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

[rcgldr]

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
v_a = v₁ + v₀ = final velocity of mass 1 wrt frame
v_b = v₂ + v₀ = final velocity of mass 2 wrt frame

It's a closed system so momentum is conserved

m₁ v_a + m₂ v_b = (m₁ + m₂) v₀

total kinetic energy change in system

ΔE = 1/2 m₁ v_a² + 1/2 m₂ v_b² - 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 v_a + m₂ v_b - (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

 

[kuruman]

You showed that this works for the specific case of two masses connected with a compressed spring. You need to show that if $\Delta U +\Delta K =0$ in inertial frame of reference $O$ is true, then it is also true in inertial frame $O'$ that is moving with constant velocity $u$ relative to $O$. This is how to do it in one dimension; it can be easily extended to three dimensions. The Galilean transformations are$$x'=x-ut~\rightarrow~dx'=dx-udt;~~v'=v-ut.$$Let $F_c$ be the sum of all conservative forces acting on the mass. The change in potential energy is the negative of the work done by the conservative forces. In the unprimed frame we have, $$\Delta U=-\int F_c dx~~\mathrm{and} ~~\Delta K=\frac{1}{2} m \left( v_f^2-v_i^2 \right)$$so that$$0=\Delta U +\Delta K=-\int F_c dx+\frac{1}{2} m \left( v_f^2-v_i^2 \right).$$Similarly,$$\Delta U'=-\int F_c dx'=-\int F_c (dx-udt)=-\int F_c dx+u\int F_cdt=\Delta U+uJ~~~~~(1)$$where $J=\int F_cdt$ is the impulse delivered to the mass. Also,$$\Delta K'=\frac{1}{2}m\left({v'}_f^2-{v'}_i^2 \right)=\frac{1}{2}m\left[(v_f-u)^2-(v_i-u)^2 \right]=\frac{1}{2}m\left[v_f^2-v_i^2-2u(v_f-v_i) \right]$$ $$\Delta K'=\frac{1}{2}m\left(v_f^2-v_i^2 \right) -mu\Delta v=\Delta K-uJ~~~~~(2)$$Adding equations (1) and (2) gives the desired result,$$\Delta U' +\Delta K' =\Delta U +\Delta K=0.$$