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    (Original post by chloe-jessica)
    You want the distance from O to 5m above the ground, which will be 5m less than the distance you just calculated, so 17.3-5=12.3m. Let this equal s, they gave you u in the question, you know a=g as it's free fall, and you want v. So using suvat equation v2=u2 + 2as, v2=4.22+2*9.81*12.3. V should come out as 16.1ms-1 if I've done that calculation correctly.
    Ah understood. What about June 09 5i.
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    (Original post by Sonnyjimisgod)
    I suck at m1.....so bad, feel like im gonna get a C. Should I do an all nighter, considering I have six exams this week
    Nope. I don't suggest an all nighter! Try as much as you can till about 11pm. Then sleep.
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    (Original post by giantbecky)
    mass R has started to descend so P won't be on the ground anymore
    so no reaction force...remove it and that's correct!

    Good luck to you too!
    Thank you so much, I can't believe I couldn't grasp something so simple! :rolleyes:

    I will give you rep!
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    (Original post by Super199)
    Ah understood. What about June 09 5i.
    Hi.

    Were given the direction of motion and mass of the particles. We can use the momemtum equation.

    So we take going to the right as positive.

    (m x 0) + (0.5 x -6) = (m x -(v+1)) + (0.5 x -v)

    ==>

    -3 = -mv - m - 0.5v

    move the mv and 0.5v over to the left and the -3 to the right.

    mv + 0.5v = 3-m

    factorise the v.

    v(m+0.5) = -m + 3

    Hope you got that!
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    (Original post by Jon1234321)
    Thank you so much, I can't believe I couldn't grasp something so simple! :rolleyes:

    I will give you rep!
    Don't worry! I find pulley questions the hardest!

    Thankyou!
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    Anyone have a copy of the 2014 paper?
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    (Original post by MsFahima)
    Hi.

    Were given the direction of motion and mass of the particles. We can use the momemtum equation.

    So we take going to the right as positive.

    (m x 0) + (0.5 x -6) = (m x -(v+1)) + (0.5 x -v)

    ==>

    -3 = -mv - m - 0.5v

    move the mv and 0.5v over to the left and the -3 to the right.

    mv + 0.5v = 3-m

    factorise the v.

    v(m+0.5) = -m + 3

    Hope you got that!
    Gosh I didn't read the bit about Q being faster than p by 1ms^-1. Thanks for your help

    How would you answer January 2010 Q2?
    The first one I was thinking cosine rule. Not sure about the second bit?
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    (Original post by Issy :))
    Anyone have a copy of the 2014 paper?
    http://www.dotmaths.com/ocr/
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    (Original post by Super199)
    Gosh I didn't read the bit about Q being faster than p by 1ms^-1. Thanks for your help

    How would you answer January 2010 Q2?
    The first one I was thinking cosine rule. Not sure about the second bit?
    Draw a closed triangle of the 12N and 19N with resultant. Then use sine rule

    sinX/19 = sin120/27.07

    27.07 was from previous part of question, you do use cosine there
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    (Original post by Lilli1997)
    Draw a closed triangle of the 12N and 19N with resultant. Then use sine rule

    sinX/19 = sin120/27.07

    27.07 was from previous part of question, you do use cosine there
    How come it is 120 and not 60? Do you mind sending a sketch by any chance?
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    (Original post by Super199)
    Gosh I didn't read the bit about Q being faster than p by 1ms^-1. Thanks for your help

    How would you answer January 2010 Q2?
    The first one I was thinking cosine rule. Not sure about the second bit?
    Well I had to look at the mark scheme for that. You have to have two components of the 19N. One in the horizontal and one in the vertical. The question is so badly worded.

    So for the first part

    R^2 = sqrt [(12 + 19cos60)^2 + (19sin60)^2 ]

    and you get R as 27.1 N

    Part b

    You have a triangle. The adjacent is (12 + 19cos60) and the opposite is (19sin60.)

    So you do tan theta = opp/adj
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    (Original post by Super199)
    How come it is 120 and not 60? Do you mind sending a sketch by any chance?
    Hope this makes sense Name:  image.jpg
Views: 74
Size:  497.0 KB

    angle on first one should be 60, can show you how I used cosine rule for first one if you want?
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    (Original post by Super199)
    Ah understood. What about June 09 5i.
    Sorry, I was concentrating on my economics essay. I think someone already answered this, but I'm doing mechanics for a bit now if you need help with anything else, let me know
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    (Original post by Lilli1997)
    Hope this makes sense Name:  image.jpg
Views: 74
Size:  497.0 KB

    angle on first one should be 60, can show you how I used cosine rule for first one if you want?
    Yeah please
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    Can someone explain how you work out change in momentum

    For 3ii. Jan 2010.

    Thanks
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    (Original post by Lilli1997)
    Hope this makes sense Name:  image.jpg
Views: 74
Size:  497.0 KB

    angle on first one should be 60, can show you how I used cosine rule for first one if you want?
    Yes please
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    (Original post by Super199)
    Can someone explain how you work out change in momentum

    For 3ii. Jan 2010.

    Thanks
    Change in momentum = momentum before - momentum after = 9*0.096 - -3.5*0.096 = 1.2kgms-1. Remember to account for direction, as the particle changes its direction of motion, it changes to a negative momentum.
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    (Original post by chloe-jessica)
    Change in momentum = momentum before - momentum after = 9*0.096 - -3.5*0.096 = 1.2kgms-1. Remember to account for direction, as the particle changes its direction of motion, it changes to a negative momentum.
    I don't understand question 4a on the same paper

    I just got 0.4gsin60 and 0.4gcos60. Where does the other 0.3gsin60 and 0.3gcos60 come from? I thought it was only considered particle p.
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    Lets all hope theres no tractive force questions and we should all be ok tomorrow lmao (hopefully)
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    (Original post by Super199)
    Yes please
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