Dynamics · Chapter 5 of 10 · Intermediate
Impulse, Momentum, and Impact
When a force acts over a short time, integrate it over time instead of position. Momentum is the natural currency of collisions, and it is conserved when nothing external pushes.
Readiness check
This is the third way to use ΣF = ma, integrated over time. Tick only what you can do closed-notes.
- Write ΣF = ma and identify external forces.
- Integrate a force over time.
- Track vector signs along a line.
- Compute ½mv² for kinetic energy.
- Recognise an isolated system.
The core idea
The impulse of the net force over a time interval equals the change in momentum; with no external impulse, total momentum is conserved.
∫F dt = mv₂ − mv₁Σmv = constante = (v₂′ − v₁′)/(v₁ − v₂)Integrating ΣF = ma over time gives the impulse-momentum principle. It is the method of choice when forces act for a known time, when forces are large and brief (impacts), or when interest is in velocity changes rather than positions. For collisions, momentum is conserved and the coefficient of restitution supplies the second equation.
The skills, taught in order
This chapter completes the trio of solution methods. Five skills cover linear and angular momentum, conservation, impact, and when to reach for each method.
5.1 Linear impulse and momentum
Momentum is G = mv. Integrating the equation of motion over time gives ∫F dt = mv₂ − mv₁: the impulse equals the change in momentum. The impulse is the area under the force-time curve, so a known average force over a known time changes momentum by Favg·Δt.
5.2 Conservation of linear momentum
If the net external impulse on a system is zero, its total momentum does not change: Σmv before equals Σmv after. Internal forces (the two bodies pushing on each other) cancel in pairs, so collisions and explosions conserve momentum even when energy is lost.
5.3 Angular impulse and momentum
About a point O, angular momentum is HO = r × mv, and the angular impulse of the moments equals its change: ∫ΣMO dt = ΔHO. When the net moment about O is zero, angular momentum is conserved, the principle behind a spinning skater pulling in their arms.
5.4 Impact and the coefficient of restitution
In a direct central impact, momentum is conserved, but kinetic energy generally is not. The coefficient of restitution e relates the relative separation speed to the relative approach speed, giving the needed second equation.
| Coefficient e | Impact type | Kinetic energy |
|---|---|---|
| e = 1 | perfectly elastic | conserved |
| 0 < e < 1 | real (partially elastic) | partly lost |
| e = 0 | perfectly plastic | maximum loss; bodies stick |
5.5 Choosing among the three methods
Chapters 3, 4, and 5 give three ways to use Newton's law. Picking the right one is half the skill.
| Method | Key equation | Best when you want |
|---|---|---|
| Force (Ch 3) | ΣF = ma | acceleration or force at an instant |
| Work-energy (Ch 4) | U = ΔT | speed as a function of distance |
| Impulse-momentum (Ch 5) | ∫F dt = Δ(mv) | velocity over time, or a collision |
Engineering connection: crash safety and airbags, pile driving, jet and rocket thrust, and the analysis of any collision or short, intense force.
Worked example 1: the force of a struck baseball
A 0.15 kg baseball arrives at 40 m/s and leaves at 50 m/s in the opposite direction after a bat contact lasting 0.005 s. Find the impulse and the average force on the ball.
- ProblemFind the impulse and average force on the ball in Figure 1.
- Given / findm = 0.15 kg, v₁ = −40 m/s (incoming), v₂ = +50 m/s (outgoing), Δt = 0.005 s. Find impulse and Favg.
- AssumptionsStraight-line (one-dimensional) impact; gravity negligible over 5 ms; take the outgoing direction as positive.
- ModelApply the linear impulse-momentum principle along the line of motion.
- Equationsimpulse = mv₂ − mv₁ Favg = impulse/Δt
- Solveimpulse = 0.15(50 − (−40)) = 0.15 × 90 = 13.5 kg·m/s. Favg = 13.5/0.005 = 2700 N.
- Check2700 N on a 1.5 N ball is about 1800 times its weight, which is why gravity was safely ignored during contact. The reversal (not just slowing) is what makes the velocity change 90, not 10, m/s.
- ConclusionBrief impacts generate enormous forces. The same impulse over a longer contact (a softer glove or a longer bat follow-through) would lower the peak force, the logic behind all cushioning.
Worked example 2: a partially elastic collision
A 2 kg puck moving at 6 m/s strikes a 3 kg puck moving toward it at 2 m/s on a frictionless surface. With a coefficient of restitution e = 0.7, find the velocity of each puck after impact.
- ProblemFind both post-impact velocities for the pucks in Figure 2.
- Given / findm₁ = 2 kg, u₁ = +6 m/s; m₂ = 3 kg, u₂ = −2 m/s; e = 0.7. Find v₁ and v₂.
- AssumptionsDirect central impact on a frictionless line; no external impulse, so momentum is conserved.
- ModelTwo equations: conservation of momentum and the restitution relation, solved together.
- Equationsm₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ v₂ − v₁ = e(u₁ − u₂)
- SolveMomentum: 2(6) + 3(−2) = 6 = 2v₁ + 3v₂. Restitution: v₂ − v₁ = 0.7(6 − (−2)) = 5.6. Solving, v₁ = −2.16 m/s (rebounds backward) and v₂ = +3.44 m/s (driven forward).
- CheckMomentum after = 2(−2.16) + 3(3.44) = 6.0, matching before. Kinetic energy fell from 42 J to 22.4 J; the 19.6 J lost is consistent with e < 1, an inelastic impact.
- ConclusionMomentum gives one equation, restitution the other. Setting e = 1 would conserve energy (elastic); e = 0 would make the pucks move together. The lost energy went to deformation, heat, and sound.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Assuming energy is conserved in impact | Restitution ignored; wrong velocities | "Is the impact elastic (e = 1)?" | Only e = 1 conserves energy; otherwise use momentum plus restitution. |
| Dropping a sign | Approaching speeds added wrongly | "Did I assign a positive direction?" | Pick one positive direction and apply it to every velocity. |
| Ignoring an external impulse | Momentum "conserved" against a wall | "Does something outside exert an impulse?" | A wall or ground adds external impulse; momentum is not conserved then. |
| Confusing impulse and force | Units of N used for impulse | "Is this N or N·s?" | Impulse is force times time (N·s = kg·m/s); divide by Δt for force. |
Practice ladder
A constant 200 N force acts on a 5 kg cart at rest for 3 s. Find its final speed.
Show answer
Impulse = FΔt = 200 × 3 = 600 N·s = Δ(mv) = 5v, so v = 120 m/s. Impulse-momentum gives velocity over time directly, with no need for acceleration.
For the Worked Example 1 ball, how much would the average force drop if a softer mitt doubled the contact time to 0.010 s?
Show answer
The impulse is fixed at 13.5 N·s by the velocity change, so Favg = 13.5/0.010 = 1350 N, exactly half. Doubling the contact time halves the peak force, which is how padding protects.
A 50 kg skater throws a 4 kg ball at 8 m/s while standing at rest on frictionless ice. Find the skater's recoil speed.
Show answer
Momentum is conserved from zero: 0 = 50v + 4(8), so v = −0.64 m/s. The skater glides backward at 0.64 m/s, a small recoil because they are much heavier than the ball.
Choose a real impact or thrust event (a car crash test, a pile driver, a water jet). Set up impulse-momentum or conservation, estimate a force or velocity, and state where energy goes if the impact is inelastic.
What good work looks like
A clear before-and-after momentum statement, the correct second equation (restitution or a constraint), a positive direction defined, and an energy-loss note for inelastic cases.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Analyse one real collision or thrust event with impulse-momentum, give the before-and-after momentum, and quantify the kinetic energy lost if it is inelastic.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the linear impulse-momentum principle.
∫F dt = mv₂ − mv₁: the impulse equals the change in momentum.
2. When is momentum conserved?
When the net external impulse on the system is zero.
3. Define the coefficient of restitution.
e = (relative separation speed)/(relative approach speed); 1 is elastic, 0 is plastic.
4. Which quantity is conserved in every collision, energy or momentum?
Momentum; kinetic energy is conserved only when e = 1.
5. When do you choose impulse-momentum over the other two methods?
For forces acting over time, brief impacts, and collisions, when velocity changes matter.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Meriam and Kraige, Engineering Mechanics: Dynamics (7th ed), Chapter 3, Sections C and D (Impulse-Momentum, Impact) |
| Cross-reference | Hibbeler, Dynamics, Ch. 15 · Beer and Johnston, Ch. 13 |
| Core topics | 5.1 Linear impulse-momentum · 5.2 Conservation · 5.3 Angular impulse-momentum · 5.4 Impact and restitution · 5.5 Choosing a method |
| Engineering connection | Crash safety, pile driving, jet and rocket thrust, and collision analysis. |
| Read next | Chapter 6: Systems of Particles and Variable Mass. |