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

1. (Original post by Jm098)
I got radius as 2.89?
Same here
2. Can someone explain how the got a quadratic for tantheta in terms of u? I just got 1/2u..
3. (Original post by maisym00)
Same here
Same
4. (Original post by KittyKat97x)
Anyone get like 105° for centre of mass angle?
The angle couldnt have been greater than 90°. I got 16.5°
5. Definitely ****ed that up. Must have gotten at best a C. Goodbye forever, KCL.
6. 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?
7. What did you get for differential equation in for v
8. (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?
They wanted it in terms of U G L and M
9. (Original post by Jm098)
I got radius as 2.89?
I got something like that too
10. Unoficial Mark Scheme: http://www.thestudentroom.co.uk/show....php?t=4188831
11. (Original post by maisym00)
What did you get for differential equation in for v
(g-(e^(-lambda T))(g-lambda U))
___________________________
Lambda
12. Best recollection of my answers:
1a. 9.6J
b. 24.3J
c. 12.7ms^1
d. Air resistance negligible
2a. a = (8-4t^3)i - 18e^-3tj
b. force = a * 2
c. Magnitude of force =8.2N
d. Time to travel south = 1.26s
e. Integrate v to find r
3a. Line of symmetry
b. 10.4cm from top line
c.16.5 degrees from vertical
4a. Tension = 78.4N
b. Theta = 41.4 degrees
c. Radius = 2.89
5a. V = sqrt(2gl(1-cos30)-u^2) which is wrong, right?b. Find tension at T
c. Tension somewhere
d. Min value = 2sqrt(gl)
6a. Show that dv/dt = something
b v = (u - g/lambda)e^-(lambda* t) + g/lambda (I think this is also wrong)
7. Tan theta = (1-2u^2)/4u
8. Don't remember
13. (Original post by ashwinderk)
The angle couldnt have been greater than 90°. I got 16.5°
Whoops I think I got confused as to where the vertical was
14. (Original post by ashwinderk)
I got tan(theta)=(1-2u^2)/4u
Same

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15. So Im pretty sure I did exactly the right method for the last question with the springs.
But I forgot to square the two extensions.
How many marks do you reckon I will lose?
16. The answers i got

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)
17. I got 0.898m for Q8

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18. (Original post by XxswagmasterxX)
Best recollection of my answers:
1a. 9.6J
b. 24.3J
c. 12.7ms^1
d. Air resistance negligible
2a. a = (8-4t^3)i - 18e^-3tj
b. force = a * 2
c. Magnitude of force =8.2N
d. Time to travel south = 1.26s
e. Integrate v to find r
3a. Line of symmetry
b. 10.4cm from top line
c.16.5 degrees from vertical
4a. Tension = 78.4N
b. Theta = 41.4 degrees
c. Radius = 2.89
5a. V = sqrt(2gl(1-cos30)-u^2) which is wrong, right?b. Find tension at T
c. Tension somewhere
d. Min value = 2sqrt(gl)
6a. Show that dv/dt = something
b v = (u - g/lambda)e^-(lambda* t) + g/lambda (I think this is also wrong)
7. Tan theta = (1-2u^2)/4u
8. Don't remember
i agree with all other than 5 and 5a
19. I got 2root(gl) as well... Everyone is getting like root(5/2gl) instead though
20. (Original post by MrMagic12345)
Can someone explain how the got a quadratic for tantheta in terms of u? I just got 1/2u..
I got that aswell

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