Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Worked examples

1.4.8 · D3 · Physics › Momentum & Collisions › Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

Yeh page parent note ke liye workout gym hai. Hum formula pehle se jaante hain: $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{speed of separation}{speed of approach}$ Yahaan hum har tarah ke collision problems dhundhte hain jo yeh topic de sakta hai aur har ek ko puri tarah solve karte hain. Usse pehle, hum ek map banate hain taaki koi bhi aisi situation kabhi na mile jise humne practice na ki ho.

Sign convention (ek baar fix karo, poore page ke liye)

Neeche har velocity signed hai, aur hum SIRF EK rule use karte hain har jagah.

Dono master equations (aur closed form)

Har 1-D restitution problem exactly do facts use karti hai:

Momentum conservation (ek equation): $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ .
Restitution law (doosri equation): $v_{2} - v_{1} = e (u_{1} - u_{2})$ .

Do linear equations, do unknowns $v_{1}, v_{2}$ → ek canonical closed-form answer milta hai. Chalte hain yeh pair ek baar, generally, solve karte hain, taaki neeche ke har worked example mein sirf yahi formula numbers ke saath use ho.

Har worked example ya toh is closed form mein plug in karta hai ya ise invert karta hai ( $e$ solve karne ke liye).

Scenario matrix — aur yeh exhaustive kyun hai

Koi bhi collision chaar ingredients se fix hoti hai: do initial velocities $u_{1}, u_{2}$ (unke signs aur order), $e$ ki value ( $0$ se $1$ tak), aur yeh ki koi ek mass finite hai ya infinite (ek wall). Har possibility neeche diye branches mein se kisi ek mein aati hai — yeh ek genuine case split hai, wishlist nahi.

Intuition Yeh das cells sab kuch kyun cover karti hain

Decision tree ek baar walk karo:

Kya collision ho rahi hai? Collision ke liye ek positive approach speed $u_{1} - u_{2} > 0$ chahiye. Agar $u_{1} = u_{2}$ (ya $u_{1} < u_{2}$ , body 1 peeche pad rahi hai) toh koi impact nahi → yeh degenerate branch hai (Cell H).
Agar haan, toh $u_{1}, u_{2}$ ke signs kya hain? Ya toh dono same sign ke hain — ek chase (Cell A) — ya opposite signs — ek head-on (Cell B). (Poori picture ke sign-flips sirf left/right relabel karte hain, toh yeh dono geometry exhaust kar dete hain.)
$e$ kya hai? Endpoints $e = 1$ (Cell C) aur $e = 0$ (Cell D) do physical limits hain; har $0 < e < 1$ ek same closed form se handle hota hai (Cells A, B isko demonstrate karte hain).
Kya koi mass infinite hai? Tab woh ek fixed wall hai, $u_{2} = v_{2} = 0$ (Cell F).
Tumse kya poochha ja raha hai? Forward (find $v$ 's: A–D, F), ya inverse (find $e$ from velocities (Cell E), heights se (Cell G), ya energy loss se (Cell J)).
Kya yeh 1-D hai ya 2-D? Ek surface se 2-D bounce perpendicular axis par 1-D reduce ho jaata hai (Cell I). Is tree ka har leaf ek cell hai — toh kuch bhi chhuta nahi.

Cell	Case class	Kya special hai	Example
A	Same-direction chase ( $u_{1}, u_{2}$ dono $+$ , $u_{1} > u_{2}$ )	dono velocities positive	Ex 1
B	Head-on, opposite signs ( $u_{1} > 0$ , $u_{2} < 0$ )	ek velocity negative → approach speed add hoti hai	Ex 2
C	$e = 1$ elastic limit	KE bhi conserved hai; equal mass ke liye velocities swap hoti hain	Ex 3
D	$e = 0$ perfectly inelastic	bodies chipak jaate hain, ek common velocity	Ex 4
E	Data se $e$ find karo (inverse problem)	$e$ solve karo, $v$ 's ke liye nahi	Ex 5
F	Fixed wall / infinite mass	$u_{2} = v_{2} = 0$ , ball bas reverse hoti hai	Ex 6
G	Bouncing ball, height aur multiple bounces	$h \propto v^{2}$ , chahiye; $e^{2 n}$	Ex 7
H	Degenerate: equal initial speeds ( $u_{1} = u_{2}$ )	approach $= 0$ , koi collision nahi — formula toot jaata hai	Ex 8
I	Real-world word problem (2-D bounce off ground)	vertical component $e$ maanta hai, horizontal unchanged	Ex 9
J	Exam twist: $e$ back-solve from KE loss	$KE lost \propto (1 - e^{2})$ use karta hai	Ex 10

Prerequisites jo tumhare paas open ho sakte hain: Conservation of Linear Momentum, Elastic Collisions, Perfectly Inelastic Collisions, Kinetic Energy Loss in Collisions, Projectile Motion.

Cell A — same-direction chase

Worked example Ex 1 — ek fast ball ek slow ball ko pakad leti hai

$m_{1} = 3$ kg at $u_{1} = + 8$ m/s, $m_{2} = 1$ kg ko $u_{2} = + 2$ m/s par pakad leta hai. Restitution $e = 0.5$ . $v_{1}, v_{2}$ find karo.

Forecast: bhaari fast ball halki slow ball se takraati hai; halki ball aage zyada tez flung ho jaegi, bhaari ball thodi slow hogi lekin chalti rahegi. Guess karo $v_{2} > v_{1}$ , dono abhi bhi positive.

Approach speed $= u_{1} - u_{2} = 8 - 2 = 6$ m/s. Yeh step kyun? Dono right jaate hain, toh woh kitni tezi se close in karte hain woh unki speeds ka difference hai, sum nahi.
Restitution equation: $v_{2} - v_{1} = e (u_{1} - u_{2}) = 0.5 \times 6 = 3$ . Kyun? Newton's law: separation speed, approach speed ka $e$ times hoti hai.
Momentum: $3 (8) + 1 (2) = 3 v_{1} + 1 v_{2} \Rightarrow 26 = 3 v_{1} + v_{2}$ . Kyun? Koi external push nahi, toh total momentum unchanged hai — yeh hamari doosri equation hai.
Pair solve karo. Step 2 se, $v_{2} = v_{1} + 3$ . Substitute karo: $26 = 3 v_{1} + (v_{1} + 3) = 4 v_{1} + 3 \Rightarrow v_{1} = \frac{23}{4} = 5.75$ , phir $v_{2} = 8.75$ . Kyun? Do equations, do unknowns → unique answer. (Upar ke closed form mein plug karne ka bhi same result aata hai.)

Verify: $v_{2} - v_{1} = 8.75 - 5.75 = 3 = 0.5 \times 6$ ✓. Momentum after $= 3 (5.75) + 1 (8.75) = 17.25 + 8.75 = 26$ ✓. Dono positive, $v_{2} > v_{1}$ ✓ — forecast ke anusaar aage flung hua.

Cell B — head-on collision (opposite signs)

Worked example Ex 2 — do balls ek doosre ki taraf daudti hain

$m_{1} = 2$ kg at $u_{1} = + 5$ m/s right ja rahi hai; $m_{2} = 2$ kg at $u_{2} = - 3$ m/s left ja rahi hai. $e = 0.6$ . Final velocities find karo.

Forecast: equal mass ke saath head-on smash; expect karo ki dono reverse hon ya strongly slow hon. Guess karo $v_{1}$ negative-ish ho jaega (bounce back), $v_{2}$ positive.

Approach speed $= u_{1} - u_{2} = 5 - (- 3) = 8$ m/s. Yeh step kyun? Yahaan $u_{2}$ negative hai kyunki woh left move karta hai. Negative ko subtract karna add karta hai — yeh bilkul sahi hai, kyunki do cheezein ek doosre ki taraf daudti hain toh unki speeds ka sum hota hai closing rate.
Restitution: $v_{2} - v_{1} = 0.6 \times 8 = 4.8$ . Kyun? Same law; signed subtraction ne geometry pehle se bake kar li hai.
Momentum: $2 (5) + 2 (- 3) = 2 v_{1} + 2 v_{2} \Rightarrow 10 - 6 = 2 v_{1} + 2 v_{2} \Rightarrow v_{1} + v_{2} = 2$ . Kyun? Signs rakho — leftward momentum negative hai.
Solve karo. $v_{2} = v_{1} + 4.8$ into $v_{1} + v_{2} = 2$ : $2 v_{1} + 4.8 = 2 \Rightarrow v_{1} = - 1.4$ , $v_{2} = 3.4$ . Kyun? Same elimination pehle jaisi.

Verify: $v_{2} - v_{1} = 3.4 - (- 1.4) = 4.8$ ✓. Momentum $= 2 (- 1.4) + 2 (3.4) = - 2.8 + 6.8 = 4$ , aur pehle $= 10 - 6 = 4$ ✓. $v_{1} < 0$ (ball 1 bounce back left) forecast ke anusaar ✓.

Cell C — elastic limit $e = 1$

Worked example Ex 3 — perfect bounce, equal masses

$m_{1} = m_{2} = 1$ kg, $u_{1} = + 6$ m/s, $u_{2} = 0$ , $e = 1$ . Final velocities find karo aur confirm karo ki KE conserved hai.

Forecast: classic Newton's-cradle case — equal masses, perfect bounce. Guess karo woh swap karte hain velocities: $v_{1} = 0$ , $v_{2} = 6$ .

Restitution ( $e = 1$ ): $v_{2} - v_{1} = 1 \times (6 - 0) = 6$ . Kyun? $e = 1$ par separation speed, approach speed ke barabar hoti hai — koi speed lost nahi.
Momentum: $6 + 0 = v_{1} + v_{2} \Rightarrow v_{1} + v_{2} = 6$ . Kyun? Equal masses equation mein nicely cancel ho jaate hain.
Solve karo: add-and-subtract → $v_{2} = 6$ , $v_{1} = 0$ . Kyun? $v_{1} + v_{2} = 6$ aur $v_{2} - v_{1} = 6$ se $v_{2} = 6, v_{1} = 0$ turant milta hai. (Closed form mein $e = 1, m_{1} = m_{2}$ set karne par exactly yahi velocity swap milta hai.)
KE check: before $= \frac{1}{2} (1) (6^{2}) = 18$ J; after $= \frac{1}{2} (1) (0^{2}) + \frac{1}{2} (1) (6^{2}) = 18$ J. Yeh step kyun? $e = 1$ ko kinetic energy bhi conserve karni chahiye — yeh elastic collision ki pehchaan hai.

Verify: velocities exactly swap hue; KE identical (18 J → 18 J) ✓.

Cell D — perfectly inelastic $e = 0$

Worked example Ex 4 — woh chipak jaate hain (KE-loss formula bhi derive hoti hai)

$m_{1} = 4$ kg at $u_{1} = + 5$ m/s, $m_{2} = 6$ kg rest par ( $u_{2} = 0$ ) se takraata hai, $e = 0$ . Final velocity aur KE lost find karo.

Forecast: $e = 0$ matlab splat — ek common velocity. Total mass 10 kg; guess final speed around $2$ m/s.

Restitution ( $e = 0$ ): $v_{2} - v_{1} = 0 \Rightarrow v_{1} = v_{2} = v_{c}$ . Kyun? Zero separation speed matlab woh ek common velocity par saath nikalte hain — perfectly inelastic ki definition.
Momentum: $4 (5) + 0 = (4 + 6) v_{c} \Rightarrow 20 = 10 v_{c} \Rightarrow v_{c} = 2$ m/s. Kyun? Ek unknown bacha, toh momentum akela kaam finish kar deta hai.
KE lost: before $= \frac{1}{2} (4) (5^{2}) = 50$ J; after $= \frac{1}{2} (10) (2^{2}) = 20$ J; lost $= 30$ J. Yeh step kyun? $e = 0$ maximum possible KE loss deta hai (saari "bounce-back" energy heat/deformation mein gayi).
General fraction kahan se aata hai. $u_{2} = 0$ aur $v_{c} = \frac{m _{1} u _{1}}{m _{1} + m _{2}}$ ke saath: before $= \frac{1}{2} m_{1} u_{1}^{2}$ ; after $= \frac{1}{2} (m_{1} + m_{2}) v_{c}^{2} = \frac{1}{2} \frac{m _{1}^{2} u _{1}^{2}}{m _{1} + m _{2}}$ . After ko before se divide karo: $\frac{after}{before} = \frac{m _{1}}{m _{1} + m _{2}}$ , toh lost fraction $= 1 - \frac{m _{1}}{m _{1} + m _{2}} = \frac{m _{2}}{m _{1} + m _{2}}$ . Yeh step kyun? Yeh dikhata hai ki " $m_{2} / (m_{1} + m_{2})$ " rule magic nahi hai — yeh exactly steps 2–3 hain numbers ki jagah symbols ke saath, jo Kinetic Energy Loss in Collisions se link karta hai.

Verify: momentum after $= 10 \times 2 = 20 =$ before ✓. Numeric fraction $= 30/50 = 0.6$ ; symbolic rule $\frac{m _{2}}{m _{1} + m _{2}} = \frac{6}{10} = 0.6$ ✓ — dono agree karte hain.

Cell E — inverse problem, $e$ find karo

Worked example Ex 5 — before/after velocities se

$e$ measure karo Cart A ( $u_{1} = + 7$ m/s) cart B ( $u_{2} = + 1$ m/s) se same direction mein takraata hai. After: $v_{1} = + 3$ m/s, $v_{2} = + 6$ m/s. $e$ find karo.

Forecast: clearly bounce apart hue (B, A se aage nikal gaya). Splat nahi, perfect nahi — guess karo $e$ around $0.5$ .

Approach $= u_{1} - u_{2} = 7 - 1 = 6$ m/s. Kyun? Same-direction chase → difference.
Separation $= v_{2} - v_{1} = 6 - 3 = 3$ m/s. Kyun? Note karo swapped indices: collision ke baad B faster/leading hai, toh $v_{2} - v_{1}$ positive separation hai.
Ratio: $e = 3/6 = 0.5$ . Yeh step kyun? $e$ define hota hai separation ÷ approach — inverse problem ke liye momentum ki zaroorat nahi.

Verify: $0 \leq 0.5 \leq 1$ ✓ (physically legal value). Approach aur separation dono positive nikle, index order confirm karta hai.

Cell F — ball off a fixed wall

Worked example Ex 6 — immovable wall se bounce

Ek ball ek fixed vertical wall se $u_{1} = + 9$ m/s par takraati hai (wall ki taraf). Wall "body 2" hai jisme $u_{2} = v_{2} = 0$ . Agar $e = 0.7$ , rebound speed find karo.

Forecast: wall nahi hilti; ball sirf slower wapas aati hai. Rebound $\approx 0.7 \times 9 \approx 6.3$ m/s, lekin doosri taraf travel karegi (negative).

Restitution: $v_{2} - v_{1} = e (u_{1} - u_{2}) \Rightarrow 0 - v_{1} = 0.7 (9 - 0)$ . Kyun? Ek "infinitely heavy" wall rest par rehti hai, toh $u_{2} = v_{2} = 0$ .
Solve karo: $- v_{1} = 6.3 \Rightarrow v_{1} = - 6.3$ m/s. Kyun? Minus sign physics hai jo hume bata raha hai ki ball ab opposite direction mein ja rahi hai — woh bounce back hui.
Rebound speed (magnitude) $= 6.3$ m/s. Yeh step kyun? Speed, velocity ka size hai; sign ne direction bataya, magnitude ne kitna fast bataya.

Verify: $∣ v_{1} ∣/ u_{1} = 6.3/9 = 0.7 = e$ ✓. Fixed wall ke liye $e$ simply rebound speed aur impact speed ka ratio hai.

Cell G — bouncing ball, heights aur repeated bounces

Yahaan geometry matter karti hai, toh hum ise draw karte hain. Neeche figure mein, horizontal axis ball ki position hai jaise woh rightward hop karta hai aur vertical axis uski height metres mein hai. Ball seedha neeche left par drop ki gayi hai (dotted white line height $h_{0}$ se), aur har successive bounce arc color-coded peak tak uthta hai — blue peak 1 ke liye, yellow peak 2 ke liye, pink peak 3 ke liye. Left side ke dashed guide lines un peak heights ko mark karti hain. Figure ka poora point: har peak bilkul $e^{2}$ times us peak ki height par baith ta hai jo uske pehle thi, toh peaks ek fixed multiplying factor se neeche march karti hain.

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

Worked example Ex 7 — drop, rebound, aur

$n$ -th bounce Ek ball $h_{0} = 2.0$ m se drop ki gayi aur $h_{1} = 1.25$ m tak rebound karti hai. (a) $e$ find karo. (b) 3rd bounce ke baad kahan pahunchi?

Forecast: har bounce mein kuch height khoti hai; $e$ noticeably below 1. 3 bounces ke baad kaafi low hona chahiye, shayad aadha metre.

Impact & rebound speeds. Impact speed $= 2 g h_{0}$ , rebound speed $= 2 g h_{1}$ (definition box se). Yeh step kyun? Measured heights ko un speeds mein convert karo jo $e$ actually compare karta hai.
Height ratio → $e$ . $e = \frac{2 g h _{1}}{2 g h _{0}} = \frac{h _{1}}{h _{0}} = \frac{1.25}{2.0} = 0.625 = 0.7906 \dots$ . kyun? Kyunki $h \propto v^{2}$ , toh heights ka ratio $e^{2}$ hai; square ko undo karne ke liye square root chahiye. Figure mein, "peak 1" (blue) $e^{2} h_{0}$ par baith ta hai — blue marker ko left-hand dashed guide ke against padho.
$n$ bounces ke baad: $h_{n} = e^{2 n} h_{0}$ . Kyun? Har bounce height ko fixed factor $e^{2}$ se multiply karta hai (figure mein yellow aur pink markers same ratio repeat karte hain); teen bounces $(e^{2})^{3} = e^{6}$ se multiply karte hain.
$n = 3$ plug karo: $h_{3} = e^{6} h_{0}$ . Kyunki $e^{2} = h_{1} / h_{0} = 0.625$ , toh $e^{6} = (0.625)^{3} = 0.2441 \dots$ , isliye $h_{3} = 0.2441 \times 2.0 = 0.488$ m. Yeh step kyun? Question specifically $h_{3}$ pooch raha hai — $h_{n}$ ki definition se directly $h_{n} = e^{2 n} h_{0}$ mein $n = 3$ substitute karna sabse seedha tarika hai, aur $e^{6} = (e^{2})^{3}$ likhne se hum pehle se known $e^{2} = 0.625$ reuse kar sakte hain bina $e$ recompute kiye.

Verify: $e^{2} = 0.790 6^{2} = 0.625$ ✓; $h_{3} = (0.625)^{3} \times 2.0 = 0.4883$ m ✓ — $h_{1}$ se lower, forecast ke anusaar.

Cell H — degenerate case (koi collision nahi)

Worked example Ex 8 — agar dono same velocity par move karen?

$m_{1} = 2$ kg aur $m_{2} = 3$ kg dono $u_{1} = u_{2} = + 4$ m/s par travel karte hain. Restitution formula kya deta hai?

Forecast: woh kabhi touch nahi karte (dono step mein move kar rahe hain, koi closing nahi). Guess: formula blow up / undefined ho jaata hai.

Approach speed $= u_{1} - u_{2} = 4 - 4 = 0$ . Yeh step kyun? Unke beech ka gap shrink nahi kar raha — koi kisi ko catch up nahi kar raha — toh collide karne ke liye kuch nahi hai.
Formula check: $e = \frac{v _{2} - v _{1}}{0}$ — division by zero, undefined. Kyun? Restitution law sirf ek actual collision describe karta hai; zero approach speed ke saath koi impact nahi hai jo characterise ki ja sake.
Physical resolution: koi collision nahi → velocities unchanged hain, $v_{1} = 4$ , $v_{2} = 4$ . Kyun? Kisi bhi body par koi impulse act nahi karta.

Verify: approach $= 0$ makes $e$ undefined ✓ (yeh genuinely degenerate input hai — formula ka denominator zero ho jaata hai). Momentum trivially conserved kyunki kuch badla nahi: before $= 2 (4) + 3 (4) = 20$ , after $= 2 (4) + 3 (4) = 20$ ✓.

Cell I — real-world 2-D bounce

Hum ab velocity arrows draw karte hain bounce se pehle aur baad mein. Figure page ke top se 2-D sign convention use karta hai ( $+ x$ right, $+ y$ up). Blue arrow incoming velocity hai (down-right point karta hai, floor mein); pink arrow outgoing velocity hai (up-right); yellow horizontal arrow shared horizontal part hai jo change nahi hota; dashed white vertical arrow outgoing vertical part hai; pink arc rebound angle $θ$ mark karta hai. Dekho yellow horizontal arrow full-length rehta hai jabki vertical part shrink hota hai.

Worked example Ex 9 — ek ball ground se angle par bounce karti hai

Ek ball floor se horizontal component $u_{x} = + 6$ m/s aur vertical component $u_{y} = - 8$ m/s ke saath hit karti hai (negative kyunki woh neeche ja rahi hai, floor mein — hamara $+ y$ up hai). Floor–ball restitution $e = 0.75$ hai. Bounce ke turant baad velocity aur rebound angle above horizontal find karo.

Forecast: floor sirf upar push kar sakta hai, toh sirf vertical part affect hoga; horizontal speed same rahegi. Vertical $e$ factor se shrink hoti hai, toh ball aane se flatter nikalegi.

Components mein split karo. Collision horizontal floor ke saath hai, toh restitution sirf vertical ( $y$ ) component par apply hoti hai. Yeh step kyun? Floor ka push seedha $+ y$ ke along upar hai; woh koi horizontal force nahi laata, toh $u_{x}$ untouched rehta hai. Figure mein, yellow horizontal arrow apni length rakhta hai jabki vertical arrow shrink hota hai.
Horizontal after: $v_{x} = u_{x} = + 6$ m/s. Kyun? Koi horizontal impulse nahi → koi change nahi (yeh Cell F ke fixed-wall logic ko mirror karta hai, vertical axis ki jagah apply kiya).
Vertical after: floor ko $y$ -axis ke along fixed wall treat karo. Iska magnitude $e$ se scale hota hai: $∣ v_{y} ∣ = e ∣ u_{y} ∣ = 0.75 \times 8 = 6$ m/s. Kyunki ab woh hamari convention mein upar move kar raha hai, $v_{y} = + 6$ m/s. Kyun? Same Cell F reasoning: perpendicular velocity sign reverse karti hai ( $- y$ se $+ y$ ) aur $e$ se scale hoti hai.
Rebound angle above horizontal: $tan θ = \frac{∣ v _{y} ∣}{∣ v _{x} ∣} = \frac{6}{6} = 1 \Rightarrow θ = 4 5^{\circ}$ . $tan$ kyun? Figure mein horizontal ( $v_{x}$ ) aur vertical ( $v_{y}$ ) arrows se bane right triangle par, $tan θ = \frac{opposite (vertical)}{adjacent (horizontal)}$ — woh ratio jo launch steepness encode karta hai.

Verify: incoming angle mein $tan θ_{in} = 8/6 = 1.333$ ( $θ_{in} = 53.1 3^{\circ}$ ); outgoing $tan θ_{out} = 6/6 = 1$ ( $θ_{out} = 4 5^{\circ}$ ) — rebound flatter hai, exactly forecast ke anusaar ✓. Speed after $= 6^{2} + 6^{2} = 8.49$ m/s, aur $∣ v_{y} ∣/∣ u_{y} ∣ = 6/8 = 0.75 = e$ ✓.

Cell J — exam twist: KE loss se $e$ back-solve karo

Worked example Ex 10 — KE lost ke fraction se

$e$ find karo Ek moving ball $m_{1} = 1$ kg at $u_{1} = + 10$ m/s, identical ball $m_{2} = 1$ kg ko rest par ( $u_{2} = 0$ ) strike karti hai. Collision total kinetic energy ka 28% lose karti hai. $e$ find karo.

Forecast: modest energy loss → $e$ kaafi high, guess around $0.8$ .

Equal masses, target at rest ke liye KE-loss formula. $m_{1} = m_{2} = m$ , $u_{2} = 0$ ke liye: KE lost ka fraction $\frac{1 - e ^{2}}{2}$ hai. Yeh step kyun? KE lost $(1 - e^{2})$ ke proportion mein hota hai; extra $\frac{1}{2}$ equal-mass geometry hai. Yahaan hum energy use karte hain kyunki energy — velocities nahi — woh hai jo problem hume deta hai.
Data ke equal set karo: $\frac{1 - e ^{2}}{2} = 0.28 \Rightarrow 1 - e^{2} = 0.56 \Rightarrow e^{2} = 0.44$ . Kyun? Single unknown $e$ ke liye algebraically solve karo.
Root lo: $e = 0.44 = 0.663$ . kyun? Humhe $e$ chahiye, aur sirf $e^{2}$ appear hua.

Verify (full recompute). $e = 0.663$ ke saath: restitution $v_{2} - v_{1} = 0.663 \times 10 = 6.63$ ; momentum $v_{1} + v_{2} = 10$ ; toh $v_{2} = 8.317$ , $v_{1} = 1.683$ . KE before $= \frac{1}{2} (1 0^{2}) = 50$ J; after $= \frac{1}{2} (1.68 3^{2}) + \frac{1}{2} (8.31 7^{2}) = 1.416 + 34.585 = 36.0$ J. Loss $= 14$ J $= 28%$ of 50 J ✓ — given data se match karta hai.

Active recall

Recall Yeh kaunsi cell hai?

Ek ball wall par phenko aur woh seedha slower wapas aati hai — kaunsi cell, aur kya special hai? ::: Cell F: fixed wall, $u_{2} = v_{2} = 0$ , toh $e =$ rebound speed ÷ impact speed aur velocity sign flip hoti hai.

Recall Sign trap

Do balls head-on $+ 4$ aur $- 4$ m/s par approach karti hain. Approach speed kya hai? ::: $u_{1} - u_{2} = 4 - (- 4) = 8$ m/s — negative subtract karna speeds ko add karta hai.

Recall Heights ke liye square root kyun?

Ek ball $h_{0}$ se drop hoti hai aur $h_{1}$ tak rebound karti hai. $e = h_{1} / h_{0}$ kyun hai aur $h_{1} / h_{0}$ kyun nahi? ::: Kyunki $e$ speeds ka ratio hai aur $h \propto v^{2}$ , toh $h_{1} / h_{0} = e^{2}$ ; square undo karne ke liye root chahiye.

Recall Degenerate input

Restitution formula kab undefined ho jaata hai, aur iska matlab kya hai? ::: Jab approach speed $u_{1} - u_{2} = 0$ (equal velocities) — division by zero. Iska matlab hai koi collision nahi hoti.

Recall Closed form

$v_{1}$ ko $m_{1}, m_{2}, u_{1}, u_{2}, e$ ke terms mein likho. ::: $v_{1} = \frac{( m _{1} - e m _{2} ) u _{1} + ( 1 + e ) m _{2} u _{2}}{m _{1} + m _{2}}$ .

Connections

Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂) (index 1.4.8) — woh parent jise yeh page drill karta hai.
Conservation of Linear Momentum — har 2-unknown problem mein partner equation.
Elastic Collisions — Cell C, $e = 1$ .
Perfectly Inelastic Collisions — Cell D, $e = 0$ .
Kinetic Energy Loss in Collisions — Cells D aur J ko power deta hai.
Projectile Motion — Cells G aur I ke liye free-fall speeds.
Impulse and Momentum — $e$ ka deeper $J_{r} / J_{d}$ view.

General final velocity

v_{1}

v_{1} = \frac{( m _{1} - e m _{2} ) u _{1} + ( 1 + e ) m _{2} u _{2}}{m _{1} + m _{2}}

General final velocity

v_{2}

v_{2} = \frac{( 1 + e ) m _{1} u _{1} + ( m _{2} - e m _{1} ) u _{2}}{m _{1} + m _{2}}

Approach speed for head-on balls at

+ 4

and

- 4

m/s

u_{1} - u_{2} = 4 - (- 4) = 8

m/s

Fixed-wall restitution

e =

rebound speed ÷ impact speed, with velocity sign reversed

When is the

e

formula undefined?

When approach speed

u_{1} - u_{2} = 0

(equal velocities, no collision)

Height after

n

bounces

h_{n} = e^{2 n} h_{0}

2-D floor bounce: which component obeys

e

Only the vertical (perpendicular) component; horizontal is unchanged

Equal-mass, target-at-rest KE loss fraction

(1 - e^{2}) /2

Perfectly inelastic (

u_{2} = 0

) KE loss fraction

m_{2} / (m_{1} + m_{2})