Collisions
Elastic and inelastic collisions, coefficient of restitution
💥 Collisions
Types of Collisions
Collisions are classified based on what happens to kinetic energy:
| Type | Momentum Conserved? | Kinetic Energy Conserved? | What Happens | |------|-------------------|------------------------|--------------| | Elastic | ✅ Yes | ✅ Yes | Objects bounce off, no deformation | | Inelastic | ✅ Yes | ❌ No | Some KE lost to heat, sound, deformation | | Perfectly Inelastic | ✅ Yes | ❌ No (max loss) | Objects stick together |
💡 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:
Kinetic Energy:
Two equations, two unknowns (if we know initial velocities and masses)
Special Case: Equal Masses (Elastic)
If and object 2 is initially at rest:
Before: Object 1 moving at , object 2 at rest
After:
- Object 1 comes to rest ()
- Object 2 moves with original velocity ()
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
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:
Momentum conservation:
Only one unknown! Much simpler than elastic collisions.
Coefficient of Restitution
The coefficient of restitution measures "bounciness":
Values:
- : Perfectly elastic (ideal bounce)
- : Inelastic (some energy lost)
- : Perfectly inelastic (stick together, no bounce)
For a ball bouncing:
where is drop height, is bounce height.
Problem-Solving Strategy
For Elastic Collisions:
- Write momentum conservation:
- Write energy conservation:
- Solve system of equations (algebra intensive!)
- Check: KE should be equal before and after
For Perfectly Inelastic Collisions:
- Use simplified equation:
- Solve for
- 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:
y-direction:
For elastic collisions, also conserve energy:
Need to use components: ,
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 , velocity ) hitting object 2 (mass , at rest):
Elastic:
- Object 1 stops:
- Object 2 moves:
- No energy lost
Perfectly Inelastic:
- Both move together:
- 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:
Energy is ADDED to system (from chemical energy, spring, etc.)
Key Formulas Summary
| Type | Momentum | Kinetic Energy | Formula | |------|----------|----------------|---------| | Elastic | Conserved | Conserved | Two conservation equations | | Inelastic | Conserved | Not conserved | Momentum equation only | | Perfectly Inelastic | Conserved | Not conserved (max loss) | | | Coefficient of restitution | - | - | |
📚 Practice Problems
1Problem 1easy
❓ Question:
Two identical 2 kg balls collide elastically. Ball 1 is moving at 6 m/s and ball 2 is at rest. Find the final velocities of both balls.
💡 Show Solution
Given Information:
- Ball 1: kg, m/s
- Ball 2: kg, m/s (at rest)
- Elastic collision (momentum AND energy conserved)
Find: and
Step 1: Apply conservation of momentum
Step 2: Apply conservation of kinetic energy
Step 3: Use special case result
For elastic collision with equal masses where one is at rest:
The velocities are exchanged!
Step 4: Verify with equations
Check Equation 1:
Check Equation 2:
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!
2Problem 2medium
❓ Question:
A 3.0 kg ball moving at 8.0 m/s collides head-on with a 5.0 kg ball at rest. The collision is elastic. (a) Find the final velocity of each ball. (b) Verify that kinetic energy is conserved.
💡 Show Solution
Solution:
Given: m₁ = 3.0 kg, v₁ᵢ = 8.0 m/s, m₂ = 5.0 kg, v₂ᵢ = 0
For elastic collision with one object initially at rest:
v₁f = ((m₁ - m₂)/(m₁ + m₂))v₁ᵢ = ((3.0 - 5.0)/(3.0 + 5.0))(8.0) v₁f = (-2.0/8.0)(8.0) = -2.0 m/s (bounces back)
v₂f = (2m₁/(m₁ + m₂))v₁ᵢ = (2(3.0)/(8.0))(8.0) v₂f = (6.0/8.0)(8.0) = 6.0 m/s
(b) Check KE conservation: KE_i = ½m₁v₁ᵢ² = ½(3.0)(8.0)² = 96 J
KE_f = ½m₁v₁f² + ½m₂v₂f² KE_f = ½(3.0)(2.0)² + ½(5.0)(6.0)² KE_f = 6.0 + 90 = 96 J ✓
Kinetic energy is conserved!
3Problem 3medium
❓ Question:
A 1500 kg car traveling at 20 m/s rear-ends a 1000 kg car traveling at 15 m/s in the same direction. The cars stick together after the collision. (a) Find their common velocity after the collision. (b) How much kinetic energy is lost?
💡 Show Solution
Given Information:
- Car 1: kg, m/s
- Car 2: kg, m/s (same direction)
- Perfectly inelastic collision (stick together)
(a) Find common velocity after collision
Step 1: Apply conservation of momentum
Answer (a): Common velocity after collision is 18 m/s
(b) Find kinetic energy lost
Step 2: Calculate initial kinetic energy
Step 3: Calculate final kinetic energy
Step 4: Calculate energy lost
Answer (b): Kinetic energy lost = 7,500 J
Percentage lost:
Note: This energy goes into crumpling metal, heat, and sound. Even though both cars were moving in the same direction and the speed difference was only 5 m/s, significant energy was dissipated!
4Problem 4hard
❓ Question:
A 0.20 kg ball moving at 5.0 m/s collides with a 0.30 kg ball initially at rest. After the collision, the 0.20 kg ball moves at 2.0 m/s at 60° to its original direction. (a) Find the velocity of the 0.30 kg ball after collision. (b) Is this collision elastic or inelastic?
💡 Show Solution
Solution:
Given: m₁ = 0.20 kg, v₁ᵢ = 5.0 m/s, m₂ = 0.30 kg, v₂ᵢ = 0 After: v₁f = 2.0 m/s at 60°
(a) Velocity of m₂: Conservation of momentum (x-direction): m₁v₁ᵢ = m₁v₁f cos 60° + m₂v₂f,x 0.20(5.0) = 0.20(2.0)(0.5) + 0.30v₂f,x 1.0 = 0.20 + 0.30v₂f,x v₂f,x = 2.67 m/s
Conservation of momentum (y-direction): 0 = m₁v₁f sin 60° + m₂v₂f,y 0 = 0.20(2.0)(0.866) + 0.30v₂f,y v₂f,y = -1.15 m/s
Magnitude: v₂f = √(2.67² + 1.15²) = 2.9 m/s Direction: θ = tan⁻¹(1.15/2.67) = 23° below original direction
(b) Elastic or inelastic? KE_i = ½(0.20)(5.0)² = 2.5 J KE_f = ½(0.20)(2.0)² + ½(0.30)(2.9)² = 0.40 + 1.26 = 1.66 J
KE lost = 2.5 - 1.66 = 0.84 J Inelastic collision (KE not conserved)
5Problem 5hard
❓ Question:
A 0.2 kg ball is dropped from a height of 2 m and bounces to a height of 1.2 m. (a) Find the coefficient of restitution. (b) If the ball was in contact with the ground for 0.01 s, what average force did the ground exert on the ball?
💡 Show Solution
Given Information:
- Mass: kg
- Drop height: m
- Bounce height: m
- Contact time: s
(a) Find coefficient of restitution
Step 1: Calculate velocity just before hitting ground
Using energy conservation:
Step 2: Calculate velocity just after leaving ground
Using energy conservation:
Step 3: Calculate coefficient of restitution
Alternative formula using heights:
Answer (a): Coefficient of restitution = 0.775
This means the collision is inelastic (not perfectly elastic where ).
(b) Find average force from ground
Step 4: Set up velocity directions
Taking upward as positive:
- Velocity before collision: m/s (downward)
- Velocity after collision: m/s (upward)
Step 5: Calculate change in momentum
Step 6: Apply impulse-momentum theorem
Step 7: Find force from ground
Net force = Normal force from ground - Weight
Answer (b): Average force from ground = 224 N
Note: This is about 114 times the ball's weight! The ground must exert a large force because:
- It must stop the ball's downward motion
- It must accelerate the ball upward
- All in a very short time (0.01 s)
If the contact time were longer (softer surface), the force would be smaller.
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