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    I swear for question to the weight was in kg and the acceleration was in ms-2 therefore the answer was R=15450
    as you have to convert the masses into grams
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    (Original post by CaiusMartius)
    Taken from an almost identical question from June 2013.

    "Direction clearly marked on a diagram, with an arrow, and 45o (oe)marked"
    I knew I remembered a similar question, but I couldn't remember what form they wanted. Did that mark scheme accept other forms?

    I ended up writing four answers for the direction (or at least, the same answer in four different ways). I had a diagram of the pulley with the string's tension acting on it, an arrow for the resultant force (labelled "resultant force") and North marked on it (directly upwards). I then wrote four answers: "bearing of 225 degrees", "due southwest", drew an arrow in the right direction and "with direction vector -i-j". There is no way they can reasonably not give me the mark.
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    How did everyone do F2? Can someone show the working here
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    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
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    (Original post by Jellymath)
    Fairly easy test this time around. Here are my answers which should be correct:

    1. 104 degrees, 640i+640j
    2. T=20.6N , R=15.5N
    3. Impulse = 3Ns
    4. Graph with one point of intersection, T = 8.75s
    5. Moments qn: d=1.2m M=42kg
    6. Coefficient of friction = 0.727
    7. 2.5i+2.5j, 13 ms-1
    8. T=11.8N, F=16.6N bearing 225 degrees
    Yeah I got exactly the same, was kindof unsure about 2b but pretty sure its right.


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    Anyone know how many marks the moments question was?
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    (Original post by Jellymath)
    Fairly easy test this time around. Here are my answers which should be correct:

    1. 104 degrees, 640i+640j
    2. T=20.6N , R=15.5N
    3. Impulse = 3Ns
    4. Graph with one point of intersection, T = 8.75s
    5. Moments qn: d=1.2m M=42kg
    6. Coefficient of friction = 0.727
    7. 2.5i+2.5j, 13 ms-1
    8. T=11.8N, F=16.6N bearing 225 degrees
    Cheers matey should be it. Just the R part of 2 I'm skeptical. The question asked the Force of the Brick on the Pan. So surely that would be the weight of the Brick, not the Reaction?

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    (Original post by Reda2)
    And guys I have a very important question this is what I will either get me a nearly full UMS or complete fail. I did the graph IN PENCIL and I forgot I was using pencil and just carried on with the paper! Realising it on the last question, this won't lose me marks right??? I even wrote on the pages I used pencil "Sorry I forgot I was using pencil" !! anyone please??
    I remember watching a documentary video on YouTube on Edexcel marking. They have a specific marker who marks papers that were done in coloured ink, used highlighters or who's papers won't scan properly. So basically if it isn't visible when scanned then a person will mark the physical paper (not on the computer) so you should be ok.
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    How many marks was the impulse question worth?
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    (Original post by dj3k)
    I remember watching a documentary video on YouTube on Edexcel marking. They have a specific marker who marks papers that were done in coloured ink, used highlighters or who's papers won't scan properly. So basically if it isn't visible when scanned then a person will mark the physical paper (not on the computer) so you should be ok.
    Thank you for this beautiful answer, I can sleep well tonight then, but omg I was so worried when I found out.
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    Would I get a mark for writing force acts at 45 degrees below horizontal for last question?
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    (Original post by Formless)
    ^^^ This is 100% correct, Any idea the marks for the impulse questions
    It was worth 7 marks I think. You'll probably get 1 for frictional force after rebound, 2 for working out acceleration with f=ma, 1 for working out speed that it rebounded (suvat), and 2 for the method of calculating the impulse, and 1 for the answer. Not sure though
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    If you got 0.2 on the Impulse question, like me, then we were wrong.

    0.2 is the answer you get when you calculate the Impulse but forget to make the velocity (4) negative.

    The mass of the particle was 0.4g and the velocity AFTER it hit the wall that you worked out was 7/2 (which should get method marks if you got that answer)

    So then the Impulse was

    I = 0.4(-3.5) - 0.4(4) = -3Ns
    NOT
    I= 0.4(3.5) - 0.4(4) = -0.2Ns

    We will get some method marks though.
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    (Original post by eloised123)
    wheres the mark scheme !!!!!!!
    would u really think itll come out this quickly??
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    (Original post by hello11223344)
    I swear for question to the weight was in kg and the acceleration was in ms-2 therefore the answer was R=15450
    as you have to convert the masses into grams
    Why would you convert mass to grams?
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    (Original post by carrm123)
    How many marks was the impulse question worth?
    Impulse was 7 marks I think.
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    (Original post by Jellymath)
    Fairly easy test this time around. Here are my answers which should be correct:

    1. 104 degrees, 640i+640j
    2. T=20.6N , R=15.5N
    3. Impulse = 3Ns
    4. Graph with one point of intersection, T = 8.75s
    5. Moments qn: d=1.2m M=42kg
    6. Coefficient of friction = 0.727
    7. 2.5i+2.5j, 13 ms-1
    8. T=11.8N, F=16.6N bearing 225 degrees
    A very nice concise round up of answers. Now I don't need to worry Thanks
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    Can someone explain why I got 1.7N something for F2.

    I am 100% my trigonometry was right so I used the sine rule to find F2 and I got 1.7 on one side. What did I have to do with this 1.7?????
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    Can anyone set a poll for grade boundaries? Will be good to get an idea of how people found it

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    (Original post by Reda2)
    And guys I have a very important question this is what I will either get me a nearly full UMS or complete fail. I did the graph IN PENCIL and I forgot I was using pencil and just carried on with the paper! Realising it on the last question, this won't lose me marks right??? I even wrote on the pages I used pencil "Sorry I forgot I was using pencil" !! anyone please??
    On the front of the paper, it states that you can use pencil for diagrams, so you're all good.
 
 
 
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