Lesson 7: Momentum & Collisions

Week 7 — The language of impacts

🎯 Learning Objectives

🎯 Intuition First: What Survives a Crash?

When two cars crash, energy is destroyed (crumpled metal, heat). But something else is always conserved: momentum. Even in the messiest collisions, the total momentum before equals the total after.

💡 Analogy:

Momentum is like money in a closed economy. Money can change form (cash → bank → stock), but the total amount never changes. In collisions, kinetic energy can "evaporate" as heat, but momentum is never lost.

📋 Momentum

Momentum: p = mv
Units: kg·m/s
Momentum is a VECTOR (direction matters!)

📋 Impulse

Impulse: J = FΔt = Δp
Units: N·s or kg·m/s (same as momentum)

Impulse is the change in momentum caused by a force acting over a time interval.

📝 Safety Connection:

Airbags and crumple zones increase the time of impact (Δt), which reduces the average force (F = Δp/Δt). Same momentum change, but spread over a longer time = gentler force on you!

📋 Conservation of Momentum

If Fnet,external = 0:   pinitial = pfinal
or: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f

📋 Types of Collisions

Collision TypeMomentum Conserved?KE Conserved?Example
Elastic✅ Yes✅ YesBilliard balls (nearly)
Inelastic✅ Yes❌ No (some lost)Car crash (some crumpling)
Perfectly Inelastic✅ Yes❌ No (maximum loss)Bullet into block (stick together)

📝 Worked Example: Inelastic Collision

A 1500-kg car moving at 20 m/s hits a stationary 1200-kg car. They lock together. What is their speed after?

📝 Step-by-Step Solution:

Conservation of momentum:

m₁v₁ + m₂v₂ = (m₁ + m₂)vf
(1500)(20) + (1200)(0) = (1500 + 1200)vf
30,000 = 2700vf
vf = 30,000/2700 = 11.1 m/s

✅ Speed after collision = 11.1 m/s

Note: Check KE: Initial = ½(1500)(20²) = 300,000 J. Final = ½(2700)(11.1²) = 166,500 J. About 133,500 J lost to crumpling and heat!

📝 Worked Example: Elastic Collision (1D)

A 2-kg ball moving at 4 m/s collides elastically with a 1-kg ball at rest. Find final velocities.

Two equations:

Momentum: m₁v₁ᵢ + m₂v₂ᵢ = m₁v₁f + m₂v₂f
KE: ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²

Shortcut for 1D elastic: v₁ᵢ − v₂ᵢ = −(v₁f − v₂f) (relative velocity reverses)

With m₁ = 2, m₂ = 1, v₁ᵢ = 4, v₂ᵢ = 0:

Momentum: 2(4) + 0 = 2v₁f + 1v₂f → 8 = 2v₁f + v₂f
Relative: 4 − 0 = −(v₁f − v₂f) → v₂f − v₁f = 4

Solving:

v₂f = v₁f + 4
8 = 2v₁f + (v₁f + 4) = 3v₁f + 4
v₁f = 4/3 = 1.33 m/s
v₂f = 1.33 + 4 = 5.33 m/s

✅ After collision: m₁ moves at 1.33 m/s, m₂ moves at 5.33 m/s

✏️ Practice Problems

Problem 1: A 0.5-kg ball moving at 10 m/s has momentum of:

Problem 2: A 50-kg skater throws a 2-kg ball at 15 m/s (from rest). What is the skater's recoil velocity?

Problem 3: In a perfectly inelastic collision, what fraction of initial KE is lost if equal masses collide head-on with equal and opposite velocities?

✅ Check Your Understanding

Q1: Can momentum be conserved in one direction but not another?

Click to reveal answer

Yes! Momentum is conserved only in directions where the net external force is zero. For example, a projectile's horizontal momentum is conserved (no horizontal forces), but vertical momentum is not (gravity acts downward).

Q2 (mini-problem): A 0.01-kg bullet traveling at 500 m/s embeds itself in a 2-kg block at rest. What is the speed of the combined system immediately after?

Click to reveal answer

Conservation of momentum: mbvb + mBvB = (mb + mB)vf

(0.01)(500) + 0 = (0.01 + 2)vf
5 = 2.01vf
vf = 5/2.01 = 2.49 m/s

Answer: 2.49 m/s