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# STEP Maths I, II, III 1993 Solutions Tweet

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1. Re: STEP Maths I, II, III 1993 Solutions
(Original post by generalebriety)
STEP III Q4:
(iii)

Alternative way:

Let's study
Note that
Now S-(2/3)S=(1/3)S=
This is almost a GP, so add (1/3) to this and we have

Thus (1/3)S=3-(1/3)=8/3 i.e. S=8 and we're looking for (1/2)S, which is 4.
2. Re: STEP Maths I, II, III 1993 Solutions
I'll post my alternatives for question 4 too.

(i) Note that

Letting x = 1/2 and using the logarithmic form of arctanh(x) gives

(iii) -- the former is the sum of a geometric series and the latter is its derivative. Let x = 2/3 and we get
. To find the sum from r = 2 to infinity we need to subtract the terms for r = 1 and 0 which gives 9/2 - 1/2 = 4.
3. Re: STEP Maths I, II, III 1993 Solutions
III/14

The particle will undergo circular motion in a vertical circle. Let the velocity when it starts/rebounds be v. At a height h, it gains mgh in potential energy. At this height, let it have velocity x. Change in kinetic = Change in potential
So
The radial acceleration will equate to the tension in the string plus the weight acting radially inwards, so , but is clearly h/c. Combining and rearranging these two equations gives so when the string is slack . Clearly for , h will not lie inside the circle. So we keep halving this value (as the coefficient of restitution is 1/2) until giving h = ; the first valid value of h as it lies inside the semicircle. This happens on the way from A to B.

Now it gets messy. As , x = . The vertical angle will be , so using our kinematic equation one wishes to solve for the larger value of t, giving t = , and multiplying this by the horizontal component () gives a distance travelled of . By geometry, the distance from A when the string is slack is giving the distance from A when it particle rebounds as , so it's closer to B. The algebra for the second part was horrible (though the concept simple), and I wonder if I've gone wrong.
4. Re: STEP Maths I, II, III 1993 Solutions
(Original post by Speleo)
I/4

I like pencil vv

N.B. these are rough solutions, I'd write much more explanation of what was happening in the real exam.
Hey could you explain why, for the second integral, you used limits and instead of and 0 as the question asks?
5. Re: STEP Maths I, II, III 1993 Solutions
STEP I Question 8 - gen's solution is STEP III Q8.

edit
Attached Thumbnails

Last edited by jj193; 28-02-2010 at 15:34.
6. Re: STEP Maths I, II, III 1993 Solutions
I just did STEP II Q7, got the 'infinite descent' part but I don't really understand infinite descent (yet I know/think it is this). Why should it be that if I can keep dividing a,b,c by 5 - forever - that a,b,c are 0. I geuss 0/5=0 etc. and zero is still in the intergers. I've just convinced myself it's ok, I still don't like it.

It seems like a huge jump.

Would this be acceptable as an ending:
If an interger can be divided infinitely many times, then the interger must be zero because if it is not it must be infinite which isn't an interger.
7. Re: STEP Maths I, II, III 1993 Solutions
I think a neater way (although I don't see anything wrong with your method) is to say,

Assume a is non zero, and an integer
Then there exists some a that contains infinitely many factors of 5 (this bit is crucial. 0 is definitely in the integers, and has infinitely many factors)
No such integer exists

Similarly for b,c
Therefore only solutions when a=b=c=0
8. Re: STEP Maths I, II, III 1993 Solutions
(Original post by Rabite)
I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

 Here it is anyway.

By the product/sum formulae that no one remembers.

But if m=±n, one of the fractions explodes. So in that case the question is:

If m=n=0, the integral turns to .

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt

Ignoring the +c for now

Which you can rewrite using the log form of arsinh.
Alternatively for the second integral:

Let:

Now consider the differential of

By the product rule

Thus

So we now have two simultaneous equation where we can cancel the ugly integral:

The method with the hyperbolic functions is much nicer and removes a horrific sec cubed term but i thought i would have a go regardless using good old trg functions and i believe this is a good method to do it if you have no hyperbolic background.
I wish my STEP exam this year will have a question like this
9. Re: STEP Maths I, II, III 1993 Solutions
(Original post by SimonM)
(Updated as far as #213) SimonM - 11.05.2009
...
STEP II Q14
Consider the general collision between a particle of mass m and a fixed surface of coefficient of restitution :
speed before = speed after=
From which it follows that the kinetic energy after such collisions is:

Applying this to the problem:
let be the kinetic energy of the particle just before the nth impact with the ceiling.
By conservation of energy:

and, more generally:

Conjecture:

letting n=1:
and therefore holds for n=1.
Similarly, let n=2:
and so holds for n=2 as well.
Now let's assume the result:
.
Using this result, let n=x+1:

Therefore, by mathematical induction, (the required result) holds

Next part:
the maximum number of times the ball can hit the ceiling is the n that satisfies:

so:

So, taking logs of both sides:

As required.
Last edited by ben-smith; 28-05-2011 at 17:46.
10. Re: STEP Maths I, II, III 1993 Solutions
STEP number 9

11. Re: STEP Maths I, II, III 1993 Solutions
Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.
12. Re: STEP Maths I, II, III 1993 Solutions
STEP III Question 11

Spoiler:
Show
It is an easy calculation to see that the COM is the midpoint of BC, 1/4 of the length "away from it".

So the angle of dangle which it makes hanging from A is:

From B it is

So the change in angle for B is

I can't get the given answer either
Last edited by SimonM; 12-06-2011 at 11:23.
13. Re: STEP Maths I, II, III 1993 Solutions
(Original post by brianeverit)
..
STEP 1993 Q11: Note - done without a good diagram - might be a minor direction mistake somewhere.

Choose a coord frame s.t. C = (0,0), D = (1,0), A = (0, 1), B = (1, 1).
Then AB has CM (0.5, 1), BC has CM (1, 0.5) and CD has CM (0.5, 0).
It follows that the CM of the entire wire is [ (0.5, 1) + (1, 0.5) + (0.5, 0)] / 3 = (2/3, 1/2).

If we hang from A, then BC lies below the horizontal and makes an angle arctan((1/2)/(2/3)) = arctan(3/4) with the horizontal.
If we hang from B, then BC lies below the horizontal and makes an angle arctan((1/2)/(1/3)) = arctan(3/2) with the horizontal (going the other way).

The tan of the angle between these is

tan(arctan(3/4)+arctan(3/2)) = .

Since this is negative, it's the oblique angle between the lines; the acute angle will be arctan(18) as required.
Last edited by DFranklin; 12-06-2011 at 11:41.
14. Re: STEP Maths I, II, III 1993 Solutions
Ah, my diagram was crap.
15. Re: STEP Maths I, II, III 1993 Solutions
guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.
16. Re: STEP Maths I, II, III 1993 Solutions
(Original post by ben-smith)
guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.
Thanks. I knew about the mislabelling, but thought (incorrectly) it was 1992 and earlier.

(Original post by Simon M)
..
Care to try again? Not sure I'll have time.
17. Re: STEP Maths I, II, III 1993 Solutions
(Original post by brianeverit)
Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.

I think you can just take moments about each axis. It's easy to make mistakes, but after 3 attempts, I got their answer using this method.

I think you can also define OA = i, OB = j, OC = k and then simply find where r is the position of a point on the line of force and F is the force itself. (Note that this tallies with the intuitive result that forces through O are irrelevant).
18. Re: STEP Maths I, II, III 1993 Solutions
(Original post by DFranklin)

I think you can just take moments about each axis. It's easy to make mistakes, but after 3 attempts, I got their answer using this method.

I think you can also define OA = i, OB = j, OC = k and then simply find where r is the position of a point on the line of force and F is the force itself. (Note that this tallies with the intuitive result that forces through O are irrelevant).
It's reassuring that you had to try 3 times because I keep getting an annoyingly similar answer. I keep getting (1,2,1) as opposed to (1,1,2). Did you get that? If so, where did you go wrong?
EDIT: can I just swap the 2 and the 1 around by just redefining the axes or is that cheating?
Last edited by ben-smith; 12-06-2011 at 13:16.
19. Re: STEP Maths I, II, III 1993 Solutions
(Original post by ben-smith)
It's reassuring that you had to try 3 times because I keep getting an annoyingly similar answer. I keep getting (1,2,1) as opposed to (1,1,2). Did you get that? If so, where did you go wrong?
EDIT: can I just swap the 2 and the 1 around by just redefining the axes or is that cheating?
You can't swap the axes because you'd also swap the forces, and nothing would change. What I kept getting was (1, -1, 2) - sign errors.

As I calculate it, taking moments anticlockwise about OB you have: -1 (from the force O'A) and that's it.

Taking moments anticlockwise about OC you have: -1 (from AC'), -2 (from C'B) and 1 (from O'A) for a total of -2.

Taking moments anticlockwise about OA you have -1 (from A'C).

I think.

So I get (-1, -1, -2) which is obviously the same axis as (1,1,2).
20. Re: STEP Maths I, II, III 1993 Solutions
1993 STEP III number 13