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Co-efficient of restitution

  1. Jul 2, 2011 #1
    Suppose that an object is dropped from a height 'H'.The coefficient of restitution be C.
    This v_1 is the speed with which it bounces.Then,it strikes the floor again.
    And again,C=v_2/v_1

    v_2= C.v_1 = C^2 u
    Similarly,v_3= C^3u
    v_4=C^4u and so on.

    Suppose that the ball comes to rest after bouncing up n times.
    v_n = 0
    C^n.u = 0
    And this is only possible when n tends to infinity.In other words,it never comes to rest.But,practically,this is not the case.How come?
  2. jcsd
  3. Jul 2, 2011 #2
    First of all coefficient of restitution is
    C=v/u simply in case of bouncing of a rigid ball on a rigid surface.Not , C=(v-1)/u.
    Then, also this paradox arises as you mentioned .
    we must not forget that we apply newton's laws in some idealized cases. And this application may seem to be miraculous some time. But as we go into detail, not neglecting many phenomenons, then gradually we come close to physical reality.
    In this case of bouncing object , we assumed it to be point particle. If we consider it to big enough ,then also bouncing height will never be equal to zero mathematically.But, the height achieved after several bouncing will be so small that we will not be able to distinguish it from
    atomic motions.
  4. Jul 2, 2011 #3
    No,its no v-1, its v_1.I dont know how to use subscripts on mobile.v_1 is the velocity with which the ball bounces up after striking the floor 1st time.
    v_2 2nd time
    v_n nth time
  5. Jul 2, 2011 #4
    Then also logic is same,isn't it.
  6. Jul 2, 2011 #5
    Yeah,just wanna make it clear!
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