AP® Physics C: Mechanics · Unit 4 · Topic 4.1

Momentum and Newton's Second Law

Newton's second law is really F = dp/dt: force is the rate of change of momentum, not just mass times acceleration. F = ma is only the special case where mass never changes.

Newton did not write F = ma. He wrote that force equals the rate of change of momentum, F = dp/dt, where p = mv. Momentum is the quantity that tracks both how much mass is moving and how fast. Writing the law this way makes it work even when mass itself is changing, not just velocity. That generality is the whole point: force is whatever changes motion, and momentum is the full measure of motion.

When mass is constant, dp/dt = d(mv)/dt = m(dv/dt) = ma, since the derivative only has to act on v. The m slides out front untouched. That is why F = ma feels like the whole story: in nearly every intro problem, mass is fixed, so the two forms agree perfectly. But F = dp/dt is the deeper statement, and it is the one that still works when mass is not fixed, like a rocket burning fuel or a raindrop growing as it falls through a cloud.

This sets up the rest of the unit. Impulse, J = integral of F dt = delta p, is just this same law integrated over time, and it is how you handle forces that vary or act briefly, like a bat hitting a ball. Momentum conservation in collisions follows from the same root idea: with no external force, dp/dt = 0 for the whole system, so total momentum cannot change even while it shuffles between colliding objects.

Key terms

Momentum (p)
The product of mass and velocity, p = mv, measuring an object's quantity of motion.
F = dp/dt
Newton's second law in its original form: force equals the time rate of change of momentum.
Impulse
The integral of force over time, equal to the change in momentum an object experiences.
Variable-mass system
A system like a rocket where mass itself changes over time, so F = ma no longer applies directly.

Watch it explained

  1. How it worksHow does F = dp/dt collapse down to F = ma whenever the mass stays constant?
  2. Worked exampleFind the force on an object whose momentum is p(t) = 4t2 kg m/s at the instant t = 2 s.
  3. Common mistakeWhy doesn't a rocket need something solid to push against in order to accelerate?
  4. Going furtherWhat changes about F = dp/dt for a rocket that is throwing away its own mass as it goes?

Check yourself

Your turn1 of 2
Find the force on an object whose momentum is p(t) = 7t2 kg m/s at the instant t = 4 s.

End of topic.