Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂): Visual walkthrough

1.4.8 · D2 · Physics › Momentum & Collisions › Coefficient of restitution e = (v₂ − v₁) - (u₁ − u₂)

Shuru karne se pehle hume sirf ek idea chahiye: velocity ek number line par ek arrow hai. Lamba arrow = tez; arrow ki direction = kis taraf move kar raha hai. Bus itna hi. Baaki sab hum banate hain.

Step 1 — Do blocks, ek line, char arrows

KYA. Do blocks imagine karo, block 1 (left) aur block 2 (right), ek smooth horizontal line par slide kar rahe hain. Hum ek direction ko positive choose karte hain — maan lo rightward positive hai. Ab har velocity sirf ek signed number hai: positive matlab "right ja raha hai", negative matlab "left ja raha hai".

KYUN. Is note mein collisions one-dimensional hain: sab kuch ek hi line ke along hota hai. Ek positive direction choose karne se har velocity ek single number ban jaati hai jise hum add aur subtract kar sakte hain. Bina fixed sign convention ke, "faster" aur "which way" ulajh jaate hain — isliye hum pehle ise fix karte hain.

PICTURE. Figure dekho. Crash se pehle blocks ki velocities $u_{1}$ aur $u_{2}$ hain (letter $u$ = "before", socho "u come first"). Collision hone ke liye, block 1 ko block 2 ko pakadna zaruri hai, isliye uska arrow lamba hai: $u_{1} > u_{2}$ .

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

Step 2 — "Approach speed": gap kitni tezi se band hota hai

KYA. Do blocks ek dusre ki taraf aa rahe hain. Kitni tezi se? Sirf $u_{1}$ nahi, sirf $u_{2}$ nahi — jo matter karta hai woh hai ki unke beech ka gap kitni tezi se shrink hota hai. Woh difference hai $u_{1} - u_{2}$ .

KYUN difference, aur kyun is order mein. Imagine karo tum block 2 par baithe ho. Tumhari seat se, block 2 khada hua hai, aur block 1 tumhari taraf $u_{1} - u_{2}$ speed se aa raha hai. $u_{2}$ subtract karna exactly "block 2 ka viewpoint switch karna" hai. Hum $u_{1} - u_{2}$ likhte hain (na ki $u_{2} - u_{1}$ ) kyunki block 1 peeche se tez wala hai — isliye yeh difference positive aata hai, jo ek "closing speed" ko hona chahiye.

PICTURE. Figure mein amber bracket closing gap hai. Amber arrow jis par $u_{1} - u_{2}$ likha hai, woh hai jo block 2 par sawaar observer unki taraf aata hua dekhta hai.

Step 3 — Crash, aur "separation speed"

KYA. Woh collide karte hain, deform hote hain, aur alag ho jaate hain. Baad mein block 2 aage hona chahiye — right ki taraf tez — warna block 1 usse guzar jaata. Toh ab $v_{2} > v_{1}$ hai, aur jis speed se gap dobara khulta hai woh $v_{2} - v_{1}$ hai.

KYUN order flip hota hai ( $v_{2} - v_{1}$ , na ki $v_{1} - v_{2}$ ). Crash se pehle, block 1 tez wala tha, isliye "faster minus slower" $u_{1} - u_{2}$ tha. Crash ke baad, block 2 tez wala hai, isliye "faster minus slower" ab $v_{2} - v_{1}$ hai. Indices swap hote hain kyunki "faster block" ki role swap hoti hai. Yahi ek swap parent formula mein upar aur neeche alag index orders hone ki puri wajah hai — aur yeh dono quantities ko positive rakhta hai.

PICTURE. Wahi viewpoint trick: crash ke baad block 1 par baitho aur block 2 ko $v_{2} - v_{1}$ speed (cyan arrow) se bhagte dekho.

Step 4 — Definition seedha nikalta hai: separation ÷ approach

KYA. Ab hamare paas do positive speeds hain. Newton ki experimental discovery yeh thi ki unka ratio kisi bhi material pair ke liye ek fixed number hota hai. Use $e$ kaho: $e = \frac{separation}{approach} = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} .$

KYUN ratio aur difference nahi. Ratio poochta hai "closing speed ka kitna fraction bacha?" — yeh ek pure number hai jiska koi unit nahi (m/s ÷ m/s cancel ho jaata hai), isliye yeh sirf blocks kis cheez se bane hain par depend karta hai, na ki tumne unhe kitni tezi se phenka. Exactly yahi clean, reusable quantity hai jo physics ko pasand hai.

PICTURE. Figure Steps 2 aur 3 ke do arrows stack karta hai aur shrink dikhata hai: separation arrow (usually) approach arrow se chhota hota hai, aur $e$ literally kitna chhota hai.

Is same number ki deeper $e = J_{r} / J_{d}$ reading ke liye Impulse and Momentum dekho.

Step 5 — Bouncy end: $e = 1$

KYA. Agar blocks utni hi tezi se alag hote hain jitni tezi se aaye the, toh $v_{2} - v_{1} = u_{1} - u_{2}$ , isliye $e = 1$ . Kuch bhi lose nahi hua.

KYUN yeh ceiling hai. Separation kabhi approach se zyada nahi ho sakti, kyunki iska matlab hoga ki collision ne kinetic energy kahi se create ki. Collision se jo zyada ho sakta hai woh hai sab kuch wapas dena. Isliye $e = 1$ maximum hai — perfectly elastic case, jahan kinetic energy bhi conserve hoti hai.

PICTURE. Separation arrow approach arrow jitna hi lamba hai — ek mirror-image bounce.

Step 6 — Splat end: $e = 0$

KYA. Agar blocks chipak jaate hain aur saath chale jaate hain, toh unki ek common velocity $v_{c}$ hoti hai: toh $v_{1} = v_{2} = v_{c}$ . Phir $v_{2} - v_{1} = 0$ , jisse $e = 0$ milta hai.

KYUN yeh floor hai. Zero separation matlab gap kabhi nahi khulta — perfectly inelastic crash, ek clay-lump splat. Kyunki separation speed negative nahi ho sakti (block 1 block 2 mein se tunnel nahi kar sakta), $e = 0$ sabse chhoti possible value hai. Isliye har real collision $0 \leq e \leq 1$ mein hota hai.

PICTURE. Do "after" arrows same length ke hain aur same taraf point karte hain — blocks lock ho jaate hain. Separation arrow ki length zero hai.

Recall Do extremes check karo

$e = 1$ ka matlab hai ::: separation = approach, perfect bounce, kinetic energy conserved. $e = 0$ ka matlab hai ::: blocks chipak jaate hain, ek velocity $v_{c}$ share karte hain, separation zero hai, maximum KE lose hoti hai.

Step 7 — Degenerate case: ek block wall se bounce karna

KYA. Maan lo block 2 ek fixed floor ya wall hai — infinitely heavy, isliye kabhi nahi hilta: $u_{2} = 0$ aur $v_{2} = 0$ . Phir formula collapse ho jaata hai: $e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}} = \frac{0 - v _{1}}{u _{1} - 0} = \frac{- v _{1}}{u _{1}} .$

KYUN minus sign yahan acha hai. Impact se pehle ball wall ki taraf move karti hai — maan lo neeche, isliye $u_{1}$ hamare "up is positive" convention mein negative hai... lekin seedha padho: ball $∣ u_{1} ∣$ speed se hit karti hai aur $∣ v_{1} ∣$ speed se opposite direction mein rebound karti hai, isliye $v_{1}$ ka sign $u_{1}$ se opposite hai. Woh opposite sign $- v_{1} / u_{1}$ ko positive banata hai — exactly rebound speed aur impact speed ka ratio: $e = \frac{rebound speed}{impact speed} .$

PICTURE. Ball neeche aati hai (lamba arrow), wall se upar bounce karti hai (chhota arrow). $e$ hai kitna arrow wall ke baad bacha.

Step 8 — Worked example, picture ke against verify kiya

KYA. Do equal masses, $m_{1} = m_{2} = 2$ kg. Block 1 $u_{1} = 4$ m/s par ek still block 2 ( $u_{2} = 0$ ) se $e = 0.5$ ke saath takrata hai. After-velocities nikalo.

KYUN do equations. Momentum ek equation deta hai; $e$ doosri deta hai. Do equations, do unknowns — ab solve ho sakta hai.

$Momentum: 2 (4) + 0 = 2 v_{1} + 2 v_{2} \Rightarrow v_{1} + v_{2} = 4$ $Restitution: v_{2} - v_{1} = e (u_{1} - u_{2}) = 0.5 (4) = 2$

Do lines add karo: $2 v_{2} = 6 \Rightarrow v_{2} = 3$ m/s, phir $v_{1} = 1$ m/s.

PICTURE / check. Separation $= v_{2} - v_{1} = 3 - 1 = 2$ m/s, approach $= 4$ m/s, ratio $= 2/4 = 0.5$ ✓. Dono positive, $v_{2} > v_{1}$ — block 2 baad mein aage hai, exactly jaisa Step 3 mein demand kiya tha.

Ek-picture summary

Sab kuch ek frame mein: approach arrow (amber, pehle), crash, separation arrow (cyan, baad mein, indices swapped), aur $e$ unki lengths ka ratio — splat ( $e = 0$ ) se perfect bounce ( $e = 1$ ) tak sliding.

Recall Pure walkthrough ki Feynman retelling

Do blocks ek line par slide karte hain. Hum rightward positive kehte hain taki har speed sirf ek number ho. Crash se pehle, peeche wala block tez hai aur pakad leta hai — jis speed se gap band hota hai woh hai "faster minus slower", $u_{1} - u_{2}$ . Woh collide karte hain, squash hote hain, aur alag ho jaate hain; ab front block tez hai, isliye jis speed se gap dobara khulta hai woh hai "faster minus slower phir se", lekin tez block swap ho gaya hai, isliye yeh $v_{2} - v_{1}$ hai. Newton ne discover kiya ki reopening speed ko closing speed se divide karne par kisi bhi material pair ke liye hamesha same number milta hai — woh number $e$ hai. Yeh $1$ se zyada nahi ho sakta (tum closing speed se tez alag nahi ho sakte, woh energy kahi se banayega) aur $0$ se kam nahi ho sakta (tum through tunnel nahi kar sakte). Ek perfect rubber bounce saari speed rakhta hai, $e = 1$ ; ek clay splat kuch nahi rakhta, $e = 0$ ; ek wall wahi kahani hai jisme ek block nailed down hai, aur kyunki bounce height speed squared par depend karti hai, $e = h_{2} / h_{1}$ .

Connections

Coefficient of Restitution — parent topic jise yeh page visualise karta hai.
Conservation of Linear Momentum — pehli equation; $e$ doosri supply karta hai.
Elastic Collisions — Step 5, $e = 1$ ceiling.
Perfectly Inelastic Collisions — Step 6, $e = 0$ floor.
Impulse and Momentum — deeper $e = J_{r} / J_{d}$ meaning.
Kinetic Energy Loss in Collisions — shrunk arrow ne kitni energy lose ki.
Projectile Motion — Step 7 ke peeche $v = 2 g h$ link.