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Elastic and Inelastic Collisions

Lesson ~10 min read 8 MCQs

In simple terms: In simple terms, collisions are either elastic or inelastic. The difference depends on whether the total kinetic energy of the objects is the same before and after they hit.

Why this matters

Imagine you're at a bowling alley. You roll the ball, it strikes the pins, and there's a huge crash. Pins fly everywhere. Now, picture a game of pool. The cue ball silently strikes the 8-ball. They click, and both roll away smoothly.

Both scenarios are collisions. But they feel completely different, don't they? One is loud and messy, the other is clean and quiet. In physics, we have a precise way to describe this difference. It all comes down to energy.

Today, we're going to learn how to classify any collision as either elastic or inelastic. We'll explore what happens to the kinetic energy in each case and why momentum is the one rule that always holds true.

Concept overview

flowchart TD
    A[Collision Occurs in a Closed System] --> B{Is Momentum Conserved?};
    B -- Yes, Always --> C{Is Kinetic Energy Conserved?};
    C -- Yes --> D[Elastic Collision];
    D -- Example --> E[Ideal Billiard Balls];
    C -- No --> F[Inelastic Collision];
    F -- KE is transformed --> G[into heat, sound, deformation];
    F --> H{Do objects stick together?};
    H -- Yes --> I[Perfectly Inelastic Collision];
    I -- Example --> J[Clay balls sticking together];
    H -- No --> K[Inelastic Collision];
    K -- Example --> L[A car crash where cars bounce apart];
A flowchart shows how to classify collisions. It starts with a collision, confirms momentum is always conserved, then asks if kinetic energy is conserved. A 'yes' leads to an elastic collision. A 'no' leads to an inelastic collision, which is further divided by asking if the objects stick together, leading to either a perfectly inelastic or a regular inelastic collision.

Core explanation

When we analyze collisions in physics, we have one unshakable foundation: conservation of momentum. In a closed system (meaning no external forces like friction are interfering), the total momentum of all objects before the collision is equal to the total momentum of all objects after the collision.

p_initial_total = p_final_total

This is our starting point for every single collision problem. But to truly understand what happened, we need to ask a second question: What happened to the kinetic energy?

The answer to that question is how we classify collisions into two main families: elastic and inelastic.

Elastic Collisions: The Perfect Bounce

An elastic collision is one where the total kinetic energy of the system is also conserved.

KE_initial_total = KE_final_total

Think of it as a "perfectly bouncy" collision. The classic example is two billiard balls hitting each other. They click, and bounce off without any loss of speed due to deformation. No energy is wasted on making a loud crunching sound or generating heat. All the initial kinetic energy is returned as final kinetic energy.

Now, this is where a common point of confusion comes up. Does this mean each individual object has the same kinetic energy it started with? No, not at all.

Imagine a moving cue ball hitting a stationary 8-ball. Before the collision, the cue ball has all the kinetic energy, and the 8-ball has zero. After they collide, the cue ball might stop completely, transferring all its kinetic energy to the 8-ball, which now rolls away. The kinetic energy of each ball changed dramatically, but the total sum of their kinetic energies before and after the impact remained the same.

In the real world, perfectly elastic collisions are rare. Even billiard balls make a small sound and generate a tiny bit of heat. But for many AP Physics problems, we'll treat collisions between hard, bouncy objects (like atoms or ideal billiard balls) as elastic.

Inelastic Collisions: The Real World

An inelastic collision is one where the total kinetic energy of the system decreases.

KE_initial_total > KE_final_total

This is far more common in our everyday lives. Think about a car crash. When two cars collide, you hear a loud noise, see sparks (light and heat), and the metal bodies get bent and deformed. All of these things—sound, heat, light, and bending metal—require energy. Where did that energy come from? It was stolen from the initial kinetic energy of the cars.

Because some of the kinetic energy was transformed into other forms, the total kinetic energy of the cars after the crash is less than it was before. The collision is inelastic.

Perfectly Inelastic Collisions: Maximum Stickiness

This is a special, extreme case of an inelastic collision. A perfectly inelastic collision is one where the objects stick together after colliding and move as a single mass with a common final velocity.

This type of collision involves the maximum possible loss of kinetic energy. It's as "un-bouncy" as you can get.

Imagine two balls of wet clay flying towards each other. They collide, squish, and stick together, moving as one combined lump. Or think of a football running back tackling a safety; they collide and move forward as a single unit.

Because they move together, we can treat them as a single object with a combined mass (m₁ + m₂) after the collision. This simplifies the momentum conservation equation beautifully:

m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f

Here, v_f is the single final velocity of the combined object. This equation is your go-to tool for any problem where things stick together.


To summarize:

  1. 1
    Always start with momentum
    It's conserved in every collision in a closed system.
  2. 2
    Check the kinetic energy
    • If KE_initial = KE_final, it's elastic.
    • If KE_initial > KE_final, it's inelastic.
  3. 3
    If they stick together
    , it's a special case called perfectly inelastic, which features the maximum loss of KE.

Worked examples

Example 1

The Perfectly Inelastic Tackle

Problem: In a football game, a 110 kg running back, Marcus, is moving at 8.0 m/s. He's tackled by a 90 kg cornerback, Jordan, who was running in the same direction at 5.0 m/s. After the tackle, they stick together. What is their velocity immediately after the collision?

Solution:

  1. 1
    Identify the collision type
    The problem states they "stick together." This is the key phrase for a perfectly inelastic collision.
  2. 2
    Set up the principle
    For any collision in a closed system, momentum is conserved. For a perfectly inelastic collision, we use the special form of the momentum equation. m₁v₁ᵢ + m₂v₂ᵢ = (m₁ + m₂)v_f
  3. 3
    List your knowns
    • Marcus (m₁): 110 kg, v₁ᵢ = 8.0 m/s
    • Jordan (m₂): 90 kg, v₂ᵢ = 5.0 m/s
    • We need to find v_f.
  4. 4
    Plug in the values
    (110 kg)(8.0 m/s) + (90 kg)(5.0 m/s) = (110 kg + 90 kg)v_f 880 kg·m/s + 450 kg·m/s = (200 kg)v_f 1330 kg·m/s = (200 kg)v_f
  5. 5

    Solve for the final velocity (v_f): v_f = 1330 kg·m/s / 200 kg v_f = 6.65 m/s

Example 2

Was It Elastic?

Problem: A 0.50 kg cart moving at 2.0 m/s on a frictionless track collides with a 1.0 kg cart that is initially at rest. After the collision, the 0.50 kg cart is moving at -0.67 m/s (it bounced backward), and the 1.0 kg cart is moving at 1.33 m/s. Is this collision elastic or inelastic?

Solution:

  1. 1
    Identify the goal
    We need to classify the collision. This means we must compare the total kinetic energy before and after.
  2. 2
    Calculate Initial Kinetic Energy (KEᵢ)
    Only the first cart is moving initially. KEᵢ = KE₁ᵢ + KE₂ᵢ KEᵢ = ½m₁v₁ᵢ² + ½m₂v₂ᵢ² KEᵢ = ½(0.50 kg)(2.0 m/s)² + ½(1.0 kg)(0 m/s)² KEᵢ = ½(0.50 kg)(4.0 m²/s²) + 0 KEᵢ = 1.0 J
  3. 3
    Calculate Final Kinetic Energy (KEf)
    Both carts are moving after the collision. KEf = KE₁f + KE₂f KEf = ½m₁v₁f² + ½m₂v₂f² KEf = ½(0.50 kg)(-0.67 m/s)² + ½(1.0 kg)(1.33 m/s)²
  4. 4
    Compare and conclude
    • Initial KE ≈ 1.0 J
    • Final KE ≈ 1.1 J

    Wait, the final kinetic energy is greater than the initial? This is a huge red flag. In a real-world physical system, this is impossible without an internal energy source (like a spring or an explosion). In the context of an AP problem, this result almost certainly means the numbers were chosen to be very close to an elastic collision, and the difference is due to rounding in the problem statement. Let's re-check the numbers. If the final velocities were exactly -2/3 m/s and 4/3 m/s, the math would work out perfectly to 1.0 J. Given the values, we can conclude this is intended to represent an elastic collision.

    Teacher's Note: On the AP Exam, if you calculate KE_f > KE_i, double-check your math. If your math is correct, the collision is elastic (assuming KE_f ≈ KE_i) or the problem involves released potential energy (like a spring-loaded cart). For this topic, we assume the former. If KE_f was clearly less, like 0.8 J, it would be inelastic.

Try it yourself

Ready to try one on your own?

Problem 1: A 10,000 kg railroad freight car is rolling at 3.0 m/s when it collides with and couples to a stationary 15,000 kg freight car on a level track. a) What is the new speed of the two-car system? b) How much kinetic energy was "lost" in the collision? What type of collision was this?

Hints:

  • "Couples to" is another way of saying "sticks together." What formula does that point to?
  • To find the lost KE, calculate the total KE before the collision and the total KE after. The difference is the energy that was transformed.

Problem 2: A billiard ball moving at 5 m/s strikes an identical, stationary billiard ball. After the collision, the first ball moves at 4 m/s at an angle, and the second moves off at 3 m/s at another angle. Is this collision elastic? (Assume the mass of each ball is m).