In short

An inelastic collision is one where momentum is conserved but kinetic energy is not — some energy goes into deformation, sound, or heat. In the extreme case (perfectly inelastic), the two bodies stick together and the final velocity is v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}. The coefficient of restitution e classifies every collision: e = 0 is perfectly inelastic, e = 1 is perfectly elastic, and real collisions fall somewhere between.

A fast bowler sends a delivery at 140 km/h. The wicket-keeper dives and catches it clean — the ball thuds into the gloves and stops. For an instant, the keeper drifts backward. Their palms sting.

That sting is kinetic energy leaving the system. Before the catch, the ball carried over 120 joules of kinetic energy. After the catch, almost all of it is gone — converted into the compression of glove padding, the warmth of leather, the thwack that reached the boundary, and the ache in the keeper's fingers. The ball's momentum transferred to the keeper's hands — momentum is always conserved, no collision escapes Newton's third law. But the kinetic energy? Over 90% of it was destroyed.

This is an inelastic collision. Unlike the idealised elastic collisions where kinetic energy bounces back perfectly, inelastic collisions are what actually happen every day — and the physics of how much energy survives is what this article is about.

What makes a collision "inelastic"

Every collision conserves momentum. That is non-negotiable — it follows directly from Newton's third law, and no internal force can change the total momentum of an isolated system. The question that separates collision types is: what happens to the kinetic energy?

In an elastic collision, kinetic energy is also conserved — every joule that existed before the collision still exists after, just redistributed. A carrom striker transferring its speed perfectly to a coin is close to elastic.

In an inelastic collision, kinetic energy is not conserved. Some of it converts into:

  • Deformation — the crumpling of metal in a car crash, the squishing of a ball of clay
  • Heat — the warmth you feel in a cricket ball that has just been caught
  • Sound — the thwack of a catch, the bang of vehicles colliding
  • Internal vibration — oscillations within the colliding bodies that eventually dissipate

The momentum equation still holds:

m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2

But the kinetic energy equation becomes an inequality:

\tfrac{1}{2}m_1 v_1^2 + \tfrac{1}{2}m_2 v_2^2 \;<\; \tfrac{1}{2}m_1 u_1^2 + \tfrac{1}{2}m_2 u_2^2

With only one equation (momentum) and two unknowns (v_1 and v_2), you cannot solve the problem without additional information. That information comes from the coefficient of restitution — which you will meet shortly. But first, the simplest and most extreme case.

The perfectly inelastic collision — they stick together

The most extreme inelastic collision is when the two objects stick together after impact and move as a single body. A ball of wet clay hitting a wooden block. A bullet embedding in a sandbag. A wicket-keeper catching a delivery. Two coupled railway carriages locking bumpers.

When the objects stick, there is only one final velocity — the velocity of the combined mass.

Step 1. Write conservation of momentum. Take m_2 at rest (u_2 = 0) for the standard case.

m_1 u_1 = (m_1 + m_2)\,v \tag{1}

Why: after the collision, both masses move together with the same velocity v. The total mass is m_1 + m_2. Momentum before equals momentum after.

Step 2. Solve for the final velocity.

\boxed{v = \frac{m_1}{m_1 + m_2}\,u_1} \tag{2}

Why: divide both sides by (m_1 + m_2). The combined mass is always larger than m_1 alone, so v < u_1 always — the merged body moves slower than the incoming one. The heavier the target m_2, the slower the result.

That is the entire calculation. One equation, one unknown, one line of algebra. The simplicity is the point — when objects stick together, you do not need the energy equation at all. Momentum alone gives you the answer.

Animated: perfectly inelastic collision between a 2 kg and a 3 kg ball Ball A (2 kg) moves right at 3 m/s and hits Ball B (3 kg, stationary) at t = 2 s. They stick together and move at 1.2 m/s. The dramatic slowdown shows kinetic energy loss.
Ball A (red, 2 kg) approaches at 3 m/s. At the collision, both balls stick and crawl together at just 1.2 m/s — less than half the original speed. The trails show the dramatic deceleration. Click replay to watch again.

Watch the trails. Before the collision, the red trail grows quickly — Ball A covers ground fast. After the collision, the combined mass barely moves. The speed dropped from 3 m/s to 1.2 m/s. Where did the kinetic energy go?

How much energy is lost?

This is the central question for inelastic collisions. The momentum equation tells you the final velocity; now you want to know how much kinetic energy was destroyed.

Step 1. Write the kinetic energy before the collision (u_2 = 0).

KE_{\text{before}} = \tfrac{1}{2}m_1 u_1^2 \tag{3}

Step 2. Write the kinetic energy after (for a perfectly inelastic collision).

KE_{\text{after}} = \tfrac{1}{2}(m_1 + m_2)\,v^2 = \tfrac{1}{2}(m_1 + m_2)\left(\frac{m_1 u_1}{m_1 + m_2}\right)^2
KE_{\text{after}} = \frac{m_1^2\, u_1^2}{2(m_1 + m_2)} \tag{4}

Why: substitute v from equation (2) and simplify. One factor of (m_1 + m_2) cancels between the numerator and denominator.

Step 3. Compute the energy lost.

\Delta KE = KE_{\text{before}} - KE_{\text{after}} = \tfrac{1}{2}m_1 u_1^2 - \frac{m_1^2\, u_1^2}{2(m_1 + m_2)}
\Delta KE = \frac{m_1 u_1^2}{2}\left(1 - \frac{m_1}{m_1 + m_2}\right) = \frac{m_1 u_1^2}{2} \cdot \frac{m_2}{m_1 + m_2}
\boxed{\Delta KE = \frac{m_1\, m_2}{2(m_1 + m_2)}\,u_1^2} \tag{5}

Why: factor out \frac{m_1 u_1^2}{2} and simplify. The fraction m_2/(m_1 + m_2) is the fraction of kinetic energy lost — it depends only on the mass ratio.

Step 4. Express the fractional energy loss.

\frac{\Delta KE}{KE_{\text{before}}} = \frac{m_2}{m_1 + m_2} \tag{6}

Why: divide equation (5) by equation (3). The m_1 u_1^2/2 cancels, leaving a pure mass ratio. The heavier the target relative to the projectile, the more energy is destroyed.

This result is worth pausing over. If a 160 g cricket ball embeds in a 2 kg block, the fraction of energy lost is 2/(0.16 + 2) = 0.926 — over 92%. If two equal masses collide and stick, the fraction is m/(m + m) = 1/2 — exactly half.

The perfectly inelastic collision always loses the maximum possible kinetic energy consistent with momentum conservation. Here is why: for a system with total momentum p, the minimum kinetic energy is p^2/(2M_{\text{total}}) — achieved when all the mass moves at the same velocity. Any other distribution of the same momentum between separately-moving bodies gives higher kinetic energy (splitting the momentum always costs more energy). So the stuck-together case, where both bodies share one velocity, is the minimum-KE state — and therefore the maximum-energy-loss state.

The coefficient of restitution — measuring the bounce

Real collisions fall on a spectrum between perfectly elastic and perfectly inelastic. A steel ball bearing bouncing off granite is close to elastic. A ball of wet clay hitting a floor is close to perfectly inelastic. A cricket ball bouncing off a pitch is somewhere in between. The quantity that locates a collision on this spectrum is the coefficient of restitution, e.

Coefficient of Restitution

e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{v_2 - v_1}{u_1 - u_2}

where u_1, u_2 are the velocities before the collision and v_1, v_2 are the velocities after. The value satisfies 0 \le e \le 1.

The physical meaning is direct. Think of e as a measure of how well the collision preserves relative motion. Before the collision, the two objects approach each other at some relative speed. After, they separate at some (usually lower) relative speed. The ratio is e.

  • e = 1: the objects separate just as fast as they approached. No kinetic energy lost. Perfectly elastic.
  • e = 0: the objects do not separate at all. They stick together. Maximum kinetic energy lost. Perfectly inelastic.
  • 0 < e < 1: the objects separate, but slower than they approached. Some kinetic energy lost. Every real collision.
The coefficient of restitution spectrum A horizontal scale from e = 0 (perfectly inelastic) to e = 1 (perfectly elastic) with real-world collisions placed at their approximate e values. 0 perfectly inelastic 1 perfectly elastic wet clay car crash cricket on pitch tennis ball steel bearing
Where real collisions fall on the $e$ spectrum. Most everyday collisions cluster in the range $e = 0.2$ to $0.8$.
Collision Approximate e
Ball of wet clay on floor 0.05–0.1
Automobile crash (crumple zones active) 0.2–0.3
Cricket ball bouncing on a pitch 0.4–0.6
Tennis ball on a racquet 0.7–0.8
Rubber ball on concrete 0.8–0.9
Hardened steel ball bearing on steel plate 0.95+

Solving any collision — momentum plus restitution

The coefficient of restitution gives you the missing equation. With momentum conservation and the restitution relation, you can solve any one-dimensional collision — elastic, inelastic, or anything in between.

Take m_1 moving at u_1, m_2 at rest (u_2 = 0).

Step 1. Write conservation of momentum.

m_1 u_1 = m_1 v_1 + m_2 v_2 \tag{7}

Step 2. Write the restitution relation.

v_2 - v_1 = e\,u_1 \tag{8}

Why: with u_2 = 0, the relative approach speed is u_1 - 0 = u_1, and the relative separation speed is v_2 - v_1. Their ratio is e by definition.

Step 3. From equation (8): v_2 = v_1 + e\,u_1. Substitute into equation (7).

m_1 u_1 = m_1 v_1 + m_2(v_1 + e\,u_1)
m_1 u_1 = v_1(m_1 + m_2) + e\,m_2\,u_1
v_1(m_1 + m_2) = u_1(m_1 - e\,m_2)

Why: expand, collect all v_1 terms on one side and all known terms on the other. Now v_1 is the only unknown.

Step 4. Solve for v_1.

\boxed{v_1 = \frac{m_1 - e\,m_2}{m_1 + m_2}\,u_1} \tag{9}

Why: divide both sides by (m_1 + m_2). This is the general formula — it contains both the elastic result (e = 1 gives v_1 = \frac{m_1 - m_2}{m_1 + m_2}\,u_1) and the perfectly inelastic result (e = 0 gives v_1 = \frac{m_1}{m_1 + m_2}\,u_1) as special cases.

Step 5. Substitute back to get v_2.

v_2 = v_1 + e\,u_1 = \frac{m_1 - e\,m_2}{m_1 + m_2}\,u_1 + e\,u_1
v_2 = u_1\left(\frac{m_1 - e\,m_2 + e\,m_1 + e\,m_2}{m_1 + m_2}\right) = u_1 \cdot \frac{m_1(1 + e)}{m_1 + m_2}
\boxed{v_2 = \frac{m_1(1 + e)}{m_1 + m_2}\,u_1} \tag{10}

Why: write e\,u_1 as \frac{e(m_1 + m_2)}{m_1 + m_2}\,u_1 and combine the fractions. The e\,m_2 terms cancel in the numerator, leaving m_1(1 + e). When e = 1, this gives 2m_1 u_1/(m_1 + m_2) — the elastic formula. When e = 0, it gives m_1 u_1/(m_1 + m_2) — the stuck-together formula.

Step 6. Energy loss for the general case.

After substituting equations (9) and (10) into the kinetic energy expressions and simplifying (the algebra involves expanding two squares and combining terms):

\boxed{\Delta KE = \frac{m_1\,m_2\,(1 - e^2)}{2(m_1 + m_2)}\,u_1^2} \tag{11}

Why: the factor (1 - e^2) captures the energy loss cleanly. For e = 1: \Delta KE = 0 — elastic, no loss. For e = 0: the formula reduces to equation (5) — maximum loss. The quantity \mu = m_1 m_2/(m_1 + m_2) is the reduced mass of the system; it appears naturally in all two-body collision problems.

Equations (9), (10), and (11) are the master formulas for all one-dimensional collisions.

Explore the spectrum yourself

The interactive figure below uses equal masses (m_1 = m_2) with the incoming speed normalised to u_1 = 1 m/s. Drag e from 0 (perfectly inelastic) to 1 (perfectly elastic) and watch three things change: the final velocity of the incoming body, the final velocity of the target, and the fraction of kinetic energy lost.

For equal masses the formulas simplify to: v_1/u_1 = (1 - e)/2, \;v_2/u_1 = (1 + e)/2, \;\Delta KE/KE_0 = (1 - e^2)/2.

Interactive: how final velocities and energy loss depend on the coefficient of restitution Three curves plotted against e from 0 to 1 for equal masses. v1 over u1 drops linearly from 0.5 to 0. v2 over u1 rises linearly from 0.5 to 1. Energy loss fraction is a downward parabola from 0.5 to 0. A draggable point lets the reader explore any e value. coefficient of restitution e ratio (velocities or energy fraction) 0 0.5 1 0 0.5 1 v₁/u₁ v₂/u₁ ΔKE/KE₀ drag the red dot along the axis
Drag the red point to change $e$. At $e = 0$ (perfectly inelastic), both bodies share the velocity equally and half the energy is lost. At $e = 1$ (perfectly elastic), the incoming body stops and the target takes all the velocity — zero energy lost. The dashed curve shows the energy loss fraction, which falls as $e$ increases.

At e = 0: both objects move at u_1/2, and half the kinetic energy is destroyed. At e = 1: the first object stops, the second takes all the velocity, and no energy is lost — this is the carrom-striker result from the elastic collisions article. Between these extremes, every real collision lives.

Worked examples

Example 1: Mud ball on a block

During monsoon, a child throws a ball of wet clay (mass 200 g) at 5 m/s. It hits a wooden block (mass 300 g) resting on a smooth table and sticks to it. Find the velocity of the combined system after impact and the energy lost.

Before and after diagram for a perfectly inelastic clay–block collision Left: clay ball moving right at 5 m/s toward a stationary block. Right: clay and block stuck together, moving right at 2 m/s. Before 200 g 5 m/s 300 g at rest After 500 g 2 m/s
Before: the clay ball approaches. After: clay and block move together at the lower speed.

Step 1. Identify the knowns and the collision type.

m_1 = 0.2 kg (clay), m_2 = 0.3 kg (block), u_1 = 5 m/s, u_2 = 0. The clay sticks to the block — this is perfectly inelastic (e = 0).

Why: "sticks to it" is the defining feature of a perfectly inelastic collision. One equation (momentum) is enough.

Step 2. Apply the perfectly inelastic formula (equation 2).

v = \frac{m_1 u_1}{m_1 + m_2} = \frac{0.2 \times 5}{0.2 + 0.3} = \frac{1.0}{0.5} = 2 \text{ m/s}

Why: the combined mass (0.5 kg) must carry the same momentum the clay had (1.0 kg·m/s). A heavier combined mass means a lower velocity.

Step 3. Compute the kinetic energies.

KE_{\text{before}} = \tfrac{1}{2} \times 0.2 \times 5^2 = 2.5 \text{ J}
KE_{\text{after}} = \tfrac{1}{2} \times 0.5 \times 2^2 = 1.0 \text{ J}

Why: the velocity dropped from 5 to 2 m/s while the mass increased from 0.2 to 0.5 kg. Since kinetic energy depends on v^2, the squared drop in speed dominates — the final KE is less than half the initial.

Step 4. Find the energy lost.

\Delta KE = 2.5 - 1.0 = 1.5 \text{ J}
\text{Fraction lost} = \frac{1.5}{2.5} = 0.60 = 60\%

Why: cross-check with equation (6): m_2/(m_1 + m_2) = 0.3/0.5 = 0.6. Confirmed — 60% of the kinetic energy went into deforming the clay, heating the surfaces, and producing the thud of impact.

Result: The clay-block system moves at 2 m/s. Of the original 2.5 J, 1.5 J (60%) is lost to deformation and heat.

The diagram confirms the key physics: momentum is conserved (0.2 \times 5 = 0.5 \times 2 = 1.0 kg·m/s), but the combined system crawls at less than half the original speed.

Example 2: Autorickshaw fender-bender

An autorickshaw A (mass 400 kg including the driver) moving at 36 km/h rear-ends autorickshaw B (mass 400 kg, stationary at a red light). The coefficient of restitution for this collision is e = 0.3. Find each auto's velocity after impact and the energy dissipated.

Before and after diagram for an autorickshaw collision Left: Auto A at 36 km/h approaching stationary Auto B. Right: Auto A at 12.6 km/h, Auto B at 23.4 km/h, both moving right. Before 400 kg 36 km/h 400 kg at rest After 12.6 km/h 23.4 km/h
Auto A slows sharply; Auto B lurches forward. Both move in the same direction, but Auto B is now faster.

Step 1. Convert units and identify the knowns.

m_1 = m_2 = 400 kg. \;u_1 = 36 km/h = 10 m/s. \;u_2 = 0. \;e = 0.3.

Why: divide km/h by 3.6 to get m/s. Equal masses and a known e — this is a partially inelastic collision that you solve with equations (9) and (10).

Step 2. Compute v_1 using equation (9).

v_1 = \frac{m_1 - e\,m_2}{m_1 + m_2}\,u_1 = \frac{400 - 0.3 \times 400}{800} \times 10 = \frac{280}{800} \times 10 = 3.5 \text{ m/s}

Why: for equal masses, v_1 = (1 - e)\,u_1/2 = 0.7 \times 10/2 = 3.5 m/s. Auto A continues forward at 3.5 m/s = 12.6 km/h — barely a third of its original speed.

Step 3. Compute v_2 using equation (10).

v_2 = \frac{m_1(1 + e)}{m_1 + m_2}\,u_1 = \frac{400 \times 1.3}{800} \times 10 = 0.65 \times 10 = 6.5 \text{ m/s}

Why: Auto B, which was stationary, now moves at 6.5 m/s = 23.4 km/h. It received more velocity than Auto A retained.

Step 4. Verify momentum conservation.

Before: 400 \times 10 = 4{,}000 kg·m/s

After: 400 \times 3.5 + 400 \times 6.5 = 1{,}400 + 2{,}600 = 4{,}000 kg·m/s ✓

Why: momentum is always conserved — this sanity check confirms the arithmetic.

Step 5. Compute the energy dissipated.

KE_{\text{before}} = \tfrac{1}{2} \times 400 \times 10^2 = 20{,}000 \text{ J} = 20 \text{ kJ}
KE_{\text{after}} = \tfrac{1}{2} \times 400 \times 3.5^2 + \tfrac{1}{2} \times 400 \times 6.5^2 = 2{,}450 + 8{,}450 = 10{,}900 \text{ J}
\Delta KE = 20{,}000 - 10{,}900 = 9{,}100 \text{ J} = 9.1 \text{ kJ}

Why: cross-check with equation (11): \frac{400 \times 400 \times (1 - 0.09)}{2 \times 800} \times 100 = 200 \times 0.91 \times 100/2 = 9{,}100 J. ✓ That 9.1 kJ went into crumpling bumpers, cracking headlights, and the bang of the crash.

Result: Auto A slows from 36 km/h to 12.6 km/h. Auto B lurches forward at 23.4 km/h. Of the original 20 kJ, 9.1 kJ (45.5%) is absorbed by the collision — crumpled metal, broken glass, sound, and heat.

The diagram shows the velocity redistribution: A had all the speed before; afterward, B carries more speed than A. But unlike a perfectly inelastic collision, they do not move together — they separate, with a relative speed of 6.5 - 3.5 = 3 m/s, which is 30% of the original approach speed (e = 0.3, confirmed).

Common confusions

  • "Inelastic means the objects stick together." Not always. Perfectly inelastic means they stick (e = 0). Plain "inelastic" means kinetic energy is not conserved — the objects can still bounce apart, just not as energetically as in an elastic collision. A cricket ball bouncing off a pitch with e = 0.5 is inelastic but obviously does not stick.

  • "Momentum is not conserved in inelastic collisions." Wrong — momentum is conserved in every collision: elastic, inelastic, or perfectly inelastic. Momentum conservation comes from Newton's third law and cannot be violated by any internal force. What is not conserved in an inelastic collision is kinetic energy.

  • "A perfectly inelastic collision loses all kinetic energy." Not all — just the maximum amount consistent with momentum conservation. If m_1 = m_2 and u_2 = 0, a perfectly inelastic collision loses exactly 50%, not 100%. The other 50% survives as kinetic energy of the combined mass. Only when m_2 \to \infty (hitting a wall) does nearly all the KE disappear.

  • "Higher e means more energy transferred to the target." Not exactly. Higher e means less total kinetic energy is lost. The distribution between the two objects depends on the mass ratio. For equal masses, the target does get more velocity at higher e — from u_1/2 at e = 0 to u_1 at e = 1 — but the incoming body simultaneously changes from u_1/2 to 0. The energy is redistributed, not simply "more transferred."

  • "e is a constant for a material." Approximately, but not exactly. The coefficient of restitution depends primarily on the materials, but it also decreases at very high collision speeds (materials deform more at higher impact energies). For most textbook problems, treating e as constant is a good approximation.

If you came here to understand inelastic collisions, use the formulas, and solve problems, you have what you need. What follows is for readers who want the ballistic pendulum, the energy picture in the centre-of-mass frame, and the general case with both objects moving.

The ballistic pendulum — measuring bullet speed

The ballistic pendulum is one of the most elegant applications of the perfectly inelastic collision. It was used for over a century to measure the speed of bullets before electronic sensors existed, and it still appears in Indian physics textbooks and JEE problems.

Setup: A bullet of mass m is fired horizontally into a large wooden block of mass M suspended by strings. The bullet embeds in the block (perfectly inelastic collision), and the block-plus-bullet swings upward to a height h.

Step 1. Apply conservation of momentum during the collision.

m\,v_{\text{bullet}} = (m + M)\,v_{\text{after}}

Why: the collision is fast (milliseconds), so the block barely moves during impact. Gravity has no time to change the momentum significantly. Momentum is conserved during the collision itself.

Step 2. Apply conservation of energy during the swing (after the collision).

\tfrac{1}{2}(m + M)\,v_{\text{after}}^2 = (m + M)\,g\,h
v_{\text{after}} = \sqrt{2gh}

Why: after the collision, no external horizontal force acts (the strings are vertical at the lowest point). The kinetic energy converts entirely to gravitational potential energy at the highest point of the swing.

Step 3. Combine to find the bullet speed.

v_{\text{bullet}} = \frac{m + M}{m}\,\sqrt{2gh}

Why: substitute v_{\text{after}} from Step 2 into Step 1 and solve for v_{\text{bullet}}. You measure h (the height the block swings to), know m and M, and compute the bullet speed from those three numbers alone.

For example, a 10 g bullet into a 2 kg block that swings up by 5 cm:

v_{\text{bullet}} = \frac{2.01}{0.01} \times \sqrt{2 \times 9.8 \times 0.05} = 201 \times 0.99 \approx 199 \text{ m/s}

The ballistic pendulum uses the perfectly inelastic collision as a feature: the bullet sticks to the block, giving you the one-equation momentum formula. If the bullet bounced off, you would need to know e as well.

Energy loss in the centre-of-mass frame

In the elastic collisions going-deeper section, you saw that an elastic collision is trivial in the CM frame — both velocities simply reverse. For an inelastic collision, the CM frame reveals exactly where the energy goes.

In the CM frame, the total momentum is zero both before and after. The kinetic energy in the CM frame before the collision is:

KE_{\text{CM}} = \tfrac{1}{2}\mu\,u_{\text{rel}}^2

where \mu = m_1 m_2/(m_1 + m_2) is the reduced mass and u_{\text{rel}} = u_1 - u_2 is the relative velocity.

After the collision, the relative velocity becomes v_{\text{rel}} = e \cdot u_{\text{rel}} (by definition of e). So:

KE_{\text{CM, after}} = \tfrac{1}{2}\mu\,(e\,u_{\text{rel}})^2 = e^2 \cdot KE_{\text{CM}}

The fraction of CM-frame kinetic energy that survives is e^2. The fraction lost is (1 - e^2).

\Delta KE = (1 - e^2) \cdot \tfrac{1}{2}\mu\,u_{\text{rel}}^2 = \frac{m_1\,m_2\,(1 - e^2)}{2(m_1 + m_2)}\,u_{\text{rel}}^2

This matches equation (11) exactly (with u_{\text{rel}} = u_1 when u_2 = 0). The CM frame makes the physics transparent: the energy that disappears is the fraction (1 - e^2) of the kinetic energy of relative motion. The kinetic energy of the system's overall drift (the CM motion) is never affected by the collision — only the relative motion carries energy that can be lost.

Both objects moving — the general formulas

When both objects are moving before the collision (u_2 \neq 0), the formulas become:

v_1 = \frac{m_1 u_1 + m_2 u_2 - e\,m_2(u_1 - u_2)}{m_1 + m_2}
v_2 = \frac{m_1 u_1 + m_2 u_2 + e\,m_1(u_1 - u_2)}{m_1 + m_2}

These are derived from the same two equations — momentum conservation and the restitution definition e = (v_2 - v_1)/(u_1 - u_2) — with u_2 kept general. They reduce to equations (9) and (10) when u_2 = 0, and to the full elastic collision formulas when e = 1.

The energy loss generalises to:

\Delta KE = \frac{m_1\,m_2\,(1 - e^2)}{2(m_1 + m_2)}\,(u_1 - u_2)^2

The factor (u_1 - u_2)^2 confirms that it is the relative velocity that determines energy loss, not the individual velocities. Two cars moving at 100 km/h and 99 km/h have a gentle 1 km/h relative collision — barely any energy lost. Two cars approaching head-on at 50 km/h each have a 100 km/h relative collision — devastating. The CM frame makes this obvious: only relative motion carries convertible energy.

Where this leads next