Linear momentum

Momentum is the product of mass and velocity, $p = mv$, measured in kg·m/s. Both factors carry equal weight, so neither one ranks momentum alone: a $1000$ kg truck creeping at $2$ m/s carries $2000$ kg·m/s while a $1$ kg ball flying at $40$ m/s carries just $40$. The slow truck wins by a factor of fifty.

The same product logic runs in every direction. Equal speeds do not mean equal momenta, because the masses still differ: two balls at $5$ m/s with masses $6$ kg and $2$ kg carry $30$ and $10$ kg·m/s. And equal momenta do not mean equal speeds: solve $p = mv$ for the factor you need, $v = p/m$ or $m = p/v$, and the lighter object must move faster to carry the same momentum.

The trap is to read the momentum as the speed with new units, to rank by whichever factor stands out, or to assume mass and speed always trade off evenly. They only balance when the products actually match; multiply it out.

Three habits cost points: ranking momentum by one factor (this section), piling up magnitudes when directions oppose (§2 and §3), and treating momentum and kinetic energy as the same amount of motion (§4).

Momentum points in the direction of the velocity. In one dimension that is a sign convention: pick a positive direction, and a $4$ kg cart moving the other way at $3$ m/s carries $p = -12$ kg·m/s. The sign is the direction; momentum is not a bare magnitude, and motion in the negative direction still carries momentum.

Because direction is part of the quantity, momentum can change while the speed holds still. A car rounding a curve at a constant $15$ m/s keeps the same magnitude $mv$, but the direction of motion turns, so the momentum changes the whole way around. Constant speed is not constant momentum once the path bends.

The same idea covers a ball thrown straight up: when it falls back through its launch height at the same speed, the momentum has the same magnitude and the opposite sign. The speed matches; the momentum does not.

The trap is to treat momentum as a positive number that only cares about how fast. Check the direction first, then the size.

The momentum of a system is the vector sum of the momenta of its parts. In one dimension: pick a positive direction, write each momentum with its sign, and add.

For example, a $3$ kg cart moves right at $4$ m/s while a $2$ kg cart moves left at $6$ m/s. Rightward positive, the momenta are $+12$ and $-12$ kg·m/s, so the system carries zero total momentum, even though $24$ units of magnitude are on the track. Equal-and-opposite momenta cancel completely; momenta of $+18$ and $-8$ cancel only partly, leaving $+10$.

The cancellation is exactly as deep as the smaller momentum, no more and no less. Opposite momenta do not refuse to cancel, and they do not always wipe each other out; the signed sum says how much survives. With three carts at $+6$, $+4$, and $-10$ kg·m/s, the system again sums to zero.

The trap is the magnitude pile: adding sizes no matter which way things move. The pile describes how much motion is in play; the signed sum is the system's momentum, and the two agree only when everything moves the same way.

Momentum and kinetic energy are built from the same ingredients and keep different books. Substitute $v = p/m$ into $K = \tfrac{1}{2}mv^2$ and the kinetic energy comes out as $K = \tfrac{1}{2}m(p/m)^2 = p^2/2m$: one line, worth memorizing.

The formula says everything the trap gets wrong. A $2$ kg cart at $6$ m/s and a $6$ kg cart at $2$ m/s carry the same momentum, $12$ kg·m/s each, but the kinetic energies are $36$ J and $12$ J: at equal momenta, the lighter, faster object holds more energy, because the mass sits downstairs in $p^2/2m$. Run it the other way and equal kinetic energies give the heavier object more momentum.

Scaling tells the same story. Doubling the speed doubles $p = mv$ but quadruples $K = \tfrac{1}{2}mv^2$: momentum is linear in speed, kinetic energy is quadratic. The two quantities never move in lockstep.

This distinction is the whole game when collisions arrive later in the unit: momentum can be conserved while kinetic energy is not. Keep two ledgers now and the collision topics get much easier.

A momentum question almost always asks you to rank by the product $mv$, sum contributions with their signs, or keep the momentum and energy ledgers apart.

Rank by the product. Compute $mv$ for each object before comparing; neither the speed nor the mass decides alone. To recover a factor, divide: $v = p/m$, $m = p/v$. Equal momenta force the lighter object to move faster.

Sum with signs. Pick a positive direction first, then add the signed momenta. Same direction: the sizes add. Opposite directions: they cancel as far as the smaller one reaches. Never pile up magnitudes across opposing motion.

Keep two ledgers. Momentum is linear in speed, kinetic energy is quadratic, and at fixed momentum $K = p^2/2m$ hands the lighter object more energy. Same $p$ never guarantees same $K$; compute each from its own definition.

Setup. A $4$ kg cart moves right at $3$ m/s and a $2$ kg cart moves left at $6$ m/s on the same track. Taking rightward as positive: (a) find each cart's momentum, (b) find the total momentum of the two-cart system, and (c) find each cart's kinetic energy.

(a) The products, with signs. The $4$ kg cart carries $p = mv = (4)(+3) = +12$ kg·m/s. The $2$ kg cart moves in the negative direction, so $p = (2)(-6) = -12$ kg·m/s. Equal magnitudes from very different carts: the small one makes up in speed what it lacks in mass.

(b) The signed sum. The system's momentum is $+12 + (-12) = 0$. The magnitude pile, $24$ kg·m/s, describes how much motion is on the track, but the system's momentum is the vector sum, and here it vanishes.

(c) The energy split. $K = \tfrac{1}{2}mv^2$ gives $\tfrac{1}{2}(4)(3^2) = 18$ J and $\tfrac{1}{2}(2)(6^2) = 36$ J. Same size of momentum, double the kinetic energy for the lighter cart, exactly what $K = p^2/2m$ predicts with half the mass downstairs. Note the kinetic energies do not cancel: energy has no direction.

The traps to avoid. Do not report the system's momentum as $24$ (signs first, then add), do not read the equal momentum magnitudes as equal kinetic energies (the lighter cart carries double), and do not let the zero total suggest nothing is moving (plenty is; the vectors just cancel).