Spring-mass systems — collision problems: Foundations

1.3.13 · D1 · Physics › Work, Energy & Power › Spring-mass systems — collision problems

Yeh Spring–Mass Collision Problems ka toolbox page hai. Hum tools ko usi order mein assemble karte hain jis order mein unhe use kiya jaata hai.

1. Mass $m$ — "dhakka dena kitna mushkil hai"

Picture: do boxes socho, ek chhota aur ek mota. Dono ko ek jaisi hand-force se push karo — mota wala dheeray speed up hoga. Woh speed up karne ki reluctance hi mass hai.

Topic ko iske zaroorat kyun hai: collisions poori tarah se is baat par hain ki do masses motion kaise share karte hain. Badi mass barely change hoti hai; chhoti mass flung ho jaati hai. Letters $m_{1}$ (block 1 ki mass) aur $m_{2}$ (block 2) sirf do boxes pe name-tags hain.

2. Velocity $u$ aur $v$ — "speed WITH direction"

Koi bhi number likhne se pehle hame ek direction choose karni hoti hai jise positive bolein. Parent note kehta hai "rightward positive lo." Woh ek choice hi velocity ko negative hone deti hai — negative velocity ka matlab sirf "left ki taraf move karna" hai.

Figure — Spring-mass systems — collision problems

3. Momentum $p = m v$ — "motion ki quantity"

Picture: imagine karo ki har block apne peeche ek bar kheench raha hai jiska length $m$ hai aur jiska direction velocity hai. Momentum woh bar hai. Jab do blocks collide karte hain, saari bars ki total length, unke signs ke saath add karke, change nahi hoti — yahi law hai jis par hum lean karte hain.

$m_{1} u_{1} + m_{2} u_{2}$ kyun, $m_{1} u_{1} - m_{2} u_{2}$ kyun nahi? Tum kabhi bhi purpose se subtract nahi karte — sign har ek velocity ke andar rehta hai. Agar block 2 left move kar raha hai, toh $u_{2}$ already negative hai, toh $m_{2} u_{2}$ already negative hai. Hamesha add karo; signs ko kaam karne do.

4. Impulse — "spring hit ke dauran invisible kyun hai"

Picture: momentum mein change ek bucket bharne jaisa hai. Ek firehose (huge force, short time) aur ek dripping tap (tiny force, long time) ek hi bucket bhar sakte hain. Collision ke dauran contact force ek firehose hai — enormous lekin ek blink mein khatam.

Yahan yeh kyun matter karta hai: spring force $k x$ us blink ke dauran ek dripping tap hai — finite force, near-zero time → near-zero impulse → woh barely momentum change karta hai. Yahi woh poori wajah hai ki hum collision ke dauran energy ki jagah momentum kyun use kar sakte hain. Yeh samajh lo aur "MEME" mnemonic sense karega: Momentum during the Moment.

5. Kinetic energy $\frac{1}{2} m v^{2}$ — "motion-fund"

$\frac{1}{2}$ aur square kyun? Speed double karne par KE chaar guna ho jaati hai (yeh square ki wajah se hai); $\frac{1}{2}$ woh exact bookkeeping factor hai jo energy ko force-times-distance ke saath balance karta hai. Tumhe yahan ise derive karne ki zaroorat nahi — bas itna trust karo ki $\frac{1}{2} m v^{2}$ "motion-fund ka size" hai.

6. Spring stiffness $k$ aur elastic PE $\frac{1}{2} k x^{2}$ — "energy bank"

Spring $k x$ force se push back karta hai — jitna zyada squish karo, utna zyada dhakka deta hai. Woh force finite hai (kabhi firehose nahi), aur yahi exact reason hai ki collision instant ke dauran ise ignore kiya ja sakta hai.

Picture: spring ek piggy bank hai. Kinetic energy andar jaati hai jab yeh squish hota hai ( $x$ barhta hai), phir bahar aati hai jab yeh un-squish hota hai. Ek ideal spring ek piggy bank hai jisme neeche koi hole nahi — woh har joule wapas karta hai.

7. Common velocity $v_{c m}$ — "woh moment jab dono ki speed match hoti hai"

Kyunki momentum just-after-contact se us instant tak conserved hai: $m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v \Rightarrow v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$

Formula ko words mein padho: total momentum divided by total mass = poore system ke balance-point ki velocity. Yahi wajah hai ise centre-of-mass velocity kehte hain — dekho Centre of Mass Frame.

8. Reduced mass $μ$ — "GAP ki effective mass"

Number ki feel: $μ$ hamesha dono masses se chhota hota hai. Agar ek block ek wall hai (infinite mass), toh $μ \to$ doosre block ki mass. Agar dono equal $m$ hain, toh $μ = m /2$ .

Iska apna place kyun earn kiya: spring kitni energy store kar sakta hai yeh sirf approach speed $(u_{1} - u_{2})$ aur is effective gap-mass $μ$ par depend karta hai: $spring mein stored \frac{1}{2} k x_{ma x}^{2} = relative-motion KE \frac{1}{2} μ (u_{1} - u_{2})^{2} \Rightarrow x_{ma x} = (u_{1} - u_{2}) \frac{μ}{k}$

$\frac{1}{2} (m_{1} + m_{2}) v^{2}$ wala chunk (poora system $v$ par drift kar raha hai) kabhi store nahi ho sakta — system abhi bhi ek lump ki tarah move kar raha hai. Sirf approaching wala part convert hota hai. Yeh parent note mein sabse zyada miss ki jaane wali idea hai.

Main ek block ka aur do blocks ka momentum likh sakta hoon

p = m v

; total

= m_{1} u_{1} + m_{2} u_{2}

, hamesha add karo signs baked in ke saath.

Main explain kar sakta hoon ki spring collision instant ke dauran "invisible" kyun hai

Iska force finite hai lekin time near-zero hai, toh iska impulse (force × time) ≈ 0 hai aur momentum unchanged rehta hai.

Main kinetic energy likh sakta hoon aur bata sakta hoon yeh kabhi negative kyun nahi hoti

K E = \frac{1}{2} m v^{2}

; velocity squared hai toh direction cancel ho jaata hai.

Main elastic PE likh sakta hoon aur spring ko bank ki tarah describe kar sakta hoon

P E = \frac{1}{2} k x^{2}

; ek ideal spring har joule store karta hai aur sab wapas karta hai.

Main bata sakta hoon ki compression maximum kab hoti hai

Jab dono blocks ek velocity

v

share karte hain (relative velocity

= 0

Main common velocity compute kar sakta hoon

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

— total momentum over total mass.

Main reduced mass compute kar sakta hoon aur bata sakta hoon ise kyun use karte hain

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

; yeh relative motion ki effective mass hai jise spring actually store karta hai.

Main

x_{ma x}

square root use karke nikaal sakta hoon

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

se, root lo:

x_{ma x} = (u_{1} - u_{2}) μ / k

Foundations — Spring-mass systems — collision problems

1. Mass $m$ — "dhakka dena kitna mushkil hai"

2. Velocity $u$ aur $v$ — "speed WITH direction"

3. Momentum $p = m v$ — "motion ki quantity"

4. Impulse — "spring hit ke dauran invisible kyun hai"

5. Kinetic energy $\frac{1}{2} m v^{2}$ — "motion-fund"

6. Spring stiffness $k$ aur elastic PE $\frac{1}{2} k x^{2}$ — "energy bank"

7. Common velocity $v_{c m}$ — "woh moment jab dono ki speed match hoti hai"

8. Reduced mass $μ$ — "GAP ki effective mass"

9. Square-root — "square ko undo karo"

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