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Show PE to KE change in closed system is independent of initial velocity

  1. Sep 30, 2011 #1

    rcgldr

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    Because of the kinetic energy and frames of reference thread:

    https://www.physicsforums.com/showthread.php?t=534883

    I was wondering how to show that a change from potential to kinetic energy in a closed system is independent of the (inertial) frame of reference. I think the math below demonstrates this.

    Example closed system: a compressed massless spring with potential energy ΔE and two masses. The compressed spring is released and accelerates the two masses, increasing the mechanical energy of the closed system by ΔE.

    From frame of reference of the center of mass

    m1 = mass 1
    m2 = mass 2
    v1 = final velocity of mass 1 wrt center of mass
    v2 = final velocity of mass 2 wrt center of mass

    It's a closed system so momentum is conserved

    m1 v1 + m2 v2 = 0

    total kinetic energy change in system

    ΔE = 1/2 m1 v12 + 1/2 m2 v22

    With the center of mass moving with respect to some inertial frame of reference:

    v0 = velocity center of mass wrt frame
    va = v1 + v0 = final velocity of mass 1 wrt frame
    vb = v2 + v0 = final velocity of mass 2 wrt frame

    It's a closed system so momentum is conserved

    m1 va + m2 vb = (m1 + m2) v0

    total kinetic energy change in system

    ΔE = 1/2 m1 va2 + 1/2 m2 vb2 - 1/2 (m1 + m2) v02

    ΔE = 1/2 m1 (v1 + v0)2 + 1/2 m2 (v2 + v0)2 - 1/2 (m1 + m2) v02

    ΔE = 1/2 m1 (v12 + 2 v1 v0 + v02) + 1/2 m2 (v22 + 2 v2 v0 + v02) - 1/2 (m1 + m2) v02

    ΔE = 1/2 m1 (v12 + 2 v1 v0) + 1/2 m2 (v22 + 2 v2 v0)

    ΔE = 1/2 m1 v12 + 1/2 m2 v22 + m1 v1 v0 + m2 v2 v0

    going back to momentum equation:

    m1 va + m2 vb - (m1 + m2) v0 = 0

    m1 (v1 + v0) + m2 (v2 + v0) - (m1 + m2) v0 = 0

    m1 v1 + m2 v2 = 0

    m1 v1 v0 + m2 v2 v0 = 0

    ΔE = 1/2 m1 v12 + 1/2 m2 v22
     
    Last edited: Sep 30, 2011
  2. jcsd
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