600 \Phi(2)- \dfrac {800}{\sqrt{2\pi}} (1- \text{e}^{-2})$
$\text {which is thus the maximum tolerable licence fee foe an extra bus (as required)}$ r

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brianeverit

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$\text {2006 STEP III question 13}$

$\text {There are two possibilities (i) both points lie on the perimeter of the semicircle (first diagram)}$
$\text {(ii) one point is on perimeter and one on diameter (second diagram)}$
$\text {(i) Let P}_1 \text { lie within the arc }(\alpha,\alpha+\delta \alpha) \text { and P}_2 \text { in the arc }(\theta, \theta + \delta \theta)$
$\text { Probability that P}_2 \text { is in this arc is } \dfrac{ \delta \theta}{\pi+2} \text { and area of triangle OP}_1 \text {P}_2 \text { where O is centre of semicircle}$
$\text {is } \dfrac{1}{2} \sin ( \alpha- \theta) \ytext { if } \theta< \alpha \text { or } \dfrac {1}{2} \sin ( \theta- \alpha) \text { if } \theta> \alpha$
$\text {so E[Area] when P}_1 \text { is in }( \alpha, \alphga + \delta \alpha) \text { is given by }$
$\displaystyle \int_0^\alpha \dfrac{ \sin (\alpha-\theta)}{2(2+\pi)} \text {d} \theta+\displaystyle \int_ \alpha^\pi \dfrac{\sin(\theta-\alpha)}{2(2+\pi)} \text {d} \theta$
$=\dfrac{1}{2(2+\pi)}\left([\cos(\alpha-\theta)]_0^\alpha+[-\cos(\theta-\alpha)]_\alpha^\pi \right)$
$=\dfrac{1}{2(2+\pi)}(1-\cos \alpha+\cos \alpha+1)= \dfrac {1}{2+\pi}$
$\text {(ii) Let P}_1 \text { lie within the range }(r,r+ \delta r) \text { measured from centre of semicircle and}$
$\text{P}_2 \text { in arc }( \theta, \theta+ \delta \theta) \text { as before then P(P}_2 \text { lies in this arc)}= \dfrac{\delta \theta}{2+\pi}$
$\text {and area of triangle OP}_1 \text {P}_2 \text { is } \dfrac {1}{2} |r| \sin \theta$
$\text {so expected value of area is } \displaystyle \int_0^\pi \dfrac{1}{2(2+\pi)}|r| \sin \theta \text {d} \theta= \left[ \dfrac{-1}{2(2+\pi)} |r| \cos \theta \right]_0^\pi= \dfrac{|r|}{2+\pi}$
$\text {If P}_1 \text { is on the arc and P}_2 \text { on the diameter then obviously the result is the same}$
$\text {hence, E[Area} is } \displaystyle \int \dfrac {1}{(2+\pi)^2} \text{d} \alpha +2 \displaystyle \int_{-1}^1 \dfrac {|r|}{(2+\pi)^2} \text {d}r= \left[ \dfrac{ alpha}{(2+\pi)^2} \right]_0^\pi+4 \left[ \dfrac {r^2}{2(2+\pi)^2} \right]_0^1= \dfrac {\pi+2}{(2+\pi)^2}= \dfrac {1}{2+\pi}$

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2006'III.13(ii).jpg (10.9 KB)

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$\text {2006 STEP III question 14 }$

$\text{E}[X_1+bX_2]= a\text{E}[X_1]+b \text{E}[X_2] \text { and E}[X_1X_2]= \text{E}[X_1]\text{E}[X_2] \text { for independent events}$
$P=2(X_1+X_2) \text { so E}[P]=2 \text{E}[X_1]+2 \text{E}[X_2]=2 \mu_1+2 \mu_2$
$\text{E}[P^2]=\text{E}[2(X_1+X_2))^2=\text{E}[4X_1^2]+\text{E}[8X_1X_2]+\text{E}[4X_2^2]=4\text{E}[X_1^2]+8\text{E}[X_1] \text{E}[X_2] +4\text{E}[X_2^2]$
$\text{E}[X_1^2]= \mu_1^2+\sigma_1^2, \text{E}[X_2^2]= \mu_2^2+\sigma_2^2\text { so E}[P^2]=4(\mu_1^2+\sigma_1^2)+8\mu_1\mu_2+4(\mu_2^2+\sigma_2^2)$
$\text{Var}[P]=\text{E}[P^2]-(\text{E}[P])^2=4(\mu_1^2+\sigma_1^2)+8\mu_1\mu_2+4(\mu_2^2+\sigma_2^2)-4(\mu_1+\mu_2)^2$
$\implies \text{ standard deviation }=2 \sqrt{\sigma_1^2+\sigma_2^2}$
$\implies \text{E}[A^2]=(\mu_1^2+\sigma_1^2)(\mu_2^2+ \sigma_2^2)$
$\text {Var}[A^2]=\text{E}[A^2]-(\text{E}[A])^2=(\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2)-\mu_1^2\mu_2^2=\sigma_1^2 \mu_2^2+\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2$
$\text{E}[PA]= \text{E}[(2X_1+2X_2)X_1X_2]=2\text{E}[X_1^2]\text{E}[X_2]+2\text{E}[X_2^2] \text{E}[X_1]$
$=2(\mu_1^2+\sigma_1^2)\mu_2+2(\mu_2^2+\sigma_2^2)\mu_1$
$\text {and E}[P] \text{E}[A]=2(\mu_1+\mu_2)\mu_1\mu_2=2\mu_1^2 \mu_2+2 \mu_2^2 \mu_1$
$\text {hence E}[PA]- \text{E}[P]\text{E}[A]=2\sigmna_1^2\mu_2+2\sigma_2^2 \mu_1 \not=0 \text { since } \mu_1,\mu_2>0 \text { and } \sigma_1, \sigma_2 \not=0$
$\text { which implies that }P \text { and }A \text { are not independent}$
$\text {if }Z=P- \alpha A \text { then E}[ZA]=\text{E}[PA- \alpha A^2]=\text{E}[PA]- \alpha \text{E}[A^2]$
$=2(\mu_1^2+ \sigma_1^2) \mu_2+2(\mu_2^2+ \sigma_2^2) \mu_1- \alpha (\sigma_1^2 \mu_2^2+\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2)=0 \text { only if }$
$\alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}$
$\text { so if }\alpha \not=\dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2} \text { then }Z \text { and }A \text { are not independent}$
$\text {For the final part we need only consider the exceptional case when } \alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}$
$X1,X2=1 \text { or }3 \implies A=1,3 \text { or }9 \text { so P}(A=1)=\dfrac {1}{4}, \texct{P}(A=3)=\dfrac {1}{2} \text { and P}(A=9)= \dfrac {1}{4}$
$\text {E}[X_1]=\text{E}[X_2]=1 \times \dfrac {1}{2}+3 \times \dfrac {1}{2}=2, \text{Var}[X_1]=\text{Var}[X_2]=1 \times \dfrac {1}{2}+9 \times \dfrac {1}{2}-2^2=1$
$\text {so }\mu_1=\mu_2=2 \text { and } \sigma_1= \sigma_2=1 \implies \alpha= \dfrac{4+4}{34+4+14}= \dfrac{8}{9}$
$A=1 \implies Z=4- \dfrac {8}{9}= \dfrac {28}{9}, A=3 \implies Z=8- \dfrac {8}{3}= \dfrac {16}{3} \text { and }A=9 \implies Z=12-8=4$
$\text {Now, for example P} \left(Z= \dfrac {28}{9} \right)= \dfrac {1}{4} \text { and P} \left(Z=\dfrac {28}{9}|A=3 \right)=0$
$\text {so }Z \text { and }A \text { are not independent for any value of }\alpha$

!Phew Thank goodness that's done. I wouldn't be at all surprised if there are some errors in it. Please let me know if you spot any.

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(Original post by SimonM)
...

STEP II Q11

Solution

Note that the first particles motion can be described by $\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ut\cos \theta \\ ut\sin \theta \end{pmatrix} - \dfrac{1}{2}\begin{pmatrix} ft^2 \\ gt^2 \end{pmatrix}$

When

and $t\not =0$ ,

.

Considering the value of

under these circumstances:
$y=0 \implies t=\dfrac{2u\sin \theta}{g}$
$\implies OA = \dfrac{2u^2\sin \theta \cos \theta}{g} - \dfrac{2u^2f\sin ^2 \theta}{g^2}$
$\implies \boxed{OA =\dfrac{2u^2\sin \theta (g\cos \theta - f\sin \theta)}{g^2}}$

(i)

Note that, for the value of

that satisfies $\dot{x} =0$ , we're looking for the largest angle of projection that is smaller than the angles which satisfy

.

$\dot{x} =0 \implies t=\dfrac{u\cos \alpha}{f}$
Therefore $y>0 \implies \dfrac{u^2\cos \alpha \sin \alpha}{f} - \dfrac{u^2g\cos ^2\alpha}{2f^2} >0$

$\implies f\sin 2\alpha - g\cos ^2\alpha >0$

Since $\cos \alpha >0$ for acute $\alpha$ :

$\implies \tan \alpha > \dfrac{g}{2f}$

If follows that, since $\tan \alpha$ is an increasing function, this critical value is given by $\boxed{\alpha = \arctan \dfrac{g}{2f}}$ .

(ii)

Plugging $\theta = 45^{\circ}$ into out formula for

yields that $OB=\left(\dfrac{u}{g}\right)^2(g-f)$

Note that $OA = \dfrac{2u^2}{g^2}(\frac{1}{2}g \sin 2\theta - \frac{f}{2}(1-\cos 2\theta))$
$=\left(\dfrac{u}{g}\right)^2 (g\sin 2\theta + f\cos 2\theta - f)$

Also note that $g\sin 2\theta + f\cos 2\theta \equiv R\sin (2\theta + \phi)$
By equating coefficient of $\sin 2\theta$ and $\cos 2\theta$ , it follows that $R=\sqrt{g^2+f^2}$ and $\phi = \arctan \left(\dfrac{f}{g}\right)$

Therefore, OA has a maximum value of $\left(\dfrac{u}{g}\right)^2 (\sqrt{g^2+f^2} -f)$

It follows that $\boxed{\dfrac{OB}{OA} = \dfrac{g-f}{\sqrt{g^2+f^2}-f}}$ , as required.

Note that

means that the particle's displacement is changing at the same rate in both the horizontal and vertical directions. It will thus follow a path inclined at 45 degrees to the horizontal until it reaches it's highest point and will then return along the same path until it reaches the origin.

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