Perfectly inelastic collisions — maximum KE loss

1.4.7 · Physics › Momentum & Collisions

KYA ho raha hai?

KYU "maximum" loss hai? Momentum conservation ek aisa constraint hai jo hum tod nahi sakte. Is constraint ke allowed final states mein se, woh state jahan dono masses ek velocity share karte hain, kinetic energy "valley" ke sabse neeche hoti hai. Har doosra allowed final state zyada KE rakhta hai. Hum neeche ise prove karenge.

Kaise derive karein final velocity (scratch se)

Maano masses $m_{1}$ (velocity $u_{1}$ ) aur $m_{2}$ (velocity $u_{2}$ ) hain. Stick hone ke baad, dono $v$ se move karte hain.

Step 1 — Momentum conserve karo. Koi external horizontal force nahi ⇒ total $p$ unchanged rehta hai. $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$ Yeh step kyun? Yahi ek conservation law hai jo bachti hai; energy ko heat/sound/deformation ke roop mein leak hone diya jaata hai.

Step 2 — Common velocity ke liye solve karo. $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ Yeh step kyun? Yeh simply dono velocities ka momentum-weighted average hai — centre-of-mass velocity, jo kisi bhi collision mein kabhi nahi badlti.

Kitni KE lost hoti hai? (Derivation)

Step 1 — Pehle aur baad ki KE likho. $K E_{i} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}, K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v^{2}$

Step 2 — $v$ substitute karo aur simplify karo. Sabse clean case ke liye aur phir general case ke liye karta hoon.

KE lost hai $Δ K E = K E_{i} - K E_{f}$ Algebra ke baad ( $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ substitute karke): $Δ K E = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2}$ Yeh step kyun? Combination $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ reduced mass hai, aur $(u_{1} - u_{2})$ relative velocity hai. Lost energy exactly relative motion ki KE hai — jo tab destroy hoti hai jab bodies ek doosre ke relative move karna band kar deti hain.

Dhyaan do: $Δ K E$ sirf $(u_{1} - u_{2})^{2}$ par depend karta hai. Agar COM frame velocity kuch aur hoti, toh momentum violate hota — isliye yahi ek allowed sticking outcome hai, aur yeh saari relative-motion KE khatam kar deta hai. Isliye yeh maximum hai.

KYU yeh MAXIMUM loss hai? (key proof)

Figure — Perfectly inelastic collisions — maximum KE loss

Worked Examples

Common Mistakes

Recall Feynman: 12-saal ke bacche ko explain karo

Socho do clay ke lumps ek doosre ki taraf slide kar rahe hain. Jab woh takraate hain, toh squish hokar ek bade lump mein chipak jaate hain. Unhe milke ek hi speed se move karna padta hai (aage wala ek taraf aur peeche wala doosri taraf nahi ja sakta — woh chipke hue hain!). Kyunki unhe agree karna padta hai, unki bahut saari "moving energy" clay ko squish karne aur heat banana mein use ho jaati hai, motion mein rehne ki jagah. Agar woh seedhe ek doosre ki taraf equally force se aa rahe the, toh blob simply ruk jaata hai — saari motion energy ek squish mein badal gayi. Yahi "saath chipkane" wala rule hai jo unhe maximum possible motion energy lose karwata hai, jabki total "push" (momentum) same rehta hai.

Flashcards

Perfectly inelastic collision mein kya conserved hota hai?

Momentum (hamesha); kinetic energy conserved NAHI hoti.

Do masses ke chipkne ke baad final velocity kya hoti hai?

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

(centre-of-mass velocity).

Perfectly inelastic collision mein KE lost ka formula?

Δ K E = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} (u_{1} - u_{2})^{2} = \frac{1}{2} μ u_{r e l}^{2}

Perfectly inelastic collision mein KE loss maximum kyun hoti hai?

Sticking dono bodies ko COM frame mein rest mein force karti hai, saari relative-motion KE destroy kar deti hai — momentum conservation ke saath allowed minimum final KE.

Reduced mass kya hota hai?

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

Equal masses, ek rest mein, perfectly inelastic — KE lost ka fraction?

Exactly

1/2

(aadha).

Sticking collision mein 100% KE kab lost hoti hai?

Jab total momentum zero ho (COM rest mein), jaise equal aur opposite momenta.

Lost KE kahan jaati hai?

Heat, sound, aur bodies ki permanent deformation mein.

Yahan

v

find karne ke liye energy conservation kyun use nahi kar sakte?

Kyunki KE partly non-conservative work mein lost hoti hai; sirf momentum conserved hota hai.

Lost KE kis motion ki kinetic energy ke barabar hoti hai?

Dono bodies ki relative motion ki (

\frac{1}{2} μ u_{r e l}^{2}

Connections

Conservation of Linear Momentum — woh law jo har collision mein bachti hai.
Elastic collisions — opposite extreme: KE fully conserved, $e = 1$ .
Coefficient of restitution — perfectly inelastic ka matlab $e = 0$ .
Centre of mass motion — $v$ yahan COM velocity hi hai.
Reduced mass — KE-loss formula aur two-body problems mein aata hai.
Ballistic pendulum — classic application (Example 2).
Work–Energy theorem — "lost" KE actually kahan jaati hai.