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# STEP I, II, III 2002 Solutions Tweet

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1. Re: STEP I, II, III 2002 Solutions
This might be a good time for me to point out that STEP II Q6 is misattributed to me (I actually posted STEP III Q6).
2. Re: STEP I, II, III 2002 Solutions
STEP III 2002, Question 3

Spoiler:
Show

To help draw the graph, we make the following observations:

Which tells us that the curve has a turning point (minimum) at .

This minimum occurs to the right of b, as a > 1.

f'(x) tends to 0 as x tends to infinity, and - infinity as x tends to b.

The graph is ommited. It begins at , decreases until it reaches the minimum, and then increases indefinitely as it flattens off.

From the graph it is clear that for f(x) = c to have no solutions for positive c,

as required.

Similarly, for f(x) = c to have exactly one solution, or

For f(x) = c to have exactly two solutions,

i)

Therefore the equation has exactly one solution, and this solution is

ii)

Therefore the equation has exactly two solutions.

Squaring twice gives

Last edited by Daniel Freedman; 08-06-2009 at 11:45.
3. Re: STEP I, II, III 2002 Solutions
Some attached
There was a bit of a quadratic formula fail at the end of 2. But I agree, majorly boring.
4. Re: STEP I, II, III 2002 Solutions
I got my answer to be 81/40 km don't know why it's different from yours....
(Original post by tommm)
STEP II 2002 Q10

Spoiler:
Show
after t hours, the competitor has km left to run

the remainder is run at , in a time (T-t) hours

using speed = distance/time:

which rearranges to

This is a quadratic in t. Clearly t is real, therefore it has real roots, so

which rearranges to give for all values of t.

If T = 3, then

which has the (repeated) solution .

Therefore competitor one runs at 13km/h for 3/4 hours
then (29/2)km/h for the remains 9/4 hours

Therefore distance run by competitor one after time x

(The 9/8 is obtained by observing that the two expressions must agree when x = 3/4.)

Now the speed of competitor 2 after time x

we can integrate to find the distance:
distance
from considering the distance travelled initially (0) and after 3 hours (42+3/8), we find

Therefore distance =

therefore the distance between the two

Via differentiation, we find that the first expression has no maxima in the required range, but the second expression has a maximum at .

Substituting this in, we find that the maximum distance is km.

5. Re: STEP I, II, III 2002 Solutions
Question 5, I'll pick up where Dadeymi left off:

Now,

So

6. Re: STEP I, II, III 2002 Solutions
Question 10, STEP I, 2002

Using NLR with elasyic collisions, speed of seperation = speed of approach so collision with the end of the cylinder dont change the speed. This gives , which is an arithmetic progression with initial value v and common difference 2u so as required.

The diference in the distances from the piston and end of cylinder is the distance the piston moves in that time so

is the time take to reach the end of the cylinder + the time taken to reach the piston afterwards.
Giving

Using our previous result

Which was to be shown.

I the case we have

Giving
Last edited by DFranklin; 17-06-2010 at 01:28. Reason: Fixed Latex
7. Re: STEP I, II, III 2002 Solutions
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.
For question 2, a far nicer way of doing the final part is to notice that is just with r replaced by r-1. The last part of the question is then trivial.

For question 4 the (unordered) pair (1,0) is also a solution.
8. Re: STEP I, II, III 2002 Solutions
I tried to do the and hence that by expanding out , subtracting , and then adding to both sides. However, this gives me the wrong amount of s^2's: I get two, and I should be getting four.

Any pointers?

EDIT: crafty little dodge showed that what you get after multiplying out and then subtracting is zero. Hence, you can multiply by two and then add , and then factorise the resulting expression in the form given.
Last edited by around; 19-06-2009 at 20:33.
9. Re: STEP I, II, III 2002 Solutions
Some more;

Did these quite a while ago I'm afraid some may be partial solutions.

III Q4; (-1,0) & (0,1) also work
10. Re: STEP I, II, III 2002 Solutions
2002 Step1 Q9
Can't find an answer to this - and can't work it out.
I should supply a diagram but I'll try to describe.
Approach:
1. Take surface of slope and normal to slope as reference co-ordinate axes
2. Take components of forces along plane and at normal to plane and equate forces as in equilibrium
3. Take moments around lower wheel and equate them as in equilibrium

Now, what I'm not sure about is whether the friction at wheel contact with slope ( for friction at upper and lower wheel) is up the slope or down the slope (or is there any friction at equilibrium?). I'm assuming the friction is up the slope.

I end up with following equations:

(forces up slope)
(normal reaction forces)
(taking moments around lower wheel)

but under the given condition the normal reaction forces are equal then the second and third equations become the same.

I can't see how to use the third equation to show and hence the sum of the friction forces is zero ?????

Clive
11. Re: STEP I, II, III 2002 Solutions
(Original post by cliverlong)
2002 Step1 Q9
Can't find an answer to this - and can't work it out.
I should supply a diagram but I'll try to describe.
Approach:
1. Take surface of slope and normal to slope as reference co-ordinate axes
2. Take components of forces along plane and at normal to plane and equate forces as in equilibrium
3. Take moments around lower wheel and equate them as in equilibrium

Now, what I'm not sure about is whether the friction at wheel contact with slope ( for friction at upper and lower wheel) is up the slope or down the slope (or is there any friction at equilibrium?). I'm assuming the friction is up the slope.

I end up with following equations:

(forces up slope)
(normal reaction forces)
(taking moments around lower wheel)

but under the given condition the normal reaction forces are equal then the second and third equations become the same.

I can't see how to use the third equation to show and hence the sum of the friction forces is zero ?????

Clive
You haven't resolved correctly parallel to the plane; it should be: . A very simple way to do this bit would be to take moments about G.
Last edited by Unbounded; 16-08-2009 at 21:53.
12. Re: STEP I, II, III 2002 Solutions
(Original post by GHOSH-5)
You haven't resolved correctly parallel to the plane; it should be: . A very simple way to do this bit would be to take moments about G.
Yes, my slip, transferred it incorrectly from my notes

So I have three equations

(1)
(2)
(3)

Well, if I apply the condition the reaction forces
are the same then equation (2) reduces to (3) and I have

(1)
(3)

from which I'm trying to show

I will have a stab at multiplying r.h.s of (1) and (3)

and this will have to equal zero for all (it isn't , take ) to make

So, I'm stuck.

Ta,

Clive
13. Re: STEP I, II, III 2002 Solutions
STEP 1 Q9

Using the suggestion to take moments about G (thanks)

First take axes parallel to and perpendicular to the slope. I will take the positive directions as “up” the plane and “away” from the plane.
There are forces not indicated on the diagram given in the question

Reaction forces of the plane on the wheels at the lower and upper wheels:
The weight of the lorry, W, acts through the centre of gravity G
Friction forces acting at the lower and upper wheels which I assume to be in the direction "up" the plane

Taking components of forces parallel to plane

At equilibrium the sum of all forces is zero. The component of the weight is “down” the plane so is negative

(1)

Taking components of forces perpendicular to plane

At equilibrium the sum of all forces is zero. The component of the weight and force P are in opposite direction to the reaction forces.

(2)

Taking moments around the centre of gravity G

Since at equilibrium, sum of moments is zero.

(3)

Now to tackle the questions

(i) If the normal reactions are equal, , substitute into (3)

Since h is not equal to zero, then = 0, that is the sum of the friction forces between wheels and ground is zero

(ii)Let us make the assumption that as the size of the force P varies, the “limits of equilibrium” are between the lorry sitting on the plane as in the diagram and the van tilting/pivoting around the lower or upper wheel. Equivalently I assume the lorry does not slip up or down the plane as the size of force P varies.

At the point the lorry is to be pulled by P so it pivots around the upper wheel, becomes zero. So, for the lorry to remain on the plane

Bringing together the equations

(1)

(2)

(2.1)

(3)
(3.1)

Substituting (1) into (3.1)

(3.1.1)

Substituting (2.1) into (3.1.1)

Multiplying throughout by , rearranging and dividing by

Also at the point the lorry is to be pulled by a small enough P so it pivots and "falls" around the lower wheel, becomes zero. So, for the lorry to remain on the plane

And this all works through in an almost identical way to (to be honest I haven't checked the detail)

QED
Attached Thumbnails

Last edited by cliverlong; 27-08-2009 at 16:11.
14. Re: STEP I, II, III 2002 Solutions
Q12 - STEP I 2002

This is my first go at writing up a STEP answer, so forgive the badly worded/written explantations and I hope it's ok!

If we separate the square into areas which the centre of the hoof (i.e. a point) can land on when covering a particular number of squares, we can work out the probability of Harry picking a particular number by placing his hoof at random.

Spoiler:
Show

If we assume the hoof is put down randomly:

When we look at what happens when only one square is covered, we can see if the centre of the hoof can land anyway within the by square and only cover 1 square. Therefore 1/4 unit^2 area in each unit square that the hoof will land here. Therefore a 1/4 probaility, which is consistant with the observations.

When 4 squares are covered, the hoof must land in a quarter of a circle area will radius 1/4 around each corner. The area covered by these 4 quatre-circles is which is , again consistant with the suspicion that he places his hoof randomly.

To do the second part of the question, we will need to look at the cases when 2 and 3 squares are covered.

When we look at 2 squares being covered, there are 4 (1/2 * 1/4) squares around the edge of each unit square. This makes a 1/2 probaility of a hoof landing on 2 squares.

When we look at 3 squares being covered, we can see it occupies the 4x the left over area in a 1/4 by 1/4 square minus the quarter of a circle. Therefore the total area in each unit square is which is

Now we have done all the possibilities, we can check they are all right by summing them to equal 1. 1/2 + 1/4 + π/16 + ( 4-π ) /16 does indeed equal 1.

The average value of Harry's answers if he places his hoof at random is which is (about 2.2)

Now if the Harry the horse actually did the sums, the average of his answers would be 1*1/4 + 2*1/4 + 3*1/4 + 4*1/4 = 1/4(10) = 2.5.

Therefore, as the average of Harry's answers is 2 (even lower average then if he placed his hoofs randomly), it would seem he does not actually do the sums, and indeed places his hoof fairly randomly. Therefore, the answer to the question is: yes, you should get a new horse.
15. Re: STEP I, II, III 2002 Solutions
STEP I, 2002 , Q11

This question has been answered using significant help from the nrich forum

https://nrich.maths.org/discus/messa...tml?1252056493

The approach is to use conservation of linear momentum, and as the collision does not lose kinetic energy, use conservation of energy.

Define:
arrives with speed
initial speed is zero

After collision:
recedes with speed
recedes with speed

Conservation of (kinetic) energy:

(1)

(The algebra to derive the required expression is tractable if one resolves along the exit path of so that does not appear in the equation) Conservation of linear momentum

Conservation of linear momentum along exit path of :
(2)

Conservation of linear momentum at right-angles to initial path of :

(3)

Now tackle the algebra with the aim of eliminating the but retaining the

(2) squared : (2.1)

Eliminate in (2.1) using (1)

(2.2)

Eliminate in (2.2) using (3)

(2.2)

Divide through by giving required result:

Last part:

[using: and substituting into the derived equation]

Last edited by cliverlong; 04-09-2009 at 11:10.
16. Re: STEP I, II, III 2002 Solutions
(Original post by SimonM)
STEP III, Question 1

As , the volume tends to
Most likely a typo but I might as well notify you, the two lines above are missing a 2 from the second and third terms respectively.
17. Re: STEP I, II, III 2002 Solutions
STEP II, Q4, The solution (and i dont know how to post an alternative solution ) for part iii uses the assumption a/x^2 + b/x + c is a quadratic equation where the a's, b's, and c's have the same role (in the quadratic formula) as in the generic quadratic equation ( ax^2 + bx + c ) which is an incorrect assumption? I think the a's and c's swap roles while the b's stay in the same role. Can anyone verify this?

Using the a > 0 and b^2 < 4ca it can be shown c > 0.

therefore the integral of ( cx^2 + bx + a ) > 0

Using q > p > 0

we can divide the integral by x^2 and maintain the inequality
so the integral becomes ( c + b/x + a/x^2 ) > 0

using limits, q , p this becomes the required result.
Last edited by tjagger1; 02-05-2010 at 18:23.
18. Re: STEP I, II, III 2002 Solutions
f(x) = a/x^2 + b/x + c = (a+bx+cx^2)x^-2 for x in {the reals - 0}

you correctly asserted that c>0 since b^2>0 => 4ca>0 => c>0 since 4a>0
So we know it has no real roots and c>0, therefore a+bx+cx^2>0, therefore f(x) > 0, since x^2>0 (x=/=0)

so yeah its ok
19. Re: STEP I, II, III 2002 Solutions
(Original post by tjagger1)
STEP II, Q4, The solution (and i dont know how to post an alternative solution ) for part iii uses the assumption a/x^2 + b/x + c is a quadratic equation where the a's, b's, and c's have the same role (in the quadratic formula) as in the generic quadratic equation ( ax^2 + bx + c ) which is an incorrect assumption? I think the a's and c's swap roles while the b's stay in the same role. Can anyone verify this?

Using the a > 0 and b^2 < 4ca it can be shown c > 0.

therefore the integral of ( cx^2 + bx + a ) > 0

Using q > p > 0

we can divide the integral by x^2 and maintain the inequality
so the integral becomes ( c + b/x + a/x^2 ) > 0

using limits, q , p this becomes the required result.
It's just a quadratic in 1/x.
20. Re: STEP I, II, III 2002 Solutions
May just be being a bit dim but on 2002 I/13 part (ii) why is it (E(R))/2 not just E(R)? Got the rest of the question, this just confused me

Thanks Laura
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