Momentum & impulse
The idea
Mass in motion carries momentum, defined as p = mv — a vector pointing along the velocity, measured in kg·m/s. To change an object's momentum you must apply a force for some time; that product, force × time, is called impulse, and the impulse-momentum theorem says impulse equals the change in momentum: FΔt = Δp. This is really Newton's second law rearranged, since F = ma can be rewritten as F = Δ(mv)/Δt.
The power of the impulse picture is the trade-off it reveals: the same momentum change can come from a large force over a short time or a small force over a long time. That is why airbags, crumple zones, and bent knees on landing all work — they stretch out Δt, so the force needed to produce the same Δp shrinks. Because momentum is a vector, sign discipline matters: reversing direction means the momentum change is larger than either momentum alone.
A common misconception is that a heavy object always has more momentum than a light one. Momentum is the product of mass AND velocity: a 0.045 kg golf ball leaving the tee at 70 m/s carries more momentum than a 2.0 kg textbook drifting at 1 m/s. Always multiply before comparing.
Worked example
A 0.145 kg baseball arrives at a bat moving at 40 m/s toward the batter and leaves at 50 m/s straight back toward the pitcher. Bat and ball are in contact for 0.0012 s. Find the impulse on the ball and the average force the bat exerts.
- Set a positive direction first — call toward the pitcher positive. The incoming velocity is then −40 m/s and the outgoing velocity is +50 m/s.
- Compute the two momenta: initial p = 0.145 × (−40) = −5.8 kg·m/s; final p = 0.145 × 50 = +7.25 kg·m/s.
- The impulse is the change: Δp = 7.25 − (−5.8) = 13.05 kg·m/s toward the pitcher. Note it exceeds either momentum alone because the direction reversed.
- Divide by the contact time to find the average force: F = Δp/Δt = 13.05/0.0012 ≈ 10900 N, or about 1.1 × 10⁴ N toward the pitcher.
- Sanity-check the scale: that force is roughly 7700 times the ball's own weight (0.145 × 9.8 ≈ 1.4 N) — enormous, but applied for barely a millisecond, which is exactly the short-time, large-force end of the impulse trade-off.
Answer. The bat delivers an impulse of about 13.1 kg·m/s, requiring an average force of about 1.1 × 10⁴ N directed toward the pitcher.
Check your understanding
- Why does the momentum change exceed both the initial and final momentum when an object bounces straight back?
- How do airbags and crumple zones use the impulse-momentum theorem to protect passengers in a crash?
- What sign errors are easiest to make in a rebound problem, and what habit prevents them?
- If a ball is caught instead of hit back at the same speed, how does the impulse compare, and what does that imply for the catcher's hands?
Build the foundations first
Momentum & impulse builds on these concepts. If any feel shaky, start there.