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    Can someone help me with part b?

    A particle P of mass 3m is moving with speed 2u in a straight line on a smooth horizontal
    plane. The particle P collides directly with a particle Q of mass 4m moving on the plane with
    speed u in the opposite direction to P. The coefficient of restitution between P and Q is e.
    (a) Find the speed of Q immediately after the collision.

    Answer: U/7 (9e+2)


    Given that the direction of motion of P is reversed by the collision,
    (b) find the range of possible values of e.

    Would i make the final speed of P >0 or P<0 and please give me a reason why it is the one you have chosen?
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    For P to reverse, it's velocity must change sign, assuming you took P to be travelling right and right to be positive,
    Put your speed of Q back into the equation for conservation of momentum in order to find some equation in e and u for the velocity of P, make this velocity <0 and solve for e.
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    (Original post by JN17)
    For P to reverse, it's velocity must change sign, assuming you took P to be travelling right and right to be positive,
    Put your speed of Q back into the equation for conservation of momentum in order to find some equation in e and u for the velocity of P, make this velocity <0 and solve for e.
    That really helped me thank you.
 
 
 
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