💡 Key Point: Momentum is ALWAYS conserved in collisions (if system is isolated). Kinetic energy is only conserved in elastic collisions.

Elastic Collisions

In an elastic collision:

• Momentum is conserved
• Kinetic energy is conserved
• Objects bounce apart
• No energy lost to heat, sound, or deformation

Examples:

• Colliding billiard balls (nearly elastic)
• Molecules in ideal gas
• Atomic particles (fundamental particles)

Conservation Equations (1D)

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

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

Two equations, two unknowns (if we know initial velocities and masses)

Special Case: Equal Masses (Elastic)

If $m_1 = m_2$ and object 2 is initially at rest:

Before: Object 1 moving at $v$, object 2 at rest

After:

• Object 1 comes to rest ($v_{1f} = 0$)
• Object 2 moves with original velocity ($v_{2f} = v$)

They exchange velocities!

Special Case: One Object Much Heavier

Light object hits heavy object at rest:

• Light object bounces back (reverses direction)
• Heavy object barely moves

Heavy object hits light object at rest:

• Heavy object continues forward (barely slows)
• Light object shoots forward at high speed

Inelastic Collisions

In an inelastic collision:

• Momentum is conserved ✓
• Kinetic energy is NOT conserved ✗
• Some KE converted to other forms (heat, sound, deformation)
• Objects may or may not stick together

Examples:

• Car crash (crumple zones absorb energy)
• Dropped ball (loses energy on each bounce)
• Most real-world collisions

Energy Loss

$KE_{lost} = KE_i - KE_f$

$KE_{lost} = \left(\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2\right) - \left(\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2\right)$

This energy goes into:

• Heat (friction)
• Sound
• Deformation (crushing, bending)
• Internal energy

Perfectly Inelastic Collisions

In a perfectly inelastic collision:

• Objects stick together after collision
• Move with same final velocity
• Maximum possible kinetic energy loss (for given initial conditions)
• Momentum still conserved!

Examples:

• Clay balls colliding
• Train cars coupling
• Tackle in football
• Meteor impact creating crater

Simplified Equation

Since objects stick together: $v_{1f} = v_{2f} = v_f$

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

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

Only one unknown! Much simpler than elastic collisions.

Coefficient of Restitution

The coefficient of restitution $e$ measures "bounciness":

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

Values:

• $e = 1$: Perfectly elastic (ideal bounce)
• $0 < e < 1$: Inelastic (some energy lost)
• $e = 0$: Perfectly inelastic (stick together, no bounce)

For a ball bouncing: $e = \sqrt{\frac{h_f}{h_i}}$

where $h_i$ is drop height, $h_f$ is bounce height.

Problem-Solving Strategy

For Elastic Collisions:

1. Write momentum conservation: $m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
2. Write energy conservation: $\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2 = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$
3. Solve system of equations (algebra intensive!)
4. Check: KE should be equal before and after

For Perfectly Inelastic Collisions:

1. Use simplified equation: $m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$
2. Solve for $v_f$
3. Optional: Calculate energy lost

Much simpler because objects stick together!

⚠️ Common Mistakes

Mistake 1: Assuming KE is Always Conserved

❌ Wrong: KE is conserved in all collisions ✅ Right: KE is only conserved in ELASTIC collisions

Mistake 2: Perfectly Inelastic = No Motion

❌ Wrong: Perfectly inelastic means objects stop moving ✅ Right: Perfectly inelastic means objects stick together (they can still move!)

Mistake 3: Sign Errors in Momentum

Remember: velocities in opposite directions have opposite signs!

Mistake 4: Confusing Speed and Velocity

In coefficient of restitution, use relative speeds (magnitudes), but in momentum conservation use velocities (with signs).

2D Collisions

For collisions in two dimensions:

Conserve momentum in EACH direction:

x-direction: $m_1 v_{1ix} + m_2 v_{2ix} = m_1 v_{1fx} + m_2 v_{2fx}$

y-direction: $m_1 v_{1iy} + m_2 v_{2iy} = m_1 v_{1fy} + m_2 v_{2fy}$

For elastic collisions, also conserve energy: $\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$

Need to use components: $v_x = v\cos\theta$, $v_y = v\sin\theta$

Real-World Applications

Car Safety

Modern cars designed for inelastic collisions:

• Crumple zones deform (absorb energy)
• Reduces force on passengers
• Energy goes into crushing metal, not hurting people

Sports

Elastic collisions:

• Golf ball and club (nearly elastic for maximum energy transfer)
• Billiards (designed to be elastic)

Inelastic collisions:

• Football tackle (maximum energy dissipation)
• Boxing glove (padding increases collision time, decreases force)

Particle Physics

Particle accelerators create collisions:

• Analyze momentum and energy
• Discover new particles
• Test fundamental physics laws

Comparing Collision Types

Same Initial Conditions

Consider object 1 (mass $m$, velocity $v$) hitting object 2 (mass $m$, at rest):

Elastic:

• Object 1 stops: $v_{1f} = 0$
• Object 2 moves: $v_{2f} = v$
• $KE_i = KE_f = \frac{1}{2}mv^2$
• No energy lost

Perfectly Inelastic:

• Both move together: $v_f = \frac{v}{2}$
• $KE_i = \frac{1}{2}mv^2$
• $KE_f = \frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 = \frac{1}{4}mv^2$
• Half the kinetic energy lost!

Explosions (Reverse Collisions)

An explosion is like a collision in reverse:

• Start: objects together (or at rest)
• End: objects flying apart

Still conserve momentum!

If explosion from rest: $m_1\vec{v}_1 + m_2\vec{v}_2 = 0$

Energy is ADDED to system (from chemical energy, spring, etc.)

Key Formulas Summary

Find: $v_{1f}$ and $v_{2f}$

Step 1: Apply conservation of momentum

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

$(2)(6) + (2)(0) = (2)v_{1f} + (2)v_{2f}$

$12 = 2v_{1f} + 2v_{2f}$

$6 = v_{1f} + v_{2f} \quad \text{(Equation 1)}$

Step 2: Apply conservation of kinetic energy

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

$\frac{1}{2}(2)(6)^2 + 0 = \frac{1}{2}(2)v_{1f}^2 + \frac{1}{2}(2)v_{2f}^2$

$36 = v_{1f}^2 + v_{2f}^2 \quad \text{(Equation 2)}$

Step 3: Use special case result

For elastic collision with equal masses where one is at rest:

The velocities are exchanged!

$v_{1f} = 0 \text{ m/s}$ $v_{2f} = 6 \text{ m/s}$

Step 4: Verify with equations

Check Equation 1: $v_{1f} + v_{2f} = 0 + 6 = 6 \quad ✓$

Check Equation 2: $v_{1f}^2 + v_{2f}^2 = 0^2 + 6^2 = 36 \quad ✓$

Answers:

• Ball 1 final velocity: 0 m/s (comes to rest)
• Ball 2 final velocity: 6 m/s (takes on ball 1's velocity)

Physical Interpretation: Ball 1 transfers all its momentum and energy to ball 2. This is what you see in Newton's cradle!