AQA M2- June 2011

On the specimen paper why does kt turn positive after integration?

Q. 6 a

I'm not 100% sure if the answer in the question paper is right... can anyone confirm this?


But I think in the mark scheme they mean 1/(2v^2) = kt + 1/(2u^2) rather than -1/(2v^2) = kt + 1/(2u^2) that's what I got anyway in which case I just multiplied both sides by -1.

Yeah the energy at the top of the hill must equal the energy at the bottom of the loop as none is lost, yes this does still equal the energy at the top of the loop too but you don't need that to work out h. Hope that helps!

But if you integrate 1/v^3 it's like integrating v^-3.

That gives you (v^-2)/-2 , which gives you -1/(2v^2)

From what I got all the expressions were negative.

ahh cool, almost ruined my entire cofidence looking through that specimen but it's not actually to bad, although I can't get the answer to the one about the moon and do you know how to explain 5d at all sorry for all the questions!!

So did you get -1/(2v^2) = -kt + c ?

Because then you can multiply through by -1, to give

1/(2v^2) = kt + a

Well basically for 5d) you need to calculate the work done against friction in getting to the wall, if it's greater than the energy you had at the beginning (the elastic potential energy) you can't get to the wall as you don't have enough energy to overcome it. But if it's less you can, so work done = force x distance so plug in the values for those and if it's less than 16 (epe) you can get to the wall - any left over energy is kinetic energy .

And what part of the moon bit don't you understand cos I can help with that too!

nope I got -1/(2v^2) = -kt -1/(2U^2)

I must have miscalculated but I just can't for the life of me work out where.

ahh cool, totally get that now thanks!

and with the moon I know what I need to do and what numbers to put in but I just couldn't see how to get the printed answer :/

For a or b ?

A) F = mv^2 / r so v = sqr root of (Fr/m) then just plug in the values given

and b) I did it differently but in the mark scheme they used time = distance/speed so the distance is the circumference of the circle and speed is what you just calculated

No thaat's what I got sorry you've already calculate the constant of integration :P

But yeah so just times through by -1 and that's how they got kt to be positive I think in the mark scheme it's a mistake where they have -1/(2v^2) = kt + 1/(2U^2) I think it should be 1/(2v^2) = kt + 1/(2U^2) which is what you got!

Well basically for 5d) you need to calculate the work done against friction in getting to the wall, if it's greater than the energy you had at the beginning (the elastic potential energy) you can't get to the wall as you don't have enough energy to overcome it. But if it's less you can, so work done = force x distance so plug in the values for those and if it's less than 16 (epe) you can get to the wall - any left over energy is kinetic energy .

And what part of the moon bit don't you understand cos I can help with that too!

You know f =mv^2/r

Which direction is F ? Is it acting towards the centre?

right so I just tried 5d and am I doing it right, will the WD against friction be 0.6*F because for F I get it to be g so 0.6*g=5.88 but in the mark scheme it has 9.8 as the answer so I dont know if i'm doing it right

The distance travelled is 1 not 0.6 because it's being released from when it was extended

ahh gotcha, thanks once again!

It's no problem at all! Could you help do you think?

http://www.billgrisdaledesign.co.uk/DCHS_mathspapers/images/AQA-MM2B-W-QP-Jan11.pdf

Q5

I don't understand why the centripetal force is equal to the friction

I'd really appreciate it if somebody could go over 7a from http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MM2B-W-QP-JUN09.PDF

I know the exam's really close now, but no matter how many times I read the mark scheme I don't have a clue haha. Guess that's what happens when you sit this exam in year 12 without being taught half of it.