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    On the January 2006 OCR M1 Paper i don't quite understand question 6iii, please could someone explain?

    http://blogs.thegrangeschool.net/mat...oklet-2010.pdf
    The paper is on page 7 of the link above


    https://211c25b87002c8b34761c926d793...20M1%20OCR.pdf
    -Mark scheme
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    Your first link doesn't work for me however the URL says 2010 not 2006.
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    (Original post by Vikingninja)
    Your first link doesn't work for me however the URL says 2010 not 2006.
    It also says it's a "past paper booklet" so presumably means a past paper booklet till 2010?
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    Since P is zero, there is no force pulling the ring to the side, so it hangs below A, and because of the length of the string, it will be below B.
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    (Original post by Zacken)
    It also says it's a "past paper booklet" so presumably means a past paper booklet till 2010?
    Oh right didn't realise that. Never seen a booklet of several in one and content didn't work for me.
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    (Original post by ghostwalker)
    Since P is zero, there is no force pulling the ring to the side, so it hangs below A, and because of the length of the string, it will be below B.
    What I think is AR is only 0.4m so R can't be below B as AB is 0.5m so R will be 0.4m below A. Also I don't understand why the tension is 1N because the weight of R is 2N so since BR is slack, for it to hang in equilibrium Tension must be 2N instead of 1N to balance the weight of 2N?
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    (Original post by runny4)
    What I think is AR is only 0.4m so R can't be below B as AB is 0.5m so R will be 0.4m below A. Also I don't understand why the tension is 1N because the weight of R is 2N so since BR is slack, for it to hang in equilibrium Tension must be 2N instead of 1N to balance the weight of 2N?
    R is a smooth ring. It is not fixed to the string, but is free to slide. Does it make sense now?
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    (Original post by ghostwalker)
    R is a smooth ring. It is not fixed to the string, but is free to slide. Does it make sense now?
    I get the thing about it being able to slide know but I don't understand why the tension upwards isn't 2N to balance the weight of R which is 2N.
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    (Original post by runny4)
    I get the thing about it being able to slide know but I don't understand why the tension upwards isn't 2N to balance the weight of R which is 2N.
    Since it's free to slide and the length of the string is greater than the distance AB, then it hangs below B, and is held by two parts of the string - one either side of the ring.

    Tension in each part is 1N, giving a total upward force of 2N.

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    (Original post by ghostwalker)
    Since it's free to slide and the length of the string is greater than the distance AB, then it hangs below B, and is held by two parts of the string - one either side of the ring.

    Tension in each part is 1N, giving a total upward force of 2N.

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    ok thank you
 
 
 
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