Inelastic Collisions & Coefficient of Restitution

The modern car is a machine for lowering $e$ on purpose. A rigid, bouncy vehicle ($e$ high) would fling its occupants with a large momentum change and hand back crash energy as rebound; crumple zones instead make the collision as plastic as possible — the structure's deformation devours the $\tfrac12\mu u_{\rm rel}^2$ bill so the cabin doesn't — and crash-test ratings are, in accounting terms, audits of where the $(1-e^2)$ energy went. Sports run the opposite direction of the same industry: federations legislate the bounce — tennis's drop test, cricket and golf ball standards, the periodic "juiced ball" controversies in baseball are all fights over $e$ — because a few percent of restitution moves home-run counts and serve speeds measurably. Forensic engineers reconstruct highway collisions by assuming momentum conservation exactly and fitting $e$ from the wreckage separation; asteroid-impact modellers and even video-game physics engines carry a restitution coefficient per material pair. Newton's pendulum measurements have become a parameter in every simulated world.

Rerun the collision movie in slow motion. Compression: the bodies squash, and the kinetic energy of their relative motion pours into strain energy. Restitution: the stored energy pushes them apart again. The coefficient $e$ is the exchange rate between the two halves — how much of the compression deposit the rebound phase actually refunds, measured in speed. Perfect materials (ideal springs, molecules) refund in full: $e = 1$. Clay refunds nothing: the deposit is spent on permanently rearranging the material, and the bodies, still touching when the dust settles, share one velocity: $e = 0$. Steel, wood, rubber sit between, each pair of materials owning its own approximate $e$.

The crucial bookkeeping fact — the one that fails more exam candidates than any formula — is what the grade does not touch. Momentum conservation comes from Newton's third law: the contact forces are equal and opposite whatever the materials do, so the momentum books balance for clay exactly as for steel. Kinetic energy has no such protection: the missing $(1-e^2)$ share ends up as heat, sound, and permanent deformation — real energy, in inconvenient form. A sticky crash is not a violation of conservation; it is momentum conserved and energy converted. One geometric convention keeps everything honest: $e$ compares speeds along the line of impact — the common normal at contact. For head-on collisions that is the whole story; for glancing ones, the perpendicular components ignore the collision entirely (the next article's business). And one measurement gift: drop a ball on a rigid floor from height $h$ and catch the rebound height $h'$. Arrival speed $\sqrt{2gh}$, departure speed $\sqrt{2gh'}$, floor immobile — so $e = \sqrt{h'/h}$. A metre rule measures the coefficient in seconds; sports federations write exactly this test into their rulebooks.

Same 1D cast as before: $m_1, m_2$ with signed velocities $u_1, u_2 \to v_1, v_2$. The first equation never changes — momentum, exact at any $e$:

where: signed velocities, one axis, the eternal law. The second equation is the experimental input — Newton's law of restitution, the graded generalization of the elastic reversal rule:

where: $u_1 - u_2$ — approach (gap-closing) speed; $v_2 - v_1$ — separation speed; $e$ — the collision's grade, $0 \le e \le 1$ for passive collisions. At $e = 1$ this is the elastic reversal rule; at $e = 0$ it forces $v_1 = v_2$ — the stick. Two linear equations, two unknowns: every head-on collision is now solvable.

The perfectly inelastic case first ($e = 0$), because it is the most-asked and the simplest. The bodies leave with one shared velocity, which momentum alone fixes:

where: $V$ — the common final velocity, which is nothing but the system's centre-of-mass velocity (a connection the COM articles will cash properly). The kinetic energy destroyed (derivation A) has a form worth memorizing:

where: $\frac{m_1m_2}{m_1+m_2} \equiv \mu$ — the "reduced mass", smaller than either body; $(u_1-u_2)$ — the relative speed. Read it physically: only the energy of relative motion is at risk in a collision — the energy of the shared (centre-of-mass) motion is protected by momentum conservation and survives every crash. Sticking destroys the relative-motion energy entirely; that is why $e=0$ is the maximum possible loss, not "all the KE gone".

The general-$e$ closed forms (derivation B) interpolate everything:

where: $V$ is the common-velocity value from (3) — each body leaves at the centre-of-mass velocity plus a rebound share proportional to $e$. The energy bill carries the factor $(1-e^2)$: zero at $e=1$ (elastic — full refund), one at $e=0$ (the stick — maximum bill). Half the elastic bounce ($e = 0.5$) still destroys three quarters of the at-risk energy — $e$ enters squared, a favourite MCQ subtlety.

The dropped ball — $e$ measured with a metre rule

A ball dropped from height $h$ arrives at the floor at $\sqrt{2gh}$; the floor (mass: one planet) does not move, so the restitution law gives departure speed $e\sqrt{2gh}$, and the rebound height follows from kinematics:

where: the square is the same quadratic bookkeeping as always — speeds scale with $e$, heights (energies) with $e^2$. Bounce again and again: heights fall geometrically, $h, e^2h, e^4h, \dots$ — each bounce keeps the same fraction $e^2$. (The full geometric-series machinery — total distance, total time, "how many bounces until…" — is the Bouncing Ball archetype's business; this article owns the measurement.) The regulation tennis ball is a live example: the ITF drops it from $254$ cm onto concrete and demands a rebound of $135$–$147$ cm — that is $e$ between $\sqrt{135/254} \approx 0.73$ and $\sqrt{147/254} \approx 0.76$, a rulebook written in restitution coefficients.

1 · Board level — coupling carts (CBSE/ISC pattern, 3 marks)

A $3.0\ \mathrm{kg}$ cart moving at $4.0\ \mathrm{m/s}$ collides with a stationary $2.0\ \mathrm{kg}$ cart and they couple. Find the common velocity and the kinetic energy lost, and state where the lost energy went.

Step 1. Momentum: $(3.0)(4.0) = (5.0)V \Rightarrow V = \boxed{2.4\ \mathrm{m/s}}$.

Step 2. Energy audit: before $\tfrac12(3)(16) = 24$ J; after $\tfrac12(5)(5.76) = 14.4$ J. Lost: $\boxed{9.6\ \mathrm{J}}$ (40%).

Step 3. Formula check: $\Delta KE = \tfrac12\frac{(3)(2)}{5}(4)^2 = \tfrac12(1.2)(16) = 9.6$ J ✓.

Step 4. The destination sentence (its own mark): the $9.6$ J became heat and sound in the coupling and permanent deformation — momentum, protected by the third law, was conserved throughout.

2 · Board level — the metre-rule measurement (ICSE/A-Level pattern)

A ball dropped from $2.0\ \mathrm{m}$ onto a rigid floor rebounds to $1.28\ \mathrm{m}$. Find the coefficient of restitution and the height of the second bounce.

Step 1. Heights carry $e^2$: $e = \sqrt{h'/h} = \sqrt{1.28/2.0} = \sqrt{0.64} = \boxed{0.80}$.

Step 2. Second bounce: $h_2 = e^2h_1 = 0.64 \times 1.28 = \boxed{0.82\ \mathrm{m}}$ — each bounce keeps 64% of the previous height.

Step 3. State the assumptions for the last mark: the floor is effectively infinite mass (it doesn't recoil), air resistance neglected, and $e$ taken constant between bounces.

3 · Competitive level — general e, full audit (JEE-Main pattern, ~3 minutes)

A $2.0\ \mathrm{kg}$ ball moving at $6.0\ \mathrm{m/s}$ hits a stationary $4.0\ \mathrm{kg}$ ball head-on with $e = 0.5$. Find both final velocities and the kinetic energy destroyed.

Step 1. Spectator first: $V = \dfrac{(2)(6)}{6} = 2.0$ m/s (the centre-of-mass velocity, unchanged by any $e$).

Step 2. Closed forms (5): $v_1 = 2 - \dfrac{(4)(0.5)(6)}{6} = 2 - 2 = \boxed{0}$; $\;v_2 = 2 + \dfrac{(2)(0.5)(6)}{6} = 2 + 1 = \boxed{3.0\ \mathrm{m/s}}$.

Step 3. Restitution audit: separation $= 3.0 - 0 = 3.0 = 0.5 \times 6.0$ ✓. Momentum audit: $12 = 0 + 12$ ✓.

Step 4. Energy destroyed: $\Delta KE = \tfrac12 \cdot \dfrac{(2)(4)}{6}(1 - 0.25)(6)^2 = \tfrac12 \cdot \tfrac43 \cdot 0.75 \cdot 36 = \boxed{18\ \mathrm{J}}$ — half the original 36 J. Direct check: after $= 0 + \tfrac12(4)(9) = 18$ J ✓. Timing note: ~3 min with the closed forms; ~4.5 solving the pair from scratch. Efficiency habit: compute $V$ first (it is also the answer to "what if they had stuck"), then bolt on the $\pm e$-shares. Note the engineered curiosity: the projectile stops dead exactly when $e = m_1/m_2$ (here $0.5 = 2/4$ ✓) — a setter's favourite; never assume the projectile stops unless the algebra says so.

4 · Competitive level — the capture fraction (NEET/JEE pattern, ~2.5 minutes)

A $10\ \mathrm{g}$ bullet moving at $500\ \mathrm{m/s}$ embeds in a stationary $2.49\ \mathrm{kg}$ block on a smooth surface. Find the common velocity and the percentage of kinetic energy that survives.

Step 1. Embedding = perfectly inelastic: $V = \dfrac{(0.010)(500)}{2.50} = \boxed{2.0\ \mathrm{m/s}}$.

Step 2. KE before: $\tfrac12(0.010)(500)^2 = 1250$ J. After: $\tfrac12(2.50)(4.0) = 5.0$ J. Surviving fraction $= \boxed{0.4\%}$.

Step 3. The general pattern — worth memorizing: for a moving $m$ captured by a resting $M$, the surviving fraction is $\dfrac{m}{m+M}$ (here $0.01/2.50 = 0.4\%$ ✓), and the destroyed fraction $\dfrac{M}{m+M}$. Light bullet, heavy block: almost everything burns.

Step 4. Where it burns: heat and splintering along the bullet's channel — which is precisely why momentum, not energy, must bridge the embedding stage of ballistic-pendulum problems (that archetype is two modules ahead). Timing note: ~2.5 min. The fraction $m/(m+M)$ is a ten-second answer where first-principles takes two minutes — and it doubles as the audit for any capture problem: surviving KE must be TINY when a light body embeds in a heavy one. An answer like "60% survives" for a bullet-and-block fails the sniff test instantly.

Writing the KE equation for an inelastic collision. One illegal line, zero marks. The energy books are open only at $e = 1$; below that, energy appears only as the audit of what was destroyed.

$(1-e)$ instead of $(1-e^2)$. Speeds scale with $e$; energies with $e^2$. Half-elastic ($e = 0.5$) destroys 75% of the at-risk energy, not 50%. Same square in the dropped ball: $h' = e^2h$, so $e = \sqrt{h'/h}$ — forgetting the square root is the classic heights error.

"Perfectly inelastic = all KE lost." Only the relative-motion energy $\tfrac12\mu u_{\rm rel}^2$ dies; the centre-of-mass share survives, protected by momentum. Everything stops only if the total momentum was zero to begin with.

Restitution applied to the wrong speeds. $e$ compares relative speeds (approach and separation), not the projectile's before-and-after. For a moving target, approach $= u_1 - u_2$ with signs — head-on movers close at $u_1 + |u_2|$.

Signs, forever. All the closed forms take signed velocities. Head-on means opposite signs; feeding in two positive speeds silently converts the problem into a chase.

$e$ outside its box. For passive collisions $0 \le e \le 1$. A computed $e = 1.3$ means an arithmetic slip (or an explosive collision, which the problem would announce loudly). Use it as a self-check.

The line of impact. $e$ (and the 1D machinery) governs velocity components along the common normal. A glancing collision needs the components split first — next article. Applying head-on formulas to oblique geometry is a structural error, not a rounding one.

Momentum "reduced by e". Never. $e$ rescales relative motion; total momentum doesn't feel it. Any option showing $P_{\rm after} = eP_{\rm before}$ is instantly dead.

Texts. NCERT Physics Class 11, Ch. 6 (collisions) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 (restitution problems) · Newton's Principia, scholium to the Laws (the original measurements) · any JEE archive under "coefficient of restitution".

Provenance. All results are standard. A verifier should recompute: the loss formula (4) via the expansion in derivation A; the closed forms and $(1-e^2)$ factor in derivation B; Example 1 ($V = 2.4$, loss $9.6$ J $= 40\%$); Example 2 ($e = 0.80$, $h_2 = 0.82$ m); Example 3 ($v_1 = 0$, $v_2 = 3.0$, loss $18$ J with both routes agreeing); Example 4 ($V = 2.0$ m/s, $0.4\%$ surviving $= m/(m+M)$); the ITF window $e \approx 0.73$–$0.76$ from $\sqrt{135/254}$–$\sqrt{147/254}$.

Prerequisites. Conservation of Linear Momentum · Elastic Collisions in 1D. Leads to. The Ballistic Pendulum (archetype) · Bullet & Block (archetype) · The Bouncing Ball (archetype) · Collisions in Two Dimensions · Inelastic Collisions (UG deep dive).