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    I'm on the Heinemann book, page 80, Qu 7 (b) How is the answer of 1.5 metres reached, please?
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    Not everyone has that book, could you post the question?
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    (Original post by Snagglepuss)
    I'm on the Heinemann book, page 80, Qu 7 (b) How is the answer of 1.5 metres reached, please?
    im doing it... lets c ...
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    (Original post by Snagglepuss)
    I'm on the Heinemann book, page 80, Qu 7 (b) How is the answer of 1.5 metres reached, please?
    what did u get... is ur answer 0.5 ?
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    No, the total distance moved by A is 1.5 metres. (But your 0.5 may be the figure on top of the metre already travelled, of course). But how did you work out the 0.5?
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    Thanks PM, bit it requires a diagram, and I don't know how to enter one here.

    How are you getting on, d999? I have just tried a mehtod which gives a total movement of A from its initial positon of 1 plus 1.5 = 2.5 metres.
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    Can anyone help with this, please?
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    sorry...didnt c ur reply... quote me ...
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    okay so first of all you have to calculate the speed when the string breaks... using this formula: V^2=u^2+2as

    2.
    my reference point where mgh= 0 is the point where the string breaks the so at that point it has K.E. only. At the point where the particle comes to rest it has P.E. only
    i assume u know how to calculate the K.E.
    For the P.E you've got the value of sin theta which is 0.6 .... as you already know that it’s the opp/hyp ... in this case the opp is h and the hyp is d (the distance travelled)
    so u write the P.E in terms of d to solve it... m x g x 0.6 x d (0.6= h/d ... h=0.6 x d)

    3.
    So the K.E. is equal to the P.E. + the work done against friction
    1/2mv^2 = mg0.6d + Fr x d
    1/2m2.8^2 = (1.96m)d + mg0.6d
    3.92m = (1.96m+5.88m)d
    d is about 0.51 m
    and then u just add 1m ... i hope that’s correct because im doing M2 on my own...so somebody better have a look at it
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    I agree with your working, and I get the same result using a slightly different method:

    KE at point of string snap is 3.92m Joules. (We don't know the value of m.)

    Resistance to the line of motion is 7.84m Newtons.

    Work = force x distance, so the KE of 3.92 Joules will be exhausted after the particle has travelled a distance s against a force of 7.84m.

    3.92m = 7.84ms. Therefore s = 3.92/7.84 = 0.5.

    So we agree, and we get the correct answer when we add the 0.5 to the metre already covered.

    BUT:

    We are asked to use the principle of conservation of energy, and both of us have used SUVAT formulae as well. I strongly suspect there is a simpler solution which does not involve SUVAT.

    By the way, I'm doing M2 on my own, too, so we need to stick together, maybe!
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    Has anyone any further thoughts on this, please? Is there a way of doing this which does not involve SUVAT formulae?
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    (Original post by Snagglepuss)
    Has anyone any further thoughts on this, please? Is there a way of doing this which does not involve SUVAT formulae?
    hmm... we didnt use the suvat formula ... did we? ... we just used the K.E and P.E at particular points ... and solved part b with energy cons...
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    Hate to disagree with you, old fellow, but we DO use SUVAT. As you mention yourself, we use V^2=u^2+2as to calculate the speed of particle m, and its KE, when the string breaks. Many thanks for your help, but I suspect there is a cleaner, simpler method than both of our efforts.
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    Can anyone see a neater way of doing this, please, that does not require the use of SUVAT in the first place - i.e. to get the speed of particle A when the string breaks? Using cons of energy ONLY to get the result.
 
 
 
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