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1. Re: STEP 2006 Solutions Thread
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?
2. Re: STEP 2006 Solutions Thread
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).
3. Re: STEP 2006 Solutions Thread
(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,
4. Re: STEP 2006 Solutions Thread
(Original post by SimonM)
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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):

Conservation of linear momentum (CoLM):

Also note that this implies that

Considering first B-C collision:
NLoR:

CoLM:

if :

Since , because we can't have negative masses, for two collision to occur between A and B,

(ii)
By subbing into and :
, and .

Using , the time taken by B to reach C's initial position following the first A-B collision =

In that time, A travels a distance of
Therefore, the distance between A and B after B has collided with C =

Note that the speed of approach of A w.r.t. B after B has first collided with C is given by: .
Therefore, the time taken for A to reach B after B has collided with C is given by:

So total time between the two collisions of A and B is given by as required.
5. Re: STEP 2006 Solutions Thread
(Original post by SimonM)
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STEP I Q12
First part
First notice that the road with the single block-able section (the third road) MUST be blocked.

Let

Therefore,

as required.

Second part
Note that:

And

Therefore:

As , 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.
Last edited by Farhan.Hanif93; 07-03-2011 at 18:40.
6. Re: STEP 2006 Solutions Thread
I 2006 q1

hence
satisfies

if , then

so ,
satisfies
where

as and are prime
or

so

hence other value of n+m is , and
Last edited by kiddey; 14-03-2011 at 19:13.
7. Re: STEP 2006 Solutions Thread
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)=
(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):

take logs of both sides:

Now differentiate both sides:

To find stationary points, let

. Since f(n) can not be 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)=
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
Last edited by ben-smith; 21-03-2011 at 19:02.
8. Re: STEP 2006 Solutions Thread
(Original post by SimonM)
...
STEP II Q14

Solution
Find the graph by noting that as , and considering the behaviour as . 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 .
It should look something like this:
Graph

(i)
Considering the case :
Note that

(ii)
Since , we don't have to worry about integrating across the discontinuity at . It follows that:

, as required.

(iii)
Considering the case :
Note that , therefore we still do not have to worry about the discontinuity. It follows that:

Setting in the Maclaurin expansion of gives:

Therefore , as required.

(iv)
Note that, in the case , from our answer to part (ii).

Also note that and that . This means that, since our PDF is only non-zero for , the range lies outside the interval for which the probability is non-zero. Therefore, it follows that

Last edited by Farhan.Hanif93; 29-03-2011 at 02:00.
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Last edited by brianeverit; 27-07-2011 at 19:43.
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!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.
Last edited by brianeverit; 27-07-2011 at 08:54.
20. Re: STEP 2006 Solutions Thread
(Original post by SimonM)
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STEP II Q11
Solution
Note that the first particles motion can be described by

When and , .

Considering the value of under these circumstances:

(i)
Note that, for the value of that satisfies , we're looking for the largest angle of projection that is smaller than the angles which satisfy .

Therefore

Since for acute :

If follows that, since is an increasing function, this critical value is given by .

(ii)
Plugging into out formula for yields that

Note that

Also note that
By equating coefficient of and , it follows that and

Therefore, OA has a maximum value of

It follows that , 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.