Elastic collisions — 1D - solve for final velocities: Worked examples

1.4.5 · D3 · Physics › Momentum & Collisions › Elastic collisions — 1D - solve for final velocities

Yahan $m_{1}, m_{2}$ do masses hain, $u_{1}, u_{2}$ unki velocities before hain, $v_{1}, v_{2}$ unki velocities after hain. Subscript ek naam-tag ki tarah hai: "1" matlab left-hand object, "2" matlab right-hand object. Rightward positive hai; negative number matlab "left ja raha hai". Yeh baat dimaag mein ghusaa lo — aadhi galtiyan sirf minus sign bhoolne se hoti hain.

Scenario matrix

Har 1D elastic problem in cells mein se kisi ek mein aata hai. Neeche ke examples cell ke naam se label kiye gaye hain taaki dekh sako ki poora space cover ho gaya hai.

Cell	Kya vary karta hai	Example
A. Dono at rest / trivial	$u_{1} = u_{2} = 0$ — degenerate	Ex 1
B. Equal masses, target still	$m_{1} = m_{2}, u_{2} = 0$	Ex 2
C. Equal masses, dono chal rahe hain (opposite signs)	$m_{1} = m_{2}$ , ek negative	Ex 3
D. Unequal masses, target at rest	general $m$ , $u_{2} = 0$	Ex 4
E. Dono moving, dono nonzero, unequal masses	full general signs	Ex 5
F. Heavy → light limit	$m_{1} ≫ m_{2}$	Ex 6
G. Light → heavy wall limit	$m_{1} ≪ m_{2}$	Ex 7
H. Real-world word problem	pool/skater story	Ex 8
I. Exam twist — arrow ulta karo	$v$ 's diye hain, $u$ 's nikalo	Ex 9

Ex 1 — Cell A · kuch bhi nahi chal raha (degenerate)

Worked example

$m_{1} = 2$ kg, $m_{2} = 5$ kg, $u_{1} = u_{2} = 0$ . Forecast: Padhne se pehle guess karo. Agar koi chal hi nahi raha, toh kya "collision" kuch badal bhi sakti hai?

Step 1 — momentum. $m_{1} u_{1} + m_{2} u_{2} = 0 + 0 = 0$ . Yeh step kyun? Total momentum hamara pehla fixed quantity hai; yahan yeh zero se shuru hota hai.

Step 2 — relative-velocity rule. $u_{1} - u_{2} = 0 - 0 = 0$ , toh $v_{2} - v_{1} = 0 \Rightarrow v_{1} = v_{2}$ . Yeh step kyun? Approach speed zero hai, toh separation speed bhi zero honi chahiye — yeh kuch nahi se alag nahi ud sakte.

Step 3 — combine karo. $2 v_{1} + 5 v_{2} = 0$ with $v_{1} = v_{2}$ gives $7 v_{1} = 0 \Rightarrow v_{1} = v_{2} = 0$ . Yeh step kyun? Do equations hain, aur woh dono velocities ko zero hone par majboor karti hain.

Verify: KE before $= 0$ , KE after $= 0$ . ✓ Kuch hota hi nahi — koi collision thi hi nahi. Yeh sanity ka sabse basic level hai: zero in, zero out.

Ex 2 — Cell B · equal masses, target at rest (the swap)

Ex 3 — Cell C · equal masses, head-on (dono chal rahe hain, opposite signs)

Worked example

$m_{1} = m_{2} = 3$ kg, $u_{1} = + 4$ m/s (right), $u_{2} = - 2$ m/s (left). Forecast: Yeh ek doosre ki taraf aa rahe hain. Kya yeh seedha bounce back karte hain? $v_{1}, v_{2}$ ke signs guess karo.

Step 1 — relative rule. $u_{1} - u_{2} = 4 - (- 2) = 6 = v_{2} - v_{1}$ . Yeh step kyun? Double minus pe dhyaan do: leftward velocity subtract karna add karna hai. Yahi #1 sign error hai.

Step 2 — momentum. $3 (4) + 3 (- 2) = 12 - 6 = 6 = 3 (v_{1} + v_{2}) \Rightarrow v_{1} + v_{2} = 2$ . Yeh step kyun? Doosri equation; common mass divide kar do.

Step 3 — solve karo. Add karo: $2 v_{2} = 6 + 2 = 8 \Rightarrow v_{2} = 4$ ; phir $v_{1} = 2 - 4 = - 2$ . Yeh step kyun? Equal masses hamesha velocities swap karte hain — aur sach mein $v_{1} = u_{2}$ , $v_{2} = u_{1}$ .

Verify: KE before $= \frac{1}{2} (3) (16) + \frac{1}{2} (3) (4) = 24 + 6 = 30$ J. After $= \frac{1}{2} (3) (4) + \frac{1}{2} (3) (16) = 6 + 24 = 30$ J. ✓ Figure dekho: red (object 1) fast-right aa raha tha, slow-left jaata hai; violet (object 2) slow-left aa raha tha, fast-right jaata hai. Unhone velocities, signs samait, trade kar liye.

Ex 4 — Cell D · unequal masses, target at rest (full formula use karo)

Worked example

$m_{1} = 3$ kg, $m_{2} = 1$ kg, $u_{1} = 8$ m/s, $u_{2} = 0$ . Forecast: Heavy ball lighter pe lagti hai. Kya heavy ball aage jaati rehti hai? Kya light ball aage nikal jaati hai?

Step 1 — $v_{1}$ formula plug karo. $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} = \frac{3 - 1}{3 + 1} (8) = \frac{2}{4} (8) = 4$ m/s. Yeh step kyun? Jab $u_{2} = 0$ hota hai toh har formula ka doosra term khatam ho jaata hai, toh formula ek hi term ka reh jaata hai — sabse fast route.

Step 2 — $v_{2}$ formula plug karo. $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{6}{4} (8) = 12$ m/s. Yeh step kyun? Target ke liye wahi shortcut.

Step 3 — relative rule se sanity check. Separation $v_{2} - v_{1} = 12 - 4 = 8 = u_{1} - u_{2}$ . ✓ Yeh step kyun? Cross-check karta hai ki hamare do numbers approach = separation law satisfy karte hain.

Verify: Momentum before $= 3 (8) = 24$ ; after $= 3 (4) + 1 (12) = 12 + 12 = 24$ . ✓ KE before $= \frac{1}{2} (3) (64) = 96$ J; after $= \frac{1}{2} (3) (16) + \frac{1}{2} (1) (144) = 24 + 72 = 96$ J. ✓ Heavy ball aage jaati rehti hai (positive) lekin slower; light ball heavy ball ke aane se bhi zyada speed se aage nikal jaati hai.

Ex 5 — Cell E · dono chal rahe hain, unequal masses, full signs

Worked example

$m_{1} = 4$ kg at $u_{1} = + 5$ m/s, $m_{2} = 2$ kg at $u_{2} = + 2$ m/s (dono rightward, fast catches slow). Forecast: Ek fast rear cart slow front cart ko pakad leti hai. Kya rear cart ulti direction mein ja sakti hai? Guess karo.

Step 1 — relative rule. $u_{1} - u_{2} = 5 - 2 = 3 \Rightarrow v_{2} - v_{1} = 3 \Rightarrow v_{2} = v_{1} + 3$ . Yeh step kyun? Do unknowns ko ek substitution mein badal deta hai.

Step 2 — momentum. $4 (5) + 2 (2) = 20 + 4 = 24 = 4 v_{1} + 2 v_{2}$ . Yeh step kyun? Total $p$ hamara anchor hai; baad mein bhi 24 hona chahiye.

Step 3 — substitute & solve karo. $4 v_{1} + 2 (v_{1} + 3) = 24 \Rightarrow 6 v_{1} + 6 = 24 \Rightarrow v_{1} = 3$ ; phir $v_{2} = 6$ . Yeh step kyun? Substitute karne ke baad $v_{1}$ mein ek linear equation reh jaati hai.

Verify: Momentum after $= 4 (3) + 2 (6) = 12 + 12 = 24$ . ✓ KE before $= \frac{1}{2} (4) (25) + \frac{1}{2} (2) (4) = 50 + 4 = 54$ J. After $= \frac{1}{2} (4) (9) + \frac{1}{2} (2) (36) = 18 + 36 = 54$ J. ✓ Rear cart 5 se 3 pe aa jaati hai (abhi bhi forward, reverse nahi hoti — kyunki yeh heavier hai), front cart 2 se 6 pe chali jaati hai.

Ex 6 — Cell F · heavy hits light (limiting behaviour)

Worked example

$m_{1} = 1000$ kg at $u_{1} = 3$ m/s hits $m_{2} = 1$ kg at rest ( $u_{2} = 0$ ). Forecast: Parent note kehta hai light ball ~double speed pe jaati hai. $v_{1}$ aur $v_{2}$ guess karo, phir dekho $2 u_{1}$ ke kitna close hai.

Step 1 — $v_{1}$ formula. $v_{1} = \frac{1000 - 1}{1000 + 1} (3) = \frac{999}{1001} (3) = \frac{2997}{1001} \approx 2.994$ m/s. Yeh step kyun? Hum exact value chahte hain taaki dekh sakein ki slow-down kitna tiny hai — parent ne warn kiya tha ki $v_{1} \approx u_{1}$ sirf approximation hai.

Step 2 — $v_{2}$ formula. $v_{2} = \frac{2 ( 1000 )}{1001} (3) = \frac{6000}{1001} \approx 5.994$ m/s. Yeh step kyun? "Double" prediction $2 u_{1} = 6$ se compare karo.

Step 3 — interpret karo. $v_{1} \approx 2.994$ (heavy barely slows), $v_{2} \approx 5.994 \approx 2 u_{1}$ . Yeh step kyun? Limit confirm hoti hai: jaise $m_{1} \to \infty$ , $v_{1} \to u_{1}$ aur $v_{2} \to 2 u_{1}$ , lekin finite mass ke liye yeh thoda kam hota hai.

Verify: Momentum before $= 1000 (3) = 3000$ ; after $= 1000 (2.994...) + 1 (5.994...) = 2994.006 + 5.994 = 3000$ . ✓ Bilkul double-speed limit, finite ratio se thoda naram hua — isliye " $v_{1} = u_{1}$ exactly" wali galti galat hai.

Ex 7 — Cell G · light ball off a heavy wall (limiting behaviour)

Worked example

$m_{1} = 0.5$ kg at $u_{1} = 7$ m/s hits $m_{2} = 500$ kg at rest. Forecast: Ek patthar ek boulder se takraata hai. Kya woh seedha wapas bounce karta hai? Kitni speed pe?

Step 1 — $v_{1}$ formula. $v_{1} = \frac{0.5 - 500}{0.5 + 500} (7) = \frac{- 499.5}{500.5} (7) \approx - 6.986$ m/s. Yeh step kyun? Numerator negative hai kyunki $m_{2} > m_{1}$ , toh $v_{1}$ sign flip karta hai — patthar wapas ho jaata hai.

Step 2 — $v_{2}$ formula. $v_{2} = \frac{2 ( 0.5 )}{500.5} (7) = \frac{7}{500.5} \approx 0.0140$ m/s. Yeh step kyun? Dikhata hai ki wall bahut kam hilti hai — tiny lekin nonzero (ek real wall thoda sa recoil karti hai).

Step 3 — interpret karo. $v_{1} \approx - 7$ (almost same speed pe wapas bounce karta hai), $v_{2} \approx 0$ . Yeh step kyun? Wall limit confirm hoti hai: $m_{2} \to \infty$ pe $v_{1} \to - u_{1}$ , $v_{2} \to 0$ .

Verify: Momentum before $= 0.5 (7) = 3.5$ ; after $= 0.5 (- 6.986...) + 500 (0.01398...) = - 3.493 + 6.993 = 3.5$ . ✓ Patthar almost saari energy le jaata hai, wall almost kuch nahi — elastic "wall bounce".

Ex 8 — Cell H · real-world word problem

Worked example Ek

$0.17$ kg cue ball $2$ m/s pe roll karti hai aur ek stationary $0.16$ kg billiard ball se head-on takraati hai. Perfectly elastic hit maano. Dono ki speeds after nikalo. Forecast: Masses almost equal hain. Guess karo: kya yeh almost swap karte hain?

Step 1 — symbols naam karo. $m_{1} = 0.17, u_{1} = 2, m_{2} = 0.16, u_{2} = 0$ . Yeh step kyun? Story ko $m, u$ tags mein translate karna asli exam skill hai — yeh pehle karo, har baar.

Step 2 — $v_{1}$ formula. $v_{1} = \frac{0.17 - 0.16}{0.33} (2) = \frac{0.01}{0.33} (2) = \frac{0.02}{0.33} \approx 0.0606$ m/s. Yeh step kyun? Nearly-equal masses ⇒ cue ball ke liye nearly-zero, "almost swap" forecast se match karta hai.

Step 3 — $v_{2}$ formula. $v_{2} = \frac{2 ( 0.17 )}{0.33} (2) = \frac{0.68}{0.33} \approx 2.0606$ m/s. Yeh step kyun? Target almost full speed le leta hai — thoda $2$ se zyada.

Verify: Momentum before $= 0.17 (2) = 0.34$ ; after $= 0.17 (0.0606) + 0.16 (2.0606) = 0.0103 + 0.3297 = 0.34$ . ✓ Units: (kg)(m/s) throughout — momentum kg·m/s mein, velocities m/s mein. Cue ball almost ruk jaati hai, target thoda $2$ m/s se zyada pe nikal jaati hai: real pool wala behaviour.

Ex 9 — Cell I · exam twist, seedha ulta chalaao

Worked example Ek elastic collision ke baad do carts (

$m_{1} = 1$ kg, $m_{2} = 3$ kg) $v_{1} = - 3$ m/s aur $v_{2} = 1$ m/s pe chalti dikhi deti hain. Diya gaya hai ki cart 2 rest pe thi ( $u_{2} = 0$ ), toh initial velocity $u_{1}$ nikalo. Forecast: Kya hum collision undo kar sakte hain? Laws dono directions mein kaam karte hain — guess karo ki ek equation kaafi hai ya nahi.

Step 1 — momentum, unknown ke liye solve karo. $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ , yaani $1 \cdot u_{1} + 0 = 1 (- 3) + 3 (1) = 0$ . Yeh step kyun? $u_{1}$ ke siwa sab kuch maloom hai; momentum ek equation hai ek unknown mein.

Step 2 — padhlo. $u_{1} = 0$ ?? Yeh ho nahi sakta — ek still cart collision cause nahi kar sakti. Kuch inconsistent hai, toh relative-velocity rule se cross-check karo. Yeh step kyun? Agar diya gaya "after" state genuinely elastic hai, toh dono laws agree karni chahiye. Test karo.

Step 3 — relative-rule test. $u_{2} = 0$ ke saath: approach $= u_{1} - 0 = u_{1}$ , separation $= v_{2} - v_{1} = 1 - (- 3) = 4$ . Elastic demand karta hai $u_{1} = 4$ . Lekin momentum ne demand kiya $u_{1} = 0$ . Contradiction ⇒ in masses ke saath elastic collision mein stated after-state impossible hai. Yeh step kyun? Yeh exam trap hai: woh numbers deta hai jo theek lagte hain lekin ek law violate karte hain. Hamesha dono laws test karo.

Verify: Consistent scenario hai $u_{1} = 4$ . Phir momentum kehta hai $m_{1} (4) = 1 \cdot v_{1} + 3 \cdot v_{2}$ , yaani $v_{1} + 3 v_{2} = 4$ , aur rule kehta hai $v_{2} - v_{1} = 4$ . Solve karo: formulas use karo: $v_{1} = \frac{1 - 3}{4} (4) = - 2$ , $v_{2} = \frac{2 ( 1 )}{4} (4) = 2$ . Check karo momentum $1 (- 2) + 3 (2) = 4$ ✓ aur KE $\frac{1}{2} (1) (16) = 8$ before, $\frac{1}{2} (1) (4) + \frac{1}{2} (3) (4) = 2 + 6 = 8$ after ✓. Lesson: ek valid elastic outcome DONO laws satisfy karta hai; printed $(- 3, 1)$ nahi karta tha.

Recall Kaun sa cell kaun sa tha?

Ex1 — dono at rest (degenerate) ::: kuch nahi hota, sab zero. Ex2 — equal masses target still ::: velocities swap hoti hain, cue ball ruk jaati hai. Ex3 — equal masses head-on ::: signs samait swap, $v_{1} = u_{2}, v_{2} = u_{1}$ . Ex4 — unequal, target at rest ::: $\frac{m _{1} - m _{2}}{m _{1} + m _{2}}$ form use karo. Ex5 — dono moving, unequal ::: full signed momentum + relative rule. Ex6 — heavy hits light ::: light ball $\to 2 u_{1}$ , heavy barely slows. Ex7 — light hits heavy ::: patthar $\approx - u_{1}$ pe reverse, wall $\approx 0$ . Ex8 — pool balls ::: near-equal masses almost swap. Ex9 — reverse twist ::: valid state DONO laws obey karta hai warna impossible hai.

Connections

Parent topic — yahan har example mein use kiye gaye do tools ki derivation.
Conservation of Momentum — woh equation jo upar har step 1 ya 2 mein use hui.
Kinetic Energy — woh quantity jo hum har example ke end mein verify karte hain.
Coefficient of Restitution — Ex 9 ka "kya yeh dono laws obey karta hai?" check $e = 1$ test hai.
Center of Mass Frame — Ex 6 aur Ex 7 limits ke liye ek slick alternate route.
Newton's Cradle — Ex 2 aur Ex 3 bilkul yahi hain.
Inelastic Collisions — 1D — contrast: wahan KE conserve nahi hoti, toh Ex 9-style checks purpose se fail hote hain.