Topic 4.1 Linear Momentum: define linear momentum as the vector product of mass and velocity, p = mv, and distinguish it from kinetic energy.

What this topic is asking

The College Board (Topic 4.1) wants you to define linear momentum as the product of an object's mass and velocity, $\vec{p} = m\vec{v}$, to treat it as a vector with the same direction as the velocity, and to distinguish it from kinetic energy. Momentum is the quantity conserved in collisions, so getting its vector nature right is the foundation of the whole unit.

The definition of momentum

Momentum captures "how much motion" an object has in a way that combines how massive it is with how fast it moves. A slow-moving truck and a fast bullet can carry similar momentum. There is no special unit name for momentum; it is simply kg$\cdot$m/s.

Momentum is a vector

This vector nature is the single most important fact about momentum and the source of most mistakes. When two objects approach each other, their momenta partly or fully cancel; when they move together, they add. Always assign directions before adding.

How momentum differs from kinetic energy

Momentum and kinetic energy both describe motion, but they are fundamentally different quantities, and the exam tests the contrast directly:

• Momentum $\vec{p} = m\vec{v}$ is a vector, linear in speed. Doubling the speed doubles the momentum.
• Kinetic energy $K = \tfrac{1}{2}mv^2$ is a scalar, quadratic in speed. Doubling the speed quadruples the kinetic energy.

This difference matters in collisions. Momentum is conserved in every collision (when no external force acts), but kinetic energy is only conserved in elastic collisions. A perfectly inelastic collision conserves momentum while losing kinetic energy to heat and deformation. Keeping the two ideas distinct, one a conserved vector, the other a sometimes-conserved scalar, is what lets you analyze collisions correctly in Topic 4.4. The deeper reason momentum earns its own unit is Newton's second law in its momentum form: a net force is the rate of change of momentum, $F_{net} = \Delta p/\Delta t$, which is the launching point for impulse in Topic 4.2 and conservation in Topic 4.3.

Try this

Q1. A $1200$ kg car travels at $15$ m/s. Calculate its momentum. [2 points]

• Cue. $p = mv = (1200)(15) = 18{,}000$ kg$\cdot$m/s in the direction of motion.

Q2. A $0.40$ kg ball moves north at $5.0$ m/s and a $0.60$ kg ball moves south at $2.0$ m/s. Taking north as positive, calculate the total momentum. [2 points]

• Cue. $p_{total} = (0.40)(+5.0) + (0.60)(-2.0) = 2.0 - 1.2 = +0.80$ kg$\cdot$m/s (north).