NCERT Solutions for Class 11 Physics Chapter 4: Laws of Motion (Session 2026–27)

These Class 11 Physics Chapter 4 solutions cover Laws of Motion from the NCERT textbook (updated for 2026–27). Every numbered question of the end-of-chapter Exercises (4.1–4.23) is reproduced exactly as in the book and solved step by step, with each numerical worked out fully and verified with units. The chapter builds on the description of motion (Chapters 2 and 3) to answer the deeper question: what causes motion to change? The answer is Newton’s three laws of motion.

Class: 11 Subject: Physics Chapter: 4 Chapter Name: Laws of Motion Exercises: 4.1 – 4.23 Session: 2026–27
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Class 11 Physics Chapter 4 – Overview

Chapter 4, Laws of Motion, lays the foundation of mechanics. It begins with Aristotle’s fallacy (the wrong idea that a force is needed to keep a body moving) and corrects it with Galileo’s law of inertia. Newton then framed three laws: the first law (a body keeps its state of rest or uniform motion unless a net external force acts), the second law (F = dp/dt = ma, relating net force to acceleration through momentum), and the third law (every action has an equal and opposite reaction). From these follow the law of conservation of momentum, the idea of impulse, the equilibrium of a particle, common forces such as friction (static, kinetic and rolling), and the dynamics of circular motion including motion of a car on level and banked roads. The chapter closes with a systematic method for solving mechanics problems using free-body diagrams.

Key Concepts & Definitions

Inertia: the property by which a body resists any change in its state of rest or of uniform motion. Mass is the measure of inertia.

Newton’s First Law: if the net external force on a body is zero, its acceleration is zero — a body at rest stays at rest and a body in motion keeps moving with uniform velocity.

Momentum (p): the product of mass and velocity, p = mv. It is a vector quantity with SI unit kg m s−1.

Newton’s Second Law: the rate of change of momentum equals the net external force, F = dp/dt = ma (for constant mass). SI unit of force is the newton: 1 N = 1 kg m s−2.

Impulse: force × time duration = change in momentum. Useful when a large force acts for a very short time.

Newton’s Third Law: to every action there is an equal and opposite reaction; the two forces act on different bodies and so never cancel.

Conservation of momentum: the total momentum of an isolated system (no external force) stays constant.

Friction: the component of contact force parallel to the surfaces, opposing relative motion. Static friction fs ≤ μsN; kinetic friction fk = μkN, with μk < μs.

Centripetal force: the net inward force (mv2/R) required to keep a body moving in a circle; it is provided by tension, gravity, friction, etc.

Important Formulas

Momentum: p = m v  •  Second law: F = dp/dt = m a (constant mass).

Equations of motion (constant a): v = u + a t  •  s = u t + ½ a t2  •  v2 = u2 + 2 a s.

Impulse: J = F · Δt = Δp = m(v − u).

Friction: (fs)max = μs N  •  fk = μk N  •  angle of repose: tanθmax = μs.

Circular motion: centripetal force fc = m v2/R  •  level road vmax = √(μs R g)  •  banked road (no friction) vo = √(R g tanθ).

Resultant of two perpendicular forces: R = √(F12 + F22).

NCERT Solutions for Class 11 Physics Chapter 4 – Exercises (4.1–4.23)

For simplicity in numerical calculations, take g = 10 m s−2 (as stated in the NCERT exercises). Questions are reproduced verbatim from NCERT; all answers are original and step-by-step.

4.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a 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 a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

ANSWER In each of (a), (b), (c) and (d) the body moves with constant velocity (or stays at rest), so its acceleration is zero. By Newton’s first law the net force is zero in all four cases — the various forces (gravity, buoyancy, tension, friction, air resistance, normal reaction) exactly balance one another. (e) The electron is far from all material objects and free of electric and magnetic fields, so no agency exerts any force on it. The net force is zero.

4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

ANSWER Once the pebble leaves the hand, the only force acting on it (air resistance ignored) is gravity: F = mg = 0.05 × 10 = 0.5 N, directed vertically downward. (a) During upward motion: 0.5 N vertically downward. (b) During downward motion: 0.5 N vertically downward. (c) At the highest point: 0.5 N vertically downward. The force is the same throughout. If thrown at 45°, the answers do not change — the net force is still 0.5 N vertically downward. However, at the highest point the pebble is now not truly at rest: it retains a constant horizontal velocity component, while only the vertical velocity is momentarily zero.

4.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c) just after it is dropped from the window of a train accelerating with 1 m s−2, (d) lying on the floor of a train which is accelerating with 1 m s−2, the stone being at rest relative to the train. Neglect air resistance throughout.

ANSWER Weight of the stone: mg = 0.1 × 10 = 1 N. (a) Stationary train: the only force is gravity → 1 N vertically downward. (b) Train at constant velocity: once dropped, the stone is acted on only by gravity (the train’s uniform motion adds no horizontal force) → 1 N vertically downward, same as (a). (c) Accelerating train: the moment the stone is dropped, it loses contact with the train, so the train’s acceleration no longer affects it. Force at an instant depends on the situation at that instant, not on history → 1 N vertically downward, same as (a). (d) Stone resting on the floor and moving with the train (a = 1 m s−2): horizontally, the net force = ma = 0.1 × 1 = 0.1 N in the direction of motion of the train (provided by friction). Vertically, weight and normal reaction cancel.

4.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T − mv2/l, (iii) T + mv2/l, (iv) 0  — T is the tension in the string. [Choose the correct alternative].

ANSWER The string lies along the radius, so the tension T is the only horizontal force on the particle and it points towards the centre. The net force directed towards the centre is therefore exactly the tension. The correct alternative is (i) T. (Note: mv2/l is the required centripetal force, which is provided by T, i.e. T = mv2/l; it is not a separate force to be added or subtracted.)

4.5 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−1. How long does the body take to stop?

ANSWER Retardation: a = F/m = −50 / 20 = −2.5 m s−2. Using v = u + at with v = 0, u = 15 m s−1:   0 = 15 + (−2.5)t → t = 15 / 2.5 = 6.0 s.

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. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

ANSWER Acceleration: a = (v − u)/t = (3.5 − 2.0)/25 = 1.5/25 = 0.06 m s−2. Force: F = ma = 3.0 × 0.06 = 0.18 N. Since the speed increases in the same direction, the force acts in the direction of motion.

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 of the body.

ANSWER Resultant of two perpendicular forces: R = √(82 + 62) = √(64 + 36) = √100 = 10 N. Direction with the 8 N force: tanθ = 6/8 = 0.75 → θ = tan−1(3/4) = 37°. Acceleration: a = R/m = 10/5 = 2 m s−2, directed along the resultant force, i.e. at 37° with the 8 N force.

4.8 The driver of a three-wheeler moving with a speed of 36 km/h sees a child standing in the middle of the road and brings his vehicle to rest in 4.0 s just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is 400 kg and the mass of the driver is 65 kg.

ANSWER Initial speed: u = 36 km/h = 36 × (5/18) = 10 m s−1; final v = 0; t = 4.0 s. Acceleration: a = (v − u)/t = (0 − 10)/4 = −2.5 m s−2. Total mass = 400 + 65 = 465 kg. Retarding force = |ma| = 465 × 2.5 = 1162.5 N ≈ 1.2 × 103 N (opposite to motion).

4.9 A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s−2. Calculate the initial thrust (force) of the blast.

ANSWER At lift-off the thrust F must support the weight and also provide the upward acceleration: F − mg = ma → F = m(g + a) = 20000 × (10 + 5.0) = 20000 × 15 = 300000 N = 3.0 × 105 N.

4.10 A body of mass 0.40 kg moving initially with a constant speed of 10 m s−1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = −5 s, 25 s, 100 s.

ANSWER Take north as positive. u = +10 m s−1. The force acts only for 0 ≤ t ≤ 30 s, giving a = F/m = −8.0/0.40 = −20 m s−2 during that interval. At t = −5 s (before the force): the body moves uniformly, so x = u t = 10 × (−5) = −50 m (i.e. 50 m to the south of origin). At t = 25 s (within 0–30 s): x = u t + ½ a t2 = 10 × 25 + ½ × (−20) × 252 = 250 − 6250 = −6000 m = −6 km. At t = 100 s: the force acts only up to t = 30 s. Up to 30 s: x1 = 10 × 30 + ½ × (−20) × 302 = 300 − 9000 = −8700 m. Velocity at 30 s: v = u + at = 10 + (−20)(30) = −590 m s−1. From 30 s to 100 s (70 s, no force, uniform velocity): x2 = −590 × 70 = −41300 m. Total: x = x1 + x2 = −8700 + (−41300) = −50000 m = −50 km.

4.11 A truck starts from rest and accelerates uniformly at 2.0 m s−2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the (a) velocity, and (b) acceleration of the stone at t = 11 s? (Neglect air resistance.)

ANSWER Horizontal velocity of the truck (and hence of the stone) at t = 10 s: vx = u + at = 0 + 2.0 × 10 = 20 m s−1. Once dropped, no horizontal force acts, so by the first law vx stays 20 m s−1. After being dropped, the stone falls for (11 − 10) = 1 s. Vertical velocity: vy = g t = 10 × 1 = 10 m s−1 (downward). (a) Resultant speed: v = √(vx2 + vy2) = √(202 + 102) = √500 = 22.4 m s−1, at angle θ = tan−1(vy/vx) = tan−1(10/20) = tan−1(½) ≈ 26.6° below the horizontal. (b) After release the only force is gravity, so the acceleration is 10 m s−2 vertically downward.

4.12 A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s−1. What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

ANSWER (a) At an extreme position the bob is momentarily at rest (speed = 0). When the string is cut, with no initial velocity the bob simply falls vertically downward under gravity (a straight vertical path). (b) At the mean position the bob has a horizontal velocity of 1 m s−1. When the string is cut, it becomes a projectile — horizontal velocity plus vertical fall — so it follows a parabolic path.

4.13 A man of mass 70 kg stands on a weighing scale in a lift which is moving (a) upwards with a uniform speed of 10 m s−1, (b) downwards with a uniform acceleration of 5 m s−2, (c) upwards with a uniform acceleration of 5 m s−2. What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

ANSWER The scale reads the normal force N it exerts on the man (equal and opposite to the man’s force on it). m = 70 kg, g = 10 m s−2. (a) Uniform speed → a = 0: N − mg = 0 → N = mg = 70 × 10 = 700 N. Reading = 700/10 = 70 kg. (b) Downward acceleration 5 m s−2: mg − N = ma → N = m(g − a) = 70 × (10 − 5) = 350 N. Reading = 35 kg. (c) Upward acceleration 5 m s−2: N − mg = ma → N = m(g + a) = 70 × (10 + 5) = 1050 N. Reading = 105 kg. (d) Free fall: a = g downward, so N = m(g − g) = 0. The scale reads zero (apparent weightlessness).

4.14 Figure 4.16 shows the position-time graph of a particle of mass 4 kg. What is the (a) force on the particle for t < 0, t > 4 s, 0 < t < 4 s? (b) impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)

ANSWER In the NCERT graph the particle is at x = 0 for t < 0, moves along a straight line (uniform velocity) from x = 0 to x = 3 m during 0 to 4 s, and then stays at x = 3 m for t > 4 s. (a) In each interval the position-time graph is straight, so the velocity is constant and the acceleration is zero. Hence the force is zero in all three intervals (t < 0, 0 < t < 4 s, and t > 4 s). (b) Velocity during 0–4 s = 3 m / 4 s = 0.75 m s−1. Impulse at t = 0 = change in momentum = m(v − 0) = 4 × 0.75 = 3 kg m s−1 (the particle starts moving). Impulse at t = 4 s = m(0 − v) = 4 × (−0.75) = −3 kg m s−1 (the particle stops).

4.15 Two bodies of masses 10 kg and 20 kg respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force F = 600 N is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

ANSWER Let A = 10 kg, B = 20 kg. The surface is smooth, so both masses move together with the same acceleration a = F/(mA + mB) = 600/30 = 20 m s−2. (i) Force applied to A (10 kg): the string pulls B (20 kg). Tension provides B’s acceleration: T = mB a = 20 × 20 = 400 N. (ii) Force applied to B (20 kg): the string pulls A (10 kg): T = mA a = 10 × 20 = 200 N. (Check (i): F − T = mA a → 600 − 400 = 200 = 10 × 20 ✓.)

4.16 Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

ANSWER For the heavier mass (12 kg, moving down): 12g − T = 12a. For the lighter mass (8 kg, moving up): T − 8g = 8a. Adding: (12 − 8)g = (12 + 8)a → 4 × 10 = 20 a → a = 40/20 = 2 m s−2. Tension: T = 8(g + a) = 8 × (10 + 2) = 96 N. (Check with heavier mass: 12(g − a) = 12 × 8 = 96 N ✓.)

4.17 A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

ANSWER The nucleus is at rest, so its initial momentum is zero. No external force acts during disintegration, so by conservation of momentum the total final momentum is also zero: p1 + p2 = 0, i.e. p2 = −p1. Two non-zero momentum vectors can add to zero only if they are equal in magnitude and opposite in direction. Therefore the two product nuclei must move in exactly opposite directions (with momenta of equal magnitude).

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

ANSWER Consider one ball moving at +6 m s−1; after rebound it moves at −6 m s−1. Impulse = change in momentum = m(v − u) = 0.05 × (−6 − 6) = 0.05 × (−12) = −0.6 kg m s−1. So the magnitude of impulse on each ball = 0.6 kg m s−1; the two impulses are equal in magnitude and opposite in direction.

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

ANSWER Initially both are at rest, so total momentum = 0. By conservation of momentum: mgun V + mshell v = 0. 100 × V = 0.020 × 80 (in magnitude) → V = 1.6/100 = 0.016 m s−1 = 1.6 cm s−1, in the direction opposite to the shell (recoil).

4.20 A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 54 km/h. What is the impulse imparted to the ball? (Mass of the ball is 0.15 kg.)

ANSWER Speed: u = 54 km/h = 54 × (5/18) = 15 m s−1 (unchanged in magnitude). The ball is deflected through 45°. The impulse lies along the bisector of the incoming and outgoing directions. Its magnitude is the change in momentum: |Δp| = 2 m u cos(θ/2), where θ = 45°. |Δp| = 2 × 0.15 × 15 × cos 22.5° = 4.5 × 0.9239 = 4.2 kg m s−1, directed along the bisector of the initial and final directions.

4.21 A stone of mass 0.25 kg tied to the end of a string is whirled round in a circle of radius 1.5 m with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of 200 N?

ANSWER Frequency: 40 rev/min = 40/60 = 2/3 rev s−1. Linear speed: v = 2πR × (40/60) = 2π × 1.5 × (2/3) = 2π = 6.283 m s−1. Tension provides the centripetal force: T = m v2/R = 0.25 × (2π)2/1.5 = 0.25 × 39.48/1.5 = 6.6 N. Maximum speed: Tmax = m vmax2/R → vmax = √(Tmax R / m) = √(200 × 1.5 / 0.25) = √1200 = 34.6 m s−1 ≈ 35 m s−1.

4.22 If, in Exercise 4.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks: (a) the stone moves radially outwards, (b) the stone flies off tangentially from the instant the string breaks, (c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle?

ANSWER When the string breaks, no force acts on the stone (in the horizontal plane), so by Newton’s first law it continues to move in a straight line with the velocity it had at that instant. In circular motion the instantaneous velocity is always tangent to the circle. Therefore the correct option is (b) the stone flies off tangentially from the instant the string breaks.

4.23 Explain why (a) a horse cannot pull a cart and run in empty space, (b) passengers are thrown forward from their seats 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.

ANSWER (a) The horse pushes the ground backward; the ground pushes the horse-cart system forward (the reaction). In empty space there is no ground to push against, and the mutual forces between horse and cart are internal and cancel (third law), so the system cannot accelerate. (b) When the bus stops suddenly, the seat (and lower body) stop with it, but the upper body tends to keep moving forward due to inertia of motion, so passengers are thrown forward. (c) When pushing at an angle, the downward component of the applied force adds to the weight, increasing the normal reaction N and hence the friction (f ∝ N). When pulling at an angle, the upward component reduces N and friction. Less friction means less effort, so pulling is easier. (d) By drawing his hands back, the cricketer increases the time of contact Δt over which the ball’s momentum is reduced to zero. Since F = Δp/Δt, a larger Δt means a smaller force on the hands, so they do not get hurt.

Extra Practice Questions

Short Answer Type Questions

Q1. State Newton’s second law of motion and derive F = ma from it.

ANSWERThe rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force. Thus F ∝ dp/dt, so F = k dp/dt. For constant mass, dp/dt = d(mv)/dt = m dv/dt = ma. Choosing k = 1 (which defines the newton), we get F = ma.

Q2. Define impulse and state its SI unit. Why is it useful?

ANSWERImpulse = force × time of action = change in momentum (J = FΔt = Δp). Its SI unit is N s (or kg m s−1). It is useful when a large force acts for a very short time (as in collisions), where F and Δt are hard to measure separately but their product, the change in momentum, is easily found.

Q3. What is the angle of repose? How is it related to the coefficient of static friction?

ANSWERThe angle of repose θmax is the maximum angle of an inclined plane at which a body just begins to slide down on its own. At this angle, mg sinθ = (fs)max = μs mg cosθ, giving tanθmax = μs. It is independent of the mass of the body.

Q4. A force of 5 N gives a mass m1 an acceleration of 10 m s−2 and a mass m2 an acceleration of 20 m s−2. What acceleration would it give if both masses were tied together?

ANSWERm1 = F/a1 = 5/10 = 0.5 kg; m2 = 5/20 = 0.25 kg. Combined mass = 0.75 kg. Acceleration a = F/(m1 + m2) = 5/0.75 = 6.67 m s−2.

Q5. Why does a gun recoil when a bullet is fired from it?

ANSWERBefore firing, the total momentum of the gun-bullet system is zero. When the bullet is fired forward with momentum pb, conservation of momentum requires the gun to gain an equal and opposite momentum pg = −pb, so the gun moves backward (recoils). Because the gun is much heavier, its recoil speed is small.

Long Answer Type Questions

Q1. State and prove the law of conservation of linear momentum using Newton’s laws of motion.

ANSWERThe law states that the total momentum of an isolated system (one with no net external force) remains constant. Consider two bodies A and B colliding. During contact, A exerts a force FBA on B and B exerts FAB on A. By the third law, FAB = −FBA. By the second law, FABΔt = p′A − pA and FBAΔt = p′B − pB. Adding and using FAB = −FBA gives (p′A − pA) + (p′B − pB) = 0, i.e. p′A + p′B = pA + pB. Thus the total momentum after collision equals the total momentum before, regardless of whether the collision is elastic or inelastic.

Q2. Derive an expression for the maximum safe speed of a car on a banked road of angle θ with coefficient of friction μs.

ANSWEROn a banked road the forces are weight mg, normal reaction N and friction f (≤ μsN, directed down the slope at maximum speed). Vertically: N cosθ = mg + f sinθ. Horizontally (centripetal): N sinθ + f cosθ = mv2/R. Putting f = μsN and eliminating N gives vmax = √[Rg(μs + tanθ)/(1 − μs tanθ)]. For μs = 0, this reduces to the optimum (no-friction) speed vo = √(Rg tanθ), at which the tyres suffer least wear.

Q3. Explain, with the help of a free-body diagram, the apparent weight of a person standing in a lift accelerating upward, accelerating downward, and in free fall.

ANSWERThe apparent weight equals the normal force N from the floor (what a scale reads). The forces on the person are weight mg (down) and N (up). (1) Lift accelerating up with a: N − mg = ma → N = m(g + a) — apparent weight is greater than true weight. (2) Lift accelerating down with a: mg − N = ma → N = m(g − a) — apparent weight is less. (3) Free fall (a = g): N = m(g − g) = 0 — the person feels weightless. At constant velocity (a = 0), N = mg and apparent weight equals true weight.

MCQs & Assertion–Reason

1. The SI unit of force, the newton, is equivalent to:

(a) kg m s−1    (b) kg m s−2    (c) kg m2 s−2    (d) kg m2 s−1

2. A body of mass 2 kg moving at 5 m s−1 is brought to rest by a force in 2 s. The average force is:

(a) 2.5 N    (b) 5 N    (c) 10 N    (d) 20 N

3. Newton’s first law of motion gives the concept of:

(a) energy    (b) momentum    (c) inertia    (d) acceleration

4. Action and reaction forces in Newton’s third law:

(a) act on the same body    (b) act on different bodies    (c) are unequal    (d) cancel each other out

5. The impulse imparted to a body equals its change in:

(a) velocity    (b) acceleration    (c) momentum    (d) kinetic energy

6. For a body on the verge of sliding on a horizontal surface, the maximum static friction equals:

(a) μk N    (b) μs N    (c) mg    (d) zero

7. The maximum speed of a car on an unbanked circular road of radius R is:

(a) √(μs R g)    (b) μs R g    (c) √(R g tanθ)    (d) √(R g/μs)

8. When a man jumps out of a boat to the shore, the boat moves backward. This illustrates:

(a) inertia of rest    (b) Newton’s first law    (c) conservation of momentum    (d) conservation of energy

9. In a lift moving down with acceleration equal to g, the apparent weight of a person is:

(a) equal to mg    (b) greater than mg    (c) zero    (d) 2mg

10. A stone whirled in a horizontal circle flies off, when the string breaks, along:

(a) the radius outward    (b) the radius inward    (c) the tangent at that point    (d) a parabola

Answer key: 1-(b), 2-(b), 3-(c), 4-(b), 5-(c), 6-(b), 7-(a), 8-(c), 9-(c), 10-(c).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: A body in uniform circular motion has a net non-zero force acting on it.

Reason: Uniform circular motion involves a continuous change in the direction of velocity, which requires a centripetal force.

A-R 2. Assertion: Action and reaction forces cancel each other, so no body can ever accelerate.

Reason: Action and reaction are equal in magnitude and opposite in direction.

A-R 3. Assertion: A passenger in a lift falling freely under gravity feels weightless.

Reason: In free fall the normal force on the passenger by the floor becomes zero.

A-R 4. Assertion: It is easier to pull a lawn roller than to push it.

Reason: Pulling reduces the normal reaction and hence the force of friction, while pushing increases it.

A-R 5. Assertion: The momentum of an isolated system of particles is conserved.

Reason: An external force is always required to change the total momentum of a system.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

  • Treating mv2/R (centripetal force) or “ma” as an extra force in the free-body diagram — they are the required net force, provided by real forces like tension, friction or gravity.
  • Thinking action and reaction cancel — they act on different bodies, so they never cancel on the same body.
  • Writing fs = μsN in every case — static friction is self-adjusting (fs ≤ μsN); use the equality only at the point of impending sliding.
  • Forgetting unit conversion: 36 km/h = 10 m s−1, 54 km/h = 15 m s−1 (multiply km/h by 5/18).
  • Assuming a body momentarily at rest (e.g. at the top of its path) has zero force — gravity still acts, so a = g there.
  • Ignoring the driver’s mass or other parts when finding the total mass in F = ma (see Ex. 4.8).

Exam Tips

How to score full marks in this chapter

Always start a numerical with a clear free-body diagram and resolve forces along convenient axes. Convert all data to SI units first, and use g = 10 m s−2 when the exercise says so. Write the working step by step with units at every stage and box the final answer with its unit and direction. For conceptual “explain why” questions (4.1–4.4, 4.17, 4.22, 4.23), name the exact law used — first law (inertia), second law (F = ma), third law, or conservation of momentum. Cross-check answers with a second method where possible (e.g. compute tension from the other mass in a pulley problem).

Frequently Asked Questions

What is Class 11 Physics Chapter 4 Laws of Motion about?

Chapter 4 introduces Newton’s three laws of motion. It corrects Aristotle’s fallacy with Galileo’s law of inertia, defines momentum and impulse, and develops the second law F = ma and the third law (action–reaction). It also covers conservation of momentum, equilibrium of a particle, friction (static, kinetic, rolling) and circular motion on level and banked roads, finishing with free-body-diagram problem solving.

How many exercises are there in Class 11 Physics Chapter 4?

The NCERT (rationalised, 2026–27) Exercises for Chapter 4 contain 23 numbered questions, 4.1 to 4.23, including conceptual questions and numericals. All of them are solved step by step on this page.

Why does a cricketer pull his hands back while catching a ball?

Drawing the hands back increases the time Δt over which the ball’s momentum is reduced to zero. Since force = change in momentum / time (F = Δp/Δt), a longer time gives a smaller force on the hands, so the catch does not hurt.

Are these Class 11 Physics Chapter 4 solutions free?

Yes. All ClearStudy NCERT Solutions for Class 11 Physics are free and follow the official NCERT textbook for the 2026–27 session, with every exercise solved and verified.

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