System with external forces — conditions for conservation: Worked examples

1.4.4 · D3 · Physics › Momentum & Collisions › System with external forces — conditions for conservation

Kuch bhi shuru karne se pehle, teen symbols jinhe hum baar baar use karenge. Agar koi shaky lage, toh yeh page chalte chalte unhe dobara explain kar degi.

Parent ka master equation inhe ek saath jodhta hai: $\frac{d P}{d t} = F_{e x t}^{n e t} ⟺ Δ P = J_{e x t}$

Isse aise padho: "group ke total momentum ko sirf ek bahari impulse hi change kar sakta hai."

Scenario matrix

Yeh topic jo bhi problem throw karta hai woh in cells mein se ek (ya mixture) hoti hai. Neeche ke worked examples mein bataya gaya hai ki woh kaun sa cell cover karte hain.

Cell	Isse kya alag banata hai	Yeh trap kya chupaata hai
A. Zero total, opposite signs	System rest se start karta hai; parts opposite-sign velocities ke saath alag hote hain	Recoiling piece par minus sign bhuul jaana
B. One-axis conservation	External force sirf ek axis par act karta hai (launch mein gravity)	Yeh claim karna ki saara momentum conserved hai jab sirf $P_{x}$ hai
C. Boundary choice flips the answer	Same physics, lekin momentum conserved hai ya nahi yeh depend karta hai tum system kahan draw karte ho	Yeh sochna ki "conserved" nature ki property hai, tumhari choice ki nahi
D. Impulsive approximation	Collision + contact ke dauran ek background force (gravity, friction) act kar rahi hai	Impulses ke bajaye forces test karna
E. Degenerate / limiting mass	Ek mass $\to \infty$ (ek wall) ya $\to 0$	Yeh assume karna ki jab partner ek wall ho toh momentum "vanish" ho jaata hai
F. 2-D angled collision	Velocities angles par point karti hain; $x$ aur $y$ mein split karna zaroori hai	Speeds ko vector components ki jagah scalars ki tarah add karna
G. Momentum yes, energy no	Inelastic collision — dono laws alag se check karo	Yeh assume karna ki KE conserved hai kyunki $P$ hai
H. Exam twist: continuous mass — stream target se takraati hai, aur ek rocket mass kho raha hai	Mass, velocity nahi, woh hai jo change ho raha hai	Jab mass change ho raha ho tab $F = ma$ use karna

Ab hum aath cells mein chalte hain (Cell H ke do examples hain: ek inflow stream aur ek mass-losing rocket).

Example 1 — Cell A: zero total, opposite signs (explosion)

Worked example Rest mein ek firecracker do tukdon mein toot jaata hai

Ek stationary firecracker jiska mass $5$ kg hai, frictionless table par do tukdon mein phoot jaata hai. Ek $2$ kg ka tukda $30$ m/s par daayein ud jaata hai. $3$ kg ke tukde ki velocity nikalo.

Forecast: Doosra tukda baayi taraf jaana chahiye. $30$ m/s se tez ya dheema? (Woh bhaari hai, toh guess dheema.)

Step 1 — System choose karo aur external force check karo. System = dono tukde. Phodhne wali force tukdon ke beech hai ⇒ internal. Table frictionless hai; gravity aur normal vertical hain aur cancel kar dete hain. Toh horizontal $F_{e x t} = 0$ . Yeh step kyun? Sirf $F_{e x t, x} = 0$ confirm karne ke baad hi hum $P_{x}$ = constant likhne ke haqdar hain.

Step 2 — Initial aur final $P_{x}$ likho. Pehle rest mein: $P_{x, i} = 0$ . Daayein $= +$ lo. $0 = (2) (+ 30) + (3) (v)$ Yeh step kyun? Conservation ka matlab hai "pehle = baad mein"; hum unhe equal set karte hain.

Step 3 — Solve karo. $v = \frac{- 60}{3} = - 20 m/s \Rightarrow 20 m/s baayi taraf .$ Yeh step kyun? Minus sign physics hai — yeh humein direction batata hai, forecast se match karta hua.

Verify karo: Total after $= (2) (30) + (3) (- 20) = 60 - 60 = 0 =$ total before. ✓ Units: kg·m/s throughout. Bhaari tukda dheema (20 < 30), forecast ke anusaar. ✓

Example 2 — Cell B: sirf ek axis par conservation

Worked example Zameen par cannon (revisited, dono axes check kiye gaye)

Ek cannon ( $M = 400$ kg) ek ball ( $m = 4$ kg) ko $200$ m/s par horizontally fire karta hai. Recoil speed nikalo, aur explicitly dikhao ki kaun sa axis conserved hai aur kaun sa nahi.

Forecast: Horizontal safe hai (bahar se kuch bhi sideways push nahi karta). Vertical nahi (gravity, normal). Guess recoil $\approx 2$ m/s.

Step 1 — Forces ko axis se split karo. Firing force = internal. External: gravity (neeche) aur ground normal (upar) — dono vertical. Toh $F_{e x t, x} = 0$ lekin $F_{e x t, y} \neq = 0$ generally. Yeh step kyun? Newton's law ek equation per axis hai; hum $x$ aur $y$ ko alag treat karte hain.

Step 2 — Sirf $P_{x}$ conserve karo. Daayein $= +$ , pehle rest mein: $0 = M (- V) + m v \Rightarrow V = \frac{m v}{M} = \frac{4 \cdot 200}{400} = 2 m/s .$ Yeh step kyun? Hum $P_{y}$ conservation line likhne se forbidden hain — ground vertical impulse supply karta hai. Humein simply $y$ ki zaroorat nahi.

Verify karo: $P_{x, after} = 400 (- 2) + 4 (200) = - 800 + 800 = 0$ . ✓ Figure mein, pink (ball) aur blue (cannon) horizontal arrows equal-and-opposite hain; green vertical ground arrow dikhata hai ki $P_{y}$ conserved kyun nahi hai. Units: kg·m/s. ✓

Example 3 — Cell C: boundary choice answer flip kar deta hai

Worked example Ek ball zameen se bounce karti hai

Ek $0.5$ kg ki ball zameen se $6$ m/s neeche ki taraf jaate hue takraati hai aur $4$ m/s upar ki taraf rebound karti hai. (a) Kya uski momentum conserved hai? (b) Zameen ne kaunsa impulse deliver kiya? (c) Kis system ke liye momentum conserved hai?

Forecast: Ball akele ke liye, momentum clearly change hua (woh reverse ho gaya) — toh conserved nahi. Zameen ne impulse supply kiya hoga.

Step 1 — System = ball akela. Upar $= +$ lo. Pehle: $p_{i} = 0.5 (- 6) = - 3$ kg·m/s. Baad mein: $p_{f} = 0.5 (+ 4) = + 2$ kg·m/s. Yeh step kyun? Ball akele ke saath, gravity aur zameen ka normal dono external hain — toh hum expect karte hain ki $P$ change hoga, aur hum compute karte hain kitna.

Step 2 — Change se Impulse. $J = Δ p = p_{f} - p_{i} = 2 - (- 3) = + 5 N\cdots (upar ki taraf) .$ Yeh step kyun? $Δ P = J_{e x t}$ — change hi impulse hai. Yeh Impulse–Momentum Theorem kaam kar raha hai.

Step 3 — System ko ball + Earth tak bada karo. Ab zameen ka push aur gravity internal hain (Earth andar hai). Newton's 3rd law se woh cancel ho jaate hain ⇒ total $P$ (ball + Earth) conserved hai. Earth ek unmeasurably tiny amount se recoil karta hai. Yeh step kyun? Punchline dikhata hai: "conserved" tumhari chosen boundary ke baare mein ek statement hai, nature ke baare mein nahi.

Verify karo: $J = + 5$ N·s positive hai (upar ki taraf), ek aise zameen se consistent jo upar push karti hai. Magnitude sanity: $∣ J ∣ = 5 > ∣ p_{i} ∣ = 3$ kyunki ball sirf ruki nahi ( $+ 3$ ) balki upar bhi throw hui ( $+ 2$ ). ✓

Example 4 — Cell D: impulsive approximation (gravity ke neeche collision)

Worked example Do pucks mid-air mein collide karte hain

Do pucks (combined mass $2$ kg) ek $Δ t = 2$ ms tak chalne wale contact ke dauran collide karte hain. Is hit mein ek typical contact impulse lagbhag $5$ N·s hai. Dikhao ki gravity ko collision ke dauran ignore kiya ja sakta hai, aur tiny error quantify karo.

Forecast: Gravity contact ke dauran sirf $2$ ms tak act karti hai — uska impulse $5$ N·s ke saamne negligible hona chahiye.

Step 1 — Contact ke dauran gravity ka impulse compute karo. $J_{g} = m g Δ t = 2 \times 9.8 \times 0.002 = 0.0392 N\cdots .$ Yeh step kyun? Sahi test impulses compare karta hai, forces nahi — impulse hi woh hai jo $P$ change karta hai.

Step 2 — Ratio banao. $\frac{J _{g}}{J _{c}} = \frac{0.0392}{5} \approx 0.0078 = 0.78%.$ Yeh step kyun? $1%$ se neeche ka ratio matlab hai ki impact ke through $P$ conserved treat karne se ek percent se kam error aata hai.

Step 3 — Two-phase strategy. $P$ collision ke through conserve karo (gravity ignore karo), phir free flight mein pehle aur baad ke liye gravity ko wapas include karo. Yeh step kyun? Approximation actually aise hi use hoti hai: yeh collision ko ek instantaneous event ke roop mein isolate karta hai.

Verify karo: $J_{g} = 0.0392$ N·s aur ratio $= 0.00784 < 0.01$ . ✓ Units: (kg)(m/s²)(s) = kg·m/s = N·s. ✓

Example 5 — Cell E: degenerate mass (wall se bounce karna)

Worked example Ek ball elastically ek fixed wall se bounce karti hai

Ek $0.2$ kg ki ball ek rigid wall se $10$ m/s par takraati hai aur $10$ m/s par rebound karti hai (elastic). Kya ball ki momentum conserved hai? Kya ball + wall ke liye momentum conserved hai? Limiting picture kya hai jab wall ka mass $\to \infty$ ?

Forecast: Ball ka momentum sign flip karta hai — ball akele ke liye conserved nahi. Wall (Earth se attached) impulse absorb karta hai.

Step 1 — Ball akela. Incoming direction $= +$ lo. Pehle $p_{i} = 0.2 (+ 10) = + 2$ ; baad mein $p_{f} = 0.2 (- 10) = - 2$ . $Δ p = - 2 - 2 = - 4 N\cdots .$ Yeh step kyun? Wall ka normal force ball ke liye external hai, toh nonzero $Δ p$ expected hai.

Step 2 — Limit "wall = infinite mass". Ball+wall ka momentum conservation kehta hai wall $+ 4$ N·s gain karta hai. Uska velocity change hai $Δ V = J / M_{w a l l} = 4/ M_{w a l l}$ . Jaise $M_{w a l l} \to \infty$ , $Δ V \to 0$ : wall still raha phir bhi momentum carry karta hai. Yeh step kyun? Yeh degenerate case hai — "wall" actually itna massive object hai ki uska velocity change negligible hai, bilkul Example 3 mein Earth ki tarah.

Verify karo: $lim_{M \to \infty} 4/ M = 0$ . ✓ Ball ki KE pehle $= \frac{1}{2} (0.2) (1 0^{2}) = 10$ J, baad mein $= 10$ J — elastic, KE conserved, speed unchanged se consistent. ✓

Example 6 — Cell F: 2-D angled collision (components resolve karne zaroori hain)

Worked example Frictionless table par glancing collision

Ball A ( $1$ kg) $4$ m/s par daayein move karta hai aur stationary ball B ( $1$ kg) se takraata hai. Impact ke baad, A original line se $3 0^{\circ}$ upar ki taraf $23$ m/s par move karta hai. B ka velocity vector nikalo.

Forecast: $P_{y} = 0$ rakhnae ke liye B ko "missing" sideways momentum neeche carry karna hoga, aur forward momentum share karna hoga.

Step 1 — External force check karo, axes pick karo. Frictionless table ⇒ koi horizontal external force nahi ⇒ dono $P_{x}$ aur $P_{y}$ conserved hain. $x$ = A ki original direction lo. Yeh step kyun? Horizontal frictionless table par poora momentum vector conserved hota hai — dono axes valid hain.

Step 2 — $P_{x}$ conserve karo. Pehle: $(1) (4) + 0 = 4$ . $4 = (1) (23 cos 3 0^{\circ}) + (1) v_{B x} = 23 \cdot \frac{3}{2} + v_{B x} = 3 + v_{B x} \Rightarrow v_{B x} = 1.$ Yeh step kyun? $x$ -components resolve karne ke baad ordinary numbers ki tarah add hote hain.

Step 3 — $P_{y}$ conserve karo. Pehle: $0$ . $0 = (1) (23 sin 3 0^{\circ}) + (1) v_{B y} = 23 \cdot \frac{1}{2} + v_{B y} = 3 + v_{B y} \Rightarrow v_{B y} = - 3 .$ Yeh step kyun? Initial sideways momentum zero tha, toh dono final sideways parts cancel karne chahiye — B neeche jaata hai.

Step 4 — Assemble karo. $v_{B} = (1, - 3)$ , speed $= 1 + 3 = 2$ m/s angle $arctan (3 /1) = 6 0^{\circ}$ line se neeche par. Yeh step kyun? Components ko wapas magnitude-and-direction answer mein convert karo.

Verify karo: $P_{x}$ : $3 + 1 = 4$ ✓. $P_{y}$ : $3 - 3 = 0$ ✓. Figure mein pink (A') aur blue (B') arrows tip-to-tail original yellow arrow rebuild karte hain — yeh momentum conservation ek vector triangle ke roop mein draw hua hai. Note karein A' $3 0^{\circ}$ par aur B' $6 0^{\circ}$ par perpendicular hain — equal-mass elastic hit ki pehchaan. ✓

Example 7 — Cell G: momentum yes, kinetic energy no (perfectly inelastic)

Worked example Do clay lumps ek saath chipak jaate hain

Ek $3$ kg ka lump $4$ m/s par daayein jaata hua ek $1$ kg ke lump se takraata hai jo $4$ m/s par baayi taraf ja raha hai. Woh chipak jaate hain. Common velocity nikalo, phir kinetic energy check karo.

Forecast: Bhaari lump jeetta hai ⇒ woh daayein move karte hain, lekin $4$ se dheema. KE girega (clay deform hoga).

Step 1 — $P_{x}$ conserve karo (frictionless, koi external horizontal force nahi). Daayein $= +$ : $3 (4) + 1 (- 4) = (3 + 1) v_{f} \Rightarrow 12 - 4 = 4 v_{f} \Rightarrow v_{f} = 2 m/s (daayein) .$ Yeh step kyun? Chipakna ek internal force hai, toh $P$ conserved hai chahe kuch bhi ho.

Step 2 — Kinetic energy pehle/baad compare karo. $K E_{i} = \frac{1}{2} (3) (4^{2}) + \frac{1}{2} (1) (4^{2}) = 24 + 8 = 32 J, K E_{f} = \frac{1}{2} (4) (2^{2}) = 8 J .$ Yeh step kyun? Momentum aur energy ke alag conditions hain — hum energy alag se test karte hain (dekho Elastic vs Inelastic Collisions).

Step 3 — Loss interpret karo. $Δ K E = 8 - 32 = - 24$ J, heat aur deformation mein gaya. $P$ phir bhi conserved; KE nahi. Yeh step kyun? Parent ki steel-manned galti "momentum aur energy ek saath conserved ya nahi" ko directly refute karta hai.

Verify karo: $P_{i} = 12 - 4 = 8 = P_{f} = 4 (2) = 8$ ✓. $K E_{f} = 8 < 32 = K E_{i}$ ✓. KE ka fraction lost $= 24/32 = 75%$ . ✓

Example 8 — Cell H exam twist (part 1): continuous inflow (ek stream ek plate se takraati hai)

Is example se pehle, ek naya symbol.

Worked example Paani ki ek stream plate se takraati hai

Paani ek hose se $20$ m/s par nikalti hai aur ek wall se horizontally $3$ kg/s ki rate se takraati hai, phir seedha neeche gir jaati hai (koi rebound nahi). Wall par horizontal force nikalo.

Forecast: Force = per second destroy hone wale momentum ki rate. Guess lagbhag $3 \times 20 = 60$ N.

Step 1 — Per second aane wala momentum. Ek second mein, $\overset{m}{˙} = 3$ kg paani horizontal velocity $20$ m/s carry karte hue aata hai: $\frac{Δ p _{x}}{Δ t} = \overset{m}{˙} v = 3 \times 20 = 60 kg\cdotm/s per s .$ Yeh step kyun? Jab mass flow karta hai, sahi tool hai $F = \frac{d p}{d t} = \overset{m}{˙} v$ , na ki $F = ma$ (mass fixed nahi hai).

Step 2 — Saara horizontal momentum destroy ho jaata hai. Paani zero horizontal velocity se nikalta hai (woh bas neeche slide karta hai), toh uska horizontal momentum $\overset{m}{˙} v$ se $0$ per second drop karta hai. $F_{on water} = \frac{Δ p _{x}}{Δ t} = 0 - 60 = - 60 N .$ Yeh step kyun? Question horizontal force ke baare mein poochhta hai, toh hum sirf horizontal momentum track karte hain; wall ko exactly utna impulse per second supply karna hoga jo incoming horizontal momentum ko khatam kare. Minus sign kehta hai wall paani ko peeche push karta hai (uski motion ke against).

Step 3 — Newton's 3rd law: wall par force. Wall paani ko $- 60$ N se push karta hai; Newton's Third Law se paani wall ko opposite force se push karta hai: $F_{on wall} = + 60 N (stream ki direction mein) .$ Yeh step kyun? Question wall par force ke baare mein poochha tha, jo paani par force ka reaction hai — toh hum sign flip karte hain.

Verify karo: $∣ F_{on wall} ∣ = \overset{m}{˙} v = 3 (20) = 60$ N, forecast "lagbhag 60 N" se match karta hai. Sign positive hai — wall uss taraf push hota hai jis taraf paani ja raha tha, bilkul jaise ek hose visibly ek plate ko shove karta hai. Units: (kg/s)(m/s) = kg·m/s² = N. ✓

Example 9 — Cell H exam twist (part 2): continuous outflow (ek rocket mass khota hai)

Stream example ne ek target mein mass add kiya. Mirror case ek rocket hai: yeh khud ko aage push karne ke liye mass door phenkta hai. Same tool ( $F = \overset{m}{˙} v$ ), opposite bookkeeping.

Worked example Ek hovering rocket

Ek rocket $\overset{m}{˙}_{ex} = 4$ kg/s par fuel jaalaata hai, exhaust ko apne relative $u = 500$ m/s par eject karta hai. (a) Thrust nikalo. (b) Is waqt rocket + unburnt fuel ka mass $180$ kg hai. Kya woh lift off kar sakta hai ( $g = 9.8$ m/s²)? (c) Uska instantaneous upward acceleration kya hai?

Forecast: Thrust $= 4 \times 500 = 2000$ N. Weight $\approx 180 \times 9.8 \approx 1764$ N. Thrust weight se zyaada hai, toh haan — lift hoga, modest acceleration ke saath.

Step 1 — Outflow rule se Thrust. $F_{thrust} = u \overset{m}{˙}_{ex} = 500 \times 4 = 2000 N (upar ki taraf) .$ Yeh step kyun? Exhaust mein momentum $u \overset{m}{˙}_{ex}$ rate se daala ja raha hai; Newton's 3rd law se us hi size ka reaction rocket drive karta hai. Hum $F = \overset{m}{˙} v$ use karte hain, na ki $F = ma$ , kyunki rocket ka mass gir raha hai.

Step 2 — Thrust ko weight (yahan external force) se compare karo. $W = M g = 180 \times 9.8 = 1764 N (neeche) .$ Rocket body par net upward force: $F_{net} = 2000 - 1764 = 236 N$ . Yeh step kyun? Gravity rocket system par sachchi external force hai; thrust internal ejection reaction hai. Lift-off ke liye thrust > weight chahiye.

Step 3 — Instantaneous acceleration. Is instant par $F_{net} = M a$ apply karo (mass abhi $180$ kg hai): $a = \frac{F _{net}}{M} = \frac{236}{180} \approx 1.31 m/s^{2} upar ki taraf .$ Yeh step kyun? Ek baar hamein net force mil gayi, ordinary Newton's 2nd law us instant ka acceleration deta hai — mass-change already thrust term ke andar account ho chuka hai.

Verify karo: Thrust $2000$ N $>$ weight $1764$ N ⇒ lift off. ✓ $a = 236/180 = 1.3 \overline{1}$ m/s² $> 0$ , rise karne se consistent. Units: thrust (m/s)(kg/s) = kg·m/s² = N ✓; $a$ N/kg = m/s² mein ✓. Sanity: exhaust ( $500$ m/s) yahan kisi bhi rocket speed se kaafi zyaada hai, toh $u$ ko ejection speed treat karna valid hai.

Recall Matrix ke across quick self-test

Rest se explosion — doosre piece ki velocity ka sign? ::: Pehle ke opposite (total zero rehta hai). Cannon horizontally fire karta hai — kaun sa momentum component conserved hai? ::: Sirf horizontal; vertical gravity aur normal se toot jaata hai. Ball floor se bounce karti hai — ball akele ke liye conserved? ::: Nahi; ball + Earth ke liye haan. Collision mein gravity ignore karne ka sahi test? ::: Impulses compare karo, forces nahi. Fixed wall se ball jab $M_{w a l l} \to \infty$ — wall ka velocity change? ::: Zero, phir bhi woh momentum carry karta hai. Sticking (inelastic) collision — kya KE conserved hai? ::: Nahi; sirf momentum hai. Wall se takraati water stream — force ke liye kaun sa formula? ::: $F = \overset{m}{˙} v$ , $F = ma$ nahi. Exhaust eject karne se rocket thrust — formula? ::: $F_{thrust} = u \overset{m}{˙}_{ex}$ (exhaust speed × burn rate).

Connections

Parent topic — woh rule jise yeh examples exercise karte hain.
Conservation of Linear Momentum — $F_{e x t} = 0$ special case (Examples 1, 6, 7).
Impulse–Momentum Theorem — Examples 3, 4, 8 ko power karta hai.
Newton's Third Law — kyun internal/stream/exhaust forces pair up karte hain (Examples 1, 8, 9).
Elastic vs Inelastic Collisions — Example 7 mein separate KE test.
Center of Mass Motion — same master equation, restated.