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Which is more effective in this case, elastic or inelastic collision?

  1. Aug 11, 2015 #1
    1. The problem statement, all variables and given/known data

    A 2.9 ton weight falling through a distance of 6.5 ft drives a 0.5 ton pile 1.5 in. into the ground
    A) Assuming that the weight-pile collision is completely inelastic, find the average force of resistance exerted by ground
    B) Assuming that the resistive force in (A) remains constant, how far into the ground would the pile be driven if the collision were elastic?
    C) Which is more effective in this case, elastic or inelastic collision?
    2. Relevant equations
    P.E of 2.9 ton mass =Work done by resistive force

    3. The attempt at a solution
    To find resistive force in inelastic collision, the K.E imparted due to P.E of 2.9 ton mass= Work done by resistive forces
    => m*g*h=F_r*(0.0381m(1.5 inch))
    => F_r = 1.47 x 10^6
    (Is it right?)
    B) No idea because it says Elastic collision with resistive force constant ( Don't understand how to solve)
    C) No idea (Depends on (b))
     
  2. jcsd
  3. Aug 11, 2015 #2

    mfb

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    Those are two different things. First you have an inelastic collision between weight and pile, afterwards the combination of two moves down. You have to consider the processes separately.
    The same applies to (B).
     
  4. Aug 12, 2015 #3
    But how I am going to write two equations for Inelastic collision. It is understood that all the P.E is dissipated in work done against resistive force
     
  5. Aug 12, 2015 #4

    mfb

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    Why do you need two equations? One is sufficient.
    No, some parts are converted to heat (or something else) in the inelastic collision.
     
  6. Aug 12, 2015 #5
    But that is ignored in solving such problems.
    Please any equation for part B
     
  7. Aug 12, 2015 #6

    mfb

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    The details of the energy loss are ignored, but the energy loss itself is not. Otherwise the inelastic collision would be an elastic collision.

    Start with the inelastic collision, otherwise this thread won't lead anywhere.

    You should have formulas for elastic collisions. If not, you can derive them from energy and momentum conservation.
     
  8. Aug 12, 2015 #7
    A) m1v1 + m2v2= (m1+m2)V
    V= Common velocity, v2=0
    Then, 1/2(m1+m2)V=Fr*d
    Fr= Resistive force
    d= Depth of penetration
    (Is that right)?
    B) v1f= (m1-m2)*v1/(m1+m2)+2m2*v2(m1+m2)
    v1f (velocity after collision)
    v2f=(m2-m1)*v2/(m1+m2)+2m1/(m1+m2)
    Then, 1/2m*v1f+1/2m*v2f= Fr*d
    (Is that right)?
     
  9. Aug 12, 2015 #8

    SammyS

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    You didn't square the velocity, V . Maybe a typo.
     
  10. Aug 13, 2015 #9

    mfb

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    There is a v1 missing.
    m1 won't participate in the ramming process any more, and I think there is a square missing again.
     
  11. Aug 13, 2015 #10
    After the inelastic collision, do you need to account for the potential energy available as the combined mass drops the 1.5"?
     
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