Elastic and inelastic collisions

A collision is a brief, forceful interaction between objects. Two conserved quantities are on the table, and the whole topic is keeping them straight. Momentum is conserved in every collision of an isolated system. Kinetic energy is conserved only when the collision is elastic.

That difference is the whole point. Momentum always balances, so it is your workhorse for finding final velocities. Kinetic energy balances only in the elastic case, so use it as a second equation only when the problem says the collision is elastic. Classifying a collision is therefore an energy question, not a question about whether the objects bounced or stuck.

For students arriving from algebra-based Physics 1, almost all of this is familiar: the definitions and the one-dimensional algebra are the same. There is no new calculus in this topic. The Physics C addition is the two-dimensional work, which leans on resolving momentum into components and writing one conservation equation per axis.

The three labels are defined by what happens to kinetic energy, and by nothing else.

Elastic. The total kinetic energy is the same before and after. The objects rebound with no energy lost to heat, sound, or permanent deformation. Two hard balls in a nearly silent, nearly frictionless collision approximate this.

Inelastic. Some kinetic energy is lost. The objects can still bounce apart; inelastic only means the energy ledger does not balance. Most real collisions are inelastic.

Perfectly inelastic. The objects stick together and move off with one shared velocity. This is the extreme case, losing the most kinetic energy that momentum conservation allows. Sticking is enough to make a collision perfectly inelastic, but it is not the meaning of inelastic in general.

A compact way to capture the whole spectrum is the coefficient of restitution $e$, the ratio of the relative speed of separation to the relative speed of approach. Elastic is $e = 1$, perfectly inelastic is $e = 0$, and partially inelastic collisions fall in between. A collision with $e = 0.4$ still separates, yet it is inelastic, because kinetic energy was lost.

Elastic, inelastic, and perfectly inelastic are energy categories. Compare the total kinetic energy before and after; do not classify by bouncing, sticking, or what the objects are made of.

During a collision the two objects push on each other. By Newton's third law those forces are equal in size and opposite in direction at every instant. The impulse one object delivers is exactly the negative of the impulse it receives, so the total momentum of the pair does not change.

In the language of Physics C, the net external force on the isolated pair is zero during the brief contact, so $\frac{d\vec{p}}{dt} = 0$ for the system and the total momentum $\vec{p}$ stays constant. Equivalently, the internal impulses cancel: $\int \vec{F}_{12}\,dt + \int \vec{F}_{21}\,dt = 0$. None of this depends on whether energy is conserved.

The practical rule: write total momentum before equals total momentum after for any collision, elastic or not. In one dimension that is a single signed equation. The common error is to assume that because energy was lost the momentum was lost too. It was not; the missing energy became heat and deformation, which carry no net momentum.

Momentum conservation depends on the system being isolated, not on the collision being elastic. Total momentum before equals total momentum after, every time.

Kinetic energy, $K = \frac{1}{2}mv^2$ summed over the objects, is conserved only in an elastic collision. In any inelastic collision some of it is converted to heat, sound, and permanent deformation. In a perfectly inelastic collision the loss is the largest possible while momentum is still conserved.

Two consequences for solving problems. First, never set the kinetic energy before equal to the kinetic energy after unless the problem states the collision is elastic; doing so in an inelastic collision is using the wrong law and gives a wrong speed. Second, an elastic collision hands you two equations, momentum and kinetic energy, which is exactly what you need to solve for two unknown final velocities.

A classic two-phase problem is the ballistic pendulum: a bullet embeds in a hanging block, then the block swings upward. The embedding is perfectly inelastic, so only momentum is conserved through it; use $m v = (m + M)V$ to find the speed just after impact. The swing that follows conserves mechanical energy, so $\frac{1}{2}(m + M)V^2 = (m + M)g h$. Trying to conserve kinetic energy through the embedding itself is the signature mistake.

Kinetic energy is conserved only when the collision is elastic. Reach for an energy equation only after the problem tells you the collision is elastic, and keep collision and post-collision phases separate.

Momentum is a vector, so in two dimensions conservation is two statements at once: the $\hat{\text{ı}}$ component of the total momentum is conserved, and the $\hat{\text{ȷ}}$ component is conserved, each on its own. Resolve every velocity into components, write one conservation equation per axis, solve, and only at the end recombine into a magnitude or a direction.

Concretely, for two objects, $m_1 v_{1x} + m_2 v_{2x}$ is the same before and after, and $m_1 v_{1y} + m_2 v_{2y}$ is the same before and after. An explosion is the same idea run backward: a firecracker at rest has zero total momentum, so its fragments fly off with momentum components that sum to zero along $\hat{\text{ı}}$ and to zero along $\hat{\text{ȷ}}$.

The trap is treating momentum like a scalar: conserving the total speed, or the magnitude of the total momentum, or adding momentum magnitudes directly. Perpendicular momenta of $3$ and $4$ combine to a magnitude of $\sqrt{3^2 + 4^2} = 5$, not $7$. Conserve the components, and compute magnitudes last.

In two dimensions, conserve the $\hat{\text{ı}}$ and $\hat{\text{ȷ}}$ components of momentum separately. Recombine with Pythagoras only at the end; never conserve a single speed or a bare magnitude.

Each one is a collision problem breaking down at a different step: reading the collision type off how the objects move, reaching for the wrong conservation law, or flattening a two-dimensional impact into a single number.