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

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1. (Original post by Salty carpet)
I got 2root(gl) as well... Everyone is getting like root(5/2gl) instead though
I got the same as you. Wasn't it just 1/2mu^2 > 2mgl?
2. (Original post by TheLifelessRobot)
I got the same as you. Wasn't it just 1/2mu^2 > 2mgl?
Yes
3. (Original post by TheLifelessRobot)
I got the same as you. Wasn't it just 1/2mu^2 > 2mgl?
I got 2√gl too

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4. = Gain EPE + Work Done by Friction?

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5. Pretty sure it was 2root(gl) or root(4gl) either one was fine.
Attached Images

6. Grade boundaries should be standard

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7. (Original post by TheLifelessRobot)
Pretty sure it was 2root(gl) or root(4gl) either one was fine.
It was worth 4 marks... 2sqr(gl) takes like one line of working..
8. (Original post by MrMagic12345)
It was worth 4 marks... 2sqr(gl) takes like one line of working..
Not really. Conservation of energy (1), correct equation (2), putting v>0 (3) Re-arranging to get u = 2root(gl) (4)

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9. I think it was a similar level of difficulty to last year? Maybe slightly harder because of the lack of numbers in the questions. It felt like an exam in using the Greek alphabet with that differential equation.
10. yeah that's what I got! thank god!

(Original post by Pentaquark)
Anyone get root 5ag for the 4 marker least value of mu.

Grade boundaries for an A* anyone?
65 last year. I think similar to 2015 maybe a higher ??
11. How can tension in the horizontal circular motion question be 78.4. Doesn't Tcostheta = 78.4?

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12. I got 7.05m
(Original post by ashwinderk)
I got 7.58
13. (Original post by MrMagic12345)
Can someone explain how the got a quadratic for tantheta in terms of u? I just got 1/2u..
My friend did that, but its because he drew his frictional force from the wall downwards instead of upwards
14. (Original post by TheLifelessRobot)
Pretty sure it was 2root(gl) or root(4gl) either one was fine.
Sound. I got too.
15. (Original post by Yo12345)
How can tension in the horizontal circular motion question be 78.4. Doesn't Tcostheta = 78.4?

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The tension acts directly in the opposite direction to the weight of the 8kg particle. Remember the 8kg particle was along the same line as the tension as it was perfectly vertical.
16. yes
(Original post by MrMagic12345)
And also, for the tension in string when particle is at 30degrees, do you have to use the weight component against the tension for mv^2/r I.e does mv^2/r=T-mgcos30?
17. I got the same answers except for q8 , I got like 7.05m I think my method was correct so I hopefully only loose 2 marks or so
(Original post by Busted838)

1a. 9.6J (2)
b. 24.3J (3)
c. 12.7ms_1 (2)
d. no air resistance (1)

2a. find v (2)
b. Force 8.2N (3)
c. t= cube root of 2 (2)
d. a=3 (4)
b=-3

3a. symetry (1)
b.10.4cm (4)
c.16.5 degrees (3)
d. grav force or weight acts through center of lamina (1)
.

4a. 78.4N (1)
b, theta=41.4 (3)
c. r=2.9m (4)

5a. v^2=u^2-2gl(1-cosx) (3)
b. T=mu^2/l - 2mg + 3mgcos(30) (3)
c. T=mg + mu^2/l (2)
d.u=root(5gl) (4)

6a. show that (2)
b. v=(g+(YV-g)e^-Yt )/Y C=ln(Yu-g)/Y (6)

7a. show that (2)
b. tan theta =((1/2)-u^2)/2u (7)

8. (pretty sure this is wrong) 4.17 (8)
18. (Original post by PiTheta97)
Was the energy equation for Q8:

Loss EPE + mgsinx = Gain EPE + Work Done by Friction?

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I did EPE(pr) = GPElost + work done by friction + remaining EPE(pr) + EPE(pq)

I think those letters are correct lol
19. This paper seemed to combine the 'hard' questions from a few other papers from different years with a new twist on them (question 8 similar to june 2014 paper q8 but with two elastic strings, question 5 somewhat similar to june 2011 q8 but more difficult etc.). I think grade boundaries will be around 62/63 for 90 UMS on this paper.
20. I think so, the way I out it was . energy at A=energy at distance x
therefore...GPE+EPE=GPE+EPE+WD
(Original post by PiTheta97)
Was the energy equation for Q8:

Loss EPE + mgsinx = Gain EPE + Work Done by Friction?

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