Spring-mass systems — collision problems: Worked examples

1.3.13 · D3 · Physics › Work, Energy & Power › Spring-mass systems — collision problems

Yeh page drill floor hai. Parent note ne do-phase idea build kiya tha (momentum collision ke moment mein, energy compression ke liye). Yahaan hum har tarah ki situation dhundhte hain jo yeh topic de sakta hai — velocity ke har sign, wall aur free blocks dono, equal-mass surprise, huge-mass limit, ek latched (locked) spring, aur ek real-world word problem — aur har ek ko line one se solve karte hain.

Koi bhi formula aane se pehle, yeh vocabulary hai jo hum use karenge, plain words mein:

Scenario matrix

Is topic ka har problem is grid ka ek cell hai. Neeche ke worked examples cell ke naam se label hain.

Cell	Kya change hota hai	Kaun sa law(s)	Example
A	Ek block wall-fixed spring se takraata hai	Sirf Energy (wall = external force)	Ex 1
B	Do free blocks, target rest mein ( $u_{2} = 0$ )	Momentum + Energy	Ex 2
C	Do free blocks, head-on ( $u_{1} > 0$ , $u_{2} < 0$ )	Momentum + Energy, signs dhyan se	Ex 3
D	Do free blocks, same direction chase ( $u_{1}, u_{2}$ dono $+$ )	Momentum + Energy	Ex 4
E	Equal masses $m_{1} = m_{2}$ (degenerate: velocities swap ho jaati hain)	Elastic-collision limit	Ex 5
F	Huge target $m_{2} ≫ m_{1}$ (limit → wall case)	Limit check	Ex 6
G	Spring latched at max compression (inelastic outcome)	Momentum, energy trapped	Ex 7
H	Real-world word problem (bumper cars)	Full toolkit	Ex 8

Hum cover karte hain: har velocity ke dono signs, zero input ( $u_{2} = 0$ ), degenerate equal-mass case, $m_{2} \to \infty$ limit, elastic vs inelastic outcome branch, aur ek word problem.

Cell A — Block into a wall-fixed spring

Forecast: Kya block rukega aur seedha same speed se bounce karega, ya slower? Padhne se pehle guess karo.

Kaun sa law? Yeh step kyun? Wall fixed hai — woh spring par koi bhi force laga sakti hai. Yeh block par ek external horizontal force hai, isliye momentum conserved nahi hai. Lekin kuch bhi energy waste nahi karta (frictionless, ideal spring), isliye mechanical energy conserved hai. Isliye Cell A odd one out hai: no momentum, sirf energy.
Energy before = energy at max compression set karo. Max compression par block momentarily rest mein hota hai (wall move nahi karti, isliye "common velocity" zero hai). Yeh step kyun? Saari kinetic energy ko spring ke alawa kahin jaana nahi hai: $\frac{1}{2} m u^{2} = \frac{1}{2} k x_{ma x}^{2} \Rightarrow x_{ma x} = u \frac{m}{k}$
Plug in karo. $x_{ma x} = 3 \frac{2}{600} = 3 \times 0.05774 = 0.173 m$
Release velocity. Yeh step kyun? Ek ideal spring 100% stored energy wapas karta hai, isliye block same speed se lekin reversed direction mein nikalta hai: $v^{'} = - 3 m/s$ .

Verify karo: $u m / k$ ke units: $\frac{m}{s} \frac{kg}{N/m} = \frac{m}{s} \frac{kg \cdot m}{kg \cdot m/s ^{2}} = \frac{m}{s} \cdot s = m$ . ✓ Length, accha. Energy check: $\frac{1}{2} (2) (3^{2}) = 9 J$ in, $\frac{1}{2} (600) (0.17 3^{2}) = 9.0 J$ stored. ✓

Cell B — Do free blocks, target at rest

Forecast: Kya light block bounce back karega, ya forward slower chalta rahega?

(a) Common velocity. Yeh step kyun? Max compression tab hoti hai jab dono saath move kar rahe hote hain — Phase 2 mein momentum conserved hai (ab koi wall nahi): $v = \frac{2 ( 6 ) + 4 ( 0 )}{2 + 4} = \frac{12}{6} = 2 m/s$
(b) Max compression. Reduced mass kyun? Sirf relative-motion kinetic energy store hoti hai; poora system $v$ par glide karta rehta hai. Woh relative KE hai $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ : $μ = \frac{2 \times 4}{6} = \frac{4}{3}, x_{ma x} = ∣6 - 0∣ \frac{4/3}{1200} = 6 \times 0.03333 = 0.200 m$
(c) Final velocities. Elastic kyun? Ideal spring har joule wapas deta hai → poora encounter ek perfectly elastic collision hai (dekho Elastic and Inelastic Collisions): $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} = \frac{- 2}{6} (6) = - 2 m/s, v_{2}^{'} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{4}{6} (6) = 4 m/s$

Figure — Spring-mass systems — collision problems

Verify karo: Momentum: before $2 (6) = 12$ ; after $2 (- 2) + 4 (4) = - 4 + 16 = 12$ . ✓ Energy: before $\frac{1}{2} (2) (36) = 36 J$ ; after $\frac{1}{2} (2) (4) + \frac{1}{2} (4) (16) = 4 + 32 = 36 J$ . ✓ Light block bounce back karta hai — Forecast test!

Cell C — Head-on collision (opposite signs)

Forecast: Max squeeze par combined system kaunsi taraf drift kar raha hoga — left ya right?

Pehle signs fix karo. Yeh step kyun? Yahaan $\pm$ discipline matter karti hai. Rightward $= +$ , isliye $u_{1} = + 4$ , $u_{2} = - 6$ . Approach speed gap hai $u_{1} - u_{2} = 4 - (- 6) = 10 m/s$ — woh $10 m/s$ par close in kar rahe hain, dono individual speeds se zyada tez. Isliye hamne $∣ u_{1} - u_{2} ∣$ likha tha: ek negative ko subtract karna add karta hai.
Common velocity. $v = \frac{3 ( 4 ) + 1 ( - 6 )}{3 + 1} = \frac{12 - 6}{4} = \frac{6}{4} = 1.5 m/s$ Positive → pair max compression par right ki taraf drift karta hai (heavier, rightward block tug jeetta hai).
Max compression. $μ = \frac{3 \times 1}{4} = 0.75, x_{ma x} = ∣4 - (- 6) ∣ \frac{0.75}{500} = 10 \times 0.03873 = 0.387 m$

Verify karo: Momentum: before $3 (4) + 1 (- 6) = 6$ ; at max compression $(3 + 1) (1.5) = 6$ . ✓ Stored energy $\frac{1}{2} μ (u_{1} - u_{2})^{2} = \frac{1}{2} (0.75) (100) = 37.5 J$ ; spring PE $\frac{1}{2} k x^{2} = \frac{1}{2} (500) (0.38 7^{2}) = 37.5 J$ . ✓ Drift rightward hai — Forecast!

Cell D — Same-direction chase (dono positive)

Forecast: Jo gap yeh close karte hain woh sirf $8 - 3 = 5$ hai, $8$ nahi. Kya spring zyada ya kam squish hoga agar $m_{2}$ rest mein hota?

Common velocity. $v = \frac{2 ( 8 ) + 3 ( 3 )}{2 + 3} = \frac{16 + 9}{5} = \frac{25}{5} = 5 m/s$
Approach speed difference hai, sum nahi. Yeh step kyun? Dono right move kar rahe hain, isliye spring ko sirf relative closing speed $∣8 - 3∣ = 5 m/s$ "feel" hoti hai. Ek chasing collision same absolute speeds par head-on se kam squish karta hai — yeh Cell C ke saath key contrast hai. $μ = \frac{2 \times 3}{5} = 1.2, x_{ma x} = 5 \frac{1.2}{400} = 5 \times 0.05477 = 0.274 m$

Verify karo: Momentum: before $2 (8) + 3 (3) = 25$ ; after $(5) (5) = 25$ . ✓ Stored energy $\frac{1}{2} (1.2) (5^{2}) = 15 J$ ; spring PE $\frac{1}{2} (400) (0.27 4^{2}) = 15.0 J$ . ✓

Cell E — Equal masses (degenerate case)

Forecast: Kuch famous hota hai jab equal masses elastically collide karte hain. Dono final speeds guess karo.

Final velocities — dekho formula kaise collapse karta hai. Yeh step kyun? Elastic results mein $m_{1} = m_{2} = m$ plug karo: $v_{1}^{'} = \frac{m - m}{2 m} u_{1} = 0, v_{2}^{'} = \frac{2 m}{2 m} u_{1} = u_{1} = 5 m/s$ Moving block bilkul ruk jaata hai; target incoming speed se nikalta hai. Woh velocities swap karte hain — equal-mass elastic collision ki pehchaan.
Max compression. $μ = \frac{2 \times 2}{4} = 1, x_{ma x} = 5 \frac{1}{800} = 5 \times 0.03536 = 0.177 m$

Verify karo: Momentum: before $2 (5) = 10$ ; after $2 (0) + 2 (5) = 10$ . ✓ Energy: before $\frac{1}{2} (2) (25) = 25 J$ ; after $\frac{1}{2} (2) (25) = 25 J$ . ✓ Velocity swap confirmed — Forecast!

Cell F — Huge target (limit back to a wall)

Forecast: Agar $m_{2}$ essentially immovable hai, toh light block kya karega — aur yeh wall se takraane ke kitna close hoga?

Common velocity ≈ 0. Yeh step kyun? Ek tiny mass jo ek mountain ke saath momentum share kare, mountain ko barely move karti hai: $v = \frac{1 ( 4 ) + 1000 ( 0 )}{1001} = \frac{4}{1001} = 0.003996 m/s \approx 0$ Isliye max compression par dono essentially rest mein hain — bilkul wall picture jaisa.
Reduced mass ≈ $m_{1}$ . Yeh step kyun? $μ = \frac{1 \times 1000}{1001} = 0.999 kg \approx m_{1}$ . Jab $m_{2} \to \infty$ , $μ \to m_{1}$ , isliye $x_{ma x} \to u_{1} m_{1} / k$ — Cell A wall formula. $x_{ma x} = 4 \frac{0.999}{250} = 4 \times 0.06322 = 0.2529 m$
Rebound velocity ≈ $- u_{1}$ . $v_{1}^{'} = \frac{1 - 1000}{1001} (4) = \frac{- 999}{1001} (4) = - 3.992 m/s \approx - 4 m/s$ Yeh apni incoming speed se (almost) bounce back karta hai — bilkul wall se jaisa.

Verify karo: Wall-formula prediction $u_{1} m_{1} / k = 4 1/250 = 0.2530 m$ ; hamara $0.2529 m$ $< 0.1%$ se match karta hai. ✓ $v_{1}^{'} \approx - 4$ ✓.

Cell G — Latched spring (inelastic outcome)

Forecast: Max compression ki maths Cell B se identical hai — lekin kya light block phir bhi bounce back karega?

Same $x_{ma x}$ . Yeh step kyun? Squeezing phase max compression ke instant tak unchanged hai, isliye $x_{ma x} = 0.200 m$ bilkul Cell B jaisa. Sirf aage kya hota hai woh change hota hai.
Woh saath rehte hain. Yeh step kyun? Spring locked hone ke saath, blocks ab ek single rigid object hain jo common velocity par move kar rahe hain — yahi perfectly inelastic collision ki definition hai: $v_{final} = v = 2 m/s (dono blocks, hamesha ke liye)$
Trapped energy. Yeh step kyun? Relative-motion KE jo Cell B ne wapas ki thi woh ab compressed spring mein permanently held hai: $E_{trapped} = \frac{1}{2} μ (u_{1} - u_{2})^{2} = \frac{1}{2} (\frac{4}{3}) (6^{2}) = 24 J$

Verify karo: KE before $= \frac{1}{2} (2) (36) = 36 J$ . KE after (moving together) $= \frac{1}{2} (6) (2^{2}) = 12 J$ . Lost $= 36 - 12 = 24 J = E_{trapped}$ . ✓ Momentum still conserved: $12 = (6) (2)$ . ✓ Koi bounce-back nahi — Forecast contrast confirmed.

Cell H — Real-world word problem (bumper cars)

Forecast: Kya car A rukegi, forward roll karti rahegi, ya reverse ho jaayegi? Uska mass B se bada hai.

Common velocity at deepest squeeze. $v = \frac{150 ( 2.0 ) + 100 ( 0 )}{250} = \frac{300}{250} = 1.2 m/s$
Reduced mass aur compression. $μ = \frac{150 \times 100}{250} = 60 kg, x_{ma x} = ∣2.0 - 0∣ \frac{60}{30000} = 2.0 \times 0.04472 = 0.0894 m$ Toh bumper lagbhag 8.9 cm squish hota hai — ek believable, visible dent depth.
Separation ke baad speeds (elastic bumper). $v_{1}^{'} = \frac{150 - 100}{250} (2.0) = \frac{50}{250} (2.0) = 0.4 m/s, v_{2}^{'} = \frac{2 ( 150 )}{250} (2.0) = \frac{300}{250} (2.0) = 2.4 m/s$ Car A slowly forward drift karta rehta hai ( $+ 0.4$ ); car B $2.4 m/s$ pe aage shoot karta hai.

Verify karo: Momentum: before $150 (2.0) = 300$ ; after $150 (0.4) + 100 (2.4) = 60 + 240 = 300$ . ✓ Energy: before $\frac{1}{2} (150) (4) = 300 J$ ; after $\frac{1}{2} (150) (0.16) + \frac{1}{2} (100) (5.76) = 12 + 288 = 300 J$ . ✓ Car A reverse nahi hua (B se heavier hai) — Forecast!

Recall Main kaun se cell mein hoon? (quick decision list)

Kya koi wall involved hai? ::: Cell A — sirf energy, no momentum, $x = u m / k$ . Do free blocks — squeeze ke dauran kaun sa law? ::: $v$ ke liye Momentum, $x_{ma x}$ ke liye energy; hamesha dono. Blocks ek doosre ki taraf move kar rahe hain (opposite signs)? ::: Cell C — approach speed = $∣ u_{1} - u_{2} ∣$ magnitudes ko add karta hai. Blocks chasing kar rahe hain (same sign)? ::: Cell D — approach speed = difference; kam squish hota hai. Equal masses, elastic? ::: Cell E — velocities swap ho jaati hain; incoming block ruk jaata hai. Ek mass enormous? ::: Cell F — wall case mein reduce ho jaata hai; $μ \to m_{1}$ , block $- u_{1}$ par rebound karta hai. Spring latched ho jaaye? ::: Cell G — same $x_{ma x}$ , lekin outcome perfectly inelastic hai; energy $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ trapped ho jaati hai.

Worked examples — Spring-mass systems — collision problems