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Conservation of Energy vs Conservation of Momentum problem

  1. Dec 2, 2015 #1

    TheDemx27

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    1. The problem statement, all variables and given/known data
    We are trying to find out whether we can hit a weight ##m=3## kg up to a given ##h=15.24## meters with a hammer with a head of ##m=6## kg swung at ##v=10## m/s.

    2. Relevant equations
    KE=0.5mv^2
    PE=mgh
    p=mv

    3. The attempt at a solution
    One solution is to even see if the energy put into the system is equal to the energy needed to get to our ##h##. For the energy put into the system, we have 0.5*6*10^2=300 joules. The energy required to attain a potential energy large enough for an ##h=15.24## is 3*9.8*15.24=448. Since our energy put into the system is not enough to raise the weight up to our required ##h##, we conclude no, we cannot raise the weight to the required height.

    Another way to do it, conservation of momentum. The weight's velocity due to the hammer is 6*10/3, and the distance it will travel is ##v^2/(2*a)=s##=20^2/(2*9.8)=20.4>15.24. So now we would say that we can raise the weight to the required height.

    Which way is correct?
     
  2. jcsd
  3. Dec 2, 2015 #2
    The energy calculation is the one to use. There is no way that the full momentum of the 6Kg hammer can be transferred to the 3Kg weight without additional energy. Imagine if we split the hammer into two 3KG pieces and put a spring between them. Releasing the spring could add 10m/s to one and subtract 10m/s to the other - leaving one stationary and the other with the momentum yo wanted for your 3Kg weight.

    The energy from that spring is what you are missing.
     
  4. Dec 2, 2015 #3

    TheDemx27

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    So I solved a system of equations and got 3.33 m/s for the hammer and 13.33 m/s for the weight. Is that the standard way to do it, with systems of equations for energy and momentum?
     

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  5. Dec 2, 2015 #4

    haruspex

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    For the purpose of finding the max height actually reached, neither works.

    You certainly cannot use conservation of energy unless you have some reason to suppose it is an elastic collision. This will depend on the materials involved. The result will be very different for a mass of putty and a rubber ball.
    Even if it is perfectly elastic, the hammer will not come to a dead stop. By using both conservation of work and conservation of momentum you can find how much velocity the hammer retains and how much momentum and KE passes to the mass.
    That said, if assuming (wrongly) that all of the KE gets transferred to the mass still doesn't get the mass to the desired height then you can be sure it won't get there.

    Similarly, your conservation of momentum calculation makes the wrong assumption that all the momentum is transferred to the mass. That would mean you gain KE.
     
  6. Dec 2, 2015 #5

    haruspex

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    I couldn't read your working, but those are the right velocities if work is conserved.
     
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