💎 Elastic Collisions

Part 1 of 7 — Collisions

In an elastic collision, both momentum AND kinetic energy are conserved. These are special collisions where no energy is converted to heat, sound, or deformation. While perfectly elastic collisions are an idealization, collisions between hard objects like billiard balls and atomic/molecular collisions come very close.

What Makes a Collision Elastic?

An elastic collision satisfies TWO conservation laws simultaneously:

1. Conservation of Momentum (always)

$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

2. Conservation of Kinetic Energy (elastic only)

$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

Comparison of Collision Types

Special Cases of Elastic Collisions

Case 1: Equal Masses ($m_1 = m_2$), One at Rest

The moving object stops and the stationary object moves with the original velocity:

$v_{1f} = 0, \quad v_{2f} = v_{1i}$

This is seen clearly in Newton's cradle!

Case 2: Heavy Hits Light ($m_1 \gg m_2$), Light at Rest

The heavy object barely changes speed; the light object flies off at nearly $2v_{1i}$:

$v_{1f} \approx v_{1i}, \quad v_{2f} \approx 2v_{1i}$

Case 3: Light Hits Heavy ($m_1 \ll m_2$), Heavy at Rest

The light object bounces back at nearly its original speed; the heavy object barely moves:

$v_{1f} \approx -v_{1i}, \quad v_{2f} \approx 0$

Think of a tennis ball bouncing off a bowling ball.

Real-World Examples

Nearly Elastic

• Billiard balls — very hard, minimal deformation
• Atomic collisions — ideal gas molecules
• Newton's cradle — steel balls transfer energy with minimal loss
• Superball bouncing — coefficient of restitution near 1

NOT Elastic

• Car crashes — significant deformation (perfectly inelastic if cars lock together)
• Ball of clay — sticks on impact
• Football tackle — players move together
• Meteor impact — enormous energy converted to heat and crater formation

How to Tell

If you can calculate $KE_i$ and $KE_f$ and they're equal → elastic. If $KE_f < KE_i$ → inelastic.

Concept Check — Elastic Collisions 🎯

Elastic Collision Checks 🧮

A 2 kg ball at +6 m/s hits a 2 kg ball at rest. After collision: Ball 1 stops, Ball 2 moves at +6 m/s.

1. What is $KE_i$? (in J)

2. What is $KE_f$? (in J)

3. Is this collision elastic? (type "yes" or "no")

Elastic Collision Properties 🔍

Exit Quiz — Elastic Collisions ✅

🧲 Perfectly Inelastic Collisions

Part 2 of 7 — Collisions

A perfectly inelastic collision is one where the colliding objects stick together and move as a single unit afterward. These are the simplest collision problems to solve because there's only one unknown (the final velocity), but they also represent the maximum possible kinetic energy loss.

The Physics

In a perfectly inelastic collision:

$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f$

Solving for $v_f$:

$v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}$

Properties

Kinetic Energy Loss

The kinetic energy lost in a perfectly inelastic collision:

$\Delta KE = KE_i - KE_f$

Special Case: Target at Rest

For $m_1$ hitting stationary $m_2$:

$v_f = \frac{m_1}{m_1 + m_2} v_{1i}$

$KE_f = \frac{1}{2}(m_1 + m_2)v_f^2 = \frac{m_1^2}{2(m_1 + m_2)} v_{1i}^2$

$\text{Fraction retained} = \frac{KE_f}{KE_i} = \frac{m_1}{m_1 + m_2}$

$\text{Fraction lost} = \frac{m_2}{m_1 + m_2}$

Example

A 2 kg ball hits a 8 kg ball at rest:

• Fraction of KE retained: $2/(2+8) = 0.20$ or 20%
• Fraction of KE lost: $8/(2+8) = 0.80$ or 80% lost!

Maximum Loss

When $m_2 \gg m_1$: nearly ALL KE is lost (object stops).

When $m_1 = m_2$: exactly 50% of KE is lost.

Head-On Perfectly Inelastic Collisions

When objects move toward each other:

$v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}$

The sign of $v_f$ tells you which direction the combined mass moves.

Example

Car A (1200 kg, +20 m/s) hits Car B (800 kg, −15 m/s) head-on:

$v_f = \frac{(1200)(20) + (800)(-15)}{1200 + 800} = \frac{24000 - 12000}{2000} = \frac{12000}{2000} = +6 \text{ m/s}$

The wreckage moves in the direction of Car A (the heavier/faster car).

Energy Lost

$KE_i = \frac{1}{2}(1200)(400) + \frac{1}{2}(800)(225) = 240{,}000 + 90{,}000 = 330{,}000 \text{ J}$

$KE_f = \frac{1}{2}(2000)(36) = 36{,}000 \text{ J}$

$\text{Lost} = 330{,}000 - 36{,}000 = 294{,}000 \text{ J} \approx 89\% \text{ of initial KE}$

Concept Check — Perfectly Inelastic Collisions 🎯

Perfectly Inelastic Collision Practice 🧮

1. A 3 kg cart at +8 m/s collides with a 5 kg cart at rest and sticks. What is the final velocity? (in m/s)

2. What is the kinetic energy lost in this collision? (in J)

3. Two identical 4 kg balls moving toward each other at +6 m/s and −2 m/s collide and stick. What is the final velocity? (in m/s)

Perfectly Inelastic Collision Concepts 🔍

Exit Quiz — Perfectly Inelastic ✅

🔄 Inelastic Collisions — Between the Extremes

Part 3 of 7 — Collisions

Most real-world collisions are neither perfectly elastic nor perfectly inelastic — they're somewhere in between. In a general inelastic collision, momentum is conserved but kinetic energy is NOT conserved. The objects separate after collision but with less total KE than before.

The Collision Spectrum

The Key Insight

Momentum is ALWAYS conserved (if no net external force), regardless of collision type. The distinction between collision types is about kinetic energy.

Coefficient of Restitution

The coefficient of restitution ($e$) quantifies how "bouncy" a collision is:

$e = \frac{|v_{2f} - v_{1f}|}{|v_{1i} - v_{2i}|} = \frac{\text{relative speed after}}{\text{relative speed before}}$

Where Does the Lost Energy Go?

In inelastic collisions, the "lost" kinetic energy is converted to other forms:

Total Energy Conservation

$KE_i = KE_f + \Delta E_{\text{thermal}} + E_{\text{sound}} + E_{\text{deformation}} + \cdots$

Total energy is always conserved — it's only the kinetic energy portion that decreases.

Determining Collision Type from Data

Given initial and final velocities for both objects, calculate:

1. $KE_i = \frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2$
2. $KE_f = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$
3. If $KE_f = KE_i$ → elastic. If $KE_f < KE_i$ → inelastic.

Example: Is This Collision Elastic?

A 3 kg ball at +8 m/s hits a 1 kg ball at rest. After: Ball 1 moves at +2 m/s, Ball 2 at +18 m/s.

Check momentum:

• Before: $(3)(8) + (1)(0) = 24$ kg·m/s
• After: $(3)(2) + (1)(18) = 6 + 18 = 24$ kg·m/s ✅

Check KE:

• Before: $\frac{1}{2}(3)(64) + 0 = 96$ J
• After: $\frac{1}{2}(3)(4) + \frac{1}{2}(1)(324) = 6 + 162 = 168$ J ❌

Wait — $KE_f > KE_i$? That violates conservation of energy! This collision is impossible. The problem data must be wrong.

Lesson

Always verify that:

1. Momentum is conserved ✅
2. $KE_f \leq KE_i$ ✅
3. Both must be true for physically possible results

Concept Check — Inelastic Collisions 🎯

Collision Analysis 🧮

A 4 kg ball at +5 m/s hits a 2 kg ball at rest. After collision, the 4 kg ball moves at +3 m/s.

1. What is the velocity of the 2 kg ball after collision? (in m/s)

2. What is the total $KE$ before the collision? (in J)

3. What is the total $KE$ after the collision? (in J)

Collision Classification 🔍

Exit Quiz — Inelastic Collisions ✅

📐 1D Elastic Collision Formulas

Part 4 of 7 — Collisions

For elastic collisions in one dimension, we can derive exact formulas for the final velocities by solving the conservation of momentum and conservation of kinetic energy equations simultaneously. These formulas are powerful and save significant computation time.

The Two Equations

For a 1D elastic collision:

Momentum: $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$ ... (1)

KE: $\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$ ... (2)

A Useful Trick

Rearranging equation (2):

$m_1(v_{1i}^2 - v_{1f}^2) = m_2(v_{2f}^2 - v_{2i}^2)$

$m_1(v_{1i} - v_{1f})(v_{1i} + v_{1f}) = m_2(v_{2f} - v_{2i})(v_{2f} + v_{2i})$

Dividing by the rearranged equation (1): $m_1(v_{1i} - v_{1f}) = m_2(v_{2f} - v_{2i})$:

$v_{1i} + v_{1f} = v_{2f} + v_{2i}$

This simplifies to:

$v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$

The relative velocity of approach equals the relative velocity of separation (with opposite sign). This is the hallmark of elastic collisions!

The General Formulas

Solving equations (1) and the relative velocity equation simultaneously:

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i} + \frac{2m_2}{m_1 + m_2} v_{2i}$

$v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i} + \frac{m_2 - m_1}{m_1 + m_2} v_{2i}$

Special Case: $v_{2i} = 0$ (Target at Rest)

$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2} v_{1i}$

$v_{2f} = \frac{2m_1}{m_1 + m_2} v_{1i}$

Verification of Special Cases

Worked Example

A 6 kg ball at $+4$ m/s collides elastically with a 2 kg ball at rest.

$v_{1f} = \frac{6 - 2}{6 + 2}(4) = \frac{4}{8}(4) = 2 \text{ m/s}$

$v_{2f} = \frac{2(6)}{6 + 2}(4) = \frac{12}{8}(4) = 6 \text{ m/s}$

Verification

Momentum: $(6)(4) = (6)(2) + (2)(6) = 12 + 12 = 24$ ✅

KE: $\frac{1}{2}(6)(16) = 48$ J. $\frac{1}{2}(6)(4) + \frac{1}{2}(2)(36) = 12 + 36 = 48$ J ✅

Relative velocity: $4 - 0 = 4$. $-(2 - 6) = 4$ ✅

Concept Check — Elastic Collision Formulas 🎯

Elastic Collision Formula Practice 🧮

Use the elastic collision formulas with $v_{2i} = 0$:

1. A 4 kg ball at +10 m/s hits a 1 kg ball at rest elastically. What is $v_{1f}$? (in m/s)

2. What is $v_{2f}$? (in m/s)

3. A 2 kg ball at +9 m/s hits a 4 kg ball at rest elastically. What is $v_{1f}$? (in m/s, include sign)

Formula Applications 🔍

Exit Quiz — 1D Elastic Formulas ✅

🎯 2D Collisions (Glancing Collisions)

Part 5 of 7 — Collisions

When objects collide at an angle rather than head-on, we have a 2D (or glancing) collision. The physics is the same — momentum is conserved — but we must apply conservation independently in the $x$ and $y$ directions.

These are among the most challenging collision problems, but the systematic approach makes them manageable.

Setting Up 2D Collision Problems

Conservation Equations

$\textbf{x: } m_1 v_{1i} = m_1 v_{1f}\cos\theta_1 + m_2 v_{2f}\cos\theta_2$

$\textbf{y: } 0 = m_1 v_{1f}\sin\theta_1 - m_2 v_{2f}\sin\theta_2$

(Assuming object 1 moves along the $x$-axis initially, and object 2 is at rest.)

Why the y-equation equals zero

If the initial motion is entirely along the $x$-axis, the total initial $y$-momentum is zero. After the collision, the $y$-components of the two objects must cancel:

$m_1 v_{1f}\sin\theta_1 = m_2 v_{2f}\sin\theta_2$

Strategy

1. Choose $x$-axis along the initial velocity
2. Write conservation in $x$ and $y$
3. If elastic: add the KE conservation equation
4. You have 2 (or 3) equations for the unknowns

Special Case: Equal Masses, Elastic, Target at Rest

This is a famous result in physics:

In a 2D elastic collision between equal masses (one at rest), the two objects always move at 90° to each other after the collision.

Proof (Summary)

From momentum conservation (vectors): $m\vec{v}_{1i} = m\vec{v}_{1f} + m\vec{v}_{2f}$ $\vec{v}_{1i} = \vec{v}_{1f} + \vec{v}_{2f}$

Squaring both sides: $v_{1i}^2 = v_{1f}^2 + v_{2f}^2 + 2\vec{v}_{1f} \cdot \vec{v}_{2f}$

From KE conservation: $v_{1i}^2 = v_{1f}^2 + v_{2f}^2$

Comparing: $2\vec{v}_{1f} \cdot \vec{v}_{2f} = 0$

Since neither velocity is zero, $\vec{v}_{1f} \perp \vec{v}_{2f}$ — they move at right angles!

Applications

This is seen in:

• Billiards: cue ball and target ball go at 90° (if equal mass and elastic)
• Nuclear physics: proton-proton scattering at 90°

Worked Example: 2D Perfectly Inelastic

A 3 kg ball at 4 m/s ($+x$) collides with a 2 kg ball at 5 m/s ($+y$). They stick together. Find the final velocity.

x-momentum: $(3)(4) + (2)(0) = (5)v_{fx}$ $v_{fx} = 12/5 = 2.4 \text{ m/s}$

y-momentum: $(3)(0) + (2)(5) = (5)v_{fy}$ $v_{fy} = 10/5 = 2.0 \text{ m/s}$

Final speed: $v_f = \sqrt{2.4^2 + 2.0^2} = \sqrt{5.76 + 4.0} = \sqrt{9.76} = 3.12 \text{ m/s}$

Direction: $\theta = \arctan\left(\frac{2.0}{2.4}\right) = \arctan(0.833) = 39.8°$

above the $x$-axis.

Concept Check — 2D Collisions 🎯

2D Collision Practice 🧮

A 5 kg ball at 6 m/s ($+x$) and a 3 kg ball at 4 m/s ($+y$) collide and stick together.

1. What is $v_{fx}$? (in m/s, to 3 significant figures)

2. What is $v_{fy}$? (in m/s)

3. What is the final speed? (in m/s, to 3 significant figures)

2D Collision Analysis 🔍

Exit Quiz — 2D Collisions ✅

🔧 Problem-Solving Workshop

Part 6 of 7 — Collisions

Time to integrate all collision concepts! In this workshop, we'll solve AP-level problems covering elastic, inelastic, and perfectly inelastic collisions in 1D and 2D. The key skill is identifying the collision type and choosing the right conservation laws.

Collision Problem Decision Tree

Step 1: Identify the Collision Type

Step 2: Choose Equations

Step 3: Solve

• Perfectly inelastic: 1 unknown, 1 equation → direct solution
• Elastic: 2 unknowns, 2 equations → use formulas or solve simultaneously
• Inelastic: Need additional information (e.g., one final velocity)

Problem 1: Collision Classification 🔍

A 2 kg ball at +8 m/s hits a 3 kg ball at rest. After: Ball 1 at $-0.8$ m/s, Ball 2 at $+5.87$ m/s.

Is this collision elastic?

Problem 2: Ballistic Pendulum Revisited 🎯

A 0.015 kg bullet at 600 m/s embeds in a 3.0 kg block at rest.

1. What is the velocity of the block+bullet right after impact? (in m/s, to 3 significant figures)

2. What fraction of the bullet's kinetic energy is lost? (as a percentage, round to 3 significant figures)

3. If the block+bullet slides on a surface with $\mu_k = 0.30$, how far does it slide? (in m, to 3 significant figures, use $g = 10$ m/s²)

Problem 3: Elastic Collision 💎

A 4 kg ball at +6 m/s collides elastically with a 2 kg ball at rest.

Problem 4: Head-On Inelastic 💥

A 5 kg ball at +4 m/s and a 3 kg ball at −6 m/s collide. The 5 kg ball moves at +1 m/s after.

1. What is the velocity of the 3 kg ball after? (in m/s, include sign)

2. What is the kinetic energy lost? (in J)

3. Is this collision elastic, inelastic, or perfectly inelastic? (type "elastic", "inelastic", or "perfectly inelastic")

Collision Problem Strategy 🔍

Exit Quiz — Problem Solving ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Collisions

This final lesson brings together all collision concepts: elastic, inelastic, perfectly inelastic, 1D and 2D, and the relationships between momentum and energy conservation. Let's review the key ideas and practice AP-level questions.

Complete Collision Summary

Conservation Laws by Collision Type

Key Formulas

All collisions: $m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}$

Perfectly inelastic: $v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2}$

Elastic (target at rest): $v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i}, \quad v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i}$

Elastic relative velocity: $v_{1i} - v_{2i} = -(v_{1f} - v_{2f})$

Key Conceptual Points

1. Momentum is always conserved (no net external force)
2. KE is conserved only in elastic collisions
3. Perfectly inelastic = max KE loss (objects stick)
4. Equal masses elastic: velocity swap (1D) or 90° deflection (2D)
5. Light hits heavy: bounces back. Heavy hits light: plows through.

AP Review — Conceptual 🎯

AP Review — Quantitative 📝

AP Calculation Practice 🧮

1. A 3 kg ball at +10 m/s collides elastically with a 1 kg ball at rest. What is $v_{1f}$? (in m/s)

2. What is $v_{2f}$? (in m/s)

3. A 2 kg ball at +6 m/s and a 4 kg ball at −3 m/s collide and stick. What is the final velocity? (in m/s)

Comprehensive Review 🔍

Final Exit Quiz — Collisions ✅