Class 11 Physics · Chapter 4
Short answer:
Class 11 Physics Chapter 4 — Laws of Motion Newton ke teen laws, linear momentum conservation, common forces (normal, tension, friction) aur circular motion (banking of roads) cover karta hai. Ye chapter Class 11 Physics NCERT solutions ki series me mechanics ka core hai — Chapter 2 (Motion in a Straight Line) aur Chapter 3 (Motion in a Plane) ke kinematics concepts ab yahan force aur cause of motion se jud jaate hain. Neeche har in-text example aur end-of-chapter exercise ka step-by-step, CBSE marking-scheme style solution hai.
Chapter 4 — Laws of Motion — poore Class 11 Physics mechanics ki neev hai. Rationalised syllabus (2026-27 session) me "Physical World" chapter fully drop ho chuka hai, aur numbering shift hone ke baad Laws of Motion ab officially Chapter 4 hai (pehle old edition me Chapter 5 tha) — Units and Measurements se shuru hoke, Motion in a Straight Line aur Motion in a Plane ke turant baad. Agar tumne pehle se class 11 physics chapter 2 motion in a straight line ncert solutions ya motion in a plane projectile motion class 11 solutions padhi hai, to ye chapter unhi kinematics equations ko "force kaise motion cause karta hai" wale sawaal se jodta hai.
Chapter Aristotle ki galat dharna se shuru hota hai ("force na lage to body ruk jaati hai" — jo galat hai), phir Galileo/Newton ka inertia ka idea aata hai. Uske baad teen laws — first law (inertia), second law (F = ma / F = dp/dt), third law (action-reaction) — aur unka sabse important application: conservation of linear momentum. Aage common forces (normal reaction, tension, static aur kinetic friction) aur circular motion (banked roads, maximum safe speed) cover hote hain.
Board exam ke liye ye chapter numerical-heavy hai — class 11 physics chapter wise weightage cbse check karne pe pata chalega ki mechanics (Chapters 2-7 combined) ka hissa sabse bada hota hai, isliye Laws of Motion ke numericals (F=ma, momentum conservation, pulley/Atwood machine, circular motion) practice karna zaroori hai. Extra practice ke liye laws of motion class 11 ncert exemplar ke HOTS questions bhi try karo. Ye chapter samajhne ke baad agla step hoga Chapter 5 (Work, Energy and Power) aur Chapter 6 — jahan system of particles and rotational motion important formulas me tum dekhoge ki torque aur angular momentum, linear momentum ke concepts ka hi extension hain. Agar tumhe saare 14 chapters ek jagah chahiye to class 11 physics all chapters pdf 2026-27 search karke rationalised syllabus wali NCERT PDF hi use karo (purani 15/16-chapter wali PDF me syllabus mismatch milega — wo class 11 physics deleted syllabus 2026-27 ka hissa hai).
Chapter 4 Summary — 5 Minute Revision
Is chapter me humne dekha: (1) Aristotle ki fallacy aur Galileo/Newton ka inertia ka concept, jisse Newton ka first law of motion nikalta hai — koi body rest ya uniform velocity me tab tak rehti hai jab tak external unbalanced force na lage. (2) Second law: net force = rate of change of momentum, F = dp/dt, jo constant mass ke liye F = ma ban jaata hai — impulse (J = FΔt = Δp) is law ka hi ek application hai. (3) Third law: har action ka ek equal aur opposite reaction hota hai, jo hamesha do alag bodies pe act karta hai. (4) In teeno se conservation of linear momentum nikalti hai — isolated system me total momentum constant rehta hai (recoil of gun, collisions, rocket propulsion isi principle pe based hain). (5) Equilibrium of a particle — jab net force zero ho. (6) Common forces — normal reaction (N), tension (T), static friction (f_s ≤ μ_s N) aur kinetic friction (f_k = μ_k N). (7) Circular motion — centripetal force koi naya force nahi, balki existing force (tension/friction/normal) hi is role me kaam karti hai; banked roads is friction ki dependency kam karte hain. Ye saara framework aage Work-Energy-Power aur Rotational Motion me directly reuse hota hai.
In-Text Questions — Solutions
Ek 2 kg ke rest wale body par 10 N ka force lagaya jaata hai. 5 s baad uski velocity aur displacement nikaalo.
Diya hai: m = 2 kg, F = 10 N, u = 0, t = 5 s.
a = F/m = 10/2 = 5 m/s²
Velocity: v = u + at = 0 + 5×5 = 25 m/s
Displacement: s = ut + ½at² = 0 + ½×5×25 = 62.5 m
0.02 kg ki bullet 200 m/s speed se lakdi ke block me ghusti hai aur 10 cm (0.1 m) penetrate karke ruk jaati hai. Average retarding force nikaalo.
v² = u² + 2as, final v = 0, u = 200 m/s, s = 0.1 m
0 = (200)² + 2a(0.1) ⟹ a = −40000/0.2 = −2×10⁵ m/s²
F = ma = 0.02 × 2×10⁵ = 4000 N (retarding, motion ki opposite direction me)
4 kg aur 6 kg ke do blocks smooth surface par string se jude hain. 6 kg block par 20 N ka force lagta hai (string ki taraf). System ka acceleration aur string me tension nikaalo.
Total mass M = 4 + 6 = 10 kg
a = F/M = 20/10 = 2 m/s²
4 kg block sirf tension se accelerate hota hai:
T = m×a = 4 × 2 = 8 N, a = 2 m/s²
500 kg (load samet) wali ek lift 2 m/s² se upar accelerate karti hai. Cable me tension nikaalo. (g = 9.8 m/s²)
Lift upward accelerate kar rahi hai, isliye T > mg
T − mg = ma ⟹ T = m(g + a)
T = 500×(9.8 + 2) = 500×11.8 = 5900 N
5 kg ki gun se 0.01 kg ki bullet 300 m/s se fire hoti hai. Gun ka recoil velocity nikaalo.
Initial total momentum = 0 (system rest me tha). Momentum conservation se:
m_bullet × v_bullet = m_gun × v_recoil
0.01 × 300 = 5 × v ⟹ v = 3/5 = 0.6 m/s (bullet ki opposite direction me)
0.15 kg ka ball 12 m/s se aata hai aur bat isse ulti direction me 20 m/s se wapas bhej deta hai. Contact time 0.01 s hai. Bat dwara ball par lagaya average force nikaalo.
Initial direction ko positive lo: v_i = +12 m/s, v_f = −20 m/s (direction reverse)
Δp = m(v_f − v_i) = 0.15×(−20−12) = 0.15×(−32) = −4.8 kg m/s
F = Δp/Δt = −4.8/0.01 = −480 N (magnitude 480 N, bat ki swing direction me)
1200 kg ki car 30 m radius wale level (unbanked) turn par jaa rahi hai. Tyre-road ke beech μ_s = 0.5 hai. Skid kiye bina maximum speed kya hogi?
Level road par centripetal force sirf friction se aati hai, jiski upper limit μ_s N hai:
v_max = √(μ_s r g) = √(0.5 × 30 × 9.8) = √147
v_max ≈ 12.1 m/s ≈ 43.6 km/h
10 kg ka block horizontal surface par rakha hai, μ_s = 0.4. Block ko move karane ke liye minimum horizontal force kitna chahiye?
Block tabhi move karega jab applied force, static friction ki maximum value ko cross kare:
f_s(max) = μ_s N = μ_s mg = 0.4 × 10 × 9.8
F_min = 39.2 N
Ek particle par teen forces lag rahi hain: 10 N (East), 10 N (North), aur F, aur particle equilibrium me hai. F ka magnitude aur direction nikaalo.
Pehle do forces ka resultant nikalo (perpendicular hain):
R = √(10² + 10²) = √200 ≈ 14.14 N, 45° North of East
Equilibrium ke liye teesra force is resultant ke exactly equal aur opposite hona chahiye:
F = 14.14 N, direction 45° South of West
5000 kg ka rocket 20 kg/s ki rate se gas eject karta hai, relative exhaust velocity 3000 m/s hai. Thrust force aur initial upward acceleration nikaalo. (g = 9.8 m/s²)
Thrust:
F_thrust = v_rel × (dm/dt) = 3000 × 20 = 60000 N
Weight ko subtract karke net force milta hai:
F_net = 60000 − (5000×9.8) = 60000 − 49000 = 11000 N
a = F_net/m = 11000/5000 = 2.2 m/s²
Poore Class 11 Physics ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q28)
Net force ka magnitude aur direction batao: (a) constant speed se gir rahi rain drop, (b) 10 g ka floating cork paani par, (c) sky me stationary kite, (d) 30 km/h constant velocity se rough road par chalti car, (e) force-free space me high-speed electron.
Sabhi cases me body ya to rest me hai ya constant velocity (uniform motion) me — Newton ke first law ke hisaab se dono situations me net force zero hota hai:
(a) Constant speed = zero acceleration ⟹ net force = 0 (gravity aur air drag balance ho jaate hain)
(b) Floating cork rest me hai, equilibrium ⟹ net force = 0
(c) Kite stationary hai (rest) ⟹ net force = 0
(d) Constant velocity ⟹ zero acceleration ⟹ net force = 0
(e) Koi field/force present nahi ⟹ net force = 0
0.05 kg ka pebble vertically upar phenka jaata hai. Upward motion, downward motion, aur highest point par net force ka magnitude/direction batao. Agar 45° angle par pheka jaaye to jawab change hoga kya? (Air resistance ignore karo)
Air resistance ignore karne par, projectile par sirf gravity kaam karti hai — chahe wo seedha upar ho ya angle par.
F = mg = 0.05 × 9.8 = 0.49 N, hamesha vertically downward
Upward motion, downward motion, aur highest point — teeno me force same rehta hai: 0.49 N downward (highest point par bhi, kyunki velocity zero hone se force zero nahi hota — force sirf gravity ki wajah se hai).
45° par phenkne se bhi jawab same rahega, kyunki horizontal direction me koi force nahi hai — sirf vertical gravity acting rehti hai.
0.1 kg ke stone par net force batao: (a) stationary train ki window se drop karte hi, (b) 36 km/h constant velocity wali train se drop karte hi, (c) 1 m/s² se accelerate karti train se drop karte hi, (d) train ke floor par pada stone jab train 1 m/s² se accelerate ho raha ho (stone train ke relative rest me hai).
Jab tak stone train ke saath contact me hai NAHI (free hai), sirf gravity kaam karti hai:
(a), (b), (c) — teeno cases me drop hote hi stone free-fall me hota hai (train ka motion irrelevant hai kyunki stone ab train se force nahi le raha), isliye net force = mg = 0.1×9.8 = 0.98 N, vertically downward — teeno case same.
(d) Yahan stone train ke floor par hai aur train ke saath hi accelerate ho raha hai (relative rest), isliye stone ko bhi 1 m/s² horizontal acceleration chahiye:
F_horizontal = ma = 0.1 × 1 = 0.1 N (train ke acceleration ki direction me, friction se aata hai)
MCQ: Smooth horizontal table par ek peg se juda hua string, jiske doosre end par mass m hai, jo circle me v speed se ghoom raha hai. Particle par net force (centre ki taraf) kya hai — (i) T (ii) T − mv²/l (iii) T + mv²/l (iv) 0 ?
Circular motion me net force hi centripetal force hoti hai, jo yahan sirf string ki tension se milti hai (koi doosra horizontal force nahi):
Answer: (i) T — tension khud hi net centripetal force hai, kyunki F_net = mv²/l = T (dono same cheez hain, T alag se subtract/add nahi hota).
20 kg ke body ko 15 m/s se moving state me 50 N ka constant retarding force lagta hai. Body ko rukne me kitna time lagega?
a = F/m = 50/20 = 2.5 m/s² (deceleration)
v = u − at ⟹ 0 = 15 − 2.5t
t = 6 s
3.0 kg ke body ki speed ek constant force se 25 s me 2.0 m/s se 3.5 m/s ho jaati hai (direction same rehti hai). Force ka magnitude aur direction batao.
a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m/s²
F = ma = 3.0 × 0.06 = 0.18 N
Direction: motion ki same direction me (kyunki speed badh rahi hai).
5 kg ke body par perpendicular directions me 8 N aur 6 N ke do forces lag rahe hain. Body ke acceleration ka magnitude aur direction batao.
Perpendicular forces ka resultant Pythagoras se:
F_net = √(8² + 6²) = √(64+36) = √100 = 10 N
a = F/m = 10/5 = 2 m/s²
Direction: θ = tan⁻¹(6/8) = 36.9° — 8 N force se (6 N ki taraf).
36 km/h se chal rahe three-wheeler ka driver bachche ko bachaane ke liye 4.0 s me vehicle rok deta hai. Three-wheeler ka mass 400 kg aur driver ka 65 kg hai. Average retarding force nikaalo.
u = 36 km/h = 10 m/s, v = 0, t = 4 s, total mass M = 400+65 = 465 kg
a = (v−u)/t = (0−10)/4 = −2.5 m/s²
F = Ma = 465 × 2.5 = 1162.5 N (retarding, motion ki opposite direction)
20,000 kg ke lift-off mass wale rocket ko 5.0 m/s² ke initial acceleration se blast kiya jaata hai. Blast ka initial thrust force nikaalo. (g = 9.8 m/s²)
Thrust ko upward acceleration DENA hai aur weight ko bhi OVERCOME karna hai:
Thrust − mg = ma ⟹ Thrust = m(g + a)
Thrust = 20000 × (9.8 + 5) = 20000 × 14.8 = 2.96 × 10⁵ N
0.40 kg ka body 10 m/s se North ki taraf move kar raha hai. t=0 pe (jaha x=0 hai) usko 8.0 N ka South-directed force 30 s ke liye lagta hai. t = −5 s, 25 s aur 100 s pe body ka position predict karo.
North ko positive x lete hain. Force South me hai, isliye 0 ≤ t ≤ 30 s ke liye:
a = F/m = 8/0.4 = 20 m/s² (South direction, i.e. −20 m/s²)
t = −5 s (force lagne se pehle, constant velocity 10 m/s North):
x = ut = 10 × (−5) = −50 m
t = 25 s (still 0-30s window, x = ut + ½at²):
x = 10(25) + ½(−20)(25)² = 250 − 6250 = −6000 m = −6 km
t = 100 s — pehle t=30s tak calculate karo (force khatam hone tak):
v(30) = 10 − 20(30) = −590 m/s; x(30) = 10(30) − 10(30)² = 300 − 9000 = −8700 m
Ab t=30s ke baad force nahi hai, constant velocity −590 m/s se 70 s aur chalta hai:
x(100) = −8700 + (−590×70) = −8700 − 41300 = −50000 m = −50 km
Ek truck rest se 2.0 m/s² se uniformly accelerate hota hai. t=10 s pe, truck ke top (6 m height) se ek stone drop kar diya jaata hai. t=11 s pe stone ki (a) velocity aur (b) acceleration kya hogi? (Air resistance neglect karo)
t=10 s pe truck ki horizontal velocity: v = at = 2×10 = 20 m/s — drop hote hi stone isi horizontal velocity ko carry karta hai (aur air resistance na hone se horizontal velocity CONSTANT rehti hai, 20 m/s).
Vertically, stone free fall me hai (0 se start, gravity ke under), 1 s baad (t=11s):
v_vertical = g×1 = 9.8 m/s (downward)
(a) Resultant velocity = √(20² + 9.8²) = √(400+96.04) ≈ 22.3 m/s, horizontal se 26.1° neeche.
(b) Acceleration = sirf gravity kaam karti hai ab (stone ab truck se attached nahi) = g = 9.8 m/s², vertically downward (2 m/s² wala truck acceleration ab apply nahi hota).
0.1 kg ka pendulum bob 2 m lambi string se latka hai, mean position par uski speed 1 m/s hai. Agar string ko (a) extreme position par, (b) mean position par cut kar diya jaaye, to bob ka trajectory kya hoga?
(a) Extreme position par cut: Extreme position par bob momentarily rest me hota hai (v=0), isliye cut hote hi sirf gravity kaam karti hai — bob vertically straight neeche girega (free fall, straight-line motion).
(b) Mean position par cut: Mean position par bob ki velocity 1 m/s horizontal (tangential) hai, isliye cut hote hi bob me horizontal velocity + vertical free-fall dono milte hain — bob parabolic path (projectile motion) follow karega.
70 kg ka aadmi weighing scale par ek lift me khada hai jo (a) 10 m/s uniform speed se upar, (b) 5 m/s² uniform acceleration se neeche, (c) 5 m/s² uniform acceleration se upar move kar raha hai. Har case me scale reading kya hogi? (d) Agar lift ka mechanism fail ho jaaye aur wo freely fall kare to reading kya hogi? (g = 10 m/s² use karo)
Scale reading = Normal force N jo scale aadmi par lagata hai.
(a) Uniform speed (a=0): N = mg = 70×10 = 700 N
(b) Downward acceleration 5 m/s²: mg − N = ma ⟹ N = m(g−a) = 70×(10−5) = 350 N
(c) Upward acceleration 5 m/s²: N − mg = ma ⟹ N = m(g+a) = 70×(10+5) = 1050 N
(d) Free fall (a=g): N = m(g−g) = 0 N (weightlessness)
4 kg ke particle ka position-time graph diya hai: t<0 ke liye particle x=0 par rest me hai; t=0 pe achanak constant velocity mil jaati hai jisse wo 4 s me x=3 m tak pahunchta hai; t>4 s ke liye wapas rest (constant x=3m). (a) Force batao t<0, 04s ke liye. (b) t=0 aur t=4 s pe impulse nikaalo.
(a) Teeno interval me velocity constant hai (rest ya uniform motion), matlab acceleration zero — isliye teeno interval me force = 0.
Poora velocity change sirf t=0 aur t=4s ke instants par achanak (impulsively) hota hai — ye finite force nahi, balki impulse hai.
(b) 0 se 4s me 3 m cover hota hai: v = 3/4 = 0.75 m/s
Impulse at t=0: J = mΔv = 4×(0.75−0) = 3 kg m/s
Impulse at t=4s: J = mΔv = 4×(0−0.75) = −3 kg m/s (magnitude 3 kg m/s, opposite direction)
10 kg (A) aur 20 kg (B) ke do bodies smooth horizontal surface par light string se jude hain. F=600 N ka horizontal force string ki direction me (i) A par, (ii) B par lagaya jaata hai. Dono cases me string me tension nikaalo.
Total mass M = 30 kg, system acceleration dono case me same:
a = F/M = 600/30 = 20 m/s²
(i) F, A par lagaya: B ko sirf tension khinchti hai:
T = m_B × a = 20 × 20 = 400 N
(ii) F, B par lagaya: A ko sirf tension khinchti hai:
T = m_A × a = 10 × 20 = 200 N
8 kg aur 12 kg ke do masses ek light, inextensible string se frictionless pulley ke upar se jude hain (Atwood machine). Release karne par masses ka acceleration aur string ki tension nikaalo.
a = (m₂−m₁)g/(m₁+m₂) = (12−8)×9.8/20 = 39.2/20 = 1.96 m/s²
Tension (8 kg wale side se check):
T = m₁(g+a) = 8×(9.8+1.96) = 8×11.76 = 94.08 N
Verify (12 kg side): T = m₂(g−a) = 12×(9.8−1.96) = 12×7.84 = 94.08 N ✓
Ek nucleus laboratory frame me rest me hai. Agar wo do chhote nuclei me disintegrate ho jaaye, to dikhao ki products opposite directions me hi move karenge.
Nucleus rest me tha, isliye initial total momentum = 0.
Momentum conservation (isolated system, koi external force nahi):
p₁ + p₂ = 0 ⟹ p₁ = −p₂
Iska matlab dono products ka momentum magnitude me equal hai par direction bilkul opposite — isliye wo hamesha ek doosre ke exactly opposite directions me hi move karenge (collinear, taaki total momentum zero rahe).
0.05 kg ke do billiard balls opposite directions me 6 m/s speed se move karte hue collide karte hain aur same speed se rebound karte hain. Har ball par doosre ki wajah se lagne wale impulse ka magnitude nikaalo.
Ball A ke liye initial direction ko positive lo: v_i = +6, collision ke baad direction reverse: v_f = −6
J_A = m(v_f − v_i) = 0.05×(−6−6) = 0.05×(−12) = −0.6 kg m/s
Magnitude 0.6 N s, har ball par apni initial motion ki opposite direction me (Newton ke third law se dono balls par equal-magnitude impulse lagta hai, sign automatic opposite hai).
100 kg ki gun se 0.020 kg ka shell 80 m/s muzzle speed se fire hota hai. Gun ka recoil speed nikaalo.
Initial total momentum = 0. Momentum conservation:
m_shell × v_shell = m_gun × v_recoil
v_recoil = (0.020×80)/100 = 1.6/100 = 0.016 m/s (1.6 cm/s), shell ki opposite direction me
Ek batsman 54 km/h se aa rahi ball ko 45° angle se deflect kar deta hai bina speed change kiye. Ball ka mass 0.15 kg hai. Ball par lagaya gaya impulse nikaalo.
v = 54 km/h = 15 m/s. Standard geometry me incoming aur outgoing directions ke beech net angle change = 90° (45° har taraf, symmetric bisector ke around).
|Δp| = 2mv sin(θ/2) = 2×0.15×15×sin45°
= 4.5 × 0.707 ≈ 3.18 kg m/s
Direction: symmetry bisector ke along (bat ke perpendicular).
0.25 kg ka stone 1.5 m radius ke horizontal circle me 40 rev/min speed se ghumaya jaata hai. String me tension nikaalo. Agar string sirf 200 N tension jhel sakti hai to maximum safe speed kya hogi?
ω = 2πN/60 = 2π×40/60 = 4π/3 ≈ 4.19 rad/s
T = mω²r = 0.25×(4.19)²×1.5 ≈ 6.58 N
Max speed jab T = 200 N:
v_max = √(T_max × r/m) = √(200×1.5/0.25) = √1200 ≈ 34.6 m/s
MCQ: Pichhle question (5.21) me agar stone ki speed permissible maximum se badha di jaaye aur string achanak toot jaaye, to string tootne ke baad stone ka trajectory kya hoga — (a) radially outward move karega (b) uss instant se tangentially fly off karega (c) tangent se ek angle par fly off karega?
Answer: (b) — String toot te hi centripetal force khatam ho jaati hai. Us instant particle ki velocity direction hamesha circle ke tangent ke along hoti hai (radius ke perpendicular), aur Newton ke first law (inertia) se — koi force na hone par — particle usi tangential direction me straight line me move karta rahega.
Explain karo: (a) khaali space me ghoda cart nahi khinch sakta, (b) achanak bus rukne par passengers aage ki taraf gir jaate hain, (c) lawn mower ko push karne se pull karna aasan hai, (d) catch pakadte waqt cricketer apne hath peeche khinchta hai.
(a) Ghode ko aage badhne ke liye zameen par peeche ki taraf push karna padta hai, jiski reaction (Newton's third law) usse aage dhakelti hai. Khaali space me push karne ke liye koi ground/reaction surface nahi hai, isliye system par koi net external force nahi banta.
(b) Passenger ka body inertia (first law) ki wajah se bus ki original speed se aage move karte rehna chahta hai jab bus achanak decelerate hoti hai — isliye relative to bus, passenger aage ki taraf lurch karta hai.
(c) Push karne se applied force ka ek component neeche ki taraf jud jaata hai, jisse normal force N aur friction (f=μN) badh jaata hai. Pull karne se component upar hota hai, N kam hoti hai, friction kam — isliye pull karna aasan hai.
(d) Hath peeche khinchne se ball ka momentum zero hone me lagne wala time Δt badh jaata hai. Impulse-momentum theorem (F = Δp/Δt) ke hisaab se, same Δp ke liye zyada Δt matlab kam average force — isliye chot kam lagti hai.
1000 kg ka helicopter 15 m/s² ke vertical acceleration se upar uthta hai. Crew aur passengers ka combined mass 300 kg hai. (a) Crew dwara floor par lagaya force, (b) rotor ka surrounding air par action, (c) air dwara helicopter par force — inka magnitude/direction nikaalo. (g = 9.8 m/s²)
(a) Crew (300 kg) khud bhi 15 m/s² se accelerate ho raha hai — floor par lagne wala force = reaction of normal force N:
N = m(g+a) = 300×(9.8+15) = 300×24.8 = 7440 N, floor par neeche ki taraf
(b) aur (c) Poora system (helicopter+crew, total 1300 kg) ko upward support chahiye:
F = M(g+a) = 1300×24.8 = 32240 N
Ye force air dwara helicopter par upward (part c ka answer) lagta hai; Newton ke third law se, rotor dwara air par exactly isi magnitude ka force downward lagta hai (part b ka answer) = 32240 N.
15 m/s speed se horizontally flow karta paani ka jet, 10⁻² m² cross-section wali tube se nikal kar ek deewar se takrata hai aur rebound nahi karta. Deewar par lagne wala force nikaalo. (density of water = 1000 kg/m³)
Mass flow rate:
dm/dt = ρ × A × v = 1000 × 0.01 × 15 = 150 kg/s
Force = rate of change of momentum (final velocity = 0, no rebound):
F = (dm/dt) × Δv = 150 × 15 = 2250 N
10 ek-rupee ke coins table par ek dusre ke upar rakhe hain, har coin ka mass m hai. (a) 7th coin (neeche se ginte hue) par upar ke saare coins ka force, (b) 8th coin dwara 7th par force, (c) 6th coin ka 7th par reaction — inhe batao aur direction bhi do.
7th coin ke upar 3 coins hain (8th, 9th, 10th).
(a) In teen coins ka total weight neeche ki taraf act karta hai:
F = 3mg, downward
(b) 8th coin apna aur upar ke sabka weight 7th tak transmit karta hai (8th+9th+10th = 3 coins):
F = 3mg, downward (same interface, same value)
(c) 6th coin, 7th coin ko support karta hai — 7th aur uske upar ke sab (7th,8th,9th,10th = 4 coins) ka weight upward reaction se balance hota hai:
F = 4mg, upward
40 kg ka monkey ek rope par chadhta hai jo maximum 600 N tension jhel sakti hai. Rope kab tootegi — monkey (a) accelerating upward climb kare, (b) accelerating downward climb kare, (c) uniform speed se chadhe/utre, (d) freely rope ke saath gire? (g = 9.8 m/s²)
Monkey ka weight mg = 40×9.8 = 392 N.
(a) Upward acceleration a: T = m(g+a). Rope tootegi nahi jab tak T ≤ 600 N:
40(9.8+a) ≤ 600 ⟹ a ≤ 5.2 m/s²
Isliye a > 5.2 m/s² upward hi rope todega — yahi ek risky case hai.
(b) Downward acceleration: T = m(g−a), jo hamesha 392 N se kam hoga (safe, kabhi nahi tootegi).
(c) Uniform speed (a=0): T = mg = 392 N < 600 N — safe.
(d) Free fall: T = 0 — safe (rope slack ho jaati hai).
100 kg ka sand bag ek wire se latka hai. Isse 0.05 kg ki bullet 150 m/s speed se takrati hai aur andar hi embed ho jaati hai. Bullet ki KE ka kitna loss hua?
Perfectly inelastic collision — momentum conservation:
0.05×150 = (100+0.05)×v ⟹ v = 7.5/100.05 ≈ 0.075 m/s
KE_initial = ½×0.05×150² = 562.5 J
KE_final = ½×100.05×(0.075)² ≈ 0.28 J
Loss of KE ≈ 562.5 − 0.28 ≈ 562.2 J (~99.95% energy heat/deformation me convert ho gayi)
Important Equations — Ek Nazar Me
| Concept | Formula / Statement |
|---|---|
| Newton's First Law (Inertia) | Body rest ya uniform velocity me rehta hai jab tak koi external unbalanced force na lage |
| Linear momentum | p = mv |
| Newton's Second Law | F = dp/dt = ma (constant mass ke liye) |
| Impulse | J = F·Δt = Δp |
| Newton's Third Law | FAB = −FBA (do alag bodies pe, kabhi cancel nahi hote) |
| Conservation of linear momentum | Isolated system me total momentum before = total momentum after |
| Weight | W = mg |
| Apparent weight in lift (a upward +ve) | N = m(g ± a) |
| Static friction (maximum) | fs(max) = μs N |
| Kinetic friction | fk = μk N |
| Centripetal force | F = mv²/r = mω²r |
| Max speed on level circular road | vmax = √(μs r g) |
| Banking of road (frictionless) | tanθ = v²/(rg) |
| Banking + friction (max safe speed) | vmax = √[ rg(μs+tanθ)/(1−μstanθ) ] |
| Thrust on rocket | Fthrust = vrel(dm/dt) |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Newton's second law ko sirf F = ma likh dena bina samjhe ki ye simplified constant-mass version hai — general form F = dp/dt hai (variable-mass systems jaise rocket me galti hoti hai).
- Action-reaction pair (third law) ko ek hi body pe lagne wale do forces samajh lena — jabki ye do ALAG bodies pe act karte hain, isliye kabhi cancel nahi hote.
- Static aur kinetic friction ko same maan lena — f_s ek self-adjusting force hai (0 se μ_s N tak) jab tak body move na kare, aur f_s(max) generally f_k se zyada hota hai.
- Circular motion me centripetal force ko ek 'alag naya force' samajh lena — jabki ye sirf net force ka naam hai jo tension/friction/normal reaction jaisi real forces provide karti hain.
- Lift/elevator apparent-weight problems me acceleration ka sign galat lagana — upward ya downward decide kiye bina N = m(g+a) ki jagah m(g−a) laga dena (ya ulta).
- Momentum conservation lagate waqt vector nature bhool jana — sirf magnitudes add/subtract kar dena bina direction (sign convention) ke, jisse opposite-direction collision/recoil problems me jawab ulta aa jaata hai.
Board-Style Important Questions
- 1 mark:
MCQ: Ek book table par rakhi hai. Book ka weight aur table ka normal reaction ek Newton's third law pair hain — sahi ya galat?
Answer: Galat. Ye ek equilibrium (balanced force) pair hai, third-law pair nahi — third-law pair hamesha do ALAG bodies pe act karta hai (yahan: book-par-table force aur table-par-book force asli pair hai; weight aur normal reaction dono book ke frame me alag origin ke forces hain).
- 2 marks:
Conservation of linear momentum ka principle state karo aur ek example se explain karo.
Answer: Kisi isolated system (jispe koi external force na lage) ka total linear momentum constant rehta hai. Example — bandook se goli chalane par: fire se pehle total momentum zero tha; fire ke baad, goli ka forward momentum aur bandook ka recoil momentum equal-opposite hote hain taaki total momentum phir se zero rahe (m_bullet v_bullet = m_gun v_recoil).
- 3 marks:
25 kg ka body 20 m/s se move kar raha hai. Ek constant retarding force 100 N lagaya jaata hai. Body kitni distance tay karke rukega?
Answer:
a = F/m = 100/25 = 4 m/s² (deceleration)
v² = u² − 2as ⟹ 0 = 400 − 2(4)s
s = 50 m
- 3 marks:
Banked road par friction ke saath maximum safe speed ka expression derive karo.
Answer: Banking angle θ, friction μ_s wale road par circular motion ke liye radial aur vertical directions me force-balance likhne se (N cosθ − f sinθ = mg, aur N sinθ + f cosθ = mv²/r, jahan f = μ_s N maximum case me):
v_max = √[ rg(μ_s + tanθ)/(1 − μ_s tanθ) ]
Ye friction wale aur pure banking (μ_s=0 ⟹ tanθ=v²/rg) dono cases ko generalize karta hai.
- 5 marks:
6 kg aur 4 kg ke do masses ek light string se frictionless pulley ke upar se jude hain. Release karne par system ka acceleration aur string ki tension nikaalo. (g = 9.8 m/s²)
Answer:
a = (m₁−m₂)g/(m₁+m₂) = (6−4)×9.8/10 = 19.6/10 = 1.96 m/s²
T = m₂(g+a) = 4×(9.8+1.96) = 4×11.76 = 47.04 N
Verify: T = m₁(g−a) = 6×(9.8−1.96) = 6×7.84 = 47.04 N ✓
a = 1.96 m/s², T = 47.04 N
Aksar Poochhe Jaane Wale Sawaal
Class 11 Physics Chapter 4 Laws of Motion me kitne exercises hain?
NCERT ke end-of-chapter exercises 23 core numericals/conceptual questions cover karte hain (5.1-5.23 style), plus additional exercises jisme helicopter, water-jet, coins stack, monkey-on-rope, aur bullet-embedding-in-bag jaise HOTS problems hain. In-text me ~10 solved Examples hain jo concept-by-concept step-by-step samjhate hain — yahi ise Chapter 2 aur 3 se zyada numerical-heavy banata hai.
Board exam me Laws of Motion ka weightage kitna hai?
Exact percentage weightage har saal CBSE ke official marking scheme document me change ho sakta hai, isliye kisi fix number ka dawa karna sahi nahi — apni current session ki official CBSE sample paper/marking scheme dekho. Broadly, Laws of Motion mechanics unit (jo class 11 physics chapter wise weightage cbse me sabse bada chunk hota hai) ka core part hai, aur numericals (F=ma, momentum conservation, Atwood machine, circular motion) har saal aate hain.
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Rationalised syllabus (2026-27 session) ke hisaab se 14 chapters hote hain — Units and Measurements se Waves tak. Agar tum class 11 physics all chapters pdf 2026-27 search karoge to sirf current session ki official NCERT PDF use karo, kyunki purani (Physical World wali) 15/16-chapter edition ab class 11 physics deleted syllabus 2026-27 ka hissa ban chuki hai aur chapter numbering match nahi karegi.
Laws of Motion (Chapter 4) aur Motion in a Straight Line (Chapter 2) me kya farak hai?
Chapter 2 (class 11 physics chapter 2 motion in a straight line ncert solutions wala chapter) sirf motion KO describe karta hai — displacement, velocity, acceleration — bina ye pooche ki motion CAUSE kyun hota hai. Chapter 4 (Laws of Motion) us gap ko bharta hai — force, mass aur momentum ke through batata hai ki motion kyun hota hai ya kyun change hota hai. Ek 'description', doosra 'cause and effect'.
NCERT Exemplar se Laws of Motion me exam ke liye kya extra milega?
Laws of motion class 11 ncert exemplar me typically MCQs (multiple-correct type), assertion-reason, aur thoda tougher multi-concept numericals (jaise combined friction+circular motion, ya multi-body pulley systems) milte hain jo textbook exercises se ek level upar hote hain — competitive exam prep (JEE/NEET foundation) ke liye achha hai, lekin board exam ke liye NCERT textbook hi primary source rakho.
Laws of Motion ke concepts aage Chapter 6 (System of Particles and Rotational Motion) me kaise use hote hain?
System of particles and rotational motion important formulas dekhoge to pata chalega ki torque = r × F (Newton's second law ka rotational analogue) aur angular momentum conservation, linear momentum conservation ka hi extension hai. Isliye Laws of Motion ke force/momentum concepts pehle solid honi chahiye, tabhi rotational motion aasan lagega.
Class 11 Physics — Saare Chapters
Board exam tak sirf revision karna hai?
Class 11 Physics ke saare chapters ke colour handwritten short notes — diagrams, formulas aur important points ek jagah.
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