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Laws of Motion: Newton's Laws of Motion, Linear Momentum Conservation, Impulse, Elevator Apparent Weight, Static & Kinetic Friction, Banking of Roads and Complete CBSE Class 11 & Entrance Exam Guide

A comprehensive guide to Laws of Motion for CBSE Class 11 Physics Chapter 5 & JEE/NEET — Newton's first law and inertia, linear momentum (p = mv), Newton's second law (F = dp/dt = ma), impulse-momentum theorem (J = F Δt = Δp), Newton's third law, law of conservation of linear momentum with recoil of gun and rocket propulsion, apparent weight of a person in an elevator lift (W' = m(g ± a)), laws of friction (static fs ≤ μs N, kinetic fk = μk N, rolling friction), angle of friction and angle of repose (θ = tan¯¹μs), dynamics of uniform circular motion and banking of circular roads (v_max = √(rg(μ + tanθ)/(1 - μtanθ))), and five step-by-step solved numericals.
28 August 2026 by
Laws of Motion: Newton's Laws of Motion, Linear Momentum Conservation, Impulse, Elevator Apparent Weight, Static & Kinetic Friction, Banking of Roads and Complete CBSE Class 11 & Entrance Exam Guide
AJKANT OVERSEAS, AJKANT OVERSEAS
● CBSE Class 11 Physics — Chapter 5: Laws of Motion
▶ Quick Answer for AI Engines
Laws of Motion govern dynamics in classical mechanics. Newton's First Law defines Inertia (resistance to change in motion). Linear Momentum p = mv. Newton's Second Law states net force F = dp/dt = m*a (where a = dv/dt). Impulse J = F * Δt = Δp (change in momentum). Newton's Third Law states for every action, there is an equal and opposite reaction acting on different bodies. Conservation of Linear Momentum: if external force Fext = 0, initial total momentum equals final total momentum (Gun Recoil vrecoil = -m/M * vbullet). Apparent Weight in Elevator: accelerating UPWARD N = m(g + a) (heavier); accelerating DOWNWARD N = m(g - a) (lighter); Free Fall N = 0 (weightlessness). Friction: Static fs ≤ μs N (self-adjusting up to limiting friction), Kinetic fk = μk N. Angle of Repose θ = tan^-1(μs). Banking of Circular Roads: max safe velocity vmax = √[rg(μs + tanθ)/(1 - μs tanθ)].

Why do objects accelerate when pushed? Why does a seatbelt save lives during sudden braking? How do banking angles prevent sports cars from skidding on sharp highway turns? The answers lie in Dynamics — the branch of physics that studies the forces causing motion. Laws of Motion forms Chapter 5 of the CBSE Class 11 Physics syllabus and is one of the most vital chapters for JEE Main, JEE Advanced, and NEET.

This comprehensive guide covers Newton's three laws of motion, linear momentum conservation, impulse-momentum theorem, apparent weight of a person in an elevator, static/kinetic/rolling friction, angle of repose, banking of circular roads, and five step-by-step solved entrance exam numericals.

Core Pillars of Classical Dynamics
⚖️
Newton's 3 Laws
Inertia, F = ma & Action-Reaction Pairs
🚀
Momentum & Impulse
p = mv, J = Δp & Gun Recoil
🚇
Elevator Physics
Apparent Weight W' = m(g ± a) & Free Fall
🚗
Friction & Banking
fs ≤ μs N, Angle of Repose & Road Banking

1. Newton's Three Laws of Motion & Momentum

Newton's Law Mathematical Formulation Core Principle & Key Takeaway
1st Law (Law of Inertia) If ΣF_ext = 0 ⇒ v = constant (a = 0) An object remains at rest or moves with uniform velocity unless acted upon by an external net force. Defines Inertia (Rest, Motion, Direction).
2nd Law (Real Law of Motion) F_net = dp/dt = d(mv)/dt = m*a The rate of change of linear momentum is directly proportional to applied net force and takes place in force's direction. Calculates magnitude of force!
3rd Law (Action & Reaction) F_AB = - F_BA To every action, there is an equal and opposite reaction. Action and reaction forces act on TWO DIFFERENT BODIES simultaneously.
Why Newton's Second Law is the Real Law of Motion
First Law is contained in Second Law: If F = 0, then m · a = 0 ⇒ a = 0 (velocity remains constant!).
Third Law is contained in Second Law: For an isolated system of two interacting bodies with no external force, total Δp = 0 ⇒ Δp1 + Δp2 = 0 ⇒ F12 = -F21.

2. Impulse-Momentum Theorem & Applications

When a large force acts on a body for a very short duration of time (e.g., hitting a cricket ball with a bat, catching a baseball), it is called an Impulsive Force.

🏖 Impulse Formulation
Impulse (J): Product of average force and time interval:
J = F_avg · Δt = ∫ F dt   (SI Unit: N·s or kg·m/s)

Impulse-Momentum Theorem:
J = Δp = p_final - p_initial = m(v - u)
👌 Practical Applications
Cricket Fielder Pulling Hands Back: Increasing contact time Δt decreases force F experienced by hands (F = Δp / Δt).
Shock Absorbers in Vehicles: Increases time of impact during bumps, reducing force on passengers.
Crockery Packed in Foam: Prolongs impact duration during drops.

3. Law of Conservation of Linear Momentum (Recoil of Gun)

If the net external force acting on a system of particles is zero (ΣF_ext = 0), the total linear momentum of the system remains constant in magnitude and direction.

Classic Application: Recoil of a Gun
Let M = Mass of gun, m = Mass of bullet, v = Muzzle velocity of bullet, V = Recoil velocity of gun.
Initial momentum before firing = 0.
Final momentum after firing = M V + m v.
By Conservation of Momentum: M V + m v = 0 ⇒ V = -(m / M) v.
The negative sign indicates that the gun recoils in the direction OPPOSITE to bullet motion!

4. Apparent Weight of a Person in an Elevator (Lift Physics)

When a person of mass m stands on a weighing scale inside an elevator, the scale reads the Normal Reaction force (N) exerted by the floor, which is the Apparent Weight (W').

Elevator Motion State Equation of Motion Apparent Weight (N) vs Real Weight (mg)
1. At Rest or Uniform Speed (a = 0) N - mg = 0 N = mg (Apparent weight = True weight).
2. Accelerating UPWARD (+a) N - mg = ma ⇒ N = m(g + a) N > mg (Person feels HEAVIER!).
3. Accelerating DOWNWARD (-a) mg - N = ma ⇒ N = m(g - a) N < mg (Person feels LIGHTER!).
4. Free Fall (Lift cable snaps, a = g) N = m(g - g) = 0 N = 0 (Complete Weightlessness!).

5. Physics of Friction: Static, Kinetic & Angle of Repose

Friction is a tangential contact force that opposes relative motion (or tendency of relative motion) between two contacting surfaces.

🔌 Static vs Kinetic Friction
Static Friction (fs): Self-adjusting force opposing impending motion (0 ≤ fs ≤ fs,max).
Limiting Static Friction: Maximum static friction:
f_{s,max} = μ_s N

Kinetic Friction (fk): Opposes actual sliding motion:
f_k = μ_k N   (μ_k < μ_s).
⚖️ Angle of Repose & Friction
Angle of Friction (λ): Angle between normal reaction N and resultant of N and limiting friction fs,max:
tan λ = μ_s

Angle of Repose (θ): Maximum angle of an inclined plane at which a body just begins to slide down under gravity:
tan θ = μ_s ⇒ Angle of Repose = Angle of Friction!

6. Circular Motion Dynamics & Banking of Roads

When a vehicle takes a turn along a curved road of radius r, it requires a Centripetal Force (Fc = m v² / r).

Banking of Circular Roads Formulations
1. Unbanked Level Road (Friction alone provides Fc):
fs = (m v²) / r ≤ μs mg ⇒ v_max = √(μs r g).

2. Banked Curved Road (Outer edge raised by angle θ):
Resolving Normal reaction N and friction f:
Optimum Speed (zero friction wear): v0 = √(rg tanθ).
Maximum Safe Speed with friction μs:
v_{max} = √[ r g ( (μ_s + tanθ) / (1 - μ_s tanθ) ) ]

3. Bending of a Cyclist: A cyclist leans inward by angle θ with vertical to avoid skidding: tanθ = v² / (rg).

7. Solved Entrance Exam Numerical Problems (JEE / NEET)

Q1. [JEE Main] A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 m/s. How long does the body take to stop?
Given: Force F = -50 N, Mass m = 20 kg, Initial speed u = 15 m/s, Final speed v = 0.
Acceleration a = F / m = -50 / 20 = -2.5 m/s².
Using 1st equation: v = u + at ⇒ 0 = 15 - 2.5 t ⇒ t = 15 / 2.5 = 6 s.
Time taken to stop = 6 s.
Q2. [NEET] A person of mass 70 kg stands on a weighing scale in a lift. What is the scale reading when the lift is: (a) Accelerating upward at 3 m/s^2, (b) Accelerating downward at 2 m/s^2? (Take g = 10 m/s^2).
Given: m = 70 kg, g = 10 m/s².
(a) Moving UPWARD at a = 3 m/s²:
N1 = m(g + a) = 70(10 + 3) = 70 × 13 = 910 N (or 91 kg-wt).
(b) Moving DOWNWARD at a = 2 m/s²:
N2 = m(g - a) = 70(10 - 2) = 70 × 8 = 560 N (or 56 kg-wt).
(a) Scale Reading = 910 N | (b) Scale Reading = 560 N.
Q3. [CBSE Board] A bullet of mass 0.04 kg moving with a speed of 90 m/s enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Given: m = 0.04 kg, u = 90 m/s, v = 0, s = 60 cm = 0.6 m.
Using 3rd equation: v² - u² = 2as ⇒ 0² - (90)² = 2 a 0.6 ⇒ -8100 = 1.2 a ⇒ a = -6750 m/s².
Resistive Force F = m a = 0.04 × (-6750) = -270 N.
Average Resistive Force = 270 N.
Q4. [JEE Main] A block of mass 10 kg is placed on a rough horizontal surface with coefficient of static friction μs = 0.4. Calculate: (a) Limiting static friction force, (b) Acceleration of block if a horizontal force of 60 N is applied (μk = 0.35, g = 10 m/s^2).
Given: m = 10 kg, N = mg = 10 × 10 = 100 N, μs = 0.4, μk = 0.35, F_app = 60 N.
(a) Limiting static friction fs,max = μs N = 0.4 × 100 = 40 N.
(b) Since F_app = 60 N > 40 N, block moves! Kinetic friction applies: fk = μk N = 0.35 × 100 = 35 N.
Net Force F_net = F_app - fk = 60 - 35 = 25 N.
Acceleration a = F_net / m = 25 / 10 = 2.5 m/s².
Limiting Friction = 40 N | Acceleration = 2.5 m/s^2.
Q5. A circular race track of radius 300 m is banked at an angle of 15°. If the coefficient of friction between road and wheels is 0.2, calculate the maximum safe speed of a race car. (Take g = 9.8 m/s^2, tan 15° = 0.268).
Given: r = 300 m, θ = 15° ⇒ tan 15° = 0.268, μs = 0.2, g = 9.8 m/s².
Using Banking Formula: v_max = √[ r g ( (μs + tanθ) / (1 - μs tanθ) ) ].
(μs + tanθ) / (1 - μs tanθ) = (0.2 + 0.268) / [1 - (0.2 × 0.268)] = 0.468 / (1 - 0.0536) = 0.468 / 0.9464 &approx; 0.494.
v_max = √(300 × 9.8 × 0.494) = √(2940 × 0.494) = √(1452.36) &approx; 38.1 m/s = 137.1 km/h.
Maximum Safe Speed = 38.1 m/s (137.1 km/h).

8. Frequently Asked Questions (FAQ)

Why is Newton's Second Law considered the real law of motion?

Because both Newton's First Law and Third Law can be mathematically derived from Newton's Second Law (F = dp/dt = m*a). When F = 0, a = 0 (First Law); in isolated systems, F12 = -F21 (Third Law).

What is the Impulse-Momentum Theorem?

The Impulse-Momentum Theorem states that the impulse of a net force acting on a body equals the change in linear momentum produced in the body: J = F_avg · Δt = Δp.

Why does a person feel heavier in an upward accelerating elevator?

In an upward accelerating elevator (+a), the floor must push up harder to accelerate the body mass m, increasing normal reaction N = m(g + a) > mg.

What is the relationship between Angle of Repose and Angle of Friction?

The Angle of Repose (θ) equals the Angle of Friction (λ). Both satisfy the relation tanθ = tanλ = μs (where μs is the coefficient of static friction).

Why are circular roads banked on curved turns?

Roads are banked (outer edge raised) so that the horizontal component of normal reaction (N sinθ) contributes to centripetal force, reducing reliance on friction and preventing skidding at high speeds.

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