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

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    (Original post by Jm098)
    I got radius as 2.89?
    Same here
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    Can someone explain how the got a quadratic for tantheta in terms of u? I just got 1/2u..
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    (Original post by maisym00)
    Same here
    Same
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    (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°
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    Definitely ****ed that up. Must have gotten at best a C. Goodbye forever, KCL.
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    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?
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    What did you get for differential equation in for v
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    (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
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    (Original post by Jm098)
    I got radius as 2.89?
    I got something like that too
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    Unoficial Mark Scheme: http://www.thestudentroom.co.uk/show....php?t=4188831
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    (Original post by maisym00)
    What did you get for differential equation in for v
    (g-(e^(-lambda T))(g-lambda U))
    ___________________________
    Lambda
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    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
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    (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 :facepalm:
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    (Original post by ashwinderk)
    I got tan(theta)=(1-2u^2)/4u
    Same


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    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?
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    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)
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    I got 0.898m for Q8


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    (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
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    I got 2root(gl) as well... Everyone is getting like root(5/2gl) instead though
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    (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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