Momentum
What is conserved when objects interact — the 'quantity of motion.'
Momentum is the product of mass and velocity: p = mv. It is a vector quantity — direction = direction of velocity. SI unit: kg·m/s (equivalent to N·s). A large, slow object can have the same momentum as a small, fast one. Momentum quantifies how difficult it is to stop an object.
Impulse is the product of force and the time for which it acts: J = FΔt. The impulse-momentum theorem states that impulse equals change in momentum: FΔt = Δp = mv − mu. The area under a F–t graph equals the impulse. A large force over a short time (like a bat hitting a ball) delivers the same impulse as a small force over a longer time.
In a closed system (no net external force), the total momentum before an interaction equals the total momentum after: Σp_before = Σp_after. This applies to all collisions and explosions. Choose a positive direction first, then assign signs to velocities accordingly. Works in all directions independently.
Both momentum AND kinetic energy are conserved. Objects bounce off each other. Total KE before = Total KE after. This is an idealised case (perfectly elastic). In practice, truly elastic collisions occur between atomic/subatomic particles. At the macroscopic level, some energy is always lost.
Momentum is conserved but kinetic energy is NOT conserved — some KE is lost as heat, sound, or deformation. Perfectly inelastic: the two objects stick together after collision (maximum KE loss). To check: calculate total KE before and after. If they differ, the collision is inelastic.
An explosion is the reverse of a perfectly inelastic collision. One object splits into two or more parts. If the system starts at rest (total momentum = 0), the parts must fly off with equal and opposite momenta: m₁v₁ = −m₂v₂. Kinetic energy is gained (from chemical or other potential energy).
For a perfectly elastic head-on collision: v₁' = (m₁−m₂)/(m₁+m₂) × u₁ and v₂' = 2m₁/(m₁+m₂) × u₁ (assuming u₂ = 0). Special case — EQUAL masses (m₁ = m₂): v₁' = 0, v₂' = u₁. The first object stops completely and the second moves at the original speed. This is seen in Newton's cradle and billiard balls. Special case — very heavy target (m₂ >> m₁): v₁' ≈ −u₁ (ball bounces back), v₂' ≈ 0. Both formulas and these special cases may appear in exams.
Impulse J = FΔt = Δp. Since Δp must equal a fixed value (the required change in momentum), increasing Δt reduces F. Applications: Crumple zones in cars — extend collision time → lower peak force on passengers. Air bags — longer deceleration time → lower force on head. Catching a ball — move hands back to increase time → reduce pain. Bungee cords — stretch over longer time → reduce peak force. In all cases: same momentum change, but smaller force over longer time.
F–t Graph (Force vs. Time)
area: Area = impulse = Δp (kg·m/s or N·s)Area under the F–t graph = impulse = change in momentum. This is especially useful when the force varies over time (a non-rectangular shape). The impulse is the total area, calculated by integration or geometry.
p–t Graph (Momentum vs. Time)
slope: Slope = net force (N)The slope of a p–t graph equals the net force. A horizontal line means no net force (constant momentum). A steep slope means a large net force.
Newton's 2nd law in its full form is F = Δp/Δt. Momentum is the deeper version of F = ma — it works even when mass changes (e.g. rockets).
💡Exam tip: Use momentum conservation when time is unknown. Use F = ma when both force and time are given.
Click any formula to see symbol definitions.
All formulas for this topic are in BINAS BINAS 35A4. In the exam you don't need to memorise the equations — but you must know which table to open and what every symbol means.