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Remedial kinetic energy equivalency question

  1. Mar 16, 2009 #1
    I'm working on a demonstration of where I want to show the equivalent result of impacts of two separate masses. I'd like to make sure I am understanding these concepts correctly (it's been more than a few years since my college level physics class).

    An object weighing 100kg moving at a velocity of 20 m/s has a KE of 20,000J

    To get the equivalent KE from an object that weighs 10kg, I've calculated a velocity of 63.25m/s

    Now, here's my uncertainty: does this mean that an impact of the 100kg object moving at 20m/w would create the same amount of damage as the 10kg object moving at 63.25m/s? Assuming that the materials are the same, so the impact distance and rebound would be equivalent. Or are there other factors involved?

    Thanks in advance.
  2. jcsd
  3. Mar 17, 2009 #2
    cross sectional area might factor in , they have the same kinetic energy but
    the smaller cross sectional area would focus that energy to a smaller point.
  4. Mar 17, 2009 #3
    First, assume that the coefficient of restitution is 1, so that there is no "damage" (energy loss). In this case, both momentum and energy are conserved (the first mass always recoils unless M1 = M2). You have two equations in two unknowns.
    Second, solve the same problem where the coefficient of restitution is zero.
    Third, solve the same problem with an arbitrary coefficient of restitution.
    Beware of using rolling billiard balls, because 2/7 of the total kinetic energy is rotational energy (I = (2/5) m R^2) and is not easily transferred during a collision.
  5. Mar 17, 2009 #4
    Compare these two drop weight impacts:

    Impact 1
    Mass = 1 kg
    Drop height = 2 m
    Speed on impact = [tex]\sqrt {2gh} = 6.26 ms^{-1}[/tex]
    Kinetic energy at impact = [tex]\frac {1}{2} m v^{2} = 19.62 J[/tex]

    Impact 2
    Mass = 2 kg
    Drop height = 1 m
    Speed on impact = [tex]\sqrt {2gh} = 4.43 ms^{-1}[/tex]
    Kinetic energy at impact = [tex]\frac {1}{2} m v^{2} = 19.62 J[/tex]

    Let's assume that all the kinetic energy of the impacting object (KE) is perfectly converted to elastic strain energy (W) by purely compressing a sample of the same material and dimensions:

    [tex]W=\int^{\epsilon=\epsilon_{1}}_{\epsilon=0} \sigma d \epsilon = KE[/tex]

    If your sample has a compressive stress-strain profile that does not vary with strain rate, the strain epsilon_1 at which this is achieved (and the subsequent stress that this is achieved at) will be the same for both impacts.

    However, if your sample shows strong rate dependency (for instance most polymers) then the fact that one impact occurs at a greater initial velocity means that the material response will generally be stronger and stiffer. That's something to bear in mind when you talk about two equal-energy impacts with the same contact area - rate sensitivity in materials means you will probably see a difference in 'impact severity'.

    Also, when you begin to change the contact area between collisions, you will begin to see greater differences in 'impact damage'. The large stresses created by the relatively sharp point of a bullet will create more damage in a structure than a relatively bluff ball bearing of the same mass and impact speed.
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