Physics for ME · Module 7 of 16
Momentum, Impulse, and Collisions
During an impact, forces are huge and unknown. Momentum sidesteps them: what goes in must come out, however violent the middle.
Readiness check
From Modules 4 and 6. Tick only what you can do closed-notes.
- Apply F = ma and read its time-integral meaning.
- Compute kinetic energies quickly.
- Work with signed velocities along one axis.
- Distinguish internal from external forces on a system.
- Evaluate ∫F dt conceptually (Math Module 5).
The core idea
If no external force acts, total momentum cannot change. Collisions become bookkeeping.
p = mvJ = ∫F dt = ΔpImpulse is force accumulated over time; it equals the momentum change. In a collision the contact forces are internal to the pair, so the pair's total momentum is untouched. Energy, by contrast, is free to leak into deformation: that difference classifies collisions.
The skills, taught in order
Momentum is the law that survives the chaos of an impact, when forces are huge, brief, and unknown. These seven skills let you read a collision from its before and after alone.
7.1 Momentum
Momentum is mass in motion, p = mv, a vector pointing the way the body moves, in units of kg·m/s. A system's total momentum is the vector sum of every part's mv.
7.2 Impulse and the impulse-momentum theorem
Impulse is force accumulated over time, J = ∫F dt (or FavgΔt), in units of N·s. The impulse-momentum theorem says it equals the change in momentum: J = Δp = mv₂ − mv₁. It is Newton's second law written for an interval rather than an instant.
Micro-example. A 0.05 kg ball hitting a wall at 10 m/s and rebounding at 8 m/s changes momentum by 0.05(10 + 8) = 0.9 N·s, the impulse the wall delivered.
7.3 Conservation of momentum
The forces two bodies exert on each other are equal and opposite (Newton's third law), so they cancel inside the system. If no net external force acts, the total momentum cannot change. Choose the system so the violent forces are internal and the brief impact's external impulses are negligible.
7.4 Perfectly inelastic collisions
The bodies lock together and move as one: m₁v₁ + m₂v₂ = (m₁ + m₂)v. Momentum is conserved, but the maximum possible kinetic energy is lost to deformation and heat. Worked Example 1 is exactly this case.
7.5 Elastic collisions
An idealised elastic collision conserves kinetic energy as well as momentum, giving two equations. A useful consequence: the relative speed of approach equals the relative speed of separation. Worked Example 2 uses both laws together.
7.6 Collisions in two dimensions
Momentum is conserved separately along each axis. A 2D collision is two 1D momentum balances sharing the same masses, exactly as projectile motion split into independent axes in Module 3.
7.7 Impulse management
In an unavoidable impact the momentum change Δp is fixed, so the only control is time: Favg = Δp/Δt. Stretching the stop over a longer time lowers the peak force, which is the whole job of crumple zones, air bags, helmets, and packaging foam.
Micro-example. Doubling an impact's duration halves the average force for the same change in momentum.
| Collision type | Momentum | Kinetic energy | Outcome |
|---|---|---|---|
| Elastic | conserved | conserved | bodies bounce apart |
| Inelastic | conserved | partly lost | some deformation |
| Perfectly inelastic | conserved | maximum lost | bodies move off together |
Engineering connection: crashworthiness, impact tooling, and jet and rocket thrust (which is momentum flowing out of the nozzle). The rotational twin, angular momentum, arrives in Module 9.
Worked example 1: the rear-end collision
A 1500 kg car at 20 m/s rear-ends a 1000 kg car moving at 10 m/s in the same direction; the bumpers lock. Find the common speed after impact, the kinetic energy lost, and the average force on the lighter car if the impact lasts 0.15 s.
- ProblemFind the post-impact speed, energy loss, and average impact force in Figure 1.
- Given / findm₁ = 1500 kg at 20 m/s; m₂ = 1000 kg at 10 m/s; lock together; impact 0.15 s.
- AssumptionsStraight-line impact; road friction negligible during the brief contact; perfectly inelastic.
- ModelSystem = both cars: the crash forces are internal, so momentum is conserved. Energy is not.
- Equationsm₁v₁ + m₂v₂ = (m₁ + m₂)v J = Δp = FavgΔt
- Solvev = (30 000 + 10 000)/2500 = 16 m/s. KE before = 300 + 50 = 350 kJ; after = ½ × 2500 × 256 = 320 kJ: 30 kJ lost to deformation. Impulse on car 2: Δp = 1000 × (16 − 10) = 6000 N·s, so Favg = 6000/0.15 = 40 kN.
- Checkv lies between 10 and 20, nearer the heavier car. Newton's third law check: car 1 loses Δp = 1500 × (20 − 16) = 6000 N·s, exactly what car 2 gained.
- ConclusionMomentum found the outcome without ever knowing the crash force history; impulse then exposed the 40 kN average that the structures and occupants must survive. Doubling the impact time (softer crumple) would halve that force: crashworthiness in one line.
Worked example 2: an elastic collision
A 0.20 kg steel ball moving at 3.0 m/s strikes a stationary 0.30 kg steel ball head-on. The hardened balls collide almost elastically. Find each ball's velocity afterward, and confirm that both momentum and kinetic energy are conserved.
- ProblemFind the two velocities after the elastic head-on collision in Figure 2.
- Given / findm₁ = 0.20 kg at v₁ = 3.0 m/s; m₂ = 0.30 kg at rest; elastic. Find v₁′ and v₂′.
- AssumptionsStraight-line head-on impact, no friction, perfectly elastic so kinetic energy is conserved.
- ModelTwo conservation laws at once: momentum, plus the elastic rule that relative speed is unchanged (which replaces the kinetic-energy equation).
- Equationsm₁v₁ = m₁v₁′ + m₂v₂′ v₁ − v₂ = −(v₁′ − v₂′)
- SolveWith v₂ = 0 these give v₁′ = ((m₁ − m₂)/(m₁ + m₂))v₁ = (−0.1/0.5)(3.0) = −0.6 m/s and v₂′ = (2m₁/(m₁ + m₂))v₁ = (0.4/0.5)(3.0) = 2.4 m/s.
- CheckMomentum: 0.20(3.0) = 0.60 kg·m/s before; 0.20(−0.6) + 0.30(2.4) = 0.60 after. Kinetic energy: ½(0.2)(3.0²) = 0.90 J before; ½(0.2)(0.6²) + ½(0.3)(2.4²) = 0.90 J after. Both hold.
- ConclusionThe lighter ball rebounds off the heavier one while the heavier moves away more slowly: the same reason a ball bounces back off a wall but a bowling ball barely notices a pin. Elastic collisions waste no energy, which is why hardened bearings and billiard balls approximate them.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Kinetic energy conserved in every collision | Audits that balance through a crunch | "Did anything permanently deform, heat, or make noise?" | Only idealized elastic collisions keep KE. Momentum is the law that always holds. |
| Momentum treated as a scalar | Head-on momenta added instead of subtracted | "What are the signs of the velocities?" | Momentum is a vector: declare a positive direction and keep the signs. |
| System chosen so the big force is external | "Momentum not conserved" in a simple impact | "Is the violent force inside my system boundary?" | Put both colliding bodies inside; then their forces are internal and cancel. |
| Impulse confused with force | N·s reported in newtons | "Amount of push, or push times time?" | Impulse is F·Δt with units N·s; the same impulse can be gentle-and-long or brutal-and-short. |
Practice ladder
A 0.5 kg hammer head moving at 8 m/s stops in 0.002 s against a nail. Find the impulse and the average force.
Show answer
J = 0.5 × 8 = 4 N·s; F = 4/0.002 = 2000 N. A small mass, briefly stopped, beats a large steady push: that is why hammers exist.
A 75 kg skater at rest throws a 5 kg toolbox forward at 6 m/s. Find the skater's recoil speed and name the principle.
Show answer
Total momentum stays zero: 75v = 5 × 6, v = 0.4 m/s backward. Conservation with an internal launch: the rocket principle in miniature.
In the worked example, suppose the cars do not lock but the rear car slows to 14 m/s. Find the front car's new speed and check whether this collision was elastic.
Show answer
Momentum: 1500(20) + 1000(10) = 1500(14) + 1000v₂, so v₂ = 19 m/s. KE after = ½(1500)(196) + ½(1000)(361) = 147 + 180.5 = 327.5 kJ < 350 kJ: still inelastic, 22.5 kJ lost. Real bumpers always sit between the extremes.
Analyze one impulse-management device (bike helmet, crash barrier, machine bumper, packaging foam): estimate the momentum change it must absorb and the time it stretches the impact over, and compute the force reduction versus a rigid stop.
What good work looks like
A momentum estimate with stated mass and speed, two impact durations (rigid versus cushioned), the two average forces compared, and one sentence on the design implication.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Film a simple collision (two rolling bottles, marbles, toy cars) with a phone. Extract speeds before and after from frames, test momentum conservation numerically, and report the kinetic-energy fate.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Define momentum and impulse, with units.
p = mv (kg·m/s); impulse J = ∫F dt (N·s); and J = Δp.
2. State exactly when momentum is conserved.
When the net external impulse on the chosen system is zero (or negligible over the interval): internal forces cancel by the third law.
3. Elastic versus perfectly inelastic: what distinguishes them?
Elastic keeps kinetic energy; perfectly inelastic loses the most possible, with the bodies locking together. Momentum holds in both.
4. Why do crumple zones and helmets work, in impulse language?
The momentum change is fixed; stretching Δt shrinks the average force F = Δp/Δt.
5. How does a rocket accelerate with nothing to push on?
It throws mass backward; conservation hands the body equal forward momentum. Thrust is momentum flow.