Visual walkthrough — Conservation of linear momentum — derivation from Newton's third law

1.4.3 · D2 · Physics › Momentum & Collisions › Conservation of linear momentum — derivation from Newton's t

Hum bilkul base se shuru karte hain: ek arrow ka kya matlab hai, "momentum" ek arrow ke roop mein kya hota hai, "rate of change" kaisa dikhta hai, aur tabhi kyun total stable rehta hai.

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Step 1 — Ek arrow (ek vector) kya hota hai, aur momentum kya hai?

Arrows ko push karne se pehle, hume un do arrows ki zaroorat hai jo yahan matter karte hain.

• $\vec{v}$ — velocity arrow: point karta hai jahan object move karta hai, length = kitna fast (metres per second).
• $m$ — mass: ek plain number (kilograms), koi direction nahi. Yeh kehta hai ki object ko dhakkelna kitna mushkil hai.

WHAT humne kiya: momentum ko ek stretched velocity arrow ke roop mein define kiya. WHY: momentum, velocity nahi, woh quantity hai jo conserved hoti hai — kyunki Newton's laws likhte hain ki momentum kaise change hota hai. PICTURE: neeche, thin velocity arrow $\vec{v}$ thick momentum arrow $\vec{p}=m\vec{v}$ mein stretch ho jaata hai.

Yahan $m\vec{v}$ ka matlab hai: $\vec{v}$ ke direction mein same, length $m$ se multiply hui. Symbol $m$ stretching kar raha hai; arrow $\vec{v}$ pointing kar raha hai.

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Step 2 — Do particles jo sirf ek doosre ko push karte hain

Naye symbols, har ek earn kiya hua:

• ${\vec{F}}_{12}$ — woh force 1 par, 2 se (subscript padho "on–from").
• ${\vec{F}}_{21}$ — woh force 2 par, 1 se.

WHAT: humne scene set kiya — do objects, unke beech do forces. WHY: momentum conservation ek closed room ke baare mein ek statement hai. Agar koi bahar se andar aata hai, to story badal jaati hai (Step 8 mein handle karenge). PICTURE: do balls, har ek ke paas ek force arrow jo doosre ki taraf point karta hai.

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Step 3 — Newton's 2nd law: force = momentum arrow kitni fast change hota hai

WHY yeh tool, plain $\vec{F}=m\vec{a}$ nahi? Kyunki hum momentum $\vec{p}$ ke baare mein ek statement chahte hain, aur yeh form kehta hai ki force directly hai $\vec{p}$ ki rate of change. Yeh exactly woh quantity hai jise hum track karne ki koshish kar rahe hain. (Dekho Newton's Second Law (momentum form).)

WHAT: humne har push ko har object ke momentum change se link kiya. PICTURE: 1 par force arrow bilkul "kidhar 1 ka momentum grow karta hai" arrow ke upar baithta hai — same arrow.

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Step 4 — Newton's 3rd law: do pushes mirror images hain

WHAT: humne kaha ki do forces equal length, opposite direction hain. WHY: yeh poore proof ka engine hai. Iske bina, do forces cancel nahi hote aur momentum drift kar sakta tha. (Dekho Newton's Third Law.) PICTURE: do arrows ek jaise length ke, bilkul opposite point karte hue. Agar aap ek ko doosre ke upar tip-to-tail slide karo, aap wapas wahin pahunch jaoge jahan se shuru kiya tha — woh kuch bhi sum nahi karte.

Minus sign literally padho: "${\vec{F}}_{21}$ woh ${\vec{F}}_{12}$ hai jo ghuma diya gaya hai."

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Step 5 — Do equations add karo taaki twin forces milein

Hum Step 3 ki do equations stack karte hain aur unke left sides aur right sides add karte hain:

$\underset{\text{do pushes}}{\underbrace{{\vec{F}}_{12}+{\vec{F}}_{21}}}=\underset{\text{do momentum-change rates}}{\underbrace{\frac{d{\vec{p}}_1}{dt}+\frac{d{\vec{p}}_2}{dt}}}$

WHAT: humne do equations ko ek mein chipkaya, aur do momenta ko ek single total arrow $\vec{P}={\vec{p}}_1+{\vec{p}}_2$ mein bundle kiya. WHY: action–reaction twins ko left par side by side laane ke liye, cancel hone ke liye ready. PICTURE: tip-to-tail addition — ${\vec{p}}_1$ phir ${\vec{p}}_2$ — ek total arrow $\vec{P}$ deta hai; aur alag se do force arrows saath laaye gaye.

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Step 6 — Forces cancel hote hain; total still rehta hai

Step 4 ko Step 5 ke left side mein daalo:

${\vec{F}}_{12}+{\vec{F}}_{21}={\vec{F}}_{12}+(-{\vec{F}}_{12})=\vec{0}$

• ${\vec{F}}_{12}$ aur $-{\vec{F}}_{12}$ — same length, opposite way.
• $\vec{0}$ — zero arrow: koi length nahi, koi direction nahi. Kuch nahi.

To right side bhi zero hona chahiye:

$\frac{d}{dt}({\vec{p}}_1+{\vec{p}}_2)=\vec{0}$

$\boxed{{\vec{p}}_1+{\vec{p}}_2=\vec{P}=\text{constant}}$

WHAT: twin forces annihilate ho gaye, total-momentum change-rate ko zero force kiya. WHY: zero change-rate "same rehta hai" ka mathematical fingerprint hai. PICTURE: do force arrows ek dot mein collapse ho jaate hain ($\vec{0}$); total momentum arrow $\vec{P}$ pehle aur baad mein unchanged baithta hai.

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Step 7 — Ek real trade mein dekho: collision

Numbers se concrete hota hai. Cart 1 ($3$ kg at $4$ m/s) stationary cart 2 ($1$ kg) se takraata hai aur woh stick ho jaate hain (ek perfectly inelastic collision, dekho Elastic vs Inelastic Collisions).

$m_1{v}_1+m_2{v}_2=(m_1+m_2)v^{\prime }$ $(3)(4)+(1)(0)=(4)v^{\prime }\Rightarrow v^{\prime }=3\text{m/s}$

• Pehle: total momentum arrow = $12\text{kg⋅m/s}$ right ki taraf (sab cart 1 mein).
• Baad: $12\text{kg⋅m/s}$ abhi bhi, ab $4$ kg pair mein shared.

WHAT: momentum redistribute hua — cart 1 ne cart 2 ko kuch diya — lekin total arrow ki length untouched hai. WHY: exactly Step 6 action mein. Contact forces ek internal action–reaction pair hain. PICTURE: total arrow pehle = total arrow baad mein, chahe pieces ne kaise bhi regroup kiya ho.

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Step 8 — Edge cases: agar room closed nahi hai, ya kuch move nahi karta?

Hume har scenario cover karna hai taaki aap kabhi koi unseen na milein.

Case A — ek bahari push exist karta hai. Tab har object ek extra external force ${\vec{F}}^{\text{ext}}$ feel karta hai. Internal pair abhi bhi cancel karta hai, lekin leftovers nahi karte: $\frac{d\vec{P}}{dt}={\vec{F}}_{\text{net}}^{\text{ext}}$ Agar ${\vec{F}}_{\text{net}}^{\text{ext}}\ne \vec{0}$, to total arrow grow ya turn karta hai. Momentum conserved nahi hota. (Yeh Impulse–Momentum Theorem view hai.)

Case B — sab kuch shuru mein rest par hai. Tab $\vec{P}=\vec{0}$ pehle. Conservation force karta hai $\vec{P}=\vec{0}$ baad mein bhi — to jo bhi pieces fly apart karte hain unke momentum arrows cancel hone chahiye (ek gun ki bullet vs. uska recoil; ek exploding shell). Equal-and-opposite, guaranteed.

Case C — 2D motion. Ek arrow $x$-part aur $y$-part mein split hota hai. Kyunki $\vec{P}$ ek arrow ke roop mein constant hai, har part alag alag constant hai. $x$ aur $y$ ko independently conserve karo.

WHAT: humne teen tarike sweep kiye jisme reality clean proof se alag hoti hai. WHY: aapko ek aisa rule dene ke liye jo kabhi break na ho: pehle check karo ki bahari push zero hai ya nahi. PICTURE: teen mini-panels — (A) ek bahari arrow total ko tilt karta hai; (B) rest start se do opposite arrows; (C) ek slanted arrow $x$ aur $y$ mein toot ke har ek hold kiya gaya.

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Ek-picture summary

Upar sab kuch ek diagram mein compress kiya: equal-and-opposite pushes ⇒ woh zero arrow mein add ho jaate hain ⇒ total momentum arrow frozen hai.

Total-momentum arrow $\vec{P}$ pehle aur baad mein same length aur direction ka hai, chahe pieces kaise bhi rearrange ho jaayein — kyunki room ke andar ke sirf forces cancelling pairs mein aaye. Link back: $\vec{P}=M{\vec{v}}_{cm}$, to Centre of Mass Motion unchanged glide karta rehta hai.

Recall Feynman retelling — plain words mein poora walkthrough

Do kids ko ice par imagine karo, haath milaye. Har kid ka "motion-money" unka momentum arrow hai: woh kitne bhaare hain times kitna fast slide karte hain. Jab woh shove off karte hain, third law kehta hai har ek doosre ko exactly utna hi hard, exactly opposite way shove karta hai — mirror arrows. Newton's second law kehta hai ek shove woh speed hai jis par aapka motion-money change hota hai. Dono kids ki books add karo: unke do shoves mirror images hain, to milke woh kuch nahi add karte. Aur agar total ki change-rate kuch nahi hai, to total frozen hai. To kids alag fly karte hain, ek left ek right, lekin do motion-monies add up karke wahi hain jaise pehle tha — zero agar woh still shuru kiye. Grand total change karne ka ek hi tarika hai ice ke bahar se push — ek wall, ek rope, gravity. Bahar push nahi, change nahi. Yahi poori story hai, drawn.

Recall Newton's 3rd law exactly kahan use hua?

Step 4 (mirror arrows) Step 6 mein gaya, ${\vec{F}}_{12}+{\vec{F}}_{21}$ ko zero arrow mein badal diya.

Recall Momentum conserved hai claim karne se pehle aapko kya ek condition check karni chahiye?

Kya us interval mein net external force zero hai? (Step 8, Case A.)

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Connections

• Newton's Third Law — mirror-arrow engine (Step 4)
• Newton's Second Law (momentum form) — push = momentum change-rate (Step 3)
• Elastic vs Inelastic Collisions — Step 7 ka energy caveat
• Impulse–Momentum Theorem — Step 8 Case A, integral view
• Centre of Mass Motion — frozen $\vec{P}$ as steady ${\vec{v}}_{cm}$
• Rocket Equation — Step 8 Case B with changing mass