Spring-mass systems — collision problems: Question bank

1.3.13 · D5 · Physics › Work, Energy & Power › Spring-mass systems — collision problems

Shuru karne se pehle, har symbol ko dobara anchor karte hain jo hum use karte hain, taaki kuch bhi abrupt na lage. Yeh us poore page ka setup hai — frictionless ice par do blocks ke beech ek spring:

Figure — Spring-mass systems — collision problems

Definition Symbols ki cast

$m_{1}, m_{2}$ — do masses (kg): ice par do blocks.
$u_{1}, u_{2}$ — encounter se pehle velocities; prime ( $u_{1}^{'}, u_{2}^{'}$ ) matlab poore encounter ke baad ki velocity. Rightward positive hai, isliye minus sign matlab "left taraf ja raha hai."
$k$ — spring constant (N/m): spring kitni stiff hai. Bada $k$ matlab spring usi squeeze ke liye zyada push back karti hai, toh compression ko zyada resist karti hai. Yeh stored energy $\frac{1}{2} k x^{2}$ mein aata hai aur isliye har compression formula ke andar bhi.
$v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ — centre-of-mass velocity: total momentum ko total mass se divide karo. Ise poore do-block system ka "average drift" samjho, jo tab kabhi nahi badalta jab koi bahari push na ho.
$μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ — reduced mass: woh ek "effective mass" jo decide karti hai ki do blocks ek-doosre ke relative kaise move karte hain. Dekho Reduced Mass and Two-Body Problems.
$x$ — spring kitni squish hui hai (m); $\frac{1}{2} k x^{2}$ uski stored elastic energy hai — dekho Elastic Potential Energy of a Spring.

Formula Is poore page ka ek formula — aur

kyun usme $μ$ hai Maximum compression par dono blocks $v_{c m}$ par saath drift karte hain, toh sirf woh KE store ho sakti hai jo drift KE se aage ho. Energy conservation deta hai: $\frac{1}{2} k x_{ma x}^{2} = start KE \frac{1}{2} (m_{1} u_{1}^{2} + m_{2} u_{2}^{2}) - drift KE at peak \frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2} .$ Right side $μ$ par kyun collapse hoti hai — $v_{c m} = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ substitute karo aur expand karo. Woh messy square deta hai, common denominator $2 (m_{1} + m_{2})$ par: $\frac{( m _{1} u _{1}^{2} + m _{2} u _{2}^{2} ) ( m _{1} + m _{2} ) - ( m _{1} u _{1} + m _{2} u _{2} ) ^{2}}{2 ( m _{1} + m _{2} )} .$ Numerator multiply out karo: $m_{1}^{2} u_{1}^{2}$ aur $m_{2}^{2} u_{2}^{2}$ terms square ke andar waale terms se cancel ho jaate hain, sirf cross terms bachte hain $m_{1} m_{2} u_{1}^{2} + m_{1} m_{2} u_{2}^{2} - 2 m_{1} m_{2} u_{1} u_{2} = m_{1} m_{2} (u_{1} - u_{2})^{2}$ . Toh poora cheez ban jaata hai: $\frac{m _{1} m _{2}}{2 ( m _{1} + m _{2} )} (u_{1} - u_{2})^{2} = \frac{1}{2} μ (u_{1} - u_{2})^{2} .$ Reduced mass $μ$ precisely isliye aata hai kyunki do-body system ki relative motion ek single body ki tarah behave karti hai jiska mass $μ$ ho. (Positive) compression depth ke liye solve karo: $x_{ma x} = ∣ u_{1} - u_{2} ∣ \frac{μ}{k} .$ Absolute value $x_{ma x} \geq 0$ rakhta hai chahe koi bhi block faster ho; yeh exactly zero hota hai jab blocks kabhi approach hi na karein. Neeche ke har TF aur edge-case is line par tikta hai; ise saamne rakho.

True or false — justify

TF1. Spring collision ke instant of contact mein kinetic energy conserved hoti hai.

Yeh ek general claim ke roop mein False hai — spring force hi contact force hai, aur bilkul pehle instant mein spring muskil se squish hoti hai, toh woh force

k x

(jab

x \approx 0

) aur uska impulse tiny hota hai; almost koi momentum ya energy exchange nahi hua. Tum encounter ko momentum se track karte ho, aur KE sirf poore encounter mein hi conserved hoti hai kyunki ideal spring baad mein sab kuch return karti hai jo usne store kiya tha.

TF2. Spring maximum compressed hoti hai theek jab incoming block momentarily rest mein ho.

False — yeh maximum tab hota hai jab dono blocks same velocity (

v_{c m}

) share kar rahe hon, yani relative velocity zero ho. Incoming block tab usually abhi bhi move kar raha hota hai.

TF3. Ideal spring se joined do free blocks ke liye, overall outcome ek perfectly elastic collision hai.

True — ek ideal spring KE store karti hai phir uska 100% return karti hai, toh before-and-after KE match karti hai, jo elastic ki definition hai (dekho Elastic and Inelastic Collisions).

TF4. Momentum conserved hoti hai jab ek block wall se bolted spring se takraata hai.

False — wall ek external horizontal force supply karta hai, isliye momentum conserved nahi hoti; sirf mechanical energy hoti hai.

TF5. Maximum compression par system ki kinetic energy zero hoti hai.

False — poora system abhi bhi

v_{c m}

par drift karta hai, isliye drift KE

\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}

bani rehti hai. Sirf relative-motion KE,

\frac{1}{2} μ (u_{1} - u_{2})^{2}

, spring mein stored hoti hai.

TF6. Agar ek clip spring ko maximum compression par latch kar le, toh encounter perfectly inelastic ban jaata hai.

True — blocks

v_{c m}

par saath lock ho jaate hain, aur stored

\frac{1}{2} μ (u_{1} - u_{2})^{2}

hamesha ke liye trap ho jaati hai, jo exactly inelastic ki pehchaan hai.

TF7.

m_{1}

aur

m_{2}

ke roles reverse karne se maximum compression change hoti hai.

False —

x_{ma x} = ∣ u_{1} - u_{2} ∣ μ / k

relative speed aur symmetric

μ

par depend karta hai (

m_{1} \leftrightarrow m_{2}

swap karne se

μ

unchanged rehta hai), isliye depth same hoti hai.

TF8. Agar dono blocks same velocity se start karein, toh spring phir bhi thoda compress hogi phir push back karegi.

False — shuru se zero relative velocity ke saath, gap kabhi nahi shrinkta, isliye

x_{ma x} = ∣ u_{1} - u_{2} ∣ μ / k = 0

. Kuch bhi compress nahi hota.

Spot the error

SE1. "Spring ideal hai, isliye energy hamesha conserved hai — toh collision mein main

\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} k x^{2}

use karunga."

Error timing ka hai: tum initial KE ko spring PE se first contact par equate nahi kar sakte, jahan

x \approx 0

hota hai aur almost kuch exchange nahi hua. Energy sirf tab clean hoti hai jab tum poori compression mein stored PE ko sahi track karo aur drift KE subtract karo.

SE2. "Do free blocks: saari KE spring mein jaati hai, isliye

\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} k x^{2}

Galat — system abhi bhi

v_{c m}

par move karta hai, drift KE carry karta hai jo store nahi ho sakti. Sirf CM-frame part

\frac{1}{2} μ (u_{1} - u_{2})^{2}

spring PE banta hai; tumhe

\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}

subtract karna hoga.

SE3. "Max compression tab hoti hai jab target block apni top speed reach kare."

Target ki speed max compression ke baad peak karti hai, jab spring usse forward push kar rahi hoti hai. Compression equal velocities par peak karti hai, spring re-expand hone se pehle.

SE4. "Block wall-spring se zyada speed ke saath bounce karta hai jitni se aaya tha, kyunki spring usse push karti hai."

Nahi — ek ideal spring exactly utni hi energy return karti hai jitni usne store ki thi, isliye block उसी speed se nikalta hai jिससे aaya tha (reversed direction mein). Speed gain karna energy conservation violate karega.

SE5. "Wall case mein (ek single block of mass

m

at speed

u

jo wall-fixed spring se takraata hai) mujhe phir bhi common velocity

v

dhundhne ke liye

m u = m v

likhna chahiye."

Koi doosra free body nahi hai jiske saath momentum share ho, aur wall momentum ko non-conserved banata hai, isliye woh equation yahan meaningless hai. Sahi tool energy hai:

\frac{1}{2} m u^{2} = \frac{1}{2} k x^{2}

SE6. "Kyunki

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

dono masses se chhota hai, expected se kam energy store hoti hai — iska matlab energy lost hui."

Koi energy lost nahi hoti;

μ

sirf relative motion ke liye correct effective mass hai. "Missing" KE destroy nahi hoti — yeh drift KE

\frac{1}{2} (m_{1} + m_{2}) v_{c m}^{2}

hai jo kabhi store hone ke liye available hi nahi thi.

SE7. "Ek light block (

m_{1} < m_{2}

) jo heavier block se takraata hai woh kabhi backward nahi ja sakta, isliye negative final velocity

u_{1}^{'}

ek galti hogi."

Ek light block heavier block se genuinely rebound karta hai; elastic result

u_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}

negative hota hai jab

m_{1} < m_{2}

, aur minus sign sirf leftward matlab rakhta hai. Momentum aur energy dono check out karte hain, isliye rebound real hai.

Why questions

WQ1. Hum problem ko "collision phase" aur "compression phase" mein split kyun karte hain?

Kyunki alag forces alag timescales par dominate karti hain — huge, brief contact force ko momentum chahiye (impulse view), jabki slow, finite spring force ko energy chahiye. Ek law per phase, us force ke hisaab se choose karo jo negligible hai.

WQ2. Reduced mass

μ

kyun

x_{ma x}

mein aata hai lekin

v_{c m}

mein kabhi nahi?

v_{c m}

poore system ke drift ke baare mein hai, isliye total mass use karta hai;

x_{ma x}

blocks ki relative squeezing ke baare mein hai, aur relative motion ek single body ki tarah behave karti hai jiska mass

μ

ho. Dekho Reduced Mass and Two-Body Problems.

WQ3. Hum poore spring encounter ko elastic kyun treat kar sakte hain bina kabhi coefficient of restitution compute kiye?

Kyunki "ideal spring saari stored energy return karti hai" hi restitution

= 1

ka physical content hai; spring mechanism elastic result ko construction se guarantee karta hai.

WQ4. Maximum compression ka moment momentum conservation apply karne ke liye natural jagah kyun hai?

Us instant par dono blocks ek velocity share karte hain,

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

ko ek single-unknown equation mein badal deta hai — problem ka sabse clean snapshot.

WQ5. Energy conservation ek ordinary (non-spring) inelastic collision ke dauran fail kyun hoti hai lekin poore spring encounter mein succeed karti hai?

Inelastic collisions KE ko heat/deformation mein convert karti hain (unrecoverable); ek ideal spring sirf KE borrow karti hai recoverable elastic PE ke roop mein aur sab wapas kar deti hai.

WQ6. Centre of Mass Frame mein problem dekhne se stored energy obvious kyun ho jaati hai?

Us frame mein total momentum zero hota hai, isliye max compression par sab kuch momentarily still hota hai aur frame ki saari KE — exactly

\frac{1}{2} μ (u_{1} - u_{2})^{2}

— spring mein baith jaati hai.

WQ7. Max-compression formula same kyun hai chahe spring baad mein release ho ya latch ho jaaye?

Turning point tak compression phase physically dono cases mein identical hai; latching ya releasing sirf decide karta hai ki peak ke baad kya hoga, na ki squeeze kitna gehra gaya.

Edge cases

EC1. Agar

u_{1} = u_{2}

ho (dono blocks saath move kar rahe hon) toh

x_{ma x}

kya hai?

Zero — relative approach speed

∣ u_{1} - u_{2} ∣ = 0

, isliye

x_{ma x} = ∣ u_{1} - u_{2} ∣ μ / k = 0

; woh simply saath coast karte hain.

EC2. Spring stiffness

k \to \infty

hone par

x_{ma x}

ka kya hota hai?

x_{ma x} = ∣ u_{1} - u_{2} ∣ μ / k \to 0

— infinitely stiff spring muskil se compress hoti hai, aur encounter ek instantaneous elastic bounce jaisi lagti hai.

EC3. Agar

m_{2} \to \infty

ho (target immovable, jaise wall) toh kya hoga?

μ \to m_{1}

aur

v_{c m} \to 0

, wall case recover hota hai:

x_{ma x} = ∣ u_{1} ∣ m_{1} / k

jisme block 1 ki saari KE stored hoti hai.

EC4. Agar two-free-block elastic result mein

m_{1} = m_{2}

ho toh kya hoga?

Blocks velocities swap karte hain — incoming block dead stop ho jaata hai aur target

u_{1}

par move off karta hai, kyunki

u_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} = 0

EC5. Do blocks jisme spring block 2 ke front face se attached hai (toh contact sirf tab hoga jab block 1 catch up kare): agar woh apart start ho rahe hain,

u_{1} < u_{2}

, toh kya hoga?

Kyunki block 1 peeche reh raha hai, uska face spring ke free end ko kabhi press nahi karta, isliye koi compression nahi hoti aur blocks simply unchanged drift apart ho jaate hain. (Agar spring dono blocks se glued hoti, toh apart move karna usse stretch karta — ek alag, tension problem jo is note se bahar hai.)

EC6.

k \to 0

hone par (spring vanishingly soft),

x_{ma x} = ∣ u_{1} - u_{2} ∣ μ / k

kya predict karta hai, aur kya yeh physical hai?

Mathematically

x_{ma x} \to \infty

, lekin yeh ek idealisation hai: ek real spring ki finite length hoti hai aur coils eventually touch karti hain (ya blocks collide karte hain), isliye formula sirf tab valid hai jab spring ideally behave kare. Practically, ek bahut soft spring matlab ek huge lekin bounded squeeze aur ek bahut gentle, slow interaction.

EC7. Agar blocks ki equal masses hon aur equal-but-opposite velocities hon (

u_{1} = - u_{2}

) toh

v_{c m}

kya hai?

Zero — total momentum cancel hoti hai, isliye max compression par dono momentarily rest mein hote hain aur saari initial KE,

\frac{1}{2} μ (2 u_{1})^{2}

, stored hoti hai (Centre of Mass Frame aur lab frame yahan coincide karte hain).

Recall One-line survival kit

MEME + Max squish = Same swish. Momentum during the Moment; Energy for the Expansion; aur deepest squeeze tab hota hai jab dono blocks same speed par swish karein. Is poore page ki har cheez inn do lines ka consequence hai.

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