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# STEP 2005 Solutions Thread Tweet

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1. Re: STEP 2005 Solutions Thread
(Original post by The Bigtime)
I also think that the solution to the final part of I question 7 can be simplified further by expanding the sine function.
Indeed it can:

I'll edit it now.
2. Re: STEP 2005 Solutions Thread
(Original post by Sk1lLz)
STEP I 2005 Question 10:

First part
Spoiler:
Show

From A colliding with B:

Let represent the velocity of A after colliding with A.

1. NEL:

2. CLM:

Substitute into 1.:

3.

From B colliding with C:

Let represent the velocity of B after colliding with C.

4. NEL:

5. CLM:

Substitute into 4.:

6.

Now, substitute 3. into 6.:

i)
Spoiler:
Show

Use 6.:

For v, use 3.:

ii)
Spoiler:
Show

Use 6.:

Use 3.:

I don't agree with this. I think you've misread the question. It asks you to find the final velocities of each of the three particles but haven't you just found the final velocity of C and the initial velocity of B following the collision with A?
From your solution, the final velocities of A and B are and but you didn't solve for them. I'll post the remainder of the solution after college.
3. Re: STEP 2005 Solutions Thread
(Original post by SimonM)
...
STEP II, Q9

(i)
Diagram

Consider the particles individually when they are at the point of slipping upwards.

Considering A:
Resolving perpendicular to the slope:

Resolving parallel to the slope: by subbing in from above and rearranging.

Considering B:
Resolving perpendicular to the slope:

Resolving parallel to the slope: by subbing in from above and rearranging.

Expressing in the form gives it to be equivalent to .

Note that this implies that , which is minimised when is maximised i.e. when .

Therefore and P must act at an angle of to the horizontal i.e. parallel to the slope.

(ii)
If the particles are not sliding down the slope then they are either on the point of slipping down it or moving up it. We've considered the latter in part (i) so we need to find the smallest value of P for which the system is at the point of slipping downwards. If you were to draw a diagram for this, it would be the same as the one for part (i) except the frictional forces will be acting in the opposite direction.

Considering A:
Resolving perpendicular to the slope:

Resolving parallel to the slope:

Considering B:
Resolving perpendicular to the slope:

Resolving parallel to the slope:

Note that .

Therefore in this case. This is smaller than the answer to part (i), therefore is the correct answer.
Last edited by Farhan.Hanif93; 07-11-2010 at 02:05.
4. Re: STEP 2005 Solutions Thread
(Original post by SimonM)
...
That was a long one!
STEP II, Q11
First part
Diagram

Where is the tension in the string, A is the object on the slope and B is the hanging object. (I chose not to use the conventional because this letter is assigned to something else later on)

Considering B:
Resolving downwards:

Considering A
Resolving perpendicular to the slope:

Resolving parallel to the slope:

Note that:
Useful triangle

Using SOHCAHTOA on this triangle yields that:

Therefore using these results and subbing into :
as required.

Second part
Considering the velocity, , of A at the point where the string breaks:

Therefore, using :

.

In this case, A can be modelled as follows:
Diagram

Where is the acceleration of A after the string breaks.

Resolving perpendicular to the slope:

Resolving down the slope:

Considering the motion of A between the time where the string breaks and the particle reaches it's maximum height (let this time be denoted by :

Therefore using :

Note that the time taken (after release) to reach maximum height is given by:

Third part
Let the maximum displacement of the particle be denoted by ; let the distance between the initial position and the position at which the string breaks be and let the distance between the string's breaking position and the maximum height of motion be such that

Considering upwards motion for the distance :

,

Therefore, using :

Considering upwards motion from to :

Therefore, using :

Note that, by resolving forces of the object when it is at it's maximum height, the acceleration down the slope is .

Considering motion down the slope from max height to initial position:

Therefore, using :

Last edited by Farhan.Hanif93; 07-11-2010 at 04:56.
5. Re: STEP 2005 Solutions Thread
(Original post by SimonM)
...
STEP I Q14

solution
Note that , where is the PDF for the distribution of X for X>0.

Differentiating the above:

Hence, the PDF for X is given by:

(i)
Note that

(ii)
Note that

Integrating the second term by parts with :

, as required.

(iii)
Variance
Note that

Integrating the second term by parts with :

In part (ii), we had shown that , hence:

Median
Let be the median of the distribution of X, it follows that:

(By using the CDF given at the start)

(iv)

Using a substitution of for the integral:

Given that :

Last edited by Farhan.Hanif93; 10-04-2011 at 13:09.
6. Re: STEP 2005 Solutions Thread
STEP II Q7

Spoiler:
Show
i) Both P and Q perform circles in a counterclockwise direction. P about the k-axis, radius 1 in the i-j plane, Q about a different axis, radius 3

ii)

as required.

iii) Sketching this, the answer is clear.
7. Re: STEP 2005 Solutions Thread
2005 STEP I question 13

8. Re: STEP 2005 Solutions Thread
2005 STEP II question 13

9. Re: STEP 2005 Solutions Thread
2005 STEP II question 14

Attached Thumbnails

10. Re: STEP 2005 Solutions Thread
2005 STEP III question 10

Last edited by brianeverit; 21-07-2011 at 19:45.
11. Re: STEP 2005 Solutions Thread

12. Re: STEP 2005 Solutions Thread

\times \dfrac{1}{1-\frac{w}{w+1})^2=w [/latex]

13. Re: STEP 2005 Solutions Thread
(Original post by sonofdot)
STEP II 2005 Question 8

[expand=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:
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
14. Re: STEP 2005 Solutions Thread
(Original post by rath90)
Hi there how you work out the first few terms of the expansion (1+(3/x))^-1 ?
Standard binomial expansion
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