Newton's third law — action-reaction, common misconceptions: Worked examples

1.2.3 · D3 · Physics › Newton's Laws & Dynamics › Newton's third law — action-reaction, common misconceptions

Yeh Newton's Third Law ke liye ek worked-example dojo hai. Parent note ne idea build kiya; yahan hum ise har tarah ki situation ke against drill karte hain jo yeh law throw kar sakti hai — taaki jab exam ya koi real problem aaye, toh tumne already uski shape dekhi ho.

Scenario matrix

Har third-law problem is table ke ek cell mein rehta hai socho. Hamara goal: sabko cover karna.

Cell	Case class	Tricky kya hai	Covered by
C1	Static equilibrium (kuch nahi hil raha)	Equal-opposite forces jo pair NAHI hain	Ex 1
C2	Two free bodies push apart (symmetric)	Same force, different accelerations	Ex 2
C3	Coupled bodies accelerating together	Internal pair vs. external force	Ex 3
C4	Degenerate: zero force / no contact	Release ke instant par kya hota hai	Ex 4
C5	Limiting case: mass ratio → 0 ya → ∞	Truck-vs-fly, Earth-vs-you	Ex 5
C6	Continuous mass ejection (rocket)	Partner exhaust gas hai, air nahi	Ex 6
C7	Real-world word problem (walking / swimming)	Kaunsi surface tumhe push karti hai?	Ex 7
C8	Exam twist: do blocks ke beech contact force ek train mein	Internal pair dhundhne ke liye ek block isolate karo	Ex 8

Neeche har numeric answer ke saath uska apna Verify line hai, aur har ek machine-checked hai.

Ex 1 — Table par static book (Cell C1: equilibrium, fake pair)

Figure — Newton's third law — action-reaction, common misconceptions

Figure — Ex 1 (alt text): Left panel mein ek single black dot hai jis par "book" likha hai, jisme do arrows hain — ek green arrow upar ki taraf hai jis par "N (table→book)" likha hai aur ek red arrow neeche ki taraf hai jis par "W (Earth→book)" likha hai; caption note karta hai ki yeh sirf equilibrium se equal hain aur 3rd-law pair NAHI hain. Right panel do genuine pairs dikhata hai, har ek do alag dots span karta hai: book↔Earth red gravity arrows se juda hai, aur book↔table green normal arrows se juda hai.

Step 1 — Book par forces list karo. Weight $W = m g = 2 \times 9.8 = 19.6 N$ downward, aur normal $N$ upward. Yeh step kyun? Pairs name karne se pehle hume pata hona chahiye ki kaun se forces book ko touch bhi karte hain. Figure ke left tower mein, dono arrows ek hi dot (book) par hain.

Step 2 — $N$ get karne ke liye equilibrium use karo. Book accelerate nahi karti, isliye Newton's Second Law se net force zero hai: $N - W = 0 ⟹ N = 19.6 N .$ Yeh step kyun? $N = W$ yahi cheez logon ko tempt karti hai ki yeh pair hain. Lekin yeh equality equilibrium se aati hai, third law se nahi — equilibrium hata do (book gira do) aur yeh turant alag ho jaate hain.

Step 3 — True partners dhundho (labels swap karo).

"Earth pulls book down" ka partner $=$ "book pulls Earth up" (gravity ↔ gravity).
"table pushes book up" ka partner $=$ "book pushes table down" (normal ↔ normal). Figure ke right half mein, har real pair do alag dots span karta hai (book aur Earth; book aur table). Yeh step kyun? Partners ka rule — $A \leftrightarrow B$ swap karo, type same rakho (upar test 4).

Step 4 — (c) ka jawab. Nahi. Weight aur normal dono ek hi body (book) par act karte hain aur different types ke hain → yeh upar define kiye gaye paanch pair-tests mein test 3 (same type) aur test 5 (do alag bodies) dono fail karte hain.

Verify: $N = 19.6 N$ upward bilkul $W = 19.6 N$ down ko balance karta hai → net $= 0 N$ , "at rest" ke saath consistent. Units: $kg \cdot m/s^{2} = N$ . ✓

Ex 2 — Do skaters push apart (Cell C2: symmetric free bodies)

Step 1 — Third-law pair identify karo. $F_{A \to B} = - F_{B \to A}$ , dono ki magnitude $F = 120 N$ . Yeh step kyun? Har skater par ek hi horizontal force hai dusre ki push — ek clean pair.

Step 2 — Impulse ke zariye force ko velocity mein convert karo. Impulse $=$ force $\times$ time $=$ momentum mein change. Yeh Impulse–Momentum Theorem hai (Newton's Second Law ka direct time-integral, momentum ka conservation nahi): $F t = m v .$ Yeh step kyun? Force akela speed nahi de sakta; hume chahiye ki yeh kitni der tak act karta hai. $t$ bridge hai.

Step 3 — Har skater solve karo. $v_{A} = \frac{F t}{m _{A}} = \frac{120 \times 0.50}{60} = 1.0 m/s, v_{B} = \frac{120 \times 0.50}{40} = 1.5 m/s .$ Yeh step kyun? Upar same $F t$ , neeche chota mass ⇒ badi speed. Lighter skater faster hota hai.

Step 4 — Ratio. $\frac{v _{B}}{v _{A}} = \frac{F t / m _{B}}{F t / m _{A}} = \frac{m _{A}}{m _{B}} = \frac{60}{40} = 1.5.$ Yeh step kyun? Ratio lene se shared impulse $F t$ top-and-bottom cancel ho jaata hai, sirf inverse mass ratio bachta hai. Yeh sabse clean proof hai ki "equal force ≠ equal speed": outcome purely is baat par depend karta hai ki masses kaise compare hote hain, $F$ ya $t$ par bilkul nahi.

Verify (via conservation of momentum): do skaters par equal-opposite impulses guarantee karte hain ki total momentum zero rahega (rest se shuru hua) — yahan Conservation of Momentum enter karta hai. $m_{A} v_{A} - m_{B} v_{B} = 60 (1.0) - 40 (1.5) = 60 - 60 = 0$ . ✓ Same force, unequal accelerations — bilkul "equal force ≠ equal effect" ka lesson.

Ex 3 — Do blocks ek saath push kiye gaye (Cell C3: internal vs external)

Figure — Ex 3 (alt text): Do touching blocks gray floor par rest kar rahe hain. Left block blue hai, "m1 3 kg" label hai; right block orange hai, "m2 2 kg" label hai. Ek red arrow "F=10 N" label ke saath left block ko left se push kar raha hai. Ek green arrow "C=4 N" label ke saath block m2 ke top par right ki taraf point kar raha hai (contact push jo ise feel hoti hai). Ek black arrow "a=2 m/s²" shared acceleration dikhata hai, aur caption padhta hai "Isolated m2: only C acts → C = m2·a = 4 N".

Step 1 — (a) ka jawab: dono blocks ko EK system treat karo. Total mass $= 3 + 2 = 5 kg$ . Ek hi external horizontal force hai tumhara $10 N$ : $a = \frac{F}{m _{1} + m _{2}} = \frac{10}{5} = 2 m/s^{2} .$ Yeh step kyun? Blocks ke beech contact force internal hai — yeh ek third-law pair hai aur whole-system view mein cancel ho jaata hai. Isliye yeh shared $a$ ko affect nahi kar sakta. Yeh Center of Mass Motion logic hai.

Step 2 — (b) ka jawab: contact force dhundhne ke liye $m_{2}$ isolate karo. $m_{2}$ par ek hi force hai push $C$ (contact magnitude) $m_{1}$ se: $C = m_{2} a = 2 \times 2 = 4 N .$ Yeh step kyun? Contact force do-block system ke liye internal hai, isliye Step 1 mein kabhi nahi dikhा। Ise reveal karne ke liye hume system tod na hoga aur $m_{2}$ ko akele draw karna hoga (Free Body Diagrams): ek baar isolate hone ke baad, $C$ sirf horizontal force hai $m_{2}$ par, isliye Newton's second law $C = m_{2} a$ seedha de deta hai. Figure mein yeh right block hai apne single green arrow ke saath.

Step 3 — (b) interpret karo: kya contact force $10 N$ hai? Nahi. Contact force $4 N$ hai, tumhare $10 N$ se kam — kyunki $m_{1}$ khud ko accelerate karne ke liye kuch push apne paas rakhta hai.

Step 4 — (c) ka jawab: pair verify karo. Third law se, $m_{2}$ wapas $m_{1}$ par $4 N$ se push karta hai. $m_{1}$ akele check karo: $F - C = m_{1} a ⟹ 10 - 4 = 3 \times 2 = 6 N . ✓$

Verify: whole system $F = (m_{1} + m_{2}) a = 5 \times 2 = 10 N$ ✓. $4 N$ pair internal aur equal-opposite hai; yeh system equation mein kabhi nahi aata.

Ex 4 — Release ka instant (Cell C4: degenerate / zero force)

Step 1 — (a) ka jawab: contact toot gaya ⇒ normal force zero hai. $N = 0 N .$ Yeh step kyun? Normal force ek contact force hai; no contact, no force. Yeh degenerate case hai — ek input zero ho jaata hai.

Step 2 — (b) ka jawab: uska partner bhi usi waqt vanish ho jaata hai. "table pushes book up" aur "book pushes table down" ka pair dono ek hi instant par $0 N$ par drop karte hain. Ek pair hamesha saath paida hota hai aur saath hi maarta hai — koi time delay nahi hota. Yeh step kyun? Misconception khatam karta hai "reaction lags action". Zero aur zero, ek saath.

Step 3 — (c) ka jawab: sirf gravity bacha rehta hai. $a = \frac{W}{m} = \frac{m g}{m} = g = 9.8 m/s^{2} (downward) .$ Yeh step kyun? Gravity ka partner ("book pulls Earth up") abhi bhi exist karta hai — gravity contact force nahi hai. Lekin yeh Earth par act karta hai, book par nahi, isliye fall slow nahi hoti.

Verify: $N = 0$ ke saath, book par net force $= W = 19.6 N$ down, toh $a = 19.6/2 = 9.8 m/s^{2} = g$ . ✓ Free fall, jaise expect tha.

Ex 5 — Truck vs. fly (Cell C5: extreme mass ratio limit)

Step 1 — Forces bilkul equal hain (third law). $F_{truck \to fly} = F_{fly \to truck} = 5 N .$ Yeh step kyun? Mass chahe kuch bhi ho, pair magnitude mein identical hai. "Bada object zyada zyaada se maarta hai" — yeh galat hai.

Step 2 — Accelerations $a = F / m$ se. $a_{fly} = \frac{5}{0.001} = 5000 m/s^{2}, a_{truck} = \frac{5}{2000} = 0.0025 m/s^{2} .$ Yeh step kyun? Mass denominator mein hai, isliye tiny fly ko monstrous acceleration milti hai.

Step 3 — Ratio (limiting behaviour). $\frac{a _{fly}}{a _{truck}} = \frac{F / m _{fly}}{F / m _{truck}} = \frac{m _{truck}}{m _{fly}} = \frac{2000}{0.001} = 2000 000.$ Yeh step kyun? Ratio lene se shared force $F$ cancel ho jaata hai, sirf inverse mass ratio bachta hai. Yeh physical lesson ko exact banata hai: jaise mass ratio $\to \infty$ , acceleration ratio bhi $\to \infty$ , toh heavy body almost notice nahi karta jabki light body annihilate ho jaata hai — same force, opposite fates. (Isliye "tum Earth ko utna hi upar push karte ho jitna woh tumhe neeche pull karti hai", phir bhi Earth visibly move nahi karti.)

Verify: $a_{fly} / a_{truck} = 5000/0.0025 = 2000000$ ✓, $m_{truck} / m_{fly}$ ke equal. Units: $N / kg = m/s^{2}$ ✓.

Ex 6 — Rocket in vacuum (Cell C6: continuous ejection)

Step 1 — (a) ka jawab: thrust = exhaust ko har second diya gaya momentum. $T = \overset{m}{˙} u = 5 \times 2000 = 10000 N .$ Yeh step kyun? Har second rocket $\overset{m}{˙}$ gas ko speed $u$ par backward fling karta hai; Conservation of Momentum se, rocket ko equal forward momentum milta hai. Yeh Rocket Propulsion & Variable Mass hai.

Step 2 — (b) ka jawab: acceleration. $a = \frac{T}{m} = \frac{10 000}{1000} = 10 m/s^{2} .$ Yeh step kyun? Rocket par Newton's second law, thrust ko forward force ke roop mein use karte hue.

Step 3 — (c) ka jawab: partner ka naam batao. $F_{rocket \to gas} = - F_{gas \to rocket}$ . Partner exhaust gas hai, surrounding air NAHI. Yeh step kyun? "Vacuum mein rocket kya push karta hai?" — yeh sawaal sirf tab confuse karta hai jab tum koi external medium dhundho. Third law iska saaf jawab deta hai: reaction partner woh gas hai jo rocket khud eject karta hai, isliye koi air zaroori nahi — yahi reason hai ki rockets space mein kaam karte hain.

Verify: har second momentum bookkeeping: gas $\overset{m}{˙} u = 5 \times 2000 = 10000 kg\cdotm/s$ backward le jaata hai; rocket ko $10000 kg\cdotm/s$ forward milta hai → thrust $10000 N$ ✓. $a = 10 000/1000 = 10 m/s^{2}$ ✓.

Ex 7 — Walking / swimming (Cell C7: real-world word problem)

Figure — Ex 7 (alt text): Ek stick person gray ground par khada hai. Paaon par, ek red arrow backward point karta hai, jis par "you on ground 140 N (back)" likha hai. Body par, ek blue arrow forward point karta hai, jis par "ground on you 140 N (forward)" — friction third-law pair. Caption padhta hai "a = 140/70 = 2 m/s² forward".

Step 1 — (a) ka jawab: forward force ground ki reaction hai. Paaon ground ko backward $140 N$ se push karta hai ⇒ ground person ko forward $140 N$ se push karta hai (friction pair, Normal Force and Friction). Figure mein, red arrow (you on ground) backward point karta hai, blue arrow (ground on you) forward point karta hai. Yeh step kyun? Tum kabhi directly khud ko propel nahi karte; ground tumhara engine hai. Frictionless ice par tum push karte aur kahin nahi jaate.

Step 2 — (b) ka jawab: person ki acceleration. $a = \frac{F}{m} = \frac{140}{70} = 2 m/s^{2} forward .$ Yeh step kyun? Person par horizontally sirf forward ground-force act karta hai.

Step 3 — (c) ka jawab: kya Earth move karta hai? (phir se limiting case). Haan, ek immeasurable amount se: $a_{Earth} = 140/ (6 \times 1 0^{24}) \approx 2.3 \times 1 0^{- 23} m/s^{2}$ — disguise mein C5 mass-ratio limit.

Verify: $a = 140/70 = 2 m/s^{2}$ ✓. Same-magnitude pair $140 N$ ; backward force ground par hai, forward force tumhare par — different bodies, koi cancellation nahi.

Ex 8 — Blocks ki "train" mein contact force (Cell C8: exam twist)

Step 1 — System acceleration. $a = \frac{F}{m _{1} + m _{2} + m _{3}} = \frac{12}{1 + 2 + 3} = \frac{12}{6} = 2 m/s^{2} .$ Yeh step kyun? Saare between-block contact forces internal third-law pairs hain; har pair equal-and-opposite hai, isliye poori train par sum karne par cancel ho jaate hain. Iska matlab sirf external $F$ total $6 kg$ drive karta hai — isliye teeno blocks ko ek body treat karne par shared $a$ ek line mein milti hai.

Step 2 — $m_{3}$ isolate karo. $m_{3}$ par sirf ek horizontal force hai contact push $C_{23}$ (contact magnitude) $m_{2}$ se: $C_{23} = m_{3} a = 3 \times 2 = 6 N .$ Yeh step kyun? $C_{23}$ poori train ke liye internal tha, isliye Step 1 mein kabhi show nahi ho sakta tha. $m_{3}$ ko akele draw karna ise expose karta hai: horizontally $m_{3}$ ko aur kuch touch nahi kar raha, $C_{23}$ hi ek force hai jo ise accelerate karta hai, isliye Newton's second law $C_{23} = m_{3} a$ turant de deta hai. General rule: ek contact exactly uske aage wale mass ko carry karta hai.

Step 3 — Third-law check. $m_{3}$ wapas $m_{2}$ par same $6 N$ se push karta hai: ek genuine equal-opposite contact pair, do alag blocks par act karta hua.

Verify: $m_{1}, m_{2}$ ke beech contact $(m_{2} + m_{3})$ ko push karna chahiye: $C_{12} = (2 + 3) \times 2 = 10 N$ . $m_{1}$ par consistency: $F - C_{12} = 12 - 10 = 2 = m_{1} a = 1 \times 2$ ✓. Aur $C_{23} = 3 \times 2 = 6 N$ ✓.

Recall Har example kaun se cell mein tha?

Ex1 → C1 (equilibrium fake-pair). Ex2 → C2 (symmetric push). Ex3 → C3 (internal vs external). Ex4 → C4 (zero/degenerate). Ex5 → C5 (mass-ratio limit). Ex6 → C6 (ejection). Ex7 → C7 (word problem). Ex8 → C8 (exam twist).

Rapid recall

Book & table normal force — kya yeh gravity ki reaction hai?

Nahi; dono book par act karte hain aur different types ke hain. Gravity ka partner hai "book pulls Earth up".

Do skaters 60 & 40 kg push off karte hain — speed ratio

v_{40} / v_{60}

60/40 = 1.5

Truck (2000 kg) vs fly (0.001 kg), acceleration ratio?

2000000

— equal force, mass ratio decide karta hai.

Rocket

\overset{m}{˙} = 5 kg/s

u = 2000 m/s

— thrust?

10000 N

Teen blocks 1,2,3 kg,

F = 12 N

— 2 aur 3 ke beech contact?

6 N

(mass ahead

= 3

\times a = 2

Jab contact toot jaata hai, normal force kitni jaldi vanish ho jaati hai?

Instantly, aur uska partner bhi simultaneously vanish ho jaata hai.

Connections

Newton's Third Law — Action & Reaction — parent law jise yeh examples exercise karte hain.
Newton's Second Law — $a = F / m$ equal forces ko unequal accelerations mein convert karta hai (Ex 2, 5).
Conservation of Momentum — Ex 2 aur Ex 6 ke peeche ka check.
Free Body Diagrams — ek body isolate karna internal pairs reveal karne ke liye (Ex 3, 8).
Normal Force and Friction — woh pair jo tumhe walk karne deta hai (Ex 7).
Rocket Propulsion & Variable Mass — continuous-ejection thrust (Ex 6).
Center of Mass Motion — kyun internal pairs system acceleration ko affect nahi karte (Ex 3, 8).