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    (Original post by bcmh)
    Anyone agree or disagree with these answers?
    Vectors
    1a) 104 degrees
    C) 640i+640j

    Pan on cable
    2a) 20.6N
    B) 15.5N

    Particle into a wall
    3) 1.4Ns

    Speed-time graph
    8.75s

    Moments
    M=42
    D=1.2

    Slope question
    Coefficient = 0.73

    Vectors (again)
    F2= 2.5i+2.5j
    Speed=13

    Connected Particles
    T=11.76
    Resultant force = 16.6 bearing 45 degrees



    Posted from TSR Mobile
    1.a) I got 284... anyone else?
    c) same

    2. same

    3. same
    Slope question - sameI didn't get the other vectors question and I don't remember what i got for the last one but it might have been the same as you.
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    Anyone mind giving their opinion on a quick question...

    For 7b.) With the vector of velocity (3i - 22j) and acceleration of (3i + 9j) i think it was.

    I done it right and got 13, but then i stupidly crossed it out and found the magnitude of the velocity (√493) and magnitude of the acceleration (3√10). And so after 3 seconds of acceleration, i got √493 + 9√10 = 50.7 ms-1.

    Do you think i will get 0 marks? I'm thinking yes...
    Also, does anyone know how many marks 7b was worth?

    Any replies are appreciated
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    (Original post by marshmellow:))
    I can confirm that the coefficient of friction was 0.7271966072
    can you leave g answers to 3sf?

    that paper was easier than past papers
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    (Original post by gunners98)
    No worries. Yeah I agree - although I didn't find it too bad I think a few of the questions were tougher than other years. I reckon
    73 - 100ums
    66 - 90ums
    60 - 80ums
    I disagree. A lot of the questions have been seen before but with different numbers. The moments question was probably the only unique one there.

    I agree with your UMS predictions though.
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    (Original post by Reda2)
    Yes you will.
    cheers
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    (Original post by stephencurtis)
    cheers
    How many marks will I lose if I forgot to mutiple by the mass for the impulse question. I got v--u = 7.5 and wrote the formula flr impukse, just stupidly forgot to multiply by 0.4?

    Posted from TSR Mobile
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    What's 9+10
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    (Original post by DW4)
    1.a) I got 284... anyone else?
    c) same

    2. same

    3. same
    Slope question - sameI didn't get the other vectors question and I don't remember what i got for the last one but it might have been the same as you.
    Sorry to burst your bubble on the first one.
    The vector was 20i - 5j, if you draw it you'll see the bearing would be 90° + arctan(5/20) = 104°.
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    so is the general gist for T 8.7 ish seconds?

    also, how many marks will I lose for drawing the graph without one point of intersection?
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    This paper was harder than last year's and last year paper was 59 for an A so i'm thinking it will be 58 for an A this year
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    (Original post by JLegion)
    I disagree. A lot of the questions have been seen before but with different numbers. The moments question was probably the only unique one there.

    I agree with your UMS predictions though.
    I haven't even nearly done all the papers so I'll take your word for it, but I thought the impulse question was harder than just a standard collision question, the second vectors seemed harder than usual, and like you say the moments question too. But I agree there was nothing ridiculous in this paper.
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    Did we need a bearing for 8b? I only put down the resultant force.
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    (Original post by TercioOfParma)
    Forgot to do the bearing, but got the force.

    2/4 sound like correct number of marks?

    Posted from TSR Mobile
    Same here :/ I'm hoping that it's 3 because the 45 degree angle has been clearly recognised, but unfortunately I think it's most likely going to be 2.
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    (Original post by LukeB98)
    June exams only:
    2010: 75
    2011: 75
    2012: 75
    2013: 68
    2014: 70
    2015: 71

    According to their grade boundaries spreadsheet.
    Thanks. It was a bit similar to June 2013, I'd say this year it'll be around 70
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    For 8b, didn't the question ask for the Force,so why does it have the bearings and how many marks will i lose for working out only the Force
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    I got all the working out for questions as I did the paper again in the allocated time, anyone need to see it? I made the corrections for the ones I realised I did wrong after the exam
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    Does anyone have the answer for the final part of question 1?
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    (Original post by Pradesh)
    For 8b, didn't the question ask for the Force,so why does it have the bearings and how many marks will i lose for working out only the Force
    The question asked for magnitude and direction.
    Since it was 4 marks you'll probably lose 2 marks for not listing the direction.
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    Guys for the moments question, I basically got it all right up to the simultaneous equations, where after that I solved them incorrectly...Lol. Would I still get 5/7 you reckon? So like lose the final two A1 marks?
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    Could someone explain to me why the force exerted by the brick is 15.45N and not just its weight, 14.7N. I understand it's accelerating upwards and so there must be a net upwards force greater than the weight. The difference between the force and the weight is the force required to make it accelerate at 0.5m/s^2, so net upwards =1.5g + 0.5x1.5=15.45. Surely though, this is the force that the pan exerts on the brick and not what the brick exerts on the pan? Or am I correct and have I just messed up when reading the question and it's actually asked me to find the force of the pan on the brick?

    Also for 7. Is F2 = 2.5i+2.5j or -3.5i+-3.5j. The two answers are pending on which way you put the ratio to find your answer ie 3(-1+x)=2+x and -1+x=3(2+x). Which one is right? I put the 2.5 one but I now think -3.5 was right. How many marks would I lose if it's wrong?

    Lastly, for the diagram, I did something stupid. I drew all of it correctly (so correct height/speed, time they decelerated and time stopped) except I didn't make their lines intersect when they were decelerating. Meaning M was above N for the entirety of the graph except for the stop point. Thus meaning they traveled different distances, which we know is wrong. Do I just lose one mark?
 
 
 
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