Impulse-momentum theorem — derivation: Worked examples

1.4.2 · D3 · Physics › Momentum & Collisions › Impulse-momentum theorem — derivation

Ye parent theorem — Impulse–Momentum ka problem gym hai. Parent note ne kyun $J = Δ p$ hota hai, ye prove kiya tha. Yahan hum us theorem ko har tarah ki situation mein daalenge, taaki koi bhi exam case tumhe surprise na kare. (Hindi-English prefer karte ho? Hinglish version dekho.)

Shuru karne se pehle, un do tools ka ek reminder jo hum baar baar use karenge:

Definition Impulse vs. momentum — same unit, alag roles

Ye dono same quantity nahi hain, chahe inke units match karte hon:

Momentum $p = m v$ ek state hai — object ko abhi kitna motion hai.
Impulse $J$ ek change hai — ek interval mein deliver hua total push, jo us state ke change ke barabar hota hai, $J = Δ p = p_{f} - p_{i}$ .

Inke units match karte hain kyunki momentum ka change usi unit mein measure hota hai jisme momentum hota hai: $N\cdots = kg\cdotm/s$ Ye equal kyun hain: $1 N = 1 kg\cdotm/s^{2}$ , isliye $1 N\cdots = 1 kg\cdotm/s^{2} \times s = 1 kg\cdotm/s$ . Is page par: jab hum $J = Δ p$ masses aur velocities se compute karte hain to naturally $kg\cdotm/s$ likhte hain; jab force × time se compute karte hain to naturally $N\cdots$ likhte hain. Ye identical numerical cheez ko label karte hain — jahan se numbers aaye, wahi wala choose karo.

Neeche sab kuch in teeno relations ka hi application hai, har baar ek alag "shape" ki problem par.

Scenario matrix

Momentum problem kuch independent knobs ke saath vary karti hai. Neeche har cell ek alag cheez hai jo galat ho sakti hai ya nayi lag sakti hai. Worked examples un cells ke saath label kiye hain jo wo cover karte hain, aur milke poora table fill karte hain.

Cell	Kya change hota hai	Example(s)
A. Stop — object rest par khatam hota hai	$v_{f} = 0$ , $Δ p = - m v_{i}$	Ex 1
B. Bounce — velocity ka sign reverse hota hai	$v_{f} = - v_{i}$ , $Δ p = - 2 m v_{i}$	Ex 2
C. Partial bounce — reverse hota hai par slower	$	v_f
D. Speed-up (same direction)	$v_{f}, v_{i}$ same sign, $	v_f
E. Varying force — integrate karna padega	$F = F (t)$ , curve ke neeche area	Ex 5
F. Zero / degenerate — koi net impulse nahi	$Δ p = 0$ chahe force act kare	Ex 6
G. Limiting case — $Δ t \to 0$	time shrink hone par force blow up hoti hai	Ex 7
H. Real-world word problem — 2D	vector components, $+$ axes choose karo	Ex 8
I. Exam twist — graph ko invert karo, back-solve velocity	given $J$ , theorem ko reverse mein chalao	Ex 9

Sabmein ek common trap hai: impulse ek vector hai, isliye signs (ya components) hi poora game hain. Hum har baar ek $+$ direction fix karte hain.

Ex 1 — Cell A: clean stop

Worked example Sandbag catch

Ek $0.40 kg$ ka beanbag $12 m/s$ se aata hai aur $0.30 s$ mein pakad liya jaata hai (rest par la diya jaata hai). Uske upar impulse aur average force nikalo.

Forecast: Agar hum incoming direction ko $+$ kahein to kya impulse positive hoga ya negative? Aage padhne se pehle sign guess karo.

Step 1 — Ek $+$ direction fix karo. Beanbag ki incoming motion ko $+$ maano. To $v_{i} = + 12$ , $v_{f} = 0$ . Ye step kyun? Impulse ek vector hai; axis choose kiye bina "the force" ka koi sign nahi hoga aur hum nonsense result paayenge.

Step 2 — Momentum mein change. $Δ p = m v_{f} - m v_{i} = 0.40 (0) - 0.40 (12) = - 4.8 kg\cdotm/s$ Ye step kyun? Theorem kehta hai $J = Δ p$ , to $Δ p$ compute karna hi impulse compute karna hai. Minus physics hume bata rahi hai ki force backwards point karti hai, incoming beanbag ko oppose karti hai. ( $kg\cdotm/s$ mein likha kyunki mass × velocity use ki; ye $- 4.8 N\cdots$ ke barabar hai.)

Step 3 — Average force. $\overset{ˉ}{F} = \frac{Δ p}{Δ t} = \frac{- 4.8}{0.30} = - 16 N$ Ye step kyun? $\overset{ˉ}{F} Δ t = Δ p$ rearrange kiya. Sign phir se "backwards" bol raha hai.

Verify: Units $\frac{kg\cdotm/s}{s} = kg\cdotm/s^{2} = N$ ✓. Sign negative hai (motion ko oppose karta hai) jaise forecast kiya tha. Magnitude $16 N$ ≈ ek $1.6 kg$ object ka weight — gentle catch ke liye plausible hai.

Ex 2 — Cell B: perfect bounce (doubling)

Worked example Ball off a wall

Ek $0.25 kg$ ki ball wall se $8 m/s$ par takraati hai aur same speed $8 m/s$ par wapas aati hai. Wall se impulse?

Forecast: Kya ye impulse ball ko sirf rokne se zyada hai, kam hai, ya barabar hai? Kitna?

Step 1 — Fix $+$ = wall ki taraf. To $v_{i} = + 8$ , $v_{f} = - 8$ . Ye step kyun? Ball opposite direction mein jaati hai, isliye uski final velocity mein minus hona chahiye. Ye flip hi bouncing ko special banata hai.

Step 2 — Momentum mein change. $Δ p = 0.25 (- 8) - 0.25 (8) = - 2 - 2 = - 4 kg\cdotm/s = - 4 N\cdots$ Ye step kyun? Kyunki $v_{f}$ aur $v_{i}$ ke opposite signs hain, dono terms magnitude mein add hote hain, partly cancel nahi karte. (Same quantity, koi bhi unit spelling.)

Step 3 — Stop se compare karo. Pure stop deta $Δ p = 0.25 (0) - 0.25 (8) = - 2 kg\cdotm/s$ . Ye step kyun? "Doubling" ko explicitly dekhne ke liye: $- 4$ vs $- 2$ .

Verify: $- 4 = 2 \times (- 2)$ ✓. Bounce karne ke liye rokne se double impulse chahiye — isliye ek bouncing hailstone equal mass aur speed wali splatting raindrop se roof ko zyada dent karta hai.

Ex 3 — Cell C: partial bounce (dono signs, unequal)

Worked example Squishy ball

Ek $0.30 kg$ ki ball floor par neeche $10 m/s$ se girti hai aur sirf $5 m/s$ se upar wapas uchhalti hai (energy lost hoti hai). Floor se impulse nikalo.

Forecast: Magnitude ek stop ( $3 kg\cdotm/s$ ) ke kareeb hoga ya full bounce ( $6 kg\cdotm/s$ ) ke?

Step 1 — Fix $+$ = upar. To incoming velocity neeche hai: $v_{i} = - 10$ . Rebound upar hai: $v_{f} = + 5$ . Ye step kyun? Up ko $+$ choose karna arbitrary hai par consistent rehna chahiye; incoming ball to minus carry karti hai.

Step 2 — Momentum mein change. $Δ p = 0.30 (+ 5) - 0.30 (- 10) = 1.5 + 3.0 = + 4.5 kg\cdotm/s$ Ye step kyun? Negative $v_{i}$ ko subtract karna total mein add karta hai — floor ne fall ko bhi roka aur usse wapas upar bhi launch kiya.

Step 3 — $+$ sign interpret karo. Positive matlab floor ka impulse upar point karta hai. Ye step kyun? Sanity: floor ko upar push karna chahiye taaki neeche girte ball ko reverse kar sake.

Verify: $4.5$ bilkul beech mein hai $3.0$ (stop) aur $6.0$ (perfect bounce) ke — bilkul jaise forecast kiya, kyunki bounce partial hai ✓. Sign upar hai ✓.

Ex 4 — Cell D: same direction mein speed-up

Worked example Rocket sled kick

Ek $50 kg$ ki cart already $4 m/s$ se right taraf roll kar rahi hai, use ek booster milta hai jo use $2.0 s$ mein $9 m/s$ tak (abhi bhi right) le jaata hai. Impulse aur average force nikalo.

Forecast: Dono velocities same way point karti hain. Kya momentum terms yahan add hoti hain ya subtract?

Step 1 — Fix $+$ = right. $v_{i} = + 4$ , $v_{f} = + 9$ — same sign. Ye step kyun? Yahan koi reversal nahi, isliye ye wo case hai jahan signs flip nahi hote. Hum galti se minus introduce nahi kar sakte.

Step 2 — Momentum mein change. $Δ p = 50 (9) - 50 (4) = 450 - 200 = + 250 kg\cdotm/s$ Ye step kyun? Same-sign velocities matlab hum genuine difference lete hain, sum nahi — bounce cases se contrast karo.

Step 3 — Average force. $\overset{ˉ}{F} = \frac{250}{2.0} = + 125 N$ Ye step kyun? $\overset{ˉ}{F} = Δ p /Δ t$ ; $+$ bolta hai push motion ki direction mein hai (speed up karta hai).

Verify: Units → N ✓. Pehle se forward object par positive force → speed up hoti hai ✓, $v_{f} > v_{i}$ ke saath consistent.

Ex 5 — Cell E: varying force (integrate karna padega)

Worked example Triangular push, graph se

Ek block rest par shuru hota hai. Ek force (jo $+$ direction mein push karti hai, jise hum motion ki direction ke roop mein lete hain) $t = 0$ se $t = 4 s$ tak linearly $0 N$ se $30 N$ tak ramp karti hai, force–time graph par ek triangle banati hai (figure dekho). Block ka mass $3 kg$ hai. Uski final speed nikalo.

Figure mein red shaded triangle hi impulse hai: uska area

\int F d t

ke barabar hai. Dhyan do ki force kabhi bhi apne peak par nahi rehti — isliye naive

F Δ t

(dashed rectangle at

30 N

) overshoot karta hai.

Forecast: Force constant nahi hai. Kya $J = F Δ t$ with $F = 30$ yahan legal hai? (Nahi — par true answer ka ballpark guess karo.)

Step 0 — $+$ direction fix karo. Jis direction mein force push karti hai use $+$ lo. To $F (t) > 0$ throughout, isliye impulse positive aata hai aur block $+$ direction mein speed up karta hai. Ye step kyun? "1D" integration mein bhi sign convention chahiye, warna area ka sign meaningless hai.

Step 1 — Ramp ka slope nikalo. $(0, 0)$ se $(4 s, 30 N)$ tak straight line ka slope: $slope = \frac{30 - 0}{4 - 0} = 7.5 N/s, so F (t) = 7.5 t .$ Ye step kyun? Integrate karne se pehle actual function $F (t)$ chahiye; slope = rise over run do endpoints se deta hai.

Step 2 — Impulse = curve ke neeche area. Red triangle ka base $4 s$ aur height $30 N$ hai: $J = \frac{1}{2} \times base \times height = \frac{1}{2} (4) (30) = 60 N\cdots$ Ye step kyun? $J = \int F d t$ hi $F (t)$ ke neeche ka area hai. Straight ramp ke liye area ek triangle hai — koi calculus nahi chahiye, geometry kaam aati hai.

Step 3 — Integral se check karo. Step 1 se $F (t) = 7.5 t$ use karke: $J = \int_{0}^{4} 7.5 t d t = [3.75 t^{2}]_{0}^{4} = 3.75 (16) = 60 N\cdots$ Ye step kyun? "Area = integral" claim ko prove karne ke liye, picture par trust karne ke bajaye.

Step 4 — Theorem final speed deta hai. $J = m v_{f} - 0 \Rightarrow 60 = 3 v_{f} \Rightarrow v_{f} = 20 m/s$ Ye step kyun? Jo area humne nikala wo hi momentum change hai; velocity ke liye mass se divide karo.

Verify: Agar hum galti se $F Δ t = 30 \times 4 = 120$ use karte, to $v_{f} = 40$ milta — double truth, kyunki average force sirf $15 N$ hai, $30$ nahi. $\overset{ˉ}{F} Δ t = 15 (4) = 60$ ✓ match karta hai. Isliye hum integrate karte hain.

Ex 6 — Cell F: zero net impulse (degenerate)

Worked example Full round trip, wapas start par

Ek $2 kg$ ka puck east mein $5 m/s$ se move kar raha hai. Phir use push kiya jaata hai (kuch forces use slow karti hain, rokti hain, reverse karti hain, phir wapas speed up karti hain) taaki poore sequence ke baad wo phir se east mein $5 m/s$ se move kar raha ho, bilkul waise hi jaisa shuru mein tha. Poore sequence mein deliver kiya gaya net impulse kya hai?

Forecast: Forces clearly act kiye (unhone puck ko reverse karke wapas laaya!). Kya net impulse phir bhi zero ho sakta hai?

Step 1 — Fix $+$ = east. Sirf start aur end states identify karo. Start: $v_{i} = + 5$ . Poore sequence ka end: $v_{f} = + 5$ (start jaisa hi). Ye step kyun? Theorem sirf endpoints ki parwah karta hai, beech ki messy path ki nahi — isliye hum intermediate stopping aur reversing bilkul ignore karte hain.

Step 2 — Net impulse. $J_{net} = m v_{f} - m v_{i} = 2 (5) - 2 (5) = 0 N\cdots$ Ye step kyun? Same start aur end velocity → $Δ p = 0$ → zero net impulse, chahe raaste mein individual forces badi rahi hon.

Step 3 — Intuition se reconcile karo. Badi forces huiyin, par wo canceling pairs mein aaiyin: ek westward impulse jo puck ko reverse karta hai, phir ek eastward impulse jo usse wapas restore karta hai. Vectors ke roop mein ye zero sum hote hain. Ye step kyun? Ye dikhane ke liye ki "no net impulse" ≠ "no forces" — matlab hai impulses ka vector sum cancel ho jaata hai.

Verify: $Δ p = 0$ ✓. Ye Conservation of Momentum ka seed idea hai: jab net external impulse zero hota hai, momentum unchanged rehta hai.

Ex 7 — Cell G: limiting case ( $Δ t \to 0$ )

Worked example Hard floors kyun hurt karte hain

Ek $0.5 kg$ ka phone girta hai aur $6 m/s$ par hit karta hai, phir ruk jaata hai. Do stopping times ke liye average force compute karo: (a) foam par, $Δ t = 0.10 s$ ; (b) concrete par, $Δ t = 0.002 s$ . $Δ t \to 0$ hone par kya hota hai?

Forecast: Momentum change dono baar same hai. To "damage" mein difference kya carry karta hai?

Step 1 — Dono ke liye same $Δ p$ . $+$ = impact ki downward direction lo; $v_{i} = + 6$ , $v_{f} = 0$ : $Δ p = 0.5 (0) - 0.5 (6) = - 3 kg\cdotm/s$ Ye step kyun? Floor material phone ka wo momentum change nahi badal sakti jo use lose karna hai — sirf kitne time mein hota hai wo badal sakti hai.

Step 2 — Foam. $\overset{ˉ}{F} = \frac{- 3}{0.10} = - 30 N$ . Step 3 — Concrete. $\overset{ˉ}{F} = \frac{- 3}{0.002} = - 1500 N$ . Ye steps kyun? Same numerator $Δ p$ , tiny denominator $Δ t$ → fraction explode karta hai.

Step 4 — The limit. $Δ p$ fixed rakhe aur $Δ t \to 0$ karo: $\overset{ˉ}{F} = \frac{Δ p}{Δ t} ⟶ \infty$ Ye step kyun? General law dikhata hai: instantaneous stop matlab infinite force. Real materials hume $Δ t$ stretch karke bachate hain (crumple zones, airbags, knees bend karna).

Verify: Concrete force foam force ka $\frac{0.10}{0.002} = 50 \times$ hai ✓ ( $30 \times 50 = 1500$ ). Same $Δ p$ , $50 \times$ chhota time, $50 \times$ badi force — ek clean inverse relationship.

Ex 8 — Cell H: 2D real-world word problem (components)

Worked example Hockey puck deflected

Ek $0.16 kg$ ka puck east mein $15 m/s$ se slide karta hai. Ek stick use strike karti hai to wo $15 m/s$ se jaata hai par ab due north move karta hai. Stick se impulse vector (magnitude aur direction) nikalo.

Figure mein, do black arrows initial (east) aur final (north) momenta hain;

p_{i}

ki tip se

p_{f}

ki tip tak jaane wala red arrow impulse

J = p_{f} - p_{i}

hai. Impulse literally momenta ka triangle close karta hai.

Forecast: Speed nahi baddhi, sirf direction baddhi. Kya impulse zero hai? (Dhyan raho — ye ek vector hai!)

Step 1 — Axes set karo. $+ x$ = east, $+ y$ = north. Initial: $p_{i} = (0.16 \times 15, 0) = (2.4, 0) kg\cdotm/s$ . Final: $p_{f} = (0, 0.16 \times 15) = (0, 2.4) kg\cdotm/s$ . Ye step kyun? 2D mein har axis apni 1D problem hai; hum $x$ aur $y$ alag alag handle karte hain.

Step 2 — Har component mein Impulse. $J_{x} = 0 - 2.4 = - 2.4 kg\cdotm/s, J_{y} = 2.4 - 0 = + 2.4 kg\cdotm/s$ Ye step kyun? $J = Δ p$ har direction mein independently apply karo. $J_{x}$ negative hai (stick ne eastward motion khatam ki); $J_{y}$ positive hai (stick ne northward motion create ki).

Step 3 — Magnitude. $∣ J ∣ = J_{x}^{2} + J_{y}^{2} = (- 2.4)^{2} + (2.4)^{2} = 2.4 2 \approx 3.39 N\cdots$ Ye step kyun? Total impulse uske perpendicular parts ka vector sum hai (figure mein arrows par Pythagoras).

Step 4 — Direction. Ye northwest point karta hai — north se west ki taraf $4 5^{\circ}$ par (yaani east se counter-clockwise $13 5^{\circ}$ measure karke), kyunki $J_{x} < 0$ , $J_{y} > 0$ aur wo size mein equal hain. Ye step kyun? Equal-and-opposite-sign components → arrow second quadrant mein bisect karta hai.

Verify: Constant speed ke bawajood, $J \neq = 0$ — direction change karna ek momentum change hai ✓. $2.4 2 \approx 3.394$ ✓, aur impulse incoming east arrow se door, outgoing north arrow ki taraf point karta hai, jaise figure dikhata hai.

Ex 9 — Cell I: exam twist (theorem ko invert karo, back-solve velocity)

Worked example Theorem ko ulta padhna

$0.020 s$ ke collision mein ek sensor record karta hai ki deliver kiya gaya impulse $J = 12 N\cdots$ tha. Strike kiye gaye object ka mass $0.60 kg$ hai aur wo $v_{i} = - 8 m/s$ se move kar raha tha (rightward ko $+$ maanke). (a) Average force, (b) final velocity nikalo.

Forecast: Hume impulse diya gaya hai aur velocity get karne ke liye theorem ko reverse mein chalana hai. Kya object right ya left move karega?

Step 1 — Fix $+$ = right (already given). $v_{i} = - 8$ matlab wo left move kar raha tha; recorded $J = + 12$ ek rightward impulse hai. Ye step kyun? Signs hi poora game hain; kuch bhi invert karne se pehle convention lock karte hain.

Step 2 — Impulse se average force. $\overset{ˉ}{F} = \frac{J}{Δ t} = \frac{12}{0.020} = 600 N$ Ye step kyun? $J = \overset{ˉ}{F} Δ t$ ko $\overset{ˉ}{F}$ ke liye solve kiya; ye equivalent rectangle ki height hai jiska area recorded impulse ke barabar hai (Ex 5 ke "find the area" ka inverse).

Step 3 — Theorem se final velocity. $J = m v_{f} - m v_{i} \Rightarrow 12 = 0.60 v_{f} - 0.60 (- 8)$ $12 = 0.60 v_{f} + 4.8 \Rightarrow 0.60 v_{f} = 7.2 \Rightarrow v_{f} = + 12 m/s$ Ye step kyun? Hume $J$ aur initial state pata hai, to theorem final state pin down karta hai — har pehle example ka inverse, jahan hum velocities jaante the aur $J$ nikalte the.

Step 4 — Interpret karo. Object $- 8$ (leftward) par aaya aur $+ 12$ (rightward) par gaya: bade rightward impulse se wo reverse bhi hua aur speed up bhi hua. Ye step kyun? Sign check: ek bada $+$ impulse ek leftward object ko rightward motion mein push karna chahiye, aur karta hai.

Verify: Plug back karo: $Δ p = 0.60 (12) - 0.60 (- 8) = 7.2 + 4.8 = 12 = J$ ✓ (loop close hota hai). Aur $\overset{ˉ}{F} Δ t = 600 (0.020) = 12 = J$ ✓. Units: $\frac{N\cdots}{s} = N$ ✓.

Recall

Recall Kaun sa cell kaun sa hai?

Stop karne par $Δ p = ?$ ::: $- m v_{i}$ (final velocity zero). Perfect bounce karne par $Δ p = ?$ ::: $- 2 m v_{i}$ — stop se double, kyunki velocity sign reverse karti hai. Jab force time ke saath vary kare, impulse barabar hota hai ::: force–time curve ke neeche area ke, $\int F d t$ . Kya net impulse zero ho sakta hai chahe badi forces act karein? ::: Haan — agar start aur end velocities match karein, $Δ p = 0$ (impulses cancel ho jaate hain). Fixed $Δ p$ ke saath stopping time $Δ t \to 0$ hone par, force ::: infinity ki taraf blow up hoti hai, kyunki $\overset{ˉ}{F} = Δ p /Δ t$ . 2D mein, kya impulse zero hota hai jab sirf direction change ho? ::: Nahi — direction change bhi ek momentum change hai; components ko Pythagoras se combine karo. Kya impulse momentum ke same quantity hai? ::: Nahi — impulse momentum ka change hai, $J = Δ p$ ; ye sirf ek unit share karte hain.

Connections

Newton's Second Law — har $\overset{ˉ}{F} = Δ p /Δ t$ step $F = d p / d t$ averaged hai; theorem ka origin.
Hinglish exposition — same derivation Hindi-English mein.
Force–Time Graphs — Ex 5 aur Ex 9 pure "area under the curve" problems hain.
Collisions — Elastic and Inelastic — bounce vs stop (Ex 2, 3) collision types ka doorway hai.
Conservation of Momentum — Ex 6 ka zero-net-impulse case conservation ka seed hai.
Work–Energy Theorem — "force over distance" wala cousin; yahan force over time se contrast karo.
Center of Mass Motion — external impulse poore system ka total momentum move karta hai.