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# AQA Mechanics 2(B) Jan 20th watch

1. I thought this was a challenging, but quite a good paper in all. I especially liked the part of the last question asking you to find the max velocity. Range of approaches acceptable and nice link with the 'Core' modules (of which I think there should be more).
2. im sure youll get the b or above, seems like most of our answers come out the same (:
ive seen easier papers before, this one was quite hard
3. (Original post by ajay_ganger)
i missed out that complete the square ? at the end time run out!!!!! i got t = 1 for that minimum acceleration ?
I got t=1 as well.

Edit: I didn't think this was much harder than the past paper Qs except for the final question. I knew if that (the 6 marker) type of question came up I hadn't learnt it, so that was a shame. The vertical circular motion was a nice easy one though, because they can be really tough. If you have access, try Jan '05; the last question is a killer.
4. I got t=0 for minimum acceleration. I differentiated acceleration, which came to 24ti, and put it equal to zero to find the minimum point of the acceleration, which was 0.
5. what did you guys get for the max distance on end question? 54 or something like that?

for the 31.8 one,
i just differentiated it and equaled it to zero to get the max value
i hate completing the square

i also got t=1 for the acceleration q
6. (Original post by maxfire)
I got t=0 for minimum acceleration. I differentiated acceleration, which came to 24ti, and put it equal to zero to find the minimum point of the acceleration, which was 0.
t could = 0 or 1, so I took 1 because 0 was a given.

(Original post by jahan218)
what did you guys get for the max distance on end question? 54 or something like that?

i just differentiated it and equaled it to zero to get the max value
i hate completing the square

i also got t=1 for the acceleration q
I think I got 54 ish as well. I just used my equation solver on my calc. I wonder if I'll get all the method marks therefore. I just wrote "use quadratic formula" next to it, lol :|
7. i just pretended to look like i used the quadratic formula when i used my calc aswell, youll be alright if you stated it i would think
8. What did people get for the maximum velocity of the jumper? I think I ****** that up and got 2800 or something, which doesn't sound remotely correct.
9. (Original post by maxfire)
What did people get for the maximum velocity of the jumper? I think I ****** that up and got 2800 or something, which doesn't sound remotely correct.
i just subbed in 31.8 into the equation i think, i got 29 or someting like that
10. I did the same. hmmmmmm
11. we have to COMPLETE THE SQUARE on the last question??? whaaat... i just used the quadratic equation .....and i managed to get (u^2+2ag)/(a^2+2ag) for the angle pheta question

this is bad...... i thought this paper was alot harder than the past papers, especially with the amount of core maths involved, and i totally ignored core cos our exam is in june, now im totally regretting it
12. (Original post by bunzy)
we have to COMPLETE THE SQUARE on the last question??? whaaat... i just used the quadratic equation .....and i managed to get (u^2+2ag)/(a^2+2ag) for the angle pheta question

this is bad...... i thought this paper was alot harder than the past papers, especially with the amount of core maths involved, and i totally ignored core cos our exam is in june, now im totally regretting it
i dont think we had to complete the square, it didnt say so
which question are we talking about here, i cant really remember,

there was, find the maximum x, where you could just use the quadratic formula

then there was, show that the maxspeed is at the point x=31.8 where you could complete the square or use calculus

and then it was, find this speed, just subbing 31.8 into the equatoin
13. (Original post by jahan218)
i dont think we had to complete the square, it didnt say so
which question are we talking about here, i cant really remember,

there was, find the maximum x, where you could just use the quadratic formula

then there was, show that the maxspeed is at the point x=31.8 where you could complete the square or use calculus

and then it was, find this speed, just subbing 31.8 into the equatoin
oh right okay, i think i just differentiated for the 31.8 one, ah i can't remember any questions or my answers which is making me panic slightly, not that there's anything i can do except to hope until march time

coming on tsr after exams is definetely adding to the feeling of dooom and failure
14. In case anyone is interested:

mga(1-cosx) = 0.5mv^2 - 0.5mu^2
mga(1-cosx) + 0.5mu^2 = 0.5mv^2
2ga(1-cosx) + u^2 = v^2
v = sqrt[2ga(1-cosx) + u^2]

mgcosx - R = (mv^2)/r R = 0, r = a
gacosx = 2ga(1-cosx) + u ^2
gacosx = 2ga - 2gacosx + u^2
3gacosx = 2ga + u^2
cosx = (2ga + u^2)/3ga
15. dats what i got

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Updated: January 20, 2010
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