Collisions in Two Dimensions

In 1925, Patrick Blackett photographed something exam candidates now prove in four lines: an alpha particle ploughing through helium gas struck a helium nucleus, and the two tracks after the collision opened out at almost exactly ninety degrees. No protractor consultation with theory needed — equal masses colliding elastically must separate at a right angle. The cloud-chamber photograph was a momentum diagram drawn by nature, and it certified that the alpha particle and the helium nucleus weigh the same. Snooker players have known the practical version for two centuries: play a clean "cut" shot and the cue ball departs perpendicular to the object ball's path — every plant and safety shot is built on it. Both facts live in the physics this article opens: collisions where the bodies do not meet head-on, so the action spreads into a plane and momentum conservation splits into two independent component equations. The machinery is less new than it looks. One geometric idea — resolve everything along and perpendicular to the line of impact — reduces an oblique collision to a 1D collision (last two articles' machinery) running along the normal, while the perpendicular components sail through as spectators. Add the vector bookkeeping for sticky 2D crashes (cars at an intersection, the forensic classic) and the celebrated equal-mass 90° theorem, and the plane holds no further terrors. ISC asks the oblique geometry directly; JEE and NEET test the 90° result and its inelastic exceptions relentlessly.

Key takeaways

In 2D, momentum conservation is two independent equations — one per axis: $\sum p_x$ and $\sum p_y$ each balance separately.
The organizing geometry: resolve velocities along the line of impact (the line joining the centres at contact) and perpendicular to it.
For smooth bodies, the collision force acts only along the line of impact: perpendicular components pass through unchanged; the along-line components obey the 1D machinery (momentum + restitution).
Equal masses + elastic + target at rest ⇒ the two bodies separate at exactly 90° — the billiards theorem, proved by Pythagoras on the momentum triangle.
If the collision is inelastic (e < 1), the opening angle closes below 90°; perfectly inelastic bodies leave together at 0°.
Sticky 2D crashes: add momenta as vectors, $\vec V = (m_1\vec u_1 + m_2\vec u_2)/(m_1+m_2)$ — magnitude by Pythagoras, direction by $\tan\theta$ .
Key idea: a glancing collision is a 1D collision along the line of impact with two spectator components riding along — and for equal masses colliding elastically, energy conservation forces the momentum triangle to be right-angled.

The core insight

Momentum conservation in a plane holds component by component, and the collision impulse acts along the line of impact — so oblique collisions decompose into an untouched perpendicular part plus a 1D collision along the normal, with the signature result that equal-mass elastic collisions on a stationary target always open at ninety degrees.

Where this lives today

Angle-reading is the daily bread of particle physics. Detectors at the LHC are giant protractors: momentum conservation per component, applied to sprays of debris, identifies what collided and what was created — and a violation of the component books (the "missing transverse momentum" of the conservation article) is the discovery channel for invisible particles. At human scale, forensic crash reconstruction runs worked example 1 in reverse: from skid directions and rest positions, investigators recover the pre-impact velocity vectors — testimony written in momentum components, admissible in court. Sports keep the theorem in daily use: snooker and pool professionals aim entirely by the line of centres (the "ghost ball" method) and bank on the cue ball's perpendicular escape to avoid the in-off; air-hockey and curling strategy is applied smooth-sphere obliquity. And molecular-beam chemists watch reaction products scatter at characteristic angles to reconstruct how molecules hit — Rutherford's method, re-run at chemical energies, a century on.

The physical picture

What makes a collision two-dimensional? Not the bodies — the aim. Slide one coin dead-centre at another and everything stays on a line. Aim slightly off-centre and the contact happens on a slant: the struck coin departs sideways, the striker deflects the other way, and the motion now needs a plane to describe it. The off-centredness is measured by where the centres sit at the moment of contact, and the line joining them — the line of impact — is the single most important line in the problem. Why? Because for smooth (frictionless) bodies, the contact force can only push along that line — a smooth sphere cannot grip another sideways any more than a smooth wall could grip a ball in the impulse article. That one observation dismembers the problem: velocity components along the line of impact feel the entire collision (and obey the whole 1D apparatus — momentum transfer, restitution, the elastic swap); components perpendicular to it feel nothing and pass through the encounter unchanged. An oblique collision is a head-on collision wearing a rotation.

Figure

Hero — the line of impact dismembers a glancing collision into 1D + spectators

Two immediate consequences organize half the exam questions. First, a struck stationary smooth ball always departs along the line of impact — it had no perpendicular velocity to keep, and the only push it received points along that line. This is why a snooker player controls the object ball's direction purely by choosing the contact point. Second, the deflection of the striker is a tug-of-war between its kept perpendicular component and whatever the 1D collision leaves of its along-line component — for equal masses and a perfect refund, the 1D part is completely surrendered (the equal-mass swap), leaving the striker only its perpendicular motion: exactly perpendicular to the target's departure. The 90° theorem is the swap theorem, rotated. For sticky 2D collisions no decomposition is needed at all: the bodies leave as one, and vector momentum conservation alone fixes the outcome — add the momentum vectors tip-to-tail, divide by the total mass. Cars meeting at an intersection, a defender wrapping up a running attacker: Pythagoras and one arctangent.

The math: components, the oblique recipe, and the right angle

The conservation law itself needs no upgrade — only bookkeeping. With no external impulse during the collision, each component of the total momentum balances independently:

\sum m u_x = \sum m v_x, \qquad \sum m u_y = \sum m v_y \qquad (1)

where: the axes are yours to choose — and the choice is the skill. For sticky crashes, use the compass directions the problem hands you. For glancing contact between smooth bodies, use the line of impact and its perpendicular, because in those axes the second physics fact becomes usable: the impulse $\vec J$ lies along the line of impact, so

v_{\perp} = u_{\perp} \;\text{(each body)}, \qquad \text{along-line: 1D machinery with } e \qquad (2)

where: $\perp$ marks components perpendicular to the line of impact. Equation (2) is the whole oblique method: two of the four unknown components are free (the spectators), and the other two come from the 1D momentum + restitution pair of the last article. For the perfectly inelastic 2D crash, conservation alone finishes the job:

\vec V = \frac{m_1\vec u_1 + m_2\vec u_2}{m_1 + m_2} \qquad (3)

where: the numerator is a genuine vector sum — for perpendicular initial paths, magnitude by Pythagoras and direction by $\tan\theta = p_y/p_x$ . (This $\vec V$ is again the centre-of-mass velocity — the momentum diagram's fixed point, as the next two articles will make structural.) And the headline theorem, proved in derivation A: for equal masses, an elastic collision with a stationary target opens at exactly 90°,

\theta_1 + \theta_2 = 90° \qquad \text{(equal masses, elastic, target initially at rest)} \qquad (4)

where: $\theta_1, \theta_2$ — the angles of the two outgoing paths on either side of the original line of motion. The speeds then follow from right-triangle trigonometry alone: $v_1 = u\cos\theta_1$ and $v_2 = u\sin\theta_1 = u\cos\theta_2$ — each outgoing speed is a projection of the incoming one.

situation	tool
stick in 2D (cars, tackles)	vector sum (3): Pythagoras + $\tan\theta$
smooth glancing contact	resolve on the line of impact; spectators + 1D core (2)
equal masses, elastic, target at rest	90° opening (4); speeds $u\cos\theta$ , $u\sin\theta$
same but inelastic	angle < 90°; components + $e$ along the normal
"explodes into three pieces" (2D)	closure diagram from the conservation article
asked for energy lost	audit KEs after solving — never assume elastic

Derivation AThe equal-mass right angle — Pythagoras on the momentum triangle

We derive the 90° theorem (4): an elastic collision between equal masses, one initially at rest, sends the two bodies off at right angles. The strategy: write momentum conservation as a vector triangle and kinetic-energy conservation as a scalar statement about the triangle's side lengths; Pythagoras' converse then forces the right angle. Watch both hypotheses earn their keep — equal masses let the masses cancel out of both statements, and elasticity supplies the Pythagorean side-length relation.

Step 1. Momentum, as vectors, masses equal (cancel $m$ ):

\vec u = \vec v_1 + \vec v_2

Geometrically: $\vec v_1$ and $\vec v_2$ placed tip-to-tail must close on $\vec u$ — the three velocities form a triangle with $\vec u$ as one side.

Step 2. Kinetic energy, elastic promise, cancel $\tfrac12 m$ :

u^2 = v_1^2 + v_2^2

Step 3. Read the two statements together: a triangle whose sides $v_1, v_2$ and $u$ satisfy $u^2 = v_1^2 + v_2^2$ . By the converse of Pythagoras, the angle between $\vec v_1$ and $\vec v_2$ is a right angle. Why the algebraic twin says the same: square Step 1 — $u^2 = v_1^2 + v_2^2 + 2\,\vec v_1\!\cdot\!\vec v_2$ — and subtract Step 2, leaving $\vec v_1\!\cdot\!\vec v_2 = 0$ : perpendicular outgoing velocities (or one of them zero — the head-on case, where the striker stops and "the angle" degenerates).

Step 4. The trigonometric harvest: with the right angle in place, each outgoing speed is a projection of $u$ onto its own direction: $v_1 = u\cos\theta_1$ , $v_2 = u\cos\theta_2 = u\sin\theta_1$ . Verified: energy audit for any $\theta_1$ : $u^2\cos^2\theta_1 + u^2\sin^2\theta_1 = u^2$ ✓. Reality checks: Blackett's 1925 alpha-on-helium cloud-chamber tracks open at ~90° (equal masses confirmed); billiard "cut" shots send the cue ball perpendicular to the object ball. Head-on limit $\theta_1 \to 90°$ ... rather $\theta_2 \to 0$ , $v_1 \to 0$ : the 1D swap regained ✓.

Go deeper · Derivation B · 1 minThe oblique smooth-sphere recipe — 1D machinery inside a rotated frame

Derivation BThe oblique smooth-sphere recipe — 1D machinery inside a rotated frame

We derive the complete outcome of a glancing elastic collision between identical smooth spheres, one at rest — the geometry ISC names "oblique collision" — and confirm it reproduces the 90° theorem with the target moving along the line of impact. The strategy: set axes on the line of impact (at angle $\alpha$ to the incoming velocity), let the perpendicular components spectate, and run the equal-mass 1D swap along the line. Watch the entire collision reduce to ONE exchange: the along-line component changes hands, nothing else moves.

Step 1. Geometry: at contact, the line of impact makes angle $\alpha$ with the striker's velocity $\vec u$ (set by the aim — the "impact parameter"). Resolve: along-line $u\cos\alpha$ , perpendicular $u\sin\alpha$ . The target has $0$ and $0$ .

Step 2. Perpendicular spectators (smooth spheres — no sideways force): striker keeps $u\sin\alpha$ ; target keeps $0$ .

Step 3. Along the line: a 1D elastic collision between equal masses at speeds $u\cos\alpha$ and $0$ — the swap: striker's along-line component becomes $0$ , target's becomes $u\cos\alpha$ .

Step 4. Reassemble. Target: velocity $u\cos\alpha$ purely along the line of impact — it departs along the line of centres at speed $u\cos\alpha$ . Striker: velocity $u\sin\alpha$ purely perpendicular to that line — speed $u\sin\alpha$ , at right angles to the target's path. Opening angle: $90°$ , reproducing the theorem, with the speeds automatically the projections of Step 4 of derivation A. Why the recipe generalizes: unequal masses or $e<1$ change only Step 3 — run the general 1D closed forms along the line with the same spectators. The perpendicular components NEVER participate for smooth bodies; that part of the recipe survives every variant. *Verified: energy — $u^2\cos^2\alpha + u^2\sin^2\alpha = u^2$ ✓. Limits: $\alpha = 0$ (dead centre) gives the full swap ✓; $\alpha \to 90°$ (a graze) leaves the striker untouched and the target unmoved ✓. Numbers for worked example 3: $u = 10$ , $\alpha = 60°$ : target $5.0$ m/s along the line, striker $8.66$ m/s perpendicular to it.*

Figure

The momentum triangle — elasticity makes it right-angled, energy loss closes it

Limiting cases

case	result	meaning
dead-centre aim ( $\alpha = 0$ )	1D collision regained	the whole previous machinery is the special case
grazing aim ( $\alpha \to 90°$ )	nothing happens	no along-line component, no collision
equal masses, elastic, target at rest	opening angle exactly 90°	the billiards/cloud-chamber theorem
same but $e < 1$	angle closes below 90°	separation slows along the normal only
perfectly inelastic ( $e=0$ )	angle 0 — they travel together	the sticky crash, direction from the vector sum
striker much heavier	both fly forward in a narrow cone	a truck is barely deflected by a football

Common trapAny two bodies that collide fly apart at ninety degrees

The 90° theorem is a three-hypothesis result, and each hypothesis is a place to lose (or harvest) marks: equal masses, perfectly elastic, target initially at rest. Break any one and the angle moves. Unequal masses: a heavy striker barrels on, squeezing the opening angle below 90° (both fragments thrown forward); a light striker can ricochet backward, opening it above. Inelastic ( $e<1$ ): the along-normal separation is throttled by $e$ while the perpendicular spectators are not — the angle closes below 90°, reaching zero at the perfect stick. Moving target: all bets off; go back to components. The exam versions: "two identical trolleys collide and move off at 60° and 45° — is the collision elastic?" (No: $60+45 \ne 90$ , and since the angles close, kinetic energy was lost.) And the trick option "90° holds for all elastic collisions" dies on unequal masses. Quote the theorem with its certificate: equal masses + elastic + stationary target, and the right angle is guaranteed; otherwise compute.

Worked examples

1 · Board level — the intersection crash (ISC pattern, 3 marks)

A $1000\ \mathrm{kg}$ car travelling east at $20\ \mathrm{m/s}$ collides at an intersection with a $1000\ \mathrm{kg}$ car travelling north at $20\ \mathrm{m/s}$ . They lock together. Find their common velocity just after impact, and the fraction of kinetic energy destroyed.

Step 1. Momentum components: $p_x = 1000 \times 20 = 20\,000$ (east); $p_y = 20\,000$ (north). Each is conserved through the crash.

Step 2. Total mass $2000$ kg: $V_x = V_y = 10$ m/s, so $V = \sqrt{10^2 + 10^2} = \boxed{14.1\ \mathrm{m/s}}$ at $\boxed{45°}$ north of east ( $\tan\theta = 1$ ).

Step 3. Energy audit: before $2 \times \tfrac12(1000)(400) = 400$ kJ; after $\tfrac12(2000)(200) = 200$ kJ. Destroyed: $\boxed{50\%}$ — crumpled metal and heat.

Step 4. State the licence: friction from the road exists, but over the ~0.1 s crash its impulse is negligible beside the collision impulses — conservation across the bang, per component.

2 · Board level — the cut shot by the 90° theorem (ISC/NEET pattern)

A ball moving at $4.0\ \mathrm{m/s}$ strikes an identical stationary ball in a perfectly elastic glancing collision. The striker is deflected $30°$ from its original line. Find both final speeds and the target's direction.

Step 1. Certificate check: equal masses ✓ elastic ✓ target at rest ✓ — the opening angle is $90°$ , so the target departs at $\boxed{60°}$ on the other side of the original line.

Step 2. Speeds are projections of the incoming velocity: striker $v_1 = u\cos 30° = 4.0 \times 0.866 = \boxed{3.46\ \mathrm{m/s}}$ ; target $v_2 = u\cos 60° = \boxed{2.0\ \mathrm{m/s}}$ .

Step 3. Energy audit: $3.46^2 + 2.0^2 = 12 + 4 = 16 = 4.0^2$ ✓ — the Pythagorean triple confirming elasticity.

3 · Competitive level — the oblique recipe in full (JEE-Main pattern, ~3 minutes)

A smooth sphere moving at $10\ \mathrm{m/s}$ strikes an identical stationary smooth sphere; at contact, the line joining the centres makes $60°$ with the striker's velocity. The collision is elastic. Find each sphere's final velocity (magnitude and direction).

Step 1. Resolve on the line of impact: along-line $10\cos 60° = 5.0$ m/s; perpendicular $10\sin 60° = 8.66$ m/s.

Step 2. Spectators: the perpendicular $8.66$ stays with the striker; the target keeps its perpendicular $0$ .

Step 3. Along the line, equal-mass elastic swap: striker's $5.0 \to 0$ ; target's $0 \to 5.0$ .

Step 4. Reassemble: target $\boxed{5.0\ \mathrm{m/s}}$ along the line of centres; striker $\boxed{8.66\ \mathrm{m/s}}$ perpendicular to it (i.e. deflected $30°$ from the original path — the complement of $60°$ ). Angle between them $90°$ ✓; energy $5^2 + 8.66^2 = 100$ ✓. Timing note: ~3 min. The impact-parameter phrase "line of centres makes angle α" IS the instruction to rotate axes — resolve there, swap the along-line components, reassemble. Never attempt the lab-frame simultaneous equations; the rotated frame does in three lines what raw components do in fifteen.

4 · Competitive level — is it elastic? Reading angles as evidence (NEET/JEE-Adv flavour, ~2.5 minutes)

Two identical pucks collide on smooth ice, one initially at rest. After the collision their paths make $35°$ and $40°$ with the original line of motion, on opposite sides. (a) Is the collision elastic? (b) The same experiment with a different target (at rest, unknown mass) yields $30°$ and $60°$ with the striker's speed dropping from $6.0$ to $3\sqrt3\ \mathrm{m/s}$ … wait — that IS consistent with 90°. What else must be checked before certifying elasticity?

Step 1. (a) Equal masses + target at rest: elastic would demand $\theta_1 + \theta_2 = 90°$ . Here $35° + 40° = 75° < 90°$ — the opening angle has closed, so kinetic energy was lost: $\boxed{\text{inelastic}}$ (and since they didn't stick, $0 < e < 1$ ).

Step 2. (b) A right angle alone is necessary for equal-mass elasticity, not sufficient in general — for unknown masses, 90° can occur inelastically too. Certify by the energy audit: striker's projection at $30°$ would be $6\cos 30° = 3\sqrt3 \approx 5.2$ m/s ✓ matches the given speed, and the partner's should be $6\cos 60° = 3.0$ m/s — check that measurement, and check the masses are equal, before signing "elastic".

Step 3. The MCQ pattern to keep: angle-sum $< 90°$ (equal masses, rest target) ⇒ inelastic, always. Angle-sum $= 90°$ ⇒ consistent with elastic; the projections $u\cos\theta$ seal it. Timing note: ~2.5 min. This is an evidence-reading question, not a computation — the 90° theorem run in reverse, exactly how Blackett used it (his measured 89–90° openings certified the α-particle's mass against helium's). "Angles close below 90°" = "energy was lost" is worth memorizing as a sentence.

Exam traps

One momentum equation where two are owed. In 2D, $x$ and $y$ balance separately. Any solution that adds momentum magnitudes without components has failed before it began.

1D closed forms on oblique geometry. The velocity-exchange formulas of the last two articles govern components along the line of impact only, after the perpendicular spectators are set aside. Applying them to full speeds in a glancing collision is structural error number one.

The 90° theorem without its certificate. Equal masses AND elastic AND target at rest. Unequal masses or $e<1$ move the angle (below 90° for inelastic equal masses — itself a tested fact).

Restitution applied off the normal. $e$ compares relative speeds along the line of impact. The perpendicular components never hear about $e$ — for smooth bodies they are conserved per body, not just in total.

Angle bookkeeping. " $30°$ with the original direction", "with the line of centres", "between the final paths" — three different angles. Draw the diagram; label $\alpha$ (aim) and $\theta_1, \theta_2$ (deflections) before any trigonometry.

Assuming the struck ball goes "forward". It departs along the line of centres — which points wherever the aim put it, up to $90°$ off the striker's path. Only a dead-centre hit sends it straight ahead.

Energy assumed conserved. A 2D problem that never says "elastic" gets the momentum components and nothing else; compute the KE change if asked, and expect a loss. The sticky intersection crash burns half the energy at equal masses and right angles — a favourite "how much was destroyed" follow-up.

Real billiards imported too literally. Spin and friction make real cue balls curve and follow through; exam spheres are smooth and spinless. Answer the idealized question — the 90° opening — not the club-room experience.

Origins · 1 minErnest Rutherford and how this was discovered

Ernest Rutherford

1871–1937 · McGill, Manchester & the Cavendish

Rutherford turned collision geometry into the microscope of the twentieth century. His 1911 analysis of alpha particles scattering off gold — some bouncing nearly straight back, "as if you had fired a 15-inch shell at tissue paper and it came back and hit you" — read nuclear structure out of deflection angles, founding the art of learning what things are by watching what angles they emerge at. The equal-mass right angle became one of its calibration marks: when Blackett photographed alpha particles colliding with helium nuclei in a cloud chamber in 1925, the tracks opened at ninety degrees — momentum and energy conservation certifying, in silver iodide, that the alpha is a helium nucleus. Rutherford's Manchester and Cavendish groups made angle-reading routine: Chadwick weighed the neutron from recoil directions, and every collider experiment since — up to the LHC identifying particles by the geometry of their debris — runs on the method this article teaches with billiard balls.

History

1668

— Huygens' collision rules already handle the 2D case by symmetry arguments and moving frames.

1835

— Coriolis' treatise on billiards mathematics analyses the cut shot the modern way, line of impact first.

1911

— Rutherford reads the nuclear atom out of scattering angles.

1925

— Blackett's cloud-chamber photographs show alpha–helium tracks opening at 90°: the theorem, photographed.

1932

— Chadwick's neutron: mass measured from recoil angles and speeds — 2D collision analysis as an instrument.

Going further

Texts. NCERT Physics Class 11, Ch. 6 (collisions in two dimensions) · HC Verma, Concepts of Physics Vol. 1, Ch. 9 (oblique collisions) · Kleppner & Kolenkow, Ch. 4 — and Blackett's 1925 cloud-chamber paper for the photograph that is this module's theorem.

Provenance. All results are standard. A verifier should recompute: the 90° proof via the squared momentum triangle ( $\vec v_1 \cdot \vec v_2 = 0$ ); the oblique recipe's reassembly (target along the line at $u\cos\alpha$ , striker perpendicular at $u\sin\alpha$ ); Example 1 ( $14.1$ m/s at $45°$ , 50% KE destroyed); Example 2 ( $3.46$ and $2.0$ m/s, $16 = 16$ energy audit); Example 3 ( $5.0$ along, $8.66$ perpendicular); Example 4's evidence logic ( $75° < 90°$ ⇒ inelastic).

Prerequisites. Elastic Collisions in 1D · Inelastic Collisions & Coefficient of Restitution · Components & Resolution of Vectors. Leads to. Locating the Centre of Mass · Motion of the Centre of Mass · Centre-of-Mass Frame Collisions (UG deep dive).

How do I find the line of impact in a problem?

It is the line through the two centres at the instant of contact — for spheres and discs, automatically perpendicular to the touching surfaces. Problems hand it to you in one of three costumes: explicitly ("the line of centres makes 60° with the velocity"), through the geometry of a figure, or through the impact parameter (the sideways offset of the aim — offset $b$ on spheres of combined radius $R$ gives $\sin\alpha = b/R$ ). In the sticky-crash family you never need it at all: when the bodies leave together, vector momentum conservation alone fixes everything.

Why does the collision force act only along the line of impact?

Because "smooth" means the surfaces cannot exert friction — and the only force a frictionless contact can transmit is the normal push, perpendicular to the touching surfaces, which for spheres points through both centres. It is the same statement as the smooth wall in the impulse article, curved. The fine print is real, though: rough bodies (real billiard balls with cloth friction and spin, gripping tyres) do transmit tangential impulses, which is why real cue balls can curve and follow through. Exam bodies are smooth unless the problem says otherwise, and the phrase "smooth spheres" is the licence for the spectator rule.

What exactly breaks the 90° result when the collision is inelastic?

Only the along-normal part of the motion. The perpendicular components are spectators at any $e$ , but the separation along the line of impact gets throttled by the factor $e$ — the bodies come apart more slowly along the normal than elasticity would demand. The striker keeps a bit of forward along-line motion it would have surrendered (equal masses: the swap is incomplete), tilting both final velocities toward each other. The angle closes smoothly from 90° at $e=1$ to 0° at $e=0$ , where they travel together. Hence the evidence rule of worked example 4: opening angle below 90° (equal masses, rest target) is a measurement of energy loss.

Does the 90° theorem work if the target is moving?

Not as stated — "target at rest" is a genuine hypothesis, because the proof needs the momentum triangle's sides to be exactly $u, v_1, v_2$ with no extra vector. But there is a professional rescue: view the collision from a frame riding with the target (subtract its velocity from everything). In that frame the target IS at rest, the theorem holds, and you transform back at the end. That frame-hopping move — Huygens' boat, three and a half centuries old — is the gateway to the centre-of-mass frame treatment that the UG deep dive builds, where every elastic collision becomes a simple rotation.

In the sticky 2D crash, why don't I need the line of impact?

Because the question you're asked has only one unknown vector. "They move off together" means one shared final velocity — two unknown components — and momentum conservation supplies exactly two equations. The system is closed with no information about the contact geometry, the forces or the deformation; the line of impact influenced how the crash unfolded internally, but not the only thing you're computing. This is conservation at its most powerful: total ignorance of the mechanism, complete knowledge of the outcome.

The striker in billiards sometimes follows the target or even screws back. Doesn't that contradict the perpendicular rule?

It confirms how good the smooth-sphere idealization is at exposing what's extra in real play. A real cue ball carries spin, and the cloth and the ball-ball contact are not frictionless: topspin drives the cue ball forward after contact ("follow"), backspin drags it back ("screw"), and side imparts swerve. All of these are tangential-impulse effects — exactly the term the smooth model sets to zero. The stun shot (no spin at contact) is the model's clean realization, and it does behave as the theorem says: dead-centre stun stops the cue ball; a cut stun sends it perpendicular to the object ball. Players effectively choose how much to depart from this article's physics.