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

Ball par do forces act karte hain:

1. Gravity, Earth ball ko neeche ki taraf kheench raha hai: ${\vec{F}}_{E\to b}$.
2. Normal force, floor ball ko upar ki taraf push kar raha hai: ${\vec{N}}_{f\to b}$.

Reaction partner hamesha do bodies ko swap karta hai aur same type ki force hoti hai:

• "Earth ball ko neeche kheenchta hai" (${\vec{F}}_{E\to b}$) ka partner hai "ball Earth ko upar kheenchta hai" ${\vec{F}}_{b\to E}$ — dono gravity.
• "Floor ball ko upar push karta hai" (${\vec{N}}_{f\to b}$) ka partner hai "ball floor ko neeche push karta hai" ${\vec{N}}_{b\to f}$ — dono normal/contact.

Note karo ki partners Earth aur floor par act karte hain, ball par nahi. Ye rule 5 hai.

False. Weight aur normal force dono yahan ek hi object (ball) par act karte hain aur alag types ki force hain (gravity vs. contact). Ek real 3rd-law pair same type ka hona chahiye aur do alag bodies par act karna chahiye. Ye sirf isliye equal hain kyunki ball equilibrium mein hai — ye Newton's Second Law hai ($\sum F=0$), third law nahi.

Koi external force act nahi karta, isliye total momentum conserved hai (ye Conservation of Momentum hai). Rest par start: total $p=0$. $0=m_w{v}_w+m_a{v}_a$ "Wrench direction" ko positive lo. $m_w=2$, $v_w=+6$, $m_a=70$: $0=(2)(6)+(70)v_a\Rightarrow v_a=-\frac{12}{70}=-0.171\text{m/s}$ Minus ka matlab hai backward (wrench ke opposite). Speed $\approx 0.171\text{m/s}$. Third-law view: astronaut ne wrench ko short time $t$ ke liye force $F$ se push kiya; wrench ne equally push back kiya. Dono par same impulse $Ft$, isliye $m_a|v_a|=m_w|v_w|$ — heavier astronaut slower move karta hai.

Third law se wrench haath ko exactly $30\text{N}$, opposite direction mein push karta hai. Astronaut ka acceleration Newton's Second Law se: $a=F/m=30/70=0.4286{\text{m/s}}^2$, backward. (Wrench ka acceleration $30/2=15{\text{m/s}}^2$ hai — same force, much larger acceleration kyunki uski mass tiny hai.)

(a) Dono blocks ko $m_1+m_2=5\text{kg}$ mass ka ek system maano. Horizontally sirf $F$ external hai: $a=\frac{F}{m_1+m_2}=\frac{20}{5}=4{\text{m/s}}^2.$ (b) Ab block 2 (mass 2 kg) ko isolate karo. Iske upar horizontal mein sirf block 1 ki contact push hai, ise $C$ kaho (block 1 → block 2): $C=m_{2}a=(2)(4)=8\text{N}\text{(forward)}.$ Third law se, block 2, block 1 par 8 N backward push karta hai. Block 1 ko directly check karo: $m_{1}a=F-C=20-8=12=(3)(4)=12✓$ Do contact forces ($1\to 2$ forward, $2\to 1$ backward) ka size equal 8 N hai, opposite directions hain, same type (contact/normal) hain, swapped labels hain, alag blocks par hain — ye ek genuine 3rd-law pair hai. Figure mein red arrows dekho: wo alag free-body diagrams par rehte hain (isliye wo kabhi cancel nahi hote — Free Body Diagrams).

Pair phir bhi $a=20/5=4{\text{m/s}}^2$ par accelerate karta hai (ab leftward). Block 1 (3 kg) ko isolate karo, jis par sirf block 2 ki contact push hai horizontally: $C^{\prime }=m_{1}a=(3)(4)=12\text{N}.$ Contact force change ho gaya (12 N vs. pehle ka 8 N) chahe total force aur total mass identical hain! Kyun: contact force ko us block ko accelerate karna hai jo push ke peeche hai. Left se push karne par, ye 2 kg block ko drive karta hai (→ 8 N); right se push karne par, ye 3 kg block ko drive karta hai (→ 12 N).

(a) Pehle momentum = 0 (sab rest mein). Positive = ball ki direction. $0=m_{\text{ball}}{v}_{\text{ball}}+m_{\text{cannon}}{v}_{\text{cannon}}=(4)(80)+50v_c$ $v_c=-\frac{320}{50}=-6.4\text{m/s}\text{(backward)}.$ (b) Ball par impulse: $J=\Delta p=(4)(80)=320\text{kg⋅m/s}$ over $t=0.02\text{s}$: $F=\frac{J}{t}=\frac{320}{0.02}=16000\text{N}.$ Third law se ball cannon ko wahi 16000 N se push back karta hai. Cannon ka impulse check karo: $Ft=16000\times 0.02=320=50\times 6.4✓$. (c) Total momentum after $=(4)(80)+(50)(-6.4)=320-320=0$, pehle jaisa — conserved (Conservation of Momentum). Centre of mass: total momentum 0 hai, isliye centre of mass wahin rehta hai — wo fire karne se pehle ya baad kabhi move nahi karta (Center of Mass Motion). Internal explosion (ek 3rd-law pair) ise shift nahi kar sakta.

Thrust = rate of momentum jo exhaust le jaata hai (ye Rocket Propulsion & Variable Mass result hai, khud ek 3rd-law statement — gas peeche push hua, rocket aage push hua): $F_{\text{thrust}}=\left(\frac{\Delta m}{\Delta t}\right)v_{\text{exhaust}}=\left(\frac{5\text{kg}}{1\text{s}}\right)(2000\text{m/s})=10000\text{N}.$ Instantaneous acceleration: $a=\frac{F_{\text{thrust}}}{m_{\text{rocket}}}=\frac{10000}{1000}=10{\text{m/s}}^2.$ Reaction partner exhaust gas hai, koi surrounding medium nahi — yahi wajah hai ki rockets vacuum mein kaam karte hain.

Positive = rightward. $A$ se $B$ par force forward hai, $B$ ko impulse $+J=+4$ deta hai; third law se $B$ se $A$ par force backward hai, $A$ par impulse $-J=-4$. (Impulse = force × time = $\Delta p$; har instant par equal-and-opposite forces integrate hokar equal-and-opposite impulses dete hain.)

Block $B$ ($m_B=1$, rest se shuru): $v_B^{\prime }=0+\frac{J}{m_B}=0+\frac{4}{1}=4\text{m/s}.$ Block $A$ ($m_A=2$, $3$ se shuru): $v_A^{\prime }=3+\frac{-J}{m_A}=3-\frac{4}{2}=1\text{m/s}.$ Impulse check: $A$ par ye $-4$ hai, $B$ par $+4$ — exact negatives, jaisa third law demand karta hai. Momentum check: pehle $=(2)(3)+(1)(0)=6$. Baad mein $=(2)(1)+(1)(4)=2+4=6✓$. Conserved.

Child apne pairo se raft ko backward (left) push karta hai; raft child ko forward (right) push karta hai — ek 3rd-law pair, child+raft system ke internal. Koi external horizontal force nahi, isliye centre of mass fixed rehta hai (Center of Mass Motion).

Maano raft distance $d$ se left slide karta hai. Tab child (jo raft ke along 6 m chalta hai) water ke relative right mein $6-d$ move karta hai. Centre of mass ko fixed rakhne ke liye, do mass-weighted displacements cancel hone chahiye: $m_{\text{child}}(6-d)=m_{\text{raft}}d$ $40(6-d)=120d\Rightarrow 240-40d=120d\Rightarrow 240=160d\Rightarrow d=1.5\text{m}.$ Raft 1.5 m left ki taraf slide karta hai, aur child apne start ke $6-1.5=4.5\text{m}$ right end up karta hai (water ke relative). Check: $40(4.5)=180=120(1.5)=180✓$.

L1: ek 3rd-law pair bodies ko swap karta hai, same type — sirf "equal aur opposite" nahi. L2–L4: same force ⇒ unequal accelerations, aur reaction partner ejected mass hai, koi medium nahi. L5: internal 3rd-law pairs centre of mass ko fixed rakhte hain, isliye lighter parts zyada door jaate hain.