1.2 Momentum, impulse, and collisions

Momentum is the quantity Newton’s second law is really about. This lesson develops its two most useful consequences: the impulse delivered by a brief force, and the conservation law that fixes the outcome of a collision.

Impulse: the integrated second law

Integrate $\mathbf F = d\mathbf p/dt$ over a finite time interval:

\int_{t_1}^{t_2} \mathbf{F}\, dt \;=\; \Delta\mathbf{p}.

The left side is the impulse. For brief, intense events — a hammer strike, a struck drum, a molecule rebounding off a wall — the detailed time-course of the force is unknown and irrelevant; only its time-integral matters, and that integral is the change in momentum.

impulse J1.200 N·s

Δp = J1.200 kg·m/s

Δv = J/m0.600 m/s

Peak force F_peak = 8.0 N Pulse duration τ = 0.30 s Mass m = 2.00 kg

The impulse is the time-integrated force, the shaded area under F(t). By Newton's second law it equals the change in momentum Δp. Doubling τ at fixed F_peak doubles Δp; halving m for fixed J doubles Δv. This is exactly the bookkeeping kinetic theory uses to compute pressure from molecular collisions.

Slide the pulse amplitude $F_\text{peak}$ and duration $\tau$ . The shaded area under $F(t)$ is the impulse $J = \tfrac12 F_\text{peak}\,\tau$ , and the resulting velocity change is $\Delta v = J/m$ . A small force applied for a long time and a large force applied briefly deliver the same momentum if their areas match.

Collisions and the energy ledger

In a collision with no external force, total momentum is conserved:

m_1 v_1^i + m_2 v_2^i \;=\; m_1 v_1^f + m_2 v_2^f.

Kinetic energy, by contrast, need not be. Whether it is conserved is set by a single number, the coefficient of restitution $e$ , which fixes the relative velocity after the collision in terms of the relative velocity before:

v_2^f - v_1^f \;=\; -e\,(v_2^i - v_1^i).

Together with momentum conservation, this determines both final velocities completely. The extreme cases bound the behaviour: $e = 1$ is elastic — kinetic energy is conserved; $e = 0$ is perfectly inelastic — the bodies coalesce and the maximum allowed kinetic energy is lost to heat and deformation. Real macroscopic collisions sit between. The idealised molecular collisions of kinetic theory are taken to be perfectly elastic, $e = 1$ , which is what lets a gas conserve its kinetic energy indefinitely.

v₁ᶠ-2.333

v₂ᶠ1.667

p (before / after)1.00 / 1.00

K (before / after)5.50 / 5.50

m₁ = 1.0 kg m₂ = 2.0 kg v₁ⁱ = 3.00 m/s v₂ⁱ = -1.00 m/s Restitution e = 1.00 (elastic)

Momentum is conserved for any restitution (no external force in the horizontal direction). Kinetic energy is conserved only when e = 1 (elastic). The lost K at e < 1 goes into heat, deformation, or sound — bookkeeping for the work-energy theorem in dissipative settings.

Set the two masses, their initial velocities, and the restitution. The momentum readout is conserved at every $e$ ; the kinetic-energy readout shows the fraction lost to heat and deformation when $e < 1$ .

The centre-of-mass frame

A collision is simplest in the frame moving with the centre of mass, where the total momentum is zero. There the two bodies approach with equal and opposite momenta, and an elastic collision simply reverses each velocity; an inelastic one scales it down by $e$ . Transforming back to the laboratory frame then adds the centre-of-mass velocity to both. This split — internal collision in the centre-of-mass frame, plus the uniform drift of the centre of mass — is the cleanest way to see that the centre-of-mass motion is untouched by the collision, exactly as the previous lesson’s momentum theorem requires.