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    (Original post by Jacob2215)
    How many marks would I lose for 3sfgs on the coefficient of friction for q5??? i.e 0.727
    none
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    45 degrees isn't a direction, just a size, a bit like a scalar
    (Original post by SirRaza97)
    Yo I think the very last question was not a bearing. So it would be 45 degrees
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    (Original post by Michal-sweaty-g)
    45 degrees isn't a direction, just a size, a bit like a scalar
    Well 45 degress to the vertical/horizontal axis. That is a direction. I just drew an arrow in the diagram and put 45 degrees.
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    (Original post by mk_98)
    I said force acts at 45 degrees below horizontal. Would that be enough?
    no because you need to include the direction, which can be shown in a diagram or say pointing to south west direction, so i think you lost a mark for that
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    (Original post by SirRaza97)
    Well 45 degress to the vertical/horizontal axis. That is a direction. I just drew an arrow in the diagram and put 45 degrees.
    "45 degrees from the horizontal table" can be south west or north east, i think the examiners want to see "south west" or a diagram with an arrow
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    (Original post by suhaylpatel786)
    ...
    Agreed.
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    (Original post by KloppOClock)
    i thought it was an okay paper, maybe 64 for an A?

    what do you guys think?



    also, does anyone remember the mark distribution of the questions?
    I think it was along the lines of ;

    1a)
    b)
    c)

    2a) 3marks
    b) 3marks

    3) 8marks

    4a) 4marks
    b) 8marks

    5) 10marks

    6)

    7a) 7marks
    b) 4marks

    8) 8marks
    b) 4marks

    feel free to add and improve (not totally sure about all).
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    (Original post by ChilliLemon)
    For question 2) (b)

    I wrote the reaction force is equal to 15.45N.
    Buth then i wrote the force exerted on the steel pan is 15.45 - 1.5g, so i took away the weight of the brick, and got 0.75N.

    Will i stil get full marks because i wrote dwon that R=15.45N, though this was not my final answer?
    no i am sure that you wont get full mark for that, because you havent rounded up to 2 or 3 sf.
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    For the one where you had to calculate friction (mu), I put 0.727 to 3 s.f. (i think). Is that right guys?
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    (Original post by thelegend99)
    For the one where you had to calculate friction (mu), I put 0.727 to 3 s.f. (i think). Is that right guys?
    yes you are correct.
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    As far as I can tell I got it all right except the very last q in which I put southeast as the direction instead of southwest.. 73/75?
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    For 2b) did it ask for the answer to 3sf?
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    Please can someone post full working for the moments question?
    Im trying to work our where I messed up as I got d=2 and M=90
    Thank you
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    1a) find bearing 3
    b) equations 3
    c) position vector 4

    2a) tension in string 3
    b) reaction on thing inside the pan 3

    3) particle question, 7

    4a) sketch speed time graph, 4
    b) solve for T, 8

    5) Find coefficient of friction,10

    6) moments, 7

    7a) find F2, 7
    b) find speed, 4

    8) find T, 8
    b) resultant on pulley, 4
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    (Original post by geniuses)
    You and me Bro, we will rule the 7th dimension.
    yeah you can use those 2 inchers to have a sword fight over allegedly getting 100% ;D
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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
    At 8, Did you have to write down the bearing? All I did was work out the Tension and the second part I used 2Tcos(45)
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    What do u think the boundary will be for an A?

    I think 59-60
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    (Original post by examhater44)
    What do u think the boundary will be for an A?

    I think 59-60
    yeah between 59-62 for an A, was easier than last year in my opinion but the second vectors question might have caused a few problems. Maybe 72/73 for 100 UMS
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    Can some one remember what question 2 was especially the working out for 2b?
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    (Original post by KloppOClock)
    i thought it was an okay paper, maybe 64 for an A?

    what do you guys think?



    also, does anyone remember the mark distribution of the questions?
    I would like to think it would be lower than you have predicted based on the last 3 years' papers where an A was about 59.
 
 
 
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