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# AQA A2 MM2B Mechanics 2 – 27th June 2016 [Exam Discussion Thread] Watch

1. (Original post by Martin9808)
Yes it's 2root gl definitely
I'm genuinely tired of people strongly asserting that incorrect corrects are 'definitely' correct. Did you even consider why it might be wrong? It's wrong, so no idea what your definition of 'definitely' is. Look up conditions for full revolutions and be amazed.
2. (Original post by tanyapotter)
for the expression of r in the earlier question, was c= 5i-5j?
wasn't it c = 3i - 3j
3. (Original post by C0balt)
I swear it was root5lg, there was a similar question to that in one of past papers, I think it was after finding expression for T and we had to make T=0
It was . Dw about the people that disagree.
4. Did everyone get some weird stuff for the vertical circular motion before the 2rootlg or root5lg question? The tension ones
5. (Original post by forestspirit)
wasn't it c = 3i - 3j
Yeah
6. (Original post by C0balt)
Did everyone get some weird stuff for the vertical circular motion before the 2rootlg or root5lg question? The tension ones
Yeah some strange answers 😂 Can't imagine how frustrated the examiners must be marking it
7. (Original post by Jm098)
Yeah some strange answers 😂 Can't imagine how frustrated the examiners must be marking it
Those m,g,u,l expressions were ugly as hell lol.
8. (Original post by forestspirit)
wasn't it c = 3i - 3j
oh yeah i think i wrote that, i just remember them both being the same and one being positive and the other negative

when t=0, what did it say the displacement was? do you remember?
9. (Original post by IrrationalRoot)
I'm genuinely tired of people strongly asserting that incorrect corrects are 'definitely' correct. Did you even consider why it might be wrong? It's wrong, so no idea what your definition of 'definitely' is. Look up conditions for full revolutions and be amazed.
Lool mate , it's 2 root gl I'd love to see the look on your face when you realize this
10. (Original post by tanyapotter)
oh yeah i think i wrote that, i just remember them both being the same and one being positive and the other negative

when t=0, what did it say the displacement was? do you remember?
I think 3i - 5j... pretty certain
11. Thought it was 2root(ag) as well, it's definitely (I think) root(5ag) cause if you put 2root(ag) into the equation for centripetal force at the top, it gives you a negative tension which is impossible, realising this leads you to root(5ag).
12. (Original post by tanyapotter)
oh yeah i think i wrote that, i just remember them both being the same and one being positive and the other negative

when t=0, what did it say the displacement was? do you remember?
8.20 metres rings a bell
13. Obligatory Mechanics post

14. Is this the method that anyone else used for Q8? Not sure where I went wrong.

No KEs to consider first of all. Now by conservation of energy,

GPE+EPE before = GPE+EPE+Work after

Taking x to be the distance from A when next at rest, height for GPE is x/2. (xsin30.)
This is taking the 0 GPE level to be at the point when next at rest.

EPE is for one string, since extension 0 for one string, extension 5 for the other so EPE from that string.

Because of our choice of GPE level, GPE=0 after.

EPE after is total of the two EPEs for both strings, so with extension of x in one of them and extension of x-6 in another.

Work found by x multiplied by coefficient of friction multiplied by reaction force which is mgcos30.

Then I got two answers, chose the larger one, added 4 (got 7.05m).
15. (Original post by steph-carys)
I think 3i - 5j... pretty certain
that was the initial position, so when you integrate the velocity to get the displacement and set t=0, you get 3i - 5j = -2j + c, so c = 3i - 3j
16. (Original post by IrrationalRoot)
I'm genuinely tired of people strongly asserting that incorrect corrects are 'definitely' correct. Did you even consider why it might be wrong? It's wrong, so no idea what your definition of 'definitely' is. Look up conditions for full revolutions and be amazed.
Just ignore it, people are bound to believe that what they have done is correct. I got in the exam but have since realised that this expression is wrong as the tension at the top of the circle could be negative.
17. (Original post by Jm098)
Oh **** I think they'd ecf it though
You think? Well, that'd be nice but given the way the mark schemes work I'm likely not to get any marks . Hoping for some sympathy from the examiner...
18. (Original post by IrrationalRoot)
Is this the method that anyone else used for Q8? Not sure where I went wrong.

No KEs to consider first of all. Now by conservation of energy,

GPE+EPE before = GPE+EPE+Work after

Taking x to be the distance from A when next at rest, height for GPE is x/2. (xsin30.)
This is taking the 0 GPE level to be at the point when next at rest.

EPE is for one string, since extension 0 for one string, extension 5 for the other so EPE from that string.

Because of our choice of GPE level, GPE=0 after.

EPE after is total of the two EPEs for both strings, so with extension of x in one of them and extension of x-6 in another.

Work found by x multiplied by coefficient of friction multiplied by reaction force which is mgcos30.

Then I got two answers, chose the larger one, added 4 (got 7.05m).
Extension would be 6-X in the other string, but shouldn't affect your answer I don't think.

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19. (Original post by Martin9808)
Lool mate , it's 2 root gl I'd love to see the look on your face when you realize this
Yep, you're one of those people. Think you're always right even in the face of a clear proof that you're incorrect. I'm expecting you to mysteriously stop posting when you understand the correct method .
20. (Original post by jjsnyder)
Extension would be 6-X in the other string, but shouldn't affect your answer I don't think.

Posted from TSR Mobile
Yep shouldn't because of the square.

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