Types of collisions (elastic and inelastic)

Two steel balls click together on a desk and spring apart, each carrying away almost exactly the energy it arrived with. Drop a lump of wet clay onto the floor and it just splats — momentum is passed on, but the clay and the floor together are left warmer, not bouncier. Both events are collisions. Both obey the same iron rule of momentum conservation. Yet they feel completely different, and physics gives that difference a name: elastic collisions conserve kinetic energy, inelastic collisions do not. This distinction is not a footnote — it is the single fact that tells you which equation you are allowed to write down when two bodies collide. Momentum conservation, $m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2$ , is always true for an isolated system, elastic or not. Kinetic energy conservation is a bonus that only elastic collisions give you, and reaching for it in an inelastic problem is one of the most common ways marks are lost in this chapter. This article builds the classification from the ground up: what "elastic" and "inelastic" actually mean physically, why a perfectly inelastic collision is the cleanest case to compute, how the two conservation laws combine to predict the surprising "velocity swap" for equal masses colliding elastically, and how much kinetic energy genuinely disappears — and where it goes — when two bodies stick. Two worked numbers, a $2\ \text{kg}$ -on- $2\ \text{kg}$ elastic collision and a $4\ \text{kg}$ -into- $6\ \text{kg}$ perfectly inelastic collision, are carried through in full so you can see exactly which conservation law does which job.

Key takeaways

Momentum is conserved in every collision — elastic or inelastic — as long as no net external force acts during the brief impact; this follows directly from Newton's third law on the internal collision forces.
Elastic collisions also conserve kinetic energy. $\tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2$ holds only for this special class of collision.
Inelastic collisions lose kinetic energy to heat, sound and permanent deformation; momentum is still conserved, but you cannot write a KE-conservation equation.
Perfectly inelastic is the extreme case: the bodies stick and move together at one common velocity $v = \dfrac{m_1u_1+m_2u_2}{m_1+m_2}$ , found from momentum conservation alone.
Equal-mass elastic head-on collisions swap velocities — a moving ball hitting an identical stationary ball stops dead, handing over its entire speed; this is a derived special case, not a separate rule to memorise.
The coefficient of restitution (qualitative for Class 11) is a single number between 0 and 1 that measures how elastic a real collision is — 1 for perfectly elastic, 0 for perfectly inelastic.
Key idea: momentum conservation applies to every collision, but only elastic collisions additionally conserve kinetic energy, and the size of the kinetic-energy gap between "before" and "after" is exactly what separates elastic from inelastic and measures how much energy left the mechanical system as heat, sound and deformation.

The core insight

Every collision conserves total momentum because the internal impact forces are equal and opposite by Newton's third law, but only an elastic collision also conserves total kinetic energy, while an inelastic collision — most sharply, a perfectly inelastic one where the bodies stick and share a common velocity $v=(m_1u_1+m_2u_2)/(m_1+m_2)$ — converts part of that kinetic energy irreversibly into heat, sound and deformation.

Where it fits in the syllabus

This topic sits inside the CBSE Class 11 unit Work, Energy and Power, in the prescribed line "elastic and inelastic collisions in one and two dimensions; momentum conservation in collisions." It is the conceptual gate you must pass through before the separate one-dimensional and two-dimensional collision-solving nodes: those nodes hand you ready-made velocity formulas, but those formulas are only valid once you have correctly classified the collision as elastic or inelastic. The 2025-26 CBSE prescription explicitly keeps the coefficient of restitution at a qualitative level only — you should know what it means and its limiting values of 0 and 1, but you are not expected to derive it as a formal law, and no examiner should ask you to.

The physical picture

Picture any collision as a very short, violent conversation between two bodies through a mutual contact force. Whatever momentum one body loses, Newton's third law guarantees the other gains, so long as no outside force (friction, gravity along the line of motion, a wall) intrudes during the brief contact time. That gives the one law that never fails:

m_1\vec u_1 + m_2\vec u_2 = m_1\vec v_1 + m_2\vec v_2

where: $m_1, m_2$ are the two masses, $u_1, u_2$ their velocities just before collision, and $v_1, v_2$ their velocities just after. Read it as a simple accounting statement — total momentum in equals total momentum out — and notice it says nothing at all about energy. Whether energy is also conserved depends on what happens at the point of contact. If the colliding surfaces behave like ideal springs — compressing elastically and pushing back with all the energy they stored — the collision is elastic and a second law becomes available:

\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

where: the left side is the total kinetic energy just before impact and the right side is the total kinetic energy just after. No real macroscopic object is a perfect spring — some sound always escapes, some material always warms — but hardened steel or glass spheres colliding briefly, and the elastic scattering of atoms and molecules, come close enough that this equation is an excellent working model. If instead the surfaces deform permanently, crumple, or the materials grip and stick, the collision is inelastic: momentum conservation still holds, but kinetic energy conservation does not, because energy has genuinely left the mechanical system as heat, sound and permanent strain in the material. The extreme inelastic case is a perfectly inelastic collision, where the two bodies move off glued together with one shared velocity $v$ , and momentum conservation alone becomes:

m_1u_1 + m_2u_2 = (m_1+m_2)v

This single equation is solvable immediately for $v$ because there is now only one unknown — you have traded away the KE equation but gained the constraint that $v_1=v_2=v$ .

Figure

Two collision panels side by side, the top panel showing two balls approaching and then separating after an elastic collision with equal total kinetic energy before and after, the bottom panel showing two blocks approaching and then moving off stuck together after a perfectly inelastic collision with less kinetic energy after than before

The physical picture, continued: kinetic energy loss

For a perfectly inelastic collision the kinetic energy lost is not a mystery quantity — it can be computed directly as the difference between the kinetic energy the system had before impact and the (necessarily smaller) kinetic energy of the single combined mass afterward:

\Delta K = \left(\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2\right) - \frac{1}{2}(m_1+m_2)v^2

where: $\Delta K$ is always positive for a genuine inelastic collision (kinetic energy is lost, never gained, in a collision with no external energy input), and $v$ is the common final velocity found from momentum conservation first. Physically, $\Delta K$ is the energy that went into sound waves radiating away, heat generated by internal friction as the materials deform, and the permanent strain energy locked into the crumpled or fused shape — none of which shows up as macroscopic kinetic energy of the combined body any more.

Derivation AWhy equal masses swap velocities in an elastic head-on collision

The goal here is to show that the "swap" result you see numerically in Worked Example 1 below — a moving ball hitting an identical stationary ball and handing over its entire velocity — is not a lucky coincidence of the particular numbers $5\ \text{m/s}$ and $2\ \text{kg}$ , but a general algebraic consequence of combining momentum conservation with kinetic energy conservation for equal masses. We derive the general one-dimensional elastic collision result for a moving mass $m_1$ striking a stationary mass $m_2$ , using only algebraic factoring, and then set $m_1=m_2$ to recover the swap.

Step 1. Write the two conservation laws for $u_2 = 0$ (mass 2 initially at rest), which is the case both worked examples in this article use. Momentum: $m_1u_1 = m_1v_1 + m_2v_2$ , which rearranges to $m_1(u_1-v_1) = m_2v_2$ . Kinetic energy: $\tfrac12 m_1u_1^2 = \tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2$ , which rearranges to $m_1(u_1^2-v_1^2) = m_2v_2^2$ .

Step 2. Factor the energy equation using the difference of squares, $u_1^2-v_1^2=(u_1-v_1)(u_1+v_1)$ , so the energy equation becomes $m_1(u_1-v_1)(u_1+v_1) = m_2v_2^2$ .

Step 3. Divide the Step 2 result by the Step 1 momentum result (both sides have a common factor $m_1(u_1-v_1)=m_2v_2$ , assuming $u_1\neq v_1$ , i.e. an actual collision occurred). This cancels the $m_2v_2$ on the right and the $m_1(u_1-v_1)$ on the left, leaving simply $u_1+v_1 = v_2$ .

Step 4. Substitute $v_2 = u_1+v_1$ back into the momentum equation $m_1u_1 = m_1v_1+m_2v_2$ : $m_1u_1 = m_1v_1 + m_2(u_1+v_1)$ , so $m_1u_1 - m_2u_1 = m_1v_1+m_2v_1$ , giving $v_1 = \dfrac{(m_1-m_2)}{(m_1+m_2)}u_1$ and then $v_2 = u_1+v_1 = \dfrac{2m_1}{m_1+m_2}u_1$ .

Step 5. Set $m_1=m_2=m$ (equal masses). Then $v_1 = \dfrac{0}{2m}u_1 = 0$ and $v_2 = \dfrac{2m}{2m}u_1 = u_1$ . The incoming ball stops completely and the target ball leaves with exactly the incoming velocity — a full velocity swap, proven in general rather than observed only in one numerical case. The same formulas also show two other limits worth remembering: if $m_1 \gg m_2$ (a bowling ball striking a marble), $v_1 \approx u_1$ and $v_2 \approx 2u_1$ — the heavy body barely slows and the light one shoots off at roughly double the incoming speed; if $m_1 \ll m_2$ (a marble striking a bowling ball), $v_1 \approx -u_1$ — the light body bounces straight back.

Worked examples

Board level - equal-mass elastic head-on collision

A ball of mass $2\ \text{kg}$ moving at $5\ \text{m/s}$ collides elastically head-on with an identical stationary ball of mass $2\ \text{kg}$ . Find the velocities of both balls after the collision.

Step 1. Write momentum conservation for the two equal masses $m$ . $mu_1 = mv_1+mv_2$ , and the common mass cancels: $5 = v_1+v_2$ .

Step 2. Write kinetic energy conservation, and cancel the common factor $\tfrac12 m$ the same way. $u_1^2 = v_1^2+v_2^2$ , so $25 = v_1^2+v_2^2$ .

Step 3. Solve the two equations together. From Step 1, $v_2 = 5-v_1$ . Substituting into Step 2: $25 = v_1^2+(5-v_1)^2 = 2v_1^2-10v_1+25$ , which simplifies to $0 = 2v_1^2-10v_1 = 2v_1(v_1-5)$ , giving $v_1=0$ or $v_1=5$ .

Step 4. Reject the physically impossible root. $v_1=5$ would mean the first ball passed straight through the second untouched, which a real collision cannot do, so $v_1=0$ and $v_2=5-0=5\ \text{m/s}$ . Answer: the moving ball stops dead ( $v_1=0$ ) and the previously stationary ball moves off at $5\ \text{m/s}$ — exactly the general "equal-mass swap" result proved in the derivation above, here recovered by direct substitution of the given numbers.

Competitive level - perfectly inelastic collision with kinetic energy loss

A $4\ \text{kg}$ block moving at $3\ \text{m/s}$ collides with a stationary $6\ \text{kg}$ block and the two stick together (perfectly inelastic). Find (a) their common final velocity and (b) the kinetic energy lost in the collision, and (c) express the loss as a percentage of the initial kinetic energy.

Step 1. Apply momentum conservation with the sticking condition $v_1=v_2=v$ . $m_1u_1+m_2u_2 = (m_1+m_2)v$ , so $4(3)+6(0) = (4+6)v$ , giving $12 = 10v$ and $v = 1.2\ \text{m/s}$ .

Step 2. Compute the initial kinetic energy, using only the moving block since the second block starts at rest. $K_i = \tfrac12(4)(3)^2 = \tfrac12(4)(9) = 18\ \text{J}$ .

Step 3. Compute the final kinetic energy of the combined $10\ \text{kg}$ mass moving at $1.2\ \text{m/s}$ . $K_f = \tfrac12(10)(1.2)^2 = \tfrac12(10)(1.44) = 7.2\ \text{J}$ .

Step 4. Subtract to find the loss, then express it as a percentage. $\Delta K = K_i-K_f = 18-7.2 = 10.8\ \text{J}$ , and the percentage lost is $\dfrac{10.8}{18}\times100 = 60\%$ . Answer: the blocks move off together at $v=1.2\ \text{m/s}$ , losing $10.8\ \text{J}$ of kinetic energy — a full $60\%$ of what the system started with — to heat, sound and deformation as they crumple and fuse. Timing note: this is a two-line problem once you recognise the sticking condition — do not waste time hunting for a second (kinetic energy) equation, because none exists for an inelastic collision. Compute $v$ from momentum first, then compute $K_i$ and $K_f$ separately from the definition $\tfrac12 mv^2$ before subtracting; trying to shortcut this into a single formula under exam pressure is where most arithmetic slips happen. Budget under two minutes for the velocity and under a further minute for the two energy values.

Common traps

Reaching for kinetic energy conservation in an inelastic problem. It is only available in elastic collisions. If the problem says "stick together," "get embedded," "coalesce," or gives you data that would leave two unknowns after using momentum alone, the collision is inelastic and you need the sticking condition $v_1=v_2$ , not a second energy equation.
Thinking momentum is "lost" alongside kinetic energy. It is not. Momentum conservation is a consequence of Newton's third law on the internal contact forces and holds regardless of how the collision deforms the bodies; only kinetic energy is sensitive to whether the deformation is recoverable.
Assuming "inelastic" always means "stuck together." Any collision with $\Delta K > 0$ is inelastic; sticking together is only the extreme (perfectly inelastic) case. Two balls that separate but leave with slightly less total KE than they started with are still inelastic, just not perfectly so.
Applying the equal-mass velocity swap to unequal masses or to inelastic collisions. The derivation above shows the swap is a special case of $v_1=\frac{m_1-m_2}{m_1+m_2}u_1$ and $v_2=\frac{2m_1}{m_1+m_2}u_1$ that only reduces to a clean swap when $m_1=m_2$ and the collision is elastic and it is head-on with $u_2=0$ .
Sign errors when a body moves opposite to the chosen positive direction. Momentum is a vector; a body approaching from the "wrong side" must carry a negative velocity in the conservation equation, or the answer comes out wrong even with perfectly correct algebra otherwise.

Common trapKinetic energy conservation is not a universal law — only momentum is

The single most expensive mistake in this topic is treating $\tfrac12 m_1u_1^2+\tfrac12 m_2u_2^2 = \tfrac12 m_1v_1^2+\tfrac12 m_2v_2^2$ as if it were as automatic as momentum conservation. It is not. Momentum conservation follows from Newton's third law and needs nothing else — it holds whether the balls bounce apart cleanly or the blocks crumple into one lump. Kinetic energy conservation is a special extra property that only holds when no mechanical energy escapes as heat, sound or permanent deformation during the collision, which is only ever approximately true (for hard, elastic-like bodies) and never true at all once the bodies stick. Before writing down any energy equation in a collision problem, first ask whether the problem has told you (or implied, through words like "stick," "embed," or "come to rest together") that the collision is inelastic — if so, momentum conservation is your only tool, and any apparent "extra" unknown is removed by the sticking condition $v_1=v_2$ , not by inventing a kinetic energy equation that does not hold.

You should be able to…

State, in one sentence each, what makes a collision elastic versus inelastic versus perfectly inelastic.
Explain why momentum conservation applies to every collision but kinetic energy conservation applies only to elastic ones.
Derive the equal-mass elastic-collision velocity swap from the two conservation laws, rather than merely quoting it.
Solve for the common final velocity of a perfectly inelastic collision and compute the kinetic energy lost, including as a percentage.
Recognise, from the wording of a problem ("stick," "embed," "bounce back elastically"), which conservation equations are legally available to use.
Explain qualitatively what the coefficient of restitution measures and its two limiting values, without needing to derive it.

Ask the tutor: Want to see how these same two conservation laws extend to a collision where both bodies are moving before impact, or where the collision happens at an angle in two dimensions — ask the AI tutor to walk through the general oblique-collision case next.

If kinetic energy just disappears in an inelastic collision, doesn't that break the law of conservation of energy?

No — total energy (in every form) is always conserved; only mechanical kinetic energy of the visible, bulk motion is not. In an inelastic collision the "missing" kinetic energy has not vanished; it has been converted into other forms of energy that are just as real but harder to see directly — heat in the deformed material (the atoms jiggling faster, which you could measure as a temperature rise), sound waves radiating away as the loud bang or crunch, and permanent strain energy locked into whatever dent or crumple is left behind. Add all of those up along with the remaining kinetic energy and the total is exactly conserved, as the overall law of conservation of energy demands. What is special about kinetic energy conservation in collisions is that it is a stronger, extra condition that only elastic collisions satisfy — it says none of the mechanical energy escaped into those other forms at all.

Why do real collisions between everyday objects always lose some kinetic energy, even hard ones like billiard balls?

Every real material has some internal friction and imperfect elasticity — when two bodies touch and deform even slightly, the compression and rebound is never perfectly reversible at the atomic level, and a little energy always leaks out as sound (the click you hear) and a tiny amount of heat. Billiard balls, hardened steel spheres and similar very hard, smooth objects lose only a small fraction of their kinetic energy this way, which is why treating them as "elastic" is an excellent approximation for problem-solving even though it is never exactly true. At the other extreme, soft or sticky materials like clay or putty convert almost all of the kinetic energy into heat and permanent deformation, which is why they are the standard textbook example of a perfectly (or nearly perfectly) inelastic collision.

Is there a shortcut formula for the final velocities in a general elastic collision, or do I always have to solve the two equations from scratch?

For the general case of an elastic head-on collision where the second body starts at rest, the derivation in this article gives you the shortcut directly: $v_1 = \frac{m_1-m_2}{m_1+m_2}u_1$ for the incoming body and $v_2 = \frac{2m_1}{m_1+m_2}u_1$ for the initially stationary body. These are correct and useful to remember for quick checks, but the fully general one-dimensional elastic collision formulas — including the case where both bodies are moving before impact — are developed in more detail in the dedicated one-dimensional collisions node that follows this one, since deriving and applying those in full is a slightly bigger task than this classification-focused article covers. For exam purposes, it is safest to know how to re-derive the equal-mass swap from the two conservation laws (as shown above) rather than only memorising the final formula, since examiners often ask you to show the working.

How is the coefficient of restitution related to elastic and inelastic collisions, and do I need to calculate it for my board exam?

The coefficient of restitution, usually written $e$ , is a number that compares how fast the two bodies separate after collision to how fast they approached before it — roughly, $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}}$ . It equals exactly $1$ for a perfectly elastic collision (they separate exactly as fast as they approached) and exactly $0$ for a perfectly inelastic collision (they do not separate at all — they move off together). Real collisions have $0 < e < 1$ . The current CBSE Class 11 prescription keeps this at a qualitative, conceptual level only — you should be able to state what $e=1$ and $e=0$ mean and place a given collision roughly on that scale, but you are not expected to derive $e$ as a formal law or use it as a computational tool in numerical problems at this stage.