Exercises — Spring-mass systems — collision problems

• Momentum conservation $m_1{u}_1+m_2{u}_2=(m_1+m_2)v$ — collision ke dauran use hota hai (spring impulse ≈ 0 chhote contact time mein) aur jab bhi koi external horizontal push nahi hota. Dekho Conservation of Linear Momentum.
• Energy conservation $\frac{1}{2}m{u}^2$ → spring PE $\frac{1}{2}k{x}^2$ — collision ke baad use hota hai jab ideal spring squish hoti hai, kyunki kuch bhi energy waste nahi karta. Dekho Elastic Potential Energy of a Spring.
• Reduced mass $\mu =\frac{m_1{m}_2}{m_1+m_2}$ — do bodies ki relative motion ka "effective mass". Sirf relative-motion kinetic energy $\frac{1}{2}\mu (u_1-u_2{)}^2$ hi spring mein store ho sakti hai. Dekho Reduced Mass and Two-Body Problems.

(a) Max compression par velocity → momentum conservation. Hum kya karte hain: total momentum before = total momentum at max compression set karo, jisse common velocity milti hai $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}$. Kyun: dono free blocks par koi external horizontal force nahi lagta, isliye unka total momentum change nahi ho sakta; max squish par wo saath move karte hain (relative velocity zero), toh ek common $v$ sara momentum carry karta hai. (b) Maximum compression → energy conservation. Hum kya karte hain: kinetic energy before = poore system ki kinetic energy at $v$ plus stored spring PE $\frac{1}{2}k{x}^2$ set karo. Kyun: ideal spring kuch bhi waste nahi karta, isliye total mechanical energy conserved rahti hai; $x$ ke liye solve karne par compression mil jaata hai. Answer: (a) momentum, (b) energy — "MEME" split.

Nahi. Wall spring ke through ek real, external horizontal force se push back karta hai, isliye block ka momentum conserved nahi hota (max compression par zero ho jaata hai). Energy zaroor conserved hoti hai — koi friction nahi, ideal spring — toh hum $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2$ use karte hain.

Unki velocities equal hoti hain (relative velocity = 0). Dono ke beech gap tabhi chhota hota hai jab ek doosre ko pakad raha hota hai; jis moment unki speeds match hoti hain gap chhona band ho jaata hai, toh spring apni maximum compression par hoti hai. "Max squish = same swish."

Wall = external force → sirf energy. Saari kinetic energy spring PE ban jaati hai: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2\Rightarrow x=u\sqrt{m/k}=3\sqrt{2/600}=3(0.05774)=\boxed{0.173\text{m}}.$

Max compression ⇔ same velocity, aur momentum conserved hai: $v=\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=\frac{2(6)+4(0)}{6}=\boxed{2\text{m/s}}.$

Sirf relative-motion KE store hoti hai, effective mass $\mu$ ke saath: $\mu =\frac{2\cdot 4}{6}=\frac{4}{3}\text{kg},x=(u_1-u_2)\sqrt{\mu /k}=6\sqrt{\frac{4/3}{1200}}=6(0.03333)=\boxed{0.200\text{m}}.$ Picture check: spring store karta hai $\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}(4/3)(36)=24\text{J}$; aur $\frac{1}{2}k{x}^2=\frac{1}{2}(1200)(0.2{)}^2=24\text{J}$. ✓

Ek ideal spring apni stored energy ka 100% return karta hai, isliye poora encounter ek perfectly elastic collision hai (dekho Elastic and Inelastic Collisions). $u_2=0$ ke saath standard 1-D elastic results use karo: $v_1^{\prime }=\frac{m_1-m_2}{m_1+m_2}{u}_1=\frac{2-4}{6}(6)=\boxed{-2\text{m/s}}(\text{bounce back karta hai}),$ $v_2^{\prime }=\frac{2m_1}{m_1+m_2}{u}_1=\frac{4}{6}(6)=\boxed{4\text{m/s}}.$ Sanity check: momentum $2(-2)+4(4)=12=2(6)$ ✓; energy $\frac{1}{2}2(4)+\frac{1}{2}4(16)=36$ ✓.

Spring lock hone se blocks hamesha saath move karte hain → perfectly inelastic. Final velocity wohi $v_{cm}$ hai: $v=2\text{m/s},E_{\text{trapped}}=\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}\cdot \frac{4}{3}\cdot 36=\boxed{24\text{J}}.$ Baat ye hai: max-compression ka maths bilkul L2.3 jaisa hai; sirf ending alag hai — released (elastic) vs trapped (inelastic).

(a) $v=\frac{3(4)+1(-2)}{4}=\frac{10}{4}=\boxed{2.5\text{m/s}}.$ $u_2$ ka sign matter karta hai — isko apne minus sign ke saath plug karo. (b) $\mu =\frac{3\cdot 1}{4}=0.75\text{kg}$, relative speed $u_1-u_2=4-(-2)=6\text{m/s}$: $x=(u_1-u_2)\sqrt{\mu /k}=6\sqrt{0.75/500}=6(0.03873)=\boxed{0.232\text{m}}.$ Kyun relative speed = 6, 2 nahi: wo ek doosre ki taraf approach kar rahe hain, isliye unki closing speed magnitudes ka sum hai; formula $u_1-u_2$ automatically yahi capture karta hai jab signs sahi hon.

(i) ${\text{KE}}_0=\frac{1}{2}(3)(4^2)+\frac{1}{2}(1)(2^2)=24+2=26\text{J}$. (ii) bulk KE $=\frac{1}{2}(4)(2.5^2)=\frac{1}{2}(4)(6.25)=12.5\text{J}$. (iii) stored PE $=\frac{1}{2}k{x}^2=\frac{1}{2}(500)(0.232^2)=\frac{1}{2}(500)(0.0540)=13.5\text{J}$. Check: $26-12.5=13.5\text{J}$ ✓, aur $\frac{1}{2}\mu (u_1-u_2{)}^2=\frac{1}{2}(0.75)(36)=13.5\text{J}$ ✓. Teeno agree karte hain — spring exactly CM-frame KE hold karta hai.

Wall case → sirf energy: $\frac{1}{2}m{u}^2=\frac{1}{2}k{x}^2\Rightarrow k=\frac{m{u}^2}{x^2}$. $k=\frac{5(4^2)}{0.10^2}=\frac{80}{0.01}=\boxed{8000\text{N/m}}.$ "Minimum" kyun: zyada stiff spring use $0.10\text{m}$ se kam mein rok leti hai, jo theek hai; softer spring $0.10\text{m}$ se zyada squish karti hai, limit violate karti hai. Equality borderline value deta hai.

Compression formula invert karo. $\mu =\frac{2\cdot 3}{5}=1.2\text{kg}$, aur $x=(u_1-0)\sqrt{\mu /k}$ toh $u_1=\frac{x}{\sqrt{\mu /k}}=x\sqrt{k/\mu }=0.20\sqrt{750/1.2}=0.20\sqrt{625}=0.20(25)=\boxed{5\text{m/s}}.$ Kyun kaam karta hai: relative approach speed stored energy set karti hai, aur yahan $u_2=0$ toh relative speed hi $u_1$ hai.

Jab $m_2\to \infty$: $\mu =\frac{m_1{m}_2}{m_1+m_2}=\frac{m_1}{1+m_1/m_2}\to m_1$. Aur relative speed $u_1-u_2=u_1$. Toh $x\to u_1\sqrt{m_1/k},$ jo exactly wall formula hai $m=m_1$, $u=u_1$ ke saath. Kyun sense banata hai: infinitely heavy $m_2$ kabhi move nahi karta — wo hi ek wall hai. Aur $v_{cm}=\frac{m_1{u}_1}{m_1+m_2}\to 0$, toh koi bulk KE nahi bachti, saari spring mein jaati hai, wall case se match karta hai. Dekho Centre of Mass Frame.

Stored energy $=\frac{1}{2}\mu (u_1-u_2{)}^2$ relative speed par depend karti hai, jo total momentum se independent hai. Toh ek hi $p$ wale do setups wildly different energies store kar sakte hain — split matter karta hai. Extreme case CM frame ka apna frame hai: ek given launch ke liye maximum relative speed. $m_1=m_2=m$, $u_1=+u$, $u_2=-u$ ke liye: $\mu =\frac{m\cdot m}{2m}=\frac{m}{2}$, relative speed $u_1-u_2=2u$, toh $E_{\text{stored}}=\frac{1}{2}\cdot \frac{m}{2}\cdot (2u{)}^2=\frac{1}{2}\cdot \frac{m}{2}\cdot 4u^2=\boxed{m{u}^2}.$ Aur $v_{cm}=0$ (total momentum zero) — toh saari KE $=2\cdot \frac{1}{2}m{u}^2=m{u}^2$ store hoti hai, jo Simple Harmonic Motion se links ke saath consistent hai jahan compressed spring turning point hai.

Maano $x_1,x_2$ positions hain; har ek par spring force $\pm k(\text{stretch})$ hai. Har ek ke liye Newton's law likhne aur relative coordinate $s=x_1-x_2$ form karne par milta hai $\mu \ddot{s}=-ks$ — SHM equation effective mass $\mu$ ke saath. Isliye $\omega =\sqrt{k/\mu },T=2\pi \sqrt{\mu /k}.$ $\mu =\frac{2\cdot 3}{5}=1.2\text{kg}$ ke saath: $T=2\pi \sqrt{1.2/750}=2\pi \sqrt{0.0016}=2\pi (0.04)=\boxed{0.251\text{s}}$. Kyun phir $\mu$: two-body relative motion ek body jaisi behave karti hai mass $\mu$ ke saath spring par — wohi reduced-mass idea jo $x_{max}$ govern karta tha.