Impulse-momentum theorem — derivation: Exercises

1.4.2 · D4 · Physics › Momentum & Collisions › Impulse-momentum theorem — derivation

Ek hi law hai jis par hum poore raaste rely karte hain: $J = \int_{t_{1}}^{t_{2}} F d t = Δ p = m v_{f} - m v_{i}, \overset{ˉ}{F} = \frac{Δ p}{Δ t}$

Neeche jo bhi hai woh isi ek idea ke teen alag faces hain. Agar tumhe derivation chahiye, toh parent note par wapas jao.

Level 1 — Recognition

Goal: spot karo ki kaunsa formula apply hota hai aur cleanly plug in karo.

L1·Q1 — Units check

Impulse $N\cdots$ mein measure hota hai. Dikhao ki yeh $kg\cdotm/s$ , momentum ki unit, ke barabar hai, aur ek sentence mein batao ki yeh match kyun karna chahiye.

Recall Solution

Ek newton $kg\cdotm/s^{2}$ hai (mass × acceleration). Seconds se multiply karo: $N\cdots = \frac{kg\cdotm}{s ^{2}} \cdot s = \frac{kg\cdotm}{s} = kg\cdotm/s .$ Yeh zaroor match karna chahiye kyunki theorem kehta hai $J = Δ p$ — ek quantity doosri ke equal tab hi ho sakti hai jab unki units agree karti hon.

L1·Q2 — Constant force, direct plug-in

$8 N$ ki ek constant force ek cart ko $3 s$ tak push karti hai. Woh kitna impulse deliver karti hai?

Recall Solution

Force constant hai, isliye $J = F Δ t$ use karna sahi hai: $J = 8 \times 3 = 24 N\cdots .$

L1·Q3 — Momentum change from velocities

Ek $4 kg$ ka trolley ek straight line mein $2 m/s$ se $5 m/s$ tak speed up karta hai. Uske momentum mein change find karo.

Recall Solution

$Δ p = m v_{f} - m v_{i} = 4 (5) - 4 (2) = 20 - 8 = 12 kg\cdotm/s .$ Theorem ke zariye yeh us par act karne wala impulse bhi hai: $J = 12 N\cdots$ .

Level 2 — Application

Goal: theorem ko rearrange karo jo bhi missing ho usse solve karne ke liye.

L2·Q1 — Catching a ball (average force)

Ek $0.15 kg$ ki ball $20 m/s$ par aati hai aur $0.05 s$ mein rest par aa jaati hai. Tumhara glove jo average force lagata hai woh find karo.

Recall Solution

Positive = ball ki incoming direction. $v_{i} = + 20$ , $v_{f} = 0$ . $Δ p = 0.15 (0) - 0.15 (20) = - 3 kg\cdotm/s .$ $\overset{ˉ}{F} = \frac{Δ p}{Δ t} = \frac{- 3}{0.05} = - 60 N .$ Minus sign bata'ta hai ki force backward point karti hai, ball ko oppose karti hai — exactly wahi jo ek glove karta hai.

L2·Q2 — Softening the blow

Same ball, same speed, lekin ab tum apne haath peeche kheenchte ho toh woh $0.20 s$ mein ruk jaati hai. Nayi average force find karo aur compare karo.

Recall Solution

$Δ p$ unchanged rehta hai ( $- 3 kg\cdotm/s$ ) kyunki ball ko phir bhi $20$ se $0$ tak jaana hai. $\overset{ˉ}{F} = \frac{- 3}{0.20} = - 15 N .$ Chaar guna time → ek chauthai force. Exactly yahi wajah hai ki airbags, crumple zones aur ghutne modhna kaam karta hai.

L2·Q3 — Finding the stopping time

Ek $1200 kg$ ki car $30 m/s$ se chal rahi hai aur $9000 N$ ki average braking force se rok di jaati hai. Braking kitne time mein hoti hai?

Recall Solution

$Δ p = 1200 (0) - 1200 (30) = - 36000 kg\cdotm/s$ . Braking force motion ko oppose karti hai, isliye $\overset{ˉ}{F} = - 9000 N$ . $Δ t = \frac{Δ p}{F ˉ} = \frac{- 36000}{- 9000} = 4 s .$

Level 3 — Analysis

Goal: bounces, graphs, aur time-varying forces handle karo jahan naive plugging fail ho jaati hai.

L3·Q1 — Bounce vs stop

Ek $0.2 kg$ ki ball $5 m/s$ se wall se takraati hai aur $5 m/s$ se rebound karti hai. (a) Wall se impulse find karo. (b) Isse compare karo agar ball simply wall se chipak jaati.

Recall Solution

Positive = wall ke andar. $v_{i} = + 5$ , $v_{f} = - 5$ (reversed). (a) Bounce: $J = m (v_{f} - v_{i}) = 0.2 (- 5 - 5) = 0.2 (- 10) = - 2 N\cdots .$ (b) Stick: $v_{f} = 0$ , isliye $J = 0.2 (0 - 5) = - 1 N\cdots .$ Bouncing mein stopping se double impulse chahiye, kyunki wall ko pehle incoming motion cancel karni hai aur phir ball ko wapas throw karna hai.

L3·Q2 — Area under a triangular pulse

Ek kick ke dauran, ek $0.45 kg$ football par force $0.008 s$ mein linearly $0$ se $900 N$ tak badhti hai, phir agle $0.008 s$ mein linearly $0$ par wapas aa jaati hai. Ball rest se start karti hai. Uski launch speed find karo.

Figure dekho: impulse triangle ka area hai, na force × total time.

Figure — Impulse-momentum theorem — derivation

Recall Solution

Pulse ek triangle hai: base $= 0.008 + 0.008 = 0.016 s$ , peak height $= 900 N$ . $J = area = \frac{1}{2} (base) (height) = \frac{1}{2} (0.016) (900) = 7.2 N\cdots .$ Phir $J = Δ p = m v_{f}$ : $v_{f} = \frac{7.2}{0.45} = 16 m/s .$ Note: peak force ko total time se multiply karne par $900 \times 0.016 = 14.4 N\cdots$ milega — exactly double sach se, kyunki triangle apne bounding rectangle ka aadha hota hai.

L3·Q3 — Time-varying force, integrate

Ek force $F (t) = 12 t N$ ek $2 kg$ ke block par (initially at rest) $t = 0$ se $t = 3 s$ tak act karti hai. Final speed find karo.

Figure dikhata hai ki hum integrate kyun karte hain: yahan "curve" ek straight ramp hai, aur impulse uske neeche shaded area hai.

Recall Solution

Force time ke saath change hoti hai, isliye $F Δ t$ illegal hai — integrate karo: $J = \int_{0}^{3} 12 t d t = [6 t^{2}]_{0}^{3} = 6 (9) - 0 = 54 N\cdots .$ Phir $J = Δ p = m v_{f}$ : $v_{f} = \frac{54}{2} = 27 m/s .$ Geometrically ramp $F (3) = 36 N$ tak pahunchta hai; shaded triangle ka area $\frac{1}{2} (3) (36) = 54$ hai — integral aur area agree karte hain.

Level 4 — Synthesis

Goal: impulse ko momentum conservation, kinematics, ya two-stage motion ke saath combine karo.

L4·Q1 — Two-stage impulse

Ek $3 kg$ ka puck right ki taraf $4 m/s$ se move kar raha hai, use $9 N\cdots$ ka rightward impulse milta hai, aur baad mein $15 N\cdots$ ka leftward impulse. Uski final velocity find karo (direction bhi batao).

Recall Solution

Positive = right. Initial momentum $p_{i} = 3 (4) = 12 kg\cdotm/s$ . Total impulse $= + 9 - 15 = - 6 N\cdots$ . $p_{f} = p_{i} + J_{total} = 12 + (- 6) = 6 kg\cdotm/s .$ $v_{f} = \frac{6}{3} = 2 m/s (to the right) .$ Impulses vectors ki tarah add hote hain; net wahi hai jo momentum ko shift karta hai.

L4·Q2 — Recoil (impulse pair)

Ek $60 kg$ ka skater khada hua hai aur ek $2 kg$ ki ball $0.30 s$ mein $9 m/s$ aage throw karta hai. (a) Skater jo average force ball par lagata hai woh find karo. (b) Skater ki recoil speed find karo, aur Newton's third law se connect karo.

Recall Solution

(a) Ball: $Δ p = 2 (9) - 0 = 18 kg\cdotm/s$ . $\overset{ˉ}{F} = \frac{18}{0.30} = 60 N (forward on the ball) .$ (b) Newton's third law ke zariye ball skater par $- 60 N$ se same $0.30 s$ tak push back karti hai, isliye skater ka impulse $- 18 N\cdots$ hai: $v_{skater} = \frac{- 18}{60} = - 0.30 m/s,$ yaani $0.30 m/s$ backward. Equal-and-opposite impulses cancel hote hain, isliye total momentum zero rehta hai — yeh impulse ke zariye dekha gaya Conservation of Momentum hai.

L4·Q3 — Impulse from a graph, then a collision

Ek $0.5 kg$ ka cart rest par hai, use ek force push karti hai jiska graph $0.4 s$ ke liye $10 N$ ka rectangle hai, uske turant baad $0.6 s$ mein $10 N$ tak uthne aur $0$ par wapas aane wala ek triangle hai. Cart ki final speed find karo.

Recall Solution

Area ko do shapes mein split karo. Rectangle: $10 \times 0.4 = 4 N\cdots$ . Triangle: $\frac{1}{2} (0.6) (10) = 3 N\cdots$ . Total $J = 4 + 3 = 7 N\cdots$ . $v_{f} = \frac{J}{m} = \frac{7}{0.5} = 14 m/s .$

Level 5 — Mastery

Goal: full multi-step problems jahan tumhe khud method choose karna hai.

L5·Q1 — Water jet on a wall

Ek hose $20 m/s$ se horizontally wall par paani spray karta hai, $3 kg$ paani per second deliver karta hai. Paani bounce nahi karta (woh wall par neeche beh jaata hai, zero horizontal velocity ke saath khatam hota hai). Wall par average horizontal force find karo.

Recall Solution

Ek second mein, mass $Δ m = 3 kg$ $+ 20 m/s$ par aati hai aur $0$ par khatam hoti hai. Momentum destroyed per second: $\frac{Δ p}{Δ t} = \frac{Δ m ( v _{f} - v _{i} )}{Δ t} = \frac{3 ( 0 - 20 )}{1} = - 60 N .$ Yeh woh force hai jo wall paani par lagati hai; Newton's third law ke zariye paani wall ko $+ 60 N$ se push karta hai. Isliye wall par average force $60 N$ hai.

L5·Q2 — Jet that splashes straight back

Same hose, same $20 m/s$ aur $3 kg/s$ , lekin ab paani seedha $20 m/s$ par wapas rebound karta hai. Wall par force find karo aur do ka factor explain karo.

Recall Solution

Ab $v_{i} = + 20$ , $v_{f} = - 20$ . $\frac{Δ p}{Δ t} = \frac{3 ( - 20 - 20 )}{1} = \frac{3 ( - 40 )}{1} = - 120 N,$ isliye paani wall ko $120 N$ se push karta hai — sticking case se double. Flow ko reverse karne ke liye wall ko pehle paani rok na hai aur phir usse wapas phenkna hai, exactly wahi bounce-vs-stop doubling jo L3·Q1 mein thi, ab continuously ho rahi hai.

L5·Q3 — Impulse, then projectile

Ek $0.25 kg$ ki ball zameen par resting hai, use strike kiya jaata hai toh force graph ek triangle form karta hai jo $500 N$ peak karta hai total contact time $0.010 s$ mein, use vertically launch karta hai. Tiny contact ke dauran gravity ignore karo, find karo (a) launch speed aur (b) maximum height ( $g = 10 m/s^{2}$ ).

Recall Solution

(a) Impulse = triangle area: $J = \frac{1}{2} (0.010) (500) = 2.5 N\cdots .$ $v = \frac{J}{m} = \frac{2.5}{0.25} = 10 m/s upward .$ (b) Ab kinematics use karo (energy ya $v^{2} = u^{2} - 2 g h$ with $v = 0$ at top): $h = \frac{u ^{2}}{2 g} = \frac{1 0 ^{2}}{2 ( 10 )} = 5 m .$ Do tools, har ek apne kaam ke liye: impulse messy contact ko ek clean launch speed mein convert karta hai; kinematics phir smooth flight handle karta hai.

L5·Q4 — Design an airbag

Ek $70 kg$ ka crash-test dummy $15 m/s$ se chal raha hai aur use rokna hai. Safety average force ko $7000 N$ tak limit karti hai. (a) Minimum stopping time kitna chahiye? (b) Airbag ke bina dummy $0.05 s$ mein steering wheel se ruk jaata hai — woh force kya hai, aur airbag kis factor se help karta hai?

Recall Solution

$Δ p = 70 (0) - 70 (15) = - 1050 kg\cdotm/s$ ; magnitude $1050 N\cdots$ . (a) $Δ t_{m i n} = \frac{∣Δ p ∣}{F _{m a x}} = \frac{1050}{7000} = 0.15 s .$ (b) Hard stop: $\overset{ˉ}{F} = \frac{1050}{0.05} = 21000 N .$ Ratio $= \frac{21000}{7000} = 3$ . Airbag stopping time ko teen guna stretch karta hai, peak force ko ek tehai kar deta hai — same $Δ p$ , gentle landing.

Connections

Newton's Second Law — woh source law jise upar har solution quietly use karta hai.
Force–Time Graphs — L3 aur L4 area se seedha impulse padhte hain.
Conservation of Momentum — recoil (L4·Q2) aur jet (L5) problems impulse ke zariye conservation hain.
Collisions — Elastic and Inelastic — bounce-vs-stick (L3·Q1) collision analysis ka seed hai.
Work–Energy Theorem — L5·Q3 ka flight phase distance analogue par switch karta hai.
Center of Mass Motion — external impulse poore system ka momentum shift karta hai.

Recall Self-test checklist

Kya har L1–L2 answer mein sahi sign tha? ::: Haan — pehle positive direction choose ki, phir signs ke saath $v_{f} - v_{i}$ . Tumne $F Δ t$ ki jagah integrate kab kiya? ::: Jab bhi force time ke saath vary karti thi (L3·Q2, L3·Q3) — impulse curve ke neeche ka area hai. Bounce mein impulse double kyun ho jaata hai? ::: Velocity reverse hoti hai, isliye $Δ v = v_{f} - v_{i}$ $- v$ ki jagah $- 2 v$ ban jaata hai. Stream force kaise ban jaati hai? ::: $F = (Δ m /Δ t) Δ v$ — momentum change per second.