Perfectly inelastic collisions — maximum KE loss: Exercises

1.4.7 · D4 · Physics › Momentum & Collisions › Perfectly inelastic collisions — maximum KE loss

Is page par sirf do master results use hote hain jo parent parent topic se aate hain:

Signs ke baare mein ek reminder, kyunki in problems mein se aadhe signs ki wajah se hi bante ya bigte hain:

Level 1 — Recognition

L1.1

Putty ke do lumps collide ke baad ek saath stick ho jaate hain. Kaun si quantity conserved hoti hai: momentum, kinetic energy, dono, ya koi nahi?

Recall Solution

Sirf Momentum. Perfectly inelastic collision mein koi external horizontal force nahi hoti, isliye total momentum unchanged rehta hai (yeh Conservation of Linear Momentum hai). Kinetic energy conserved nahi hoti — kuch putty ko squish karne mein aur heat/sound banane mein kharach ho jaati hai. Answer: momentum conserved hai, KE nahi.

L1.2

$3 kg$ ka ek mass $4 m/s$ par $1 kg$ ke stationary mass se takrake stick ho jaata hai. Common velocity $v$ symbolically likho, phir calculate karo.

Recall Solution

Dono diye hue numbers lo aur $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ mein daal do. $v = \frac{( 3 ) ( 4 ) + ( 1 ) ( 0 )}{3 + 1} = \frac{12}{4} = 3 m/s$ Answer: $v = 3 m/s$ (moving mass ke same direction mein).

L1.3

L1.2 wali same collision ke liye reduced mass $μ$ kya hai?

Recall Solution

$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} = \frac{( 3 ) ( 1 )}{3 + 1} = \frac{3}{4} = 0.75 kg$ Answer: $μ = 0.75 kg$ . Dhyaan do ki $μ$ hamesha dono individual masses se chhota hota hai — isliye ise "reduced" kehte hain.

Level 2 — Application

L2.1

$0.02 kg$ ki ek bullet $500 m/s$ par travel kar rahi hai aur stationary $3.98 kg$ ke wooden block mein embed ho jaati hai. Find karo (a) common velocity, (b) KE lost, (c) initial KE ka percentage lost.

Recall Solution

(a) Pehle Momentum. Total mass $= 0.02 + 3.98 = 4.00 kg$ . $v = \frac{( 0.02 ) ( 500 ) + 0}{4.00} = \frac{10}{4.00} = 2.5 m/s$ (b) KE lost — $\frac{1}{2} μ u_{r e l}^{2}$ use karo. Relative speed $u_{r e l} = 500 - 0 = 500 m/s$ . $μ = \frac{( 0.02 ) ( 3.98 )}{4.00} = \frac{0.0796}{4.00} = 0.0199 kg$ Yeh value exact hai, rounded nahi: $3.98 \times 0.02 = 0.0796$ aur $0.0796/4.00 = 0.0199$ cleanly terminate hoti hai, isliye hum confidence ke saath chaaron significant figures carry karte hain. $Δ K E = \frac{1}{2} (0.0199) (500)^{2} = \frac{1}{2} (0.0199) (250000) = 2487.5 J$ (c) Percentage. $K E_{i} = \frac{1}{2} (0.02) (500)^{2} = 2500 J$ . $\frac{Δ K E}{K E _{i}} = \frac{2487.5}{2500} = 0.995 = 99.5%$ Answers: $v = 2.5 m/s$ , $Δ K E = 2487.5 J$ , $99.5%$ lost. Ek chhota mass jo ek bade mass ke saath momentum share karne par majboor ho, woh slow ho jaata hai, isliye almost saari energy deformation mein dump ho jaati hai — Ballistic pendulum bilkul isi fact par kaam karta hai.

L2.2

Do identical railway cars, har ek $m = 8000 kg$ , same track par aate hain: ek $+ 2 m/s$ par, doosra $+ 0.5 m/s$ par (teez wala pakad kar dheeme wale se couple ho jaata hai). $v$ aur $Δ K E$ find karo.

Recall Solution

Dono same direction mein move kar rahe hain, isliye dono $+$ hain. $v = \frac{8000 ( 2 ) + 8000 ( 0.5 )}{16000} = \frac{16000 + 4000}{16000} = \frac{20000}{16000} = 1.25 m/s$ Relative speed $u_{r e l} = 2 - 0.5 = 1.5 m/s$ , aur $μ = \frac{8000 \cdot 8000}{16000} = 4000 kg$ . $Δ K E = \frac{1}{2} (4000) (1.5)^{2} = \frac{1}{2} (4000) (2.25) = 4500 J$ Answers: $v = 1.25 m/s$ , $Δ K E = 4500 J$ .

L2.3

$2 kg$ ki ek ball right mein $6 m/s$ par chal rahi hai jo left mein $3 m/s$ par chal rahi $4 kg$ ki ball se head-on collide karke stick ho jaati hai. $v$ (direction batao) aur $Δ K E$ find karo.

Recall Solution

Right $= +$ . Toh $u_{1} = + 6$ , $u_{2} = - 3$ . $v = \frac{2 ( 6 ) + 4 ( - 3 )}{6} = \frac{12 - 12}{6} = \frac{0}{6} = 0 m/s$ Blob momentarily at rest hai! $K E_{i} = \frac{1}{2} (2) (6)^{2} + \frac{1}{2} (4) (3)^{2} = 36 + 18 = 54 J, K E_{f} = 0$ $Δ K E = 54 J = 100% lost$ $\frac{1}{2} μ u_{r e l}^{2}$ se cross-check: $u_{r e l} = 6 - (- 3) = 9$ , $μ = \frac{2 \cdot 4}{6} = \frac{4}{3}$ , toh $Δ K E = \frac{1}{2} \cdot \frac{4}{3} \cdot 81 = 54 J$ ✓. Answers: $v = 0$ , $Δ K E = 54 J$ (saari ki saari).

Level 3 — Analysis

L3.1

Prove karo ki jab moving mass $m_{1}$ (speed $u$ ) ek equal stationary mass $m_{2} = m_{1}$ se stick hoti hai, initial KE ki exactly half lost hoti hai — $m_{1}$ ya $u$ ki value se independent.

Recall Solution

Maano har mass $= m$ , $u_{1} = u$ , $u_{2} = 0$ . $v = \frac{m u + 0}{2 m} = \frac{u}{2}$ $K E_{i} = \frac{1}{2} m u^{2}, K E_{f} = \frac{1}{2} (2 m) (\frac{u}{2})^{2} = \frac{1}{2} (2 m) \frac{u ^{2}}{4} = \frac{1}{4} m u^{2}$ $\frac{Δ K E}{K E _{i}} = \frac{\frac{1}{2} m u ^{2} - \frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{\frac{1}{4} m u ^{2}}{\frac{1}{2} m u ^{2}} = \frac{1}{2}$ $m$ aur $u$ completely cancel ho jaate hain, isliye fraction hamesha $1/2$ hota hai. Proven. Geometry yeh hai: moving mass apni aadhi speed apne partner ko de deta hai, aur speed aadhi hone par jabki mass double hone par KE quarter hoti hai — aadhi nahi — jo ek clean one-half deficit chhod deta hai.

L3.2

Do masses collide karke stick ho jaate hain. Coefficient of restitution $e$ ke terms mein, perfectly inelastic collision mein $e = 0$ hota hai. Explain karo kyun $e = 0$ objects ko ek velocity share karne par majboor karta hai, aur ise "maximum KE loss" se connect karo.

Recall Solution

Pehle labels pin down karo. Dono bodies ko $+$ direction mein ek line par rakhte hain. Body 1 woh hai jo peeche hai (velocity $u_{1}$ ) aur body 2 woh hai jo aage hai (velocity $u_{2}$ ), aur choose karo ki body 1 pakad rahi hai: $u_{1} > u_{2}$ , yani approach difference $u_{1} - u_{2} > 0$ . Unki final velocities $v_{1}^{'}$ (body 1) aur $v_{2}^{'}$ (body 2) hain. Coefficient of restitution dono speeds ka ratio hai — jo magnitudes hain, isliye hum absolute-value bars use karte hain taaki sign conventions ki parwah na ho: $e = \frac{speed of separation}{speed of approach} = \frac{∣ v _{2}^{'} - v _{1}^{'} ∣}{∣ u _{1} - u _{2} ∣}$ $e = 0$ set karne se numerator zero ho jaata hai: $∣ v_{2}^{'} - v_{1}^{'} ∣ = 0$ , yani $v_{2}^{'} = v_{1}^{'}$ . Yahi precisely "woh ek common velocity ke saath saath move karte hain" hai — aur yeh shared value exactly woh bare $v$ hai jo hum har jagah compute karte hain: $v_{1}^{'} = v_{2}^{'} = v$ . Kyunki woh zero speed par separate hote hain, saari relative-motion KE ( $\frac{1}{2} μ u_{r e l}^{2}$ wala chunk) destroy ho jaati hai — yeh maximum hai jo momentum-conserving collision lose kar sakti hai. Koi bhi $e > 0$ kuch separation speed chhod deta hai, isliye kuch relative KE survive karti hai, isliye less lost hoti hai. Answer: $e = 0 \Rightarrow v_{1}^{'} = v_{2}^{'} = v$ , poori relative-motion KE khatam — maximum loss.

L3.3

$5 kg$ ka ek mass $u_{1}$ par (unknown, right mein move karta hua) $5 kg$ ke stationary mass se stick hota hai, aur collision $250 J$ lose karti hai. $u_{1}$ find karo.

Recall Solution

Equal masses, ek at rest. Reduced mass $μ = \frac{5 \cdot 5}{10} = 2.5 kg$ , relative speed $= u_{1}$ . $Δ K E = \frac{1}{2} μ u_{1}^{2} \Rightarrow 250 = \frac{1}{2} (2.5) u_{1}^{2} = 1.25 u_{1}^{2}$ $u_{1}^{2} = \frac{250}{1.25} = 200 \Rightarrow u_{1} = \pm 200 = \pm 102$ Algebra do roots deta hai. Humari sign convention ne "right mein move karna" $+$ fix kiya, aur problem state karti hai ki mass right mein move karta hai, isliye hum negative root $- 102$ ko reject karte hain (woh ek leftward-moving mass describe karta, jo given ke against hai). Positive wala rakhte hain: $u_{1} = 102 \approx 14.14 m/s$ Answer: $u_{1} = 102 \approx 14.14 m/s$ . Humne energy formula invert ki kyunki yeh directly loss ko ek unknown speed se link karti hai — $v$ find karne ki koi zarurat nahi.

Level 4 — Synthesis

L4.1 — Ballistic pendulum

$m = 0.05 kg$ ki ek bullet horizontally fire ki jaati hai aur strings se latke $M = 4.95 kg$ ke block mein embed ho jaati hai. Block+bullet phir $h = 0.20 m$ tak swing up karta hai. $g = 10 m/s^{2}$ lo. Bullet ki original speed $u$ find karo.

Neeche di figure two-phase story dikhati hai. Left (Phase 1): magenta arrow mass $m$ ki bullet hai jo speed $u$ par violet block $M$ ki taraf fly kar rahi hai jo do vertical strings se rest mein latka hua hai; is fast impact ke dauran momentum conserved hota hai (magenta mein label kiya). Right (Phase 2): stuck lump $M + m$ (orange) dashed orange arc ke saath swing karta hai, vertical height $h$ (navy double-headed arrow) tak rise karta hai; lowest point par chhota magenta arrow $v$ sticking ke baad common velocity hai, aur is slow rise ke dauran energy conserved hoti hai (orange mein label kiya). Dhyaan do ki dono phases do alag laws follow karte hain.

Figure — Perfectly inelastic collisions — maximum KE loss

Recall Solution

Pehle masses combine karo (aur precision ka dhyaan rakho). $M + m = 4.95 + 0.05 = 5.00 kg$ exactly — dono decimals clean $5.00$ mein add hote hain, isliye koi rounding nahi aur hum poore time $5.00 kg$ teen significant figures rakhte hain. Pehle Phase 2 (swing) — energy use karo. Sticking ke baad, combined mass $M + m$ speed $v$ par move karta hai, phir height $h$ tak rise karta hai, saari KE ko potential energy mein convert karta hai (yeh gravity ke saath Work–Energy theorem hai): $\frac{1}{2} (M + m) v^{2} = (M + m) g h \Rightarrow v = 2 g h = 2 (10) (0.20) = 4 = 2 m/s$ Phase 1 (collision) — momentum use karo. Bullet+block stick ho jaate hain, isliye: $m u = (M + m) v \Rightarrow u = \frac{( M + m ) v}{m} = \frac{( 5.00 ) ( 2 )}{0.05} = \frac{10}{0.05} = 200 m/s$ Answer: $u = 200 m/s$ . Do laws kyun? Collision ke dauran, energy deformation mein leak hoti hai — isliye KE conservation wahan forbidden hai; sirf momentum survive karta hai. Swing ke dauran, kuch stick ya squish nahi hota, isliye energy conservation wahan valid hai. Sahi law sahi phase mein use karna — yahi poora trick hai.

L4.2 — Lost fraction sirf mass ratio par depend karta hai

Prove karo ki jab mass $m_{1}$ stationary mass $m_{2}$ se takrata hai, lost KE ki fraction $\frac{m _{2}}{m _{1} + m _{2}}$ hoti hai. Phir ise ek bullet ( $m_{1} = 0.01$ ) aur block ( $m_{2} = 1.99$ ) ke liye evaluate karo.

Recall Solution

$u_{2} = 0$ ke saath: $u_{r e l} = u_{1}$ , aur $K E_{i} = \frac{1}{2} m_{1} u_{1}^{2}$ . $\frac{Δ K E}{K E _{i}} = \frac{\frac{1}{2} \frac{m _{1} m _{2}}{m _{1} + m _{2}} u _{1}^{2}}{\frac{1}{2} m _{1} u _{1}^{2}} = \frac{\frac{m _{1} m _{2}}{m _{1} + m _{2}}}{m _{1}} = \frac{m _{2}}{m _{1} + m _{2}}$ $\frac{1}{2}$ , $m_{1}$ , aur $u_{1}^{2}$ sab cancel ho jaate hain — fraction masses ki pure geometry hai. Bullet/block ke liye: $\frac{1.99}{0.01 + 1.99} = \frac{1.99}{2.00} = 0.995 = 99.5%$ . Answer: fraction $= \frac{m _{2}}{m _{1} + m _{2}}$ ; numerically $99.5%$ . Isliye ek light projectile jo heavy target se takrata hai almost sab kuch lose kar deta hai — surviving KE fraction $\frac{m _{1}}{m _{1} + m _{2}}$ hoti hai, jo $m_{1} ≪ m_{2}$ hone par bahut chhoti hoti hai.

Level 5 — Mastery

L5.1 — Movable second mass ke saath loss minimise karna

$m_{1} = 2 kg$ ka ek block $u_{1} = 9 m/s$ par move karta hua $m_{2}$ ke doosre block se stick hoga jo rest mein hai. Tum $m_{2}$ choose kar sakte ho. (a) Jab $m_{2} \to 0$ , KE lost ki fraction kya hai? (b) Jab $m_{2} \to \infty$ ? (c) $m_{2} = 2 kg$ ke liye, $Δ K E$ compute karo. Trend interpret karo.

Recall Solution

L4.2 ka result use karo: fraction lost $= \frac{m _{2}}{m _{1} + m _{2}}$ . (a) $m_{2} \to 0$ : fraction $\to \frac{0}{2 + 0} = 0$ . Share karne ke liye almost koi mass nahi, isliye almost koi KE lost nahi — collision barely hoti hai. (b) $m_{2} \to \infty$ : fraction $\to \frac{m _{2}}{2 + m _{2}} \to 1$ , yani $100%$ . Ek immovable wall se takra ke stick ho jaana teri saari KE waste kar deta hai. (c) $m_{2} = 2 kg$ : equal masses, isliye fraction $= \frac{1}{2}$ . Actual loss compute karo: $μ = \frac{2 \cdot 2}{4} = 1 kg, Δ K E = \frac{1}{2} (1) (9)^{2} = 40.5 J$ (Check: $K E_{i} = \frac{1}{2} (2) (9^{2}) = 81 J$ ; half hai $40.5 J$ ✓.) Answers: (a) $0%$ , (b) $100%$ , (c) $Δ K E = 40.5 J$ . Loss fraction monotonically $0$ se $1$ tak climb karta hai jaise target bada hota jaata hai.

L5.2 — Teen-body sequential sticking

Ek frictionless line par, $m_{A} = 1 kg$ right mein $12 m/s$ par move karta hai aur $m_{B} = 2 kg$ se jo rest mein hai stick ho jaata hai, ek lump banata hai. Woh lump phir continue karta hai aur $m_{C} = 3 kg$ se jo rest mein hai stick ho jaata hai. Teeno ki final velocity aur dono collisions mein total KE lost find karo.

Recall Solution

Collision 1 ( $A$ into $B$ ): $v_{1} = \frac{( 1 ) ( 12 ) + ( 2 ) ( 0 )}{1 + 2} = \frac{12}{3} = 4 m/s (lump mass 3 kg)$ Collision 2 (lump into $C$ ): $v_{2} = \frac{( 3 ) ( 4 ) + ( 3 ) ( 0 )}{3 + 3} = \frac{12}{6} = 2 m/s (final, mass 6 kg)$ Total KE lost. Initial KE saari $A$ mein thi: $K E_{i} = \frac{1}{2} (1) (12)^{2} = 72 J$ Final KE: $K E_{f} = \frac{1}{2} (6) (2)^{2} = 12 J$ $Δ K E_{total} = 72 - 12 = 60 J$ Answers: final velocity $= 2 m/s$ ; total KE lost $= 60 J$ . Do stage-losses sum karke sanity check: stage 1 loses $\frac{1}{2} μ_{1} u_{r e l 1}^{2} = \frac{1}{2} \cdot \frac{1 \cdot 2}{3} \cdot 1 2^{2} = 48 J$ ; stage 2 loses $\frac{1}{2} \cdot \frac{3 \cdot 3}{6} \cdot 4^{2} = \frac{1}{2} \cdot 1.5 \cdot 16 = 12 J$ ; total $48 + 12 = 60 J$ ✓. Dhyaan do: momentum throughout conserved raha ( $1 \times 12 = 6 \times 2 = 12 kg\cdotm/s$ ), lekin KE steadily har stick par bleed hoti rahi.

L5.3 — Sticking collision exactly half KE kab lose karta hai?

Prove karo: ek moving mass $m_{1}$ jo stationary mass $m_{2}$ se stick hoti hai, initial KE ki exactly half lose karti hai iff $m_{1} = m_{2}$ .

Recall Solution

L4.2 se, fraction lost $= \frac{m _{2}}{m _{1} + m _{2}}$ . Ise $\frac{1}{2}$ ke equal set karo: $\frac{m _{2}}{m _{1} + m _{2}} = \frac{1}{2} \Rightarrow 2 m_{2} = m_{1} + m_{2} \Rightarrow m_{1} = m_{2}$ Yeh step reversible hai (har implication ek equivalence hai), isliye yeh if and only if masses equal hone par hold karta hai. Proven. Yahi deep reason hai ki L3.1 ka answer $m$ aur $u$ se independent tha: equal masses hi woh condition hai jo fraction ko exactly one-half par pin karti hai.

Recall Poori ladder ki one-line summary

Momentum speed decide karta hai ( $v = \frac{\sum m u}{\sum m}$ ); reduced mass aur relative speed loss decide karte hain ( $\frac{1}{2} μ u_{r e l}^{2}$ ); jab target rest mein ho toh lost fraction pure mass geometry hai ( $\frac{m _{2}}{m _{1} + m _{2}}$ ), $100%$ sirf tab reach hota hai jab total momentum zero ho.

Connections

Parent topic — woh theory jo yeh exercises drill karti hain.
Conservation of Linear Momentum — har single problem mein use hoti hai.
Ballistic pendulum — L4.1 classic application hai.
Work–Energy theorem — L4.1 ka swing phase.
Coefficient of restitution — L3.2 $e = 0$ ko sticking se link karta hai.
Reduced mass — loss formula mein $μ$ .
Elastic collisions — contrast ke liye $e = 1$ opposite extreme.