STEP Prep Thread 2016

Original post by tiny hobbit

It means exactly North. This matters because of the wind direction. Think 3D to translate this question. The angle theta is above the horizontal, which is nothing to do with the compass direction.

Either I've gone completely out of my mind or I'm just being thick. How does this question have anything to do with 3D? It's a two dimensional question and Farhan's solution agrees with mine.

Reply 704

tiny hobbit

15

Original post by Zacken

Either I've gone completely out of my mind or I'm just being thick. How does this question have anything to do with 3D? It's a two dimensional question and Farhan's solution agrees with mine.

I said this because the conversation seemed to be mixing the angle theta with the compass direction. Your posting talked about the projectile going "a bit north".

The OP needed to realise that the projectile is fired exactly northwards and the wind is blowing north to south.

(edited 8 years ago)

Reply 705

Original post by rcmehta

This?

Thank you!

Reply 706

demigawdz

Original post by tiny hobbit

I said this because the conversation seemed to be mixing the angle theta with the compass direction. Your posting talked about the projectile going "a bit north".

The OP needed to realise that the projectile is fired exactly northwards and the wind is blowing north to south.

yeah thanks bro, they really should of just used plain english really
rofl

Reply 707

STEP I, 1995, Q10, part about showing

v^2 + \frac{\lambda x^2}{ml} = u_0^2.

I resolve downwards to get

\displaystyle m\ddot{x} = mg - T = mg - \frac{\lambda(x+e)}{l} = mg - \frac{\lambda x}{l} - \frac{\lambda e}{l} = mg - \frac{\lambda x}{l} = mg - \frac{\lambda x}{l} - mg

So,

\displaystyle m\ddot{x} = -\frac{\lambda x}{l} \implies \omega^2 = \frac{\lambda x}{ml}

... which is correct. Wow! Looks like I do maths better on TSR and Latex than on paper. Oh well, just gonna post this now, given that I've taken the time to type it up. :tongue:

Reply 708

username1162972

18

Original post by Zacken

STEP I, 1995, Q10, part about showing

v^2 + \frac{\lambda x^2}{ml} = u_0^2.

I resolve downwards to get

\displaystyle m\ddot{x} = mg - T = mg - \frac{\lambda(x+e)}{l} = mg - \frac{\lambda x}{l} - \frac{\lambda e}{l} = mg - \frac{\lambda x}{l} = mg - \frac{\lambda x}{l} - mg

So,

\displaystyle m\ddot{x} = -\frac{\lambda x}{l} \implies \omega^2 = \frac{\lambda x}{ml}

... which is correct. Wow! Looks like I do maths better on TSR and Latex than on paper. Oh well, just gonna post this now, given that I've taken the time to type it up. :tongue:

Multiplying theough the result by 1/2m and rearranging is interesting also. Just an idea

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Reply 709

Original post by physicsmaths

Multiplying theough the result by 1/2m and rearranging is interesting also. Just an idea

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Thanks, I'll give that a go too. I think I found another way of doing this part without SHM. Just letting

\ddot{x} = v \frac{dv}{dx}

then solving the resulting differential equation.

Reply 710

username1763791

For those who did all of this last year for Camb, what would you say was the best interview prep you did?

Reply 711

username1162972

18

Original post by Jordan\

For those who did all of this last year for Camb, what would you say was the best interview prep you did?

Probably the questions i put up. Weirdly i think Mat is pretty useful u know. Step I aswell but the problems in step are on the longer side for an interview(bar some ofcourse)

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Reply 712

Original post by physicsmaths

Probably the questions i put up. Weirdly i think Mat is pretty useful u know. Step I aswell but the problems in step are on the longer side for an interview(bar some ofcourse)

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What questions did you put up?

How easy is the MAT?

Reply 713

130398

3

There is this question in the "Advanced problems in mathematics"booklet and the last part asks so show that integrating 1/1+x^2 from 0 to 1 is equal to π/4. The solutions use something called the shell method I believe (not quite sure how you're suppose to know that from the C4 integration but maybe that's just me). What is wanted to ask is, if you see that that is the integral from fp3 of arctanx could you use that?

Reply 714

username1763791

Original post by 130398

There is this question in the "Advanced problems in mathematics"booklet and the last part asks so show that integrating 1/1+x^2 from 0 to 1 is equal to π/4. The solutions use something called the shell method I believe (not quite sure how you're suppose to know that from the C4 integration but maybe that's just me). What is wanted to ask is, if you see that that is the integral from fp3 of arctanx could you use that?

Normally I would say yes, especially on the pre-2000 papers when you do them as its used quite a bit there. However, in the most recent exams I probably wouldn't, and if the whole point of the question was to integrate that expression then using the formula would make the question trivial, possibly ruining the point off the whole question.

In general, if your own method that is off-syllabus renders the question pointless and trivial, I would use the intended method/ one on syllabus to be safe!

Reply 715

username1763791