Linear momentum p = mv: Worked examples — Hinglish deep dive

1.4.1 · D3 · Physics › Momentum & Collisions › Linear momentum p = mv

Yeh page Linear momentum $p=mv$ ki practice arena hai. Parent note ne bataya tha ki momentum kya hota hai; yahan hum har tarah ke sawaal ko drill karte hain jo yeh topic pooch sakta hai — har sign, dono directions, zero aur infinite limits, ek real-world word problem, aur ek exam-style trap. Koi naya assumption nahi hai: agar koi symbol aata hai, toh hum use yahan hi dobara establish karenge.

Recall Woh ek formula jis par sab kuch tika hai

$p = m v$ $m$ mass hai (hamesha ek positive number, kilograms mein measured, $kg$ ). $v$ velocity hai — speed with ek direction. $p$ par chhota arrow ka matlab hai ki momentum ek vector hai: iska ek size bhi hai aur yeh kisi direction mein point karta hai. Iska unit $kg\cdotm/s$ hai.

Scenario matrix

Kuch bhi solve karne se pehle, chalte hain har distinct tarah ki situation list karte hain jo yeh topic aap par throw kar sakta hai. Neeche ek worked example har cell ko cover karta hai.

#	Cell (case class)	Ise alag kya banata hai	Example jo ise cover karta hai
A	Plain 1D, ek object, positive velocity	Bare formula, koi signs ki chinta nahi	Ex 1
B	1D negative velocity	Momentum negative ho sakta hai — direction sign ke roop mein encode hoti hai	Ex 2
C	1D, do objects, opposite directions	Cancellation ke saath vector sum	Ex 3
D	Zero / degenerate input	Object at rest, ya massless idealisation — $p$ ka kya hoga?	Ex 4
E	2D, perpendicular components	Right angles par add karna → Pythagoras + angle	Ex 5
F	2D, general angle (dono components signed)	Ek component "galat" taraf point karta hai	Ex 6
G	Limiting behaviour	Same $p$ , mass $\to$ huge / speed $\to$ huge	Ex 7
H	Real-world word problem	Messy English ko $m$ aur $v$ mein translate karna	Ex 8
I	Exam-style twist (units + $K E$ link)	Mixed units, aur $K E = p^{2} /2 m$	Ex 9

Hum cells A–I ko nau examples se cover karte hain. Chalte hain.

Apni direction ki picture set karna

Neeche sab kuch ek sign convention use karta hai: ek direction ko positive choose karo. Hum hamesha right = positive $(+)$ aur left = negative $(-)$ draw karenge. Yeh akela choice "direction" ko "ek plus ya minus sign" mein badal deta hai, jo woh poora trick hai jo 1D momentum ko easy banata hai.

Cell A — Plain 1D, positive velocity

Worked example Ex 1 · Ek rolling bowling ball

Ek $7.0 kg$ ka bowling ball right ki taraf $4.0 m/s$ par roll karta hai. Uska momentum nikalo.

Forecast: Aage padhne se pehle number aur uska sign guess karo. $10$ se bada? Positive ya negative?

Pieces identify karo. $m = 7.0 kg$ , $v = + 4.0 m/s$ (right = positive). Yeh step kyun? Momentum ko exactly do inputs chahiye, mass aur velocity — pehle unhe pin down karo.
$p = m v$ apply karo. $p = (7.0) (+ 4.0) = + 28 kg\cdotm/s$ Yeh step kyun? Definition ka direct use; ek object ke liye 1D mein kuch aur zaroorat nahi.
Sign padho. $+ 28$ matlab "28 units of oomph, right ki taraf point karta hua." Yeh step kyun? Sign HI direction hai — yeh answer ka part hai, decoration nahi.

Verify: Units: $kg \times m/s = kg\cdotm/s$ ✓. Magnitude $28$ , ek $10 kg\cdotm/s$ wale football se heavier×faster hai, jo "bowling ball = bada oomph" se match karta hai. ✓

Cell B — 1D negative velocity

Worked example Ex 2 · Ball leftward fire kiya gaya

Ek $0.30 kg$ ka ball left ki taraf $5.0 m/s$ par move karta hai. Uska momentum nikalo.

Forecast: Pehle jaisi hi size? Is baar sign kya hoga?

"Left" ko sign mein translate karo. Right $+$ hai, toh left $-$ hai: $v = - 5.0 m/s$ . Yeh step kyun? Hamaari convention (Fig s01) ek direction word ko ek signed number mein convert karti hai.
$p = m v$ apply karo. $p = (0.30) (- 5.0) = - 1.5 kg\cdotm/s$ Yeh step kyun? Mass hamesha positive hota hai, toh velocity ka sign seedha $p$ mein carry hota hai.
Interpret karo. Minus sign kehta hai momentum left ki taraf point karta hai. Iska magnitude (upar define kiye gaye bars ke saath uska size) $∣ p ∣ = 1.5 kg\cdotm/s$ hai. Yeh step kyun? Magnitude sign drop karta hai; sign direction ke baare mein alag information hai.

Verify: Ek negative $p$ ka matlab kabhi "zero se kam oomph" nahi — matlab hai oomph doosri taraf aimed hai. Agar yeh Ex 1 ke ball se head-on takraata, toh yeh $- 1.5$ us ball ke $+ 28$ mein se subtract karta. Yahi reason hai ki hum signs rakhte hain. ✓

Cell C — 1D, opposite directions, total momentum

Worked example Ex 3 · Do carts approach kar rahe hain

Cart A: $2.0 kg$ at $+ 4.0 m/s$ . Cart B: $3.0 kg$ at $- 2.0 m/s$ (left ki taraf move karta hua). System ka total momentum nikalo.

Forecast: Kya total positive hoga (net rightward), negative, ya zero?

Har cart ka momentum. $p_{A} = (2.0) (+ 4.0) = + 8.0 kg\cdotm/s, p_{B} = (3.0) (- 2.0) = - 6.0 kg\cdotm/s$ Yeh step kyun? Ek system ka total momentum parts ka vector sum hota hai, isliye pehle har part chahiye.
Signs ke saath add karo. $p_{total} = (+ 8.0) + (- 6.0) = + 2.0 kg\cdotm/s$ Yeh step kyun? Ek positive aur ek negative add karne se opposite directions partly cancel ho jaati hain — yeh ek line par vector sum hai.

Verify: Fig s02 mein green net arrow right ki taraf point karta hai aur chhota hai, $+ 2.0$ se match karta hua. Sanity check: A ka rightward oomph ( $8$ ) B ke leftward oomph ( $6$ ) se zyada hai, isliye bacha hua right ki taraf jhukta hai — consistent. ✓

Cell D — Zero aur degenerate inputs

Worked example Ex 4 · At rest, aur massless limit

(a) Ek $12 kg$ ka box still baitha hai. (b) Ek idealised "massless" tracer speck $10 m/s$ par drift karta hai. Dono ka momentum?

Forecast: Kya koi moving cheez ka zero momentum ho sakta hai? Kya koi ruki hui cheez ka?

Box at rest: $v = 0$ , toh $p = (12) (0) = 0 kg\cdotm/s .$ Yeh step kyun? Zero velocity product ko kill kar deta hai chahe box kitna bhi heavy kyun na ho — koi motion nahi, koi oomph nahi.
Massless speck: idealization mein $m \to 0$ , toh $p = (0) (10) = 0 kg\cdotm/s .$ Yeh step kyun? Zero mass bhi product ko kill kar deta hai, chahe woh kitni bhi tez kyun na jaaye.

Verify: Dono degenerate inputs ( $v = 0$ ya $m = 0$ ) $p = 0$ dete hain. Yeh structure check karta hai: $p = m v$ ek product hai, isliye yeh us waqt vanish karta hai jab koi bhi factor zero ho. Ek parked truck aur ek hypothetical massless dust mote dono zero momentum carry karte hain — opposite reasons ke liye. ✓

Cell E — 2D, perpendicular components

Ab motion line chhod deti hai, isliye ek akela $\pm$ sign ab kaafi nahi. Hum components chahiye: velocity ko split karo kitna east ( $x$ ) jaata hai aur kitna north ( $y$ ) jaata hai, dono ko alag handle karo, phir rebuild karo.

$p_{x} = m v_{x}, p_{y} = m v_{y}, ∣ p ∣ = p_{x}^{2} + p_{y}^{2}$

Worked example Ex 5 · Air table par ek puck

Ek $0.20 kg$ puck ka $v_{x} = 3.0 m/s$ (east) aur $v_{y} = 4.0 m/s$ (north) hai. $∣ p ∣$ aur uski direction nikalo.

Forecast: Ek tempting-but-wrong guess hai do component speeds simply add karna, $v_{x} + v_{y} = 3.0 + 4.0 = 7.0$ , aur mass se multiply karna $0.20 \times 7 = 1.4$ paane ke liye. Kya tumhe lagta hai true magnitude us $1.4$ se badi hai ya chhoti? (Perpendicular cheezein seedhi add nahi hoti — woh dhyan mein rakho.)

Component momenta. $p_{x} = (0.20) (3.0) = 0.60, p_{y} = (0.20) (4.0) = 0.80 (kg\cdotm/s)$ Yeh step kyun? Mass har direction ko independently scale karta hai — momentum velocity ke components inherit karta hai.
Pythagoras se combine karo. $∣ p ∣ = 0.6 0^{2} + 0.8 0^{2} = 0.36 + 0.64 = 1.00 = 1.0 kg\cdotm/s$ Yeh step kyun? Legs $0.6$ aur $0.8$ perpendicular hain, isliye diagonal hypotenuse hai.
Direction (east-of-north angle). Angle $θ$ satisfy karta hai $tan θ = p_{y} / p_{x}$ . Hum $tan$ = opposite/adjacent use karte hain kyunki yeh do legs ka ratio hai, jo tilt ko pin karta hai. $θ = arctan (\frac{0.80}{0.60}) = arctan (1.333 \dots) \approx 53. 1^{\circ}$ Yeh step kyun? $arctan$ answer karta hai "kaun sa angle is leg-ratio se correspond karta hai?" — yeh tan ko undo kar ke direction wapas deta hai.

Verify: $0.6, 0.8, 1.0$ ek scaled $3$ - $4$ - $5$ triangle hai — ek classic exact right triangle. ✓ Aur $1.0 < 1.4$ (naive straight-add guess), confirm karta hai ki perpendicular parts ek straight sum se chhote add hote hain. ✓

Charon quadrants mein angles ke liye ek consistent rule

Cell F signs mix kare usse pehle, chalte hain ek direction name karne ka ek tarika fix karte hain taaki hum kabhi na bhatakein. Hum har angle ko standard angle $ϕ$ ke roop mein measure karte hain: positive $x$ -axis (east) se start karo aur counter-clockwise sweep karo. East $= 0^{\circ}$ , north $= 9 0^{\circ}$ , west $= 18 0^{\circ}$ , south $= 27 0^{\circ}$ .

Formula Quadrant recipe (har baar yeh karo)

Reference angle $α = arctan (\frac{∣ p _{y} ∣}{∣ p _{x} ∣})$ compute karo — $arctan$ ko hamesha positive leg-ratio do, toh $α$ ek clean acute angle hoga $0^{\circ}$ aur $9 0^{\circ}$ ke beech.
Quadrant dhoondne ke liye $p_{x}, p_{y}$ ke signs padho, phir convert karo:

Quadrant	$p_{x}$ ka sign	$p_{y}$ ka sign	Standard angle $ϕ$
I (NE)	$+$	$+$	$ϕ = α$
II (NW)	$-$	$+$	$ϕ = 18 0^{\circ} - α$
III (SW)	$-$	$-$	$ϕ = 18 0^{\circ} + α$
IV (SE)	$+$	$-$	$ϕ = 36 0^{\circ} - α$

Recipe kyun, raw $arctan$ kyun nahi? Plain $arctan$ button sirf $- 9 0^{\circ}$ aur $+ 9 0^{\circ}$ ke beech answers return karta hai (woh NW aur SE mein farq nahi kar sakta — dono same ratio dete hain). Sign table hi hai jo charon quadrants ko disambiguate karta hai. Cell E mein puck Quadrant I mein tha, isliye $ϕ = α = 53. 1^{\circ}$ directly.

Cell F — 2D, ek component "galat" taraf point karta hua

Worked example Ex 6 · Westward drift ke saath puck

Ek $0.50 kg$ puck ka $v_{x} = - 6.0 m/s$ (matlab west) aur $v_{y} = + 8.0 m/s$ (north) hai. $∣ p ∣$ aur uska standard-angle direction nikalo.

Forecast: Kya minus sign total momentum ko shrink karta hai? Arrow kaun se quadrant mein point karta hai?

Component momenta — sign rakho. $p_{x} = (0.50) (- 6.0) = - 3.0, p_{y} = (0.50) (+ 8.0) = + 4.0 (kg\cdotm/s)$ Yeh step kyun? 2D mein har axis apna sign rakhta hai; $x$ -component west ki taraf point karta hai, isliye yeh negative rehta hai.
Magnitude — squares sign erase kar dete hain. $∣ p ∣ = (- 3.0)^{2} + (4.0)^{2} = 9 + 16 = 25 = 5.0 kg\cdotm/s$ Yeh step kyun? $(- 3.0)$ ko square karne par $+ 9$ milta hai: size ko direction ki parwah nahi, isliye ek negative component length mein utna hi add karta hai jitna ek positive karta.
Direction — quadrant recipe use karo. Pehle reference angle: $α = arctan (\frac{∣ p _{y} ∣}{∣ p _{x} ∣}) = arctan (\frac{4.0}{3.0}) \approx 53. 1^{\circ} .$ Signs hain $p_{x} < 0$ , $p_{y} > 0$ ⇒ Quadrant II (north-west), toh table se $ϕ = 18 0^{\circ} - α \approx 18 0^{\circ} - 53. 1^{\circ} = 126. 9^{\circ} .$ Yeh step kyun? Sirf magnitude direction lose kar deta hai; signs quadrant pick karte hain, aur recipe acute reference angle ko true standard angle mein convert karti hai.

Verify: Phir $3$ - $4$ - $5$ ⇒ length $5.0$ . ✓ Standard angle $126. 9^{\circ}$ , $9 0^{\circ}$ (north) aur $18 0^{\circ}$ (west) ke beech hai, matlab upper-left — Fig s04 aur negative $p_{x}$ se match karta hua. ✓

Cell G — Limiting behaviour (same momentum, extreme mass/speed)

Worked example Ex 7 · Machhar aur glacier

Do objects mein se har ek exactly $p = 12 kg\cdotm/s$ carry karta hai. Object 1 ek light dart hai, $m_{1} = 0.010 kg$ . Object 2 ek slow slab hai, $v_{2} = 0.0020 m/s$ . $v_{1}$ aur $m_{2}$ nikalo, aur limits par comment karo.

Forecast: Jab fixed $p$ ke liye mass zero ki taraf shrink ho, toh speed kya karna chahiye?

$p = m v$ ko missing factor ke liye solve karo. Rearrange karke, $v = p / m$ aur $m = p / v$ . Yeh step kyun? Jo bhi factor unknown ho, fixed $p$ ko jaane wale se divide karo.
Dart ki speed. $v_{1} = \frac{p}{m _{1}} = \frac{12}{0.010} = 1200 m/s .$ Yeh step kyun? Tiny mass mein sama oomph ek huge speed force karta hai.
Slab ki mass. $m_{2} = \frac{p}{v _{2}} = \frac{12}{0.0020} = 6000 kg .$ Yeh step kyun? Crawling speed par sama oomph ek enormous mass force karta hai.
Limits state karo. Fixed $p$ ke liye: jab $m \to 0$ , $v = p / m \to \infty$ ; jab $v \to 0$ , $m = p / v \to \infty$ . Yeh step kyun? $p$ ek product hai, isliye ek factor shrink karna doosre ko blow up kar deta hai taaki product constant rahe.

Verify: $0.010 \times 1200 = 12$ ✓ aur $6000 \times 0.0020 = 12$ ✓. Same momentum, wildly different mass aur speed — exactly parent note se "equal oomph, unequal everything else" idea. ✓

Cell H — Real-world word problem

Worked example Ex 8 · Ek recoiling cannon

Ek $500 kg$ ka cannon, initially at rest, ek $5.0 kg$ ki ball ko forward $60 m/s$ par fire karta hai. Momentum conservation use karke, cannon ki recoil velocity nikalo. (Total momentum before = total after, kyunki firing forces internal hain — dekho Conservation of Linear Momentum.)

Forecast: Compute karne se pehle — kya cannon tez ya dheere move karega, aur kaun si taraf?

Pehle total momentum. Sab kuch at rest hai: $p_{before} = 0.$ Yeh step kyun? Koi motion nahi ⇒ zero oomph (phir Cell D).
Total momentum after bhi $0$ hona chahiye. Cannon ki velocity $V$ hone do aur ball ki direction $+$ hone do. $m_{cannon} = 500 kg$ aur $m_{ball} = 5.0 kg$ likhte hue: $0 = m_{ball} v_{ball} + m_{cannon} V = (5.0) (+ 60) + (500) V .$ Yeh step kyun? Internal forces system ka total momentum change nahi kar sakte, isliye "after" "before" $= 0$ ke barabar hai.
$V$ ke liye solve karo. $500 V = - 300 \Rightarrow V = - 0.60 m/s .$ Yeh step kyun? Balance rearrange karna; minus sign matlab cannon backward move karta hai, ball ke opposite.

Verify: Ball momentum $+ 300$ , cannon momentum $500 \times (- 0.60) = - 300$ ; sum $= 0$ ✓, initial $0$ se match karta hua. Heavy cannon dheere recoil karta hai ( $0.60 m/s$ ) jabki light ball tez fly karta hai — heavy ⇒ equal $∣ p ∣$ ke liye chhota $v$ , Ex 7 se consistent. ✓

Cell I — Exam twist: mixed units + kinetic-energy link

Worked example Ex 9 · Grams, km/h, aur energy

Ek $200 g$ ki ball $36 km/h$ par fly karti hai. (a) SI units mein $p$ nikalo. (b) $K E = p^{2} /2 m$ use karke uski kinetic energy nikalo.

Forecast: Students ko kya zyada trip karta hai — grams ya km/h? Pehle dono convert karo.

SI mein convert karo. $200 g = 0.200 kg$ ( $1000$ se divide karo). $36 km/h = 36 \times \frac{1000}{3600} = 10 m/s$ . Yeh step kyun? $p = m v$ sirf $kg\cdotm/s$ deta hai agar $m$ $kg$ mein ho aur $v$ $m/s$ mein ho; mixed units classic trap hai.
Momentum. $p = (0.200) (10) = 2.0 kg\cdotm/s .$ Yeh step kyun? Units saaf hone ke baad plain formula.
$K E = p^{2} /2 m$ se kinetic energy. Hum yeh form use karte hain (instead of $\frac{1}{2} m v^{2}$ ) taaki parent note se momentum–energy bridge practice karen. $K E = \frac{p ^{2}}{2 m} = \frac{( 2.0 ) ^{2}}{2 ( 0.200 )} = \frac{4.0}{0.40} = 10 J .$ Yeh step kyun? $K E = p^{2} /2 m$ answer karta hai "itne oomph aur mass ke liye kitni work-capacity?" — yeh do quantities ko directly link karta hai.

Verify: $\frac{1}{2} m v^{2} = \frac{1}{2} (0.200) (10)^{2} = \frac{1}{2} (0.200) (100) = 10 J$ ✓ se cross-check — dono routes agree karte hain. Units: $\frac{( kg\cdotm/s ) ^{2}}{kg} = \frac{kg ^{2} m ^{2} / s ^{2}}{kg} = kg\cdotm^{2} / s^{2} = J$ ✓. Dekho Kinetic Energy. ✓

Recall

Recall Kaun si cell kaun si hai?

Positive 1D single object ::: Ex 1 (Cell A) 1D mein direction kahan "rehti" hai? ::: $v$ ke $\pm$ sign mein, isliye $p$ mein bhi. $m = 0$ aur $v = 0$ dono dete hain ::: $p = 0$ , kyunki $p = m v$ ek product hai. Perpendicular components kaise combine hote hain? ::: Pythagoras se, $∣ p ∣ = p_{x}^{2} + p_{y}^{2}$ . Ek negative component magnitude ko kaise affect karta hai? ::: Bilkul nahi — squaring uska sign erase karta hai; yeh sirf quadrant set karta hai. Kisi bhi quadrant mein true direction kaise milti hai? ::: $α = arctan (∣ p_{y} ∣/∣ p_{x} ∣)$ lo, phir sign table use karo (I: $α$ , II: $180 - α$ , III: $180 + α$ , IV: $360 - α$ ). Fixed $p$ ke liye, jab $m \to 0$ toh speed ::: $\to \infty$ (aur jab $v \to 0$ , $m \to \infty$ ). Cannon recoil ki direction ::: Ball ke opposite, taaki total $p$ $0$ rahe. $36 km/h$ ko m/s mein convert karo ::: $36 \times 1000/3600 = 10 m/s$ . $K E$ momentum se ::: $K E = p^{2} /2 m$ .

Connections

Parent: Linear momentum $p=mv$ — woh definitions jo yeh examples exercise karti hain.
Conservation of Linear Momentum — Ex 8 (cannon recoil) ke peeche ka rule.
Impulse–Momentum Theorem — kaise ek stationary wall (Ex 4) phir bhi ek car rok leti hai.
Kinetic Energy — Ex 9 mein use kiya gaya $K E = p^{2} /2 m$ bridge.
Elastic and Inelastic Collisions — jahan signed 1D sums (Ex 3) constantly use hote hain.
Centre of Mass — total momentum $= M v_{c m}$ , Ex 3 aur Ex 8 ka system view.