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

1. (Original post by IrrationalRoot)
Yes there is the correct answer is and that is certain.
I certainly hope so since that's what I put!
2. (Original post by IrrationalRoot)
I needed 0UMS too XD. Still really disappointing that I got 80s in this paper, worst mark I've got in an M2 paper by far and that's just from two tiny mistakes.
Me too! Made a couple silly errors, but I don't mind as I had only done a couple past papers and was never taught the module, actually just read the M2 book on a train XD
3. (Original post by LiesandAlibis)
Also, is there a general consensus on 2(root)gl versus (root)5gl?
In my opinion:

The answer is (root)5gl because this is the answer found with the assumption that the tension in the string must be greater than or equal to zero at the top.

An answer of 2(root)gl is yielded when assuming that the velocity is greater than or equal to zero at the top.

This is true for a constrained circular motion, e.g. a bead on a wire where it cannot go anywhere else other than along the wire, but for situations where the particle is unconstrained, the tension must also be greater than or equal to zero. This is an additional condition which stops the particle from taking a decay path before reaching the top.
4. (Original post by jjsnyder)
Me too! Made a couple silly errors, but I don't mind as I had only done a couple past papers and was never taught the module, actually just read the M2 book on a train XD
Wow fair play to you then, I was taught the whole module. Also I did quite a few past papers (10 or so). So for me, no excuses for stupid mistakes lol.
5. (Original post by MrCahooner)
In my opinion:

The answer is (root)5gl because this is the answer found with the assumption that the tension in the string must be greater than or equal to zero at the top.

An answer of 2(root)gl is yielded when assuming that the velocity is greater than or equal to zero at the top.

This is true for a constrained circular motion, e.g. a bead on a wire where it cannot go anywhere else other than along the wire, but for situations where the particle is unconstrained, the tension must also be greater than or equal to zero. This is an additional condition which stops the particle from taking a decay path before reaching the top.
Yep this is correct reasoning. Bead on a wire can have a reaction force upwards but tension will not act upwards.
6. do people think this paper was comparable to june 2014, or was it easier/harder?
7. (Original post by tanyapotter)
do people think this paper was comparable to june 2014, or was it easier/harder?
Really similar but I think there were little more things u could trip up on so I think slightly Lower gb that 2014
8. How many marks was the last question?
9. 1-u/2 for the ladder question? 7.15m last question?
10. (Original post by mcride98)
How many marks was the last question?
8 i think
11. for the last question, i have a strong opinion that the final answer is 5.46m. Both extensions where taken from the distance from q where the next point of rest is. one of the extensions was 4+x and the other i think was 11-x.

12. Working for question 8.
13. (Original post by Andrew Brockbank)

Working for question 8.
you have used the conservation of energy, but i used forces, the next point of rest will mean that the forces are in equilibrium from each string.
14. (Original post by higginsy7)
you have used the conservation of energy, but i used forces, the next point of rest will mean that the forces are in equilibrium from each string.
The forces dont need to be in equilibrium. Where the forces are in equilibrium will be the final place the mass comes to rest, it simply asks for the next postition the mass comes to rest. It will only be breifly at 6.78m before going up again towards Q.
15. (Original post by Andrew Brockbank)

Working for question 8.
You messed up loss in gpe you said it was dsin30 but it's mgdsin30
16. Scenes, so I did 😶
(Original post by C0balt)
You messed up loss in gpe you said it was dsin30 but it's mgdsin30
17. (Original post by C0balt)
You messed up loss in gpe you said it was dsin30 but it's mgdsin30
stared at that for so long trying to find the error lmao fml

did you get 7.58?
18. (Original post by C0balt)
You messed up loss in gpe you said it was dsin30 but it's mgdsin30
What did you get c0balt?

Posted from TSR Mobile
19. (Original post by Yo12345)
What did you get c0balt?

Posted from TSR Mobile
7.58
stared at that for so long trying to find the error lmao fml

did you get 7.58?
Yeah I did

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