Elastic Collisions

A carrom striker slides across the board and hits a stationary coin dead-on. The striker stops. The coin shoots forward at roughly the speed the striker had. If you have played carrom, you have seen this a hundred times and never thought twice about it — but something remarkable just happened. The striker transferred all of its motion to the coin: its speed, its energy, everything. Nothing was lost to sound or heat. The coin left with exactly what the striker brought in.

This is an elastic collision — the physicist's ideal. In the real world, collisions always lose a little energy to sound, deformation, or heat. But some collisions come close enough to the ideal that treating them as perfectly elastic gives you the right answer. A steel ball bearing bouncing off another steel ball bearing. Two billiard balls clicking together on a table. A cricket ball hitting the sweet spot of a bat (close enough for a first approximation). These are the collisions where the physics is cleanest and the algebra gives you everything.

What makes a collision "elastic"

Every collision obeys conservation of momentum — that is Newton's third law at work, and no collision can escape it. But not every collision conserves kinetic energy. When two lumps of clay slam together and stick, momentum is conserved but kinetic energy is not — some of it went into deforming the clay. That is an inelastic collision.

An elastic collision is one where both conservation laws hold simultaneously:

Here u_1, u_2 are the velocities before the collision and v_1, v_2 are the velocities after. Two equations, two unknowns (v_1 and v_2) — this is a system you can solve completely.

The physical content is this: no energy goes into sound, heat, deformation, or light. Every joule of kinetic energy that existed before the collision still exists after it, just redistributed between the two bodies. That is a strong constraint, and it is what makes elastic collisions special.

Deriving the final velocities — head-on elastic collision

Take the simplest and most important case: mass m_1 moves at velocity u_1 and hits mass m_2, which is initially at rest (u_2 = 0). This is the carrom striker hitting a coin, the cue ball hitting the eight ball, a cricket ball hitting a stationary stump.

Assumptions: The collision is head-on (one-dimensional — both objects move along the same straight line). There are no external forces during the collision. Both momentum and kinetic energy are conserved.

Step 1. Write the conservation of momentum.

Why: the total momentum before (m_1 u_1 + m_2 \cdot 0) equals the total momentum after. This is Newton's third law — the force m_1 exerts on m_2 is equal and opposite to the force m_2 exerts on m_1, so the total momentum cannot change.

Step 2. Write the conservation of kinetic energy.

Why: elastic means no energy is lost. Every joule of kinetic energy in the system before the collision is still kinetic energy after.

Step 3. Rearrange equation (1) to isolate terms by mass.

Why: move the m_1 v_1 term to the left side so that the left has only m_1 terms and the right has only m_2 terms. This grouping will be useful in a moment.

Step 4. Rearrange equation (2) similarly. Cancel the \frac{1}{2} on both sides first.

Why: same grouping strategy — m_1 terms on the left, m_2 on the right. The left side is a difference of squares, which factors nicely.

Step 5. Factor the left side of (4) using a^2 - b^2 = (a+b)(a-b).

Why: factoring the difference of squares creates a factor of (u_1 - v_1) that also appears in equation (3). This is the algebraic trick that makes the system solvable without the quadratic formula.

Step 6. Divide equation (5) by equation (3).

Why: dividing removes m_1(u_1 - v_1) from the left and m_2 v_2 from the right. This is valid as long as u_1 \neq v_1 and v_2 \neq 0 — that is, as long as a collision actually happens. The result is beautifully simple: the separation velocity after the collision equals the approach velocity before it.

Step 7. Substitute v_2 = u_1 + v_1 from equation (6) into equation (1).

Why: now there is only one unknown, v_1. Collect all v_1 terms on one side and all known terms on the other.

Step 8. Solve for v_1.

Why: factor u_1 out of the left side. The final velocity of m_1 depends only on the mass ratio and its initial speed.

Step 9. Substitute v_1 back into equation (6) to get v_2.

Why: combine the fractions by writing 1 as \frac{m_1 + m_2}{m_1 + m_2}. The m_2 terms cancel in the numerator, leaving 2m_1. The target mass m_2 always moves forward (positive v_2) — it gets pushed in the direction the incoming mass was moving.

These two formulas — equations (7) and (8) — are the complete solution to the head-on elastic collision with one body initially at rest. Everything that follows is a special case.

Special cases — the physics hides in the mass ratio

The formulas look algebraic, but the physics is in the ratio m_1/m_2. Three cases reveal what is really going on.

Case 1: Equal masses — the velocity exchange

Set m_1 = m_2. The formulas give:

The incoming mass stops dead. The target mass moves off at exactly the speed the first mass had. This is a complete transfer of motion — the carrom striker stopping while the coin shoots forward, two identical billiard balls exchanging velocities on the table.

This is why Newton's cradle works. When the left ball swings down and strikes the row, the impact transfers through the chain of equal-mass contacts until the ball on the far right swings up. Each intermediate transfer is an equal-mass elastic collision — full velocity exchange.

Case 2: Heavy hits light — the bowling ball

Set m_1 = 5 kg, m_2 = 1 kg, u_1 = 3 m/s.

The heavy ball barely slows down (3 m/s to 2 m/s), but the light ball flies off at 5 m/s — faster than the heavy ball was originally moving. Think of a fast bowler's delivery hitting a bail: the ball barely slows, but the bail goes flying.

In the extreme case where m_1 \gg m_2, the formulas approach v_1 \approx u_1 (the heavy mass is barely affected) and v_2 \approx 2u_1 (the light mass bounces off at nearly twice the incoming speed).

Case 3: Light hits heavy — the bouncing ball

Set m_1 = 1 kg, m_2 = 10 kg, u_1 = 4 m/s.

The light ball bounces back — v_1 is negative. It returns almost as fast as it came, losing barely any speed. The heavy ball nudges forward slightly. Think of a tennis ball hitting a wall (the wall being effectively infinite mass): the ball bounces back at nearly the same speed, and the wall does not move.

In the extreme case where m_2 \gg m_1: v_1 \approx -u_1 (bounces back at the same speed) and v_2 \approx 0 (the heavy mass barely moves).

Explore the mass ratio yourself

The three cases above are snapshots. What happens between them? The interactive figure below lets you drag the mass ratio m_2/m_1 from 0.1 to 10 and watch how the final velocities change. The incoming speed u_1 is fixed at 1 m/s so you can read the formulas directly as fractions.

Worked examples

Common confusions

• "Elastic means bouncy." Not quite. "Elastic" means kinetic energy is conserved — it has nothing to do with whether the objects bounce or stick. A superball bouncing off a floor is close to elastic because very little energy is lost. Two billiard balls clicking together are close to elastic. But a ball of rubber putty bouncing off a wall might look "bouncy" yet lose significant kinetic energy to deformation — that is not elastic.

• "Momentum conservation is special to elastic collisions." No — momentum is conserved in every collision, elastic or inelastic. What is special about elastic collisions is that kinetic energy is also conserved. Momentum conservation is Newton's third law; it cannot be violated by any internal force.

• "If the collision is elastic, nothing changes." Wrong interpretation. The total kinetic energy is conserved, but the distribution of energy between the two objects can change dramatically. In the equal-mass case, 100% of the energy transfers from one object to the other. In the heavy-on-light case, a small ball can fly off at nearly twice the incoming speed.

• "The formulas work for both objects moving." They do, but the formulas derived above assume u_2 = 0. For the general case where both objects are moving, use the full formulas:

These reduce to equations (7) and (8) when u_2 = 0.

• "In a real elastic collision, the objects don't touch for long." There is no requirement on the duration of contact. The collision could be instantaneous (hard spheres) or take a finite time (spring-like deformation and rebound). What matters is the end state: did all kinetic energy survive? If yes, the collision is elastic regardless of how it happened.

Where this leads next

• Inelastic Collisions — what happens when kinetic energy is not conserved, including the perfectly inelastic case where objects stick together.
• Coefficient of Restitution — the general framework for classifying collisions by how much energy they lose.
• Centre of Mass — the reference frame that makes multi-body problems simpler, and the key to understanding oblique collisions.
• Impulse and Momentum — the force-time integral that connects Newton's second law to momentum change during a collision.
• Conservation of Momentum — the foundational law underlying all collision physics.