Announcements
9 years ago
#21
STEP III, Question 3

Spoiler:
Show

Comparing coefficients, we have

Therefore if we can find suitable values for we have:

If a,b,c satisfy those equations, then we can solve simultaneously to find p,r,s and take q to be whatever value we need.

Expanding

This clearly satisfies our equation, so we're done.

so it satisfies our equation.

Therefore:

Therefore we have

Therefore
0
9 years ago
#22
STEP III, Question 4

Spoiler:
Show

Assuming no zero terms (which they let us know)

Therefore we have the recurrence: in

(induction if that's what floats your boat...)

This proves what we want, and

Therefore

and

i) Geometric , .

ii) Periodic, period 2:

iii) Period, period 4:
2
9 years ago
#23
STEP II, Question 1

Spoiler:
Show
Let

Therefore

Therefore the solutions are

We have

Therefore

Therefore

Let

Therefore comparing coefficients, we have:

Therefore

Therefore take
9
9 years ago
#24
STEP II, Question 3

Spoiler:
Show

is an increasing function, from 0 to 1 with a slight curve

i)

ii)

Since

Therefore
0
9 years ago
#25
GHOSH gave all the solutions I have for 1 2005:Q2,3,4,7
I'll try another one
2
9 years ago
#26
STEP III 2005 Question 9

Hopefully my diagram makes it clear what all the symbols I use mean
Diagram
First Part
For the first collision, the coefficient of restitution is given by

By conservation of momentum:

Solving these simultaneously for and gives
Second Part
B then collides with the wall, and, using the coefficient of restitution,

The next collision takes place at a distance from the wall. The time between the collisions is For particle A, and for B,

So we have which boils down to give
Third Part
Clearly, when this collision happens, particle A will continue in the same direction, as will particle B, which will rebound off the wall to collide with A again in a similar way. I'm going to outline a brief proof by induction that the speed of A after the nth collision is and the speed of B after the nth collision with A and after it has hit the wall, , though I don't know how much detail you'd have to go into on a STEP mechanics question...

So, assume that, going into the (r+1)th collision, the speed of A is and the speed of B is

The coefficient of restitution for this collision is given by:

And, by conservation of momentum:

Again, solving these simultaneously for and gives

And then, after B strikes the wall, you get , so our hypothesis has been "proved".

Now consider , the distance from the wall at which the (n+1)th collision takes place, and , the time between the nth and (n+1)th collisions. For particle A: and for particle B:

This gives which boils down to

Clearly, the first distance is d, so the next is de, then de^2, and so on, so

The time between the (n+1)th and the nth collisions, is therefore given by:

Since d, e and v are all constant, we have shown that the times between successive collisions are equal.
1
9 years ago
#27
Ah what the hell,
STEP I 2005 Question 1

Part (i)

For a 5-digit number to add up to 43 the digits must be either:
(1): (9, 9, 9, 9, 7) [ 5 ways of rearranging ]
(2): (9, 9, 9, 8, 8) [ 5!/3!2! = 10 ways of rearranging ]

5 + 10 = 15
Part (ii)

We can classify groupings of digits by how many 9s they have and, if no 9s. Also, since since 8*4 + 7 = 39 there can't be less than four 8s if there are no 9s.

FOUR NINES:
(9, 9, 9, 9, 3) [ 5 ways of rearranging]

THREE NINES:
(9, 9, 9, 8, 4) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 7, 5) [ 5!/3! = 20 ways of rearranging ]
(9, 9, 9, 6, 6) [ 5!/3!2! = 10 ways of rearranging ]

TWO NINES:
(9, 9, 8, 8, 5) [ 5!/2!2! = 30 ways of rearranging ]
(9, 9, 8, 7, 6) [ 5!/2! = 60 ways of rearranging ]
(9, 9, 7, 7, 7) [ 5!/2!3! = 10 ways of rearranging ]

ONE NINE:
(9, 8, 8, 8, 6) [5!/3! = 20 ways of rearranging ]
(9, 8, 8, 7, 7) [5!/2!2! = 30 ways of rearranging ]

NO NINES:
(8, 8, 8, 8, 7) [5 ways of rearranging ]

Doing some good ol' addition gives you 5 + 20 + 20 + 10 + 30 + 60 + 10 + 20 + 30 + 5 = 210 ways of rearranging the numbers. And no LaTeX required!
2
9 years ago
#28
Question 5, STEP I, 2005
(i)
Case 1:

Case 2:

As

I wasn't exactly sure what to put for that last part
(ii)
With the substitution

Case 1:

Case 2:

Case 2:

1
9 years ago
#29
STEP I 2005 Question 9

Solution

The rod is non-uniform so the weight can be anywhere. I said it acted at distance from B. In the instance when there are two strings holding it up, taking moments about B gives you .

When there is a pivot (call this P) at a distance from A and a string at B, the distance from P from which the weight acts is therefore . Therefore, taking moments about P gives you . Rearranging this with the substitution gives you , as required.

We can now also simplify , to get ; this is the distance along the rod from B that the weight acts.

To find it's best just to take moments about B again. Doing so gives . Using our result from before that , this simplifies to , which simplifies to

Now we've found , we need to find the coefficient of friction. Resolving vertically to find the normal contact force gives us . Using the information we have already found, we have so this simplifies to .

In order to find we need to resolve horizontally to find the frictional force acting horizontally "inwards" (towards the rest of the rod) at B. We are told that the rod is not slipping, so the frictional force must be greater than opposing forces, so we therefore get:

0
9 years ago
#30
STEP II 2005 Question 8

First Part
for with y=1 when x=0

The RHS can be integrated by parts, with and

Substituting in x=0 and y=1 gives C=1/3, so we get as required.

For large x, we can say so

Looking at the first two terms of the expansion of gives, for large x:

Sketch One
I was lazy and didn't sketch it myself, but some points to note:

1) dy/dx is 0 only when x=0, since y cannot be 0 or the above equation blows up, so the only turning point is at (0,1)

2) The graph has an asymptote x=3, and looks like the graph of y=3-9/x for large x

Sketch Two
Note that if we separate the variables again, we will get:

So this graph is just graph of y^2 (where the above was a sketch of y)

So it is symmetrical in the x-axis (which gives the two curves requested, passing through (0,1) and (0,-1)) and it has asymptotes

1
9 years ago
#31
2
9 years ago
#32
STEP I 2005 Question 11

Part (i)

The path of the particle can be represented by parametric equations . A the origin, x = y = 0, so using the double-angle formulae:

, as required.
Part (ii)

Differentiating to find the velocity components gives us . If this is perpendicular to the displacement, then , giving us .

Using the double-angle formula for again gives us:

Part (iii)

I maybe cheated a bit here by using a matrix... there are other ways to do this, but this was by far the simplest.

If the lines are parallel, then the matrix is singular, so

That gives me

Once again using the double-angle formula for sine, we get:

Now using , we get:

Subbing these into the equations gives , which is the origin, as required.
Part (iv)

If the particle is at a maximum distance from the origin, then , so . This happens when and , which gives us , so . Subbing this back in to the displacement equations gives us , so the maximum distance is 2. [We know it's a maximum and not a minimum because we've already shown that it goes through the origin.]

The sketch looks like a figure-of-8, with the part where the loops cross at the origin, and crossing the y-axis at ... I'd draw a diagram but I have no means to do so!
To be honest, this question could just as well have been in the Pure section!
0
9 years ago
#33
Christ, I'm normally a Pure man, but these applied questions are great fun

STEP II 2005 Question 12
Part (i)

To help I drew a table that looked something like:

If it is heads, then EITHER [Anna is at the computer AND telling the truth] OR [Bella is at the computer AND telling the truth], so , as required.
Part (ii)

Either the first correspondence is true and the second is false (1) or the first is false and the second is true (2). The probability that the coin came down heads requires (1), so the probability is (1)/[(1) + (2)].

(1) can be expressed by, . Letting , this simplfies to . Similarly, (2) can be expressed as , meaning that it is the same as (1), so (1) + (2) = 2(1).

This means that the probability that the coin came down heads is , as required.
Part (iii)

In a similar fashion, either both times whoever we speak to is telling the truth (3) or both times they are lying (4).

(3):
(4):

So, the probability is . Since we are given that , this simplifies to
0
9 years ago
#34
Question 6, STEP I, 2005
(i)

Or:
(ii)
Let

Comparing coefficients of different terms of with , the ratio between the coefficients must be the same for the path to be the same.

And similarly:

We can divide by , for the case

2
9 years ago
#35
(Original post by nuodai)
STEP I 2005 Question 11
(ii)
Not sure about the bit I've boxed. for some integer n.

edit: for (iii) I think as a different approach: the direction of the velocity and displacement will be the same when
Spoiler:
Show
So we end up solving:

which are the times when it is at the origin, as found in part (i)
0
9 years ago
#36
STEP III 2005, Question 5

Spoiler:
Show

This gives the x coordinate when the gradient of the tangent to the curve is m. The y coordinate is given by

The equation of a line is given by

as required.

From the above, the equations of the tangent with gradient m to each curve can be written:

. Equating the RHS of each equation (as these two equations describe the same tangent):

as required.

For the two curves to have exactly one common tangent, this quadratic in m must have equal roots, meaning that its discriminant must be equal to 0.

For the curves to touch each other, the quadratic in x that results from eliminating y must have equal roots ie it's discriminant must be 0.

This quadratic is , whose discriminant is the expression above equal to 0 (provided ). Hence, in the case , the two curves have exactly one common tangent if and only if they touch each other.

If , the condition on the tangent becomes

Which gives the condition (if they were equal, this would imply ie. the curves wouldn't be distinct).
0
9 years ago
#37
(Original post by DeanK22)
STEP III

Question 1

If cos(B) = Sin(A) this implies A = (4n+1)pi/2 (+/-) B

Let B = A + x for some x in R

It follows the problem is now Cos(A+x) = Sin(A) *

Epanding using the addition formulae we see that * is transformed to;

Cos(A)Cos(x) - Sin(A)Sin(x) = Sin(A)

<->
<->

Spoiler:
Show

It is relativel easy to show that if after searching for GA formula / derivation it has not been found, we use the identity sin(x) = cos(pi/2 - x) = cos(2(pi/4 - x/2)) and expand using the double angle formulae and the result follows fairly soon.

We also note that multiples of pi can be added to the tangent functions argument without altering its value to obtain;

<->

and the result

follows - although interestingly we do not have wip is obviously needed to obtain this.
Bit of an odd method?

.

From looking at the cos graph, we can see that or for some integer k, which gives

which we can write as for some integer n.
0
9 years ago
#38
(Original post by GHOSH-5)
(ii)
With the substitution

Case 1:

Case 2:

Case 2:

It might be a little neater just to do:

then the results follow immediately from part (i)
1
9 years ago
#39
III/2 - with the proviso that I think I may have missed something somewhere.

Solution of Differential Equation

Separating variables and integrating:

with

with

I don't agree with your constant after the integration You say this equals with However, substituting that B into it gives and -0.5ln(ln(1/sqrt(k))) =/ k.

I got:
Spoiler:
Show
So which also gives the desired form, with c=e^k
1
9 years ago
#40
Yes, I mean K = -1/2 ln B, don't I? I was just shifting around the constant. Sorries.
0
X

new posts
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### University open days

• Arts University Bournemouth
Art and Design Foundation Diploma Further education
Sat, 25 May '19
• SOAS University of London
Wed, 29 May '19
• University of Exeter
Thu, 30 May '19

### Poll

Join the discussion

#### How did your AQA GCSE Physics Paper 1 go?

Loved the paper - Feeling positive (351)
30.6%
The paper was reasonable (460)
40.1%
Not feeling great about that exam... (197)
17.18%
It was TERRIBLE (139)
12.12%