04 // Interactive Formula Compendium // PHYSICSJ = ∫ F(t) dt = F_avg · Δt = Δp = m(v_f − v_i)
Impulse-Momentum Theorem & Elastic/Inelastic Collisions Sheet
The Impulse-Momentum Theorem states that the impulse J = ∫ F(t)dt = F_avg Δt delivered to an object equals its vector change in linear momentum Δp = m v_f − m v_i (in N·s or kg·m/s).
Formula Reference
Main Formula
J = ∫ F(t) dt = F_avg · Δt = Δp = m(v_f − v_i)
Rearranged Forms
m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂ (Conservation of Linear Momentum)
e = (v₂ − v₁) / (u₁ − u₂) (Coefficient of Restitution, 0 ≤ e ≤ 1)
v_final = (m₁ u₁ + m₂ u₂) / (m₁ + m₂) (Perfectly Inelastic Collision, e = 0)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| F_avg | Average Impact Force (m·Δv / Δt) | N |
| m | Object Mass (m) | kg |
| dv | Velocity Change Magnitude |Δv| | m/s |
| dt | Contact Impact Duration (Δt) | s |
Inputs & Parameters
[kg]
[m/s]
[s]
Calculated Result (Average Impact Force (m·Δv / Δt))
F_avg =11,600N
Formula Notes & Background
- Write Newton's Second Law in momentum form: F(t) = dp / dt.
- Multiply by dt and integrate from t₁ to t₂: ∫_{t₁}^{t₂} F(t) dt = ∫_{p_i}^{p_f} dp = p_f − p_i.
- For an isolated two-body collision, Newton's Third Law F₁₂ = −F₂₁ implies equal and opposite impulses Δp₁ = −Δp₂, conserving total momentum p₁ + p₂.
Step-by-Step Examples
#1: A 0.145 kg baseball arriving at −38 m/s is batted back at +42 m/s (|Δv| = 80 m/s) during a 1.0 ms contact time.
m = 0.145 kgΔv = 42 − (−38) = 80 m/sΔt = 0.001 s
J = 0.145 × 80 = 11.6 N·s; F_avg = 11.6 / 0.001
→ Impulse J = 11.6 N·s, Average Force F_avg = 11,600 N (11.6 kN)
#2: How do airbags and crumple zones save lives in a Δv = 15 m/s crash for a 75 kg passenger?
Fixed impulse J = 75 × 15 = 1,125 N·s
Rigid dash (Δt = 0.01 s) ⇒ F = 112.5 kN; Airbag (Δt = 0.15 s) ⇒ F = 7.5 kN
→ Extending impact duration 15× reduces peak force on the body by 15×
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Frequently Asked Questions
Is momentum conserved even when kinetic energy is lost in a car crash?+
Yes! Total linear momentum is conserved in every isolated collision (elastic or inelastic) because internal contact forces obey Newton's Third Law.
What happens in a 1D head-on elastic collision (e = 1) between two identical billiard balls?+
When m₁ = m₂ and e = 1, the two balls completely exchange their velocities (v₁ = u₂ and v₂ = u₁).
Why must velocity signs (+ and −) be included when computing Δv = v_f − v_i?+
Because velocity and momentum are vectors, rebounding at the same speed v reverses direction, making Δv = v − (−v) = 2v rather than 0.