Perfectly inelastic collisions — maximum KE loss: Question bank

1.4.7 · D5 · Physics › Momentum & Collisions › Perfectly inelastic collisions — maximum KE loss

Definition Is page par use kiye gaye symbols aur idealizations

Poore page mein hum do point masses (saari mass ek hi point par, koi rotation nahi, koi size nahi) ko ek frictionless line par collide karte hue imagine karte hain — brief impact ke dauran koi external horizontal force nahi — ye standard idealizations hain. Symbols:

$m_{1}, m_{2}$ — do masses.
$u_{1}, u_{2}$ — impact se pehle unki velocities (vectors: sign direction batata hai).
$v$ — ek hi common velocity jab dono chipak jaate hain.
$v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ — centre-of-mass velocity: dono ki momentum-weighted average velocity. Ye collision mein kabhi nahi badlti, aur chipkane ke baad $v = v_{c m}$ . Dekho Centre of mass motion.
$u_{r e l} = u_{1} - u_{2}$ — relative velocity: body 2 par baithe hue body 1 kitni tezi se body 2 ke paas aati hai.
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ — reduced mass: ek akela effective mass jo do-body relative motion ko capture karta hai (hamesha choti mass se bhi chota hota hai). Dekho Reduced mass.
$e$ — coefficient of restitution $= \frac{relative speed after}{relative speed before}$ . Perfectly inelastic $\Rightarrow e = 0$ . Dekho Coefficient of restitution.

Formula

$Δ K E = \frac{1}{2} μ u_{r e l}^{2}$ kyun hai — COM frame mein jaake derive kiya Neeche ka diagram poori derivation ka map hai. Top row mein lab frame mein dono bodies ko velocities $u_{1}$ (blue) aur $u_{2}$ (orange) ke saath dikhaya gaya hai. Bottom-left green arrow COM drift $v_{c m}$ hai jise hum subtract karte hain. Bottom-right subtract karne ke baad jo bachta hai woh dikhata hai: do equal-and-opposite arrows — relative wobble. Red label us piece ko mark karta hai jo sticking destroy kar deti hai. Aage badhte waqt ise dekhte rehna.

Step 1 — Motion ko split karo aur COM-frame speeds dhundho. "COM frame mein jump karne" ke liye hum har velocity se $v_{c m}$ subtract karte hain. Baaki bachi velocities ko $v_{1}^{'} = u_{1} - v_{c m}$ aur $v_{2}^{'} = u_{2} - v_{c m}$ kaho (primed = "COM frame mein jaise dikhta hai"). Body 1 ke liye $v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ plug in karo: $v_{1}^{'} = u_{1} - \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}} = \frac{( m _{1} + m _{2} ) u _{1} - m _{1} u _{1} - m _{2} u _{2}}{m _{1} + m _{2}} = \frac{m _{2} ( u _{1} - u _{2} )}{m _{1} + m _{2}} = \frac{m _{2}}{m _{1} + m _{2}} u_{r e l} .$ $m_{1} u_{1}$ terms cancel ho jaate hain, sirf difference $u_{1} - u_{2} = u_{r e l}$ bachta hai — isliye COM-frame speed purely relative velocity se banti hai. Body 2 ke liye wahi subtraction karne par $v_{2}^{'} = \frac{m _{1} ( u _{2} - u _{1} )}{m _{1} + m _{2}} = - \frac{m _{1}}{m _{1} + m _{2}} u_{r e l}$ milta hai (minus sign ki wajah se diagram mein dono arrows opposite point karte hain). Dhyan do ki momenta equal and opposite hain: $m_{1} v_{1}^{'} = \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{r e l} = - m_{2} v_{2}^{'}$ , toh is frame mein total momentum zero hai, jaisa hona chahiye.

Step 2 — KE add karo aur dekho kaise collapse hoti hai. Ab un speeds se dono kinetic energies add karo: $K E_{rel} = \frac{1}{2} m_{1} v_{1}^{'2} + \frac{1}{2} m_{2} v_{2}^{'2} = \frac{1}{2} m_{1} \frac{m _{2}^{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2} + \frac{1}{2} m_{2} \frac{m _{1}^{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2} .$ Dono terms mein $\frac{m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} u_{r e l}^{2}$ common factor hai; ise bahar nikalne par andar $m_{2} + m_{1}$ bachta hai: $K E_{rel} = \frac{1}{2} \frac{m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} (m_{2} + m_{1}) u_{r e l}^{2} = \frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{r e l}^{2} = \frac{1}{2} μ u_{r e l}^{2} .$ Key cancellation yeh hai ki numerator mein $(m_{1} + m_{2})$ ka ek factor denominator ke ek factor se cancel hota hai — yehi woh cheez hai jo awkward two-mass sum ko tidy reduced mass $μ$ mein badal deti hai. Yeh relative motion ki kinetic energy hai — woh akela KE jo sirf poora system saath drift karne ki wajah se nahi hai.

Step 3 — Sticking bilkul yahi khatam kar deti hai. Chipkane ke baad, dono COM frame mein rest par hain (saath move kar rahe hain = koi relative motion nahi), toh $K E_{rel}$ ka poora hissa destroy ho jaata hai jabki drift-together KE $\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}$ bachi rehti hai. Isliye $Δ K E = \frac{1}{2} μ u_{r e l}^{2}$ .

Figure — Perfectly inelastic collisions — maximum KE loss

True or false — justify

Perfectly inelastic collision mein momentum conserved hota hai.

True. Koi external horizontal force nahi hone se, impact forces internal hain aur equal-and-opposite hain (Newton's third law), toh unka sum zero hai aur ye total momentum nahi badal sakta — chahe chipko ya nahi. Sirf kinetic energy leak hone ke liye free hai.

"Maximum KE loss" ka matlab hai ki saari kinetic energy destroy ho jaati hai.

False. Iska matlab hai momentum conservation ki ijazat se zyada se zyada loss. Agar total momentum nonzero hai toh centre of mass chalta rehna chahiye, toh drift KE

K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2} > 0

hamesha bachti hai.

Ek zyada clever collision chunke tum perfectly inelastic collision se zyada KE kho sakte ho.

False. Sticking already dono bodies ko COM frame mein rest par le aati hai (zero relative motion), jo momentum allow karta hai usme se lowest KE hai. Isse kam karne ke liye negative KE wali body chahiye, jo impossible hai.

Perfectly inelastic collision mein dono bodies ki velocity baad mein zero honi chahiye.

False. Dono ek common velocity

v = v_{c m}

share karte hain, jo zero sirf tab hoti hai jab total momentum zero ho. Generally dono saath nonzero speed par move karte hain.

Perfectly inelastic collision ke liye coefficient of restitution

e = 0

hota hai.

True. Definition se

e = \frac{relative speed after}{relative speed before}

; chipkane ka matlab after mein relative speed zero hai, toh

e = \frac{0}{u _{r e l}} = 0

Agar do objects collide karke chipak jaate hain, toh baad mein woh possibly alag nahi ho sakte.

True. "Stick" ka matlab hi hai impact ke baad zero relative velocity — alag hone ke liye

e

ka numerator positive hona chahiye, yaani

e > 0

, jo perfectly inelastic (

e = 0

) case ko contradict karta hai.

KE ka loss do bodies ki relative motion ki kinetic energy ke barabar hai.

True. Upar derive kiya gaya hai, relative-motion KE

\frac{1}{2} μ u_{r e l}^{2}

hai jahan

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

; sticking exactly woh motion destroy karta hai aur kuch nahi.

Perfectly inelastic collision hamesha KE ko sirf heat mein convert karti hai.

False. Lost KE heat, sound, aur bodies ki permanent deformation mein jaati hai — sirf heat mein nahi.

Dono initial velocity vectors ko double karne se KE loss chaar guna ho jaata hai, 1D aur 2D dono mein.

True.

Δ K E = \frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2}

; dono velocity vectors ko double karne se relative-velocity vector

u_{1} - u_{2}

double hota hai, aur uska magnitude squared chaar guna ho jaata hai. Yeh generally hold karta hai — bas dhyan yeh rakhna hai ki velocities ko vectors ki tarah combine karo, bare numbers ki tarah nahi.

Agar ek body doosri se infinitely zyada massive hai, toh essentially light body ki 100% KE lost ho jaati hai.

True. Bhari mass barely hilti hai, toh common velocity tiny hoti hai; light body ki motion energy almost poori deformation/heat mein chali jaati hai (bullet-in-block limit).

Spot the error

"KE conserved hai, toh

v

dhundhne ke liye

K E_{i} = K E_{f}

set karunga."

Error: KE yahan conserved nahi hai — kuch deformation mein leak ho jaati hai, toh

K E_{i} > K E_{f}

v

dhundhne ke liye tumhe momentum conservation,

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

, use karna hoga, phir loss calculate karo.

"Wo equal speeds se head-on collide kiye, toh saari momentum destroy ho gayi."

Error: Momentum kabhi destroy nahi hoti. Impact forces internal Newton's-third-law pairs hain, toh cancel ho jaate hain; agar momenta (

m u

) equal aur opposite hain toh unka sum zero hai — lekin woh zero total conserved hai, lost nahi.

m_{1} = 2

kg at

3

m/s hits

m_{2} = 1

kg at

6

m/s (opposite), toh

v = \frac{2 ( 3 ) + 1 ( 6 )}{3} = 4

Error: Velocity ek vector hai — doosri body opposite direction mein move kar rahi hai, toh

u_{2} = - 6

. Sahi:

v = \frac{6 - 6}{3} = 0

. Sign bhool jaana classic trap hai.

"Reduced mass

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

dono masses se bada hota hai."

Error:

μ

hamesha choti mass se bhi chota hota hai. Equal masses ke liye

μ = m /2

; jab ek mass

\to \infty

hoti hai,

μ \to

choti mass par lekin kabhi exceed nahi karta — toh woh kabhi bhi kisi bhi mass se bada nahi ho sakta.

"Chipkane ke baad, dono bodies lab frame mein rest par hain kyunki unhone saari relative KE kho di."

Error: Wo relative-motion KE kho dete hain, lekin centre-of-mass drift untouched rehti hai. Lab frame mein dono

v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

par glide karte hain jab tak woh COM velocity zero na ho.

"Kyunki objects deform hote hain, momentum deformation mein leak hoti hai."

Error: Deformation energy absorb karta hai, momentum nahi. Kyun? Bodies ke beech deformation forces ek internal equal-and-opposite pair hain (Newton's third law), toh unke impulses cancel ho jaate hain aur total momentum fixed rehti hai; energy mein aisi koi cancellation nahi hoti aur woh heat ke roop mein pile up ho sakti hai.

Why questions

Momentum conserved kyun hota hai lekin kinetic energy nahi?

Collision forces internal aur equal-and-opposite hain, toh unka net impulse zero hai aur total momentum nahi badal sakta; lekin har force apni body par negative work karti hai (woh motion ke against squeeze karti hain), aur woh work KE ko heat aur deformation mein drain kar deta hai. Dekho Work–Energy theorem.

Sticking minimum possible final KE kyun deti hai?

COM frame mein total momentum zero hota hai; "momentum zero" se consistent lowest KE state dono bodies ka rest par hona hai — jo exactly "saath move karna" hai. Koi bhi doosra outcome relative motion chhodta hai, aur relative motion hamesha positive KE

\frac{1}{2} μ u_{r e l}^{2}

carry karta hai.

Δ K E

frame ki overall velocity se independent kyun hai?

Yeh sirf

u_{r e l} = u_{1} - u_{2}

par depend karta hai, aur dono bodies mein common velocity add karne se har individual velocity (aur

v_{c m}

) badal jaati hai lekin unka difference unchanged rehta hai — toh relative motion, aur uski KE, har inertial frame mein same hai.

Ballistic pendulum mein bullet ki almost saari KE kyun lost ho jaati hai?

Ek tiny mass (

m_{1}

, bullet) apni momentum ek bhari mass (

m_{2}

, block) ke saath share karta hai, toh common speed

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

negligible hoti hai; bullet ki kinetic energy almost poori heat aur deformation mein convert ho jaati hai. Dekho Ballistic pendulum.

Ballistic pendulum ki post-collision height impact ke through energy conservation use karke kyun nahi dhundh sakte?

Impact khud inelastic hai aur KE destroy kar deta hai, toh energy conserved nahi hoti — impact ke liye momentum use karo

v

pane ke liye, phir baad ki swing ke liye energy conservation switch karo.

Δ K E

ke formula mein reduced mass

μ

naturally kyun aata hai?

Jab tum two-body relative motion ko ek effective particle ke roop mein rewrite karte ho, us particle ki mass

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

nikali hai, toh uski kinetic energy

\frac{1}{2} μ u_{r e l}^{2}

hoti hai — aur exactly wahi energy sticking destroy karta hai. Dekho Reduced mass.

Perfectly inelastic collision elastic collision ka opposite extreme kyun hai?

Elastic collisions saari KE conserve karte hain (

e = 1

, bodies same relative speed par bounce apart ho jaati hain); perfectly inelastic koi relative KE conserve nahi karta (

e = 0

, bodies chipak jaati hain). Har real collision

0 < e < 1

ke beech hoti hai. Dekho Elastic collisions.

Oblique (2D) perfectly inelastic collision bhi same tarah kaam kyun karti hai?

Momentum ek vector hai, toh yeh component by component conserved hoti hai (

x

aur

y

alag-alag); common final velocity vector

v_{c m}

hai, aur KE lost abhi bhi

\frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2}

hai relative-velocity vector ki magnitude use karke.

Edge cases

Impact se pehle total momentum exactly zero hai — kitni KE lost hogi?

100%. Zero total momentum ke saath COM rest par hai (

v_{c m} = 0

), toh sticking sab kuch rest par le aata hai, saari kinetic energy destroy kar deta hai.

Dono bodies touch karne se pehle same velocity par move kar rahi hain — agar woh chipak jaayein toh kya hoga?

Kuch nahi badlega aur zero KE lost hogi. Yahan

u_{1} = u_{2}

hai, toh

u_{r e l} = 0

aur

Δ K E = \frac{1}{2} μ \cdot 0 = 0

; destroy karne ke liye koi relative motion thi hi nahi.

Ek body ki mass zero hai — kya formula abhi bhi meaningful hai?

Degenerate case hai.

m_{2} = 0

ke saath,

μ = \frac{m _{1} \cdot 0}{m _{1}} = 0

toh

Δ K E = 0

, aur

v = u_{1}

; ek massless body koi momentum ya KE carry nahi karti, toh collision real mass par kuch nahi karta.

Ek mass doosri se bahut badi hai aur initially rest par hai — common velocity kya hogi?

Essentially zero. Jab

m_{2} \to \infty

v = \frac{m _{1} u _{1}}{m _{1} + m _{2}} \to 0

; wall-jaisi body momentum absorb kar leti hai bina visibly hiley, aur light body ki KE almost poori lost ho jaati hai.

Do bodies ki equal masses hain aur ek rest par hai — kitna KE bachta hai?

Exactly aadha bachta hai, aadha lost hota hai.

v = u /2

deta hai

K E_{f} = \frac{1}{4} m u^{2}

against

K E_{i} = \frac{1}{2} m u^{2}

, toh exactly

50%

destroy hota hai.

KE ka sabse bada possible fraction kitna lost ho sakta hai, aur kab?

100% tak, sirf tab jab total momentum zero ho. Koi bhi nonzero total momentum ek moving COM force karta hai, loss ko 100% se neeche cap karta hai kyunki drift KE

\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}

bachni chahiye.

Do bodies right angles par milti hain (oblique impact) aur chipak jaati hain — kya koi KE lost hoti hai?

Haan.

9 0^{\circ}

par bhi ek nonzero relative-velocity vector

u_{1} - u_{2}

hota hai, toh

Δ K E = \frac{1}{2} μ ∣ u_{1} - u_{2} ∣^{2} > 0

; loss sirf tab vanish hota hai jab dono velocity vectors identical hon.

Connections

Conservation of Linear Momentum — woh law jo yahan har "true" answer ke peeche hai.
Coefficient of restitution — $e = 0$ edge case jinke around ye traps orbit karte hain.
Reduced mass — kyun $μ$ aur $u_{r e l}$ loss govern karte hain.
Centre of mass motion — surviving KE yahan rehti hai.
Elastic collisions — contrast ke liye opposite extreme.
Ballistic pendulum — "almost saari KE lost" ka showcase.
Work–Energy theorem — lost KE actually kahan jaati hai.