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Find the Distance (Momentum)

  1. Nov 17, 2014 #1
    1. The problem statement, all variables and given/known data
    You are 60kg and you are holding a 10kg weight as you stand at rest on a frictionless surface. you can throw the weight 5m on ground where there is friction to keep you from moving backwards. How far will the 10kg weight be from you after you throw it on the frictionless surface?


    2. Relevant equations

    M1V1 + M2V2 = 0

    3. The attempt at a solution

    Total momentum of the system (you + weight) after the throw must be the same as it was before so it will be 0.


    M1 = 60kg
    M2= 10kg
    V1 = ?
    V2 = ?

    Please help! Thanks!
     
  2. jcsd
  3. Nov 17, 2014 #2

    haruspex

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    There isn't really enough information. To answer this you will need to make some assumptions about how arms work.
    Which of the following do you think will be approximately the same in the two cases:
    • the force exerted by the muscles
    • the distance over which the force is exerted (as far as the muscles are concerned)
    • the power delivered by the muscles
    • the time for which the power is delivered
    ?
     
  4. Nov 17, 2014 #3

    PhanthomJay

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    Yes.
    Try solving for V1 in terms of V2. Assume you the throw the weight horizontally.
     
  5. Nov 18, 2014 #4

    haruspex

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    Hi Tom2,

    Just had a private chat with PhanthomJay. His model is like firing a gun. That's rather different because the speed of the bullet would be the same with or without friction. For that to apply to a thrower, the thrower would have to absorb the recoil initially by flexing the body, so as to avoid any horizontal force on the feet until the mass has been released (with and without friction).
    I regard that as the least realistic of three models (the other two being constant force and constant power), but if the question is not intended to be much advanced then it may be what the questioner has in mind.
    The constant force model can also be analysed without calculus, but I regard constant power as the most realistic.
     
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