Visual walkthrough — System with external forces — conditions for conservation

1.4.4 · D2 · Physics › Momentum & Collisions › System with external forces — conditions for conservation

Har symbol ko pehle banate hain, phir use karte hain. Agar aapne kabhi kisi push ke liye arrow nahi dekha, Step 1 se shuru karo aur skip mat karo.

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Step 1 — Ek akele arrow ka matlab (momentum aur force as arrows)

KYA. Kisi bhi equation se pehle, hum picture pe agree karte hain. Ek vector bas ek arrow hai: uski lambai batati hai "kitna," uski direction batati hai "kidhar." Hum ek akeli chalti hui ball ke liye do tarah ke arrow draw karte hain:

• $\vec{p}$ — momentum arrow. Yeh us direction mein point karta hai jis taraf ball chal rahi hai, aur yeh lamba hota hai jab ball bhaari ya tez ho. Symbols mein $\vec{p}=m\vec{v}$, jahan $m$ mass hai (ek seedha number, "kitna stuff") aur $\vec{v}$ velocity arrow hai (kitni tez, kidhar).
• $\vec{F}$ — force arrow. Yeh ball par push ya pull hai.

YEH DONO KYO. Newton ne discover kiya ki force kisi cheez ki jagah ke baare mein nahi hai, balki is baare mein hai ki uska momentum arrow kaise badlta hai. Isliye pehle momentum ko arrow ke roop mein draw karna zaroori hai, phir force ko us cheez ke roop mein jo use khainchti ya ghumaati hai.

PICTURE. Amber ball ko dekho. Uska cyan momentum arrow $\vec{p}$ daayein point karta hai. Safed force arrow $\vec{F}$ use halka sa dhakka deta hai. Thodi si der baad momentum arrow $\vec{F}$ ki direction mein thoda sa badhh jaata hai — woh change hi dashed arrow $d\vec{p}$ hai.

Recall

$\frac{d\vec{p}}{dt}$ sahi tool kyun hai aur $\Delta \vec{p}$ kyun nahi? Kyunki forces har instant mein alag ho sakti hain. ::: Ek derivative har moment par momentum change rate deta hai; ek seedha difference $\Delta\vec p$ sirf poore interval ka total deta aur yeh nahi batata ki force moment-to-moment kaise act karti rahi.

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Step 2 — Ek system draw karna: kaafi saari balls, ek boundary

KYA. Ek system bas un objects ka group hai jinhe hum circle karna choose karte hain. Hum un balls ke around ek dashed loop — boundary — draw karte hain jinki hume parwah hai. Andar sab kuch "hum" hain; bahar sab kuch "baaki duniya" hai.

Balls ko $1,2,3,\dots$ label karte hain. Ball number $i$ ka mass $m_i$, velocity ${\vec{v}}_i$, aur momentum ${\vec{p}}_i=m_i{\vec{v}}_i$ hai.

KYO. Poora subject isi loop par tika hua hai. Koi force "internal" hai ya "external" — yeh force ki property nahi hai — yeh poori tarah depend karta hai ki dashed loop kahan draw ki gayi. Isliye hum ise pehle draw karte hain, koi bhi force judge karne se pehle.

PICTURE. Teen amber balls ek dashed cyan loop ke andar baithi hain. Total momentum teen individual arrows ka tip-to-tail sum hai — lamba safed arrow $\vec{P}$ ke roop mein dikhaya gaya hai.

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Step 3 — Har force ko sort karna: internal twins vs bahar ke pushes

KYA. Ab hum ball $i$ par push karne waale har arrow ko dekhte hain aur har ek ko ek box mein daalte hain:

• Agar push kisi doosri ball $j$ se aati hai jo loop ke andar hai, toh use internal force ${\vec{f}}_{ij}$ kehte hain ("$j$ se $i$ par force").
• Agar push loop ke bahar se aati hai (gravity, ek haath, ek deewar), toh use external force ${\vec{F}}_i^{ext}$ kehte hain.

Koi teesra box nahi. Har push ka ek source hota hai; woh source ya toh loop ke andar hai ya bahar.

KYO. Hum yeh prove karne wale hain ki andar ke pushes secretly cancel ho jaate hain aur sirf bahar wale bachte hain. Uske liye hume pehle unhe kaagaz par saaf alag karna hoga.

PICTURE. Ball $1$ par hum ek cyan internal arrow draw karte hain (ball $2$ se) aur ek amber external arrow (gravity, dashed boundary ke bahar se aata hua). Forces ko colour-code kiya gaya hai ki woh kis box se belong karte hain.

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Step 4 — Poore system par Newton ka niyam add karo

KYA. Har ball ke liye ${\vec{F}}_i=\frac{d{\vec{p}}_i}{dt}$ likho, phir un saari equations ko ek saath add karo ($i$ par sum karo).

KYO. Hum poore group ke momentum $\vec{P}$ ke baare mein ek niyam chahte hain, har ball alag nahi. Per-ball laws add karna exactly wahi hai jisse individual ${\vec{p}}_i$ mili $\vec{P}$ mein ghul jaati hain.

PICTURE. Teen stacked rows, ek per ball, har row ek equation "force = us ball ke arrow ke change ki rate." Daayein ek badi cyan brace teeno $\frac{d{\vec{p}}_i}{dt}$ ko ek akele $\frac{d\vec{P}}{dt}$ mein collect karti hai, kyunki ek sum ki derivative derivatives ka sum hoti hai.

${\sum }_i{\vec{F}}_i=\underset{\text{all internal}}{\underbrace{\sum _i\sum _{j\ne i}{\vec{f}}_{ij}}}+\underset{\text{all external}}{\underbrace{\sum _i{\vec{F}}_i^{ext}}}=\frac{d}{dt}{\sum }_i{\vec{p}}_i=\frac{d\vec{P}}{dt}$

• ${\sum }_i{\vec{F}}_i$ — saari balls par saari pushes ka total.
• ${\sum }_i{\sum }_{j\ne i}{\vec{f}}_{ij}$ — andar ki pushes ka poora dher.
• ${\sum }_i{\vec{F}}_i^{ext}$ — baahr ki pushes ka poora dher.
• $\frac{d\vec{P}}{dt}$ — group ka combined arrow kaise badlta hai. Isliye hum sum ko derivative ke andar le ja sakte hain: (arrow 1 + arrow 2 + arrow 3) ki rate-of-change equals (arrow 1 ki rate) + (arrow 2 ki rate) + ..., isliye sum aur $d/dt$ aazaadi se jagah badal sakte hain.

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Step 5 — Jaadu wali cancellation (Newton ka Third Law internal pile ko khatam karta hai)

KYA. Andar ki koi bhi pair of balls lo, maano $1$ aur $2$. Ball $2$, ball $1$ ko ${\vec{f}}_{12}$ se push karti hai. Ball $1$ wapas ball $2$ ko ${\vec{f}}_{21}$ se push karti hai. Newton's Third Law kehta hai ye lambai mein barabar, direction mein ulte hote hain: ${\vec{f}}_{12}=-{\vec{f}}_{21}$ Isliye jab dono giant internal pile mein aate hain, toh woh ek zero arrow mein add ho jaate hain. Har internal push ka ek twin hota hai; har twin use cancel kar deta hai.

KYO. Yahi woh reason hai ki internal forces — chahe kitni bhi violent ho, jaise ek explosion — group ka total arrow nahi hila sakti. Woh hamesha equal-aur-opposite pairs mein aati hain jo sum mein ek doosre ko mita deti hain.

PICTURE. Do amber balls; har ek se ek cyan arrow doosre ko push karta hua, bilkul barabar aur ulta drawn. Neeche, do arrows tip-to-tail rakh kar ek single point mein collapse ho jaate hain — zero arrow $\vec{0}$.

${\sum }_i{\sum }_{j\ne i}{\vec{f}}_{ij}=\vec{0}$

• Double sum har ordered pair par run karta hai, isliye ${\vec{f}}_{12}$ aur ${\vec{f}}_{21}$ dono usme hain.
• Twins ke roop mein group karo $({\vec{f}}_{12}+{\vec{f}}_{21})$, har twin ek zero arrow hai, isliye poori pile $\vec{0}$ hai.

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Step 6 — Master equation saamne aati hai

KYA. Internal pile khatam ho gayi ($=\vec{0}$), daayein sirf external pile khadi reh jaati hai: $\frac{d\vec{P}}{dt}={\sum }_i{\vec{F}}_i^{ext}={\vec{F}}_{ext}^{net}$

KYO. Yeh pehle paanch steps ka fayda hai. Group ka total momentum arrow exactly utni hi rate se badlta hai jitni net bahar ki push hai — koi bhi internal cheez kabhi enter nahi karti.

PICTURE. Wahi dashed loop, lekin ab sirf amber external arrows bachte hain, sab milakar ek bold amber arrow ${\vec{F}}_{ext}^{net}$ ban jaata hai, badhte safed arrow $\vec{P}$ ke bilkul paas baith ke. Internal cyan arrows ghosts mein fade ho jaate hain yeh dikhane ke liye ki woh out ho gaye.

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Step 7 — Edge case A: ek axis par jeetna doosre par haarte hue

KYA. Ek arrow equation secretly do equations hoti hain: ek horizontal ($x$) part ke liye, ek vertical ($y$) part ke liye. Woh independently chalta hai: $\frac{d{P}_x}{dt}=F_{ext,x}^{net},\frac{d{P}_y}{dt}=F_{ext,y}^{net}$

Isliye chahe bahar ki push ho bhi, agar uska horizontal part zero hai, toh $P_x$ phir bhi conserved hai.

KYO. Yahi cannon hai. Gravity seedha neeche point karti hai — uska horizontal shadow zero hai. Isliye (cannon + ball) ka sideways momentum safe hai chahe up-down momentum nahi.

PICTURE. Ek cannon ek ball ko daayein fire karta hai; recoil cannon ko baayein bhejta hai. Gravity (amber, neeche) ka koi horizontal component nahi hai (uska $x$-shadow ek dot hai). Do horizontal momentum arrows barabar aur ulte hain — unka $x$-sum pehle aur baad mein zero hi rehta hai.

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Step 8 — Edge case B: impulsive approximation (choti collisions)

KYA. Jo cheez actually momentum badlati hai woh hai impulse = force $\times$ time, likha jaata hai $\vec{J}=\int \vec{F}dt$ (Impulse–Momentum Theorem). Ek tiny $\Delta t$ lasting collision ke dauran, gravity jaise ek finite bahar ki force almost-zero impulse deliver karti hai: ${\vec{J}}_{ext}=\int {\vec{F}}_{ext}dt\approx {\vec{F}}_{ext}\Delta t\to \vec{0}(\Delta t\to 0)$

KYO. Hit ke dauran contact force bahut badi lekin brief hoti hai; gravity halki hoti hai. Unka time wahi tiny sliver hai, isliye gravity ka impulse bahut chhota pad jaata hai. Hum $\vec{P}$ ko impact ke through conserved treat karte hain, phir slow flight ke liye gravity wapas add karte hain.

PICTURE. Ek hi axis par do force-vs-time graphs: ek tall thin cyan spike (contact force, bahut bada lekin $2$ ms ke liye) aur ek chhoti flat amber line (gravity). Har ek ke neeche area impulse hai — spike ka area bada hai, gravity ka area barely visible sliver hai.

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Step 9 — Edge case C: boundary sab decide karti hai (girती ball)

KYA. Wahi girती ball, do alag loops:

• Sirf ball ke around loop: gravity (Earth se) aur floor ki push loop ke bahar hain ⇒ external ⇒ $\vec{P}$ conserved nahi (bounce par ulta ho jaata hai).
• Ball + Earth ke around loop: wahi forces ab loop ke andar hain ⇒ internal ⇒ $\vec{P}$ conserved hai (Earth recoil karta hai, bahut hi halka sa).

KYO. "External" kabhi akeli force ki property nahi hoti — yeh force ki property hai aapki loop ke relative. Loop dobara khaincho, aur wahi force boxes badal leti hai.

PICTURE. Baayi panel: ball ke around chhoti dashed loop; gravity arrow boundary ko pierce karta hai (external) — momentum arrow bounce ke baad upar flip ho jaata hai. Daayein panel: ball aur Earth dono ko andar leti badi loop; wahi force ab ek internal twin pair hai — combined arrow unchanged hai. Yahi idea hai Center of Mass Motion ki: koi external push nahi, center of mass seedha chalta rehta hai.

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Ek-picture summary

Upar sab kuch ek single flow mein compress kiya: har ball par force → internal + external mein split → sum → internal twins cancel → sirf net external bachta hai → agar woh zero hai, $\vec{P}$ frozen hai.

Recall Poore walkthrough ki Feynman-style retelling

Ek bag of marbles ki picture karo. Har marble ke paas ek chhota arrow hai jo dikhata hai ki woh kaise chal raha hai; saare arrows ko tip-to-tail add karo aur poore bag ke liye ek bada arrow milega — yahi total momentum hai. Ab marbles bag ke andar ek doosre se takraate hain: har takkar ek do-taraf ki shove hai, "main tumhe push karta hoon, tum mujhe bilkul utni hi force se wapas push karo doosri taraf." Woh cancel ho jaate hain — isliye duniya ki saari andar ki takkar bag ke bade arrow ko nahi hila sakti. Jo cheez ise hila sakti hai woh hai bahar se ek shove — ek haath, gravity, ek deewar. Isliye bada arrow exactly tab frozen rehta hai jab bahar se koi nahi dhakela. Aur teen bonus tricks: (1) agar bahar ki shove sirf sideways hai, arrow ka aage wala part phir bhi frozen hai; (2) agar ek collision ek pal mein khatam ho jaaye, gravity ko shove karne ka bahut kam time milta hai, isliye crash ke through arrow almost frozen hai; (3) koi force "bahar" maani jaati hai ya nahi yeh poori tarah depend karta hai ki aapne bag kahan khainchi — bag badi khaincho aur kal ki bahar ki shove aaj ki andar ki takkar ban jaati hai.

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Connections

• Parent topic — full note
• Newton's Third Law — Step 5 mein twin-cancellation.
• Impulse–Momentum Theorem — Step 8 mein area-under-the-curve idea.
• Center of Mass Motion — wahi master equation Step 9 mein dobara bataya gaya.
• Elastic vs Inelastic Collisions — alag energy condition.
• Conservation of Linear Momentum — special case ${\vec{F}}_{ext}^{net}=\vec{0}$.