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

1.4.8 · Physics › Momentum & Collisions

YEH HAI KYA?

Numerator $v_{2} - v_{1}$ kyun hai aur denominator $u_{1} - u_{2}$ kyun (indices ki order swap notice karo)? Kyunki actually collide karne ke liye body 1 ko body 2 ko pakad-te rehna chahiye, isliye $u_{1} > u_{2}$ → approach $= u_{1} - u_{2} > 0$ . Collision ke baad dono alag ho jaate hain, toh body 2 aage/tez ho jaati hai, $v_{2} > v_{1}$ → separation $= v_{2} - v_{1} > 0$ . Yeh swap dono quantities ko positive rakhta hai, isliye $e \geq 0$ automatically aa jaata hai.

YEH EXIST KYUN KARTA HAI? (Iske peeche ki physics)

PEHLE PRINCIPLES SE DERIVE KAISE KAREIN

Hum collision ko do phases mein model karte hain.

Phase 1 — Deformation: bodies ek dusre mein dhans jaati hain jab tak momentarily ek common velocity $v_{c}$ par nahi aa jaatein. Colliding surfaces ek dusre par push karti hain; impulse $J_{d}$ act karta hai.

Phase 2 — Restoration: daba hua material wapas spring karta hai, bodies ko alag push karta hai; impulse $J_{r}$ act karta hai.

Impulses se velocity formula derive karna. Body 1 (mass $m_{1}$ ) lo. Positive direction uski motion ki taraf maan lo.

Deformation phase ( $u_{1}$ se common $v_{c}$ tak): $- J_{d} = m_{1} (v_{c} - u_{1}) \Rightarrow J_{d} = m_{1} (u_{1} - v_{c})$

Restoration phase ( $v_{c}$ se final $v_{1}$ tak): $- J_{r} = m_{1} (v_{1} - v_{c}) \Rightarrow J_{r} = m_{1} (v_{c} - v_{1})$

Toh body 1 ke liye: $e = \frac{J _{r}}{J _{d}} = \frac{v _{c} - v _{1}}{u _{1} - v _{c}}$

Body 2 ke liye bhi yahi bookkeeping karo (forces reverse hain, isliye uske upar impulses $+ J_{d}, + J_{r}$ hain): $J_{d} = m_{2} (v_{c} - u_{2}), J_{r} = m_{2} (v_{2} - v_{c})$ $\Rightarrow e = \frac{v _{2} - v _{c}}{v _{c} - u _{2}}$

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

Special case: ball ka fixed floor se bounce karna

Floor "body 2" hai jiska infinite mass hai, isliye $u_{2} = v_{2} = 0$ . Height $h_{1}$ se giraao → floor par $u_{1} = 2 g h_{1}$ (neeche ki taraf) speed se hit karta hai. $v_{1} = 2 g h_{2}$ (upar ki taraf) speed se rebound karta hai, isliye chosen sign convention mein separation $= v_{2} - v_{1}$ rebound speed ban jaata hai.

Worked examples

Common mistakes (Steel-manned)

Active recall

Recall Cover karo aur answer do

$e$ ko velocities ke terms mein likhein aur har part ka naam batayein.
Humein $e$ ek separate equation ke roop mein kyun chahiye?
Jab $e < 1$ ho toh kya conserved hota hai aur kya lost hota hai?
Bounce-height relation $h_{2} = e^{2} h_{1}$ derive karein.
$e = 0$ physically kya matlab rakhta hai?

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho do toy cars crash kar rahi hain. Crash se pehle woh ek dusre ki taraf daud rahi theen — yeh unki "paas aane ki speed" hai. Crash ke baad woh bounce hokar alag hoti hain — yeh unki "door jaane ki speed" hai. Number $e$ bas yeh poochta hai: crash ke baad kitni speed bachi? Ek bahut bouncy rubber ball mein almost sab bachi rehti hai ( $e$ 1 ke paas). Ek clay ki dali mein kuch nahi bachta — woh bas splat hokar chipak jaati hai ( $e = 0$ ). Toh $e$ crash ki "bounciness score" hai, aur yeh depend karta hai ki cheezein kis cheez ki bani hain.

Connections

Conservation of Linear Momentum — pehli equation; restitution doosri hai.
Elastic Collisions — $e = 1$ ka case jahan KE bhi conserved hoti hai.
Perfectly Inelastic Collisions — $e = 0$ ka case, bodies saath milke chalti hain.
Impulse and Momentum — $e$ ki $J_{r} / J_{d}$ definition.
Kinetic Energy Loss in Collisions — energy lost $\propto (1 - e^{2})$ .
Projectile Motion — bouncing-ball height analysis ke liye zaroori hai.

Coefficient of restitution formula

e = \frac{v _{2} - v _{1}}{u _{1} - u _{2}}

= separation speed / approach speed

Numerator vs denominator mein indices swap kyun hain?

Taaki approach (

u_{1} > u_{2}

) aur separation (

v_{2} > v_{1}

) dono positive niklen, aur

0 \leq e \leq 1

mile

e = 1

physically kya matlab rakhta hai?

Perfectly elastic collision — separation ki speed, approach ki speed ke barabar hai, kinetic energy conserved hai

e = 0

physically kya matlab rakhta hai?

Perfectly inelastic — bodies chipak jaati hain aur common velocity se chalti hain, maximum KE loss hota hai

Kya

e < 1

par momentum conserved hota hai?

Haan, hamesha (koi external impulse nahi). Kinetic energy lost hoti hai

e

as a ratio of impulses

e = J_{r} / J_{d}

= restoration impulse / deformation impulse

Fixed floor par ball ke liye bounce height relation

e = h_{2} / h_{1}

, toh

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

Ek ball 2 m se girayi aur 1.28 m tak rebound hui;

e

nikaalein

e = 1.28/2 = 0.64 = 0.8

e

speeds ka ratio hai lekin heights

e^{2}

kyun deti hain?

Kyunki

h = v^{2} /2 g

, isliye height speed squared ke saath scale hoti hai

Derivation mein common velocity trick

Dono impulse ratios ke numerators aur denominators add karo;

v_{c}

cancel ho jaata hai, aur

e = (v_{2} - v_{1}) / (u_{1} - u_{2})

milta hai