8.4 Elastic and Inelastic Collisions

Two Collisions, One Momentum, Two Different Worlds

Two balls of equal mass collide head-on at equal speeds. In one collision, they bounce off perfectly --- each reverses direction and flies away at the same speed it arrived. In another collision, they stick together and stop dead.

The total momentum before each collision is zero. The total momentum after each collision is also zero. Momentum conservation holds in both cases --- Section 8.3 guarantees it.

But look at what happened to the kinetic energy. In the first collision, every bit of kinetic energy that existed before the collision still exists afterward --- just redistributed. In the second, all the kinetic energy is gone. Two objects that were moving are now motionless. That energy did not vanish from the universe; it went into deforming the objects, heating them up, producing sound --- but it is no longer kinetic energy.

Momentum was conserved in both cases. Kinetic energy was conserved in one and destroyed in the other. This section is about understanding that distinction, classifying collisions by what happens to kinetic energy, and knowing which equations to use in each case.

Before you read on: Two equal-mass balls collide head-on. One is moving to the right; the other is at rest. The collision is perfectly elastic (no energy loss). What happens afterward?

(a) Both move forward (to the right) at reduced speeds.

(b) The first ball stops completely, and the second ball moves to the right at the original speed.

(c) Both balls bounce backward (to the left).

Commit to your answer before continuing.

[Interactive: Predict-Then-Reveal. The student selects (a), (b), or (c). Their choice is locked in and revisited after the exploration.]

Exploration: The Collision Simulator

[Interactive: Two-Ball Collision Lab. Two balls on a frictionless track with adjustable masses ($m_1$, $m_2$) and initial velocities ($v_1$, $v_2$). A "restitution" slider controls the collision type, sweeping continuously from perfectly inelastic (coefficient of restitution $e = 0$, balls stick together) to perfectly elastic ($e = 1$, no kinetic energy loss). Before and after each collision, bar charts display:

• Total momentum (before and after) --- always equal
• Total kinetic energy (before and after) --- varies with the restitution slider
• Kinetic energy lost (shaded in red, labeled "transferred to heat, sound, deformation")

The student can: - Set specific mass values and velocities - Run the collision and watch the balls interact in real time - Toggle slow-motion to see the moment of contact - Compare the bar charts before and after

Guided prompts:]

Step 1: Set equal masses, give the left ball a velocity to the right, and leave the right ball at rest. Set the restitution slider to fully elastic ($e = 1$). Run the collision. What happens to each ball? Compare the kinetic energy bars before and after.

Step 2: Keep everything the same, but slide the restitution to perfectly inelastic ($e = 0$). Run the collision again. What is the final velocity of the combined object? How much kinetic energy was lost?

Step 3: Now try intermediate values of the slider. What fraction of kinetic energy survives? Is the momentum still conserved at every slider position?

Step 4: Change the mass ratio. Make $m_1$ much larger than $m_2$ (a bowling ball hitting a tennis ball). Run a perfectly elastic collision. What happens to each object's speed? Now reverse the ratio --- light ball hits heavy ball. What changes?

If you explored carefully, here is what you should have noticed:

1. Momentum is always conserved. No matter the slider position, no matter the mass ratio, the total momentum bar is identical before and after. This is not a coincidence --- it is the law you established in Section 8.3.

2. Kinetic energy is only conserved when $e = 1$. At every other slider position, some kinetic energy disappears from the bars. The lower the restitution, the more energy is lost.

3. The equal-mass elastic case is special. The moving ball stops; the stationary ball picks up all its speed. This is the physics behind Newton's cradle.

4. Mass ratio matters. When a heavy object hits a light one elastically, the heavy object barely slows down and the light one flies off fast. When a light object hits a heavy one, the light one bounces back and the heavy one barely moves.

The Concept: Classifying Collisions by Energy

All collisions conserve momentum (assuming the system is isolated --- no significant external forces during the brief collision). But collisions differ in what happens to kinetic energy. There are three categories.

Perfectly Elastic Collisions

A perfectly elastic collision conserves both momentum and kinetic energy. No kinetic energy is converted to other forms --- no permanent deformation, no heat, no sound.

$$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \quad \text{(momentum conservation)}$$

$$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 = \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \quad \text{(kinetic energy conservation)}$$

Two equations, two unknowns ($v_{1f}$ and $v_{2f}$). This system can be solved exactly. The algebra is somewhat involved, but the result is clean. For a one-dimensional elastic collision with object 2 initially at rest ($v_{2i} = 0$):

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

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

These formulas reward inspection. When $m_1 = m_2$: $v_{1f} = 0$ and $v_{2f} = v_{1i}$. The first ball stops, the second takes over. That is Newton's cradle --- and the answer to the prediction question is (b).

When $m_1 \gg m_2$: $v_{1f} \approx v_{1i}$ (the heavy ball barely notices) and $v_{2f} \approx 2v_{1i}$ (the light ball flies off at nearly twice the incoming speed).

When $m_1 \ll m_2$: $v_{1f} \approx -v_{1i}$ (the light ball bounces back at nearly its original speed) and $v_{2f} \approx 0$ (the heavy ball barely moves).

Perfect elasticity is an idealization. Collisions between hard steel balls or between gas molecules at moderate temperatures come close. Most everyday collisions are not perfectly elastic.

Inelastic Collisions

An inelastic collision conserves momentum but not kinetic energy. Some kinetic energy is transformed into internal energy --- deformation of the objects, heat, sound. This is the most common type of collision in everyday life.

$$m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \quad \text{(momentum conservation)}$$

$$\frac{1}{2}m_1 v_{1i}^2 + \frac{1}{2}m_2 v_{2i}^2 > \frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2 \quad \text{(kinetic energy decreases)}$$

With only one equation (momentum conservation) and two unknowns, you cannot solve for both final velocities unless you have additional information --- such as one of the final velocities, or the fraction of energy lost.

Perfectly Inelastic Collisions

A perfectly inelastic collision is the extreme case: the objects stick together after impact and move as a single combined mass. This loses the maximum kinetic energy possible while still conserving momentum.

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

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

One equation, one unknown. This is the simplest collision to solve. The final velocity is just the weighted average of the initial velocities, weighted by mass.

The kinetic energy lost is:

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

For the special case where object 2 is initially at rest:

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

This energy goes into deformation, heat, and sound. It is not destroyed --- it is transformed into forms that are no longer organized kinetic energy.

Variation: What Changes, What Stays the Same?

Understanding the difference between collision types deepens when you hold some variables fixed and vary others systematically.

Hold the masses fixed. Vary the collision type.

Consider $m_1 = m_2 = m$, with $v_{1i} = v_0$ and $v_{2i} = 0$. The total initial momentum is $mv_0$ in every case.

Collision type	$v_{1f}$	$v_{2f}$	Final $KE$	$KE$ lost
Perfectly elastic	$0$	$v_0$	$\frac{1}{2}mv_0^2$	$0$
Perfectly inelastic	$v_0/2$	$v_0/2$ (stuck together)	$\frac{1}{4}mv_0^2$	$\frac{1}{4}mv_0^2$ (50%)

In both rows, total momentum is $mv_0$. The elastic collision redistributes all the kinetic energy from one ball to the other. The perfectly inelastic collision keeps the momentum the same but destroys half the kinetic energy.

Hold the collision type fixed. Vary the mass ratio.

For perfectly elastic collisions with the target at rest:

Mass ratio $m_1/m_2$	$v_{1f}/v_{1i}$	$v_{2f}/v_{1i}$	Description
$1$	$0$	$1$	Complete transfer
$\gg 1$	$\approx 1$	$\approx 2$	Heavy hits light: light flies off fast
$\ll 1$	$\approx -1$	$\approx 0$	Light hits heavy: light bounces back

The pattern: when the masses are equal, the transfer of motion is complete. When they are mismatched, the heavier object dominates the outcome.

Why Perfectly Inelastic Collisions Lose the Most Energy

This is a structural question worth thinking through carefully. Among all collisions that conserve momentum, why does sticking together lose more kinetic energy than any other outcome?

Here is the reasoning. Momentum is conserved: $p_{\text{total}} = m_1 v_{1f} + m_2 v_{2f}$. This is a constraint on the final velocities. Among all pairs $(v_{1f}, v_{2f})$ that satisfy this constraint, the total final kinetic energy $\frac{1}{2}m_1 v_{1f}^2 + \frac{1}{2}m_2 v_{2f}^2$ varies.

When the two objects stick together, they move at the same velocity $v_f$. This is the configuration that minimizes the total kinetic energy for a given total momentum. Any other outcome --- where the objects move at different velocities --- must have more kinetic energy. The proof uses the fact that for a fixed total momentum, kinetic energy is minimized when all parts of the system move at the same velocity (there is no relative motion between them).

Think of it this way: kinetic energy can always be decomposed into the kinetic energy of the center-of-mass motion plus the kinetic energy of motion relative to the center of mass. Momentum conservation fixes the center-of-mass motion. Sticking together eliminates all relative motion, leaving only the center-of-mass kinetic energy --- the bare minimum.

Connection to Prior Sections

Energy conservation (Chapter 7) and momentum conservation (Section 8.3) are now working together. But they play different roles depending on the collision type.

Elastic collisions require both conservation laws simultaneously. Momentum conservation alone gives one equation with two unknowns. Kinetic energy conservation provides the second equation. Together, they fully determine the outcome.

Inelastic collisions can only use momentum conservation. You cannot write a kinetic energy conservation equation because kinetic energy is not conserved. If you try, the equations will be inconsistent --- they will have no solution. This is exactly the error explored in the Debug practice problem below.

Perfectly inelastic collisions need only momentum conservation, and they have only one unknown (the common final velocity). They are the most straightforward to solve.

The work-energy theorem (Section 7.3) explains where the "lost" kinetic energy goes. During an inelastic collision, internal forces do work that deforms the objects, generates heat, and produces sound. This work transforms organized kinetic energy into disorganized internal energy. The total energy of the universe is still conserved --- but the kinetic energy of the colliding objects is not.

Worked Example: Perfectly Elastic Collision

A 3 kg ball moving at 4 m/s to the right collides head-on with a 1 kg ball at rest. The collision is perfectly elastic. Find the final velocities of both balls.

Using the elastic collision formulas (with $v_{2i} = 0$):

$$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}\,v_{1i} = \frac{3 - 1}{3 + 1}(4) = \frac{2}{4}(4) = 2 \text{ m/s}$$

$$v_{2f} = \frac{2m_1}{m_1 + m_2}\,v_{1i} = \frac{2(3)}{3 + 1}(4) = \frac{6}{4}(4) = 6 \text{ m/s}$$

Check momentum: $p_i = (3)(4) + (1)(0) = 12$ kg m/s. $p_f = (3)(2) + (1)(6) = 6 + 6 = 12$ kg m/s. Conserved.

Check kinetic energy: $KE_i = \frac{1}{2}(3)(4)^2 = 24$ J. $KE_f = \frac{1}{2}(3)(2)^2 + \frac{1}{2}(1)(6)^2 = 6 + 18 = 24$ J. Conserved.

Both conservation laws hold. The heavier ball slows from 4 m/s to 2 m/s; the lighter ball shoots off at 6 m/s. Notice the lighter ball ends up moving faster than the initial speed of the heavier ball --- this is characteristic of elastic collisions with a heavy-to-light mass ratio.

Faded Example: Perfectly Inelastic Collision

A 2 kg cart moving at 5 m/s to the right collides with a 3 kg cart moving at 2 m/s to the left. The carts stick together. Find the final velocity and the kinetic energy lost.

Step 1: Define a sign convention and write the momentum conservation equation.

Check your answer

Take rightward as positive. Then $v_{1i} = +5$ m/s and $v_{2i} = -2$ m/s. $$m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2)v_f$$ $$(2)(5) + (3)(-2) = (2 + 3)v_f$$ $$10 - 6 = 5v_f$$

Step 2: Solve for $v_f$ and interpret its sign.

Check your answer

$$v_f = \frac{4}{5} = 0.8 \text{ m/s}$$ The positive sign means the combined carts move to the right. The heavier cart was moving left, but the lighter cart's higher speed gave it more momentum to the right, so the right-moving momentum wins.

Step 3: Compute the kinetic energy before and after, and find the energy lost.

Check your answer

$$KE_i = \frac{1}{2}(2)(5)^2 + \frac{1}{2}(3)(2)^2 = 25 + 6 = 31 \text{ J}$$ $$KE_f = \frac{1}{2}(5)(0.8)^2 = \frac{1}{2}(5)(0.64) = 1.6 \text{ J}$$ $$\Delta KE = 31 - 1.6 = 29.4 \text{ J lost}$$ That is 94.8% of the initial kinetic energy, lost to deformation, heat, and sound. Nearly all the kinetic energy was destroyed, even though every bit of momentum was conserved. This dramatically illustrates that momentum conservation and energy conservation are independent statements.

Spaced Retrieval

Before moving to practice, test your recall of earlier material.

Recall prompt 1: State the impulse-momentum theorem in words and in an equation. (Section 8.1)

Recall prompt 2: Under what conditions is the total momentum of a system conserved? What role does Newton's third law play? (Section 8.3)

Recall prompt 3: What is the difference between conservative and nonconservative forces? How does this relate to whether a collision is elastic or inelastic? (Section 7.4)

Practice

Layer 1: Concrete

Problem 1. A 0.5 kg ball moving at 6 m/s to the right collides head-on with a 0.5 kg ball at rest. The collision is perfectly elastic. Find the final velocity of each ball.

Check your answer

For equal masses with the target at rest, the elastic collision formulas give: $$v_{1f} = \frac{m_1 - m_2}{m_1 + m_2}v_{1i} = \frac{0.5 - 0.5}{0.5 + 0.5}(6) = 0 \text{ m/s}$$ $$v_{2f} = \frac{2m_1}{m_1 + m_2}v_{1i} = \frac{2(0.5)}{1.0}(6) = 6 \text{ m/s}$$ The first ball stops completely. The second ball moves at 6 m/s to the right. All the motion transfers from one ball to the other --- this is the Newton's cradle result.

Problem 2. A 4 kg truck moving at 3 m/s collides with a 2 kg cart at rest. The objects stick together. Find the final velocity and the kinetic energy lost.

Check your answer

Momentum conservation: $$m_1 v_{1i} = (m_1 + m_2)v_f$$ $$(4)(3) = (6)v_f$$ $$v_f = 2 \text{ m/s}$$ Kinetic energy before: $KE_i = \frac{1}{2}(4)(3)^2 = 18$ J. Kinetic energy after: $KE_f = \frac{1}{2}(6)(2)^2 = 12$ J. Energy lost: $\Delta KE = 18 - 12 = 6$ J. That is one-third of the initial kinetic energy, lost to deformation, heat, and sound. You can verify this with the formula: $\Delta KE = \frac{1}{2}\frac{m_1 m_2}{m_1 + m_2}v_{1i}^2 = \frac{1}{2}\frac{(4)(2)}{6}(3)^2 = \frac{1}{2}\frac{8}{6}(9) = 6$ J. It matches.

Layer 2: Pattern

Problem 3. In each scenario below, you are given the initial and final states of a one-dimensional collision. Classify each as elastic, inelastic, or impossible (violates a conservation law).

(a) $m_1 = 2$ kg, $m_2 = 2$ kg. Before: $v_{1i} = 3$ m/s, $v_{2i} = 0$. After: $v_{1f} = 0$, $v_{2f} = 3$ m/s.

(b) $m_1 = 2$ kg, $m_2 = 2$ kg. Before: $v_{1i} = 3$ m/s, $v_{2i} = 0$. After: $v_{1f} = 1$ m/s, $v_{2f} = 2$ m/s.

(c) $m_1 = 2$ kg, $m_2 = 2$ kg. Before: $v_{1i} = 3$ m/s, $v_{2i} = 0$. After: $v_{1f} = -1$ m/s, $v_{2f} = 3$ m/s.

(d) $m_1 = 1$ kg, $m_2 = 3$ kg. Before: $v_{1i} = 4$ m/s, $v_{2i} = 0$. After: $v_{1f} = -2$ m/s, $v_{2f} = 2$ m/s.

Check your answer

For each case, check momentum conservation first, then compare kinetic energies. **(a)** Momentum: $p_i = (2)(3) + (2)(0) = 6$. $p_f = (2)(0) + (2)(3) = 6$. Conserved. $KE_i = \frac{1}{2}(2)(9) = 9$ J. $KE_f = \frac{1}{2}(2)(9) = 9$ J. Conserved. **Elastic.** **(b)** Momentum: $p_i = 6$. $p_f = (2)(1) + (2)(2) = 2 + 4 = 6$. Conserved. $KE_i = 9$ J. $KE_f = \frac{1}{2}(2)(1) + \frac{1}{2}(2)(4) = 1 + 4 = 5$ J. Kinetic energy decreased. **Inelastic.** **(c)** Momentum: $p_i = 6$. $p_f = (2)(-1) + (2)(3) = -2 + 6 = 4$. Momentum is NOT conserved ($6 \neq 4$). **Impossible.** This collision violates momentum conservation and cannot occur in an isolated system. **(d)** Momentum: $p_i = (1)(4) + (3)(0) = 4$. $p_f = (1)(-2) + (3)(2) = -2 + 6 = 4$. Conserved. $KE_i = \frac{1}{2}(1)(16) = 8$ J. $KE_f = \frac{1}{2}(1)(4) + \frac{1}{2}(3)(4) = 2 + 6 = 8$ J. Conserved. **Elastic.** The lesson: you classify a collision by checking the numbers, not by guessing from the description. Momentum conservation is non-negotiable for isolated systems. If it fails, the scenario is impossible. If momentum is conserved, compare kinetic energies to determine the type.

Layer 3: Structure

Problem 4. Why does a perfectly inelastic collision lose the most kinetic energy of any collision type (for given initial conditions)?

Check your answer

The total kinetic energy of a system can be split into two parts: the kinetic energy of the center-of-mass motion, and the kinetic energy of motion relative to the center of mass. $$KE_{\text{total}} = KE_{\text{cm}} + KE_{\text{relative}}$$ Momentum conservation fixes the center-of-mass velocity (it is the same before and after any collision). Therefore $KE_{\text{cm}}$ is the same before and after. The only kinetic energy that *can* change is $KE_{\text{relative}}$ --- the energy of motion of the objects relative to each other. In a perfectly inelastic collision, the objects stick together and move at the same velocity. Their relative velocity is zero. Therefore $KE_{\text{relative}} = 0$ after the collision. All the relative kinetic energy has been destroyed. In any other collision type, the objects have some relative motion after impact, so $KE_{\text{relative}} > 0$ after the collision. Less kinetic energy is lost. In a perfectly elastic collision, $KE_{\text{relative}}$ is fully preserved (the relative speed is the same before and after, just reversed in direction). No kinetic energy is lost at all. The perfectly inelastic collision is the extreme that eliminates all relative motion. You cannot lose more kinetic energy than that without violating momentum conservation.

Layer 4: Debug

Problem 5. A student is solving the following problem: "A 3 kg ball moving at 4 m/s to the right collides with a 1 kg ball at rest. The objects stick together. Find the final velocity."

The student writes:

"Since the collision conserves momentum and energy:"

$$3(4) + 1(0) = 4v_f \quad \Rightarrow \quad v_f = 3 \text{ m/s}$$

$$\frac{1}{2}(3)(4)^2 + 0 = \frac{1}{2}(4)(v_f)^2 \quad \Rightarrow \quad 24 = 2v_f^2 \quad \Rightarrow \quad v_f = \sqrt{12} \approx 3.46 \text{ m/s}$$

"I got two different answers. I'll average them: $v_f \approx 3.23$ m/s."

What went wrong?

Check your answer

The student applied both momentum conservation *and* kinetic energy conservation to a perfectly inelastic collision. This is the core error. A perfectly inelastic collision does **not** conserve kinetic energy. The two conservation equations are inconsistent --- they give different values for $v_f$ --- because one of them (kinetic energy conservation) does not apply. There is no way to reconcile them, and averaging is not a valid resolution. The correct approach: use **only** momentum conservation for a perfectly inelastic collision. $$v_f = \frac{m_1 v_{1i} + m_2 v_{2i}}{m_1 + m_2} = \frac{12}{4} = 3 \text{ m/s}$$ The kinetic energy equation gives $v_f = 3.46$ m/s, but this is the speed that *would* be needed to conserve kinetic energy. Since the actual $v_f = 3$ m/s is less than 3.46 m/s, kinetic energy has decreased --- confirming that the collision is inelastic. The deeper lesson: the type of collision determines which equations you are allowed to use. Using kinetic energy conservation when it does not hold produces contradictions. If you get two different answers from momentum and energy, that is not a sign to average --- it is a sign that one equation does not apply.

Reflection

After a collision, how do you determine what kind of collision it was?

You measure the initial and final states --- the masses and velocities of each object before and after the collision --- and then you compute. First, check momentum: is $\vec{p}{\text{total}}$ the same before and after? It should be, for an isolated system. Then check kinetic energy: is $KE$ the same before and after? If yes, the collision was elastic. If kinetic energy decreased, the collision was inelastic. If the objects stuck together, it was perfectly inelastic.}

The collision type is not something you assume in advance. It is something you determine from the data. This is a critical habit: check your conservation laws against the evidence, rather than imposing them by assumption.

Looking Ahead

So far, every collision in this section has been one-dimensional --- objects moving along a single line. But real collisions rarely cooperate. A cue ball strikes an object ball off-center, and both fly off at angles. A car T-bones another at an intersection, and the wreckage slides diagonally.

In Section 8.5, you will extend collision analysis to two dimensions. Momentum conservation will give you two equations --- one for each component --- and you will see a beautiful geometric result: when two equal-mass objects collide elastically with one initially at rest, their paths after the collision are always perpendicular. The vector nature of momentum, which was easy to overlook in one dimension, becomes essential.