Module 06
Many particles
Two carts roll toward each other on a track and collide. Slow the film down and the middle of that crash is a horror of detail: bumpers flexing, metal shivering, contact forces spiking and collapsing in fractions of a millisecond. No instrument you own can follow it. Be honest and write down what you actually know about the middle: nothing.
Now do this. Before the crash, multiply each cart's mass by its velocity and add the two results. After the crash, when things are calm again, compute the same sum. The two numbers agree. Rerun the crash harder, gentler, with different bumpers; they agree every time.
You were completely ignorant of the middle and you are exactly right about the end. What kind of bookkeeping makes that possible?
The takeaway
Total momentum, the sum of mass times velocity over every body in a closed system, cannot change, because the third law cancels every internal push in matched pairs. The middle can be chaos; the total is bookkept by nature itself.
The pair rule earns its keep
Give each body a new number: its momentum, mass times velocity, written . A vector, aimed the way the body moves. Heavy-and-fast means a lot of it; light-and-slow means a little. On its own this looks like bookkeeping for its own sake. The payoff is what happens when bodies interact.
First, draw a boundary. Pick which bodies are on your list, and call the list a system: the two carts and nothing else. Every force in the world is now one of two kinds. Forces between list members are internal. Forces from outside the list are external. The boundary is yours to draw, and drawing it well is a skill worth developing, because the theorem coming next cares about nothing else.
During the crash, cart one pushes cart two, and module 05's third law insists cart two pushes back, equal and opposite, at every instant, whatever the bumpers are doing. Each push changes the other cart's momentum, and because the pushes are opposite twins, whatever momentum one cart gains the other loses, tick for tick, all the way through the chaos. Internal forces move momentum around the list. They cannot create it, and they cannot destroy it. Only an external force, a push from off the list, can change the total.
Think of momentum as a currency, and every push as a transaction: your account and mine change, the economy's total does not. The analogy breaks at the mint. A real currency has a government that prints and burns money, and an external force is exactly that, an outside mint. The theorem holds only for a closed economy, a system with no mint, and the phrase for that is a closed system.
This is the demon's first true shortcut. Through module 05 it computed by brute force: every push, every tick. Now entire stretches of the calculation collapse. The unknowable middle of the crash stopped mattering; the demon writes down one number going in and reads the constraint coming out. And the fireworks burst that ended module 05 yields to the same move: ten thousand fragments, millions of internal pulls, and the total momentum of the cloud sails on exactly as the unexploded shell would have.
The math
Restate the second law in momentum's language. For one body of fixed mass:
Force is the rate at which momentum changes. Now take two bodies. Let be the force on body 1 from body 2, and let be any outside force on body 1:
Add the two equations. The third law says , so the internal pair cancels:
If the external forces are zero, the total momentum has zero derivative: it is constant. The argument never asked what was, only that it came paired. That silence is why the middle of the crash never mattered. The same cancellation runs for any number of bodies: internal forces arrive in pairs, and pairs die in the sum.
Work the sticky crash. A 2 kg cart at 3 m/s hits a 1 kg cart at rest, and they lock together into one 3 kg body at speed :
Six units of momentum in, six out, one equation, no bumper physics.
Play with it
Two carts, your choice of masses and speeds, sticky or bouncy bumpers. Watch the per-cart momentum bars trade while the green total refuses to move. Then find a run where the bars end far from where they started and the total still has not budged.
Check yourself
Problem 1. A 2 kg cart moving at 3 m/s hits a 1 kg cart at rest, and they stick together. How fast does the pair move?
Reveal the solution
Momentum in: 2 × 3 + 1 × 0 = 6. The stuck pair has mass 3, so 3v = 6 and v = 2 m/s.
Problem 2. A 60 kg skater at rest on smooth ice throws a 0.5 kg ball forward at 10 m/s. What happens to the skater?
Reveal the solution
The system starts with zero total momentum and nothing external pushes along the ice, so it must end at zero: 0.5 × 10 + 60 × v = 0 gives v = -1/12, about 0.083 m/s backward. Throwing the ball is the transaction; the recoil is the balancing entry.
Problem 3. Two 1 kg carts, perfectly bouncy: one arrives at 4 m/s, the other rests. What comes out?
Reveal the solution
They swap: the mover stops dead and the target leaves at 4 m/s. Check the ledger: momentum 4 in, 4 out. For equal masses a perfectly elastic head-on collision always exchanges the velocities; the sim's bouncy setting will show it.
What this did to the demon
The demon learned to skip: draw the boundary, total the momentum, and the unknowable middle of any interior chaos stops being its problem. But run the carts through a valley and something else is clearly being traded, speed for height and back, and momentum's ledger has no column for it. Module 07 opens the second currency.
Energy