Class 11 Physics Chapter 4: Laws of Motion — Newton's Laws & Friction

01 / Chapter Theory

Laws of Motion

Official NCERT syllabus: Dynamics of motion, inertia, momentum, Newton's 1st, 2nd, and 3rd laws, impulse, conservation of linear momentum, equilibrium of concurrent forces, friction laws, circular motion & road banking, and systematic problem solving using Free Body Diagrams (FBD).

11 SECTIONS

Chapter Navigation

4.1-4.2Aristotle's Fallacy 4.3The Law of Inertia 4.4Newton's 1st Law 4.5Newton's 2nd Law & Impulse 4.6Newton's 3rd Law 4.7Momentum Conservation 4.8Equilibrium of a Particle 4.9Friction (Static, Kinetic, Rolling) 4.10Circular Motion & Banking 4.11Free Body Diagrams (FBD)

4.1 - 4.3

NCERT Sections

Aristotle's Fallacy & Galileo's Law of Inertia

Historical Foundations

4.2 Aristotle's Flaw vs. Galileo's Insight

Aristotle's Fallacy: Held the view that "an external force is required to keep a body in uniform motion". This view is flawed because everyday moving bodies come to rest solely due to external opposing forces like friction and viscous drag.
Galileo's Law of Inertia: If the net external force on a body is zero, a body at rest remains at rest, and a body in uniform motion continues to move with constant velocity along a straight line. State of rest and state of uniform linear motion are physically equivalent.

Figure 4.1: Galileo's double inclined plane experiment: On a frictionless surface, a ball released from height h travels infinitely on a horizontal plane to reach its original height.

Interactive 3D Galileo's Double Inclined Plane (Law of Inertia)

Inertia Lab

Drag to change perspective

Physical Regime & Distance Traveled

Right Incline: 30° • Ball travels distance $d = h/\sin\theta$ to reach identical initial height $h$.

Right Incline Angle ($\theta$): Set to 0° for infinite horizontal motion

4.4

NCERT Section

Newton's First Law of Motion

The First Law

4.4.1 Definition & Physical Meaning

Statement: Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.
Mathematical Formulation: If ∑ F_ext = 0 ⇒ a = 0 (and vice versa).

Inertia of Rest: Inability of a body to move by itself (e.g. dust particles falling off a beaten carpet; passengers jerking backward when a bus starts suddenly).
Inertia of Motion: Inability of a body to stop or change its speed by itself (e.g. passenger thrown forward when a speeding bus stops abruptly).
Inertia of Direction: Inability to change direction without external lateral force (e.g. mud flying off tangentially from bicycle wheels).

Example 4.1 • Astronaut Outside Accelerating Spaceship

NCERT Solved

An astronaut accidentally gets separated out of his small spaceship accelerating in interstellar space at a constant rate of 100 m s⁻². What is the acceleration of the astronaut the instant after he is outside the spaceship? (Assume no nearby stars).

Answer: Acceleration = 0 m s⁻².

Reasoning: Once the astronaut is outside the spaceship, there are no contact forces and no gravitational pulls from nearby stars. By Newton's First Law, since net external force ∑ F_ext = 0, the acceleration is zero. (The astronaut continues to move with whatever constant velocity he possessed at the moment of detachment).

4.5

NCERT Section

Newton's Second Law of Motion & Impulse

Force and Momentum

4.5.1 Rate of Change of Momentum & Vector Character

Linear momentum p is the product of mass m and velocity v (p = mv):

Second Law: F = dp/dt = d(mv)/dt = m (dv/dt) = m a (for constant mass)
Component Form: F_x = m a_x | F_y = m a_y | F_z = m a_z
SI Unit of Force: 1 Newton (N) = 1 kg m s⁻²

Local Nature of the Second Law: Force F at a location at an instant t determines acceleration a at that exact location and instant. The particle carries no memory of past forces.

4.5.2 Impulse

When a large force acts for a very short time duration Δt, its time-integral is called Impulse (J):

Impulse (J): J = F_avg Δt = Δp = p_final − p_initial
SI Unit: N s or kg m s⁻¹

Interactive 3D Newton's Second Law ($\mathbf{F} = m\mathbf{a}$)

Dynamics Lab

Drag to change perspective

Applied Force, Mass & Acceleration Relation

Applied Force: 10 N | Mass: 2.0 kg ➔ Acceleration: a = F/m = 5.00 m/s²

Applied Force ($F$):

Mass ($m$):

Example 4.2 • Bullet Retardation in Wooden Block

NCERT Solved

A bullet of mass 0.04 kg moving with speed 90 m s⁻¹ enters a heavy wooden block and stops after 60 cm. What is the average resistive force exerted by the block?

Answer: Average Resistive Force = 270 N (Retardation a = −6750 m s⁻²)

Step 1: Calculate deceleration using kinematics.
v² = u² + 2as → 0 = 90² + 2a(0.60) → 1.2a = −8100 → a = −6750 m s⁻².

Step 2: Calculate resistive force.
F = ma = (0.04 kg) × (−6750 m s⁻²) = −270 N.

Example 4.3 • Position Equation and Force

NCERT Solved

The motion of a particle of mass m is described by y = ut + ½gt². Find the force acting on the particle.

Step 1: Successive time differentiation.
v = dy/dt = u + gt.
a = dv/dt = g (constant downward acceleration).

Step 2: Force.
F = ma = mg (gravitational force).

Example 4.4 • Batsman Returning Straight Drive

NCERT Solved

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s⁻¹. If mass of the ball is 0.15 kg, determine impulse imparted.

Answer: Impulse = 3.6 N s (directed towards the bowler).

Step 1: Calculate vector momentum change.
p_i = −mu = −(0.15 × 12) = −1.8 kg m s⁻¹.
p_f = +mu = +(0.15 × 12) = +1.8 kg m s⁻¹.
Impulse J = p_f − p_i = 1.8 − (−1.8) = 3.6 N s.

4.6 - 4.7

NCERT Sections

Newton's Third Law & Momentum Conservation

Action-Reaction Pairs

4.6.1 Features of Newton's Third Law

Statement: To every action, there is always an equal and opposite reaction.
Vector Relation: F_AB = −F_BA (Force on A by B = − Force on B by A)

Act on Different Bodies: Action and reaction forces never act on the same body; hence they never cancel each other out for individual bodies.
Simultaneous: Action does not cause reaction; they exist instantaneously together.
Internal Forces Cancel: For a composite system (A + B), internal action-reaction pairs sum to zero.

4.7.1 Law of Conservation of Linear Momentum

In an isolated system (where net external force ∑ F_ext = 0), total linear momentum is strictly conserved:

p_total = p₁ + p₂ + ... = Constant
Gun-Bullet Recoil: M V_recoil + m v_bullet = 0 ⇒ V_recoil = − (m / M) v_bullet

Example 4.5 • Billiard Ball Collision with Wall

NCERT Solved

Two identical balls strike a rigid wall with speed u: (a) normally, and (b) at 30° to the normal, reflecting without speed loss. Find the direction of force and the ratio of impulse magnitudes.

Answers: Force is normal to the wall in both cases • Ratio of Impulses J_a / J_b = 2 / √3 ≈ 1.15 ≈ 1.2.

Case (a) Normal Incidence:
Δp_x = −mu − (mu) = −2mu | Δp_y = 0.
Impulse magnitude J_a = 2mu (normal to wall).

Case (b) Inclined at 30° to Normal:
Δp_x = −mu cos 30° − (mu cos 30°) = −2mu cos 30° = −√3 mu.
Δp_y = −mu sin 30° − (−mu sin 30°) = 0.
Impulse magnitude J_b = 2mu cos 30° = √3 mu.

Ratio: J_a / J_b = 2 / √3 = 1.155 ≈ 1.2.

4.8

NCERT Section

Equilibrium of a Particle

Translational Equilibrium

4.8.1 Concurrent Forces & Lami's Theorem

A particle is in translational equilibrium when the vector sum of all concurrent forces acting on it is zero:

∑ F = F₁ + F₂ + F₃ = 0
∑ F_x = 0 | ∑ F_y = 0 | ∑ F_z = 0

Lami's Theorem (for 3 concurrent forces):
F₁ / sin α = F₂ / sin β = F₃ / sin γ

Example 4.6 • Suspended Mass with Midpoint Horizontal Pull

NCERT Solved

A mass of 6 kg is suspended by a rope of length 2 m from ceiling. A horizontal force of 50 N is applied at the midpoint P. What angle does the upper rope make with the vertical in equilibrium? (g = 10 m s⁻²).

Answer: θ = tan⁻¹(5/6) ≈ 39.8° ≈ 40° with vertical.

Step 1: Lower segment tension T₂.
T₂ = mg = 6 × 10 = 60 N.

Step 2: Equilibrium at point P.
Vertical: T₁ cos θ = T₂ = 60 N.
Horizontal: T₁ sin θ = 50 N.
Dividing: tan θ = 50 / 60 = 5/6 ≈ 0.833 → θ ≈ 40°.

4.9

NCERT Section

Common Forces & Friction Laws

Tribology & Contact Mechanics

4.9.1 Static, Kinetic, and Rolling Friction

Friction Type	Formula / Behavior	Key Characteristic
Static Friction (f_s)	f_s ≤ f_s,max = μ_s N	Self-adjusting; opposes impending relative motion.
Kinetic Friction (f_k)	f_k = μ_k N	Constant; opposes actual relative sliding (μ_k < μ_s).
Rolling Friction (f_r)	f_r = μ_r (N / R)	Much smaller (by 2–3 orders of magnitude); occurs due to surface deformation.

Friction & Angle of Repose $$f_\text{s} \le f_\text{s,max} = \mu_\text{s} N, \qquad f_\text{k} = \mu_\text{k} N$$ $$\tan\theta_\text{max} = \mu_\text{s} \implies \theta_\text{max} = \tan^{-1}\mu_\text{s} \quad (\text{Angle of Repose})$$ $$\text{Pulling: } N = mg - F\sin\theta \quad (\text{Lower Friction} \implies \text{Easier to pull than push})$$

Interactive 3D Static vs Kinetic Friction & Angle of Repose

Tribology Lab

Drag to change perspective

Friction State & Threshold Breakdown

Angle θ = 15° ≤ θ_repose (21.8°) • Static Equilibrium (f_s = mg sinθ)

Incline Angle ($\theta$):

Static Friction Coeff ($\mu_\text{s}$):

Example 4.7 • Maximum Train Acceleration for Stationary Box

NCERT Solved

Determine maximum acceleration of a train in which a box on its floor remains stationary ($\mu_\text{s} = 0.15$, $g = 10\text{ m s}^{-2}$).

Solution: $$ma \le f_\text{s,max} = \mu_\text{s} mg \implies a_\text{max} = \mu_\text{s} g = 0.15 \times 10 = \mathbf{1.5\text{ m s}^{-2}}$$

Example 4.8 • Angle of Repose of Incline

NCERT Solved

A mass of $4\text{ kg}$ rests on a plane. The plane is inclined until at $\theta = 15^\circ$, the mass just begins to slide. What is $\mu_\text{s}$?

Solution: $$\mu_\text{s} = \tan\theta_\text{max} = \tan 15^\circ = 2 - \sqrt{3} \approx \mathbf{0.27}$$

Example 4.9 • Block and Trolley System

NCERT Solved

A 20 kg trolley on a horizontal table is connected by a string over a pulley to a hanging 3 kg block (μ_k = 0.04, g = 10 m s⁻²). Find acceleration and tension.

Answers: Acceleration a = 0.96 m s⁻² • String Tension T = 27.1 N

Step 1: Equations of motion.
Hanging block (3 kg): 30 − T = 3a.
Trolley (20 kg): T − f_k = 20a, where f_k = 0.04 × (20 × 10) = 8 N → T − 8 = 20a.

Step 2: Solve simultaneously.
Adding equations: 30 − 8 = 23a → 22 = 23a → a = 22/23 ≈ 0.96 m s⁻².
Tension: T = 30 − 3(0.96) = 30 − 2.88 = 27.12 N ≈ 27.1 N.

4.10

NCERT Section

Circular Motion & Banking of Roads

Centripetal Force & Banking

4.10.1 Level vs. Banked Curve Dynamics

Figure 4.14: Circular turning mechanics on: (a) Flat unbanked road, (b) Banked road at angle θ.

1. Level Road Max Speed: v_max = √(μ_s R g)
2. Optimum Banked Speed (Zero Tyre Wear, μ = 0): v₀ = √(R g tan θ)
3. Max Safe Speed on Banked Road: v_max = √[ R g (μ_s + tan θ) / (1 − μ_s tan θ) ]

Interactive 3D Circular Motion & Road Banking Dynamics

Centripetal Lab

Drag to inspect bank tilt

Bank Angle & Optimum Safe Speed ($v_0 = \sqrt{Rg\tan\theta}$)

Bank Angle: 18° • Optimum Speed (Zero Tyre Wear): v₀ = √(Rg tanθ) ≈ 60.1 km/h

Bank Angle ($\theta$):

Vehicle Speed:

Example 4.10 • Cyclist on a Level Turn

NCERT Solved

A cyclist speeding at 18 km/h on a level road takes a sharp turn of radius 3 m (μ_s = 0.1, g = 9.8 m s⁻²). Will he slip?

Answer: Yes, the cyclist will slip.

Condition to prevent slipping: v² ≤ μ_sRg.
v = 18 × (5/18) = 5 m s⁻¹ → v² = 25 m² s⁻².
μ_sRg = 0.1 × 3 × 9.8 = 2.94 m² s⁻².
Since 25 > 2.94, the required centripetal force exceeds maximum static friction, so the cyclist slips.

Example 4.11 • Banked Racetrack Speeds

NCERT Solved

A circular track of radius 300 m is banked at 15° (μ_s = 0.2, g = 9.8 m s⁻²). Find: (a) optimum speed, (b) maximum permissible speed.

Answers: (a) Optimum Speed v₀ = 28.1 m s⁻¹ • (b) Maximum Speed v_max = 38.1 m s⁻¹

(a) Optimum speed:
v₀ = √(Rg tan 15°) = √(300 × 9.8 × 0.2679) = √787.6 ≈ 28.1 m s⁻¹ (approx 101 km/h).

(b) Maximum permissible speed:
v_max = √[ (300 × 9.8) × (0.2 + 0.2679) / (1 − 0.2 × 0.2679) ] = √[ 2940 × 0.4679 / 0.9464 ] = √1453.6 ≈ 38.1 m s⁻¹ (approx 137 km/h).

4.11

NCERT Section

Solving Problems in Mechanics & FBDs

Example 4.12 • Yielding Floor Action-Reaction Pairs

NCERT Solved

A 2 kg block rests on a floor. A 25 kg cylinder is placed on top. The floor yields steadily with acceleration 0.1 m s⁻² downwards. What is the action on the floor (a) before and (b) after yielding? Identify action-reaction pairs (g = 10 m s⁻²).

Answers: (a) 20 N downwards • (b) 267.3 N downwards

(a) Before yielding (at rest):
R = mg = 2 × 10 = 20 N. Action on floor is 20 N downwards.

(b) After yielding (accelerating down at 0.1 m s⁻²):
Total mass M = 27 kg. Weight = 270 N.
W − R' = Ma → 270 − R' = 27(0.1) = 2.7 → R' = 267.3 N. Action on floor is 267.3 N downwards.

Action-Reaction Pairs:
1. Gravity on mass by Earth ↔ Gravity on Earth by mass (270 N).
2. Normal push on floor by block ↔ Normal push on block by floor (267.3 N).
3. Push on block by cylinder ↔ Push on cylinder by block.

02 / Self-Assessment Lab

Concept Check Questions

Instant per-option explanations for every choice, illustrating fundamental dynamic principles and pinpointing common misconceptions.

20 QUESTIONS

Progress: 0 / 20 Answered

Score: 0

Q01UNANSWERED

If the net external force acting on a body is zero, the body:

Option A is incorrect. The body could also move with uniform velocity in a straight line.

Option B is correct. Newton's First Law states that ∑ F_ext = 0 implies a = 0 (constant velocity or rest).

Option C is incorrect. Circular motion requires a non-zero inward centripetal force.

Option D is incorrect. Slowing down requires a retarding force.

Core Rule

∑ F_ext = 0 ⇔ a = 0.

Q02UNANSWERED

A passenger standing in a bus is thrown backward when the bus accelerates forward because of:

Option A is correct. The feet move forward with the floor due to friction, while the upper body tends to remain at rest due to inertia.

Option B is incorrect. Inertia of motion acts when a moving bus stops.

Option C is incorrect. The effect is explained by inertia (First Law).

Option D is incorrect. Motion is linear, not rotating.

Core Rule

Inertia of rest opposes sudden change in state from rest to motion.

Q03UNANSWERED

Why does a cricketer draw his hands backward while catching a ball?

Option A is incorrect. Momentum change Δp = mu is fixed regardless of hand motion.

Option B is correct. Since F = Δp / Δt, increasing Δt drastically reduces the impulsive force felt by the hands.

Option C is incorrect. Impulse equals Δp, which is constant.

Option D is incorrect. The core principle here is the rate of momentum change (Second Law).

Core Rule

F_avg ∝ 1 / Δt for a given momentum change Δp.

Q04UNANSWERED

Action and reaction forces:

Option A is incorrect. They always act on two distinct interacting bodies.

Option B is correct. F_AB = −F_BA: They are simultaneous mutual interaction forces on different objects.

Option C is incorrect. There is no time lag between action and reaction.

Option D is incorrect. Newton's Third Law is universally true whether bodies are at rest or accelerating.

Core Rule

Action and reaction are equal, opposite, simultaneous, and act on different bodies.

Q05UNANSWERED

The coefficient of static friction μ_s and kinetic friction μ_k satisfy:

Option A is correct. Interlocking of microscopic surface irregularities is stronger when static than during relative sliding, so μ_s > μ_k.

Option B is incorrect. Kinetic friction is always smaller than limiting static friction.

Option C is incorrect. They are distinct constants.

Option D is incorrect. Friction is independent of apparent area of contact.

Core Rule

Limiting static friction is greater than kinetic friction (μ_s > μ_k).

Q06UNANSWERED

The optimum speed on a banked road of radius R at angle θ where tyres experience no lateral friction is:

Option A is correct. When N sin θ = mv²/R and N cos θ = mg, dividing gives tan θ = v²/(Rg), so v₀ = √(Rg tan θ).

Option B is incorrect. That is the max speed on a flat unbanked road.

Option C is incorrect. Dimensional formula is incorrect.

Option D is incorrect. Normal component uses tan θ.

Core Rule

Optimum banking speed: v₀ = √(Rg tan θ).

Q07UNANSWERED

Why is it easier to pull a lawn mower than to push it?

Option A is correct. The vertical component of pulling force acts upward, reducing normal reaction N and frictional resistance. Pushing adds F sin θ downward, increasing N.

Option B is incorrect. Mass is constant.

Option C is incorrect. It remains rolling motion in both cases.

Option D is incorrect. Friction opposes rolling motion in both.

Core Rule

Pulling decreases normal force (N = mg − F sin θ), decreasing frictional resistance.

Q08UNANSWERED

A 70 kg man in an elevator accelerating downwards at 5 m s⁻² reads on a weighing scale (g = 10 m s⁻²):

Option A is incorrect. 700 N is reading at rest or uniform velocity.

Option B is correct. Apparent weight R = m(g − a) = 70(10 − 5) = 70 × 5 = 350 N (scale reads 35 kg).

Option C is incorrect. 1050 N occurs when accelerating upwards (m(g+a)).

Option D is incorrect. Zero reading (weightlessness) occurs in free fall (a = g).

Core Rule

Lift downward acceleration: R = m(g − a).

Q09UNANSWERED

A gun of mass 100 kg fires a shell of mass 0.02 kg with muzzle speed 80 m s⁻¹. The recoil speed of the gun is:

Option A is correct. V_recoil = (m/M)v = (0.02 × 80) / 100 = 1.6 / 100 = 0.016 m s⁻¹.

Option B is incorrect. Missed dividing by 100 kg.

Option C is incorrect. Factor of 10 error.

Option D is incorrect. Incorrect calculation.

Core Rule

Recoil velocity V = (m/M)v from momentum conservation.

Q10UNANSWERED

When a stone whirled in a horizontal circle breaks its string, it flies off:

Option A is incorrect. There is no outward force; velocity at the instant of break is strictly tangential.

Option B is correct. By inertia of direction (First Law), once centripetal tension vanishes, the stone continues along its instantaneous tangential velocity.

Option C is incorrect. Tension is zero after the break.

Option D is incorrect. Straight-line inertial flight occurs.

Core Rule

Inertia of direction: Objects fly tangentially when circular constraint is removed.

Q11UNANSWERED

An astronaut accidentally gets separated from a spaceship accelerating at 100 m s⁻² in deep space. His acceleration immediately after separation is:

Option A is incorrect. The astronaut no longer touches the spaceship thrusters.

Option B is correct. With no net external contact or gravitational force in deep space, ∑ F = 0 ⇒ a = 0 by Newton's First Law.

Option C is incorrect. Deep space has no Earth gravity.

Option D is incorrect. Force drops instantaneously to zero.

Core Rule

No contact force + no field force = Zero acceleration.

Q12UNANSWERED

Static friction is called a "self-adjusting" force because:

Option A is incorrect. μ_sN is only its upper limiting ceiling.

Option B is correct. f_s adjusts exactly from 0 to μ_sN to maintain zero relative motion under applied shear force.

Option C is incorrect. Static friction applies when speed is zero.

Option D is incorrect. Friction is area-independent.

Core Rule

Static friction: 0 ≤ f_s ≤ μ_sN.

Q13UNANSWERED

Newton's Second Law F = ma is a "local" relation. This means:

Option A is incorrect. Laws of physics are universal.

Option B is correct. The particle retains no memory of past forces or past accelerations.

Option C is incorrect. Only velocity carries momentum history; acceleration is instantaneous.

Option D is incorrect. Internal forces sum to zero.

Core Rule

Local property: F(t) governs a(t) at the exact instant t.

Q14UNANSWERED

In an isolated system of two colliding particles, total linear momentum is conserved because:

Option A is incorrect. Momentum is conserved even in highly inelastic collisions where KE is lost.

Option B is correct. Δp₁ = F₁₂ Δt and Δp₂ = −F₁₂ Δt ⇒ Δp_total = 0.

Option C is incorrect. System is isolated from external forces.

Option D is incorrect. Internal friction does not change total momentum.

Core Rule

Momentum conservation is a direct mathematical consequence of 2nd and 3rd laws.

Q15UNANSWERED

The angle of repose of an incline is θ_max = 30°. The coefficient of static friction μ_s between block and incline is:

Option A is incorrect. sin 30° = 0.5, but μ_s uses tan θ.

Option B is correct. μ_s = tan θ_max = tan 30° = 1/√3 ≈ 0.577.

Option C is incorrect. tan 60° = √3.

Option D is incorrect. cos 30° = 0.866.

Core Rule

Angle of Repose: μ_s = tan θ_max.

Q16UNANSWERED

On a level unbanked road of radius R = 100 m with μ_s = 0.4 and g = 10 m s⁻², maximum safe speed is:

Option A is incorrect. 10 m/s is below limit.

Option B is correct. v_max = √(μ_s R g) = √(0.4 × 100 × 10) = √400 = 20 m s⁻¹.

Option C is incorrect. Forgot square root.

Option D is incorrect. Calculation error.

Core Rule

Level curve: v_max = √(μ_s R g).

Q17UNANSWERED

Atwood machine with masses m₁ = 3 kg and m₂ = 5 kg over frictionless pulley has acceleration (g = 10 m s⁻²):

Option A is correct. a = [(m₂ − m₁)/(m₁ + m₂)]g = [(5 − 3)/8] × 10 = (2/8) × 10 = 2.5 m s⁻².

Option B is incorrect. Incorrect mass ratio.

Option C is incorrect. Factor of 2 error.

Option D is incorrect. 10 m/s² is free fall.

Core Rule

Atwood machine: a = g(Δm / ∑m).

Q18UNANSWERED

Area under a Force versus Time (F-t) graph represents:

Option A is incorrect. Work done is area under a Force-Displacement (F-x) curve.

Option B is correct. ∫ F dt = Δp = Impulse.

Option C is incorrect. Power is dW/dt.

Option D is incorrect. Acceleration is F/m.

Core Rule

Area under F-t curve = ∫ F dt = Impulse = Δp.

Q19UNANSWERED

Which of the following is NOT a method to reduce friction?

Option A is incorrect. Lubricants separate moving surfaces, reducing friction.

Option B is incorrect. Ball bearings convert sliding friction into rolling friction.

Option C is incorrect. Air cushion minimizes solid-solid contact.

Option D is correct. Since f_s ≤ μN and f_k = μN, increasing the normal force N directly increases frictional resistance.

Core Rule

Friction is directly proportional to normal force N.

Q20UNANSWERED

A book resting on a table is in equilibrium. The upward normal reaction force N equals downward weight W because of:

Option A is incorrect. N and W act on the SAME body; they are not an action-reaction pair.

Option B is correct. Since the book is observed to have a = 0, net external force must be zero: N − W = 0 ⇒ N = W.

Option C is incorrect. Momentum is statically zero.

Option D is incorrect. Not applicable here.

Core Rule

N = W for a resting body on a table arises from First Law equilibrium, NOT Third Law.

03 / Textbook Solutions

NCERT Exercises 4.1 – 4.23

Complete stepwise solutions for every textbook exercise question in NCERT Class 11 Physics Chapter 4 Reprint 2026-27 (take g = 10 m s⁻²).

23 QUESTIONS

Ex 4.1Net Force Identification

4.1 Give the magnitude and direction of the net force acting on:

(a)

A drop of rain falling down with constant speed.

(b)

A cork of mass 10 g floating on water.

(c)

A kite skillfully held stationary in the sky.

(d)

A car moving with constant velocity of 30 km/h on a rough road.

(e)

A high-speed electron in space far from all material objects and fields.

In all five cases (a to e), the net force is ZERO.

(a) Zero: Constant speed in straight line → a = 0 → F_net = 0 (downward gravity is balanced by upward viscous drag).

(b) Zero: Floating at rest → F_net = 0 (downward weight is balanced by upward buoyant force).

(c) Zero: Held stationary → a = 0 → F_net = 0 (wind thrust and gravity balanced by string tension).

(d) Zero: Constant velocity → a = 0 → F_net = 0 (engine forward force cancels road friction).

(e) Zero: Free of all gravitational, electric, and magnetic fields in deep space → F_net = 0.

Ex 4.2Pebble Thrown Vertically Up

4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the magnitude and direction of the net force on the pebble:
(a) during upward motion, (b) during downward motion, (c) at highest point where it is momentarily at rest. Do answers change if thrown at 45°? (Ignore air resistance, g = 10 m s⁻²).

In all cases, Net Force = 0.5 N vertically downwards (No change for 45° projection).

(a), (b), (c): The only external force acting throughout flight is gravity: F = mg = (0.05 kg) × 10 m s⁻² = 0.5 N vertically downwards.

At 45° angle: The force remains unchanged (0.5 N vertically downwards) because gravity acts exclusively vertically.

Ex 4.3Stone Dropped from Train

4.3 Give magnitude and direction of net force on a stone of mass 0.1 kg:
(a) dropped from stationary train, (b) dropped from train at constant 36 km/h, (c) dropped from train accelerating at 1 m s⁻², (d) lying on floor of train accelerating at 1 m s⁻² at rest relative to train.

(a), (b), (c): 1.0 N vertically downwards • (d): 0.1 N horizontally in direction of train acceleration.

(a), (b), (c): The moment the stone leaves the hand, contact with train is lost. It experiences only gravity: F = mg = 0.1 × 10 = 1.0 N vertically downwards.

(d): Stone accelerates with the train (a = 1 m s⁻²). Vertical weight is balanced by floor normal reaction (N = mg = 1 N). Net force is horizontal static friction: F = ma = 0.1 × 1 = 0.1 N horizontally forward.

Ex 4.4Whirled Particle Net Force

4.4 One end of a string of length l is connected to a particle of mass m on a smooth horizontal table moving in circle with speed v. The net force on the particle (directed towards centre) is:
(i) T, (ii) T − mv²/l, (iii) T + mv²/l, (iv) 0.

Correct Alternative: (i) T

Reasoning: Tension T in the string is the physical force acting towards the centre that provides the necessary centripetal acceleration (T = mv²/l). mv²/l is not a separate force; it is the ma effect of tension.

Ex 4.5Stopping Time Calculation

4.5 A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with speed 15 m s⁻¹. How long does the body take to stop?

Answer: Stopping Time t = 6.0 s

Step 1: Retardation.
a = −F / m = −50 / 20 = −2.5 m s⁻².

Step 2: Time.
v = u + at → 0 = 15 − 2.5t → t = 15 / 2.5 = 6.0 s.

Ex 4.6Force from Speed Change

4.6 A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s in the same direction. Find the magnitude and direction of the force.

Answer: Force = 0.18 N along the direction of motion.

Step 1: Acceleration.
a = (v − u) / t = (3.5 − 2.0) / 25 = 1.5 / 25 = 0.06 m s⁻².

Step 2: Force.
F = ma = 3.0 × 0.06 = 0.18 N.

Ex 4.7Perpendicular Forces Acceleration

4.7 A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration.

Answer: Acceleration = 2.0 m s⁻² at θ ≈ 36.9° ≈ 37° with the 8 N force.

Step 1: Resultant force.
F = √(8² + 6²) = √(64 + 36) = √100 = 10 N.

Step 2: Acceleration & Direction.
a = F / m = 10 / 5 = 2.0 m s⁻².
tan θ = 6 / 8 = 0.75 → θ = tan⁻¹(0.75) ≈ 36.87° ≈ 37°.

Ex 4.8Three-Wheeler Braking Force

4.8 A three-wheeler (mass 400 kg + driver 65 kg = 465 kg) moving at 36 km/h stops in 4.0 s to save a child. What is the average retarding force?

Answer: Average Retarding Force = 1162.5 N ≈ 1.16 × 10³ N

Step 1: Convert speed and find deceleration.
u = 36 × (5/18) = 10 m s⁻¹.
a = (0 − 10) / 4.0 = −2.5 m s⁻².

Step 2: Retarding force.
F = M_total × |a| = 465 kg × 2.5 m s⁻² = 1162.5 N.

Ex 4.9Rocket Lift-off Thrust

4.9 A rocket with lift-off mass 20,000 kg is blasted upwards with initial acceleration 5.0 m s⁻². Calculate initial thrust (g = 10 m s⁻²).

Answer: Initial Thrust = 3.0 × 10⁵ N (300 kN)

Thrust formula:
F_thrust − mg = ma → F_thrust = m(g + a) = 20000 × (10 + 5.0) = 20000 × 15 = 300,000 N = 3.0 × 10⁵ N.

Ex 4.10Piecewise 1D Motion under Force

4.10 Body of mass 0.40 kg moving at 10 m s⁻¹ North is subjected to 8.0 N South for 30 s. Position at t = 0 is x = 0. Predict position at t = −5 s, 25 s, 100 s.

x(−5 s) = −50 m • x(25 s) = −6000 m (−6 km) • x(100 s) = −50,000 m (−50 km)

For t < 0 (uniform velocity 10 m/s North):
x(−5 s) = u × t = 10 × (−5) = −50 m.

For 0 ≤ t ≤ 30 s (acceleration a = −8.0/0.40 = −20 m s⁻²):
x(25 s) = ut + ½at² = 10(25) + ½(−20)(25)² = 250 − 6250 = −6000 m (−6.0 km).

For t = 100 s:
At t = 30 s: x(30) = 10(30) + ½(−20)(900) = 300 − 9000 = −8700 m.
Velocity at t = 30 s: v = 10 − 20(30) = −590 m/s.
From t = 30 to 100 s (70 s of uniform motion at −590 m/s):
x(100) = −8700 + (−590 × 70) = −8700 − 41300 = −50,000 m (−50 km).

Ex 4.11Stone Dropped from Accelerated Truck

4.11 A truck accelerates at 2.0 m s⁻² from rest. At t = 10 s, a stone is dropped from 6 m height. What are velocity and acceleration of stone at t = 11 s? (g = 10 m s⁻²).

Velocity = 22.36 m s⁻¹ at 26.6° below horizontal • Acceleration = 10 m s⁻² vertically downwards

Step 1: Horizontal velocity at release (t = 10 s).
v_x = at = 2.0 × 10 = 20 m s⁻¹ (constant).

Step 2: Vertical velocity after 1 s of fall (at t = 11 s).
v_y = 0 − g(1) = −10 m s⁻¹.
Speed: v = √(20² + (−10)²) = √500 = 22.36 m s⁻¹.
Direction: tan θ = −10 / 20 = −0.5 → θ = −26.6° with horizontal.

Step 3: Acceleration.
Once in air, horizontal acceleration is zero, so a = 10 m s⁻² vertically downwards.

Ex 4.12Oscillating Bob String Cut

4.12 Bob of mass 0.1 kg on 2 m string oscillates with speed 1 m s⁻¹ at mean position. What is its trajectory if string is cut at: (a) extreme position, (b) mean position?

(a) Extreme Position: At the extreme position, instantaneous velocity is zero (v = 0). When cut, the bob falls vertically straight down under gravity.

(b) Mean Position: At the mean position, the bob has purely horizontal velocity (v = 1 m s⁻¹). When cut, it traces a parabolic trajectory like a horizontally projected object.

Ex 4.13Weighing Scale in an Elevator

4.13 A 70 kg man stands on a scale in a lift. What is the reading when lift moves:
(a) upwards with uniform speed 10 m/s,
(b) downwards with uniform acceleration 5 m s⁻²,
(c) upwards with uniform acceleration 5 m s⁻²,
(d) in free fall under gravity? (g = 10 m s⁻²).

(a) 700 N (70 kg) • (b) 350 N (35 kg) • (c) 1050 N (105 kg) • (d) 0 N (Weightlessness)

(a) Uniform speed (a = 0): R = mg = 70 × 10 = 700 N (70 kg).

(b) Downward acceleration (a = 5 m s⁻²): R = m(g − a) = 70(10 − 5) = 350 N (35 kg).

(c) Upward acceleration (a = 5 m s⁻²): R = m(g + a) = 70(10 + 5) = 1050 N (105 kg).

(d) Free fall (a = g): R = m(g − g) = 0 N (0 kg, Weightlessness).

Ex 4.14Position-Time Graph Forces & Impulse

4.14 Figure 4.16 shows the x-t graph of 4 kg particle (x = 0 for t < 0; x rises linearly to 3 m at t = 4 s; x = 3 m for t > 4 s). Find: (a) force in each segment, (b) impulse at t = 0 and t = 4 s.

Figure 4.16: Position-time (x-t) graph for Exercise 4.14.

(a) Force = 0 for all intervals • (b) Impulse at t = 0 is +3 N s • Impulse at t = 4 s is −3 N s

(a) Forces:
• t < 0: At rest (v = 0) → F = 0.
• 0 < t < 4 s: Constant slope → v = 3/4 = 0.75 m/s → a = 0 → F = 0.
• t > 4 s: At rest (v = 0) → F = 0.

(b) Impulses (momentum change):
• At t = 0: J = m(v_after − v_before) = 4(0.75 − 0) = +3.0 N s (kg m s⁻¹).
• At t = 4 s: J = m(v_after − v_before) = 4(0 − 0.75) = −3.0 N s (kg m s⁻¹).

Ex 4.15Two Connected Blocks Tension

4.15 Two blocks of mass 10 kg (A) and 20 kg (B) on smooth horizontal surface are connected by a light string. Force F = 600 N is applied to: (i) A, (ii) B along string direction. Find tension in each case.

(i) Tension T = 400 N (pulling A) • (ii) Tension T = 200 N (pulling B)

System Acceleration: a = F / (m_A + m_B) = 600 / (10 + 20) = 600 / 30 = 20 m s⁻².

(i) When force is applied to A (10 kg):
The string pulls block B alone: T = m_B a = 20 kg × 20 m s⁻² = 400 N.

(ii) When force is applied to B (20 kg):
The string pulls block A alone: T = m_A a = 10 kg × 20 m s⁻² = 200 N.

Ex 4.16Atwood Machine (8 kg & 12 kg)

4.16 Two masses 8 kg (m₁) and 12 kg (m₂) are connected by light string over a frictionless pulley. Find acceleration and tension (g = 10 m s⁻²).

Acceleration a = 2.0 m s⁻² • String Tension T = 96 N

Step 1: Acceleration of Atwood machine.
a = [(m₂ − m₁) / (m₁ + m₂)] g = [(12 − 8) / (12 + 8)] × 10 = (4 / 20) × 10 = 2.0 m s⁻².

Step 2: String Tension.
T = [2 m₁ m₂ / (m₁ + m₂)] g = [2 × 8 × 12 / 20] × 10 = 192 / 2 = 96 N.

Ex 4.17Nuclear Disintegration Momentum

4.17 A nucleus at rest in lab frame disintegrates into two smaller nuclei. Show that the products must move in opposite directions.

Proof from Momentum Conservation:
Initial momentum of resting parent nucleus = 0.
By conservation of linear momentum: p₁ + p₂ = 0 ⇒ p₁ = −p₂.
m₁ v₁ = −m₂ v₂ ⇒ v₁ = −(m₂ / m₁) v₂.
The negative sign demonstrates that the two product nuclei must move in strictly opposite directions along the same straight line.

Ex 4.18Head-on Billiard Ball Rebound

4.18 Two billiard balls of mass 0.05 kg each moving in opposite directions at 6 m s⁻¹ collide head-on and rebound with the same speed. What is the impulse imparted to each?

Impulse magnitude = 0.6 N s (kg m s⁻¹)

Calculation:
Initial momentum of ball 1 = m(+u) = 0.05 × 6 = +0.3 kg m s⁻¹.
Final momentum of ball 1 = m(−u) = 0.05 × (−6) = −0.3 kg m s⁻¹.
Impulse = |Δp| = |−0.3 − (+0.3)| = |−0.6| = 0.6 N s.

Ex 4.19Gun Recoil Speed

4.19 A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If muzzle speed of the shell is 80 m s⁻¹, what is the recoil speed of the gun?

Answer: Recoil speed = 0.016 m s⁻¹ (1.6 cm s⁻¹)

Conservation of momentum:
M V + m v = 0 → V = −(m/M)v = −(0.020 × 80) / 100 = −1.6 / 100 = −0.016 m s⁻¹.

Ex 4.20Glance Deflection Impulse

4.20 A batsman deflects a ball by an angle of 45° without changing initial speed 54 km/h (15 m/s). What is the impulse imparted (ball mass = 0.15 kg)?

Answer: Impulse = 4.16 N s ≈ 4.2 N s

Step 1: Angle of incidence and reflection.
Total deflection is 45° → angle with normal to deflection bisector is θ = 45°/2 = 22.5°.

Step 2: Vector impulse formula.
J = 2 m u cos(θ/2) = 2 × (0.15) × (15) × cos 22.5° = 4.5 × 0.9239 = 4.157 N s ≈ 4.2 N s.

Ex 4.21Horizontal Whirled Stone Tension

4.21 Stone of mass 0.25 kg on 1.5 m string is whirled horizontally at 40 rev/min. What is the tension? What is the maximum speed if string can withstand 200 N?

Tension T = 6.57 N ≈ 6.6 N • Maximum permissible speed v_max = 34.64 m s⁻¹ ≈ 35 m s⁻¹

Step 1: Calculate angular velocity & tension.
ν = 40/60 = 2/3 rev/s → ω = 2πν = 2π(2/3) = 4π/3 = 4.189 rad/s.
T = m ω² R = 0.25 × (4.189)² × 1.5 = 0.375 × 17.55 = 6.58 N ≈ 6.6 N.

Step 2: Maximum speed for T_max = 200 N.
T_max = m v_max² / R → v_max² = (200 × 1.5) / 0.25 = 300 / 0.25 = 1200 → v_max = √1200 = 34.64 m s⁻¹.

Ex 4.22String Break Trajectory

4.22 If speed exceeds maximum in Ex 4.21 and string breaks, what describes the stone's path:
(a) moves radially outwards, (b) flies off tangentially, (c) flies off at an angle with tangent?

Correct Alternative: (b) The stone flies off tangentially from the instant the string breaks.

Reasoning: At every instant, the stone's velocity vector is tangential to the circle. When the centripetal tension vanishes, Newton's First Law guarantees that the stone continues straight along this tangential velocity.

Ex 4.23Conceptual Explanations

4.23 Explain why:

(a)

A horse cannot pull a cart and run in empty space.

(b)

Passengers are thrown forward when a speeding bus stops suddenly.

(c)

It is easier to pull a lawn mower than to push it.

(d)

A cricketer moves his hands backwards while holding a catch.

(a) Horse in space: To move forward, the horse must push the ground backward with its hooves; the ground reacts by pushing the horse forward. In empty space, there is no ground reaction force.

(b) Passengers thrown forward: When the bus stops, the lower body stops with the floor due to friction, but the upper body continues moving forward due to inertia of motion.

(c) Lawn mower: Pulling creates an upward vertical component reducing normal reaction N = mg − F sin θ, reducing friction. Pushing increases N = mg + F sin θ.

(d) Cricketer hands backward: Increases time of impact Δt, decreasing the average stopping force F = Δp / Δt and preventing injury.

04 / Rapid Reference

Chapter Summary & Formulas

Newton's laws of motion, conservation of momentum, friction dynamics, circular road banking, FBD shortcuts, and exam-focused mnemonics.

100% SYLLABUS

4.1 - 4.5 • Core Newton Laws

Inertia, Momentum & Newton's Three Laws

First Law (Inertia): An object remains in a state of rest or uniform motion unless acted upon by an external force. Defines force qualitatively.
Second Law (Dynamics): The rate of change of momentum is proportional to the applied force: $$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$$ If mass $m$ is constant, $\vec{F} = m\vec{a}$. If mass is variable (e.g. rocket), $\vec{F} = m\frac{d\vec{v}}{dt} + \vec{v}\frac{dm}{dt}$.
Third Law (Reciprocity): Action and reaction are equal and opposite ($\vec{F}_{AB} = -\vec{F}_{BA}$).
Crucial Condition: Action and reaction act on two different bodies. They never act on the same body and therefore **never cancel each other**.

Impulse and Linear Momentum

Linear Momentum ($\vec{p}$): $\vec{p} = m\vec{v}$ (vector quantity).

Impulse ($\vec{I}$): The product of large average force and the short time interval: $$\vec{I} = \vec{F}_{\text{avg}}\Delta t = \Delta \vec{p} = \int \vec{F} dt$$ Geometrically, Impulse is the area under the Force-time ($F-t$) graph.

Conservation of Linear Momentum: In the absence of an external net force ($\vec{F}_{\text{ext}} = 0$), the total linear momentum of an isolated system remains constant: $\sum \vec{p}_i = \sum \vec{p}_f$.
Recoil Velocity of Gun ($v_r$): $v_r = -\frac{m}{M}v$.

4.6 • Friction Mechanics

Static, Limiting & Kinetic Friction

Friction is a contact force parallel to the surfaces that opposes impending or actual relative motion.

Friction Type	Mathematical Formula	Characteristics & Conditions
Static Friction ($f_s$)	$f_s \le f_{s,\text{max}} = \mu_s N$	Self-adjusting. It exactly matches the applied force up to the limiting value.
Limiting Friction	$f_{s,\text{max}} = \mu_s N$	Maximum value of static friction before sliding begins. Dependes on surface nature.
Kinetic Friction ($f_k$)	$f_k = \mu_k N$	Acts during relative sliding. Constant magnitude, $\mu_k < \mu_s$.
Rolling Friction ($f_r$)	$f_r = \mu_r \frac{N}{R}$	Acts on rolling wheels. $\mu_r \ll \mu_k$. Much smaller than sliding.

Friction Angles & Lawn Mower Trick

Angle of Friction ($\theta$): Angle between normal reaction $N$ and resultant of limiting friction and $N$: $\mu_s = \tan\theta$.
Angle of Repose ($\alpha$): Max inclination angle of plane at which a block just stays at rest: $\mu_s = \tan\alpha$.
Pulling vs Pushing Lawn Mower: Pulling is easier than pushing. In pulling, the upward vertical component of applied force reduces normal reaction $N$ ($N = mg - F\sin\theta$), decreasing friction. In pushing, the downward component increases $N$ ($N = mg + F\sin\theta$), increasing friction.

4.7 • Circular Road Dynamics

Dynamics of Circular Motion & Banking

Centripetal force is not a new force, but a label for the net inward radial force ($F_c = \frac{mv^2}{R}$).

Road Condition	Speed Limit Formula	Limiting Variable & Notes
Flat Circular Road	$$v_{\text{max}} = \sqrt{\mu_s R g}$$	Centripetal force is provided solely by static friction.
Banked Road (Frictionless)	$$v_0 = \sqrt{R g \tan\theta}$$	Optimum speed. Normal reaction component ($N\sin\theta$) provides centripetal force. Zero wear on tires.
Banked Road (With Friction)	$$v_{\text{max}} = \sqrt{R g \left(\frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta}\right)}$$	Maximum safe speed incorporating both banking angle $\theta$ and static friction coefficient $\mu_s$.

System Dynamics • FBD Shortcuts

Free Body Diagram (FBD) Shortcuts for Connected Motion

Atwood Machine (Two masses on a pulley, $m_1 > m_2$)

System Acceleration Mnemonic: Difference in weights over sum of masses: $$a = g\left(\frac{m_1 - m_2}{m_1 + m_2}\right)$$
String Tension Mnemonic: Twice the product over sum: $$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$

Apparent Weight in a Lift

Motion of Lift	Apparent Weight ($N$)	Sensation
At rest or moving with constant velocity ($a = 0$)	$N = mg$	Normal Weight
Accelerating upward ($a$)	$N = m(g+a)$	Heavier (Overweight)
Accelerating downward ($a$)	$N = m(g-a)$	Lighter (Underweight)
Free Fall ($a = g$)	$N = 0$	Weightlessness

NCERT Official

Points to Ponder & Exam Tips

Newton's laws are valid only in inertial frames. In non-inertial (accelerating) frames, pseudo forces must be added to use $F = ma$.
If net external force is zero, velocity does not change. Zero force means zero acceleration, not zero velocity.
Normal reaction is always perpendicular to the contact surface, not necessarily opposite to gravity (e.g. on an incline $N = mg\cos\theta$).
Friction is not always a resistive force; it is the force that enables walking, driving, and sliding block systems to accelerate.
Static friction limits are limiting friction values. Never write $f_s = \mu_s N$ unless sliding is just about to start.

05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to calculate score and review answers.

24 QUESTIONS

1. SI unit of linear momentum is:

2. If a body moves with constant velocity, net force is:

3. Action and reaction forces act on:

4. The relationship between static and kinetic friction coefficients is:

5. A 20 kg mass experiencing 50 N force stops from 15 m/s in:

6. In an elevator falling freely under gravity, apparent weight is:

7. Optimum speed on a banked curve of radius R at angle θ is:

8. Impulse is equal to:

1. Atwood machine with 8 kg and 12 kg masses has acceleration (g = 10 m/s²):

2. Tension in Atwood machine string with 8 kg and 12 kg is:

3. 70 kg man in lift accelerating UP at 5 m/s² weighs:

4. Train max acceleration without slipping a box (μ_s = 0.15, g = 10) is:

5. Gun of 100 kg firing 0.02 kg shell at 80 m/s recoils at:

6. 20,000 kg rocket blasted up at 5 m/s² needs initial thrust of:

7. 5 kg body acted upon by perpendicular 8 N and 6 N has acceleration:

8. 0.04 kg bullet entering block at 90 m/s stopping in 60 cm experiences force:

1. Two blocks 10 kg (A) and 20 kg (B) pulled by 600 N applied at A has string tension:

2. If 600 N is applied at B (20 kg), string tension between A and B is:

3. 0.25 kg stone on 1.5 m string with T_max = 200 N has max speed:

4. Batsman deflecting 0.15 kg ball at 54 km/h by 45° imparts impulse of:

5. 4 kg particle with x-t rising 3 m in 4 s has impulse at t = 4 s of:

6. 20 kg trolley connected to hanging 3 kg block (μ_k = 0.04) has acceleration:

7. 2 kg block + 25 kg cylinder yielding floor downwards at 0.1 m/s² exerts action of:

8. 300 m radius track banked at 15° has optimum speed of:

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