Perfectly inelastic collisions — maximum KE loss: Worked examples

1.4.7 · D3 · Physics › Momentum & Collisions › Perfectly inelastic collisions — maximum KE loss

Shuru karne se pehle: ek symbol dictionary taaki kuch bhi bina explain ke na aaye.

Definition Symbols, simple shabdon mein

$m_{1}, m_{2}$ — do masses (kilograms mein). Hamesha positive numbers.
$u_{1}, u_{2}$ — hit se pehle ki velocities. Yeh signed hain: ek direction ko $+$ lo (jaise rightward), to leftward $-$ hoga. Velocity ek speed hai jisme direction bhi attached hai.
$v$ — single velocity jo dono share karte hain sticking ke baad.
$p$ — total momentum, defined as $p = m_{1} u_{1} + m_{2} u_{2}$ (mass times velocity, summed, signs ke saath). Yeh pehle aur baad mein same rehta hai — yahi ek law hai jo hum kabhi nahi todte. Dekho Conservation of Linear Momentum.
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ — reduced mass. Ise "relative motion ki effective mass" samjho. Dekho Reduced mass.
$u_{r e l} = u_{1} - u_{2}$ — relative velocity: object 1, object 2 ke perspective se kitni tezi se approach kar raha hai.
$K E = \frac{1}{2} m v^{2}$ — kinetic energy, "motion ki energy." Square ki wajah se, sign se koi fark nahi padta — left ya right move karo, KE same hogi.

The scenario matrix

Har perfectly inelastic collision teen haan/na questions se fix hoti hai: same direction ya opposite?, kya ek at rest hai?, equal ya unequal masses? — plus degenerate edges. Total momentum column $p = m_{1} u_{1} + m_{2} u_{2}$ use karta hai (upar define kiya). Neeche ki table mein har cell listed hai; har example tag [C#] ek row se map karta hai.

Cell	$u_{1}, u_{2}$ ka sign	Masses	Total momentum $p = m_{1} u_{1} + m_{2} u_{2}$	Kya special hai	Example
C1	dono $+$ , ek $0$ hai	equal	$> 0$	textbook baseline	Ex 1
C2	dono $+$ , ek $0$ hai	bahut unequal	$> 0$	bada KE loss $%$	Ex 2
C3	$+$ aur $-$	unequal	$= 0$	dono ruk jaate hain, 100 $%$ loss	Ex 3
C4	$+$ aur $-$	unequal	$\neq = 0$	survivors chalte rehte hain	Ex 4
C5	dono $+$	any	$> 0$	dono pehle se chal rahe, catch-up	Ex 5
C6 (degenerate)	$u_{1} = u_{2}$	any	$\neq = 0$	koi relative motion nahi ⇒ zero loss	Ex 6
C7 (degenerate)	any	$m_{2} \to 0$	$\approx m_{1} u_{1}$	limiting behaviour	Ex 7
C8 (word)	real-world	unequal	$> 0$	ballistic-pendulum height, Ballistic pendulum	Ex 8
C9 (exam twist)	fraction puchha	equal	$> 0$	"KE ka kitna fraction bachta hai?"	Ex 9

Figure — Perfectly inelastic collisions — maximum KE loss

Worked Examples

Worked example Ex 1 — Equal masses, ek at rest

[C1] $m_{1} = m_{2} = 2 kg$ , $u_{1} = + 6 m/s$ , $u_{2} = 0$ .

Forecast: padhne se pehle guess karo kitna fraction KE lost hoga. Half? Quarter?

Step 1 — Common velocity. $v = \frac{2 ( 6 ) + 2 ( 0 )}{2 + 2} = \frac{12}{4} = 3 m/s$ . Yeh step kyun? Sticking ek shared velocity force karti hai; momentum conservation hi ek law hai jo yeh deti hai.

Step 2 — KE before. $K E_{i} = \frac{1}{2} (2) (6)^{2} + 0 = 36 J$ . Yeh step kyun? Hume compare karne ke liye ek "before" chahiye.

Step 3 — KE after. $K E_{f} = \frac{1}{2} (2 + 2) (3)^{2} = 18 J$ . To $Δ K E = K E_{i} - K E_{f} = 36 - 18 = 18 J$ = aadha lost. Yeh step kyun? Equal masses hamesha exactly half lose karte hain — moving wale ka momentum double mass par spread ho jaata hai.

Verify: reduced-mass route $\frac{1}{2} μ u_{r e l}^{2} = \frac{1}{2} \cdot \frac{2 \cdot 2}{4} \cdot (6 - 0)^{2} = \frac{1}{2} (1) (36) = 18 J$ ✓. Units: $kg \cdot (m/s)^{2} = J$ ✓.

Worked example Ex 7 — Limiting case:

$m_{2} \to 0$ [C7] $m_{1} = 1 kg$ at $u_{1} = + 10 m/s$ ; ek speck $m_{2} = 0.001 kg$ at rest isse stick ho jaata hai.

Forecast: $m_{2} \to 0$ ke saath $v$ kya approach karega? $Δ K E$ kya approach karega?

Step 1 — Common velocity. $v = \frac{1 ( 10 ) + 0.001 ( 0 )}{1.001} = \frac{10}{1.001} \approx 9.990 m/s$ . Yeh step kyun? Jaise doosra mass khatam hota hai, $v \to u_{1}$ : ek massless speck tumhe slow nahi kar sakta.

Step 2 — Reduced mass limit. $μ = \frac{1 \cdot 0.001}{1.001} \approx 0.0009990 kg \to 0$ as $m_{2} \to 0$ . Yeh step kyun? Reduced mass chhote mass se dominate hoti hai, isliye $μ \to 0$ aur loss $\to 0$ .

Step 3 — KE loss. $Δ K E = \frac{1}{2} (0.0009990) (10 - 0)^{2} \approx 0.04995 J$ — vanishingly small.

Verify: $K E_{i} = \frac{1}{2} (1) (10)^{2} = 50 J$ ; $K E_{f} = \frac{1}{2} (1.001) (9.990)^{2} \approx 49.950 J$ ; difference $\approx 0.0499 J$ ✓. Jaise $m_{2} \to 0$ , $Δ K E \to 0$ .

Worked example Ex 8 — Word problem: ballistic pendulum height

[C8] $0.05 kg$ ka dart $u_{1} = + 120 m/s$ par $1.95 kg$ ke block mein embed hota hai jo rest mein hang kar raha hai. Block-plus-dart kitni height $h$ tak swing karega? $g = 10 m/s^{2}$ use karo. Dekho Ballistic pendulum.

Forecast: do energy regimes hain — ek jahan KE lost hoti hai, ek jahan conserved hoti hai. Kaun sa kaun sa hai?

Step 1 — Collision (momentum only). $v = \frac{0.05 ( 120 ) + 1.95 ( 0 )}{2} = \frac{6}{2} = 3 m/s$ . Yeh step kyun? Embedding ke dauran KE lost hoti hai — isliye hum yahan energy use nahi kar sakte, sirf momentum. Dekho Conservation of Linear Momentum.

Step 2 — Swing (energy conserved). Stick hone ke baad, aur squishing nahi, isliye blob ki KE height mein convert hoti hai: $\frac{1}{2} (2) v^{2} = (2) g h$ . Dekho Work–Energy theorem. Yeh step kyun? Post-collision koi non-conservative work nahi hai, isliye mechanical energy ab conserved hai.

Step 3 — Height solve karo. $h = \frac{v ^{2}}{2 g} = \frac{3 ^{2}}{2 ( 10 )} = \frac{9}{20} = 0.45 m$ .

Verify: $K E$ at bottom $= \frac{1}{2} (2) (3)^{2} = 9 J$ ; $P E$ at top $= 2 \cdot 10 \cdot 0.45 = 9 J$ ✓. Units: $\frac{( m/s ) ^{2}}{m/s ^{2}} = m$ ✓.

Worked example Ex 9 — Exam twist: KE ka fraction jo

bachta hai [C9] Mass $m$ speed $u$ par identical mass $m$ se rest mein stick karta hai. Initial KE ka kitna fraction bacha rehta hai? (Symbolically answer do, phir numerically check karo $m = 1, u = 4$ se.)

Forecast: Ex 1 mein humne "aadha lost" find kiya tha — to kya aadha bachta hai?

Step 1 — Momentum se common velocity. Momentum conserve karo: $m u + m (0) = (m + m) v$ , to $2 m v = m u$ , giving $v = \frac{u}{2}$ . Yeh step kyun? Hum $v$ guess nahi kar sakte; ek sticking collision mein bachne wala ek hi law momentum conservation hai, aur iske liye $v$ solve karne par $v = u /2$ milta hai. Aage sab kuch isi par depend karta hai.

Step 2 — Fraction banao. $K E_{f} = \frac{1}{2} (2 m) (u /2)^{2} = \frac{1}{4} m u^{2}$ , $K E_{i} = \frac{1}{2} m u^{2}$ , to $\frac{K E _{f}}{K E _{i}} = \frac{\frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{1}{2}$ . Yeh step kyun? $m, u$ ko symbolic rakhne se prove hota hai ki fraction universal hai — masses aur speeds cancel hokar ek pure number dete hain.

Step 3 — Interpret karo. Exactly $\frac{1}{2}$ bachta hai aur $\frac{1}{2}$ lost hota hai, kisi bhi equal-mass one-at-rest sticking collision ke liye.

Verify: $m = 1, u = 4$ ke saath: $K E_{i} = 8 J$ , $v = 2$ , $K E_{f} = \frac{1}{2} (2) (2)^{2} = 4 J$ , ratio $= 4/8 = 0.5$ ✓.

Recall Self-check: kaun sa cell kaun sa hai?

Ek blob collision ke baad rest mein aata hai — kya sach hona chahiye tha? ::: Total momentum zero tha (equal aur opposite), C3 cell — 100 $%$ KE lost. Collision zero KE loose karti hai — yeh kya batata hai? ::: Dono bodies ki velocities identical thi ( $u_{r e l} = 0$ ), degenerate C6 cell. Bullet-into-block ~99 $%$ KE loose karta hai — kyun? ::: Wildly unequal masses (C2/C7 regime); tiny mass ek huge mass ke saath momentum share karta hai isliye $v$ aur $K E_{f}$ tiny hain. Ballistic pendulum mein pure problem mein energy conservation kyun use nahi kar sakte? ::: Embedding phase (C8) KE lose karta hai; sirf sticking ke baad swing energy conserve karta hai.

Connections

Parent — full derivation
Conservation of Linear Momentum — ek hi law jo upar har cell mein use hua.
Reduced mass — woh $μ$ jo saari lost energy carry karta hai.
Ballistic pendulum — Ex 8 ki two-phase logic.
Work–Energy theorem — bachne wali KE ko swing height mein convert karta hai.
Coefficient of restitution — in saare cells mein $e = 0$ hai.
Elastic collisions — opposite extreme, $e = 1$ , zero loss.
Centre of mass motion — yahan $v$ hamesha COM velocity hoti hai.