The 2012 STEP Results Discussion Thread

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I have to say that I was until a few weeks ago and I realised how shifting coordinate systems around could be really handy. I mean by creating new unit vectors, say, orthogonal to planes that have particles colliding with them.

is that something in STEP III i presume.....I haven't hit the mechanics in step III yet.....im just killing off questions in step I and II mainly.

do send me the question your referring to.....might consider attacking it :colone:

Reply 981

Rahul.S

Original post by twig

Got it to work! Must say though that a bit of hindsight did help, as I knew (from doing the standard method first) what to approach, and otherwise it is not really obvious. But it is a lot shorter this way.

Vectors are weird for me, because I sat SI before ever looking at a c3/4 paper, so I guess I am more used to this type. My slow speed and sloppy algebra skills make up for it though!

btw what college did you apply to? :tongue:

dont think ive seen you much on TSR :eek:

Reply 982

bananarama2

Original post by Rahul.S

STEP II 1987 q 11. It's difficult. I struggled even after peeking at the answer....
came in handy in M4 too.

It's not a collision question

(edited 12 years ago)

Reply 983

Rahul.S

Original post by wcp100

STEP II 1987 q 11. It's difficult. I struggled even after peeking at the answer....
came in handy in M4 too.

thanks

ill give it a go later tonight. :colone:

Reply 984

fruktas

STEP II 2007 Q2. Perfectly lovely question, but I am struggling to understand the symmetry bit. I see that the cubic is symmetrical about the point P, thats clear, but what I do not see is how the area R can be modeled by a simple triangle, ie how can you be sure that the bits of the curve coming out of the triangle are identical to the bits missing from the triangle

Mostly based on post #41 http://www.thestudentroom.co.uk/showthread.php?t=443433&page=3&p=9663502#post9663502 p

Reply 985

Rahul.S

for some reason or another I decided to do the 2010 STEP II paper in timed conditions and I found it very good lol :colone:

I would love our II paper to be like that......on the discussion of vectors...the vector question on that paper was a GIFT!

Reply 986

safmaster

Original post by fruktas

Did this question yesterday; from what I can recall, equating the second derivative of y to zero yields x = (p+q) / 2. Therefore, at this point the rate of change of gradient is zero, hence the curve is completely symmetrical at this point of inflection, which is very obvious from a good sketch. It follows R can be modelled by a simple triangle (which lets you do do part (iv) in literally one line of working :tongue:

Reply 987

safmaster

Original post by Rahul.S

for some reason or another I decided to do the 2010 STEP II paper in timed conditions and I found it very good lol :colone:

I would love our II paper to be like that......on the discussion of vectors...the vector question on that paper was a GIFT!

I haven't seen that paper yet (saving it for later), but it must have been easy since the boundaries were sky high (79 for a 1, 105 for a S)! I'd rather have a slightly harder paper with lower boundaries :tongue:

Reply 988

number23

has anyone got any tips for the STEP papers? I did maths a2 last year and doing FM a2 this year. Finding the STEP quite difficult atm, dont feel confident in answering many of the questions... any suggestions?

Reply 989

Rahul.S

Original post by safmaster

im leaving the 2011 paper but the rest im smashing :colone:

79 for a 1 :eek:

I can see why duo :colondollar:

Reply 990

fruktas

Original post by safmaster

I look at the question properly again, it makes sense! Thanks :smile:

Reply 991

worriedParnt

For step iii 1998 qu 8 part 2 i would change the answer slightly to n.d +/- a =l because a is the radius of the sphere and hence positive. (The sphere can be tangential on either side of the plane.)

Reply 992

ian.slater

Original post by fruktas

All cubics have rotational symmetry about their point of inflexion - that can be useful in several STEP questions. The proof is left to the reader :tongue:

Reply 993

shamika

I'm becoming more and more convinced that learning a few basic facts about probability distributions (definitions of pdf, expectation, median, variance and cdf) will bag most people here an easy 20 marks on STEP II. STEP II 2010 Q12 is a perfect example - that should be easy fodder for a serious STEP student. Might be worth trying a few to see whether you can use it as insurance in case the pure turns out to be a bit of a nightmare...

Reply 994

Rahul.S

Original post by shamika

might look at bits of stats in the last month....I heard they can be quite easy to get marks in as well. :tongue:

Reply 995

Farhan.Hanif93

Original post by worriedParnt

For step iii 1998 qu 8 part 2 i would change the answer slightly to n.d +/- a =l because a is the radius of the sphere and hence positive. (The sphere can be tangential on either side of the plane.)

Notice that you can define any point with position

\mathbf{p}

on the sphere by choosing a suitable unit vector

\mathbf{n}

so that

\mathbf{p} = \mathbf{d} + a\mathbf{n}

i.e. you cover all possible points for the plane to be tangential to the sphere at. Having

\pm a

instead just describes these points twice (unless you fix

\mathbf{n}

for diametrically opposite points, which I haven't specified and thus can't be assumed. Since my solution only chose to consider one point rather than a pair of diametrically opposite points, my answer should still hold without the ±. It can be taken WLOG).

Reply 996

fruktas

What is STEP III Mechanics mostly based on? Rotations of rigid bodies about fixed (smooth) axis?

Reply 997

Exp!

Any ideas on who/what this may be?

step.cambridge

...

Please, feel free to introduce yourself.

Reply 998

Farhan.Hanif93

Original post by fruktas

What is STEP III Mechanics mostly based on? Rotations of rigid bodies about fixed (smooth) axis?

I don't think that it can be narrowed down to just one topic. That said, rotational dynamics is a huge part of the syllabus so there are quite a few questions on it. There's also quite a lot on oscillations (simple harmonic or otherwise).

Reply 999

worriedParnt

Original post by Farhan.Hanif93

Notice that you can define any point with position

\mathbf{p}

on the sphere by choosing a suitable unit vector

\mathbf{n}

so that

\mathbf{p} = \mathbf{d} + a\mathbf{n}

i.e. you cover all possible points for the plane to be tangential to the sphere at. Having

\pm a

instead just describes these points twice (unless you fix

\mathbf{n}

I may be wrong but my point is that there are two possible spheres. We start with the origin, we fix n and l which gives our plane, and now we have two places to put the sphere, "above" or "below" the plane.