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    for question 2 part b i got 15.7N because i did F=2(0.5)+1.5(9.8), how many marks is that
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    (Original post by E1Ephan7)
    I completely guessed 7a) F=2.5i and forgot to write j component. Showed no working. How many marks could i get??
    Zero
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    Any working out for the 0.74 question
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    Marks:
    Question 1 A -3, B-3, C-4
    Question 2 A-3, B-3
    Question 3 A-7
    Question 4 A-4, B-8
    Question 5 A-10
    Question 6 A-7
    Question 7 A-7, B-4
    Question 8 A-8, B-4
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    I got 0.65 for the coefficient of friction wtfff i thought i aced it
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    how many marks was 1c worth?
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    (Original post by Khaddad)
    I got 0.65 for the coefficient of friction wtfff i thought i aced it
    And I got 0.873
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    How many marks would I lose for 3sfgs on the coefficient of friction for q5??? i.e 0.727
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    (Original post by VA123)
    was μ meant to be given to 2sf or is 3sf ok?
    3is fine, i gave it to 3 as well
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    (Original post by student bang)
    And I got 0.873
    Rip. At least were not the only ones
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    (Original post by KloppOClock)
    Message me if anything is wrong.
    Also, if anyone remembers how many marks each question was worth, please comment so I can add them on too.

    Question 1:
    Spoiler:
    Show
    1(a) 104, (90+14)

    1(b) p = (400+15t)i + (20t)j
    q = (20t)i + (800-5t)j

    1(c) The 'j' vectors for both are equal, as it is due west.
    800-5t = 20t
    t = 32
    therefore q = 640i + 640j
    Question 2:
    Spoiler:
    Show
    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)
    Question 3:
    Spoiler:
    Show
    3 Ns
    Question 4:
    Spoiler:
    Show
    4(a)
    4(b) Area under graph = 975
    Area for slower car for first 25 seconds = 750
    975-750 = 225
    1/2 * b * 30 =225
    b = 15
    total time = 15+25 = 40
    so area under faster car = 975 = 1/2 * (40)(T+40)
    T = 8.75 s
    Question 5: (10 Marks)
    Spoiler:
    Show
    μ=0.73 to 2 significant figures
    Question 6: (7 Marks?)
    Spoiler:
    Show
    For 1st situation, R(T) = 0
    For 2nd situation, R(S) = 0

    M(S) => 0.5M = 30d-15
    M(T) => M = 60 - 15d
    Therefore,
    d = 1.2m
    M = 42kg
    Question 7:
    Spoiler:
    Show
    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1
    Question 8:
    Spoiler:
    Show
    8(a) Fmax = 0.3g
    acceleration = 0.6g
    Tension = 11.76N = 11.8 N (3sf)

    8(b) RF = √(11.76)^2 + (11.76)^2 = 16.6
    It has a bearing of 225

    Woo think I only lost about 5 marks,
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    (Original post by student bang)
    Good luck in life with this attitude
    Good luck in life with that name.

    And your mark.
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    What do people estimate the raw mark to be for 90 or 100 UMS
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    (Original post by Verish)
    Its an AS module, you cant get an A*?
    Probably means 90 UMS.
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    Anyone has the actual questions?
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    poll: http://www.thestudentroom.co.uk/show....php?t=4150305
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    (Original post by geniuses)
    Dear Failures,

    I have just realised i have achieved 100% in this exam.

    Good luck next year!
    Same, but I think its a bit rude to call people failures. I'm sure loads of people did really well and tried their best.
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    (Original post by Ilyas_99)
    Same, but I think its a bit rude to call people failures. I'm sure loads of people did really well and tried their best.
    You and me Bro, we will rule the 7th dimension.
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    All you had to do is 2Tcos(45), as the forces where at right angles to each other.
    (Original post by vvlp)
    For the magnitude of the force on the pulley I didn't do root T^2 + T^2 because i got confused when it said work out the resultant magnitude and it messed me up. I thought the resultant force was the root of ((T-friction)^2 + (weight - T)^2)

    Was this stupid of me? I thought 11.47^2 with an angle of 45 from the horizontal was too simple for 4 marks.
    (Original post by vvlp)
    For the magnitude of the force on the pulley I didn't do root T^2 + T^2 because i got confused when it said work out the resultant magnitude and it messed me up. I thought the resultant force was the root of ((T-friction)^2 + (weight - T)^2)

    Was this stupid of me? I thought 11.47^2 with an angle of 45 from the horizontal was too simple for 4 marks.
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    (Original post by geniuses)
    You and me Bro, we will rule the 7th dimension.
    I'm a girl
 
 
 
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