Unit 4 · Topic 4.4

4.4 Elastic and Inelastic Collisions

Every collision with no net external force conserves momentum, but kinetic energy is another story. In an elastic collision the total kinetic energy is the same before and after; in an inelastic collision some becomes thermal energy, sound or deformation; and in a perfectly inelastic collision the objects stick together and lose the most kinetic energy that momentum conservation allows.

Key terms

elastic collision
inelastic collision
perfectly inelastic collision
kinetic energy
thermal energy

Types of collisions

Type	Momentum conserved?	Kinetic energy conserved?	What happens
elastic	yes	yes	objects bounce apart with no lasting deformation, like ideal billiard balls or magnets that repel
inelastic	yes	no, some is lost	objects bounce apart but some energy becomes thermal energy, sound or deformation
perfectly inelastic	yes	no, the most is lost	objects stick together and move with one common velocity

Elastic collisions

In an elastic collision, both momentum and total kinetic energy are conserved, giving you two equations. Individual objects can still gain or lose kinetic energy; only the total stays fixed.

For a head-on elastic collision of $m_1$ moving at $v_1$ into $m_2$ at rest, the two equations give:

$v_1' = \frac{m_1 - m_2}{m_1 + m_2}v_1, \qquad v_2' = \frac{2m_1}{m_1 + m_2}v_1$

You don't need to memorize these, but they show useful patterns. Equal masses swap velocities: the moving one stops dead. A light object bouncing off a heavy one comes back at almost the same speed. A heavy object hitting a light one barely slows down. In any head-on elastic collision, the objects separate at the same relative speed they approached with.

Inelastic and perfectly inelastic collisions

Most real collisions are inelastic. Momentum is still conserved (if the external force is negligible), but kinetic energy isn't, so you have only the momentum equation unless you're given more information.

In a perfectly inelastic collision, the objects stick together, so there's only one final velocity: $m_1v_1 + m_2v_2 = (m_1 + m_2)v_f$ . That's just the center-of-mass velocity, which doesn't change. The kinetic energy left over is only the energy of the center of mass's motion; everything else is lost.

Telling them apart, and two-stage problems

To classify a collision, compute the total kinetic energy before and after. Equal means elastic; less after means inelastic; objects moving together means perfectly inelastic. (More after means stored energy was released, as in an explosion.)

Many problems chain a collision with another process. In a ballistic pendulum, a bullet embeds in a hanging block, which then swings up. Stage 1, the collision: momentum is conserved but kinetic energy isn't. Stage 2, the swing: mechanical energy is conserved but momentum isn't (gravity and tension act). Use the right law for each stage, and never use energy conservation across a sticking collision.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Ballistic pendulum (classic trap)
A 0.010 kg bullet embeds itself in a 2.0 kg wooden block hanging from strings. The block and bullet swing up 0.10 m. Find the bullet's speed before impact and the fraction of its kinetic energy lost in the collision. Use g = 9.8 m/s².
Show the solutionHide the solution
1. Step 1: Stage 2 (swing, mechanical energy conserved): $\frac{1}{2}(m + M)V^2 = (m + M)gh$ , so $V = \sqrt{2gh} = \sqrt{2(9.8)(0.10)} = 1.4$ m/s just after the collision.
2. Step 2: Stage 1 (collision, momentum conserved): $mv = (m + M)V$ , so $v = \frac{2.010}{0.010}(1.4) \approx 281$ m/s.
3. Step 3: Kinetic energy before: $\frac{1}{2}(0.010)(281.4)^2 \approx 396$ J. Just after: $\frac{1}{2}(2.010)(1.4)^2 \approx 1.97$ J. About 99.5% was lost to thermal energy and deformation.
4. Step 4: The trap is setting the bullet's kinetic energy equal to (m + M)gh. That ignores the huge energy loss in the collision and gives about 20 m/s, far too low.
Answer: About 280 m/s; about 99.5% of the kinetic energy is lost.
Example 2
A head-on elastic collision
A 1.0 kg cart moving at 4.0 m/s collides elastically and head-on with a 3.0 kg cart at rest. Find both velocities afterward and check that kinetic energy is conserved.
Show the solutionHide the solution
1. Step 1: $v_1' = \frac{1.0 - 3.0}{4.0}(4.0) = -2.0$ m/s (it bounces back) and $v_2' = \frac{2(1.0)}{4.0}(4.0) = 2.0$ m/s.
2. Step 2: Momentum check: before, (1.0)(4.0) = 4.0; after, (1.0)(−2.0) + (3.0)(2.0) = 4.0 kg·m/s. ✓
3. Step 3: Kinetic energy check: before, $\frac{1}{2}(1.0)(4.0)^2 = 8.0$ J; after, $\frac{1}{2}(1.0)(2.0)^2 + \frac{1}{2}(3.0)(2.0)^2 = 2.0 + 6.0 = 8.0$ J. ✓
4. Step 4: Relative speed check: they approach at 4.0 m/s and separate at 2.0 − (−2.0) = 4.0 m/s. ✓
Answer: The 1.0 kg cart moves at 2.0 m/s backward; the 3.0 kg cart moves at 2.0 m/s forward.
Example 3Calculator allowed
Classifying a collision
A 2.0 kg puck moving at 5.0 m/s hits a 3.0 kg puck at rest head-on. Afterward the 2.0 kg puck moves at 1.0 m/s in its original direction. Find the 3.0 kg puck's velocity and decide what kind of collision it was.
Show the solutionHide the solution
1. Step 1: Momentum: $(2.0)(5.0) = (2.0)(1.0) + (3.0)v$ , so $v = \frac{8.0}{3.0} \approx 2.67$ m/s forward.
2. Step 2: Kinetic energy before: $\frac{1}{2}(2.0)(5.0)^2 = 25$ J. After: $\frac{1}{2}(2.0)(1.0)^2 + \frac{1}{2}(3.0)(2.67)^2 \approx 1.0 + 10.7 = 11.7$ J.
3. Step 3: Kinetic energy dropped by about 13.3 J and the pucks didn't stick, so the collision was inelastic (but not perfectly inelastic).
Answer: About 2.67 m/s forward; inelastic, with about 13 J of kinetic energy lost.

Common mistakes

Using conservation of kinetic energy for a collision that isn't stated (or shown) to be elastic.
Using energy conservation across a collision where objects stick, as in the ballistic pendulum. Use momentum for the collision, energy for the swing.
Thinking momentum is lost in an inelastic collision. Kinetic energy is lost; momentum is still conserved.
Assuming each object keeps its kinetic energy in an elastic collision. Only the total is conserved.

On the exam

Expect a question asking you to classify a collision from data. Calculate the total kinetic energy before and after and compare them; a claim without the numbers rarely earns full credit.
Two-stage problems (collision then swing, slide or spring compression) are common on free response. Say which conservation law applies in each stage and why.

Connected topics

Videos

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Check yourself

4 questions on 4.4 Elastic and Inelastic Collisions. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A billiard ball moving at 2.0 m/s hits an identical ball at rest head-on, and the collision is elastic. What are the balls' velocities afterward?

A 2.0 kg cart moving at 6.0 m/s to the right collides head-on with a 4.0 kg cart at rest on a frictionless track.

Described situation

Question 2 of 4Calculator allowed

If the collision is elastic, what is the 2.0 kg cart's velocity afterward?

Question 3 of 4Calculator allowed

If instead the carts stick together, how much kinetic energy is converted to other forms?

Question 4 of 4Calculator allowed

A puck slides across frictionless ice and makes a glancing (not head-on) elastic collision with an identical puck at rest. What is the angle between the two pucks' velocities after the collision?

0 of 4 answered