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Conservation of Momentum/Linear Momentum

  1. Feb 27, 2007 #1
    1. The problem statement, all variables and given/known data

    A man (weighing 915 N) stands on a long railroad flatcar (weighing 2805 N) as it rolls at 18.0 m/s in the positive direction of an x axis, with negligible friction. Then the man runs along the flatcar in the negative x direction at 40.00 m/s relative to the flatcar. What is the resulting increase in the speed of the flatcar?

    2. Relevant equations

    Pi = Pf

    3. The attempt at a solution

    (915+2805)g(18)=(915)g(-40+18)+(2805)g(v+18)
    v=-1.3 m/s

    But this is the wrong answer...what am I doing wrong???
     
  2. jcsd
  3. Feb 27, 2007 #2

    mjsd

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    Homework Helper

    firstly, weights were given in Newton already, no needs to add an extra "g". secondly, if you do want to use momentum (somehow), you want the "mass" of the objects.
     
  4. Feb 27, 2007 #3
    usually i get ma equations in variables n thn substitute the values.........suppose the man moves at a speed w towards left and the platform recoils at a speed V towards right...hence the speed of the man relative to the platform is V+w=v...or w= v-V....
     
  5. Feb 27, 2007 #4
    taking the man n the platform to b inna system, there is no external horizontal force on the system, the linear momentum remians constant. thus
    (M+m)18 = MV- mv...now solve n check
     
  6. Feb 27, 2007 #5
    sorry, the eqn above shud hav been (M+m)18 = MV- mw
     
  7. Feb 27, 2007 #6
    ooookay...the catch is , it will recoil wit V+18 and not just V....ill just try n let u know
     
  8. Feb 27, 2007 #7
    If You look at what I posted above that is the equation that I used and it did not work...(M+m)18=M(v+18)+m(v+18) and it did not give me the right answer.
     
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