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aqa mechanics 2b june 2010 q9 Watch

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    can i have some help please?

    A particle, of mass 8kg, is attached to one end of a length of elastic string. The particle is placed on a smooth horizontal surface. The other end of the elastic string is attached to a point O fixed on the horizontal surface.

    The elastic string had natural length 1.2m and modulus of elasticity 192N.

    The particle is set in motion on the horizontal surface so that it moves in a circle, centre O, with a constant speed 3m/s.

    Find the radius of the circle.


    ive used t=(mv^2)/r where t is the centripetal force
    i thought that this might be equal to the string when its moving so i set it equal to ?x/l
    ?=192, l=1.2 x(the extension) = r-l?

    (mv^2)/r = ?x/l
    8x9/r = 192(r-1.2)/1.2
    0.45=r(r-1.2)
    r^2 -1.2r -0.45=0

    i've ended up with a quadratic which i solved and selected the positive solution
    my final answer for the radius is 1.5m can anyone confirm?
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    (Original post by hey_its_nay)
    my final answer for the radius is 1.5m can anyone confirm?
    Not check the details of your working, but I agree with that answer.
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    I think you can also check it with the mark scheme of the paper.
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    (Original post by Rishabh95)
    I think you can also check it with the mark scheme of the paper.
    its the june 2010 paper, the mark scheme hasnt been released
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    ohh, ok.
 
 
 
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