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Collision of particles; centre of mass problem

  1. Jun 4, 2012 #1
    1. The problem statement, all variables and given/known data

    One particle, mass m1, is fired vertically upwards with an initial velocity v0 from the ground (height y = 0). At the same time (t = 0), another particle, mass m2, is released from rest at a height y = h, directly above the first, where h > 0. The particles are then in free fall near the earth's surface. Air resistance and the rotation of the earth may be neglected.

    e) Assume that the particles collide and that the collision is very brief and completely inelastic. Write an expression for the position of the particles after the collision but before they hit the ground.

    f) From the results above, derive an expression for the total momentum of the two particles both before and after the collision, but before they hit the ground.

    2. Relevant equations



    3. The attempt at a solution

    for e) I had

    yx=h - 1/2 g(h/vo))2

    ie replacing t with (h/vo)

    the expression is correct, but Im having trouble with part f).

    I had

    before p= m1v1+m2v2

    = m1(vo-gt)+m2(-gt)

    and after p= (m2m2)vx

    deriving vx from part e)

    vx= -g(h/vo)

    so after p= (m1+m2)(-g×h/vo)

    but this is wrong for part f), can someone please help with why?

    Thanks.
     
  2. jcsd
  3. Jun 4, 2012 #2
    But this vx is the velocity of the particle that was dropped just before the collision, and not the total velocity of both the masses together.

    It is a completely inelastic collision, what quantity is conserved in such a collision??
     
  4. Jun 4, 2012 #3
    ohh right thanks. infinitum. momentum.
     
  5. Jun 4, 2012 #4
    Yep, that should give you the answer! :approve:
     
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