STEP 1 2008 QUESTION 10

For the initial part, notice that as g’ < g, after the particle crosses the boundary where gravity changes, the rate it is decelerating at decreases. Hence, the particle will travel a further distance before it stops in mid-air, and will of course thereafter accelerate downwards.

So, with that in mind, the sketch of the trajectory of the particle when g’ < g will be that it reaches a higher maximum displacement, and therefore a further horizontal displacement. I cannot sketch it here as I don’t know how to!

For the next part, we will consider the time the particle takes to reach the boundary where gravity changes, and then the time it takes to reach its maximum displacement under g’. As its path is symmetrical, the total time taken is simply the sum of these two multiplied by two.

Time taken to travel h, where gravity changes:

Using suvat:

[br][br]s = h[br]u = V\sin\alpha[br]v[br]a = -g[br]t = t[br]

h = Vt\sin\alpha \ – \ \frac{1}{2}gt^2

t = \dfrac{ V\sin\alpha - \sqrt{V^2 \sin^2\alpha - 2gh}}{g}

This is the time the particle takes to reach the boundary.

Now calculating the velocity of the particles as it crosses the boundary:

v = V\sin\alpha \ – \ gt

v = V\sin\alpha - V\sin\alpha + \sqrt{V^2 \sin^2\alpha -2gh}

v = \sqrt{V^2 \sin^2\alpha - 2gh}

Now, the particle is under g’ gravity.

Time taken to reach maximum displacement under g’:

Using v = u + at:

[br][br]s[br]u = \sqrt{V^2 \sin^2\alpha - 2gh}[br]v = 0[br]a = -g’[br]t = t[br]

0 = \sqrt{V^2 \sin^2\alpha - 2gh} \ – \ g’t

t = \dfrac{\sqrt{V^2 \sin^2\alpha - 2gh}}{g’}

So the total time is:

[br][br]T = 2 (\dfrac {V\sin\alpha - \sqrt{V^2 \sin^2\alpha - 2gh}}{g} + \dfrac { \sqrt{V^2 \sin^2\alpha - 2gh}}{g’})[br][br]

There is no horizontal acceleration, hence it follows that:

d = 2V\cos\alpha \ (\dfrac {V\sin\alpha - \sqrt{V^2 \sin^2\alpha - 2gh}}{g} + \dfrac { \sqrt{V^2 \sin^2\alpha - 2gh}}{g’})

\sqrt{V^2 \sin^2\alpha - 2gh} = (\sqrt{\sin^2\alpha}) \ (\sqrt{V^2 - 2gh\mathrm{cosec}^2\alpha})

\sqrt{V^2 \sin^2\alpha - 2gh} = \sin\alpha \ (\sqrt{V^2 - 2gh\mathrm{cosec}^2\alpha})

Thus:

\sqrt{V^2 \sin^2\alpha - 2gh} = V'sin\alpha

, where

V' = \sqrt{V^2 - 2gh\mathrm{cosec}^2\alpha}

Hence, taking

\sin\alpha

out, we are left with:

d = 2V\cos\alpha\sin\alpha \ (\dfrac {V - V’}{g} + \dfrac {V’}{g’})

d = V\sin2\alpha \ (\dfrac {V - V’}{g} + \dfrac {V’}{g’}) \ , \ as \ required.

(edited 13 years ago)

Reply 48

Reminisce

11

Original post by nuodai

...

STEP I Question 6:

Unparseable latex formula:

f(x) = \dfrac{e^x - 1}{e - 1} \ for \ x \ \geq 0[br] \[br]Let y = f(x)[br]y = \dfrac{e^x - 1}{e - 1} \[br]y(e - 1) = e^x - 1[br]y(e - 1) +1 = e^x[br]x = \ln (y(e - 1) + 1)[br]g(x) = \ln (x(e - 1) + 1)[br] \[br]Drawing the graph...[br] \[br]y = \dfrac{e^x - 1}{e - 1}

This is an exponential through the origin when x = 0, y = 0. And also, as

x \rightarrow \infty , y \rightarrow \infty .

We could tell also by the fact that it is basically a constant multiplied by

(e^x - 1)

and obviously has an increasing gradient. We should also notice that when x = 1, g(1) = f(1) = 1. Therefore, they intersect at (1,1) and g(x) is f(x) reflected on y = x.

Unparseable latex formula:

\int^{\frac{1}{2}}_0 f(x) \ dx[br]= \int^{\frac{1}{2}}_0 \dfrac{e^x - 1}{e - 1} \ dx[br]= \dfrac{1}{e - 1} \int^{\frac{1}{2}}_0e^x - 1 \ dx[br]= \dfrac{1}{e - 1} \left[ e^x - x\right]^{\frac{1}{2}}_0[br]= \dfrac{1}{e - 1} \left[ \sqrt e - \frac{3}{2} \right] = \dfrac{2 \sqrt e - 3}{2(e - 1)}[br]\[br]\int^k_0 g(x) \ dx = \int^k_0 \ln (x(e - 1) + 1) \ dx [br]Let \ u = x(e - 1) + 1 = ex - x + 1[br]\dfrac{du}{dx} = e - 1[br]\dfrac{1}{e - 1} du = dx

x = k \ (k = \dfrac{1}{\sqrt e + 1} ) [br]u = \dfrac{e - 1}{\sqrt e + 1} + 1[br]u = \dfrac{e - 1 + \sqrt e + 1}{\sqrt e + 1} = \dfrac{\sqrt e + e}{\sqrt e + 1}[br]u = \dfrac{\sqrt e (1 + \sqrt e)}{1 + \sqrt e)} = \sqrt e[br]x = 0, u = 1[br]\int^{\sqrt e}_1 \dfrac{1}{e - 1} \ln u \ du[br]\int \ln x \ dx = x \ln x - x + c[br]\dfrac{1}{e - 1} \left[ u \ln u - u \right]^{\sqrt e}_1 = \dfrac{1}{e - 1} \left[ \sqrt e \ln \sqrt e - \sqrt e + 1 \right][br]\dfrac{1}{e - 1} \left[ \dfrac{\sqrt e}{2} - \sqrt e + 1 \right] = \dfrac{\sqrt e - 2\sqrt e + 2}{2(e - 1)}[br]= \dfrac{2 - \sqrt e}{2(e - 1)}[br]

\int^{\frac{1}{2}}_{0} f(x) \ dx + \int^{k}_{0} g(x) \ dx = \dfrac{2 \sqrt e - 3}{2(e - 1)} + \dfrac{2 - \sqrt e}{2(e - 1)}[br]= \dfrac{\sqrt e - 1}{2(e - 1)} = \dfrac{1}{2} (\dfrac {\sqrt e - 1}{e - 1})[br]

Using the difference of two squares on the denominator:

= \dfrac{1}{2} \dfrac {\sqrt e - 1}{(\sqrt e - 1)(\sqrt e + 1)}[br]= \dfrac{1}{2 (\sqrt e + 1)}[br]

As given.

Finally, we should notice the relation that

f(\frac{1}{2}) = k \ and \ g(k) = \frac{1}{2}

. Therefore, because g(x) is the inverse of f(x), we can reflect the area given by the definite integral of f(x) on the y = x line and the area would form a rectangle with the definite integral of g(x) going k along the x axis and a half along the x axis.

The area is therefore

\frac{k}{2}

which is also the result we arrived at.

(I'll make a graph if necessary.)

(edited 12 years ago)

Reply 49

Give

nuodai

...

STEP I Q7

Spoiler

Reply 50

metaltron

11

Original post by Oh I Really Don't Care

STEP I, question 1

Introduction

If a number , x, is rational it can be expressed as the ratio of two integers p and q. That is;

x = \frac{p}{q} \; p,q \in \mathbb{Z} \; q \ne 0

. If a number x is irrational it cannot be epressed as the ratio of two integers.

A For the sake of contradiction we assume that pq is irrational but p and q are rational. That is,

p = \frac{a}{c} \; and q = \frac{b}{d} \; so \; pq = \frac{ab}{cd}

the integers are closed under multiplcation so it follows that ab/cd is the ratio of two integers: a contradiction.

B As above;

p + q = \frac{ad + bc}{dc}

and we note the integers are closed under addition and the contradiction establishes the result.

C

\pi + \left(1 - \pi\right) = 1

Labelling;

pi + e = M

pi - e = N

pi^2 - e^2 = O

pi^2 + e^2 = P

Adding M and N we realise by A and B that oone of M and N is irrational. We do the same and realise that one of O and P is irrational.

We note:

M = O/N and that N = O/M

assuming that O and N and O and M are both rational leads to a contradiction of the established results about the irrationality of either N or M so we realise that P and one of M,N may be rational

Spoiler

(edited 11 years ago)

Reply 51

und

16

Can I offer a nicely written up proof for the last part of I Q1?

STEP I 2008 Q1.pdf64.9KB

Reply 52

thorn0123

12

Original post by nuodai

STEP I Question 3

Introduction

Part (i)

Part (ii) coming when I can be bothered with tedium.

Could you finish part (i) and part (ii) please?

Reply 53

Felix Felicis

13

Original post by nuodai

x

Solution for I, Q9

Position of P

Velocity of P

Total kinetic energy

(edited 11 years ago)

II/12:

part i)

Spoiler

part ii)

Spoiler

Reply 55

brianeverit

9

2008 STEP I question 11
If friction is limiting then the total reactions at

P_1\mathrm{\ and\ }P_2 \mathrm{\ are\ at\ an\ angle\ }\lambda

to the normals and meet at C, directly above the mid-point of the rod.

Unparseable latex formula:

\angleP_1OP_2=90^{\circ},\mathrm{\ so\ } \angleP_1CP_2=90^{\circ}\mathrm{\ and\ }P_1OCP_2 \mathrm{\ is\ cyclic\ }

G is the mid-point of the rod and is obviously the centre of the circle

P_1OCP_2

Unparseable latex formula:

\mathrm{if\ }ON\mathrm{\ is\ vertical\ then\ we\ have}\angleCGO=2\lambda

(angle at centre theorem)

Unparseable latex formula:

\angleGON=2\lambda \mathrm{\ alternate\ angles\ }\angleNOP_1=\alpha \mathrm{\ alternate\ angles\ and\ }\angleGOP_1=\frac{\pi}{4}

\mathrm{so\ }2\lambda+\alpha=\frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{4}-2\lambda

\mathrm{hence\ }\tan\alpha=\frac{\tan\frac{\pi}{4}-\tan2\lambda}{1+\tan\frac{\pi}{4}\tan2\lambda}=\frac{1-\frac{2\mu}{1-\mu^2}}{1+\frac{2\mu}{1+\mu^2}} \mathrm{\ since\ }\tan\lambda=\mu

\mathrm{i.e.\ }\tan\alpha=\frac{1-\mu^2-2\mu}{1-\mu^2+2\mu} \mathrm{\ as\ required}

Unparseable latex formula:

\mathrm{if\ }\theta\mathrm{\ is\ the\ angle\ of\ the\ rod\ to \ the\ horizontal\ then\ }\angleCGO=\theta\mathrm{\ hence\ }\theta=2\lambda

Unparseable latex formula:

2\lambda+\alpha=\frac{\pi}{4} \Rightarrow 0<\lambda<22\frac{1}{2}^\circ}

2008 STEP I .11.jpg19.4KB

Reply 56

brianeverit

9

2008 STEP II Question 11.
By conservation of momentum horizontally

Unparseable latex formula:

-kmc+m(-v+u\cos\theta)=0\mathrm{\ and\ }u=at\mathrm_\ so}

v(k+1)=at\cos\theta\Rightarrow\frac{dv}{dt}=\frac{a\cos\theta}{k+1}\mathrm{\ as\ required}

\mathrm{If\ P\ appears\ to\ descend\ along\ a\ line\ at\ }45^\circ \mathrm{\ to\ the\ horizontal\ then}

Unparseable latex formula:

at\cos\theta-v=at\sin\theta \Rightarrowv=at(\cos\theta-\sin\theta)

\Rightarrow \frac{dv}{dt}=a(\cos\theta-\sin\theta) \Rightarrow\frac{a\cos\theta}{k+1}=a(\cos\theta-\sin\theta)

\Rightarrow \cos\theta=(k+1)(\cos\theta-\sin\theta)

Unparseable latex formula:

\Rightarrow \sin\theta= \frac{k\cos\theta}{k+1} \RIghtarrow\tan\theta= \frac{k}{k+1} \mathrm{\ as\ required}

k=3 \Rightarrow \tan\theta=\frac{3}{4}, \mathrm{\ so\ }\sin\theta=\frac{3}{5} \mathrm{\ and\ }\cos\theta=\frac{4}{5}\mathrm{\ so\ }\frac{dv}{dt}=\frac{a}{5}

Unparseable latex formula:

\mathrm{Resolving\ along\ the\ slope\ }mg\sintheta-\muN=m\left(-\frac{dv}{dt}\cos\theta+a\Right) \Rightarrow3mg-5\muN=5m\left(-\frac{4a}{25}+a\right)

Unparseable latex formula:

\mathrm{Resolving\ perpendicular\ to\ slope\ }mg\cosTheta-N=m\frac{dv}{dt}\sin\theta \Rightarrow4mg-5N=\frac{3am}{5}

\Rightarrow N=\frac{20mg-3am}{25}

\mathrm{so\ }3mg-\frac{(20mg-3am)\mu}{5}=\frac{21am}{5} \Rightarrow 15g-20g\mu+3a\mu=21a \Rightarrow a=\frac{(20\mu-15)g}{(3\mu-21)}

If P is released with

\tan\theta\le\mu

then total contact force on particle will be to the left of vertical (see diagram) so wedge will slide to the left but the particle will not move relative to the wedge.

2008 STEP II.11.jpg10.6KB

Reply 57

brianeverit

9

2008 STEP III Question 11.
If the resisting couple (G) is constant then using

G=I\alpha

for the second phase of the motion we have

G=\frac{I}{T}\omega_0

and rotational K.E. Used doing work against the couple is

\frac{1}{2}I\omega_0^2=2\pi n_2G

Hence,

\frac{1}{2}I\omega_0^2=2\pi n_2\times \frac{I\omega_0}{T} \Rightarrow \frac{1}{2}I\omega_0T=2\pi n_2 \mathrm{\ as\ required}

If the particle descends a distance h in the first part of the motion then

h=2\pi rn_1

and its speed will be

v=r\omega_0

so by the work-energy principle

mgh-2\pi n_1G=\frac{1}{2}I\omega_0^2+ \frac{1}{2} mv^2

\mathrm{now\ }h=2\pi rn_1,G=\frac{I\omega_0^2}{4\pi n_2},\omega_0^2=\frac{16\pi^2n_2^2}{T^2} \mathrm{\ and\ }v^2=r^2\omega_0^2

\mathrm{so\ }2\pi mgrn_1-\frac{\pi n_1I}{2\pi n_2}\times\frac{16\pi^2n_2^2}{T^2}= \frac{8I\pi^2n_2^2}{T^2}+ \frac{8mr^2\pi^2n_2^2}{T^2}

\Rightarrow \pi mgrn_1T^2-4\pi n_1I\pi n_2=4I\pi^2n_2^2+4mr^2\pi^2n_2^2

\Rightarrow I\left(4\pi^2n_1n_2+4\pi^2n_2^2 \right)=\pi mgrn_1T^2-4mr^2\pi^2n_2^2

\Rightarrow I=\frac{mgrn_1T^2-4mr^2\pi n}{4\pi n_2(n_1+n_2)}\mathrm{\ as\ required}

(edited 10 years ago)

Reply 58

brianeverit

9

2008 STEP III Question 13

\mathrm{P(1\ ring\ created\ at\ first\ step)\ }=\frac{1}{2n-1} \mathrm{\ and\ P(no\ ring\ created)\ }=\frac{2n-2}{2n-1}

\mathrm{so\ E(number\ of\ rings\ created\ at\ first\ step)\ n}=0\times\frac{2n-2}{2n-1}+1\times\frac{1}{2n-1}=\frac{1}{2n-1}

Immediately after the first step there will be 2n-2 free ends left regardless of the outcome of the first step so E(number of rings created at second step)

=\frac{1}{2n-3}

And in general, E(number created at r-th step)

=\frac{1}{2n-(2r-1)}

So, E(number of rings at end of process)

=\frac{1}{2n-1}+\frac{1}{2n-3}+\frac{1}{2n-5}+...+\frac{1}{1} \mathrm{\ or\ }\displaystyle\sum_{r=1}^n \frac{1}{2n+1-2r}

Var(number of rings at first step)

=\frac{1}{2n-1}-\left(\frac{1}{2n-1}\right)^2=\frac{2(n-1)}{(2n-1)^2}

Var(number of rings at second step)

Unparseable latex formula:

=\frac{1}{2n-3}-\left(\frac{1}{2n-3}\righ)^2=\frac{2(n-2)}{(2n-3)^2} \mathrm{\ etc.}

So since number of rings created at each step are independent.
Var(number at end of process)

= \frac{2(n-1)}{(2n-1)^2}+\frac{2(n-2)}{(2n-3)^2}+\frac{2(n-3)}{(2n-5)^2}+...+\frac{2}{3^2}

\mathrm{For\ }n=40000,

E(number of rings created)

=1+\frac{1}{3}+\frac{1}{5}+...+ \frac{1}{79999}

=1+\frac{1}{2}+ \frac{1}{3}+ \frac{1}{4}+...+ \frac{1}{80000}-\left(\frac{1}{2}+ \frac{1}{3}+ \frac{1}{5}...+ \frac{1}{80000}\right)

=\ln80000- \frac{1}{2}\ln40000=\ln\left( \frac{80000}{200}\right)=2\ln20 \approx 6

Reply 59

brianeverit

9

2008 STEP II Question 13

\mathrm {(i)\ Black\ counter\ ends\ in\ P\ if\ either\ it\ does\ not\ move\ at\ all\ or\ it\ moves\ from\ P\ to\ Q\ and\ then\ back\ from\ Q\ to\ P}

\mathrm{P(Black\ ball\ does\ not\ move\ from\ P)\ }=\frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)!k!(n-k)!}{(k-1)!(n-k)!n!}=\frac{n-k}{n}

\mathrm{P(black\ ball\ goes\ from\ P\ to\ Q)}=\frac{\binom{n-1}{k-1}}{\binom{n}{k}}=\frac{(n-1)!k!(n-k)!}{(k-1)!(-k)(n-k)!n!}=\frac{k}{n}

]\mathrm{P(black\ ball\ goes\ from\ Q\ to\ P)}=\frac{\binom{n+k-1}{k-1}}{\binom{n+1}{k}}=\frac{(n+k-1)!k!n!}{(k-1)!n!(n+k)!}=\frac{k}{n+k}

\mathrm{so\ P(black\ ball\ ends\ in\ P)}=\frac{n-k}{n}\times \frac{k}{n} \times \frac{k}{n+k}= \frac{n^2}{n(n+k)}= \frac{n}{n+k}

\mathrm{The\ probability\ is\ obviously\ maximised\ when\ }k=0 \mathrm{\ since\ it\ is\ then\ }=1

\mathrm{ii)\ The\ two\ black\ balls\ can\ end\ up\ in\ same\ bag\ in\ three\ ways:}

\mathrm{(a)\ Black\ ball\ not\ moved\ from\ P\ but\ then\ moved\ from\ Q}

\mathrm{(b)\ Black\ ball\ moved\ from\ P\ but\ not\ from\ Q}

\mathrm{(c)\ Black\ ball\ moved\ from\ P\ then\ both\ moved\ from\ Q}

]

\mathrm{P\ of\ (a)\ }=\frac{\binom{n-1}{k}}{\binom{n}{k}}\times\frac{\binom{n+k-1}{k-1}}{\binom{n+k}{k}}=\frac{(n-1)!k!(n-k)!}{k!(n-1-k)!n!}\times\frac{(n+k-1)!k!n!}{(k-1)!n!(n+k)!}=\frac{n-k}{n}\times\frac{k}{n+k}= \frac{k(n-k)}{n(n+k)}

\mathrm{P\ of\ (b)\ }=\frac{\binom{n-1}{k-1}}{\binom{n}{k}}\times\frac{ \binom{n+k-2}{k-2}}{ \binom{n+k}{k}}=\frac{(n-1)!k!(n-k)!}{(k-1)!(n-k)!n!}\times\frac{(n+k-2)!k!n!}{(k-2)!n!(n+k)!}=\frac{k}{n}\times \frac{k(k-1)}{n(n+k)}= \frac{k^2(k-1)}{n(n+k)(n+k-1)}

\mathrm{P\ of\ (c)\ }=\frac{\binom{n-1}{k-1}}{\binom{n}{k}}\times\frac{ \binom{n+k-2}{k-2}}{ \binom{n+k}{k}}=\frac{(n-1)!k!(n-k)!}{(k-1)!(n-k)!n!}\times\frac{(n+k-2)!k!n!}{k!(n-2)!(n+k)!}=\frac{k}{n}\times \frac{n(n-1)}{(n+k)(n+k-1)}= \frac{k(n-1)}{(n+k)(n+k-1)}

\mathrm{so\ probability\ that\ both\ black\ balls\ end\ up\ in\ same\ bag\ } = \frac{k(n-k)}{n(n+k)}+\frac{k^2(k-1)}{n(n+k)(n+k-1)}+\frac{k(n-1)}{(n+k)(n+k-1)}

=\frac{k(n-k)(n+k-1)+k^2(k-1)+kn(n-1)}{n(n+k)(n+k-1)}=\frac{k(n^2-k^2-n+k+k^2-k+n^2-n)}{n(n+k)(n+k-1)}

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

\mathrm{Let\ f}(k)=\frac{k}{(n+k)(n+k-1)}\mathrm{\ then\ f}'(k)=\frac{(n+k)(n+k-1)-k(n+k+n+k-1)}{(n+k)^2(n+k-1)^2}=\frac{n^2-k^2-n}{(n+k)^2(bn+k-1)^2}=0\mathrm{\ if\ }k^2=n^2-n \Rightarrow k=\sqrt{n^2-n}

\mathrm{Only\ the\ positive\ root\ considered\ since\ }k>0

\mathrm{Probability\ is\ a\ minimum\ when\ }k=0 \mathrm{\ since\ it\ is\ then\ zero\ so\ }k=\sqrt{n^2-n} \mathrm{\ gives\ maximum\ probability\ }

\mathrm{But\ }n^2-n \mathrm{\ is\ not\ a\ perfect\ square\ so\ we\ consider\ the\ integers\ on\ either\ side\ of\ }\sqrt{n^2-n}

k=n \Rightarrow p=\frac{2n(n-1)}{2n(2n-1)}=\frac{n-1}{2n-1} \mathrm{\ and\ }k=n-1 \Rightarrow p=\frac{2(n-1)^2}{(2n-1)(2n-2)}=\frac{n-1}{2n-1} \mathrm{\ so\ }k=n \mathrm{\ or\ }n-1

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