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Ball in head-on collision and being squeezed

  1. Feb 26, 2010 #1
    1. The problem statement, all variables and given/known data
    Two steel balls, each of diameter 24.9 mm, moving in opposite directions at 4 m/s, run into each other head-on and bounce apart.
    a. Compute an estimate for the time interval for which the balls are in contact.
    b One of the balls is squeezed in a vise while precise measurements are made of the resulting amount of compression. The results show that Hooke's law is a fair model of the ball's elastic behavior. For one datum, a force of 24 kN exerted by each jaw of the vise results in a 0.4 mm reduction in the ball's diameter. The diameter returns to its original value when the force is removed. Modeling the ball as a spring, find its spring constant.

    2. Relevant equations
    F = kx
    not sure about other equations


    3. The attempt at a solution
    a.
    At first I tried to use impulse but failed..
    I = F Δt, then I can't continue..

    b.
    F = kx
    24 x 103 = k * 0.4 x 10-3
    k = 6 x 107 N / m
    Do I get it right for part (b)?

    Thanks
     
  2. jcsd
  3. Feb 26, 2010 #2

    ehild

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    Gold Member

    Part b. is right. A hint for a: The deformation of the balls is elastic, so the mechanical energy is conserved. The balls are identical and move with the same speed, so the centre of mass of the two-ball system is in rest and stays so during and after the collision. When the balls are in contact, their translational energy is converted to elastic energy, as they squeeze each other. Can you determine the maximum deformation? And the maximum force of interaction?

    ehild
     
  4. Feb 26, 2010 #3
    Hi ehild

    My idea is :
    1) 1/2 mv2 = 1/2 kx2 --> I'll get x
    2) F = kx ---> I'll get F
    3) I = F Δt

    Is this right? If right, then to find the mass I have to search for the density of steel?

    Thanks
     
  5. Feb 26, 2010 #4

    ehild

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    Yes, find the density of steel.

    That F you get is the maximum, and the average force is less, but this is an estimation.

    Do not forget, that Δt is the time when the balls are squeezed, and equal time is needed to relax and move apart.

    ehild
     
  6. Feb 26, 2010 #5
    Oh you mean the time interval for which the balls are in contact is twice of Δt I found ?

    Thanks
     
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