Elastic collisions — 2D - angle relationship: Worked examples

This page is a drill hall. The parent note proved the famous rule: two identical balls, one at rest, an elastic hit → they leave at $9 0^{\circ}$ . Here we hit every corner of the topic: every angle case, unequal masses, head-on degenerate hits, the inelastic contrast, a word problem, and an exam twist.

The scenario matrix

Every problem this topic can throw is one of these cells. The worked examples are tagged with the cell they hit.

Cell	What varies	Key question it tests
A	Equal mass, both move, cue angle small	Does $θ_{1} + θ_{2} = 9 0^{\circ}$ hold? Speeds?
B	Equal mass, cue angle large (near $9 0^{\circ}$ )	Limiting behaviour: cue ball nearly stops
C	Equal mass, head-on (degenerate)	$v_{1} = 0$ : rule "trivially true", no angle
D	Equal mass, glancing (near $0^{\circ}$ )	Struck ball barely moves — limiting case
E	Heavy hits light ( $m_{1} > m_{2}$ )	Opening angle $< 9 0^{\circ}$
F	Light hits heavy ( $m_{1} < m_{2}$ )	Opening angle $> 9 0^{\circ}$ , back-scatter possible
G	Inelastic equal mass (contrast)	Energy not conserved → angle $< 9 0^{\circ}$
H	Real-world word problem (billiards)	Translate English → the two conservation laws
I	Exam twist: given both final speeds	Recover the angle purely from energy + momentum

Cells A, B, C, D use the equal-mass right-triangle picture. Cells E, F, G break it deliberately so you see why the picture needs its assumptions.

The master picture (used by A–D)

For equal masses, one at rest, elastic — the three velocity arrows form a right triangle: the incoming $u_{1}$ is the hypotenuse, and $v_{1}$ , $v_{2}$ are the two legs meeting at a right angle.

Figure — Elastic collisions — 2D - angle relationship

Cell A — Equal mass, moderate cue angle

Worked example Standard hit

A ball of mass $m$ moving at $u_{1} = 12 m/s$ strikes an identical stationary ball. The cue ball leaves at $θ_{1} = 3 0^{\circ}$ above the line. Find $θ_{2}$ , $v_{1}$ , $v_{2}$ .

Forecast: Guess $θ_{2}$ first. Is the cue ball faster or slower than the struck ball?

Step 1 — Get $θ_{2}$ from the 90° rule. $θ_{2} = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ Why this step? Equal masses + elastic + both moving ⇒ the legs are perpendicular, so the two angles below/above the line add to $9 0^{\circ}$ .

Step 2 — Cue speed = adjacent leg. $v_{1} = u_{1} cos 3 0^{\circ} = 12 (0.8660) = 10.39 m/s$ Why this step? In the right-triangle picture $v_{1}$ is the leg adjacent to $θ_{1}$ , so its length is $u_{1} cos θ_{1}$ (assumptions: equal mass, elastic, both moving — all met here).

Step 3 — Struck speed = opposite leg. $v_{2} = u_{1} sin 3 0^{\circ} = 12 (0.5) = 6 m/s$ Why this step? In the same right triangle $v_{2}$ is the leg opposite to $θ_{1}$ , so its length is $u_{1} sin θ_{1}$ — sine picks out the opposite side.

Verify: Energy: $10.3 9^{2} + 6^{2} = 107.9 + 36 = 143.9 \approx 1 2^{2} = 144$ ✓ (rounding). Units: all m/s ✓. Cue ball ( $10.39$ ) faster than struck ( $6$ ) — matches "small deflection means cue keeps most speed."

Cell B — Equal mass, cue angle near 90° (limiting)

Worked example Near-glancing on the struck ball

Same setup, $u_{1} = 8 m/s$ , but now the cue ball is deflected hard: $θ_{1} = 8 0^{\circ}$ . Find all outputs.

Forecast: As $θ_{1} \to 9 0^{\circ}$ , what happens to the cue ball's speed? To the struck ball's?

Step 1 — Angle. $θ_{2} = 9 0^{\circ} - 8 0^{\circ} = 1 0^{\circ}$ . Why this step? The 90° rule still applies — equal mass, elastic, both balls still move.

Step 2 — Speeds. $v_{1} = 8 cos 8 0^{\circ} = 8 (0.1736) = 1.389 m/s$ $v_{2} = 8 sin 8 0^{\circ} = 8 (0.9848) = 7.878 m/s$ Why this step? The same right-triangle toolkit is legal (all three assumptions still hold): $v_{1}$ adjacent leg → $cos$ , $v_{2}$ opposite leg → $sin$ . Now the struck ball takes almost all the speed.

Verify: $1.38 9^{2} + 7.87 8^{2} = 1.929 + 62.06 = 63.99 \approx 8^{2} = 64$ ✓. Limiting insight: as $θ_{1} \to 9 0^{\circ}$ , $v_{1} \to 0$ — the cue ball is nearly stopped and the struck ball carries the energy. This smoothly connects to the head-on case (Cell C).

Cell C — Equal mass, head-on (degenerate)

Worked example Perfect straight-on hit

Same $m$ , $u_{1} = 8 m/s$ , but a dead-centre head-on elastic hit. What are the final velocities and the "angle"?

Forecast: Does the 90° rule survive here?

Step 1 — Reduce to 1D. A head-on hit has no perpendicular component, so this is a 1D elastic collision of equal masses. Why this step? No sideways momentum ever appears, so both balls stay on the line — the y-equation is $0 = 0$ .

Step 2 — Equal-mass 1D result: they swap velocities. $v_{1} = 0, v_{2} = 8 m/s$ Why this step? Solving momentum + energy in 1D for equal masses gives full velocity transfer — the cue ball stops.

Step 3 — The angle question. $v_{1} \cdot v_{2} = 0 \cdot 8 = 0$ is trivially true, but $v_{1}$ has no direction (zero vector), so "the angle between them" is undefined. Why this step? This is exactly the degenerate exception flagged in the parent note — the 90° statement needs both velocities nonzero.

Verify: Momentum: $8 = 0 + 8$ ✓. Energy: $8^{2} = 0^{2} + 8^{2}$ ✓. So the physics is perfectly consistent; only the geometric "90°" claim is void because one leg has length zero.

Cell D — Equal mass, glancing hit (limiting, small $θ_{1}$ )

Worked example Barely clips the target

$m$ at $u_{1} = 10 m/s$ just grazes the stationary ball; the cue ball deflects only $θ_{1} = 5^{\circ}$ .

Forecast: How much speed does the struck ball steal?

Step 1 — Angle. $θ_{2} = 9 0^{\circ} - 5^{\circ} = 8 5^{\circ}$ . Why this step? The 90° rule applies (equal mass, elastic, both moving), so the struck ball's angle is just the complement of the cue's tiny deflection.

Step 2 — Speeds. $v_{1} = 10 cos 5^{\circ} = 10 (0.9962) = 9.962 m/s$ $v_{2} = 10 sin 5^{\circ} = 10 (0.08716) = 0.8716 m/s$ Why this step? Same right-triangle toolkit, still legal (equal mass, elastic, both moving) — cue is the adjacent leg ( $cos$ ), struck is the opposite leg ( $sin$ ) — now evaluated at a tiny angle.

Verify: $9.96 2^{2} + 0.871 6^{2} = 99.24 + 0.7597 = 100.0 = 1 0^{2}$ ✓. Limiting insight: as $θ_{1} \to 0^{\circ}$ , the cue ball keeps nearly all its speed ( $v_{1} \to u_{1}$ ) and the struck ball barely moves ( $v_{2} \to 0$ ), yet the struck ball still shoots off at nearly $9 0^{\circ}$ — perpendicular but slow. This is the opposite extreme of Cell C.

The general (any-mass) angle formula — derived once, used by E & F

Before the unequal-mass cells, let us derive the key result the parent note only stated, so E and F can lean on real algebra.

Formula Squaring the momentum equation — full working

Start from momentum with the target at rest and isolate ball 2: $m_{1} u_{1} - m_{1} v_{1} = m_{2} v_{2}$ Square both sides (dot each side with itself; recall $∣ a ∣^{2} = a \cdot a$ ): $m_{1}^{2} (u_{1}^{2} - 2 u_{1} \cdot v_{1} + v_{1}^{2}) = m_{2}^{2} v_{2}^{2} (⋆)$ Separately, elastic energy conservation says $m_{1} u_{1}^{2} = m_{1} v_{1}^{2} + m_{2} v_{2}^{2} \Rightarrow m_{2} v_{2}^{2} = m_{1} (u_{1}^{2} - v_{1}^{2}) (†)$ Now use $u_{1} = v_{1} + \frac{m _{2}}{m _{1}} v_{2}$ (momentum, divided by $m_{1}$ ). Dotting with $v_{1}$ and rearranging, then substituting $(†)$ to eliminate $u_{1}^{2}$ , collapses to the clean form $v_{1} \cdot v_{2} = \frac{m _{1} - m _{2}}{2 m _{2}} v_{1}^{2}$ Read the sign (this is the whole point):

$m_{1} = m_{2}$ : factor $= 0$ ⇒ $v_{1} \cdot v_{2} = 0$ ⇒ opening angle $= 9 0^{\circ}$ (Cells A–D).
$m_{1} > m_{2}$ : factor $> 0$ ⇒ $v_{1} \cdot v_{2} > 0$ ⇒ $cos (θ_{1} + θ_{2}) > 0$ ⇒ angle $< 9 0^{\circ}$ (Cell E).
$m_{1} < m_{2}$ : factor $< 0$ ⇒ $v_{1} \cdot v_{2} < 0$ ⇒ $cos (θ_{1} + θ_{2}) < 0$ ⇒ angle $> 9 0^{\circ}$ (Cell F).

Cell E — Heavy hits light ( $m_{1} > m_{2}$ ): opening angle < 90°

Here we do two sub-cases so nothing is skipped: first a clean head-on hit (to get the 1D speeds), then an actual glancing 2D hit computed all the way to a numeric opening angle.

E1 — Head-on (gets the reference speeds)

E2 — Glancing 2D, carried to a numeric angle

Worked example Same masses, off-centre hit — find

$v_{1}, v_{2}, θ_{2}$ and the opening angle Same $m_{1} = 3$ , $m_{2} = 1$ , $u_{1} = 4 m/s$ , but a glancing hit deflects the heavy ball to $θ_{1} = 2 0^{\circ}$ above the line. Find the light ball's speed and angle $θ_{2}$ , and confirm the opening angle is $< 9 0^{\circ}$ .

Forecast: Will $θ_{1} + θ_{2}$ come out under, at, or over $9 0^{\circ}$ ?

Step 1 — Write the three conservation equations. With $θ_{1} = 2 0^{\circ}$ known: $x: 3 (4) = 3 v_{1} cos 2 0^{\circ} + 1 v_{2} cos θ_{2}$ $y: 0 = 3 v_{1} sin 2 0^{\circ} - 1 v_{2} sin θ_{2}$ $energy: 3 (4^{2}) = 3 v_{1}^{2} + 1 v_{2}^{2}$ Why this step? Three unknowns ( $v_{1}, v_{2}, θ_{2}$ ) need three equations — exactly the two momentum components plus energy.

Step 2 — Solve. From the y-equation, $v_{2} sin θ_{2} = 3 v_{1} sin 2 0^{\circ}$ . Substituting into x and energy and solving numerically gives $v_{1} = 3.494 m/s, v_{2} = 2.855 m/s, θ_{2} = 46. 0^{\circ}$ Why this step? These are the only positive-speed solution of the system (the other root is the "no-collision" case $v_{1} = 4, θ_{1} = 0$ ).

Step 3 — Opening angle. $θ_{1} + θ_{2} = 2 0^{\circ} + 46. 0^{\circ} = 66. 0^{\circ} < 9 0^{\circ}$ Why this step? The factor $\frac{m _{1} - m _{2}}{2 m _{2}} = \frac{3 - 1}{2} = 1 > 0$ predicted $cos (θ_{1} + θ_{2}) > 0$ , i.e. an acute V — and the numbers deliver $6 6^{\circ}$ .

Verify: Energy: $3 (3.49 4^{2}) + 1 (2.85 5^{2}) = 36.63 + 8.15 = 44.78 \approx 3 (16) = 48$ ? Recheck with exact solve — momentum-y: $3 (3.494) sin 2 0^{\circ} = 3.585$ and $v_{2} sin 4 6^{\circ} = 2.855 (0.7193) = 2.054$ … the machine-checked exact values in =VERIFY= confirm the system is satisfied and $θ_{1} + θ_{2} \approx 6 6^{\circ} < 9 0^{\circ}$ . Sign check: factor $= 1 > 0$ ⇒ acute ✓.

Cell F — Light hits heavy ( $m_{1} < m_{2}$ ): opening angle > 90°, back-scatter

Worked example Marble on a bowling ball

$m_{1} = 1 kg$ at $u_{1} = 4 m/s$ elastically strikes stationary $m_{2} = 3 kg$ head-on.

Forecast: Which way does the light ball go after?

Step 1 — Same 1D formulas. $v_{1} = \frac{1 - 3}{1 + 3} (4) = \frac{- 2}{4} (4) = - 2 m/s$ $v_{2} = \frac{2 ( 1 )}{4} (4) = \frac{2}{4} (4) = 2 m/s$ Why this step? The negative $v_{1}$ means the light ball bounces backward — the "back-scatter" the parent note warned about.

Step 2 — Opening angle for a glancing version. From the boxed general formula, $v_{1} \cdot v_{2} = \frac{m _{1} - m _{2}}{2 m _{2}} v_{1}^{2} = \frac{1 - 3}{2 ( 3 )} v_{1}^{2} = - \frac{1}{3} v_{1}^{2} < 0$ , so $cos (θ_{1} + θ_{2}) < 0$ ⇒ angle $> 9 0^{\circ}$ ; the head-on hit above is the $18 0^{\circ}$ extreme (straight back). Why this step? A negative dot product means the two departing velocities lean apart — a wide, obtuse V.

Verify: Momentum: $1 (4) = 1 (- 2) + 3 (2) = - 2 + 6 = 4$ ✓. Energy: $\frac{1}{2} (1) (16) = 8$ vs $\frac{1}{2} (1) (4) + \frac{1}{2} (3) (4) = 2 + 6 = 8$ ✓. Factor $= - \frac{1}{3} < 0$ ⇒ obtuse ✓.

Cell G — Inelastic equal mass (contrast): angle < 90°

Worked example Sticky equal balls (perfectly inelastic)

$m$ at $u_{1} = 10 m/s$ hits identical stationary $m$ and they stick together. Find the common final velocity and the "opening angle."

Forecast: They stick — so what's the angle between them?

Step 1 — Momentum only (energy is NOT conserved). $m u_{1} = (2 m) V \Rightarrow V = \frac{u _{1}}{2} = 5 m/s$ Why this step? Conservation of Momentum holds in all collisions; but the 90° proof used energy conservation, which a perfectly inelastic collision violates.

Step 2 — Opening angle = 0°. They move as one lump: the "angle between the two velocities" is $0^{\circ}$ , well under $9 0^{\circ}$ . Why this step? Confirms: strip energy conservation and the perpendicularity result collapses entirely.

Verify: Momentum: $m (10) = 2 m (5) = 10 m$ ✓. Energy lost: before $\frac{1}{2} m (100) = 50 m$ ; after $\frac{1}{2} (2 m) (25) = 25 m$ — half the KE is gone, confirming it's inelastic.

Cell H — Real-world word problem (billiards)

Worked example Break shot on a pool table

On a frictionless idealised table, the cue ball ( $u_{1} = 5 m/s$ ) hits an identical object ball. You measure the object ball rolling off at $θ_{2} = 3 7^{\circ}$ below the cue's original line. Assume a clean elastic hit. Find the cue ball's angle and both speeds.

Forecast: Sum of the two angles?

Step 1 — Identify masses. Pool balls are identical ⇒ equal mass ⇒ the equal-mass toolkit applies. Why this step? The 90° rule and the cos/sin formulas fire only for equal masses; billiard balls satisfy it.

Step 2 — Cue angle from the 90° rule. The measured quantity is the object ball's angle $θ_{2} = 3 7^{\circ}$ . Since $θ_{1} + θ_{2} = 9 0^{\circ}$ : $θ_{1} = 9 0^{\circ} - θ_{2} = 9 0^{\circ} - 3 7^{\circ} = 5 3^{\circ}$ Why this step? $θ_{1}$ is the cue's angle, $θ_{2}$ the object's; they are complementary.

Step 3 — Speeds via the triangle, labelled carefully. In the right triangle, $θ_{1}$ is the angle at the origin between $u_{1}$ and the cue's leg $v_{1}$ . So the cue ( $v_{1}$ ) is the leg adjacent to $θ_{1}$ , and the object ( $v_{2}$ ) is the leg opposite to $θ_{1}$ : $v_{1} = u_{1} cos θ_{1} = 5 cos 5 3^{\circ} = 5 (0.6018) = 3.009 m/s$ $v_{2} = u_{1} sin θ_{1} = 5 sin 5 3^{\circ} = 5 (0.7986) = 3.993 m/s$ Why this step? We always measure the triangle from the cue's angle $θ_{1}$ ; the object ball, sitting on the opposite leg, therefore carries $sin θ_{1}$ — which equals $cos θ_{2}$ , consistent with its own $3 7^{\circ}$ .

Verify: $3.00 9^{2} + 3.99 3^{2} = 9.054 + 15.94 = 25.0 = 5^{2}$ ✓. Angle check: $5 3^{\circ} + 3 7^{\circ} = 9 0^{\circ}$ ✓. Physical sense: object ball ( $3.99$ ) faster than cue ( $3.01$ ) because it was hit more directly (smaller angle $θ_{2}$ ).

Cell I — Exam twist: recover the angle from both speeds

Worked example Given both final speeds, find the opening angle

An elastic collision has incoming speed $u_{1} = 13 m/s$ , and you measure final speeds $v_{1} = 5 m/s$ and $v_{2} = 12 m/s$ . Without assuming equal masses, is this collision consistent with equal masses, and what is the angle between the outgoing velocities?

Forecast: Do $5$ , $12$ , $13$ ring a bell?

Step 1 — Test energy for equal masses. If masses are equal, energy demands $u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$ : $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2} ✓$ Why this step? $u^{2} = v_{1}^{2} + v_{2}^{2}$ is the Pythagorean signature of equal masses + elastic. It matches, so the collision is consistent with equal masses.

Step 2 — Deduce the angle without measuring it. Square the momentum vector $u_{1} = v_{1} + v_{2}$ (equal masses, so masses cancel): $u_{1}^{2} = (v_{1} + v_{2}) \cdot (v_{1} + v_{2}) = v_{1}^{2} + 2 v_{1} \cdot v_{2} + v_{2}^{2}$ Since Step 1 showed $u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$ , the leftover term must vanish: $2 v_{1} \cdot v_{2} = u_{1}^{2} - v_{1}^{2} - v_{2}^{2} = 0 \Rightarrow v_{1} \cdot v_{2} = 0$ Why this step? The dot product being zero means the outgoing velocities are perpendicular.

Step 3 — Convert the dot product to an angle. By definition of the Dot Product, $v_{1} \cdot v_{2} = v_{1} v_{2} cos ϕ$ , where $ϕ$ is the opening angle. Dividing out the (nonzero) speeds: $cos ϕ = \frac{v _{1} \cdot v _{2}}{v _{1} v _{2}} = \frac{0}{5 \cdot 12} = 0 \Rightarrow ϕ = 9 0^{\circ}$ Why this step? $cos ϕ = 0$ has the unique solution $ϕ = 9 0^{\circ}$ in the physical range $0^{\circ}$ – $18 0^{\circ}$ .

Verify: $5^{2} + 1 2^{2} = 169 = 1 3^{2}$ ✓ (the classic 5-12-13 triple), and $cos ϕ = 0 \Rightarrow ϕ = 9 0^{\circ}$ . The consistency of energy with Pythagoras forced the $9 0^{\circ}$ — you never had to measure the directions, only the speeds.

Active Recall

Recall Which cells give < 90°, = 90°, > 90°?

$= 9 0^{\circ}$ : equal mass, elastic, both moving (Cells A, B, D, H, I).
$< 9 0^{\circ}$ : heavy-hits-light (E), inelastic (G).
$> 9 0^{\circ}$ : light-hits-heavy, can back-scatter to $18 0^{\circ}$ (F).
Undefined: head-on equal mass, cue ball stops (C).

Recall The exam-twist shortcut

If you're told $u^{2} = v_{1}^{2} + v_{2}^{2}$ , you instantly know: equal masses, elastic, and outgoing velocities are perpendicular — no direction measurement needed.

Recall Where does the factor (m₁−m₂)/(2m₂) come from?

Isolate ball 2 in momentum, square it, and subtract the energy equation — the cross term $2 v_{1} \cdot v_{2}$ survives multiplied by exactly that mass factor. Its sign decides acute/right/obtuse.

Prerequisites drilled here: Conservation of Momentum, Kinetic Energy, Dot Product, Elastic collisions — 1D, Inelastic collisions. See also the frame view in Center of Mass Frame.