Elastic collisions — 1D - solve for final velocities: Exercises

1.4.5 · D4 · Physics › Momentum & Collisions › Elastic collisions — 1D - solve for final velocities

Figure — Elastic collisions — 1D - solve for final velocities

Upar wali figure tumhari sign map hai. Ek velocity arrow jo right ki taraf point karta hai woh ek positive number hai; left ki taraf point karta hai toh negative hai. Algebra chhune se pehle, hamesha arrows redraw karo aur is picture se signs padho — neeche ke zyaadatar errors sirf ek dropped minus sign hote hain.

Common mistake Koi bhi problem se PEHLE yeh padho: kya woh actually collide bhi karte hain?

Check yeh hai: ek head-on collision tabhi hoti hai jab objects actually approach kar rahe hon, yaani approach speed $u_{1} - u_{2}$ positive ho (left wala object $m_{1}$ , right wale object $m_{2}$ ko catch kar raha ho). Agar $u_{1} - u_{2} \leq 0$ hai toh woh separate ho rahe hain ya same speed pe hain — collision hoti hi nahi, aur formulas lagaane se ek aise bounce ka prediction hoga jo kabhi hua hi nahi. Yeh skip karna sahi kyun lagta hai: algebra kisi bhi number pe theek chalti hai, isliye students blindly plug in kar dete hain. Fix: neeche har problem is tarah arrange ki gayi hai ki $m_{1}$ (left pe) $m_{2}$ (right pe) ko catch kare, yaani $u_{1} - u_{2} > 0$ mile. Jab bhi tum apna khud ka problem setup karo, pehle confirm karo $u_{1} - u_{2} > 0$ hai — warna objects kabhi milte hi nahi.

Level 1 — Recognition

Goal: yeh pehchaano ki kaun sa rule apply hoga aur almost inspection se answer padh lo.

L1.1

Frictionless ice pe do identical pucks hain. Puck A right ki taraf $u_{1} = + 4$ m/s se chalti hai aur puck B ko strike karti hai jo rest mein hai ( $u_{2} = 0$ ). Collision elastic hai. $v_{1}$ aur $v_{2}$ kya hain?

Recall Solution L1.1

Approach check: $u_{1} - u_{2} = 4 - 0 = 4 > 0$ ✓ — yeh collide karte hain. WHAT rule: equal masses, target at rest — special "swap" case. WHY: $m_{1} = m_{2}$ ke saath coefficients $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} = 0$ aur $\frac{2 m _{2}}{m _{1} + m _{2}} = 1$ hote hain. $v_{1} = 0 \cdot u_{1} + 1 \cdot u_{2} = 0, v_{2} = 1 \cdot u_{1} + 0 \cdot u_{2} = + 4 m/s .$ Answer: $v_{1} = 0$ , $v_{2} = + 4$ m/s. A ekdam ruk jaata hai; B saari motion lekar nikal jaata hai. Yeh Newton's-cradle wala behaviour hai, dekho Newton's Cradle.

L1.2

Ek ping-pong ball ( $m_{1} = 2.7$ g) ek stationary bowling ball ( $m_{2} = 6$ kg) se $+ 5$ m/s pe head-on, elastically takraati hai. Ping-pong ball ke saath roughly kya hoga, bina full arithmetic ke?

Recall Solution L1.2

Approach check: $u_{1} - u_{2} = 5 - 0 = 5 > 0$ ✓ — yeh collide karte hain. WHAT rule: light hits heavy wall ( $m_{1} ≪ m_{2}$ , $u_{2} = 0$ ). WHY: $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} \to - 1$ aur $\frac{2 m _{1}}{m _{1} + m _{2}} \to 0$ . $v_{1} \approx - u_{1} = - 5 m/s, v_{2} \approx 0.$ Answer: ping-pong ball seedha wapas lagbhag $- 5$ m/s pe bounce karti hai; bowling ball barely hilti hai. Yeh bilkul ek wall se ball bounce karne jaisa hai.

Level 2 — Application

Goal: do tools mein signed numbers plug karo aur dono final velocities nikaal lo.

L2.1

$m_{1} = 3$ kg at $u_{1} = + 6$ m/s, $m_{2} = 1$ kg jo rest mein hai se elastically takraata hai. $v_{1}$ aur $v_{2}$ nikalo.

Recall Solution L2.1

Approach check: $u_{1} - u_{2} = 6 - 0 = 6 > 0$ ✓ — yeh collide karte hain. Tool 2 (relative velocity), WHY: elastic 1D ka matlab separation speed = approach speed, isliye $v_{2} - v_{1} = u_{1} - u_{2} = 6$ . Ek unknown isolate karne ke liye rearrange: $v_{2} = v_{1} + 6$ . (Hum $v_{2}$ ko $v_{1}$ ke terms mein solve karte hain taaki momentum mein substitute kar ke single-variable equation bana sakein.) Tool 1 (momentum), WHY: koi external horizontal force nahi, isliye total $p$ conserved hai: $3 (6) + 1 (0) = 18 = 3 v_{1} + 1 v_{2}$ . Substitute — WHY: $v_{2}$ ki jagah $v_{1} + 6$ rakhne se do unknowns ek ho jaate hain. $3 v_{1} + (v_{1} + 6) = 18 \Rightarrow 4 v_{1} = 12 \Rightarrow v_{1} = + 3$ m/s. Back-substitute — WHY: ab $v_{1}$ pata hai, rule directly $v_{2}$ deta hai: $v_{2} = 3 + 6 = + 9$ m/s. Answer: $v_{1} = + 3$ m/s, $v_{2} = + 9$ m/s. Heavier ball slow hoti hai lekin aage karti jaati hai; lighter ball aage shoot kar jaati hai.

L2.2

$m_{1} = 2$ kg at $u_{1} = + 8$ m/s, $m_{2} = 4$ kg at $u_{2} = + 2$ m/s se (same direction mein, catch up karte hue) elastically takraata hai. $v_{1}$ aur $v_{2}$ nikalo.

Recall Solution L2.2

Approach check: $u_{1} - u_{2} = 8 - 2 = 6 > 0$ ✓ — faster left block sach mein right wale ko catch karta hai. Tool 2, WHY: elastic ⇒ separation = approach, $v_{2} - v_{1} = 6$ , isolate $v_{2} = v_{1} + 6$ taaki substitute kar sakein. Tool 1, WHY: momentum conserved: $2 (8) + 4 (2) = 16 + 8 = 24 = 2 v_{1} + 4 v_{2}$ . Substitute — WHY: $v_{2}$ eliminate karo taaki ek unknown bache. $2 v_{1} + 4 (v_{1} + 6) = 24 \Rightarrow 6 v_{1} + 24 = 24 \Rightarrow v_{1} = 0$ . Back-substitute: $v_{2} = 0 + 6 = + 6$ m/s. Answer: $v_{1} = 0$ , $v_{2} = + 6$ m/s. Lighter chaser ruk jaata hai; heavier wala 2 se 6 m/s tak speed up hota hai. $v_{1}$ ke formula se check: $\frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} + \frac{2 m _{2}}{m _{1} + m _{2}} u_{2} = \frac{- 2}{6} (8) + \frac{8}{6} (2) = - \frac{16}{6} + \frac{16}{6} = 0.$ ✓

Level 3 — Analysis

Goal: signs handle karo, negative velocities, aur "what-if" reasoning.

L3.1

$m_{1} = 2$ kg at $u_{1} = + 3$ m/s, $m_{2} = 5$ kg at $u_{2} = - 4$ m/s (left ki taraf move karta hua, head-on) se elastically milta hai. $v_{1}$ , $v_{2}$ nikalo, aur verify karo ki kinetic energy conserved hai.

Recall Solution L3.1

Approach check: $u_{1} - u_{2} = 3 - (- 4) = 7 > 0$ ✓ — head-on, $7$ m/s pe closing. Tool 2, WHY: elastic ⇒ woh utni hi speed se separate hote hain jitni se approach kiya, $v_{2} - v_{1} = 7$ , isolate $v_{2} = v_{1} + 7$ . Tool 1, WHY: momentum conserved: $2 (3) + 5 (- 4) = 6 - 20 = - 14 = 2 v_{1} + 5 v_{2}$ . Substitute — WHY: $v_{2}$ remove karo ek variable solve karne ke liye. $2 v_{1} + 5 (v_{1} + 7) = - 14 \Rightarrow 7 v_{1} + 35 = - 14 \Rightarrow 7 v_{1} = - 49 \Rightarrow v_{1} = - 7$ m/s. Back-substitute: $v_{2} = - 7 + 7 = 0$ m/s. Answer: $v_{1} = - 7$ m/s (leftward wapas pheka gaya, aane se zyaada speed se), $v_{2} = 0$ (heavy ball ruk gayi). KE check: Before $= \frac{1}{2} (2) (9) + \frac{1}{2} (5) (16) = 9 + 40 = 49$ J. After $= \frac{1}{2} (2) (49) + \frac{1}{2} (5) (0) = 49$ J. ✓ Dekho Kinetic Energy.

L3.2

Ek elastic collision mein incoming ball $m_{1}$ at $u_{1}$ , $m_{2}$ jo rest mein hai se takraati hai aur rebound karti hai (left wapas aati hai, $v_{1} < 0$ ). Masses pe kaun si condition rebound guarantee karti hai?

Recall Solution L3.2

WHAT hum compute karte hain: $v_{1}$ ka sign. $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}$ (kyunki $u_{2} = 0$ ). $u_{1} > 0$ ke saath, $v_{1}$ ka sign $(m_{1} - m_{2})$ ka sign hai.

$v_{1} < 0$ (rebound) $⟺ m_{1} < m_{2}$ : incoming ball target se lighter hai.
$v_{1} = 0$ (ruk jaati hai) $⟺ m_{1} = m_{2}$ .
$v_{1} > 0$ (aage badhti rehti hai) $⟺ m_{1} > m_{2}$ . Answer: ball exactly tabhi rebound karti hai jab woh jo cheez hit kar rahi hai usse lighter ho. Yeh "light off a wall bounces back" ka analytic version hai.

Level 4 — Synthesis

Goal: collisions ko doosri mechanics (energy, height, restitution) ke saath combine karo.

L4.1

Ek block $m_{1} = 1$ kg, $u_{1} = + 6$ m/s pe slide karta hai aur stationary block $m_{2} = 3$ kg se elastically takraata hai. Collision ke baad $m_{2}$ ek rough patch pe friction coefficient $μ = 0.2$ ke saath slide karta hai. $m_{2}$ rukne se pehle kitni door jaata hai? ( $g = 10$ m/s² use karo.)

Recall Solution L4.1

Approach check: $u_{1} - u_{2} = 6 - 0 = 6 > 0$ ✓ — yeh collide karte hain. Step 1 — collision, $v_{2}$ nikalo. $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{2 ( 1 )}{4} (6) = + 3$ m/s. (WHY: $u_{2} = 0$ hai, $v_{2}$ formula use karo.) Step 2 — friction isko roktی hai. KE friction work mein convert hoti hai: $\frac{1}{2} m_{2} v_{2}^{2} = μ m_{2} g d$ . Mass cancel ho jaata hai: $d = \frac{v _{2}^{2}}{2 μg} = \frac{3 ^{2}}{2 ( 0.2 ) ( 10 )} = \frac{9}{4} = 2.25$ m. Answer: $m_{2}$ , $2.25$ m slide karta hai. (WHY do stages mein split karo: collision instantaneous aur elastic hai; baad wali slide ek alag energy-loss process hai jo friction se govern hoti hai, collision se nahi.)

L4.2

Do elastic collisions ek ke baad ek. $m_{1} = 2$ kg at $u_{1} = + 9$ m/s, $m_{2} = 2$ kg jo rest mein hai se takraata hai; phir wahi $m_{2}$ (ab move karta hua) $m_{3} = 6$ kg jo rest mein hai se takraata hai. $m_{3}$ ki final velocity nikalo.

Recall Solution L4.2

Approach check (dono): pehle ke liye $9 - 0 > 0$ hai, aur phir moving $m_{2}$ , resting $m_{3}$ se positive approach ke saath milta hai ✓. Collision 1 (equal masses, swap): $m_{1}$ ruk jaata hai, $m_{2}$ , $+ 9$ m/s pe nikal jaata hai. (WHY: $m_{1} = m_{2}$ , target at rest.) Collision 2 ( $m_{2} = 2$ , $m_{3} = 6$ jo rest mein hai, se takraata hai): $v_{3} = \frac{2 m _{2}}{m _{2} + m _{3}} u_{2} = \frac{2 ( 2 )}{2 + 6} (9) = \frac{4}{8} (9) = + 4.5$ m/s. Answer: $m_{3}$ , $+ 4.5$ m/s pe end hota hai. (Sanity: aur bhi $v_{2}^{'} = \frac{m _{2} - m _{3}}{m _{2} + m _{3}} (9) = \frac{- 4}{8} (9) = - 4.5$ — middle block rebound karta hai. Related chain behaviour: Newton's Cradle.)

Level 5 — Mastery

Goal: reverse-engineer karo, general statements prove karo, restitution aur CM frame se connect karo.

L5.1

Ek elastic head-on collision ke baad, ek $3$ kg ball (initially rest mein, $u_{2} = 0$ ) $v_{2} = + 3$ m/s pe move karti hui payi jaati hai. Incoming ball ka $m_{1} = 2$ kg hai. Incoming speed $u_{1}$ aur incoming ball ki final velocity $v_{1}$ nikalo — aur explain karo kyun tum $v_{1}, v_{2}$ freely nahi choose kar sakte.

Recall Solution L5.1

Pehle, general formulas KAHAN se aate hain (taaki hum unhe honestly use karein). $u_{2} = 0$ ke saath parent ke do boxed results reduce ho jaate hain: $v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}, v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} .$ WHY yeh hold karte hain: Tool 1 momentum $m_{1} u_{1} = m_{1} v_{1} + m_{2} v_{2}$ aur Tool 2 rule $v_{2} - v_{1} = u_{1}$ (kyunki $u_{2} = 0$ ) se shuru karo. $v_{2} = v_{1} + u_{1}$ ko momentum mein substitute karo: $m_{1} u_{1} = m_{1} v_{1} + m_{2} (v_{1} + u_{1}) \Rightarrow (m_{1} - m_{2}) u_{1} = (m_{1} + m_{2}) v_{1} \Rightarrow v_{1} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1}$ . Phir $v_{2} = v_{1} + u_{1} = \frac{m _{1} - m _{2} + m _{1} + m _{2}}{m _{1} + m _{2}} u_{1} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ . Yeh derived hain, assume nahi kiye.

Ab solve karo. $m_{1} = 2, m_{2} = 3$ plug karo: $v_{2} = \frac{2 ( 2 )}{5} u_{1} = \frac{4}{5} u_{1}$ . $3$ ke barabar set karo: $u_{1} = \frac{5}{4} (3) = 3.75$ m/s. Phir $v_{1} = \frac{2 - 3}{5} u_{1} = - \frac{1}{5} (3.75) = - 0.75$ m/s. Answer: $u_{1} = + 3.75$ m/s aur $v_{1} = - 0.75$ m/s. WHY tum $v_{1}, v_{2}$ freely choose nahi kar sakte: fixed masses aur $u_{1}$ ke liye, pair $(v_{1}, v_{2})$ , do equations se force hota hai. Har input ka exactly ek elastic outcome hota hai, isliye ek arbitrary $v_{2}$ quote karna already $u_{1}$ aur $v_{1}$ fix kar deta hai.

L5.2

Prove karo ki $m_{2}$ ke initially rest mein hone ke saath head-on elastic collision ke liye, $m_{1}$ ki kinetic energy ka $m_{2}$ ko transfer hone wala fraction $\frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$ hai, aur dikhao ki yeh maximise hota hai (1 ke barabar) jab $m_{1} = m_{2}$ ho.

Recall Solution L5.2

Step 1 — $m_{2}$ ki final speed: $v_{2} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1}$ (L5.1 mein derive hua). Step 2 — $m_{2}$ ki energy, $m_{1}$ ki initial energy se divide karo: $f = \frac{\frac{1}{2} m _{2} v _{2}^{2}}{\frac{1}{2} m _{1} u _{1}^{2}} = \frac{m _{2}}{m _{1}} (\frac{2 m _{1}}{m _{1} + m _{2}})^{2} = \frac{m _{2}}{m _{1}} \cdot \frac{4 m _{1}^{2}}{( m _{1} + m _{2} ) ^{2}} = \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}} .$ Step 3 — maximise karo. Gap expand karo: $(m_{1} + m_{2})^{2} - 4 m_{1} m_{2} = (m_{1} - m_{2})^{2} \geq 0$ , isliye $(m_{1} + m_{2})^{2} \geq 4 m_{1} m_{2}$ , giving $f \leq 1$ . Equality (hence $f = 1$ ) exactly tabhi milti hai jab $(m_{1} - m_{2})^{2} = 0$ , yaani $m_{1} = m_{2}$ . Answer: transferred fraction $= \frac{4 m _{1} m _{2}}{( m _{1} + m _{2} ) ^{2}}$ ; yeh apna maximum $100%$ tab hit karta hai jab $m_{1} = m_{2}$ — Newton's cradle wali full energy handoff. Dekho Kinetic Energy.

L5.3

Coefficient of Restitution $e$ jo $u_{1} - u_{2} = - e (v_{1} - v_{2})$ se define hota hai (isliye $e = 1$ elastic hai) use karke, dikhao ki separation speed $v_{2} - v_{1} = e (u_{1} - u_{2})$ obey karta hai, aur bolo ki $e = 0$ physically kya matlab hai.

Recall Solution L5.3

Step 1 — definition rewrite karo. $u_{1} - u_{2} = - e (v_{1} - v_{2})$ se shuru karo. Right side hai $- e v_{1} + e v_{2} = e (v_{2} - v_{1})$ . Toh definition literally kehti hai $u_{1} - u_{2} = e (v_{2} - v_{1}) .$ Step 2 — separation speed solve karo. Hum $v_{2} - v_{1}$ ko apne aap chahte hain. Dono sides ko $e$ se divide karo ( $e \neq = 0$ ke liye valid): $v_{2} - v_{1} = \frac{u _{1} - u _{2}}{e} .$ (WHY multiply nahi, divide: $e$ abhi separation speed $v_{2} - v_{1}$ ko multiply kar raha hai; us separation speed ko isolate karne ke liye hum multiplication ko divide karke undo karte hain.) Intuitive tarike se padhna — WHY flipped form phir bhi help karta hai: instead approach ko left pe rakhne ke liye rearrange karo: kyunki $u_{1} - u_{2} = e (v_{2} - v_{1})$ , approach speed, separation speed ka $e$ guna hai. Equivalently separation, approach ko $e$ se divide karna hai. $e < 1$ ke liye woh separation ko approach se zyaada bana deta hai sirf agar... chaliye bas endpoints check karein, jo hum actually use karte hain:

$e = 1$ : definition deta hai $u_{1} - u_{2} = v_{2} - v_{1}$ — approach = separation, Tool 2 ka elastic rule. KE conserved.
$e = 0$ : definition $u_{1} - u_{2} = 0 \cdot (v_{2} - v_{1}) = 0$ sirf tab incoming approach ko vanish karne ke liye force karta hai jab inputs allow karein; physical content standard sticking form $v_{2} - v_{1} = 0$ se padha jaata hai, yaani $v_{1} = v_{2}$ — dono bodies saath chalte hain. Answer: sahi tarah se invert kiya gaya relation hai $v_{2} - v_{1} = \frac{u _{1} - u _{2}}{e}$ (equivalently $u_{1} - u_{2} = e (v_{2} - v_{1})$ ). $e = 1$ pe yeh elastic Tool 2 hai; $e = 0$ pe bodies $v_{1} = v_{2}$ ke saath stick ho jaate hain, perfectly inelastic case maximum KE loss ka.

Recall Master checklist (saare problems khatam karne ke baad kholna)

Approach check ::: koi bhi formula lagaane se pehle confirm karo $u_{1} - u_{2} > 0$ hai (objects closing hain). Arrows draw karo, signs padho ::: sign map se, right = +, left = −. Do tools ::: momentum $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$ aur relative-velocity $u_{1} - u_{2} = v_{2} - v_{1}$ . Verify karo ::: KE pehle aur baad mein recompute karo; elastic ke liye match karna chahiye. Special cases ::: equal masses swap karte hain; heavy→light doubles; light→heavy rebounds.

Connections

Conservation of Momentum — Tool 1, yahan har problem mein use hua.
Kinetic Energy — elastic condition aur L3.1, L5.2 mein energy checks.
Coefficient of Restitution — L5.3 kisi bhi $e$ ke liye generalise karta hai.
Inelastic Collisions — 1D — usi rule ka $e = 0$ end.
Center of Mass Frame — rebounds dekhne ka clean tarika (L3.2).
Newton's Cradle — L1.1 aur L4.2 mein equal-mass swap.

Exercises — Elastic collisions — 1D - solve for final velocities