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STEP Maths I,II,III 1987 Solutions

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Reply 60
Go for it, I maybe got about a third of the way through before i got stumped.
Square
Go for it, I maybe got about a third of the way through before i got stumped.
So where did you get to? (Slightly curious, as it's a two part question and you say you got a third of the way...)
Reply 62
I made the subsitution for dx and then substituted for x on the denominator whereas coffeym has substituted for x-a and x-b - which makes more sense.

I think sort of fiddled around abit and got a pretty nasty expression - it would've worked if I'd changed my cos^2 into sin^2 and such would've had all the individual alphas and betas being subtracted from each other.
Reply 63
DFranklin
No.

02tdt=021dt\int_0^2 t \, dt = \int_0^2 1\, dt but this doesn't mean t = 1.

On the other hand, you can correctly argue that the -t^2 integral is negative and the other integral is positive, so they can't be equal. But even then, all you are really saying is "these two integrals aren't the same, so something must have gone wrong". To get the marks, I think you need to identify that something.



ok, fair enough.

but your evidence that when the original expression is integrated by parts, two distinct equations are produced (and added), one of which is the second integrand and hence the two cannot be equivalent would be fine for all the marks, right?

thanks for the help with this btw, it's helped clear up some confusion i had with this kind of question :smile:
kabbers
but your evidence that when the original expression is integrated by parts, two distinct equations are produced (and added), one of which is the second integrand and hence the two cannot be equivalent would be fine for all the marks, right?No, not really. The question isn't asking you "prove this method gives the wrong result", it's asking "what is wrong with this method?".

Suppose I ask you "what is wrong with the following argument?":

[indent]Suppose x = y. Then:

x2=xyx^2 = xy

x2y2=xyy2x^2-y^2= xy-y^2

(x+y)(xy)=y(xy)(x+y)(x-y) = y(x-y)

(x+y)=y(x+y) = y

2x=y2x = y

Taking the particular choice x=y=1, we find 2 = 1.[/indent]

If you say "the argument is wrong because 212 \neq 1", you won't get any marks. Instead you would need to say something like "You can't go from (x+y)(x-y) = y(x-y) to x+y = y, because you're dividing by (x-y) which is zero".

To look at it another way, the question asks "explain why the substitution x=1/t does not show the integrals are equal" (emphasis mine). So if your explanation doesn't even mention the substitution x=1/t, then you're in trouble!

In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to 011(1+x2)2dx\int_0^1 \frac{1}{(1+x^2)^2}\, dx and 101(1+x2)2dx\int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/tx=1/t.

This is somewhat belabouring the point for this particular question - I can't think the explanation would be worth many marks, and I'd expect you do get most of them for any explanation mentioning the behaviour of 1/x at x=0.
DFranklin
In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to 011(1+x2)2dx\int_0^1 \frac{1}{(1+x^2)^2}\, dx and 111(1+x2)2dx\int_{-1}^1 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/tx=1/t.

You didn't happen to mean 101(1+x2)2dx\int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx the second time?

At least that was what I did in my working...
nota bene
You didn't happen to mean 101(1+x2)2dx\int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx the second time?Yes, thanks. Darned cut'n'paste errors...
Reply 67
DFranklin
No, not really. The question isn't asking you "prove this method gives the wrong result", it's asking "what is wrong with this method?".

Suppose I ask you "what is wrong with the following argument?":

[indent]Suppose x = y. Then:

x2=xyx^2 = xy

x2y2=xyy2x^2-y^2= xy-y^2

(x+y)(xy)=y(xy)(x+y)(x-y) = y(x-y)

(x+y)=y(x+y) = y

2x=y2x = y

Taking the particular choice x=y=1, we find 2 = 1.[/indent]

If you say "the argument is wrong because 212 \neq 1", you won't get any marks. Instead you would need to say something like "You can't go from (x+y)(x-y) = y(x-y) to x+y = y, because you're dividing by (x-y) which is zero".

To look at it another way, the question asks "explain why the substitution x=1/t does not show the integrals are equal" (emphasis mine). So if your explanation doesn't even mention the substitution x=1/t, then you're in trouble!

In terms of the original question, to understand a bit better what's going on, you might want to consider what happens to 011(1+x2)2dx\int_0^1 \frac{1}{(1+x^2)^2}\, dx and 101(1+x2)2dx\int_{-1}^0 \frac{1}{(1+x^2)^2}\, dx under the substitution x=1/tx=1/t.

This is somewhat belabouring the point for this particular question - I can't think the explanation would be worth many marks, and I'd expect you do get most of them for any explanation mentioning the behaviour of 1/x at x=0.


right, ok.. i understand :smile:
thanks!
Reply 68
Step II, Q1

(i)(xa)2+(2xb)2+(3xc)2=x22ax+a2+4x24bx+b2+9x26c+c2 (\mathrm{i}) (x-\mathrm{a})^2 + (2x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = x^2 - 2\mathrm{a}x + \mathrm{a}^2 + 4x^2 - 4\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

=14x2x(2a+4b+6c)+a2+b2+c2 = 14x^2 - x(2\mathrm{a} + 4\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

=14x2x(14x)+a2+b2+c2since7x=a+2b+3c = 14x^2 - x(14x) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 \,\, \mathrm{since} \,\,7x = \mathrm{a} + 2\mathrm{b} + 3\mathrm{c}

=a2+b2+c2 = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2



(ii)(2xa)2+(3xb)2+(3xc)2=4x24ax+a2+9x26bx+b2+9x26c+c2 (\mathrm{ii}) (2x-\mathrm{a})^2 + (3x - \mathrm{b})^2 + (3x - \mathrm{c})^2 = 4x^2 - 4\mathrm{a}x + \mathrm{a}^2 + 9x^2 - 6\mathrm{b}x + \mathrm{b}^2 + 9x^2 - 6\mathrm{c} + \mathrm{c}^2

=22x2x(4a+6b+6c)+a2+b2+c2 = 22x^2 - x(4\mathrm{a} + 6\mathrm{b} + 6\mathrm{c}) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

=22x2x(22x)+a2+b2+c2since11x=2a+3b+3c = 22x^2 - x(22x) + \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 \,\, \mathrm{since} \,\,11x = 2\mathrm{a} + 3\mathrm{b} + 3\mathrm{c}

=a2+b2+c2 = \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2

ifαa+βb+γc=α2+β2+γ22x \mathrm{if}\,\, \alpha\mathrm{a} + \beta\mathrm{b} + \gamma\mathrm{c} = \frac{\alpha^2 + \beta^2 + \gamma^2}{2}x

then a2+b2+c2=(αxa)2+(βxb)2+(γxc)2 \mathrm{a}^2 + \mathrm{b}^2 + \mathrm{c}^2 = (\alpha x - \mathrm{a})^2 + (\beta x - \mathrm{b})^2 + (\gamma x - \mathrm{c})^2

Apologies if there are errors, rather rushed it.
Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?
Reply 70
DFranklin
Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?


OK, I think this is a good idea :biggrin:
Reply 71
So do I :P

I have to pop out for abit then I'm going to try some more.
Reply 72
DFranklin
Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?


I don't think any of the new comers could do STEP III yet since most of us haven't done any of the further maths modules, especially not FP3.
Reply 73
Hey. :smile:

I'm new to STEP. I'm trying to do STEP I/12. Any help would be great!

So far, I have got some equations for the vertical collisions of the ball with the steps. But in order to find the coefficient of restitution, I'm thinking that I'd need some equations for the horizontal motion of the ball. This is what I'm struggling with. My current thinking is that surely the horizontal velocity of the ball will just remain constant.

Here is my working so far.

Let unu_{n} be the initial velocity of collision nn, vnv_{n} be the final velocity of collision nn and ee be the coefficient of restitution.

Then for the first collision eu1=v1.e u_{1} = v_{1}. However from conservation of energy u1=2ghe=v12gh.u_{1} = \sqrt{2 g h} \Rightarrow e = \frac{v_{1}}{\sqrt{2 g h}}.

For the second collision u2=v1+2gh=2gh(e+1)u_{2} = v_{1} + \sqrt{2 g h} = \sqrt{2 g h} (e + 1) and so we get v2=eu2=2gh(e)(e+1).v_{2} = e u_{2} = \sqrt{2 g h} (e) (e + 1).

Unfortunately that's all I've got up to. :redface:
I need some help in getting equations involving dd which will give me more ammo for finding e.e.
Dharma
Hey. :smile:

I'm new to STEP. I'm trying to do STEP I/12. Any help would be great!

So far, I have got some equations for the vertical collisions of the ball with the steps. But in order to find the coefficient of restitution, I'm thinking that I'd need some equations for the horizontal motion of the ball. This is what I'm struggling with. My current thinking is that surely the horizontal velocity of the ball will just remain constant.Yes, horizontal velocity is constant.

I think the obvious thing to do next is to consider how long it is between the initial 'kick' and the first bounce, and how long between the first bounce and the 2nd bounce.
Reply 75
Dharma
However from conservation of energy u1=2ghu_{1} = \sqrt{2 g h}

Not really, there's an initial kinetic energy as well. (I assume your u, v are not just the vertical components of the velocity)


For the second collision u2=v1+2ghu_{2} = v_{1} + \sqrt{2 g h}


Again, not really: what conservation of energy gives you is u22=v12+2ghu_2^2 = v_1^2 + 2gh, which is not the same thing.

(disclaimer: I have not yet solved the question)

Question to anybody: Is the coefficient of restitution, in an oblique impact, equal to the ratio of the speeds after and before impact, or is it the ratio of the vertical (or in general perpendicular to the plane tangent to both the impacting objects at the point of impact at the moment of impact) components of the velocity?

I don't know this and it seems kinda essential to solving the question *shrugs*
ukgea
Not really, there's an initial kinetic energy as well. (I assume your u, v are not just the vertical components of the velocity)I was assuming they were, in which case
what conservation of energy gives you is u22=v12+2ghu_2^2 = v_1^2 + 2gh, which is not the same thing.
is effectively the same thing - the K.E. due to the horizontal velocity is just a constant, so as you only care about the change in K.E. v.s. G.P.E. to work out the vertical velocity the end result is the same. (It's perhaps not totally obvious, but do these questions enough and you tend to get a bit lazy about showing this to be the case).

Edit: ukgea is right. Ignore me about this bit...

But yes, you are right, if u,v, aren't the vertical components, Dharma's equations are incorrect. Although I'd say the only sensible way to approach this question is to consider the vertical and horizontal motion separately.

Question to anybody: Is the coefficient of restitution, in an oblique impact, equal to the ratio of the speeds after and before impact, or is it the ratio of the vertical (or in general perpendicular to the plane tangent to both the impacting objects at the point of impact at the moment of impact) components of the velocity?
The latter. The normal (pun intended) approach is to resolve the velocity into components parallel and perpendicular to the surface normal. Only the component parallel to the normal is affected by the impact (neglecting friction, which of course you do...)
DeathAwaitsU
I don't think any of the new comers could do STEP III yet since most of us haven't done any of the further maths modules, especially not FP3.
Although true for the majority of the questions, I don't see STEP III, Q1 requires anything beyond GCSE maths, let alone FM. There are a few others that should also be reasonably accessible I think.
Reply 78
DFranklin
in which caseis effectively the same thing - the K.E. due to the horizontal velocity is just a constant, so as you only care about the change in K.E. v.s. G.P.E. to work out the vertical velocity the end result is the same. (It's perhaps not totally obvious, but do these questions enough and you tend to get a bit lazy about showing this to be the case).


This I still don't get. Perhaps I just haven't done these questions enough. But I get Conservation of energy:

u2x2+u2y2=v1x2+v1y2+2ghu_{2 x}^2 + u_{2y}^2 = v_{1 x}^2 + v_{1 y}^2 + 2gh

and since horizontal velocity is constant they cancel:

u2y2=v1y2+2ghu_{2y}^2 = v_{1y}^2 + 2gh (1)

. Now if

u2y=v1y+2ghu_{2y} = v_{1y} + \sqrt{2gh} (2)

also were to hold, wouldn't we by squaring (2) and subtracting (1) get

0=2v1y2gh0 = 2v_{1y}\sqrt{2gh}

which I can't see hold true in any case (obviously neither of the factors is zero). So what am I missing?


The latter. The normal (pun intended) approach is to resolve the velocity into components parallel and perpendicular to the surface normal. Only the component parallel to the normal is affected by the impact (neglecting friction, which of course you do...)


Okay, thanks *heads off to solve the question*
DFranklin
Looking at the replies to this thread, can I suggest us 'STEP veterans' hold off from posting entire solutions and instead try to help the newcomers through the questions?

And conversely, if you're new to STEP and you're stuck, why not post where you've got up to and see if anyone can suggest anything?

Could I just ask that, on a point of pragmatism, someone write up the solutions in LaTeX once they're done? :redface: We're currently putting the solutions onto the wiki (link will be provided as soon as the damn thing is unfrozen) and it's much more difficult if we have to essentially re-write them ourselves. :smile:

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