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Average force of impact

  1. Nov 17, 2007 #1
    1. The problem statement, all variables and given/known data

    During an autumn storm, a 0.012 kg hail stone traveling at 20 m/s made a .20 cm deep dent in the hood of a person's car. What average force did the car exert to stop the damaging hail stone?

    2. Relevant equations

    d = vt
    ft = mv

    3. The attempt at a solution

    Vav = (Vf + Vo)/2 = (0 + 20)/2 = 10 m/s

    V = 10 m/s
    d = 0.002 m

    t = d/v = 0.002/10 = 0.0002 seconds

    m = 0.012kg
    delta v = 20 m/s
    t = 0.0002 seconds

    f = mdeltav / t = (0.012 * 20) / 0.0002 = 1200 N
  2. jcsd
  3. Nov 17, 2007 #2
    Thats not right. You cant use d=vt here as there is an average force applied resulting in an average acceleration. What you have to do is compute the average acceleration using the impulse momentum theorem, then use that to calculate the time required ([tex]s=ut-\frac{1}{2}a_{av}t^2[/tex]). From there, you may calculate the average force using, again, the impulse momentum theorem. To help you along, I shall give you the answers, but you have to get there.

    [tex]t=1.267\times 10^{-6}[/tex]
    [tex]F_{av}=189.423\times 10^{3}[/tex]
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