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Transfer of momentum .help needed ?

  1. May 21, 2010 #1
    transfer of momentum.....help needed ???

    1. The problem statement, all variables and given/known data

    two particles of masses m1 & m2 are moving along a straight line with velocities u & v respectively. they collide to each other..if the collision between them is an elastic collision then show that momentum transfered from first particle to second particle is 2m1*m2(u - v)/m1 + m2....


    2. Relevant equations

    I am trying to use the condition of an elastic collision...their linear momentum & kinetic energy will be conserved..but the proof is not coming out....please help me with it.........

    if the velocity of first particle after collision is u1 & the same of the 2nd particle is v1 then we know, u1 = (m1 - m2)*u/(m1 + m2) + 2m2*v/(m1 + m2)

    v1 = 2m1*u/(m1 + m2) - (m1 - m2)*v/(m1 + m2)

    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 21, 2010 #2

    ehild

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    Re: transfer of momentum.....help needed ???

    Show your work.

    ehild
     
  4. May 22, 2010 #3
    Re: transfer of momentum.....help needed ???

    v1 = 2m1*u/(m1 +m2) - (m1 - m2)*v/(m1+m2)

    momentum of the 2nd particle after collision

    m2v1 = 2m1*m2*u/(m1+m2) - (m1-m2)*m2*v/(m1+m2)

    transfer of momemtum

    m2v1 - m2v = 2m1*m2*u/(m1+m2) - (m1-m2)*m2*v/(m1+m2) - m2v

    [2m1*m2*u - (m1-m2)*m2*v - (m1+m2)*m2v]/(m1+m2)

    [2m1*m2*u - (m1-m2)*m2*v - m1*m2*v - m2^2v]/(m1+m2)

    [2m1*m2*u - m1*m2v + m2^2v - m1*m2*v - m2^2v]/(m1+m2)

    [2m1*m2*u - 2m1*m2*v]/(m1+m2)

    2m1*m2(u-v)/(m1+m2)


    it is sloved

    BTW thanks
     
  5. May 22, 2010 #4

    ehild

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    Gold Member

    Re: transfer of momentum.....help needed ???

    Good job!

    ehild
     
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