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    (Original post by Themathgeek)
    How many would you lose if you got the magnitude right but wrote 45 degrees and drew it outwards.... read the question wrong.

    Also if you forgot to multiply inpulse by mass how maby would you lose?

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    if you drew it in the wrong direction then i would think you would lose 1 mark
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    (Original post by KloppOClock)
    maybe, 45 degrees below horizontal could be south west or south east, if you drew a diagram with an arrow then you would get it i think
    Also I would lose a mark for writing tension as 11.76 instead of 11.8 or does it not matter?
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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?
    Has it ever been that high? I reckon 60-61 for an A, a lot of my friends found it very difficult

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    Did we need to state an angle for 8b? Because i got the answer, and I drew a diagram of vector forces to help me get that answer.
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    (Original post by mk_98)
    Also I would lose a mark for writing tension as 11.76 instead of 11.8 or does it not matter?
    I think so, the front of the exam explicity says that when using g you have to round to 2 or 3 s.f. But it is only 1 accuarcy mark
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    (Original post by jacobfrosst)
    how many marks was the moments question worth?
    7 marks I think.

    (Original post by KloppOClock)
    maybe, 45 degrees below horizontal could be south west or south east, if you drew a diagram with an arrow then you would get it i think
    This was how I did it, said it was 45 degrees below the horizontal, but had drawn a diagram for part (b) with the resultant marked, R, pointing south-westerly. Still, I'd think it'd only be 1 mark out of the 4 for that part.
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    Worked iut the right time 40s for velocity questio but didnt have time to find t.....how many method marks would i get you think?
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    What about not multiplying by 0.4 for impulse?

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    (Original post by themathgeek)
    what about not multiplying by 0.4 for impulse?

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    -i = 0.4(-4-3.5)
    -i = 0.4(-7)
    -i = -3

    3=i
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    For the coefficient of friction question I think I may have got 0.72 rather than 0.73. That would only be one mark lost right?Also I drew my speed-time graph in the blank space between the lines but to the right of the question. The examiner should see that right...
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    does anyone remember how many marks each question was worth
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    (Original post by suhaylpatel786)
    These are my answers:

    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

    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)

    3(a) Acceleration = -g/8
    velocity after rebound = 3.5
    thus Impulse = 3 Ns

    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

    5) μ = 0.73

    6) 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

    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1

    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

    For question 7 i used simultaneous equations but only managed to find the magnititude of F2, how many marks would i lose for that?Also for question 3 b I also wrote 15.5N initially then crossed it out and used R - 0.5g = 0.5a If i can recall the question asked 'the force exerted on the scale pan by the brick' in other questions they always asked force exerted on the brick by the scale pan and used the mass of the brick, i thought it was the other way around because of this..How many marks would i lose if both of my answers are wrong? I hope I'd still get some from working out etcAlso I said acting downwards for the final question, it didn't ask for direction as a bearing so would my answer be wrong? :/
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    For 7B I left my answer as the velocity rather than working out the speed - how many marks do you think I'd drop for this?

    Pretty sure Q7A was 7 marks, Q7B was worth 4 & Q3 was 7 marks but I might be wrong!
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    For question 7 i used simultaneous equations but only managed to find the magnititude of F2, how many marks would i lose for that?Also for question 3 b I also wrote 15.5N initially then crossed it out and used R - 0.5g = 0.5a If i can recall the question asked 'the force exerted on the scale pan by the brick' in other questions they always asked force exerted on the brick by the scale pan and used the mass of the brick, i thought it was the other way around because of this..How many marks would i lose if both of my answers are wrong? I hope I'd still get some from working out etcAlso I said acting downwards for the final question, it didn't ask for direction as a bearing so would my answer be wrong? :/
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    I said its acts southwest would i get it?
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    My teacher reckons it will be
    A=60
    A*=66
    70=100ums
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    Can u send me model answers now please
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    Isn't the last Q:

    resultant force = 2Tcos45? So around 8. something
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    (Original post by suhaylpatel786)
    These are my answers:

    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

    2(a) T = 20.6N
    2(b) 15.45 N => 15.5 N (3sf)

    3(a) Acceleration = -g/8
    velocity after rebound = 3.5
    thus Impulse = 3 Ns

    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

    5) μ = 0.73

    6) 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

    7(a) F2 = 2.5i + 2.5j (this required simultaneous equations)
    (b) V = 12i +5j, thus speed = 13 ms-1

    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
    hey, can you just remind me what question 2 was? i cant remember....
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    For 7 a i didnt use simultaneous equations but got the right answer somehow. I was playing around with the unit vectors and i remember there being a 2j involved.
 
 
 
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