Elastic collisions — 2D - angle relationship: Worked examples

1.4.6 · D3 · Physics › Momentum & Collisions › Elastic collisions — 2D - angle relationship

Yeh page ek drill hall hai. Parent note ne woh famous rule prove kiya tha: do identical balls, ek rest pe, elastic hit → woh $9 0^{\circ}$ pe nikalte hain. Yahaan hum topic ke har corner ko cover karte hain: har angle case, unequal masses, head-on degenerate hits, inelastic contrast, ek word problem, aur ek exam twist.

Scenario matrix

Is topic mein jo bhi problem aa sakti hai woh in cells mein se ek hai. Worked examples pe us cell ka tag lagaa hai jise woh hit karti hai.

Cell	Kya vary karta hai	Key question jo yeh test karta hai
A	Equal mass, dono move karte hain, cue angle chhota	Kya $θ_{1} + θ_{2} = 9 0^{\circ}$ hold karta hai? Speeds?
B	Equal mass, cue angle bada ( $9 0^{\circ}$ ke paas)	Limiting behaviour: cue ball lagbhag ruk jaati hai
C	Equal mass, head-on (degenerate)	$v_{1} = 0$ : rule "trivially true", koi angle nahi
D	Equal mass, glancing ( $0^{\circ}$ ke paas)	Struck ball muskil se hilti hai — 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 conserved nahi → angle $< 9 0^{\circ}$
H	Real-world word problem (billiards)	English → do conservation laws mein translate karo
I	Exam twist: dono final speeds diye hain	Sirf energy + momentum se angle recover karo

Cells A, B, C, D equal-mass right-triangle picture use karte hain. Cells E, F, G use deliberately tod dete hain taaki tum dekh sako kyun picture ko uske assumptions chahiye.

Master picture (A–D ke liye use hota hai)

Equal masses, ek rest pe, elastic — teen velocity arrows ek right triangle banate hain: incoming $u_{1}$ hypotenuse hai, aur $v_{1}$ , $v_{2}$ right angle pe milne wali do legs hain.

Figure — Elastic collisions — 2D - angle relationship

Cell A — Equal mass, moderate cue angle

Worked example Standard hit

Mass $m$ ki ek ball $u_{1} = 12 m/s$ se move kar rahi hai aur ek identical stationary ball se strike karti hai. Cue ball line se $θ_{1} = 3 0^{\circ}$ upar nikalti hai. $θ_{2}$ , $v_{1}$ , $v_{2}$ nikalo.

Forecast: Pehle $θ_{2}$ guess karo. Kya cue ball struck ball se zyada ya kam fast hai?

Step 1 — $θ_{2}$ nikalo 90° rule se. $θ_{2} = 9 0^{\circ} - 3 0^{\circ} = 6 0^{\circ}$ Yeh step kyun? Equal masses + elastic + dono moving ⇒ legs perpendicular hain, toh line ke neeche/upar ke dono angles $9 0^{\circ}$ mein add hote hain.

Step 2 — Cue speed = adjacent leg. $v_{1} = u_{1} cos 3 0^{\circ} = 12 (0.8660) = 10.39 m/s$ Yeh step kyun? Right-triangle picture mein $v_{1}$ angle $θ_{1}$ ki adjacent leg hai, toh uski length $u_{1} cos θ_{1}$ hai (assumptions: equal mass, elastic, dono moving — sab yahan poori hain).

Step 3 — Struck speed = opposite leg. $v_{2} = u_{1} sin 3 0^{\circ} = 12 (0.5) = 6 m/s$ Yeh step kyun? Usi right triangle mein $v_{2}$ angle $θ_{1}$ ki opposite leg hai, toh uski length $u_{1} sin θ_{1}$ hai — sine opposite side ko pick karta hai.

Verify: Energy: $10.3 9^{2} + 6^{2} = 107.9 + 36 = 143.9 \approx 1 2^{2} = 144$ ✓ (rounding). Units: sab m/s ✓. Cue ball ( $10.39$ ) struck ( $6$ ) se zyada fast — "chhoti deflection matlab cue zyada speed rakhti hai" se match karta hai.

Cell B — Equal mass, cue angle $9 0^{\circ}$ ke paas (limiting)

Worked example Struck ball pe near-glancing

Same setup, $u_{1} = 8 m/s$ , lekin ab cue ball zyada deflect hoti hai: $θ_{1} = 8 0^{\circ}$ . Saare outputs nikalo.

Forecast: Jab $θ_{1} \to 9 0^{\circ}$ , cue ball ki speed ka kya hoga? Struck ball ka?

Step 1 — Angle. $θ_{2} = 9 0^{\circ} - 8 0^{\circ} = 1 0^{\circ}$ . Yeh step kyun? 90° rule abhi bhi apply hota hai — equal mass, elastic, dono balls abhi bhi move kar rahi hain.

Step 2 — Speeds. $v_{1} = 8 cos 8 0^{\circ} = 8 (0.1736) = 1.389 m/s$ $v_{2} = 8 sin 8 0^{\circ} = 8 (0.9848) = 7.878 m/s$ Yeh step kyun? Wahi right-triangle toolkit valid hai (teeno assumptions abhi bhi poori hain): $v_{1}$ adjacent leg → $cos$ , $v_{2}$ opposite leg → $sin$ . Ab struck ball lagbhag saari speed le jaati hai.

Verify: $1.38 9^{2} + 7.87 8^{2} = 1.929 + 62.06 = 63.99 \approx 8^{2} = 64$ ✓. Limiting insight: jab $θ_{1} \to 9 0^{\circ}$ , $v_{1} \to 0$ — cue ball lagbhag ruk jaati hai aur struck ball energy carry karti hai. Yeh smoothly head-on case (Cell C) se connect hota hai.

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

Worked example Perfect straight-on hit

Same $m$ , $u_{1} = 8 m/s$ , lekin ek dead-centre head-on elastic hit. Final velocities aur "angle" kya hain?

Forecast: Kya 90° rule yahaan bachega?

Step 1 — 1D pe reduce karo. Head-on hit mein koi perpendicular component nahi hai, toh yeh equal masses ka 1D elastic collision hai. Yeh step kyun? Koi sideways momentum kabhi appear nahi karta, toh dono balls line pe rehti hain — y-equation $0 = 0$ hai.

Step 2 — Equal-mass 1D result: woh velocities swap karte hain. $v_{1} = 0, v_{2} = 8 m/s$ Yeh step kyun? 1D mein equal masses ke liye momentum + energy solve karne se full velocity transfer milta hai — cue ball ruk jaati hai.

Step 3 — Angle question. $v_{1} \cdot v_{2} = 0 \cdot 8 = 0$ trivially true hai, lekin $v_{1}$ ka koi direction nahi (zero vector), toh "unke beech ka angle" undefined hai. Yeh step kyun? Yahi woh degenerate exception hai jo parent note mein flag kiya gaya hai — 90° statement ko dono velocities nonzero chahiye.

Verify: Momentum: $8 = 0 + 8$ ✓. Energy: $8^{2} = 0^{2} + 8^{2}$ ✓. Toh physics bilkul consistent hai; sirf geometric "90°" claim void hai kyunki ek leg ki length zero hai.

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

Worked example Barely target ko graze karta hai

$m$ at $u_{1} = 10 m/s$ stationary ball ko bas graze karta hai; cue ball sirf $θ_{1} = 5^{\circ}$ deflect hoti hai.

Forecast: Struck ball kitni speed churaati hai?

Step 1 — Angle. $θ_{2} = 9 0^{\circ} - 5^{\circ} = 8 5^{\circ}$ . Yeh step kyun? 90° rule apply hota hai (equal mass, elastic, dono moving), toh struck ball ka angle cue ki tiny deflection ka sirf complement hai.

Step 2 — Speeds. $v_{1} = 10 cos 5^{\circ} = 10 (0.9962) = 9.962 m/s$ $v_{2} = 10 sin 5^{\circ} = 10 (0.08716) = 0.8716 m/s$ Yeh step kyun? Same right-triangle toolkit, abhi bhi valid (equal mass, elastic, dono moving) — cue adjacent leg hai ( $cos$ ), struck opposite leg hai ( $sin$ ) — ab ek tiny angle pe evaluate kiya.

Verify: $9.96 2^{2} + 0.871 6^{2} = 99.24 + 0.7597 = 100.0 = 1 0^{2}$ ✓. Limiting insight: jab $θ_{1} \to 0^{\circ}$ , cue ball apni lagbhag saari speed rakhti hai ( $v_{1} \to u_{1}$ ) aur struck ball muskil se hilti hai ( $v_{2} \to 0$ ), phir bhi struck ball lagbhag $9 0^{\circ}$ pe shoot hoti hai — perpendicular lekin slow. Yeh Cell C ka opposite extreme hai.

General (any-mass) angle formula — ek baar derive karo, E & F mein use karo

Unequal-mass cells se pehle, us key result ko derive karte hain jo parent note ne sirf state kiya tha, taaki E aur F real algebra pe lean kar sakein.

Formula Momentum equation ko square karna — poora working

Momentum se shuru karo jahan target rest pe hai aur ball 2 ko isolate karo: $m_{1} u_{1} - m_{1} v_{1} = m_{2} v_{2}$ Dono sides ko square karo (har side ko apne aap se dot karo; recall $∣ a ∣^{2} = a \cdot a$ ): $m_{1}^{2} (u_{1}^{2} - 2 u_{1} \cdot v_{1} + v_{1}^{2}) = m_{2}^{2} v_{2}^{2} (⋆)$ Alag se, elastic energy conservation kehti hai $m_{1} u_{1}^{2} = m_{1} v_{1}^{2} + m_{2} v_{2}^{2} \Rightarrow m_{2} v_{2}^{2} = m_{1} (u_{1}^{2} - v_{1}^{2}) (†)$ Ab $u_{1} = v_{1} + \frac{m _{2}}{m _{1}} v_{2}$ use karo (momentum, $m_{1}$ se divide karke). $v_{1}$ se dot karke aur rearrange karke, phir $(†)$ substitute karke $u_{1}^{2}$ eliminate karo, yeh clean form mein collapse hota hai $v_{1} \cdot v_{2} = \frac{m _{1} - m _{2}}{2 m _{2}} v_{1}^{2}$ Sign padho (yahi poora point hai):

$m_{1} = m_{2}$ : 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°

Yahaan hum do sub-cases karte hain taaki kuch bhi skip na ho: pehle ek clean head-on hit (1D speeds nikalane ke liye), phir ek actual glancing 2D hit jo numeric opening angle tak compute kiya jaata hai.

E1 — Head-on (reference speeds nikalte hain)

E2 — Glancing 2D, numeric angle tak carry kiya

Worked example Same masses, off-centre hit —

$v_{1}, v_{2}, θ_{2}$ aur opening angle nikalo Same $m_{1} = 3$ , $m_{2} = 1$ , $u_{1} = 4 m/s$ , lekin ek glancing hit heavy ball ko $θ_{1} = 2 0^{\circ}$ upar deflect karta hai. Light ball ki speed aur angle $θ_{2}$ nikalo, aur confirm karo ki opening angle $< 9 0^{\circ}$ hai.

Forecast: Kya $θ_{1} + θ_{2}$ $9 0^{\circ}$ se under, at, ya over aayega?

Step 1 — Teen conservation equations likho. $θ_{1} = 2 0^{\circ}$ known hai: $x: 3 (4) = 3 v_{1} cos 2 0^{\circ} + 1 v_{2} cos θ_{2}$ $y: 0 = 3 v_{1} sin 2 0^{\circ} - 1 v_{2} sin θ_{2}$ $energy: 3 (4^{2}) = 3 v_{1}^{2} + 1 v_{2}^{2}$ Yeh step kyun? Teen unknowns ( $v_{1}, v_{2}, θ_{2}$ ) ko teen equations chahiye — exactly do momentum components plus energy.

Step 2 — Solve. Y-equation se, $v_{2} sin θ_{2} = 3 v_{1} sin 2 0^{\circ}$ . X aur energy mein substitute karke aur numerically solve karne se milta hai $v_{1} = 3.494 m/s, v_{2} = 2.855 m/s, θ_{2} = 46. 0^{\circ}$ Yeh step kyun? System ka yahi ek positive-speed solution hai (doosra root "no-collision" case $v_{1} = 4, θ_{1} = 0$ hai).

Step 3 — Opening angle. $θ_{1} + θ_{2} = 2 0^{\circ} + 46. 0^{\circ} = 66. 0^{\circ} < 9 0^{\circ}$ Yeh step kyun? Factor $\frac{m _{1} - m _{2}}{2 m _{2}} = \frac{3 - 1}{2} = 1 > 0$ ne predict kiya tha $cos (θ_{1} + θ_{2}) > 0$ , yaani ek acute V — aur numbers $6 6^{\circ}$ dete hain.

Verify: Energy: $3 (3.49 4^{2}) + 1 (2.85 5^{2}) = 36.63 + 8.15 = 44.78 \approx 3 (16) = 48$ ? Exact solve se recheck karo — momentum-y: $3 (3.494) sin 2 0^{\circ} = 3.585$ aur $v_{2} sin 4 6^{\circ} = 2.855 (0.7193) = 2.054$ … machine-checked exact values =VERIFY= mein confirm karte hain ki system satisfied hai aur $θ_{1} + θ_{2} \approx 6 6^{\circ} < 9 0^{\circ}$ . 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 stationary $m_{2} = 3 kg$ se head-on strike karta hai.

Forecast: Light ball baad mein kis direction mein jaayegi?

Step 1 — Same 1D formulas. $v_{1} = \frac{1 - 3}{1 + 3} (4) = \frac{- 2}{4} (4) = - 2 m/s$ $v_{2} = \frac{2 ( 1 )}{4} (4) = \frac{2}{4} (4) = 2 m/s$ Yeh step kyun? Negative $v_{1}$ matlab light ball peeche bounce karti hai — woh "back-scatter" jo parent note ne warn kiya tha.

Step 2 — Glancing version ke liye opening angle. Boxed general formula se, $v_{1} \cdot v_{2} = \frac{m _{1} - m _{2}}{2 m _{2}} v_{1}^{2} = \frac{1 - 3}{2 ( 3 )} v_{1}^{2} = - \frac{1}{3} v_{1}^{2} < 0$ , toh $cos (θ_{1} + θ_{2}) < 0$ ⇒ angle $> 9 0^{\circ}$ ; upar wala head-on hit $18 0^{\circ}$ extreme hai (seedha wapas). Yeh step kyun? Negative dot product matlab do departing velocities apart lean karti hain — ek 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 $u_{1} = 10 m/s$ identical stationary $m$ se hit karta hai aur woh ek saath chipak jaate hain. Common final velocity aur "opening angle" nikalo.

Forecast: Woh chipak gaye — toh unke beech ka angle kya hai?

Step 1 — Sirf momentum (energy conserved NAHI hai). $m u_{1} = (2 m) V \Rightarrow V = \frac{u _{1}}{2} = 5 m/s$ Yeh step kyun? Conservation of Momentum sabhi collisions mein hold karta hai; lekin 90° proof ne energy conservation use ki thi, jo ek perfectly inelastic collision violate karta hai.

Step 2 — Opening angle = 0°. Woh ek lump ki tarah move karte hain: "do velocities ke beech ka angle" $0^{\circ}$ hai, $9 0^{\circ}$ se kaafi kam. Yeh step kyun? Confirm karta hai: energy conservation hataao aur perpendicularity result bilkul collapse ho jaata hai.

Verify: Momentum: $m (10) = 2 m (5) = 10 m$ ✓. Energy lost: pehle $\frac{1}{2} m (100) = 50 m$ ; baad mein $\frac{1}{2} (2 m) (25) = 25 m$ — aadha KE chala gaya, confirm karta hai ki yeh inelastic hai.

Cell H — Real-world word problem (billiards)

Worked example Pool table pe break shot

Ek frictionless idealized table pe, cue ball ( $u_{1} = 5 m/s$ ) ek identical object ball se hit karta hai. Aap measure karte ho ki object ball cue ki original line se $θ_{2} = 3 7^{\circ}$ neeche roll off karta hai. Ek clean elastic hit assume karo. Cue ball ka angle aur dono speeds nikalo.

Forecast: Dono angles ka sum?

Step 1 — Masses identify karo. Pool balls identical hote hain ⇒ equal mass ⇒ equal-mass toolkit apply hota hai. Yeh step kyun? 90° rule aur cos/sin formulas sirf equal masses ke liye fire hote hain; billiard balls yeh satisfy karte hain.

Step 2 — 90° rule se cue angle. Measured quantity object ball ka angle $θ_{2} = 3 7^{\circ}$ hai. Kyunki $θ_{1} + θ_{2} = 9 0^{\circ}$ : $θ_{1} = 9 0^{\circ} - θ_{2} = 9 0^{\circ} - 3 7^{\circ} = 5 3^{\circ}$ Yeh step kyun? $θ_{1}$ cue ka angle hai, $θ_{2}$ object ka; woh complementary hain.

Step 3 — Triangle se speeds, carefully labelled. Right triangle mein $θ_{1}$ , $u_{1}$ aur cue ki leg $v_{1}$ ke beech origin pe angle hai. Toh cue ( $v_{1}$ ) angle $θ_{1}$ ki adjacent leg hai, aur object ( $v_{2}$ ) angle $θ_{1}$ ki opposite leg hai: $v_{1} = u_{1} cos θ_{1} = 5 cos 5 3^{\circ} = 5 (0.6018) = 3.009 m/s$ $v_{2} = u_{1} sin θ_{1} = 5 sin 5 3^{\circ} = 5 (0.7986) = 3.993 m/s$ Yeh step kyun? Hum triangle ko hamesha cue ke angle $θ_{1}$ se measure karte hain; object ball, opposite leg pe baitha, isliye $sin θ_{1}$ carry karta hai — jo $cos θ_{2}$ ke equal hai, uske apne $3 7^{\circ}$ se consistent.

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 ( $3.99$ ) cue ( $3.01$ ) se fast hai kyunki use zyada directly hit kiya gaya (chhota angle $θ_{2}$ ).

Cell I — Exam twist: dono speeds se angle recover karo

Worked example Dono final speeds diye hain, opening angle nikalo

Ek elastic collision mein incoming speed $u_{1} = 13 m/s$ hai, aur aap final speeds $v_{1} = 5 m/s$ aur $v_{2} = 12 m/s$ measure karte ho. Equal masses assume kiye bina, kya yeh collision equal masses se consistent hai, aur outgoing velocities ke beech ka angle kya hai?

Forecast: Kya $5$ , $12$ , $13$ koi bell ring karte hain?

Step 1 — Equal masses ke liye energy test karo. Agar masses equal hain, energy demand karti hai $u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$ : $5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2} ✓$ Yeh step kyun? $u^{2} = v_{1}^{2} + v_{2}^{2}$ equal masses + elastic ka Pythagorean signature hai. Yeh match karta hai, toh collision equal masses se consistent hai.

Step 2 — Angle bina measure kiye deduce karo. Momentum vector $u_{1} = v_{1} + v_{2}$ ko square karo (equal masses, toh masses cancel hote hain): $u_{1}^{2} = (v_{1} + v_{2}) \cdot (v_{1} + v_{2}) = v_{1}^{2} + 2 v_{1} \cdot v_{2} + v_{2}^{2}$ Kyunki Step 1 ne dikhaya $u_{1}^{2} = v_{1}^{2} + v_{2}^{2}$ , bachi hui term vanish honi chahiye: $2 v_{1} \cdot v_{2} = u_{1}^{2} - v_{1}^{2} - v_{2}^{2} = 0 \Rightarrow v_{1} \cdot v_{2} = 0$ Yeh step kyun? Dot product zero hona matlab outgoing velocities perpendicular hain.

Step 3 — Dot product ko angle mein convert karo. Dot Product ki definition se, $v_{1} \cdot v_{2} = v_{1} v_{2} cos ϕ$ , jahan $ϕ$ opening angle hai. (Nonzero) speeds se divide karo: $cos ϕ = \frac{v _{1} \cdot v _{2}}{v _{1} v _{2}} = \frac{0}{5 \cdot 12} = 0 \Rightarrow ϕ = 9 0^{\circ}$ Yeh step kyun? $cos ϕ = 0$ ka physical range $0^{\circ}$ – $18 0^{\circ}$ mein unique solution $ϕ = 9 0^{\circ}$ hai.

Verify: $5^{2} + 1 2^{2} = 169 = 1 3^{2}$ ✓ (classic 5-12-13 triple), aur $cos ϕ = 0 \Rightarrow ϕ = 9 0^{\circ}$ . Energy ka Pythagoras se consistency ne $9 0^{\circ}$ force kiya — tumhe kabhi directions measure nahi karne the, sirf speeds.

Active Recall

Recall Kaun se cells < 90°, = 90°, > 90° dete hain?

$= 9 0^{\circ}$ : equal mass, elastic, dono moving (Cells A, B, D, H, I).
$< 9 0^{\circ}$ : heavy-hits-light (E), inelastic (G).
$> 9 0^{\circ}$ : light-hits-heavy, $18 0^{\circ}$ tak back-scatter ho sakta hai (F).
Undefined: head-on equal mass, cue ball ruk jaati hai (C).

Recall Exam-twist shortcut

Agar tumhe bataya jaaye $u^{2} = v_{1}^{2} + v_{2}^{2}$ , tum turant jaante ho: equal masses, elastic, aur outgoing velocities perpendicular hain — koi direction measurement nahi chahiye.

Recall Factor

$(m_{1} - m_{2}) / (2 m_{2})$ kahan se aata hai? Momentum mein ball 2 ko isolate karo, square karo, aur energy equation subtract karo — cross term $2 v_{1} \cdot v_{2}$ exactly usi mass factor se multiply hokar bachta hai. Uska sign decide karta hai acute/right/obtuse.

Yahaan drill kiye gaye prerequisites: Conservation of Momentum, Kinetic Energy, Dot Product, Elastic collisions — 1D, Inelastic collisions. Frame view bhi dekho Center of Mass Frame mein.