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Momentum factor into the force applied to an object at rest

  1. Jul 5, 2005 #1
    How does momentum factor into the force applied to an object at rest hit by another which is moving?
  2. jcsd
  3. Jul 5, 2005 #2


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    Staff Emeritus
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    Assuming no external forces, momentum is conserved so the initial momentum of the moving object relates to that (as well as how elastic the collision is)- it determines the final speeds. However, the "force" involved in the collision will also depend upon how long the interaction takes. In order go from rest to whatever final speed the object has, it must accelerate during the collision. The longer the collision takes, the lower the acceleration and so the lower the force required.
  4. Jul 5, 2005 #3
    Lets say I have a human bone, and I'm putting weights on it until it snaps, then I have another bone but let the weights fall onto the bone. I'm assuming the second bone will break with less weight because we're letting the weights fall. It is momentum that's involved here? What equations would I be working with?
  5. Jul 5, 2005 #4


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    Staff: Mentor

    No, its energy. There is an engineering test called a CHARPY IMPACT TEST that takes into account all the factors that Halls mentioned and calls the energy required to break a sample "impact toughness". By using a heavy pendulum to break a test sample, the energy required to break it can be measured simply by using the difference in the height of the pendulum between the upstroke and the downstroke and applying the potential energy equation.
  6. Jul 6, 2005 #5
    The OP first question asked about the relationship between force and momentum, That relationship is:

    F = dp/dt where Force equals the change in momentum divided by the change in
    time (during the collision)

    This relationship is not evident in the arm breaking scenario but is best seen with an example such as a tennis player serving a ball or a golfer tee-ing off. The tricky part is in measuring the time of impact. Also, the force may not be constant during the collision, hence the need for differential calculus.
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