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M1v1=m2v2 momentum help

  1. Dec 12, 2008 #1
    1. The problem statement, all variables and given/known data
    A massless elastic cord (that obeys Hooke’s Law) will break if the tension in the cord exceeds Tmax. One end of the cord is attached to a fixed point, the other is attached to an object of mass 3m. If a second, smaller object of mass m moving at an initial speed v_0 strikes the larger mass and the two stick together, the cord will stretch and break, but the final kinetic energy of the two masses will be zero. If instead the two collide with a perfectly elastic one-dimensional collision, the cord will still break, and the larger mass will move off with a final speed of v_f. All motion occurs on a horizontal, frictionless surface.
    Find v_f/v_0

    2. Relevant equations
    m1v1=m2v2


    3. The attempt at a solution
    I always got 1/sqrt(2), which is not the answer.
     
  2. jcsd
  3. Dec 13, 2008 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    Re: momentum

    This is not an attempt at a solution. You have to show your work and explain your reasoning.

    AM
     
  4. Dec 13, 2008 #3
    Re: momentum

    oh,sorry,I know where I was wrong now,a stupid mistake.hehe
     
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