Elastic collisions — 1D - solve for final velocities: Visual walkthrough

1.4.5 · D2 · Physics › Momentum & Collisions › Elastic collisions — 1D - solve for final velocities

Step 1 — Woh picture jo hum explain karne ki koshish kar rahe hain

KIYA KYA. Do blocks ek smooth, seedhi, frictionless track par slide kar rahe hain. Hum agree karte hain ki rightward = positive — yeh ek choice hai, lekin ek baar yeh karne ke baad, har velocity ek signed number hai: left jaane wali block ki velocity negative hoti hai.

Main sab kuch naam de deta hun, taaki koi bhi symbol baad mein bina matlab ke na aaye:

KYUN. Kuch bhi conserve karne se pehle, hume exactly pata hona chahiye ki "before" aur "after" ka matlab kya hai aur positive direction kaunsi hai. Sign discipline woh jagah hai jahan zyaadatar collision mistakes paida hoti hain.

PICTURE. Do blocks, signed velocities dikhate arrows, track aur positive direction marked.

Step 2 — Pehla fact: total momentum badal nahi sakta

KIYA KYA. Ek block ka Momentum uski mass times uski velocity hai, $m v$ — ek signed number. Total momentum dono blocks ka sum hai. Hum claim karte hain ki yeh total crash se pehle aur baad mein same rehta hai:

$block 1’s push, before m_{1} u_{1} + block 2’s push, before m_{2} u_{2} = block 1’s push, after m_{1} v_{1} + block 2’s push, after m_{2} v_{2}$

Yeh law kyun? Collision ke dauran blocks ek doosre par push karte hain, lekin Newton's third law ke hisaab se woh do forces equal aur opposite hain — woh total mein cancel ho jaate hain. Koi bahari horizontal force nahi hone se, total momentum ke paas koi cheez nahi jo use badal sake. Dekho Conservation of Momentum.

PICTURE. Har block ke liye ek "momentum bar"; pehle stack ki gayi do bars baad mein stack ki gayi do bars ke equal hain, chahe split badal jaye.

Step 3 — Doosra fact: total kinetic energy bhi nahi badal sakti

KIYA KYA. Kinetic energy (KE) "motion ki energy" hai, ek block ke liye $\frac{1}{2} m v^{2}$ . Gaur karo ki velocity squared hai — isliye ek leftward block ( $v < 0$ ) ki energy phir bhi positive hoti hai. "Elastic" woh special promise hai ki total KE unchanged rehti hai:

$total KE before \frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = total KE after \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$

Kyun squared, aur kyun yeh law? Squaring hi woh cheez hai jo energy ko direction se andha banati hai — energy kitni fast ke baare mein hai, kis taraf ke baare mein nahi. Hume ek doosri equation chahiye kyunki hamare paas do unknowns hain ( $v_{1}$ aur $v_{2}$ ); momentum akela infinitely many possibilities chod deta hai. Elastic promise (koi heat nahi, koi dent nahi, koi sound nahi) exactly woh missing constraint hai. Dekho Kinetic Energy.

PICTURE. Energy ko squares ke area ki tarah draw karo (side = speed); pehle total tile-area baad mein total tile-area ke equal.

Step 4 — Masses group karo (momentum, rewritten)

KIYA KYA. Momentum equation lo aur block 1 ki sab cheez left mein aur block 2 ki sab cheez right mein dhakel do:

$m_{1} u_{1} - m_{1} v_{1} = m_{2} v_{2} - m_{2} u_{2} ⟹ momentum block 1 *lost* m_{1} (u_{1} - v_{1}) = momentum block 2 *gained* m_{2} (v_{2} - u_{2}) (A)$

KYUN. Yeh kuch physical aur beautiful kehta hai: block 1 jitna momentum chodta hai, block 2 utna uthata hai. Do grouped chunks $m_{1} (u_{1} - v_{1})$ aur $m_{2} (v_{2} - u_{2})$ poore trick ke stars hain — inki shapes yaad rakho.

PICTURE. Block 1 se momentum nikaalta ek arrow aur wahi identical arrow block 2 mein pahunchta hua.

Step 5 — Energy mein bhi same tarah group karo (difference of squares)

KIYA KYA. Energy equation mein bhi identical grouping karo. Pehle har $\frac{1}{2}$ cancel karo (dono sides ko 2 se multiply karo), phir block 1 ko left mein, block 2 ko right mein ikatha karo:

$m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2})$

Ab key algebra tool: do squares ka difference hamesha $a^{2} - b^{2} = (a - b) (a + b)$ ki tarah factor hota hai.

$same chunk as (A)! m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = same chunk as (A)! m_{2} (v_{2} - u_{2}) (v_{2} + u_{2}) (B)$

PICTURE. Area rectangle $u^{2} - v^{2}$ ko $(u - v)$ strip times $(u + v)$ length mein slice kiya hua.

Step 6 — (B) ko (A) se divide karo: sab kuch saaf ho jaata hai

KIYA KYA. Equation (B), equation (A) hi hai jisme left par ek extra factor $(u_{1} + v_{1})$ aur right par $(v_{2} + u_{2})$ hai. (B) ko (A) se divide karo — identical chunks upar aur neeche cancel ho jaate hain:

$\frac{m _{1} ( u _{1} - v _{1} ) ( u _{1} + v _{1} )}{m _{1} ( u _{1} - v _{1} )} = \frac{m _{2} ( v _{2} - u _{2} ) ( v _{2} + u _{2} )}{m _{2} ( v _{2} - u _{2} )} ⟹ u_{1} + v_{1} = v_{2} + u_{2}$

Rearrange karo taaki "before" quantities saath baithen aur "after" saath:

$u_{1} - u_{2} = v_{2} - v_{1}$

KYUN divide karte hain? Divide karna do quadratic equations ko ek linear mein badal deta hai. Linear easy hai; wahi payoff hai. (Guard note: humne $m_{1} (u_{1} - v_{1})$ se divide kiya, jo theek hai jab tak $u_{1} \neq = v_{1}$ — yani block 1 ki velocity actually badli, yani real collision hui. No-collision case $v_{1} = u_{1}, v_{2} = u_{2}$ Step 8 mein separately handle kiya gaya hai.)

PICTURE. Pehle approach arrows, baad mein separation arrows — same length.

Step 7 — Do linear equations → final formulas

KIYA KYA. Ab hamare paas do linear equations hain:

$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2} (1) v_{2} = v_{1} + (u_{1} - u_{2}) (rule)$

Rule ko (1) mein substitute karo aur $v_{1}$ ke liye solve karo, phir $v_{2}$ ke liye back-substitute karo:

KYUN. Ek substitution ek unknown khatam kar deta hai, aur ek single equation $v_{1}$ mein bach jaati hai. Yeh poora "algebra omitted for brevity" hai parent note se, open mein kiya gaya.

PICTURE. Do-line linear system jaise do crossing lines ek point $(v_{1}, v_{2})$ par milti hain — ek unique solution.

Step 8 — Har case, degenerate ones bhi include

KIYA KYA. Formulas mein special inputs dalo aur dekho woh kaise behave karte hain. Sahi formula in sab mein survive karna chahiye — kabhi reader ko koi unseen scenario nahi milna chahiye.

Case	Inputs	Result	Reading
Equal masses, target still	$m_{1} = m_{2}$ , $u_{2} = 0$	$v_{1} = 0, v_{2} = u_{1}$	Woh swap karte hain — Newton's Cradle
Heavy hits light (still)	$m_{1} ≫ m_{2}$ , $u_{2} = 0$	$v_{1} \approx u_{1}, v_{2} \approx 2 u_{1}$	Light wala double speed par ud jaata hai
Light hits heavy wall	$m_{1} ≪ m_{2}$ , $u_{2} = 0$	$v_{1} \approx - u_{1}, v_{2} \approx 0$	Ball wapas bounce karti hai
No collision (miss)	$u_{1} = u_{2}$	$v_{1} = u_{1}, v_{2} = u_{2}$	Kuch nahi hota — approach speed 0

Degenerate case kyun dikhate hain? Jab $u_{1} = u_{2}$ hota hai toh approach speed zero hai, isliye kuch nahi badlta — aur notice karo yeh exactly woh case hai jiske baare mein Step 6 ne warn kiya tha ( $u_{1} = v_{1}$ ). Formulas yahan bhi sahi answer dete hain, isliye kuch break nahi hota; hum sirf "divide by (A)" shortcut use karke derive nahi kar sakte.

PICTURE. Chaar mini-panels, har case ke liye ek, pehle/baad ke arrows.

Ek-picture summary

Yahan poora safar compress kar diya hai: do conservation laws andar, group kiye, divide kiye, aur bahar aate hain do formulas — special cases side mein lage hue.

Recall Feynman: poora walkthrough kisi dost ko batao

Hamare paas ek smooth track par do bouncy carts thi aur hum crash ke baad unki speeds jaanna chahte the. Do rules kabhi nahi toote: total push (mass × velocity, add karke) same rehta hai, aur total motion-energy (mass × velocity², half karke, add karke) same rehta hai. Maine dono rules likhe, phir cart 1 ko ek side aur cart 2 ko doosri side dhakel diya har rule mein. Push-rule ban gaya "cart 1 ne jo push khoya = cart 2 ne jo push liya." Energy-rule, difference of squares factor karne ke baad, exactly wahi cart-chunks bahar laaya jisme ek extra piece laga tha. Toh maine ek rule ko doosre se divide kiya, ugly chunks cancel ho gaye, aur jo nikla woh gorgeous tha: woh utni hi speed par alag hote hain jis par woh paas aaye the. Yeh ek clean straight-line equation hai. Ise push-rule ke saath pair karo, ek substitution karo, aur do final-velocity formulas bahar aa jaate hain. Sanity-check ke liye, main equal weights daalta hun (woh speeds trade karte hain), ek heavy cart ko light par marta hun (light wala double speed par ud jaata hai), aur ek light cart ko wall par marta hun (woh seedha wapas bounce karti hai). Chaaron cases behave karte hain — toh formula har jagah trustworthy hai.

Recall Quick self-test

Approach speed agar $u_{1} = 5, u_{2} = - 2$ ? ::: $u_{1} - u_{2} = 5 - (- 2) = 7$ , toh woh $7$ par bhi alag hote hain. Kaun sa algebra tool energy equation ko momentum ke same chunks mein convert karta hai? ::: Difference of squares, $u^{2} - v^{2} = (u - v) (u + v)$ . (B) ko (A) se divide karne ke baad kaunsi single equation aati hai? ::: $u_{1} - u_{2} = v_{2} - v_{1}$ (relative-velocity rule). Raw energy conservation ki jagah relative-velocity rule kyun prefer kiya jaata hai? ::: Yeh linear hai — koi squares nahi — toh do linear equations instantly solve ho jaate hain.

Connections

Conservation of Momentum — pehla pillar (Step 2).
Kinetic Energy — elastic promise (Step 3).
Coefficient of Restitution — Step 6 ka rule $e = 1$ case hai.
Center of Mass Frame — Step 8 ke "double speed" ke peeche slick re-derivation.
Newton's Cradle — Step 8 ka equal-mass swap.
Inelastic Collisions — 1D — kya hota hai jab KE promise toot jaati hai.