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

1. Sep 30, 2011

### rcgldr

Because of the kinetic energy and frames of reference thread:

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