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

1.4.3 · D4 · Physics › Momentum & Collisions › Conservation of linear momentum — derivation from Newton's t

Har problem ke liye ek hi tool. Kisi isolated system ke liye (net external force zero ho) total linear momentum $p = m v$ yeh law follow karta hai: $P_{before} = P_{after}, P = \sum_{i} m_{i} v_{i} .$ Kyunki $P$ ek vector hai, do dimensions mein hum $x$ -component aur $y$ -component ko alag-alag conserve karte hain — dono ki apni equation hoti hai. Signs dhyan se rakho: left direction ki velocity ek negative number hai, sirf "2 m/s" nahi.

Level 1 — Recognition

Goal: yeh decide karna ki momentum conserved hai ya nahi, aur seedha answer padhna.

Exercise 1.1

$50 kg$ ki ek skater frictionless ice par khadi hai aur $2 kg$ ki ball pakde hue hai. Woh ball ko east direction mein $6 m/s$ par throw karti hai. Kya total momentum conserved hai, aur throw se pehle yeh kitna hai?

Recall Solution 1.1

Kya conserved hai? Baahri forces sirf gravity (neeche) aur ice ka normal force (upar) hain; horizontally koi external force nahi hai (frictionless). Toh horizontal system isolated hai → momentum conserved hai. Pehle: sab kuch rest mein hai, isliye $P_{before} = 0.$ Yeh ek hi fact ("rest se shuru → total zero hi rahega") poora trick hai — skater ko west ki taraf recoil karna hoga taaki dono momenta milke zero hon.

Exercise 1.2

Ek $1200 kg$ ki car seedhi road par steady $20 m/s$ se chal rahi hai, engine on hai aur rolling friction bhi hai. Kya is driving ke ek second ke dauran car ka momentum conserved hai?

Recall Solution 1.2

Nahi. Steady speed ka matlab isolated nahi hota. Engine road ke through car ko aage push karta hai (external), friction aur drag peeche push karte hain (external). Yeh balance ho jaate hain, isliye $P$ constant rehta hai — lekin isliye nahi ki system isolated hai. Yeh cancellation ki coincidence hai, aur momentum sirf tab guaranteed constant hota hai jab net external force sach mein zero ho. Yahan woh balance se zero hai, toh $P = (1200) (20) = 24000 kg\cdotm/s$ hai aur tabhi tak wahan rehta hai jab tak woh balance kaayam rahe.

Level 2 — Application

Goal: $P_{before} = P_{after}$ mein numbers plug karna, ek dimension mein.

Exercise 2.1

$0.02 kg$ ki ek bullet $500 m/s$ par fire ki jaati hai $3 kg$ ki rifle se jo initially rest mein hai. Rifle ki recoil velocity nikalo.

Recall Solution 2.1

East ko ( $+$ ) lete hain bullet ki direction. Pehle: rest mein, $P_{i} = 0$ . $m_{b} v_{b} + m_{r} v_{r} = 0$ $(0.02) (500) + (3) v_{r} = 0 \Rightarrow v_{r} = - \frac{10}{3} = - 3.33 m/s .$ Minus ka matlab west direction hai — yahi recoil hai. Magnitude $3.33 m/s$ .

Exercise 2.2

$3 kg$ ki ek trolley right direction mein $4 m/s$ se chal rahi hai aur $2 kg$ ki ek trolley se collide karke usme chipak jaati hai jo right mein $1 m/s$ se chal rahi hai. Unki common velocity nikalo.

Recall Solution 2.2

Chipak jaate hain → ek final velocity $v^{'}$ . Right is $+$ . $m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$ $(3) (4) + (2) (1) = (5) v^{'} \Rightarrow 14 = 5 v^{'} \Rightarrow v^{'} = 2.8 m/s (right) .$

Exercise 2.3

Wahi do trolleys, lekin ab $2 kg$ wali left direction mein $1 m/s$ se ja rahi hai chipakne se pehle. Common velocity nikalo.

Recall Solution 2.3

Sirf $v_{2}$ ka sign badlega: leftward yani $v_{2} = - 1 m/s$ . $(3) (4) + (2) (- 1) = (5) v^{'} \Rightarrow 12 - 2 = 5 v^{'} \Rightarrow v^{'} = 2.0 m/s (right) .$ Dekho kaise ek sign flip hone se answer $2.8$ se $2.0$ ho gaya — signs hi physics hain.

Level 3 — Analysis

Goal: do dimensions, ya yeh reasoning ki conservation kya forbid karta hai.

Exercise 3.1

$4 kg$ ka ek shell rest mein hai aur teen pieces mein explode ho jaata hai. Piece A ( $1 kg$ ) east mein $9 m/s$ par udta hai; piece B ( $1 kg$ ) north mein $9 m/s$ par udta hai. Piece C ( $2 kg$ ) ki velocity (magnitude aur direction) nikalo.

Recall Solution 3.1

Pehle: rest mein → $P_{i} = 0$ , toh baad mein dono components ka sum zero hona chahiye. x-axis (east +): $(1) (9) + (1) (0) + (2) v_{C x} = 0 \Rightarrow v_{C x} = - 4.5 m/s$ . y-axis (north +): $(1) (0) + (1) (9) + (2) v_{C y} = 0 \Rightarrow v_{C y} = - 4.5 m/s$ . Toh C south-west direction mein jaata hai. Magnitude: $∣ v_{C} ∣ = (- 4.5)^{2} + (- 4.5)^{2} = 40.5 = 6.36 m/s .$ Direction: dono components negative → third quadrant (neeche-left), west axis se $4 5^{\circ}$ neeche, yaani exactly south-west ki taraf — jaise figure mein red arrow dikhata hai, yellow (A) aur green (B) arrows ko balance karte hue.

Exercise 3.2

$1000 kg$ ki car east mein $20 m/s$ se aa rahi hai aur $1500 kg$ ki van north mein $10 m/s$ se; woh aapas mein lock ho jaate hain. Impact ke baad ki common velocity nikalo.

Recall Solution 3.2

Locked → ek velocity $(v_{x}, v_{y})$ , total mass $2500 kg$ . Har axis conserve karo. x (east +): $(1000) (20) = 2500 v_{x} \Rightarrow v_{x} = 8 m/s$ . y (north +): $(1500) (10) = 2500 v_{y} \Rightarrow v_{y} = 6 m/s$ . $∣ v ∣ = 8^{2} + 6^{2} = 100 = 10 m/s .$ Direction: $θ = arctan (\frac{v _{y}}{v _{x}}) = arctan (\frac{6}{8}) = 36. 9^{\circ}$ north of east. Dono components positive → first quadrant, toh arctan par koi sign correction nahi chahiye.

Level 4 — Synthesis

Goal: momentum conservation ko ek doosre idea (energy, ya two-stage process) ke saath combine karna.

Exercise 4.1 — Ballistic pendulum

$0.05 kg$ ki bullet $1.95 kg$ ke block mein strike karke embed ho jaati hai jo rest mein hang kar raha hai. Block+bullet swing karke height $h = 0.20 m$ tak uthte hain. Bullet ki original speed nikalo. ( $g = 10 m/s^{2}$ use karo.)

Recall Solution 4.1

Do stages, do laws. Stage 1 (collision, momentum conserved — bahut fast, gravity kuch nahi kar paati abhi): $m_{b} v_{b} = (m_{b} + m_{b l k}) V \Rightarrow (0.05) v_{b} = (2.0) V .$ Stage 2 (swing up, energy conserved — ab koi collision nahi, sirf gravity): $\frac{1}{2} (2.0) V^{2} = (2.0) g h \Rightarrow V = 2 g h = 2 (10) (0.20) = 2 m/s .$ Back-substitute: $v_{b} = \frac{2.0 \times 2}{0.05} = 80 m/s .$ Do alag laws kyun? Embed ke dauran KE heat mein convert ho jaati hai, toh wahan energy use nahi kar sakte — lekin momentum bachta hai. Embed ke baad kuch nahi khoता, toh rise ke liye energy clean tool hai. Sahi stage mein sahi law lagana hi poora art hai.

Exercise 4.2 — Elastic 1D check

$2 kg$ ki ball $3 m/s$ se $1 kg$ ki stationary ball se elastically (KE conserved) takraati hai. Dono final velocities nikalo.

Recall Solution 4.2

Do conserved quantities → do equations. Right is $+$ . Momentum: $(2) (3) = 2 v_{1}^{'} + 1 v_{2}^{'} \Rightarrow 6 = 2 v_{1}^{'} + v_{2}^{'}$ . KE: $\frac{1}{2} (2) (3^{2}) = \frac{1}{2} (2) v_{1}^{'2} + \frac{1}{2} (1) v_{2}^{'2} \Rightarrow 18 = 2 v_{1}^{'2} + v_{2}^{'2}$ . Elastic shortcut use karo (relative speed reverse ho jaati hai): $v_{2}^{'} - v_{1}^{'} = 3$ . $v_{2}^{'} = v_{1}^{'} + 3$ ko momentum mein substitute karo: $6 = 2 v_{1}^{'} + v_{1}^{'} + 3 \Rightarrow v_{1}^{'} = 1 m/s$ , aur $v_{2}^{'} = 4 m/s$ . KE check: $2 (1)^{2} + (4)^{2} = 2 + 16 = 18$ ✓. Momentum: $2 (1) + 4 = 6$ ✓.

Level 5 — Mastery

Goal: variable-mass / centre-of-mass reasoning, ya ek multi-step vector problem.

Exercise 5.1 — Rocket kick

$1000 kg$ total mass ka ek rocket ( $200 kg$ fuel bhi include hai) deep space mein $100 m/s$ se drift kar raha hai. Woh saara $200 kg$ fuel ek quick burst mein backward eject karta hai $500 m/s$ par rocket ke relative. Rocket ki final speed nikalo. (Continuous version ke liye Rocket Equation dekho.)

Recall Solution 5.1

Isolated system → momentum conserved. Maano final rocket speed $v_{r}$ hai (forward $+$ ), burn ke baad body mass $= 800 kg$ . Fuel rocket se $500 m/s$ peeche jaata hai, toh ground velocity hai $v_{fuel} = v_{r} - 500$ . $before (1000) (100) = rocket (800) v_{r} + fuel (200) (v_{r} - 500)$ $100000 = 800 v_{r} + 200 v_{r} - 100000$ $200000 = 1000 v_{r} \Rightarrow v_{r} = 200 m/s .$ Rocket $100$ se $200 m/s$ tak speed up ho jaata hai — mass peeche phenk kar $100 m/s$ ka gain.

Exercise 5.2 — Centre of mass kabhi jhooth nahi bolta

Frictionless ice par, ek $60 kg$ ka insaan $40 kg$ , $4 m$ lambi boat ke left end par khada hai, sab rest mein. Woh insaan right end tak chalta hai. Boat kitni door move hoti hai (ice ke relative), aur kis direction mein? (Centre of Mass Motion use karta hai.)

Recall Solution 5.2

Key idea: system rest se shuru hota hai aur isolated hai, toh $P = 0$ hamesha → ice ke relative centre of mass move nahi karta. Maano boat left ki taraf $d$ distance slide karti hai. Insaan boat ke relative $4 m$ right chalta hai, toh ice ke relative insaan $(4 - d)$ right move karta hai. Centre of mass fixed rehta hai: $m_{person} (4 - d) = m_{boat} d$ $(60) (4 - d) = (40) d \Rightarrow 240 - 60 d = 40 d \Rightarrow 240 = 100 d \Rightarrow d = 2.4 m .$ Boat $2.4 m$ left slide karti hai (walk ke opposite), aur insaan actually ice ke upar $4 - 2.4 = 1.6 m$ right move karta hai.

Quick self-check ladder

Apne aap ko rate karo

L1 done = tum jaante ho kab momentum conserved hota hai; L5 done = tum ise energy, frames, aur centre of mass ke saath combine kar sakte ho.

Inelastic crash ke dauran kaun sa law?

Sirf Momentum (KE lost ho jaati hai).

Frictionless swing/rise ke baad kaun sa law?

Energy conservation.

2D collisions: kitni equations?

Ek per axis —

x

aur

y

alag-alag conserve karo.

Connections

Newton's Third Law — kyun internal forces har problem mein cancel ho jaate hain
Newton's Second Law (momentum form) — $F = d p / d t$ , derivation ki root
Elastic vs Inelastic Collisions — Exercises 2.2, 4.1 (inelastic) vs 4.2 (elastic)
Impulse–Momentum Theorem — 2.1 mein recoil ka integral view
Centre of Mass Motion — Exercise 5.2 ke peeche ka engine
Rocket Equation — Exercise 5.1 ka continuous version