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Momentum question

  1. Mar 24, 2006 #1
    My teacher showed me a question he was confused on with a picture so I don't know the original word problem, so I'll make one up: A 2000 kg hammer strikes a 400 kg stake into the ground. The velocity of the hammer is 6 meters per second and the stake goes into the ground .75 meters. What is the velocity of the stake after the hammer strikes it and what is the resistant force? This is a completely inelastic collision.

    This is what I did:

    [tex]m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}' + m_{2}v_{2}'[/tex]

    [tex]v_{1}' = v_{2}'[/tex]

    [tex]v_{2}' = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}[/tex]

    [tex]v_{2}' = \frac{2000(6) + 400(0)}{2000 + 400}[/tex]

    [tex]v_{2}' = 5 m/s[/tex]

    [tex]\Delta W = Fd[/tex]

    [tex]\Delta W = \Delta PE + \Delta KE[/tex]

    [tex]\Delta KE = \frac{1}{2}mv^2[/tex]

    [tex]\Delta KE_{hammer} = \frac{1}{2}2000*(6)^2 = 36000 J[/tex]

    [tex]\Delta PE = mgh[/tex]

    [tex]\Delta PE_{stake} = 400*9.8*.75 = 2940 J[/tex]

    [tex]\Delta W = 36000 + 2940 = 38940 J[/tex]

    [tex]F = \frac{\Delta W}{d}[/tex]

    [tex]F = \frac{38940}{.75}[/tex]

    [tex]F = 51,920 N[/tex]

    That's what I got, but it isn't the right answer, what am I doing wrong here?
     
    Last edited: Mar 24, 2006
  2. jcsd
  3. Mar 24, 2006 #2
    Oh ya I forgot to put that this is a completely inelastic collision.
     
  4. Mar 24, 2006 #3

    Hootenanny

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    If [itex]v'_{1} = v'_{2}[/itex], then this implies that the hammer and the stake combine to become one object after the collision. Therefore to calculate the kinetic energy you need to use the sum of the masses of the hammer and stake. I am also confused to why you have used the velocity of six [itex]m\cdot s^{-1}[/itex] when you've just calculated to be 5 [itex]m\cdot s^{-1}[/itex]?
     
  5. Mar 24, 2006 #4
    Oh I was using the velocity of the hammer before it hit the stake to get kinetic energy. Should I take the sum of kinetic energy of the hammer & stake after the collision plus potential energy of the stake before the collision to find the work? I probably worded my question wrong.

    Would it be correct to say that for the hammer to completely stop after the collision it would have to be going 30 m/s?
     
    Last edited: Mar 24, 2006
  6. Mar 24, 2006 #5

    Hootenanny

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    Yep, that sounds about right.
     
  7. Mar 24, 2006 #6

    Hootenanny

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    You've found the velocity of the combined body, because it stated that [itex]v'_{1} = v'_{2}[/itex]

    So you know that the kinetic energy upon collision is given by;

    [tex]E_{k} = \frac{1}{2}mv^2 = \frac{1}{2} (6000+400) v'_{2}^{2}[/tex]
     
  8. Mar 24, 2006 #7
    Ok I get 30000J for the sum of the kinetic energies and subtract 2940J potential energy. 27060/.75 = 36,080N. But that is not what the book says.

    I'm confused on the label you used: [tex]m\cdot s^{-1}[/tex] is that worded " meters in one second ". Then how would you word [tex]m\cdot s[/tex] ? I don't get why you would use " inverse seconds "
     
    Last edited: Mar 24, 2006
  9. Mar 24, 2006 #8

    Hootenanny

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    What answer does the book give? Working it out I get about 169kN.
     
  10. Mar 24, 2006 #9
    The answer in the book gives 64kN
     
  11. Mar 24, 2006 #10

    Hootenanny

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    What if you assume the hammer become stationary after the collision?

    [tex]m\cdot s^{-1} = \frac{m}{s}[/tex]

    Which is metres per second. :smile:
     
  12. Mar 24, 2006 #11
    I get 240kN if the hammer becomes stationary after collision.
     
  13. Mar 24, 2006 #12

    nrqed

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    I get 63520 N so it must be right.

    You just have to take into account the motion of both stake and hammer after the collison. In the kinetic energy and in mgh, use mass= total mass.
    Then it works out.

    Patrick
     
  14. Mar 24, 2006 #13

    Hootenanny

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    Correction, I get the same as you 63544N. I was looking at another thread in which the mass of an object was 6000kg, so I was using 6000kg instead of 2000kg :blushing: .

    Just to clarify for the OP;

    [tex]WD = \frac{1}{2}\Sigma m \cdot v'_{2}^{2} + \Sigma m \cdot gh[/tex]

    Sorry about the confusion.

    -Hoot:smile:
     
  15. Mar 24, 2006 #14

    Hootenanny

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    Do you have the correct answer now firestrider?
     
  16. Mar 24, 2006 #15
    Ok thanks for all the help. That's what I got too. But why don't I have to take in account the kinetic energy of the hammer before the collision?
     
  17. Mar 24, 2006 #16

    Hootenanny

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    You stated that the collision is totally inelastic, which means the kinetic energy of the objects is not conserved, however, momentum is always conserved. Basically, you could split the question up into three sections. You took the velocity of the hammer into account when you used momentum, you can't use energy because the collision isnt elastic.

    Does that make sense?
     
  18. Mar 24, 2006 #17

    nrqed

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    Because when the collision occurs, some energy is dissipated in heat and used to deform the objects, etc. This is an inelastic collision, so kinetic energy is not conserved (you actually have all the numbers to calculate the change of kinetic energy during the collision...that number corresponds to energy dissipated in the forms I mentioned above). You have to start from the total kinetic energy [itex] after [/itex] the collision and work from there.

    Patrick

    EDIT: After posting I saw that Hootenanny had basically said the same thing.
     
    Last edited: Mar 24, 2006
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