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    (Original post by GabbytheGreek_48)
    think i also got 1.12
    I definitely got 1.12 as well

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    (Original post by wil_is_he)
    I can do some if it... Hopeully others wilk fill in after.

    2 helicopters A and B,
    A is at origin, B is at position (maybe 50i 40j? Someone needs to help)
    A has acceleration (-0.2i + 0.1j), B: (0.2i - 0.1j)
    Cant remember inial velocity...
    Q: What is the distance A to B when their velocities are parallel. Answer=106. If others could fill in gaps thatll help. You can either do c4 method or gradient method, t=12
    Everything you put is right. Initial velocity of B was (6i + 9j). Cant remember A
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    (Original post by Peter Wilkinson)
    Did no one get around 07.6 for the bearing ? :|
    Bearing was 008 so ye think so
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    (Original post by LiesandAlibis)
    R = Tsin(6) +49
    How
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    (Original post by OturuDansay)
    for question 7, was the horizontal velocity 12cos50 or something, I can't remember what the initial figures were when setting up the question. Someone help lol
    horizontal a=0
    u=12cos(50)
    t=1.76(from previous answer)
    s= what your finding
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    Welcome Squad
    (Original post by jdtjCdiifn)
    Everything you put is right. Initial velocity of B was (6i + 9j). Cant remember A
    A was 4i and can't remember the j
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    (Original post by gcsekid)
    Attachment 554175
    Ignore the pencil line
    How are people getting 43 newtons for T
    R=49+Tsin60 not R+Tsin60=49 I think

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    (Original post by gcsekid)
    How
    The vertical component of T acts downwards along with weight.
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    (Original post by PiTheta97)
    I definitely got 1.12 as well

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    uh i must of typed in something wrong, i put v=0
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    (Original post by gcsekid)
    How
    think he means 60 instead of 6 and 49 = 5g (5*9.8)
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    (Original post by koolgurl14)
    10 more marks gone
    Sorry.
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    (Original post by LiesandAlibis)
    The vertical component of T acts downwards along with weight.
    What would I lose for that then how many marks for the minus sogn
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    (Original post by GabbytheGreek_48)
    horizontal a=0
    u=12cos(50)
    t=1.76(from previous answer)
    s= what your finding
    Could you tell me how you got 1.76 because I got like 1.18 seconds or somit and just wanna see ur method if that's alright
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    (Original post by Ano123)
    Sorry.
    think the grade boundaries will probably b low because thw whole 10 marks for one question and easy to lose marks some places
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    (Original post by matt5822)
    Full as long as you got the distance to be 106 m.
    i found the distance of each and found the difference between distances rather than finding difference between the displacements and then finding distance
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    (Original post by OturuDansay)
    for question 7, was the horizontal velocity 12cos50 or something, I can't remember what the initial figures were when setting up the question. Someone help lol
    The horizintal velocity to the start of the box was 12cos50 yes
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    its not R+tsine60=5g. its R=5g-Tsin60. the vertical component of tension acts against the reaction therefore its pushing the block towards the ground, not lifting it up as you have suggested.
    (Original post by gcsekid)
    Attachment 554175
    Ignore the pencil line
    How are people getting 43 newtons for T
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    (Original post by GabbytheGreek_48)
    think he means 60 instead of 6 and 49 = 5g (5*9.8)
    Yes, that was a typo!
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    (Original post by OturuDansay)
    Could you tell me how you got 1.76 because I got like 1.18 seconds or somit and just wanna see ur method if that's alright
    idr but i think i resolved vertically so vertical was
    s=1
    u=12sin(50)
    a=-9.8
    t=what your finding
    you put those in and eventually get a quadratic do quadratic formula and then i used the higher value
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    (Original post by postexamtalk)
    i found the distance of each and found the difference between distances rather than finding difference between the displacements and then finding distance
    You should still get most of the marks. I think the hardest part was finding the value of t.
 
 
 
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