Perfectly inelastic collisions — maximum KE loss: Visual walkthrough

1.4.7 · D2 · Physics › Momentum & Collisions › Perfectly inelastic collisions — maximum KE loss

Hum sirf yeh ideas use karte hain, aur har ek ko build karte hain:

mass — ek lump mein kitna "stuff" hai (ek number, kilograms mein);
velocity — koi cheez kitni tez aur kis direction mein move karti hai (ek arrow: length = speed, direction = jis taraf point kare);
momentum — mass times velocity, woh "push" jo ek moving lump carry karta hai;
kinetic energy — ek moving lump mein stored "motion-energy."

Baaki sab hum derive karte hain.

Step 1 — Touch karne se pehle do lumps draw karo

KYA. Do clay blobs ek seedhi line par slide karte hain. Baaye wale ki mass $m_{1}$ hai aur woh velocity $u_{1}$ se move karta hai; daaye wale ki mass $m_{2}$ hai aur velocity $u_{2}$ hai. Velocity ko ek arrow ki tarah draw kiya jaata hai: lamba arrow = tez, aur arrow us direction mein point karta hai jis taraf lump travel karta hai. Daaye ko hum $+$ kehte hain, baaye ko $-$ .

KYUN. Isse pehle ki hum conservation ki baat karein, hume ek positive direction fix karni hogi taaki har speed ek signed number ban sake. Signs ke bina, ek head-on crash aur ek rear-end bump ek jaisi dikhti hain — par woh hoti nahi hain.

PICTURE. Neeche, pale-yellow arrow ( $u_{1}$ ) blue arrow ( $u_{2}$ ) se lamba hai, isliye left blob catch up kar raha hai.

Figure — Perfectly inelastic collisions — maximum KE loss

Step 2 — Har lump ko ek momentum arrow do

KYA. Momentum hai $p = m u$ : velocity arrow lo aur use mass se stretch karo. Ek bhaari slow lump utna hi push carry kar sakta hai jitna ek halka fast lump.

KYUN. Ek hi fact har collision se bachta hai woh yeh hai ki total momentum nahi badalta (bahar se koi pair ko push nahi karta — dekho Conservation of Linear Momentum). Isliye momentum, velocity nahi, woh quantity hai jise hum crash ke through track karte hain. Hum momentum arrows par switch karte hain kyunki woh arrows impact ke baad bhi same add hote rahenge.

PICTURE. Har blob ab ek mota momentum arrow carry karta hai $p_{1} = m_{1} u_{1}$ aur $p_{2} = m_{2} u_{2}$ . Unka tip-to-tail sum (pink) hai total momentum $p_{tot}$ .

Step 3 — Woh stick karte hain: ek lump, ek velocity

KYA. Crash ke baad blobs ek single lump mein glue ho jaate hain jiska mass $m_{1} + m_{2}$ hai aur jo ek shared velocity $v$ par move karta hai. Yeh "dono ke liye ek velocity" perfectly inelastic ka poora matlab hai: front aur back ab ek hi object hain, isliye woh alag-alag speeds par move nahi kar sakte.

KYUN. Kyunki pieces glued hain, ab hamare paas do unknown final velocities nahi hain — exactly ek hai, $v$ . Uss single unknown ko momentum conservation pin down karega.

PICTURE. Do arrows collapse hokar ek bade blob par ek combined arrow $v$ ban jaate hain.

Step 4 — Pushes ko equal set karo, aur $v$ solve karo

KYA. Total momentum before = total momentum after. "Before" push hai $m_{1} u_{1} + m_{2} u_{2}$ ; "after" push hai mass $m_{1} + m_{2}$ ka ek lump $v$ par move karta hua, yaani $(m_{1} + m_{2}) v$ . Inhe equal set karke divide karne par $v$ milta hai.

KYUN. Yahi ek equation hai jo hum likhne ke allowed hain. Kinetic energy yahan off-limits hai kyunki squish ke dauran kuch secretly heat mein leak ho jaata hai — abhi hume pata nahi kitna. Momentum kahin leak nahi hota, isliye yeh safe bookkeeping quantity hai.

$push before m_{1} u_{1} + m_{2} u_{2} = glued mass (m_{1} + m_{2}) shared velocity v$

Dono sides ko $m_{1} + m_{2}$ se divide karo:

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

PICTURE. Step 2 ka pink total-momentum arrow reuse hota hai unchanged bade blob par; sirf uske neeche mass badha, isliye arrow-per-mass (velocity) shrink hoti hai. Woh shrink hi $v$ hai.

Step 5 — Har velocity ko "COM drift" + "relative wiggle" mein split karo

KYA. Centre of mass par baitho aur dekho. Wahan se aap har blob ki velocity ko do pieces mein dekhte ho: common drift $v$ (jo COM share karta hai), plus ek leftover relative motion. Relative velocity hai $u_{rel} = u_{1} - u_{2}$ — lump 1 kitni tez lump 2 ke paas aa raha hai.

KYUN. Hum aise split karte hain kyunki momentum conservation drift piece ko freeze karta hai ( $v$ nahi badal sakta) par relative piece ko protect karne ke baare mein kuch nahi kehta. Isliye jo bhi energy destroy ho sakti hai woh entirely relative wiggle mein rehti hai. Dono pieces ko alag karna hi energy loss ko obvious banata hai.

PICTURE. Har blob ka arrow (grey drift $v$ ) + (coloured relative part) ke roop mein draw kiya gaya hai. Sticking ke baad, coloured relative parts vanish ho jaate hain; sirf grey drift bachti hai.

Step 6 — Energy measure karo: before vs after

KYA. Kinetic energy ek lump ke liye $\frac{1}{2} m u^{2}$ hai. "Before" ke liye dono lumps add karo, aur "after" ke liye single glued lump use karo.

KYUN. Ab jo $v$ pata chal gaya, KE hi ek cheez bachi hai compute karne ke liye. Hum subtract karte hain loss nikalne ke liye.

$K E_{i} = \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2}, K E_{f} = \frac{1}{2} (m_{1} + m_{2}) v^{2}$

PICTURE. KE arrow ki length ke square ke saath badhti hai — isliye hum har energy ko uske velocity arrow par bane ek square ke area ke roop mein draw karte hain. Do "before" squares bade hain; single "after" square (chote $v$ par bana) chota hai. Missing area loss hai.

Step 7 — $v$ substitute karo aur dekho yeh reduced mass par collapse hota hai

KYA. $v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}$ ko $K E_{f}$ mein daalo aur $K E_{i}$ se subtract karo. Dust settle hone ke baad, ek khoobsurat cheez hoti hai: $u_{1}$ aur $u_{2}$ sirf $(u_{1} - u_{2})^{2}$ ke roop mein bachte hain, aur masses ek single combination $μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$ mein collapse ho jaate hain.

KYUN. Yeh Step 5 ka payoff hai. Algebra sirf woh confirm kar raha hai jo split ne pehle hi bataya tha: lost energy exactly relative motion ki kinetic energy hai.

$Δ K E = K E_{i} - K E_{f} = \frac{1}{2} μ = reduced mass \frac{m _{1} m _{2}}{m _{1} + m _{2}} u_{rel}^{2} (u_{1} - u_{2})^{2} = \frac{1}{2} μ u_{rel}^{2}$

PICTURE. Step 6 ka "missing square" area ek clean square ke roop mein redraw kiya gaya hai jiska side $u_{rel}$ hai aur $μ$ se scale kiya gaya hai — relative-motion energy, ab akele khadi hai.

Step 8 — Yeh MAXIMUM loss kyun hai (valley picture)

KYA. Un sab final states mein se jo total momentum fixed rakhte hain, sticking kinetic-energy valley ke ekdum bottom par hai.

KYUN. COM frame mein total momentum zero hai, isliye do final momenta equal aur opposite hone chahiye: kisi number $q$ ke liye $+ q$ aur $- q$ . Us frame mein final KE hai $\frac{q ^{2}}{2 m _{1}} + \frac{q ^{2}}{2 m _{2}}$ — $q$ mein ek bowl-shaped curve jo $q = 0$ par sabse chhoti hai. Aur $q = 0$ ka matlab hai COM frame mein dono lumps at rest = dono saath move karna = sticking. Aap isse neeche nahi ja sakte, kyunki KE kabhi negative nahi ho sakti.

PICTURE. Leftover relative momentum $q$ ke against final KE ka ek parabola; uska minimum exactly $q = 0$ par mark kiya gaya hai, sticking point.

Step 9 — Teen edge cases, draw kiye hue

KYA & PICTURE. Teen special inputs, har ek board par ek mini-scene:

Equal masses, ek at rest ( $m_{1} = m_{2} = m$ , $u_{2} = 0$ ): $v = u /2$ , aur exactly aadha KE lost hota hai. Valley floor aadhe raaste par hai.
Tiny into huge (bullet into block): $v$ nearly zero hai, isliye almost sab KE lost hoti hai — par bilkul nahi, kyunki $p_{tot} \neq = 0$ .
Equal and opposite momenta ( $m_{1} u_{1} = - m_{2} u_{2}$ ): $p_{tot} = 0$ , isliye $v = 0$ aur KE ka 100% destroy ho jaata hai — valley khud zero par baithe hai.

KYUN. Yeh har regime cover karte hain: partial loss, near-total loss, aur total loss. Kuch aur ho hi nahi sakta, kyunki $v$ aur $Δ K E$ sirf aapke diye hue signed inputs par depend karte hain.

Ek-picture summary

Sab kuch ek board par: do momentum arrows $p_{tot}$ mein add hote hain (unchanged), glued lump $v = p_{tot} / (m_{1} + m_{2})$ inherit karta hai, aur woh energy jo disappear hoti hai woh relative-motion square $\frac{1}{2} μ u_{rel}^{2}$ hai — KE valley ki depth.

Recall Feynman: poora walkthrough simple words mein

Do clay blobs ko ek doosre ki taraf slide karte hue imagine karo. Har ek ko ek arrow do ki woh kaise move karta hai, phir uski "push" ke liye ek mota arrow (mass times velocity). Unhe glue karo: ab yeh ek bada blob hai, aur universe ka sirf ek rule hai ki total push nahi badal sakta. Toh mote push-arrows bas add hote hain aur bhaari blob par baith jaate hain — aur kyunki blob bhaari hai, uski speed chhoti aati hai. Woh chhoti shared speed hi answer $v$ hai. Ab energy ke liye: motion-energy arrow ke square ki tarah badhti hai, isliye areas dekho. Har blob ki motion ko ek shared drift (jo bachta hai) aur ek leftover "closing-in" wiggle (jo nahi bachta) mein split karo. Sticking wiggle ka har bit erase kar deti hai — aur yeh pata chalta hai ki erased energy exactly $\frac{1}{2}$ times ek special combined mass $μ$ times closing speed squared hai. Yeh most kyun hai jo aap lose kar sakte ho? Kyunki agar blobs kisi bhi wiggle ko rakhne ki koshish karte, unhe phir alag hona padta — glue kehta hai nahi. Balance-point frame mein, aap jo sabse low energy reach kar sakte ho woh hai "dono at rest," aur exactly yahi sticking deta hai. Agar woh equal-and-opposite pushes ke saath ek doosre par charge kar rahe the, toh blob bas dead stop ho jaata hai aur sab motion-energy chali jaati hai.

Recall

Sticking ke baad, kaunsi ek quantity pehle jaisi unchanged rehti hai? ::: Total momentum $m_{1} u_{1} + m_{2} u_{2}$ . $v$ find karne ke liye hum KE conservation kyun nahi use kar sakte? ::: KE partly heat/sound/deformation mein leak hoti hai, isliye yeh unknown hai jab tak $v$ momentum se na nikla ho. Lost KE kis motion ki kinetic energy ke equal hai? ::: Relative motion ki, $\frac{1}{2} μ u_{rel}^{2}$ . Loss 100% kab hota hai? ::: Jab total momentum zero ho (equal-and-opposite pushes), isliye $v = 0$ .

Connections

Parent topic — woh master results jo yeh page draw out karta hai.
Conservation of Linear Momentum — Steps 2–4 mein use kiya gaya ek law.
Centre of mass motion — $v$ hi COM velocity hai (Step 4).
Reduced mass — $μ$ jo Step 7 mein appear hota hai.
Coefficient of restitution — sticking $e = 0$ end hai; Elastic collisions $e = 1$ hai.
Ballistic pendulum — bullet-into-block edge case (Step 9).
Work–Energy theorem — missing KE clay ko deform karne mein kiye gaye work ke equal hai.