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M2 question watch

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    A light inextensible string of length 4a has a particle A, of mass m, attached at one end and a particle B, of mass m, attached at the other end. The string passes through a small smooth ring which is fixed at a point O at a distance 3a above a horizontal table. The system is hanging in equilibrium with OB = 2a when a smooth bead of mass 2m, which is threaded on the string between O and B, is released from rest at O. The bead falls under gravity until it collides with and adheres to the particle B to form a composite particle C.
    Given that the string remains taut, show that the speed of C immediately after the collision is sqrt(ga).


    I can solve this, but I'm not really content with my reasoning. I did it in two ways, once considering energy and once using the standard equations of motion, but I think I'm doing something wrong... So I wanted to see how you guys would approach it.
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    hmm, is this an Edexcel question?
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    Yup. It's from the 2nd review exercise.
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    (Original post by shift3)
    Yup. It's from the 2nd review exercise.
    question no? .... ah q20 working on it
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    hehe i cant do it! Noooo... better revise that chapter... got about 2.5 hours before my exams!! so can you post wht u got so far?
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    Before C is released:
    PE=mgh=(2m)g(3a)=6mga
    KE=0

    When C joins B:
    PE=mgh=(m+m+2m)ga=4mga
    KE=(1/2)mv²=(1/2)(m+m+2m)v²=2mv²

    so....
    2mv²=2mga
    v²=sqrt(ga)
 
 
 
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Updated: January 21, 2005

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