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Sandbox Physics

M031 · Observers & invariants

Center-of-Momentum Frame Lab

Resolve an oblique two-body impact, switch observers and separate mass-centre motion from internal kinetic energy. Compare elastic exchange, normal restitution and a clean miss.

Interactive modelCenter-of-Momentum Frame Lab
Nearest valid window timeReady\text{Ready}
Estimated horizontal momentumReady\text{Ready}
Estimated kinetic energyReady\text{Ready}
Motion stateReady\text{Ready}
Selected timeReady\text{Ready}
Observer horizontal speedReady\text{Ready}
A horizontal velocity in frameReady\text{Ready}
A transverse velocityReady\text{Ready}
B horizontal velocity in frameReady\text{Ready}
B transverse velocityReady\text{Ready}
Total horizontal momentum in frameReady\text{Ready}
Total transverse momentumReady\text{Ready}
Total kinetic energy in frameReady\text{Ready}
Internal kinetic energyReady\text{Ready}
Mass-centre kinetic energy in frameReady\text{Ready}
Dissipated energy up to clockReady\text{Ready}
Energy decomposition residualReady\text{Ready}
Frame transformation residualReady\text{Ready}
Raw position framesReady\text{Ready}
Available slope estimatesReady\text{Ready}
Model contact timeReady\text{Ready}
Impulse up to clockReady\text{Ready}

Physics tutorial

Center-of-Momentum Frame Lab

BackgroundNewtonian mechanics permits equally valid constant-velocity observers.

Why it mattersResolve an oblique two-body impact, switch observers and separate mass-centre motion from internal kinetic energy. Compare elastic exchange, normal restitution and a clean miss.

Start with the essentials

Focus question
Can total momentum vanish while both bodies keep moving?
One-sentence intuition
Zero total momentum minimizes kinetic energy without eliminating internal motion.

Core mathematical model

The same event in another frame

xi′=xi−Utvx,i′=vx,i−U\begin{aligned}x_i^\prime&=x_i-Ut\\v_{x,i}^\prime&=v_{x,i}-U\end{aligned}

Only constant horizontal boosts are used. Time and transverse coordinates are unchanged.

Momentum changes with the observer

Px′=Px−MU,Py′=PyP_x^\prime=P_x-MU,\qquad P_y^\prime=P_y

Total momentum vanishes in the centre frame for this initial condition.

Energy has a minimum

K′=K−UPx+12MU2=12M∣Vcm−Ux^∣2+Kint\begin{aligned}K^\prime&=K-UP_x+\tfrac12MU^2\\&=\tfrac12M|\mathbf V_{\rm cm}-U\hat{\mathbf x}|^2+K_{\rm int}\end{aligned}

The residual is checked against the direct sum of the two body energies.

Resolve the normal impulse

J=−(1+e)μ w− ⁣⋅nQ=12μ(1−e2)(w− ⁣⋅n)2\begin{aligned}J&=-(1+e)\mu\,\mathbf w^-\!\cdot\mathbf n\\Q&=\tfrac12\mu(1-e^2)(\mathbf w^-\!\cdot\mathbf n)^2\end{aligned}

Relative velocity is B minus A; the normal points from A to B. Only closing contact receives an impulse.

Common difficulties

Separate frame choice from dynamics

Typical misconceptionChanging the observer changes the collision or removes all kinetic energy.

Better mental modelThe same physical positions and impulses receive different coordinates. Internal kinetic energy persists unless the interaction dissipates it.

Run the experiment

  1. 01

    Predict the observer

    Find an observer that makes A stationary, then compare the mass-centre observer.

    What to observe: One stationary body does not generally make total momentum vanish.
  2. 02

    Move along the energy curve

    Drag the copper handle across the observer-energy plot.

    What to observe: The minimum follows the total mass and momentum, not a preferred absolute rest frame.
  3. 03

    Resolve the encounter

    Compare oblique, elastic, zero-restitution and miss presets.

    What to observe: Impulse and dissipated energy are observer invariant.
  4. 04

    Acquire before switching

    Acquire positions, change observers and compare finite-window estimates with the reference.

    What to observe: The raw laboratory record is identical after a frame switch.