5.3 Momentum, Conservation of Momentum & Impulse

Momentum, Conservation of Momentum & Impulse

Newton’s second law can be written in terms of momentum. That form is ideal for collisions, landings, and any situation where a large force acts for a short time. Module 2 expects p = m v, conservation of momentum, and impulse = change in momentum, with clear SI units and landing-gear intuition.

Linear Momentum

Linear momentum of a particle or rigid body treated as a particle is:

p = m v

• m = mass (kg)
• v = velocity (m/s) — a vector, so p is a vector in the same direction as v
• Unit: kg·m/s, which is identical to N·s (because 1 N·s = 1 kg·m/s² × s)

A heavy slow object can have the same momentum as a light fast object. Example: 2 000 kg at 10 m/s has p = 20 000 kg·m/s; 500 kg at 40 m/s has the same magnitude of momentum.

Worked example. Aircraft mass 15 000 kg, groundspeed 60 m/s along the runway. Momentum magnitude p = 15 000 × 60 = 900 000 kg·m/s in the direction of motion. To stop, the total impulse from brakes, reverse thrust, and drag must equal −900 000 N·s (opposite direction).

Momentum is more complete than velocity alone when comparing “how hard it is to stop” different masses: stopping force and time link to Δp, not to speed only.

Newton’s Second Law in Momentum Form

Originally, Newton stated that force is proportional to the rate of change of momentum:

F_net = dp/dt

If mass is constant, dp/dt = m dv/dt = m a, recovering F = m a. If mass changes (conveyor loading bulk cargo, or idealised variable-mass systems), the momentum form is more fundamental. For Module 2, constant-mass F = ma and impulse–momentum are the working tools.

Conservation of Momentum

If the net external force on a system is zero (or negligible during a brief event), then total momentum of the system is conserved:

Σ p_initial = Σ p_final

or m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ (one dimension, two bodies).

Internal forces (action–reaction pairs between parts of the system) cancel in the total momentum balance by the third law. External forces such as friction with the ground or gravity may break perfect conservation for the subsystem you chose — always define the system carefully.

Collisions (Conceptual)

• Elastic collision: total kinetic energy conserved as well as momentum (hard smooth ideal spheres).
• Inelastic collision: momentum conserved (if no external force), but some KE converts to heat and deformation.
• Perfectly inelastic: bodies stick together; maximum KE loss consistent with momentum conservation.

Worked example — perfectly inelastic. A 1 000 kg trolley moving at 4 m/s couples to a stationary 1 500 kg trolley on a level, low-friction track. After coupling, common speed v:

(1 000 × 4) + (1 500 × 0) = (1 000 + 1 500) v
4 000 = 2 500 v
v = 1.6 m/s in the original direction

Initial KE = ½ × 1 000 × 16 = 8 000 J; final KE = ½ × 2 500 × 2.56 = 3 200 J. Momentum conserved; KE not conserved (inelastic).

Worked example — recoil (one dimension). A gun/aircraft gun or a simple model: mass m_g fires a projectile mass m_p at speed v_p relative to ground; system initially at rest. If external horizontal forces are neglected during the brief firing:

0 = m_g v_g + m_p v_p ⇒ v_g = − (m_p / m_g) v_p

The gun recoils opposite to the projectile. Propeller and jet propulsion are continuous versions of “throwing mass aft to gain forward momentum.”

Impulse

Impulse of a constant force over a time interval Δt is:

J = F Δt

For a varying force, impulse is the integral (area under the F–t graph). The impulse–momentum theorem states:

J = F_avg Δt = Δp = m v_final − m v_initial

This is simply F = ma rewritten with a = Δv/Δt: F Δt = m Δv.

Implications:

• Same change in momentum can come from a large force for a short time or a small force for a long time.
• Peak structural loads during impact scale with how short Δt is.
• Unit of impulse: N·s (= kg·m/s).

Worked example — impulse. A net force of 5 000 N acts for 0.40 s on a 1 250 kg mass initially at rest.

Impulse J = 5 000 × 0.40 = 2 000 N·s
Δp = 2 000 kg·m/s
v = Δp / m = 2 000 / 1 250 = 1.6 m/s

Worked example — stopping force. Aircraft mass 20 000 kg slows from 50 m/s to 0. Required Δp = 20 000 × (0 − 50) = −1 000 000 kg·m/s. If stopping time is 25 s, average net force F_avg = Δp / Δt = −1 000 000 / 25 = −40 000 N (40 kN opposite to motion). If the same stop were forced in 5 s, average force would be −200 kN — five times larger.

Landing Gear and Impact Intuition

Landing gear (and crashworthy seats, cargo nets, and fuselage structure) are designed with impulse–momentum in mind.

Oleo-pneumatic struts

On touchdown, the vertical velocity of the airframe must be brought toward zero. The change in vertical momentum is roughly m × v_sink (for a simple one-mass model). The average force on the gear is Δp / Δt. An oleo strut lengthens the time Δt over which the momentum changes by allowing controlled stroke: oil is forced through orifices as the strut compresses, and gas spring stores and returns energy. Longer stroke and longer Δt → lower peak force for the same sink-rate momentum change.

Without shock absorption (rigid gear), Δt is very small, peak force is huge, and structure or runway is damaged — the dynamical reason “don’t land with locked struts / inadequate damping.”

Soft versus hard landing numbers (order of magnitude)

Suppose effective mass on one main gear is 8 000 kg and vertical speed at contact is 3 m/s. Δp ≈ 8 000 × 3 = 24 000 kg·m/s (upward impulse needed).

• If deceleration distance/time corresponds to Δt ≈ 0.20 s: F_avg ≈ 24 000 / 0.20 = 120 000 N (plus weight management in a full analysis).
• If Δt ≈ 0.05 s (much stiffer): F_avg ≈ 480 000 N — four times the force.

Real multi-body gear models include tyre spring, oleo, and airframe flexibility, but the Module 2 message is robust: increase Δt to reduce peak F for a given Δp.

Horizontal landing and braking

Spin-up of wheels on touchdown is another impulse problem: the tyre must accelerate angularly; friction impulse from the runway provides the wheel’s angular momentum gain and a brief drag spike on the airframe (wheel spin-up drag). Braking later applies a sustained friction force; impulse of braking force over the ground roll equals the change in aircraft horizontal momentum.

Cargo and passenger restraint

During a crash or hard stop, occupants and cargo have large forward momentum. Restraints apply an impulse that reduces that momentum to match the decelerating cabin. Webbing that stretches slightly increases Δt and can lower peak loads compared with a perfectly rigid stop — again impulse–momentum, not magic.

System Selection Tips for Exams

1. Draw the system boundary.
2. Ask: are external impulses negligible during the short event? If yes, total p conserved.
3. If external forces matter (brakes over many seconds, sustained thrust), use impulse = ∫F dt = Δp with those forces included, or use F = ma.
4. Keep directions: assign + and − along a line; momentum and impulse are signed in 1D.
5. Do not confuse energy conservation with momentum conservation — inelastic collisions conserve momentum (no external force) but not mechanical KE.

Formula Checklist

• p = m v (vector)
• Conservation: Σ m u = Σ m v when ΣF_ext ≈ 0
• J = F Δt = Δp = m v − m u
• Longer impact time → smaller average force for same Δp
• Units: kg·m/s or N·s

Momentum and impulse connect Newton’s laws to every short-duration load event on the aircraft — from tyre spin-up to hard landings and emergency stops.