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    how the **** did i get k=24 and v=14 :/
    what do you think 65/72 will be UMS?
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    (Original post by jt663)
    I got different answers for question 7 iirc. Think T was 1.2 for me? (Instead of 0.525)
    same, and 1.12 for the final answer for me
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    Ideas as to what 66/72 will be in UMS this year?
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    (Original post by newy181818)
    something like 9.74N and 35.6 degrees
    Yes - same!
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    iit must be 13.3 for the resultant force
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    (Original post by aliah4666)
    Why would the acceleration = 0 for questions 5iii and 5iv??
    For iv, there was no sideways movement, and the force up (1.2) was less than gravity (0.4g)

    For iii, there was sideways movement, but it was less than friction, so no movement.
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    Welcome Squad
    (Original post by xsanda)
    Xsanda's solutions:

    Falling question
    1i) v=28 [2]
    1ii) s=6.384 [2]
    1iii) t=0.598 [3]

    Collision question
    2i) u=6 [4]
    2ii) m=0.2 [4]

    Cycling question
    3i) 20km, 5hrs [3]
    3ii) 7 hrs [3]
    3iii) 50km [2]

    110° tick
    4i) 9.74N, 35.4° [6]
    4ii) 10.3N [1]
    4iii) 54.6° [2]

    θ force
    5i) 1.04 [3]
    5ii) 1.06 [3]
    5iii) a=0 [3]
    5iv) a=0 [2]

    Coalescing
    6i) v=68 [4]
    6ii) s=78 [4]
    6iii) k=132 [1]
    6iv) v=7 [5]

    Prism
    7i) a=0.7 (show that), T=1.68 [5]
    7ii) v=1.26 [2]
    7iii) -a = 0.9 [2]
    7iva) T=1.2 [3]
    7ivb) Fr = 1.12 [3]
    For q5. The force was applied so those answers are wrong... that's what I think anyway.

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    7)iv)b 0.3gsin30 -fr = 0.3*-0.9

    fr = 1.47 + 0.27 = 1.74?
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    (Original post by lolkik)
    iit must be 13.3 for the resultant force
    13.3 was the wrong diagonal: the diagonal from the ends of both of the vectors, rather than the diagonal from the start to the end tip-to-tail. You needed to move one vector to the end of the other.
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    Was q5 pulling?
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    T=7 hours
    K=132

    Is deffo correct
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    (Original post by seizetoday)
    T=7 hours
    K=132

    Is deffo correct
    how did you get k to be 132? so i can see where i went wrong
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    (Original post by wat a wizard)
    7)iv)b 0.3gsin30 -fr = 0.3*-0.9

    fr = 1.47 + 0.27 = 1.74?
    You've missed out the tension I think
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    What the **** was that paper? Questions 1-3 and 6-7 were fine but 4 and 5 were abysmal! I will be lucky to have picked up 5 of those 20 marks. Urgh.

    I think 56/72 will be an A. Question 3 was a bit hard, but I got 7 hours BY GUESSING.

    I guess overall the difficulty balances to medium, as all questions were easy apart from 4 and 5.
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    Predicted grade boundaries?
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    (Original post by ChoccyPhilly)
    A lot of m friends are saying that the 1.2 force was pushing down onto the block. I thought it was pulling
    This was the strangest question in terms of wording. 'A 1.2N force is applied to P' there was never any indication for pushing or pulling so it required an assumption to be made. I assumed push, but loads assumed pull. Some are saying that it can't have been push because the angle would have to be below the horizontal. But that's a load of rubbish, since when was that a law? However, I know I got the last two parts wrong because I did a silly mistake, which I'm kicking myself for now. Just hoping that's all I got wrong, which I think it is, in which case this stupid question doesn't matter to me, I can afford to lose it.
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    i second that motion...i got k as 96 i think. how did you get 132?
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    (Original post by Sid1234)
    how did you get k to be 132? so i can see where i went wrong
    They gave you are formula which was displacement of Q= k-2t^3 I think
    Cant remember exactly, but you plug in t=3
    So you get k-54 but you equal this to what you get in your previous answer which was the displacement OP (OP=OQ at t=3)

    So K-54=78
    K=132

    Hope this helps.
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    (Original post by Sid1234)
    how the **** did i get k=24 and v=14 :/
    what do you think 65/72 will be UMS?
    i also got 14
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    (Original post by seizetoday)
    They gave you are formula which was displacement of Q= k-2t^3 I think
    Cant remember exactly, but you plug in t=3
    So you get k-54 but you equal this to what you get in your previous answer which was the displacement OP (OP=OQ at t=3)

    So K-54=78
    K=132

    Hope this helps.
    AHHH i did 78-24 FUUUUUUU
 
 
 

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