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Kinetic Energy / Momentum Problem

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Two Railway cars, m1 and m2, are moving along a track with velocities v1 and v2, respectively. The cars collide, and after the collision the velocities are v'1 and v'2. Show that the change in kinetic energy, K' - K, will be maximum if the cars couple together.
    Hint: Set d(K' - K)/dv'1 = 0 and show that v'1 = v'2.

    2. Relevant equations
    Conservation of linear momentum: m1v1 + m2v2 = m1v'1 + m2v'2.
    Kinetic energy K = 0.5mv^2
    Difference in kinetic energy: K' - K = 0.5m1v1^2 + 0.5m2v2^2 - 0.5m1v'1^2 - 0.5m2v'2^2.

    3. The attempt at a solution
    I solved the conservation of momentum equation for v1 and substituted that into the K' - K equation. This yields v'2 = v2.
    I then solved the conservation of momentum equation for v2 and substituted that into the K' - K equation. I got v'1 = (m1v1 - m2v'2) / (m1 - m2).
     
  2. jcsd
  3. Jul 17, 2011 #2
    Since the hint mentions finding the derivative of the kinetic energy difference with respect to [itex]v'_{1}[/itex], you should find the value of [itex]v'_{2}[/itex] in terms of [itex]v'_{1}[/itex]. Substitute that into the energy difference and then use the hint (Note that [itex]v_{1}[/itex] and [itex]v_{2}[/itex] are constant in that expression).
     
    Last edited: Jul 17, 2011
  4. Jul 17, 2011 #3
    Thank you for responding, Pi-Bond. If I solve the momentum equation for v2-prime in terms of the other v's, and then substitute this into the energy equation, the energy equation wil be in terms of v1-prime, v1, and v2. I will not be able to show that v1-prime equals v2-prime.
     
  5. Jul 17, 2011 #4
    If you do that, what value for [itex]v'_{1}[/itex] do you get?
     
  6. Jul 17, 2011 #5
    Your two equations are sufficient; you can get the result by using the conservation equation to find an expression [itex]m_{1} v_{1}[/itex], which can be substituted into your second equation along with your first equation.
     
  7. Jul 18, 2011 #6
    I still cannot solve this.

    When I solve the conservation of momentum equation for m2v'2 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v'2.

    When I solve the conservation of momentum equation for m1v1 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v1.

    Finally, when I solve the conservation of momentum equation for m2v2 and then substitute this into the K' - K equation and then take its derivative and set it equal to zero, I get v'1 = 0.5v2.
     
  8. Jul 19, 2011 #7
    You got the equations:

    [itex]v'_{2}=v_{2}[/itex]
    [itex]v'_{1}=\frac{m_{1} v_{1} - m_{2} v'_{2} }{m_{1}-m_{2}}[/itex]

    From conservation of momentum,

    [itex]m_{1} v_{1} = m_{1} v'_{1}+ m_{2} v'_{2} - m_{2} v_{2}[/itex]

    Substitute that above to get:

    [itex]m_{1} v'_{1} - m_{2} v'_{1} = m_{1} v'_{1}+ m_{2} v'_{2} - m_{2} v_{2} - m_{2} v'_{2}[/itex]

    Can you get the result now?
     
  9. Jul 19, 2011 #8
    OK, I finally got it. It would have gone quicker if I had just followed the advice in your first post.

    I solved the conservation of momentum equation for v'2, then substituted that into the kinetic energy equation, then set its derivative equal to zero and solved for v'1.

    I then substituted this value for v'1 back into my equation for v'2 and showed that v'2 reduced to the same expression as v'1.

    Pi-Bond, thank you for your help and patience.
     
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