Conservation of linear momentum — derivation from Newton's third law: Worked examples

1.4.3 · D3 · Physics › Momentum & Collisions › Conservation of linear momentum — derivation from Newton's t

Intuition Ye page kya hai

Parent note ne law prove kiya tha. Ye page use har us case ke against stress-test karta hai jo universe de sakti hai: dono objects move kar rahe hain, head-on vs. same-direction, stick karna vs. bounce karna, ek system jo rest se shuru hota hai, two dimensions (explosion aur scattering dono), ek wall jo rule tod deti hai, aur ek limiting mass. Agar tum ye sab kar sako, toh koi bhi exam collision tumhe surprise nahi kar sakta.

Ek hi tool hai jo hum har jagah reuse karte hain, wo hai master result: $before m_{1} v_{1} + m_{2} v_{2} = after m_{1} v_{1}^{'} + m_{2} v_{2}^{'} (when F_{net}^{ext} = 0)$ Primes ka matlab hai "after". Vector arrow ka matlab hai quantity ki ek direction hai, aur 1D mein hum us direction ko sign ke roop mein encode karte hain: rightward ke liye $+$ , leftward ke liye $-$ . Ye sign convention hi pura game hai — ek axis chuno, uspe tikay raho.

The scenario matrix

Kuch bhi work karne se pehle, aao har distinct situation list karein jo ek momentum problem mein aa sakti hai. Baad ke har example ko us cell ke saath tag kiya gaya hai jise wo fill karta hai.

#	Case class	Isme kya alag hai	Example
A	Rest se shuru, alag ho jaata hai	Total $P = 0$ before → baad mein dono momenta equal aur opposite hote hain	Ex 1
B	Dono move kar rahe hain, head-on, bounce karte hain	Before mein opposite signs; dono change ho jaate hain	Ex 2
C	Dono move kar rahe hain, same direction, catch-up	Before mein same signs; slower wala speed up hota hai	Ex 3
D	Ye stick kar jaate hain (perfectly inelastic)	Ek shared final velocity; KE girta hai	Ex 4
E1	Two dimensions — explosion	$x$ aur $y$ alag-alag conserve karo, rest se shuru	Ex 5
E2	Two dimensions — scattering	Dono pieces move off karte hain; dono component equations solve karo	Ex 6
F	External force present (ek wall)	Momentum conserved NAHI hota — ye trap case hai	Ex 7
G	Limiting mass ( $m_{2} \to \infty$ ya $m_{2} = 0$ )	Degenerate: wall se ball / feather ko truck ki takkar	Ex 8
H	Exam twist — ek unknown mass, baaki sab given	Hidden quantity ke liye rearrange karo	Ex 9

Hum A–H neeche cover karenge. Note karo ki hum jo axis choose karte hain ( $+ x$ = right) aur har velocity ka sign — yahi discipline ek sahi answer ko sign-flipped answer se alag karti hai.

Example 1 — Cell A: rest se shuru, splits apart

Worked example Firecracker on ice

Ek $4 kg$ block frictionless ice par still baitha hai. Ye do pieces mein burst hota hai: piece $P$ jiska weight $1 kg$ hai, right mein $6 m/s$ par fly karta hai. Piece $Q$ ki velocity nikalo ( $3 kg$ ).

Forecast: Burst se pehle kuch nahi hil raha, toh total momentum zero hai. Kuch right mein gaya — toh $Q$ ki direction guess karo padhne se pehle. Use left mein move karna chahiye taaki total zero rahe. Uski speed bhi guess karo.

Step 1 — Axis choose karo aur "before" likho. $+ x$ = right maano. $P_{i} = 0 (block at rest)$ Ye step kyun? Koi bhi external horizontal force nahi hai (frictionless), toh momentum conserved hai. Rest se shuru karna total ko exactly zero fix karta hai, jo possible sabse cleanest constraint hai.

Step 2 — "After" likho. $P_{f} = m_{P} v_{P} + m_{Q} v_{Q} = (1) (+ 6) + (3) v_{Q}$ Ye step kyun? Burst ke baad momentum sirf dono flying pieces carry karte hain.

Step 3 — Inhe equal set karo aur solve karo. $0 = 6 + 3 v_{Q} \Rightarrow v_{Q} = - 2 m/s$ Ye step kyun? Conservation kehta hai $P_{f} = P_{i} = 0$ . Negative sign physics hai: $Q$ left mein $2 m/s$ par move karta hai, bilkul forecast ke according.

Verify: After total $= (1) (6) + (3) (- 2) = 6 - 6 = 0$ ✓, "before" se match karta hai. Units: $kg \cdot m/s$ throughout ✓.

Ye gun-recoil example ka twin hai — same skeleton jaise Rocket Equation, jahan "flying out piece" exhaust gas hai.

Example 2 — Cell B: head-on, dono bounce

Worked example Do carts head-on collide karte hain

Cart 1 ( $2 kg$ ) right mein $3 m/s$ par move kar raha hai; cart 2 ( $1 kg$ ) left mein $4 m/s$ par move kar raha hai. Ek elastic collision ke baad cart 1 ko left mein $\frac{7}{3} m/s$ par move karte measure kiya gaya hai. Cart 2 ki final velocity nikalo.

Forecast: Ye opposite sides se approach karte hain, toh unke momenta partly cancel hote hain. Cart 2 left mein fast ja raha tha — heavier cart se takkar ke baad, guess karo: kya ye right ya left mein end up hoga?

Step 1 — Axis aur signed "before". $+ x$ = right. $v_{1} = + 3, v_{2} = - 4$ Ye step kyun? Head-on matlab opposite signs. Leftward cart ko $- 4$ likhna hi sign convention ka pura point hai.

Step 2 — Before total momentum. $P_{i} = (2) (+ 3) + (1) (- 4) = 6 - 4 = + 2 kg\cdotm/s$ Ye step kyun? Signed momenta add karne se cancellation automatically ho jaata hai; net drift gently rightward hai.

Step 3 — Conserve karo aur $v_{2}^{'}$ solve karo. Given $v_{1}^{'} = - \frac{7}{3}$ : $+ 2 = (2) (- \frac{7}{3}) + (1) v_{2}^{'} \Rightarrow v_{2}^{'} = 2 + \frac{14}{3} = \frac{20}{3} m/s$ Ye step kyun? Same conservation equation, ab unknown cart 2 hai. Positive $\Rightarrow$ cart 2 right mein $\frac{20}{3} \approx 6.67 m/s$ par end up karta hai — ise jis taraf se aaya tha us taraf wapas knock kiya gaya.

Verify (elastic ⇒ KE bhi conserved): Before: $\frac{1}{2} (2) (3^{2}) + \frac{1}{2} (1) (4^{2}) = 9 + 8 = 17 J$ . After: $\frac{1}{2} (2) (\frac{7}{3})^{2} + \frac{1}{2} (1) (\frac{20}{3})^{2} = \frac{49}{9} + \frac{200}{9} = \frac{249}{9} \approx 27.7 J$ . ⚠️ Ye match nahi karte — toh given "elastic" numbers inconsistent hain. Momentum satisfy ho jaata hai, lekin energy error flag karta hai. Lesson: momentum akela nahi bata sakta ki collision elastic hai; KE alag se check karo (dekho Elastic vs Inelastic Collisions).

Example 3 — Cell C: same direction, catch-up

Worked example Fast cart slow wale ko rear-end karta hai

Cart 1 ( $2 kg$ ) $+ 5 m/s$ par cart 2 ( $3 kg$ ) ko $+ 1 m/s$ par catch karta hai (dono right mein move kar rahe hain). Bounce ke baad, cart 1 slow ho kar $+ 2 m/s$ par aa jaata hai. $v_{2}^{'}$ nikalo.

Forecast: Dono right mein move kar rahe hain, toh dono momenta positive hain — kuch cancel nahi hota. Cart 1 ne speed khoyi, toh usne momentum forward pass kiya; cart 2 ko speed up karna chahiye. Guess karo kitna.

Step 1 — Signed before (is baar same sign). $P_{i} = (2) (+ 5) + (3) (+ 1) = 10 + 3 = + 13 kg\cdotm/s$ Ye step kyun? Same-direction motion matlab like signs add hote hain, Example 2 se contrasting jahan ye subtract karte the.

Step 2 — Conserve karo. $13 = (2) (+ 2) + (3) v_{2}^{'} \Rightarrow 3 v_{2}^{'} = 13 - 4 = 9 \Rightarrow v_{2}^{'} = + 3 m/s$ Ye step kyun? Ek conservation equation, ek unknown. Cart 2 $1$ se $3 m/s$ par aa gaya — speed up ho gaya, jaise forecast kiya tha, aur abhi bhi right mein move kar raha hai.

Verify: After $= (2) (2) + (3) (3) = 4 + 9 = 13$ ✓. Sanity: cart 1 ( $2 \to$ abhi bhi aage? $2 < 3$ ) ab cart 2 se slower hai, toh ye separate ho gaye hain — physically consistent, koi interpenetration nahi.

Example 4 — Cell D: stick kar jaate hain (perfectly inelastic)

Worked example Clay ball ek block mein jaata hai

Ek $0.5 kg$ clay ball right mein $8 m/s$ par fly karta hai ek stationary $1.5 kg$ block mein; ye stick kar jaate hain. Common speed $v^{'}$ aur kinetic energy loss nikalo.

Forecast: Puri $2 kg$ total mein se sirf $0.5$ hi move kar raha tha, toh combined lump crawl karna chahiye — $8 m/s$ se bahut kam. Fraction guess karo.

Step 1 — Ek shared final velocity. $m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$ Ye step kyun? "Stick together" matlab baad mein ek object, toh ek single $v^{'}$ — yahi is case ko algebraically sabse simple banata hai.

Step 2 — Plug in karo aur solve karo. $(0.5) (8) + (1.5) (0) = (2.0) v^{'} \Rightarrow 4 = 2 v^{'} \Rightarrow v^{'} = 2 m/s$ Ye step kyun? Momentum conserved hai (brief impact ke dauran koi external horizontal force nahi). $2 m/s$ yaani $8$ ka $\frac{1}{4}$ — ball apni motion $4 \times$ apne mass ke saath share kar raha hai.

Step 3 — Energy audit. $K E_{i} = \frac{1}{2} (0.5) (8^{2}) = 16 J, K E_{f} = \frac{1}{2} (2.0) (2^{2}) = 4 J$ Lost $= 12 J$ heat/sound/deformation mein. Ye step kyun? Inelastic collisions momentum conserve karte hain lekin KE destroy karte hain — parent ka Example 2 point, yahan quantify kiya gaya.

Verify: $P$ : before $= 4$ , after $= (2) (2) = 4$ ✓. KE ka fraction jo bacha $= 4/16 = 25% = \frac{m _{1}}{m _{1} + m _{2}}$ — ek body ke stationary wale se hit karne ka exact rule. ✓

Ek vector ko components mein decompose karna (2D ke liye tool banana)

Do-dimensional examples se pehle, hume $P_{i x}$ aur $P_{i y}$ notation earn karni hogi.

Example 5 — Cell E1: two-dimensional explosion

Worked example L-shaped explosion

Ek $5 kg$ object rest par burst ho kar teen mein banta hai: A ( $1 kg$ , $+ x$ mein $9 m/s$ par), B ( $2 kg$ , $+ y$ mein $3 m/s$ par), aur C ( $2 kg$ , unknown $v_{C}$ ). $v_{C}$ nikalo.

Figure 1 (Cell E1). Momentum-space diagram. Orange arrow: A ka momentum, horizontal (x-momentum) axis ke along

+ 9

. Teal arrow: B ka momentum, vertical (y-momentum) axis ke along

+ 6

. Dashed ink arrow: unka combined pull

(9, 6)

. Plum arrow: C ka momentum

(- 9, - 6)

, third quadrant mein point karta hai taaki us pull ko cancel kare aur teeno arrows wapas origin (rest) tak sum ho jaayein.

Forecast: Do pieces $+ x$ aur $+ y$ ke along fly out karte hain; C ko third quadrant mein fly karna chahiye (dono components negative) taaki inhe cancel kare. Figure mein orange aur teal arrows right aur up point karte hain, toh plum arrow down-left point karna chahiye.

Step 1 — Momentum ek vector hai: axes mein split karo. $P_{i x} = 0, P_{i y} = 0$ Ye step kyun? $P$ $x$ aur $y$ mein independently conserve hota hai (abhi define kiya upar). Rest se shuru karna dono totals ko zero set karta hai.

Step 2 — $x$ -equation. $0 = (1) (9) + (2) (0) + (2) v_{C x} \Rightarrow v_{C x} = - 4.5 m/s$ Ye step kyun? Sirf A ke paas $x$ -momentum hai; C ko ise cancel karna hai → negative (leftward), figure se match karta hai.

Step 3 — $y$ -equation. $0 = (1) (0) + (2) (3) + (2) v_{C y} \Rightarrow v_{C y} = - 3 m/s$ Ye step kyun? Sirf B $y$ -momentum carry karta hai; C ise cancel karta hai → downward.

Step 4 — Vector assemble karo aur uski speed nikalo. $v_{C} = (- 4.5, - 3) m/s, ∣ v_{C} ∣ = 4. 5^{2} + 3^{2} = 29.25 \approx 5.41 m/s$ Ye step kyun? Magnitude do components par Pythagorean theorem use karta hai — figure mein arrow ki length.

Verify: $x$ -total after $= 9 + 2 (- 4.5) = 0$ ✓; $y$ -total $= 6 + 2 (- 3) = 0$ ✓. Third-quadrant direction confirmed. Dekho Centre of Mass Motion: centre of mass kabhi nahi hila, exactly isliye sabhi momenta zero sum karte hain.

Example 6 — Cell E2: two-dimensional scattering

Worked example Puck ek stationary puck se takraata hai aur dono scatter ho jaate hain

Ek $2 kg$ puck right ( $+ x$ ) mein $5 m/s$ par move karta hai aur ek stationary $2 kg$ puck se glance karta hai. Hit ke baad, incoming puck $3 m/s$ par $x$ -axis ke upar $+ 3 0^{\circ}$ direction mein move karta hai. Struck puck ke velocity components nikalo.

Forecast: System purely $+ x$ ke along move kar raha tha, aur $y$ -momentum zero tha. Pehla puck ab upward drift karta hai, toh struck puck ko $y$ -momentum zero rakhne ke liye downward drift karna chahiye. $x$ ke along, struck puck ko jo pehle wale ne diya wo carry karna chahiye.

Step 1 — Components mein signed "before". Incoming puck 1, target puck 2 (at rest). $P_{i x} = (2) (5) + (2) (0) = 10, P_{i y} = 0$ Ye step kyun? Puri initial momentum $x$ ke along hai; $y$ -total exactly zero se shuru hota hai, hamaara tightest constraint.

Step 2 — Puck 1 ki final velocity ko components mein resolve karo. Speed $3$ at $3 0^{\circ}$ : $v_{1 x}^{'} = 3 cos 3 0^{\circ} = 3 \cdot \frac{3}{2} = \frac{3 3}{2} \approx 2.598, v_{1 y}^{'} = 3 sin 3 0^{\circ} = 3 \cdot \frac{1}{2} = 1.5$ Ye step kyun? Hum momenta sirf axis-by-axis add kar sakte hain, toh "speed aur angle" ke roop mein given ek arrow ko pehle apne $x$ aur $y$ shadows mein todna hoga (upar define tool).

Step 3 — Puck 2 ke liye $x$ -equation. $10 = (2) (\frac{3 3}{2}) + (2) v_{2 x}^{'} \Rightarrow 10 = 33 + 2 v_{2 x}^{'} \Rightarrow v_{2 x}^{'} = 5 - \frac{3 3}{2} \approx 2.402 m/s$ Ye step kyun? $x$ -momentum conserve karo; struck puck baaki carry karta hai — abhi bhi positive, toh right mein move karta hai.

Step 4 — Puck 2 ke liye $y$ -equation. $0 = (2) (1.5) + (2) v_{2 y}^{'} \Rightarrow v_{2 y}^{'} = - 1.5 m/s$ Ye step kyun? Total $y$ -momentum zero rehna chahiye; puck 1 upar gaya ( $+ 1.5$ per unit mass), toh puck 2 neeche jaana chahiye ( $- 1.5$ ), jaise forecast kiya tha.

Verify: $x$ -total after $= 2 (2.598) + 2 (2.402) = 5.196 + 4.804 = 10$ ✓; $y$ -total after $= 2 (1.5) + 2 (- 1.5) = 0$ ✓. Dono axes independently balance karte hain — yahi 2D conservation ka essence hai.

Example 7 — Cell F: external force, conserved NAHI hai (the trap)

Worked example Ball wall se bounce karti hai

Ek $0.2 kg$ ball wall se right mein $10 m/s$ par takraati hai aur left mein $8 m/s$ par rebound karti hai. Kya ball ka momentum conserved hai? Wall ne jo impulse deliver kiya wo nikalo.

Forecast: Wall clearly ball ko shove karta hai — ball-alone system par ek external force. Toh ball ka momentum change hona chahiye. Us change ka sign guess karo.

Step 1 — Before aur after ball momentum. $p_{i} = (0.2) (+ 10) = + 2, p_{f} = (0.2) (- 8) = - 1.6 kg\cdotm/s$ Ye step kyun? Direction reversed → sign flip ho gaya. Ye akela dikhata hai ki momentum constant nahi raha.

Step 2 — Impulse = change in momentum. $J = p_{f} - p_{i} = - 1.6 - 2 = - 3.6 kg\cdotm/s$ Ye step kyun? Impulse–Momentum Theorem kehta hai $J = Δ p$ . Wall ne leftward ( $-$ ) push kiya, isliye ball ka momentum $3.6$ units se drop ho gaya.

Verify: Magnitude $3.6 kg\cdotm/s$ leftward ball par; Newton's Third Law se ball wall (aur Earth) ko $+ 3.6$ rightward push karta hai — toh ball + Earth system does momentum conserve karta hai. Momentum tab hi "lost" hota hai jab hum system ko bahut chota cut karte hain. ✓

Example 8 — Cell G: limiting masses (degenerate cases)

Worked example Ek saath do limits

(a) Ek $0.5 kg$ ball ek immovable wall ( $m_{2} \to \infty$ ) se elastically $6 m/s$ par takraati hai. Uski rebound speed nikalo. (b) Ek $2000 kg$ truck $20 m/s$ par ek stationary $0.01 kg$ feather ko swallow karta hai ( $m_{2} \to 0$ , ye stick kar jaate hain). Truck ki new speed nikalo.

Forecast: (a) Wall infinitely heavy hai — ball ko same speed par wapas bounce karna chahiye. (b) Feather basically massless hai — truck ko kuch feel nahi hona chahiye.

Part (a) — infinite mass limit. $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1} m_{2} \to \infty \frac{- m _{2}}{m _{2}} v_{1} = - v_{1}$ Ye step kyun? Top aur bottom ko $m_{2}$ se divide karo aur ise badhne do toh har finite term vanish ho jaata hai, $- v_{1}$ bachta hai. Toh $v_{1}^{'} = - 6 m/s$ : same speed, reversed. Exactly wall bounce.

Part (b) — zero mass limit (stick case). $v^{'} = \frac{m _{1} v _{1}}{m _{1} + m _{2}} = \frac{( 2000 ) ( 20 )}{2000 + 0.01} = \frac{40000}{2000.01} \approx 19.9999 m/s$ Ye step kyun? Jaise $m_{2} \to 0$ hota hai, denominator $\to m_{1}$ ho jaata hai aur $v^{'} \to v_{1}$ — truck barely speed change karta hai, jaise forecast kiya tha.

Verify: (a) $∣ v_{1}^{'} ∣ = ∣ v_{1} ∣ = 6$ ✓, aur KE $\frac{1}{2} (0.5) (6^{2}) = 9 J$ before aur after ✓ (elastic). (b) $v^{'} < 20$ lekin sirf $\sim 5 \times 1 0^{- 5} m/s$ se ✓ — negligible, intuition se match karta hai.

Example 9 — Cell H: exam twist, hidden mass

Worked example Unknown mass nikalo

Ek unknown mass $m$ ki ball right mein $6 m/s$ par ek stationary $4 kg$ ball se takraati hai. Ye stick ho kar together $2 m/s$ par move off karte hain. $m$ nikalo.

Forecast: Pair $2 m/s$ par end hota hai, jo $6$ ka $\frac{1}{3}$ hai. Toh moving mass total ka roughly ek third hona chahiye — $4 kg$ se chhota. Guess karo.

Step 1 — $m$ ko unknown maanke conservation likho. $m (6) + (4) (0) = (m + 4) (2)$ Ye step kyun? Same stick-together equation jaise Example 4, lekin ab final velocity given hai aur mass hidden hai — algebra backward chalti hai.

Step 2 — Expand karo, $m$ terms collect karo, aur solve karo. $6 m = 2 m + 8 \Rightarrow 6 m - 2 m = 8 \Rightarrow 4 m = 8 \Rightarrow m = 2 kg$ Ye step kyun? $m$ wale har term ko ek side laana unknown isolate karta hai. Answer: $m = 2 kg$ — total mass $6 kg$ , toh moving ball uska $\frac{2}{6} = \frac{1}{3}$ hai, exactly wo $\frac{1}{3}$ speed ratio se match karta hai jo humne forecast kiya tha.

Verify: Before $= (2) (6) = 12$ ; after $= (2 + 4) (2) = (6) (2) = 12$ ✓. Units $kg\cdotm/s$ ✓.

Active recall

Recall Cell A (rest se splits) mein, do pieces ke momenta ka kya relation hai?

Magnitude mein equal, direction mein opposite — ye zero sum karte hain.

Recall Hum 2D mein

$x$ aur $y$ alag-alag kyun conserve karte hain (Cells E1, E2)? Momentum ek vector hai; $x$ aur $y$ components independent equations hain jo mix nahi hoti, kyunki perpendicular pushes ek doosre ko affect nahi karte.

Recall Cell F mein, kya ball ka momentum conserved hai? Conservation kyaa restore karta hai?

Nahi — wall ek external force hai. System mein Earth ko include karna conservation restore karta hai.

Recall Infinite-mass limit (Cell G) mein same-speed-reversal result kya hai?

$v_{1}^{'} \to - v_{1}$ : ball same speed par wapas bounce karti hai.

Recall Ek baar momentum balance ho jaane ke baad elastic aur inelastic outcome mein kya ek check distinguish karta hai?

Kinetic-energy check — momentum dono ke liye balance karta hai; sirf elastic KE conserve karta hai.

Connections

Conservation of Linear Momentum — Derivation from Newton's Third Law — parent proof jise ye examples exercise karte hain
Elastic vs Inelastic Collisions — Cells B, E2, G; KE check aur $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} v_{1}$ ki derivation
Impulse–Momentum Theorem — Cell F, $J = Δ p$
Centre of Mass Motion — Cells A & E1, kyun $P = 0$ matlab CM wahin rehta hai
Newton's Third Law — Cell F, wall reaction conservation restore karta hai
Rocket Equation — Cell A ka cousin continuous mass ejection ke saath