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    How is it that I've done all the past year papers but I still feel ill prepared I'm still making careless mistakes and such, it's frustrating.
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    I've done no revision at all yet lol , gonna do a few hours of revision and past papers now
    aiming for 100 UMS but ill end up with like 95 due to stupid mistakes as usual
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    Can someone explain Q7 to me, please?

    https://drive.google.com/file/d/0B1Z...M2V1FZTnM/view
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    Name:  3851e6e6c2ccdbb02b472a5430ede9c5.png
Views: 125
Size:  87.1 KBdidnt get chance before so thanks Zacken for helping me last time . anyways for this question (June 2014 R) for part b) i got that s = 0.3m but the mark scheme says you need to do 0.3 x 2 to get the answer and i don't understand why?
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    (Original post by KloppOClock)
    Thrust is opposite to tension.

    Tension:

    O---->-------<-----O

    Thrust:

    O<---->O
    (Original post by wenogk)
    Thank you!


    This question uses "thrust" check out examsolutions video for that question(Q6)
    http://www.examsolutions.net/a-level...e/paper.php#Q6
    Ahh right thanks :yy:

    Posted from TSR Mobile
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    Just did the specimen paper, what are the boundaries?
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    (Original post by Thomith)
    Name:  3851e6e6c2ccdbb02b472a5430ede9c5.png
Views: 125
Size:  87.1 KBdidnt get chance before so thanks Zacken for helping me last time . anyways for this question (June 2014 R) for part b) i got that s = 0.3m but the mark scheme says you need to do 0.3 x 2 to get the answer and i don't understand why?
    The particle returns to the point where the tension is lost. So it will travel back the same distance hence x 2.
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    (Original post by AlphaArgonian)
    Can someone explain Q7 to me, please?

    https://drive.google.com/file/d/0B1Z...M2V1FZTnM/view
    Arsey's way is superior imo, but hope this helps. Name:  For some guy.jpeg
Views: 117
Size:  232.5 KB
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    (Original post by AlphaArgonian)
    Can someone explain Q7 to me, please?

    https://drive.google.com/file/d/0B1Z...M2V1FZTnM/view
    I used (v-u)/t = acceleration
    So : 2i+9j – (-3i-3j) divided by t
    = (5i + 12j)/t

    Gives us the magnitude therefore:
    (5/t)^2 + (12/t)^2 = 2.6^2
    = 25/(t^2) + 144/(t^2) = 6.76

    Multiply all my t^2 to get 169=6.76t^2
    SoT= 5

    Sub this into the first part (5i +12j)/5 = 1 and 2.4 
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    (Original post by Marxist)
    Arsey's way is superior imo, but hope this helps. Name:  For some guy.jpeg
Views: 117
Size:  232.5 KB
    Got there before me

    My method is a tad different tho, shocked I got it right tbh haha.
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    (Original post by Louiseelg0rt)
    When a vector is A relative to B, is that the same as the vector BA? or is it the vector AB?
    A relative to B is BA
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    Anyone found any *really* hard questions that come up?

    And I just can't get my head around "what does inextensible show/how have you used that in your calculation" kinda questions.
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    (Original post by thelegend99)
    Anyone found any *really* hard questions that come up?

    And I just can't get my head around "what does inextensible show/how have you used that in your calculation" kinda questions.
    inextinsible string means that acceleration of both particles attached will be the same
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    (Original post by thelegend99)
    Anyone found any *really* hard questions that come up?

    And I just can't get my head around "what does inextensible show/how have you used that in your calculation" kinda questions.
    This is because as soon as one particle moves, the other one moves with it since the string can't stretch!
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    For vectors, does it matter whether we write them in i + j form, or the other form with the brackets (forgot their name!)?
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    How do we use the information that the string is light?

    Also, for questions where the string breaks, that would mean that the tension in the string is now 0 and so we would have to find the new acceleration for the particle, right?
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    (Original post by thesuperdark)
    For vectors, does it matter whether we write them in i + j form, or the other form with the brackets (forgot their name!)?
    doesn't matter i think, but i usually leave them as i and j as the final answer
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    (Original post by amelienine)
    How do we use the information that the string is light?

    Also, for questions where the string breaks, that would mean that the tension in the string is now 0 and so we would have to find the new acceleration for the particle, right?
    i think that means the tension and acceleration are the same
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    (Original post by Don Pedro K.)
    This is because as soon as one particle moves, the other one moves with it since the string can't stretch!
    There u go its the ultimate HARD GOLD TIER M1 !
    I hope u like this for practice!
    Attached Images
  1. File Type: pdf 15 Gold 5 - M1 Edexcel.pdf (198.2 KB, 67 views)
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    (Original post by KloppOClock)
    i think that means the tension and acceleration are the same
    Pretty sure the fact that tension is the same is given by "string passes over a smooth pulley" from what I remember
 
 
 
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