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A-level Physics conservation of momentum question help Watch

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    The question is divided in two parts, and I managed to work out part ii) as it was fairly straight forward. The question is:

    A particle A, of mass m, moving with an initial velocity u makes a head on collision with particle B of mass 2m, being initially at rest. In terms of u calculate the final velocity of A if the collision is i) elastic ii) inelastic. Assume that in the inelastic collision the two particles adhere.

    I am not quite sure of how to go on to solve part i) as we don't have any info regarding the final KEs or partcle B's final velocity. Am I thinking too much into it and missing something very obvious?

    PS: I have tried to solve it by simultaneous equations by calling x and y, A and B's respective final velocities into the equations for conservation of energy and momentum but ended up with the wrong answers.

    Thanks in advance for any help you may be able to offer.
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    In an elastic collision kinetic energy is conserved, in an inelastic collision it isn't.
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    (Original post by Joules911)
    The question is divided in two parts, and I managed to work out part ii) as it was fairly straight forward. The question is:

    A particle A, of mass m, moving with an initial velocity u makes a head on collision with particle B of mass 2m, being initially at rest. In terms of u calculate the final velocity of A if the collision is i) elastic ii) inelastic. Assume that in the inelastic collision the two particles adhere.

    I am not quite sure of how to go on to solve part i) as we don't have any info regarding the final KEs or partcle B's final velocity. Am I thinking too much into it and missing something very obvious?

    PS: I have tried to solve it by simultaneous equations by calling x and y, A and B's respective final velocities into the equations for conservation of energy and momentum but ended up with the wrong answers.

    Thanks in advance for any help you may be able to offer.
    In part ii) it says that the two particles adhere which means that the resultant mass becomes the sum of the individual masses. Hence the total mass becomes 3m.

    Momentum before collision = momentum after collision

    So mu = 3mv, so v=u/3

    For part i) the particles do not coalesce. Remember that for elastic collisions, the speed at which two bodies approach one another is the same as the speed at which the two bodies move away from one another. Hence, uB - uA = vA - vB

    Since uB = 0, the equation becomes -uA = vA - vB

    Therefore, uA = vB - vA



    Now, we have to use the conservation of momentum:

    muA = mvA + 2mvB

    uA = vA + 2vB

    vB = (uA - vA)/2


    Simply plug VB into equation 1 to find vA:

    uA = (uA - vA)/2 - vA

    2uA = uA - vA - 2vA

    3va = -uA

    Hence va = -uA/3

    This means that when A collides with B, it rebounds (since it has opposite direction due to -ve sign) and the speed of A after the collision is 1/3 of its original speed.

    Hopefully this helps.
 
 
 
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