Spring-mass systems — collision problems: Visual walkthrough

1.3.13 · D2 · Physics › Work, Energy & Power › Spring-mass systems — collision problems

Yeh page do-block spring collision ka result bilkul zero se rebuild karti hai, ek picture ek time. Hum har symbol ko pehle kamayenge, phir use karenge — aur end tak tumhe dikhega kyun spring ki maximum squish exactly yeh hoti hai:

$x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} .$

Agar koi bhi symbol yahan scary lage — achha hai. Hum har ek se dheere milenge. Parent note: 1.3.13 Spring-mass systems — collision problems (Hinglish).

Step 1 — Do carts draw karo aur sab kuch name karo

KYA. Do blocks frictionless ice par slide kar rahe hain. Left wala, mass $m_{1}$ , right ki taraf speed $u_{1}$ se ja raha hai. Right wala, mass $m_{2}$ , right ki taraf speed $u_{2}$ se ja raha hai. Ek spring of stiffness $k$ unme se kisi ek par glued hai.

KYUN. Kisi bhi physics se pehle, hume names aur ek direction fix karni padegi, warna baad mein ek "minus sign" ka koi matlab nahi hoga. Hum choose karte hain rightward = positive. Har velocity ek signed number hai: ek block jo left drift kar raha hai uska $u$ bas negative hoga.

PICTURE. Figure dekho. Do arrows velocities hain; unki lengths speeds hain. Kyunki $u_{1}$ ko $u_{2}$ se lamba draw kiya gaya hai, left block catch up kar raha hai — yehi ek wajah hai ki spring kabhi touch karegi.

Figure — Spring-mass systems — collision problems

Spring compress ho, iske liye left block ka right block par gain karna zaroori hai, yani approach speed $u_{1} - u_{2}$ positive honi chahiye. Is quantity ko mind mein rakho — yeh poori story ka hero hai.

Step 2 — Collision instant: kyun momentum, energy nahi

KYA. Pehli touch par, hum poochte hain: contact ke instant ke paas se kaunsi quantity safe hai carry karne ke liye?

KYUN. Spring $k x$ force se push karta hai — jahan $x$ kitna squish hua hai. Pehli touch par $x = 0$ hai, toh force essentially zero hai, aur thodi der baad bhi yeh sirf ek finite force hai jo tiny time ke liye act karta hai. Force $\times$ time ko impulse kehte hain, aur yahi ek cheez hai jo momentum change kar sakti hai. Ek finite force near-zero time mein near-zero impulse deta hai.

PICTURE. Figure mein spring force $k x$ zero se shuru hoke gently rise karta dikhta hai — kuch bhi sudden nahi. Isse compare karo hammer-hard rigid collision se (dashed spike). Spring kabhi spike nahi karta, isliye momentum bas bina chuye flow karta rehta hai.

Step 3 — "Maximum compression" ka asli matlab kya hai

KYA. Spring $x$ length se squish hota hai jo time ke saath change karta rehta hai. Hum chahte hain $x$ ki sabse badi value jo kabhi reach ho, use $x_{ma x}$ kehte hain.

KYUN. Squish $x$ tab badhta hai jab do blocks paas aate hain aur tab ghatta hai jab woh alag hote hain. Unke beech gap tab hi ghatta hai jab left block still faster ho right se: jab $v_{1} > v_{2}$ . Gap ghatta band ho jaata hai — toh $x$ bhi badhna band — exactly us instant par jab do velocities equal ho jaati hain.

PICTURE. Do velocity curves time ke against draw hain: fast block slow ho raha hai, slow block speed up ho raha hai. Jahan curves cross karti hain (same height = same velocity) wahan squish sabse deep hai. Cross se pehle, woh approach karte hain; cross ke baad, woh separate hote hain.

Khud prove karo: niche wajah reveal karo.

Max compression par velocities equal kyun honi chahiye?

Agar woh alag hote, ek block abhi bhi doosre par close kar raha hota (ya usse open ho raha hota), toh

x

abhi bhi change ho raha hota — matlab abhi maximum/turning point par nahi pahuncha.

Step 4 — Shared velocity $v$ solve karo

KYA. Max-compression instant par dono blocks same $v$ par move karte hain. Step 2 ki momentum equation mein $v_{1} = v_{2} = v$ plug karo.

KYUN. Momentum conserved tha (Step 2), aur ab hum jaante hain ki dono final speeds same number $v$ hain. Yeh do unknowns ko ek mein collapse kar deta hai, jise hum solve kar sakte hain.

PICTURE. Figure mein pair briefly ek velocity $v$ par locked dikhti hai — dono blocks par same arrow — beech mein spring apni deepest squish par hai.

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$

Left side: total incoming momentum (Step 2 se unchanged).
$(m_{1} + m_{2})$ : do blocks ab ek lump ki tarah move kar rahe hain, toh unki masses add hoti hain.
$v$ : woh ek velocity jise woh share karte hain.

Step 5 — Energy books: kinetic energy kahaan gayi?

KYA. Ab tools switch karo. Jab blocks contact mein hain aur move kar rahe hain, sirf ek conservative, frictionless spring force hai, toh mechanical energy conserved hai pehli touch se max compression tak.

KYUN. Kinetic energy (motion ki energy, $\frac{1}{2} m v^{2}$ ) vanish nahi hoti — yeh squished spring mein elastic potential energy ki tarah store ho jaati hai (dekho Elastic Potential Energy of a Spring). Likho "KE before = KE at max squish + energy locked in spring."

PICTURE. Ek bar chart: tall "before" KE bar do hisso mein split hoti hai — ek chhoti "still-moving" KE bar (poora lump $v$ par glide karta hua) aur ek orange "stored in spring" bar. Stored bar woh piece hai jo hum chahte hain.

$KE before touching \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = KE of the moving lump \frac{1}{2} (m_{1} + m_{2}) v^{2} + energy stored in the spring \frac{1}{2} k x_{ma x}^{2}$

$\frac{1}{2} m u^{2}$ — mass $m$ ki speed $u$ par kinetic energy.
$\frac{1}{2} (m_{1} + m_{2}) v^{2}$ — poora system still $v$ par glide kar raha hai; woh motion spring ko squish karne ke liye available nahi hai.
$\frac{1}{2} k x_{ma x}^{2}$ — spring $x_{ma x}$ se squish hone par jo energy hold karta hai.

Step 6 — Algebra: reduced mass $μ$ appear hota hai

KYA. Step 5 ko rearrange karo stored energy isolate karne ke liye, phir Step 4 se $v$ substitute karo.

KYUN. Hum $x_{ma x}$ chahte hain given numbers ( $m_{1}, m_{2}, u_{1}, u_{2}, k$ ) mein, derived $v$ mein nahi. Jab hum substitute karke simplify karte hain, ek khoobsurat cancellation sab kuch ek tidy grouping mein pack kar deta hai.

PICTURE. Figure mein stored-energy expression term by term shrink hoti dikhti hai, single compact block $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ mein — messy sum ek neat parcel mein fold ho jaati hai.

Stored energy isolate karo:

$\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} (m_{1} + m_{2}) v^{2} .$

$v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ substitute karke aur algebra grind karne par, sab kuch collapse ho jaata hai:

$\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} μ (u_{1} - u_{2})^{2}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}} .$

$(u_{1} - u_{2})$ — approach speed Step 1 se. Sirf relative motion hi spring ko squish kar sakti hai.
$μ$ — reduced mass: do bodies ki relative motion ke liye "effective single mass" (dekho Reduced Mass and Two-Body Problems).

Boxed line ko $x_{ma x}$ ke liye solve karo:

$x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}$

Step 7 — Sanity: har edge aur degenerate case

KYA. Formula ko extreme inputs ke against test karo. Ek achha formula kabhi break nahi hona chahiye.

KYUN. Agar koi limiting case nonsense deta hai, derivation galat hai. Har case niche ek "free experiment" hai.

PICTURE. Chaar mini-panels, ek case ke liye ek, har ek mein carts aur resulting squish dikhti hai.

Equal speeds, $u_{1} = u_{2}$ . Approach speed $= 0$ → $x_{ma x} = 0$ . Woh kabhi close in nahi karte, spring kabhi touch nahi karti. ✔
Target at rest, $u_{2} = 0$ . $x_{ma x} = u_{1} μ / k$ — standard textbook case. ✔
Bahut heavy target, $m_{2} \to \infty$ . Tab $μ \to m_{1}$ , aur $x_{ma x} \to u_{1} m_{1} / k$ — exactly wall result, kyunki infinite mass immovable wall ki tarah act karta hai. ✔ (Dekho Elastic and Inelastic Collisions.)
Blocks apart move kar rahe hain, $u_{1} < u_{2}$ . Approach speed negative hai; $(u_{1} - u_{2})^{2}$ phir bhi positive hai lekin spring pehle place mein engage hi nahi hoti, toh physically $x_{ma x} = 0$ — formula sirf tab apply hota hai jab contact shuru ho. ✔

Ek-picture summary

Sab kuch ek canvas par: do carts (Step 1) → momentum gentle spring force se flow karta hua (Step 2) → curves equal velocity $v$ par cross karti hain = deepest squish (Steps 3–4) → energy bar split hoti hai, aur stored slice $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ ke barabar hai (Steps 5–6) → $x_{ma x} = (u_{1} - u_{2}) μ / k$ milta hai.

Recall Feynman retelling — poora walkthrough simple words mein

Do carts ice par; fast wale ke paas springy bumper hai aur woh slow wale ko catch karta hai. Jab woh touch karte hain, push kuch nahi se shuru hoti hai aur gently build up hoti hai, isliye quick bump ke dauran hum bas kehte hain total momentum unchanged hai — koi sudden hammer-blow nahi jo use change kare. Ab spring squish hoti hai. Yeh tab tak squish hoti rehti hai jab tak fast cart abhi bhi ground gain kar raha hai; deepest squish us ek instant par hoti hai jab dono carts ek same speed $v$ par roll kar rahe hote hain — uske baad spring unhe alag dhakelta hai. Depth find karne ke liye, hum energy count karte hain: kuch motion energy as the whole pair glides at $v$ remain karti hai, aur bacha hua — jo exactly unke ek doosre ki taraf squeeze karne ki energy hai — wahi spring store karti hai. Woh bacha hua sirf is par depend karta hai ki woh kitni fast close in kar rahe the, $u_{1} - u_{2}$ , aur ek special combined mass $μ$ par. Us stored energy ko spring ki $\frac{1}{2} k x^{2}$ ke barabar set karo aur $x_{ma x} = (u_{1} - u_{2}) μ / k$ nikal aata hai. Har extreme case check out karta hai: same speed → no squish; infinite target → wall answer.

Connections

1.3.13 Spring-mass systems — collision problems (Hinglish)
Conservation of Linear Momentum
Elastic and Inelastic Collisions
Reduced Mass and Two-Body Problems
Elastic Potential Energy of a Spring
Centre of Mass Frame
Simple Harmonic Motion