Momentum · 5.7 · free game
Orbital Dock
Start here
Orbital Dock — dock five pods.
Latch inside the ring under 0.80 m/s
Dock 1 · First contactThe pod is parked dead ahead. Build speed, flip, and brake to a crawl before the ring.
Tap the stage to begin, then hold to thrust — every second of burn is impulse.
How the physics works
The thruster is fixed at 420 N, so holding it for a time Δt delivers an impulse J = F Δt, and impulse is change of momentum: Δv = J/m. The Δv readout on the stage is simply the fuel that remains divided by the tug’s mass — which is why the 1200 kg tug in Dock 3 turns the same tank into far less velocity change.
Nothing slows you down out here. Every m/s you buy, you keep until you burn the opposite way. Docking is judged on relative velocity, vrel = |vtug − vpod|: under 0.80 m/s inside the ring the latch holds, while a faster contact bounces both craft apart with restitution 0.45 — equal and opposite impulses, so momentum is conserved even in the failure.
A successful capture is a perfectly inelastic collision. The pair leaves at vdock = (mt vt + mp vp)/(mt + mp) — watch the docked pair drift, because that vector is the momentum you arrived with, shared over more mass. The dashed ghost line runs the same update law forward each frame, which is exactly how a game engine integrates it: apply thrust, step the velocity, step the position.
Momentum · 5.7
Momentum Model for Gaming
Fly a tug with a real thruster: every second of burn is impulse, every m/s is bought with fuel, and a gentle latch is a perfectly inelastic collision.
Build the model
Track direction as carefully as magnitude.
A convincing docking game needs three checks: the burn changes the tug's momentum, relative velocity sets impact safety, and conservation sets the joined velocity after capture.
The tug's new velocity after a thruster burn: old velocity plus the kick divided by the tug's mass.
Resolve J into x and y components
How fast the two craft close on each other — keep this small for a gentle docking.
Low relative speed protects the docking system
After docking, the joined craft move at the combined momentum divided by the combined mass.
Perfectly inelastic vector collision
State
Store each craft's mass and two velocity components before the docking event.
Rules
Apply the commanded impulse, test relative speed, then conserve x and y momentum for a successful capture.
Feedback
Show impulse budget, impact safety, final velocity error, and the exact failed constraint.
Catch the common trap
Explain before calculating.
To reduce relative speed most efficiently, which way should a corrective impulse point?
Choose an answer, then use the explanation to check your model.
Worked examples
Choose signs, balance, then check.
EasyA 500 kg tug needs a velocity change of 0.40 m/s. What impulse must its thruster deliver?
- J = mΔv = 500 × 0.40.
- J = 200 N·s.
AnswerJ = 200 N·s
MediumA 600 kg tug at +2.0 m/s captures a 400 kg cargo pod at −0.50 m/s. Find the joined velocity.
- Before capture: p = (600)(2.0) + (400)(−0.50) = 1,000 kg·m/s.
- The joined mass is 600 + 400 = 1,000 kg.
- vdock = ptotal/mtotal = 1,000/1,000.
Answervdock = +1.0 m/s
HardA 600 kg tug at (+2.0, 0) m/s fires a 900 N·s impulse along +y. Find its new velocity vector.
- Δvᵧ = 900 ÷ 600 = 1.5 m/s; vₓ unchanged.
- v = (2.0, 1.5) m/s.
- |v| = √(4 + 2.25) = 2.5 m/s at 37° above +x.
Answer(2.0, 1.5) m/s — 2.5 m/s at 37°
ChallengingDocking safety allows a relative speed of at most 0.20 m/s. A 600 kg tug at +1.80 m/s approaches a 400 kg pod at +1.50 m/s. The thruster delivers impulse in 50 N·s bursts (retrograde). How many bursts are needed before capture is safe, and what is the docked velocity?
- Relative speed = 0.30 m/s; must shed ≥ 0.10 m/s of tug speed: Δv per burst = 50/600 ≈ 0.083 m/s.
- Two bursts give 0.167 m/s reduction → tug at 1.633 m/s, relative 0.133 m/s ≤ 0.20 ✓ (one burst leaves 0.217 — unsafe).
- Docked: (600 × 1.633 + 400 × 1.50)/1000 ≈ 1.58 m/s.
AnswerTwo bursts; docked velocity ≈ +1.58 m/s