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    A particle A of mass 4m moves with speed \frac{3}{2}u collides directly with a particle B of mass m which is at rest on a smooth horizontal table. The coefficient of restitution between the particles is e. Given that there are no further collisions, find the range of possible values for e.
    Using conservation of momentum:

     4.\frac{3}{2} + 0 = 4v_A + v_B
     6 = 4v_A + v_B

    Coefficient of restitution:

     \frac{2(v_B - v_A)}{3} = e
     2v_B - 3e = 2v_A

     6 = 2(2v_B - 3e) + v_B


    Where do I go from here...?
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    (Original post by Patent Pending)
    Using conservation of momentum:

     4.\frac{3}{2} + 0 = 4v_A + v_B
     6 = 4v_A + v_B
    should be  6 <b>u</b>= 4v_A + v_B
    Coefficient of restitution:

     \frac{2(v_B - v_A)}{3} = e
     2v_B - 3e = 2v_A

     6 = 2(2v_B - 3e) + v_B
    should be 6u=...
    should be /3u
    Where do I go from here...?
    find equations for va and vb in terms of u and solve vb>va
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    Is it necessary to retain the 'u's (or the 'm's)? Normally I find that these can be omitted..
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    it will be very hard to get equations of va and vb in terms of u if you omitt u in the only two equations you have lol
 
 
 
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