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# STEP Maths II, III 2009 Solutions

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1. STEP III Q11

Solution
(i)
Conservation of momentum:

Since , it follows that and hence (discarding negative soln as )

(ii)
as constant. yields so . That is:

Hence, using to eliminate the constant of integration:

Note, by considering proportionality term-by-term, is constant iff:

Hence s.t. , as required.

Furthermore, assuming that , note . That is, tends to a constant speed as , as desired.

2. (Original post by Zacken)
...
There are a couple solutions at the end of page 2 that you missed.

STEP III Q2

Spoiler
(i)
Note, by linearity and the fact that the old constant term vanishes:

Similarly,

Explicitly, that is

(ii)

Plugging into :

Equating coefficients:

In the case
Observe that, if for some , then so

Furthermore, writing for some inductive use of yields that:

So , as required.

In the case :
By the earlier argument, we have . Considering such terms, write for and proceed inductively:

So

3. STEP II Q6:

values

(i)

Using the recurrence relation (for ): .

We then have .

So it follows immediately that as required.

Hence: by sum of a geometric progression.

A similar analysis yields that for .

So it follows immediately that .

Now, splitting off the odd and even terms which is an obvious approach once the series has been written out explicitly, we have:

(ii)

Note here that it is simple to simply sum the first few terms (since they are all positive) to get:

For the upper bound, consider:

As required.

4. (Original post by Farhan.Hanif93)
There are a couple solutions at the end of page 2 that you missed.
Thank you, for doing Q2 as well, that must have been hell to LaTeX!
5. STEP II, Q10:

solution

Consider Newton's Law of Restitution between and : .

Consider Newton's Law of Restitution between and : .

We have:

Let be the initial distance between and and be the initial distance between and .

So since the collision between occur at a time we have and similarly for : .

From these, we get and so that by subtracting these two equations we are left with:

by .
6. (Original post by Zacken)
STEP II, Q10:

solution

Consider Newton's Law of Restitution between and : .

Consider Newton's Law of Restitution between and : .

We have:

Let be the initial distance between and and be the initial distance between and .

So since the collision between occur at a time we have and similarly for : .

From these, we get and so that by subtracting these two equations we are left with:

by .
What?! Is that question really that short?
7. (Original post by Zacken)
Thank you, for doing Q2 as well, that must have been hell to LaTeX!
It was pretty painful to type, as was Q7. I assumed that's why no one had bothered.
8. (Original post by IrrationalRoot)
What?! Is that question really that short?
The 'normal' method would result in pages of algebra but there are a few concise approaches that sneakily cuts out all that work and makes it this short, yes. Here's what the examiners report says:

9. STEP III Q9

Solution
(i)
Define a cartesian coordinate system s.t. is the origin and . Suppose that the particle passes through and with speeds and respectively, and let denote the time taken to travel from to . Throughout this solution, denote the relevant trig functions by their respective initial letters (for ease of LaTeX). Note:

So

Furthermore, by :

So , as desired.

(ii)
Note, at Q:

Observe that this quadratic in has solutions as there are at most two trajectories that travel from to with the given initial speed and those two trajectories are known.

It follows that:

Hence , as required. Equivalently,

Observe further, via the compound angle formulae:

(Note: There is probably a better way to obtain the above equality via a similar quadratic, but it's late and I didn't trust myself to spot the symmetry accurately)

Hence, by :

The angles of projection are acute so that and, intuitively, the angles of reception at must be acute in magnitude (but can be negative) i.e. . Furthermore, the angles of reception must be strictly smaller than their respective projection counterparts, so that and therefore . Also noting that and , it follow that . Hence:

, as desired.

10. (Original post by Farhan.Hanif93)
STEP III Q9
That looked painful... thanks for doing it!

(mind changing the heading to STEP II Q9?)
11. (Original post by Zacken)
That looked painful... thanks for doing it!

(mind changing the heading to STEP II Q9?)
Oh, it was.

It's definitely STEP III Q9, I think you have accidentally tagged one of my other solutions to III Q9.
12. (Original post by Farhan.Hanif93)
Oh, it was.

It's definitely STEP III Q9, I think you have accidentally tagged one of my other solutions to III Q9.
Oops. My bad, should be fixed now. I'm just going to blame it on the time...
13. (Original post by Zacken)
Oops. My bad, should be fixed now. I'm just going to blame it on the time...
Almost fixed. There is no longer a bijection between the links in OP and the solutions (as it's still linked under the currently unanswered II Q9, as well).
14. (Original post by Zacken)
STEP II, Q10:

solution

Consider Newton's Law of Restitution between and : .

Consider Newton's Law of Restitution between and : .

We have:

Let be the initial distance between and and be the initial distance between and .

So since the collision between occur at a time we have and similarly for : .

From these, we get and so that by subtracting these two equations we are left with:

by .
15. STEP II Q11

Solution

Note that: from Newton's second law.

The only two forces here are the resistive force and the engines driving force. The net force is therefore , where is the driving force of the engine and is the resistive force.

as each of the trucks and the engine have resistive force .

The mass for all of the trucks and the engine.
So therefore

The engine can output a maximum power of .
We know that , where . So from this:

Eliminate the fractional v by multiplying through:
, which we rearrange for a as required:

obviously, for the train to reach a velocity it has to accelerate. For the train to accelerate, must be true.

(i)
Firstly, we want to construct a differential equation by noticing that

We can separate variables apart:

And then integrate both sides:

Now let and , and rearrange for

factoring out the denominator here yields the following:

Simplify the integral down by taking out to obtain:

. Evaluating the limits:

from taking the out.

Now note that they give the approximation of . The log on top of the equation above can therefore be simplified:

some easy cancellations are to be had here to yield the result as

Which can then be multiplied through to give the approximation

Which one should note is the equation for kinetic energy, and therefore the equation shows that work done is the change in kinetic energy. This holds for when there is no resistance acting against the train, and therefore

(ii)
Note first that Work done is force multiplied by distance. As such, the work done against the force is , where is the distance travelled by the train.
We still have trucks however, so therefore
, where is the work done against R.

Any energy not used in doing work against the resistive force must go into the kinetic energy of the train. As the only thing providing energy to the train is the engine, it's trivial to formulate an equation.

where is the engine energy output. (E=PT)

Now it's a simple rearrangement to find that

as required.

Zacken
16. STEP III Q13

Solution
(i)
Let and . Then:

Furthermore,

Note that , so i.e. uncorrelated as desired.

It's clear that and are not independent, as fixing restricts to two points as opposed to the whole of .

(ii)
Observe first that .

Furthermore, if , then (by independence) . If instead , then by (i). Hence

It follows , so uncorrelated as was to be shown.

Considering the case of large : See first that .

Assuming sufficiently large and applying the central limit theorem, we have . It follows that:

, as desired.

17. STEP II Q13

Solution
Consider a launch of the Andover: Let denote the cost of repairing the Andover if engines fail, and let denote the number of failed engines.

Consider a launch of the Basingstoke: Let denote the cost of repairing the Basingstoke if engines fail, and let denote the number of failed engines.

Assume (reasonably) that the replacement of an engine is never free. Note that, by collecting terms on one side and cancelling down, we have iff:

Hence the expected repair cost of the Andover is the expected repair cost of the Basingstoke if and only if .

18. STEP II Q12

Solution
(i)
Standard decaying exponential of amplitude restricted to the upper right quadrant of plane. Note the striking resemblance to one half of the normal pdf.

(ii)
Note that

But we see that by definition of the standard normal distribution, so:

(iii)

Integrating by parts:

(iv)
Seeking s.t. , observe (using the manipulations of part (i)):

Where is the CDF for the standard normal distribution.

Hence, noting that ,

So the median of is approximately , by the tables.

19. Ill do Q9 II.
Looks proper easy.

Posted from TSR Mobile
20. (Original post by physicsmaths)
Ill do Q9 II.
Looks proper easy.

Posted from TSR Mobile
It is. Quite tedious but still easy by step standards.

Posted from TSR Mobile

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