STEP Maths I, II, III 1994 Solutions

Reply 180

Original post

by zrancis

STEP I - Question 11

i)

To ensure the ball hits the ground, it's mass must be greater than the parallel component of the wagon:

$M>mgsin\theta$

By considering both bodies separately, there will be a tension pulling the wagon up the slope and a tension resisting the ball from falling, by applying N2L, firstly to the wagon:

$T-mgsin\theta=ma_c$

and to the ball:

$Mg-T=Ma_c$

Adding these gives:

$Mg-mgsin\theta=a_c(M+m)$

So:

$a_c=\frac{g(M-msin\theta)}{M+m}$

By taking the length of the slope to be: $s=dsin\theta$
And applying the fact the system starts from rest,
Suvat can be applied:

$v^2=u^2+2a_cs$

$v^2=0+2\frac{g(M-msin\theta)}{M+m}dsin\theta$

$v^2=\frac{2g(M-msin\theta)dsin\theta}{M+m}$

As required.

ii)

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

$v^2\le0$

hence:

$\frac{2g(M-msin\theta)dsin\theta}{M+m}\le0$

$sin\theta(M-msin\theta)\le0$

So:

Provided $sin\theta\ge0$

$M-msin\theta\le0$

QED (I hope)

I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.

Original post

by DFranklin

I don't think this is right. I think the scenario you are supposed to be considering is "ball falls towards floor, accelerating wagon. Ball hits floor, wagon carries on under it's own momentum, (but now has gravity providing -ve acceleration). Does wagon have enough momentum to hit top?"

(Incidentally, the condition $M > m sin \theta$ is basically the condition that the ball is heavy enough to accelerate the wagon in the first place).

I didnt look at momentum. I did it using the work energy principle.

I got an answer of 1-sinx>{(M-msinx)(sinx)}/(M+m)

Is this right? (Used x instead of theta as clearer to read...)

Reply 181

Adug Deguengo

Original post

by eponymous

Step I Q1

Since all planes are at the same angle \theta to the horizontal, the roof ridge is central.

The height (h) of the roof (see image 1) is equal to $p\tan\theta$ . Therefore $x$ (see diagram) = $\frac{h}{\tan\theta} = \frac{p\tan\theta}{\tan\theta} = p$ . Since the ridge is central, this will be the same at the other end of the roof.

So R (ridge length) = 2q - 2p, since $\theta$ is not present in this expression, the ridge length is independent of $\theta$ .

Volume of the roof:

Slicing up the roof, we can join the two ends to make a pyramid and leave a prism. (see image 2)

The volume of a pyramid = $\frac{1}{3} \times area of base \times height(h)$

Area of base = $p^{2}$

So its volume = $\frac{1}{3} \times p^{2} \times p\tan\theta = \frac{p^{3}}{3} \tan\theta.$

Volume of a prism = $area of face \times length$

Area of face = $\frac{1}{2} \times 2p \times h = p^{2} \tan\theta.$

So the volume of the prism = $p^{2} \tan\theta \times (2q - 2p) = 2p^{2}(q - p)\tan\theta$

So the total volume of the roof = volume of prism + volume of pyramid = $\frac{p^{3}}{3} \tan\theta + 2p^{2} (q - p)\tan\theta = \frac{p^{2}}{3} (6q - 5p) \tan\theta.$

there probably is a much simpler way to do this :redface:

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way :smile:

*

I don't know if someone has pointed this out already, but, shouldn't the volume of the prism be $\frac {4p^{3}\tan\theta}{3}$ as p = x from before (since R = 2q - 2x = 2q - 2p).This would make the base of the pyramid 2p and 2p, and the height remains $p\tan\theta$ .

Reply 182

shiny718

Original post

by SimonM

STEP III, Question 12

Spoiler

Hi there! I tried this question but I got stucked with it. For the first part, I got something like

$p_a p_b ( \frac{q_a}{1-q_a p_b} + \frac{1}{1-p_a q_b})$ based on: if A wins in two rounds when A serves first, probability is $p_a p_b$ , if player A wins in three rounds, probability is $q_a p_a p_b$ , if player A wins in four rounds, probability is $p_a q_b p_a p_b$ since player B cannot win in two consecutive rounds, and so on... so actually $p_a q_b$ and $p_b q_a$ are alternating in two separate cases which are then summed up to get probability of A winning if she serves first. But unfortunately i couldn't get the right answer. So something should be wrong with my solution.

I don't really understand how you got the first part of your solution.

How did you get $P_{\text{short}} = p_Ap_B + (p_Aq_B+q_Ap_B)P_{\text{short}}$ ? If there is a series of $( q_A p_B)$ times with $p_A p_B$ , wouldn't A already win 3 consecutive times in the first place? Please correct me if i have misunderstood the question. Thanks!

Reply 183

brianeverit

9

Original post

by shiny718

Hi there! I tried this question but I got stucked with it. For the first part, I got something like

$p_a p_b ( \frac{q_a}{1-q_a p_b} + \frac{1}{1-p_a q_b})$ based on: if A wins in two rounds when A serves first, probability is $p_a p_b$ , if player A wins in three rounds, probability is $q_a p_a p_b$ , if player A wins in four rounds, probability is $p_a q_b p_a p_b$ since player B cannot win in two consecutive rounds, and so on... so actually $p_a q_b$ and $p_b q_a$ are alternating in two separate cases which are then summed up to get probability of A winning if she serves first. But unfortunately i couldn't get the right answer. So something should be wrong with my solution.

I don't really understand how you got the first part of your solution.

How did you get $P_{\text{short}} = p_Ap_B + (p_Aq_B+q_Ap_B)P_{\text{short}}$ ? If there is a series of $( q_A p_B)$ times with $p_A p_B$ , wouldn't A already win 3 consecutive times in the first place? Please correct me if i have misunderstood the question. Thanks!

Neither player can possibly win in 3 games. The sequence of winners must be
AA ,ABAA, BAAA etc for A to have a 2 game lead over B.

Reply 184

shiny718

Original post

by brianeverit

Neither player can possibly win in 3 games. The sequence of winners must be
AA ,ABAA, BAAA etc for A to have a 2 game lead over B.

Ah I understand now! Thank you very much! :biggrin:

Reply 185

brianeverit

9

Original post

by shiny718

Ah I understand now! Thank you very much! :biggrin:

Your welcome

Reply 186

newblood

19

Original post

by DFranklin

$\displaystyle\frac{a_{2n+1}}{a_{2n-1}} = \frac{(2n-1)^2}{2n(2n+1)}$

Let $b_n$ be the corresponding coefficients for g, so that $g(x) = \sum b_n x^n$ . Then $\displaystyle\frac{b_{2n+1}}{b_{2n-1}} = \frac{2n-1}{2n+1}$

So if $c_n = a_n / b_n$ , then $\displaystyle\frac{c_{2n+1}}{c_{2n-1}} = \frac{(2n-1)^2(2n+1)}{(2n-1)2n(2n+1)} = \frac{2n-1}{2n} < 1$

So since $a_3 < b_3$ , and the $a_n,b_n$ are all non-negative, we have $a_n<b_n$ for all (odd) $n \ge 3$ .

(N.B. Am using my wife's laptop as my machine is chkdsk'ing a 250GB drive right now, so I don't have the question in front of me)

how did you get $\displaystyle\frac{a_{2n+1}}{a_{2n-1}} = \frac{(2n-1)^2}{2n(2n+1)}$

ive been trying for ages and cant for the life of me get it -.-

my formula for general term of f(x) as a power series is correct as ive checked on the internet so thats not the problem

do you mind showing a bit of working for my benefit please ,

cheers

Reply 187

Omghacklol

8

For I Q4 is it good enough to show the indefinite integral = bla bla + c by differentianting the RHS and obtaining the integrand of the LHS? Is this worthy of all the marks?

Reply 188

davros

16

Original post

by Omghacklol

For I Q4 is it good enough to show the indefinite integral = bla bla + c by differentianting the RHS and obtaining the integrand of the LHS? Is this worthy of all the marks?

Well you could, but I'm not sure you're saving much time - it's only a completing-the-square plus quoting the formula booklet integral to start with (being careful about justifying why you need |k| < 1)

Reply 189

mikelbird

8

Original post

by DFranklin

I think you can go further for the last part. We have:

$e^{i\gamma}(1+e^{i(\theta+\beta-\gamma)}+e^{i(2\theta+\alpha -\gamma)}) = 0$ . Since $e^{i\gamma} \neq 0$ we must have $1+e^{i(\theta+\beta-\gamma)}+e^{i(2\theta+\alpha -\gamma)}) = 0$ (*)

Write u, v to be the principal arguments of $e^{i(\theta+\beta-\gamma)}, e^{i(2\theta+\alpha-\gamma)}$ respectively. Considering the imaginary part of (*), we see $Im[e^{iu}] = -Im[e^{iv}]$ . Considering the real part, we see $e^{iu}, e^{iv}$ must both have real part <= 0. So we must in fact have u = -v, which in tern implies 2cos(u)= -1. So we finally deduce that $u =\theta, v=-\theta$ or vice versa.

Now $u=\theta \implies \beta-\gamma = 0$ contradicting the uniqueness of B,C. So $u=-\theta (\equiv 2\theta \pmod{2\pi})$ and so $\theta+\beta-\gamma = -\theta \pmod{2\pi} \implies \beta = \gamma + \theta \pmod{2\pi}$ .

Similarly the value for u gives $2\theta+\alpha-\gamma = \theta \pmod{2\pi} \implies \alpha = \gamma + 2\theta \pmod{2\pi}$ .

So in fact, A, B, C are the vertices of an equilateral triangle.

Just looking at this question and tried to tease a bit more out of it....

Step1994Paper2Question6.pdf78.7KB

Reply 190

Brubeckian

9

Need help for Q4 on III, there are two solutions to the differential equation, two quadratics right? I'll try to use latex to show where I am.

$(\frac{dy}{dx} )^2=4y$

$\frac{dy}{dx}=\pm2\sqrt y$

$\sqrt y = x + A$ OR $\sqrt y = -x +B$

Right, so these are the two general solutions, the question asks find the two solutions which pass through the point (a,b^2)

If I sub in b^2 into either of my equations won't I get two different constant values for EACH equation giving me in total 4 equations ? :s-smilie:

I know that's not right but honestly can't see where I'm going wrong, any help will be appreciated, thank you.

Edit; I just read something interesting about how for root(a) for example, we always take that value as positive as we're not solving for anything, is that correcT?

Reply 191

raff97

8

Original post

by Brubeckian

Need help for Q4 on III, there are two solutions to the differential equation, two quadratics right? I'll try to use latex to show where I am.

$(\frac{dy}{dx} )^2=4y$

$\frac{dy}{dx}=\pm2\sqrt y$

$\sqrt y = x + A$ OR $\sqrt y = -x +B$

Right, so these are the two general solutions, the question asks find the two solutions which pass through the point (a,b^2)

If I sub in b^2 into either of my equations won't I get two different constant values for EACH equation giving me in total 4 equations ? :s-smilie:

I know that's not right but honestly can't see where I'm going wrong, any help will be appreciated, thank you.

Edit; I just read something interesting about how for root(a) for example, we always take that value as positive as we're not solving for anything, is that correcT?

The wording on that question gave me aids

Reply 192

Brubeckian

9

Original post

by raff97

The wording on that question gave me aids

All of STEP gives me AIDS :P

Reply 193

raff97

8

Original post

by Brubeckian

All of STEP gives me AIDS :P

Ikr I'm starting to regret my life decisions. Why have I committed my life to maths?!?! I shouldve just went to a nice quiet top 20 uni for some sort of engineering and enjoyed my time at uni.........

Reply 194

IrrationalRoot

20

Original post

by zrancis

STEP I - Question 4

i)

$cos(2\alpha)=2cos^2\alpha-1$

$cos\alpha=\sqrt{\frac{1}{2}(cos(2\alpha)+1)}$

Similarily:

$sin\alpha=\sqrt{\frac{1}{2}(cos(2\alpha)-1)}$

Hence:

$tan(\frac{\alpha}{2})=\frac{\sqrt{\frac{1}{2}(cos \alpha-1)}}{\sqrt{\frac{1}{2}(cos\alpha+1)}}$

$tan(\frac{\alpha}{2})=\frac{\sqrt{\frac{1}{2}(1-cos \alpha)}}{\sqrt{\frac{1}{2}(-1-cos\alpha)}}$
Rationalising:

$tan(\frac{\alpha}{2})=\frac{\sqrt{(1-cos\alpha)^2}}{\sqrt{(-1-cos\alpha)(1-cos\alpha)}}$

$tan(\frac{\alpha}{2})=\frac{(1-cos\alpha)}{\sqrt{(sin^2\alpha)}}$

$tan(\frac{\alpha}{2})=\frac{1-cos\alpha}{sin\alpha}$

ii)

$\int^1_0\frac{sin\alpha}{1-2xcos\alpha+x^2}dx$

$\int^1_0\frac{sin\alpha}{(x-cos\alpha)^2+1-cos^2\alpha}dx$

$\int^1_0\frac{sin\alpha}{(x-cos\alpha)^2+sin^2\alpha\alpha}dx$

Using the result given:

$=[sin\alpha.\frac{1}{sin\alpha}tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0$

$=[tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0$

$=[tan^{-1}\frac{1-cos\alpha}{sin\alpha}]-[tan^{-1}\frac{-cos\alpha}{sin\alpha}]$
From part i)

$tan(\frac{\alpha}{2})=\frac{cos\alpha+1}{sin\alpha}$

$\frac{\alpha}{2}=tan^{-1}\frac{cos\alpha+1}{sin\alpha}$

By also noting that $tan^{-1}(-cot\alpha)=\alpha-\frac{\pi}{2}$

$=[tan^{-1}\frac{x-cos\alpha}{sin\alpha}]^1_0=[\frac{\alpha}{2}]-[\alpha-\frac{\pi}{2}]$

$=\frac{1}{2}(\pi-\alpha)$

QED

Lol this solution misses out the bulk of the question (the middle part).
Also the first part can be made a lot less messy and much, much quicker by simply letting say theta=1/2(alpha) and the results quickly follows from the double angle formulae.

If anyone wants an idea of the solution to the middle part look up the AQA FP2 online textbook and find the derivation of the result in Chapter 5. It's not a perfect derivation, but it somewhat suffices. Then it should be clear how to apply that derivation to the given integral in the question (complete square, sub u=x-k, etc.).

Reply 195

IrrationalRoot

20

Original post

by davros

Well you could, but I'm not sure you're saving much time - it's only a completing-the-square plus quoting the formula booklet integral to start with (being careful about justifying why you need |k| < 1)

See it's extremely unclear whether you're allowed to use the formula booklet to do this integral. It would make the whole question completely trivial; the only somewhat involved part is deriving that result.
So I did the whole derivation (I say whole, but that just means one extra substitution).
If the question did allow the use of the formula booklet result, then it's terribly worded.

Reply 196

IrrationalRoot

20

Original post

by Adug Deguengo

I don't know if someone has pointed this out already, but, shouldn't the volume of the prism be $\frac {4p^{3}\tan\theta}{3}$ as p = x from before (since R = 2q - 2x = 2q - 2p).This would make the base of the pyramid 2p and 2p, and the height remains $p\tan\theta$ .

This is correct and is what I did.
Also I got 4pqsec(theta) as the surface area if that's what anyone else got.

Reply 197

DavidOrrell

2

Original post

by zrancis

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

$v^2\le0$

hence:

$\frac{2g(M-msin\theta)dsin\theta}{M+m}\le0$

$sin\theta(M-msin\theta)\le0$

So:

Provided $sin\theta\ge0$

$M-msin\theta\le0$

QED (I hope)

I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.

I do disagree with the reasoning here. I'm new to STEP so I'm hardly an expert, so please don't take it the wrong way :smile:

That condition would indeed mean the wagon doesn't hit the pulley. However if you look back at the expression you derived for acceleration, you would notice that it would make the acceleration negative. Since we defined positive as being up the slope, this situation would have the wagon accelerating down the slope from the word go.

I think what the question is getting at is to realise that the length of the slope is greater than the distance that the ball falls. Therefore once the ball hits the ground, the wagon has a velocity and it has a surplus distance to cover before hitting the pulley. Since the cable is now slack, we are just looking at deceleration due to the weight component parallel to the slope, and we seek to find the condition for which the wagon reaches zero velocity before it reaches the pulley.

The displacement from point of slackening to pulley is $d(1-sin\theta)$ . The velocity at the point of slackening squared is $\frac{2g(M-m\sin\theta)d\sin\theta}{M+m}$ . We are looking for final velocity = 0, at a constant acceleration of $-g\sin\theta$ . Using $s = \frac{v^{2}-u^{2}}{2a}$ , we find that

$s = \frac{d(M-m\sin\theta)}{M+m}$

Finally we want the condition that the distance covered is smaller than $d(1-\sin\theta)$

Hence

$\frac{M-m\sin\theta}{M+m} < 1-\sin\theta$

$\frac{M(1+\sin\theta)}{M+m} < 1$

I think that's right? Tell me about any corrections, I probably made a million mistakes there!

Reply 198

beresford

3

Hi Guys,
Re: 1994; pII, Qs 9 & 10: my comp. cannot open the files containing these two solutions (not had this prob with filed soltions on other years' papers). Does anyone have these to hand to flash onto the sreen?
Thank you.

Beresford

Reply 199

username2452153

16

Original post

by zrancis

STEP I - Question 11

i)

To ensure the ball hits the ground, it's mass must be greater than the parallel component of the wagon:

$M>mgsin\theta$

By considering both bodies separately, there will be a tension pulling the wagon up the slope and a tension resisting the ball from falling, by applying N2L, firstly to the wagon:

$T-mgsin\theta=ma_c$

and to the ball:

$Mg-T=Ma_c$

Adding these gives:

$Mg-mgsin\theta=a_c(M+m)$

So:

$a_c=\frac{g(M-msin\theta)}{M+m}$

By taking the length of the slope to be: $s=dsin\theta$
And applying the fact the system starts from rest,
Suvat can be applied:

$v^2=u^2+2a_cs$

$v^2=0+2\frac{g(M-msin\theta)}{M+m}dsin\theta$

$v^2=\frac{2g(M-msin\theta)dsin\theta}{M+m}$

As required.

ii)

If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:

$v^2\le0$

hence:

$\frac{2g(M-msin\theta)dsin\theta}{M+m}\le0$

$sin\theta(M-msin\theta)\le0$

So:

Provided $sin\theta\ge0$

$M-msin\theta\le0$

QED (I hope)

I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great.

Certainly, the velocity of m when M hits the ground can't be 0 or less.

In the last part, when the M hits the ground, m is only d sin theta along the plane. As M hits the ground, m has same velocity as M but since M is now no longer pulling m, it undergoes deceleration. For m to not hit the pulley, the potential energy difference should be greater than the kinetic energy of m when M collides and the solution follows.
$mgd(1-sin \theta )sin \theta <m/2 \frac{2g(M-msin \theta )}{M+m} dsin \theta[br] [br]$

Alternatively, you could use DavidOrrell's method.

Original post

by DavidOrrell

I do disagree with the reasoning here. I'm new to STEP so I'm hardly an expert, so please don't take it the wrong way :smile:

That condition would indeed mean the wagon doesn't hit the pulley. However if you look back at the expression you derived for acceleration, you would notice that it would make the acceleration negative. Since we defined positive as being up the slope, this situation would have the wagon accelerating down the slope from the word go.

I think what the question is getting at is to realise that the length of the slope is greater than the distance that the ball falls. Therefore once the ball hits the ground, the wagon has a velocity and it has a surplus distance to cover before hitting the pulley. Since the cable is now slack, we are just looking at deceleration due to the weight component parallel to the slope, and we seek to find the condition for which the wagon reaches zero velocity before it reaches the pulley.

The displacement from point of slackening to pulley is $d(1-sin\theta)$ . The velocity at the point of slackening squared is $\frac{2g(M-m\sin\theta)d\sin\theta}{M+m}$ . We are looking for final velocity = 0, at a constant acceleration of $-g\sin\theta$ . Using $s = \frac{v^{2}-u^{2}}{2a}$ , we find that

$s = \frac{d(M-m\sin\theta)}{M+m}$

Finally we want the condition that the distance covered is smaller than $d(1-\sin\theta)$

Hence

$\frac{M-m\sin\theta}{M+m} < 1-\sin\theta$

$\frac{M(1+\sin\theta)}{M+m} < 1$

I think that's right? Tell me about any corrections, I probably made a million mistakes there!

(edited 7 years ago)

STEP Maths I, II, III 1994 Solutions

Related discussions

Articles for you

How The Student Room is moderated