Exercises — System with external forces — conditions for conservation

System = saare 30 fragments. Explosion ki pushes fragments ke beech hain → internal → Newton's 3rd law se pairs mein cancel ho jaati hain, toh woh $\vec{P}$ change nahi kar sakteen. Sirf external force gravity hai, jo seedha neeche point karti hai.

• Horizontal axis: koi external force nahi → $P_x$ conserved. ✅
• Vertical axis: gravity act karti hai → $P_y$ conserved nahi. ❌ Newton ka law component-wise hai, isliye tum $x$ par jeet sakte ho jab $y$ par haar rahe ho.

System = sirf ball. External forces: gravity (neeche, Earth se jo bahar hai) aur table ka normal force (upar, table bahar hai). Yahan woh equal aur opposite hain, toh ${\vec{F}}_{ext}^{net}=\vec{0}$ is instant par → $\vec{P}$ (jo zero hai, ball rest mein) zero rehta hai. Conserved hai, lekin trivially — ball bas move hi nahi kar rahi. Teaching point yeh hai: gravity aur normal yahan external hain sirf isliye kyunki Earth aur table us boundary ke bahar hain jo tumne draw ki.

System = cannon + ball. Firing force ab internal hai. External forces (gravity, normal) vertical hain, toh horizontal $P_x$ conserved hai. Fire karne se pehle, sab kuch rest mein hai: $P_{x,i}=0$. Baad mein: ball $+v$ par jaati hai, cannon $-V$ par recoil karta hai: $0=mv+M(-V)\Rightarrow V=\frac{mv}{M}=\frac{5(180)}{500}=1.8\text{m/s}$ Cannon $\boxed{1.8\text{m/s}}$ par peeche roll karta hai.

System = A + B. Rail frictionless hai, gravity/normal vertical → horizontal $P_x$ conserved. $P_i=3(4)+2(0)=12\text{kg⋅m/s}$ Saath chipke hue woh $5\text{kg}$ mass share karte hain speed $u$ par: $12=5u\Rightarrow u=2.4\text{m/s}$ Common final speed $\boxed{2.4\text{m/s}}$ (yeh perfectly inelastic collision hai — dekho Elastic vs Inelastic Collisions).

Sahi test impulses ($=$ force $\times$ time) compare karta hai, forces nahi (parent, Impulse–Momentum Theorem). Bat impulse = ball ka momentum change: $J_{bat}=m\Delta v=0.15(50)=7.5\text{N⋅s}$ Contact ke dauran gravity impulse: $J_g=mg\Delta t=0.15(9.8)(0.0012)\approx 1.76\times 10^{-3}\text{N⋅s}$ Ratio $J_g/J_{bat}\approx 2.4\times 10^{-4}$, yani gravity approximately $0.024\%$ contribute karti hai. Kyunki $J_g\ll J_{bat}$, $\vec{P}$ ko impact ke through conserved treat karo, phir flight ke liye gravity re-add karo. Impulsive approximation justified. ✅

System = dono disks. Ice frictionless hai, gravity/normal vertical → dono horizontal axes ($x$ aur $y$) conserved hain. Toh hum $P_i=P_f$ alag-alag har ek par likhte hain. Disk 1 baad mein: $v_{1x}=3\sqrt{3}\cos 30^{\circ }=3\sqrt{3}\cdot \frac{\sqrt{3}}{2}=4.5$, $v_{1y}=3\sqrt{3}\sin 30^{\circ }=3\sqrt{3}\cdot \frac{1}{2}=\frac{3\sqrt{3}}{2}\approx 2.598$. x-axis: $2(6)+2(0)=2(4.5)+2(v_{2x})$ $12=9+2v_{2x}\Rightarrow v_{2x}=1.5\text{m/s}$ y-axis: $0=2(2.598)+2(v_{2y})$ $0=5.196+2v_{2y}\Rightarrow v_{2y}=-2.598\text{m/s}$ Disk 2, $\boxed{(1.5,-2.598)\text{m/s}}$ par move karta hai, yani speed $\sqrt{1.5^2+2.598^2}=3\text{m/s}$ $x$-axis ke neeche $30^{\circ }$ par. Dono disks symmetrically alag uda jaate hain — yeh equal masses ki geometry hai.

Do phases, alag-alag valid laws ke saath. Phase 1 — embedding (impulsive): collision fast hai, toh gravity ka impulse negligible hai → momentum horizontally conserved. $m{v}_0=(m+M)V,m=0.010,M=2.00$ Phase 2 — swing up (energy): impact ke baad, koi non-conservative force swing ke along net kaam nahi karta, toh KE gravitational PE mein convert hoti hai. $\frac{1}{2}(m+M)V^2=(m+M)gh\Rightarrow V=\sqrt{2gh}=\sqrt{2(9.8)(0.20)}=1.9799\text{m/s}$ Back-substitute karo: $v_0=\frac{m+M}{m}V=\frac{2.010}{0.010}(1.9799)=398.0\text{m/s}$ Bullet speed $\boxed{v_0\approx 398\text{m/s}}$. Dono phases ke liye ek law kyun nahi? Momentum phase 1 mein conserved hai lekin KE nahi (embedding = inelastic, heat lost). Energy phase 2 mein conserved hai lekin horizontal momentum nahi (strings inward pull karti hain). Har phase woh law use karta hai jiska condition woh satisfy karta hai.

System = person + raft. Koi horizontal external force nahi → $P_x=0$ zero rehta hai → paani mein center of mass fixed rehta hai (yeh Center of Mass Motion hai: $F_{ext}=0\Rightarrow a_{cm}=0$, aur yeh rest mein shuru hua tha). Maano raft $x$ shift karta hai (walk ke opposite direction mein). Person raft ke relative $4\text{m}$ walk karta hai, toh paani ke relative person $(4-x)$ move karta hai; fixed CM enforce karo: $m_p{d}_p=m_r{d}_r,d_p+d_r=4$ $m_p=60$, $m_r=120$ ke saath: raft ka displacement hai $x=\frac{m_p}{m_p+m_r}(4)=\frac{60}{180}(4)=1.333\text{m}$ Raft person ke walk direction ke $\boxed{1.33\text{m}}$ opposite slide karta hai (person paani ke upar $2.67\text{m}$ move karta hai). CM kabhi nahi hila.

Up ko positive lo. (a) Ball alone: $\Delta p=m{v}_f-m{v}_i=1(+4)-1(-5)=+9\text{kg⋅m/s}$. Nonzero hai kyunki gravity aur floor ka normal ek-ball system ke liye external hain — floor ek real upward impulse deliver karta hai. Ball ka $\vec{P}$ conserved nahi hai. (b) Ball + Earth: ab floor ka push internal hai (floor Earth ka part hai). Koi external horizontal/vertical force nahi → total $\vec{P}$ conserved. Earth ko $-9\text{kg⋅m/s}$ absorb karna hoga: $M_{\oplus }{V}_{\oplus }=9\Rightarrow V_{\oplus }=\frac{9}{5.97\times 10^{24}}\approx 1.5\times 10^{-24}\text{m/s}$ Bilkul imperceptible, lekin nonzero — same event, opposite conservation verdict, purely iss baat se ki boundary kahan draw ki.

System = dono pieces, koi external force nahi → $P=0$ pehle aur baad mein (explosion forces internal). $0=mu+(Km)(-w)\Rightarrow w=\frac{u}{K}$ (b) Limits:

• $K\to \infty$: $w=u/K\to 0$ — ek bahut bhaari chunk barely move karta hai, jaise rifle ek light bullet ke against almost kuch nahi recoil karti. ✅
• $K\to 0$: $w=u/K\to \infty$ — vanishingly light doosra piece arbitrarily fast fling hota hai, toh saara "kick" us tiny fragment mein jaata hai. Dono limits $P=0$ ko exactly respect karte hain: heavy·slow always light·fast ko balance karta hai.

Har momentum problem par poochha jaane wala single question ::: "Meri boundary kahan hai, aur kya us axis par net external force zero hai jis par mujhe dhyan dena hai?"