Exploring Mechanics / 4. Momentum
Momentum is what force and time together leave behind — and unlike energy, it never has a "wasted" form to lose track of.
Make a prediction first: a loaded truck and a motorcycle collide with a wall at the exact same speed. Which one pushes on the wall harder, and for longer?
p = mv · In any isolated system, total momentum is conserved
Momentum answers a slightly different question than kinetic energy does. KE = ½mv² cares about speed squared and has no direction; p = mv is linear in velocity and is a genuine vector — a truck moving left and one moving right at the same speed have opposite momenta (one positive, one negative) but identical kinetic energy.
It follows directly from Newton's Third Law. In any collision between objects A and B, the force A exerts on B and the force B exerts on A are equal and opposite at every instant, so their impulses (force × time, developed fully in the next tab) are equal and opposite too — whatever momentum A loses, B gains, exactly. Add up momentum over the whole isolated system and those internal transfers cancel completely, leaving the total unchanged no matter how complicated the collision is.
Set independent masses and velocities (negative means moving left) for two carts that collide and stick together. Total momentum before always equals total momentum after — even when the carts' individual speeds change dramatically.
The two-carts-pushing-apart-from-rest version of this idea (equal and opposite forces, unequal resulting accelerations) was already built as the Push-Off Simulator in Newton's Laws → Inertia — it's the same conservation law, viewed from the force side instead of the momentum side. What kinetic energy does during a collision like this one is the subject of the next tab.
A cannon and shell start at rest — total momentum zero. Fire the shell forward and the cannon must recoil backward with exactly enough momentum to keep the total at zero.
Mvcannon + mvshell = 0 (starting momentum) gives vcannon = −mvshell/M. The recoil speed shrinks in direct proportion to how much heavier the cannon is than the shell — the same reason a rifle "kicks" noticeably but the battleship firing a much heavier shell barely shudders: mass ratio, not some difference in the underlying law.
Make a prediction first: you catch a fast ball two ways — rigidly (arm locked stiff) and by letting your arm "give," moving back with the ball as you catch it. Same ball, same speed, same final result (it stops). Which catch hurts more, and why?
J = FΔt = Δp
Stopping a given amount of momentum always takes the same impulse — but that impulse can come from a huge force acting briefly, or a small force acting longer. Every safety device in the prediction above (airbags, padded gloves, a "giving" catch) works the same way: stretch out Δt and the average force F = Δp/Δt drops in direct proportion.
For a constant force, J = FΔt is literally base × height — the area of a rectangle on a force-vs-time plot. Real collision forces aren't constant (they typically ramp up and back down over the contact time), but the same idea holds exactly: impulse is whatever area sits under the true force-time curve, and F in the simple formula is really just the average height that would give the same area.
The shaded area is fixed — it's the momentum change this object needs, which doesn't depend on how the stop happens. Stretch the stop out and watch the rectangle get wider and dramatically shorter.
Every one of these scenarios stops the same momentum change; only the time it takes to do it changes. Pick a scenario and see how violently the required force swings.
A golf or tennis follow-through extends the contact time between racket/club and ball on purpose — for a fixed desired Δp given to the ball, a longer Δt would normally mean a smaller force, so players compensate by also increasing the force itself throughout the longer stroke. The point of a follow-through isn't to reduce force at all — it's to keep applying force for as long as possible, maximizing the total impulse (and therefore the ball's speed) instead of cutting the swing short.
Same ball, same speed, same final result — only the stopping distance changes. Try the rigid-wall preset, then cupped hands, then an airbag, and watch the force readout swing by orders of magnitude.
Make a prediction first: in a Newton's cradle, you pull back two balls at one end and let them swing into the rest. Does exactly one ball fly out the far side twice as fast, or two balls fly out at the original speed?
Momentum is always conserved in a collision. Kinetic energy is conserved only in the special case of an elastic collision; a perfectly inelastic one (the objects stick together) loses the maximum amount possible while still conserving momentum. Most real collisions land somewhere between those two extremes.
Slide the elasticity from perfectly inelastic to perfectly elastic and watch momentum stay locked while kinetic energy is only rescued back to 100% right at the elastic end.
v₁′ = [(m₁−em₂)v₁ + (1+e)m₂v₂] / (m₁+m₂), and swap 1↔2 for v₂′. Setting e = 1 reduces this to the standard elastic-collision formulas; setting e = 0 makes both formulas collapse to the same shared final velocity — a perfectly inelastic collision, exactly matching the momentum-only result from the previous tab.
Choose how many balls you pull back. In this idealized (perfectly elastic, equal-mass) chain, the same number of balls always exits the far end at the same speed — the middle balls barely move at all.
One fast ball leaving with double speed would conserve momentum (2m·v either way) but would carry twice the kinetic energy that went in — impossible for an elastic chain. k balls leaving together at the original speed conserves both momentum (k times the single-ball momentum, matching what came in) and kinetic energy simultaneously, which is the only outcome consistent with both laws at once.
Hit a stationary ball off-center and watch both balls fly apart at right angles to each other — a real, testable fact about equal-mass elastic collisions, not just a billiards superstition.
A single ball's momentum has to be shared among every pin it scatters — a qualitative look at what a two-object collision looks like once there are ten objects instead of one.
Make a prediction first: a fireworks shell explodes in mid-air into dozens of glowing fragments, scattering in every direction. Does the "center" of that expanding sparkle cloud still follow the same smooth arc the unexploded shell would have?
xcom = Σmᵢxᵢ / Σmᵢ (and likewise for y)
The center of mass of a system moves exactly as if all of the system's mass were concentrated there and only the external forces acted on it — internal forces (an explosion, a diver tucking and extending, a rocket's stages separating) can throw individual pieces around violently without changing the center of mass's path at all.
Drag any of the three masses, or resize them, and watch the center of mass (★) update instantly, always pulled toward whichever mass is currently heaviest.
If all three masses were equal, the center of mass would sit at the ordinary geometric centroid of the triangle they form. Make one mass much heavier and the center of mass shifts hard toward it — mathematically because each position xᵢ in the sum is weighted by its own mass, so heavier particles simply count for more in the average.
Launch a projectile and detonate it mid-flight into two equal fragments flying apart. The ghost marker traces the path the unexploded shell would have taken — watch it stay locked to the midpoint between the two fragments the entire time.
The explosive force between the two fragments is an internal force — by Newton's Third Law it's equal and opposite on each fragment, so it contributes exactly zero to the system's total momentum. The only external force still acting on the whole system is gravity, exactly as before the explosion — so the center of mass keeps obeying the same projectile-motion equations, completely unaware that anything exploded.
Make a prediction first: a bullet is fired into a heavy wooden block hanging from strings, and the block swings up and to the side by some measurable height. Could you work out how fast the bullet was moving just from that one height measurement — without ever timing the bullet itself?
This classic device needs two different laws in sequence: momentum conservation gets you through the violent, energy-losing collision itself; energy conservation then carries you through the smooth swing afterward. Neither law alone would work for the whole process.
The bullet embedding in the block is a perfectly inelastic collision — by definition, it loses the maximum kinetic energy possible while still conserving momentum. Trying to use energy conservation across the collision itself would silently (and wrongly) assume no energy was lost. Momentum conservation is the only law guaranteed to hold across a violent, deforming impact like this; energy conservation only becomes valid again once the collision is over and the swing is smooth and lossless.
Set a "true" bullet speed below, and watch the collision-then-swing animation play out. Then check the last row: working backward from only the swing height h recovers the exact same speed you started with.
Momentum conservation across the impact: mv = (m+M)V, so V = mv/(m+M). Energy conservation across the swing: ½(m+M)V² = (m+M)gh, so V = √(2gh). Setting these equal and solving for v: v = [(m+M)/m]√(2gh) — a bullet's speed, recovered entirely from a ruler measurement of how high a block swings.