Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Visual walkthrough

We only need one idea from before we start: velocity is an arrow on a number line. Longer arrow = faster; direction of arrow = which way it moves. That's it. Everything else we build.

Step 1 — Two blocks, one line, four arrows

WHAT. Picture two blocks, block 1 (left) and block 2 (right), sliding on a smooth horizontal line. We pick one direction as positive — let's say rightward is positive. Every velocity is now just a signed number: positive means "moving right", negative means "moving left".

WHY. Collisions in this note are one-dimensional: everything happens along a single line. Choosing a positive direction turns each velocity into a single number we can add and subtract. Without a fixed sign convention, "faster" and "which way" get tangled — so we nail it down first.

PICTURE. Look at the figure. Before the crash the blocks have velocities $u_{1}$ and $u_{2}$ (the letter $u$ = "before", think "u come first"). For a collision to happen at all, block 1 must be catching up to block 2, so its arrow is longer: $u_{1} > u_{2}$ .

Figure — Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

Step 2 — "Approach speed": how fast the gap closes

WHAT. The two blocks are approaching each other. How fast? Not $u_{1}$ alone, not $u_{2}$ alone — what matters is how fast the gap between them shrinks. That is the difference $u_{1} - u_{2}$ .

WHY the difference, and why this order. Imagine you are sitting on block 2. From your seat, block 2 is standing still, and block 1 comes at you with speed $u_{1} - u_{2}$ . Subtracting $u_{2}$ is exactly "switch to block 2's viewpoint". We write $u_{1} - u_{2}$ (not $u_{2} - u_{1}$ ) because block 1 is the faster one from behind — so this difference comes out positive, which is what a "closing speed" should be.

PICTURE. In the figure the amber bracket is the closing gap. The amber arrow labelled $u_{1} - u_{2}$ is what an observer riding block 2 sees rushing toward them.

Step 3 — The crash, and "separation speed"

WHAT. They collide, deform, and push apart. Afterwards block 2 must end up ahead — faster to the right — otherwise block 1 would pass through it. So now $v_{2} > v_{1}$ , and the speed at which the gap reopens is $v_{2} - v_{1}$ .

WHY the order flips ( $v_{2} - v_{1}$ , not $v_{1} - v_{2}$ ). Before the crash, block 1 was the faster one, so "faster minus slower" was $u_{1} - u_{2}$ . After the crash, block 2 is the faster one, so "faster minus slower" is now $v_{2} - v_{1}$ . The indices swap because the roles of "faster block" swap. That single swap is the whole reason the parent formula has different index orders top and bottom — and it keeps both quantities positive.

PICTURE. Same viewpoint trick: sit on block 1 after the crash and watch block 2 flee at speed $v_{2} - v_{1}$ (cyan arrow).

Step 4 — The definition drops out: separation ÷ approach

WHAT. We now have two positive speeds. Newton's experimental discovery was that their ratio is a fixed number for a given pair of materials. Call it $e$ : $e = \frac{separation}{approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} .$

WHY a ratio and not a difference. A ratio asks "what fraction of the closing speed survived?" — it is a pure number with no units (m/s ÷ m/s cancels), so it depends only on what the blocks are made of, not on how hard you threw them. That is exactly the kind of clean, reusable quantity physics loves.

PICTURE. The figure stacks the two arrows from Steps 2 and 3 and shows the shrink: separation arrow is (usually) shorter than approach arrow, and $e$ is literally how much shorter.

See Impulse and Momentum for the deeper $e = J_{r} / J_{d}$ reading of this same number.

Step 5 — The bouncy end: $e = 1$

WHAT. If the blocks bounce apart just as fast as they came together, $v_{2} - v_{1} = u_{1} - u_{2}$ , so $e = 1$ . Nothing was lost.

WHY it's the ceiling. Separation can never exceed approach, because that would mean the collision created kinetic energy from nothing. The most a collision can do is give it all back. So $e = 1$ is the maximum — the perfectly elastic case, where kinetic energy is also conserved.

PICTURE. The separation arrow is exactly as long as the approach arrow — a mirror-image bounce.

Step 6 — The splat end: $e = 0$

WHAT. If the blocks stick and move off together, they share one common velocity $v_{c}$ : so $v_{1} = v_{2} = v_{c}$ . Then $v_{2} - v_{1} = 0$ , giving $e = 0$ .

WHY it's the floor. Zero separation means the gap never reopens — a perfectly inelastic crash, a clay-lump splat. Since a separation speed can't be negative (block 1 can't tunnel through block 2), $e = 0$ is the smallest possible value. So every real collision lives in $0 \leq e \leq 1$ .

PICTURE. The two "after" arrows are the same length and point the same way — the blocks lock together. Separation arrow has zero length.

Recall Check the two extremes

$e = 1$ means ::: separation = approach, a perfect bounce, kinetic energy conserved. $e = 0$ means ::: blocks stick, share one velocity $v_{c}$ , separation is zero, maximum KE lost.

Step 7 — The degenerate case: a block bouncing off a wall

WHAT. Let block 2 be a fixed floor or wall — infinitely heavy, so it never moves: $u_{2} = 0$ and $v_{2} = 0$ . Then the formula collapses to $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{0 - v _{1}}{u _{1} - 0} = \frac{- v _{1}}{u _{1}} .$

WHY the minus sign is good here. Before impact the ball moves toward the wall — say downward, so $u_{1}$ is negative in our "up is positive" convention... but let's read it cleanly: the ball hits at speed $∣ u_{1} ∣$ and rebounds at speed $∣ v_{1} ∣$ in the opposite direction, so $v_{1}$ has the opposite sign to $u_{1}$ . That opposite sign makes $- v_{1} / u_{1}$ come out positive — exactly the ratio of rebound speed to impact speed: $e = \frac{rebound speed}{impact speed} .$

PICTURE. Ball comes down (long arrow), bounces up (shorter arrow). $e$ is how much of the arrow survived the wall.

Step 8 — Worked example, verified against the picture

WHAT. Two equal masses, $m_{1} = m_{2} = 2$ kg. Block 1 at $u_{1} = 4$ m/s hits a still block 2 ( $u_{2} = 0$ ), with $e = 0.5$ . Find the after-velocities.

WHY two equations. Momentum gives one equation; $e$ gives the second. Two equations, two unknowns — now solvable.

$Momentum: 2 (4) + 0 = 2 v_{1} + 2 v_{2} \Rightarrow v_{1} + v_{2} = 4$ $Restitution: v_{2} - v_{1} = e (u_{1} - u_{2}) = 0.5 (4) = 2$

Add the two lines: $2 v_{2} = 6 \Rightarrow v_{2} = 3$ m/s, then $v_{1} = 1$ m/s.

PICTURE / check. Separation $= v_{2} - v_{1} = 3 - 1 = 2$ m/s, approach $= 4$ m/s, ratio $= 2/4 = 0.5$ ✓. Both positive, $v_{2} > v_{1}$ — block 2 leads afterward, exactly as Step 3 demanded.

The one-picture summary

Everything in one frame: approach arrow (amber, before), the crash, separation arrow (cyan, after, indices swapped), and $e$ as the ratio of their lengths — sliding from the splat ( $e = 0$ ) to the perfect bounce ( $e = 1$ ).

Recall Feynman retelling of the whole walkthrough

Two blocks slide on a line. We call rightward positive so every speed is just a number. Before they crash, the back block is faster and catches up — the speed at which the gap closes is "faster minus slower", $u_{1} - u_{2}$ . They collide, squash, and shove apart; now the front block is the fast one, so the speed at which the gap reopens is "faster minus slower again", but the fast block has swapped, so it's $v_{2} - v_{1}$ . Newton found that dividing the reopening speed by the closing speed always gives the same number for a given pair of materials — that number is $e$ . It can't be more than $1$ (you can't fly apart faster than you came together, that would make energy from nothing) and can't be less than $0$ (you can't tunnel through). A perfect rubber bounce keeps all the speed, $e = 1$ ; a clay splat keeps none, $e = 0$ ; a wall is just the same story with one block nailed down, and because bounce height depends on speed squared, $e = h_{2} / h_{1}$ .

Connections

Coefficient of Restitution — the parent topic this page visualises.
Conservation of Linear Momentum — the first equation; $e$ supplies the second.
Elastic Collisions — Step 5, the $e = 1$ ceiling.
Perfectly Inelastic Collisions — Step 6, the $e = 0$ floor.
Impulse and Momentum — the deeper $e = J_{r} / J_{d}$ meaning.
Kinetic Energy Loss in Collisions — how much energy the shrunken arrow lost.
Projectile Motion — the $v = 2 g h$ link behind Step 7.