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Laws of Motion — NCERT Class 11 Physics

Complete NCERT Solutions for CBSE Class 11 Physics Chapter 5 — all intext and chapter-end exercise questions on Newton's laws, friction, circular motion dynamics, free body diagrams answered clearly.

Updated 11 June 202615 min read

Exercise Solutions Available

Step-by-step answers for all textbook questions in this chapter

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How to Use This Page

Draw a free body diagram for every problem. Identify all forces, choose a coordinate system, and write Newton's second law (F = ma) along each axis. Then substitute values and compute the answer with correct units.

Key Concepts (Quick Recap)

  • Newton's First Law: A body continues in its state of rest or uniform motion unless acted upon by an external force (Law of Inertia)
  • Newton's Second Law: F = ma (net force = mass × acceleration); impulse = change in momentum
  • Newton's Third Law: Every action has an equal and opposite reaction; action-reaction act on different bodies
  • Static friction: f_s ≤ μ_s N (adjusts up to a maximum)
  • Kinetic friction: f_k = μ_k N (constant once sliding begins)
  • Rolling friction: f_r = μ_r N (smallest of the three)
  • Centripetal force: F_c = mv²/r (toward the centre of circular path)
  • Banking of roads: tan θ = v²/rg (for ideal banking with no friction)
  • Conservation of momentum: m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂' (when no external force)

Intext Questions

Page 88

Q1. 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 held stationary in the sky (d) a car moving with a constant velocity of 30 km h⁻¹ on a rough road (e) a high-speed electron in space far from all material objects and free of electric and magnetic fields.
Answer

All of these have zero net force because each is in equilibrium (constant velocity or stationary): (a) Constant speed → a = 0 → F = 0 (b) Floating → weight = buoyant force → F_net = 0 (c) Stationary → F_net = 0 (d) Constant velocity → a = 0 → F_net = 0 (e) No forces act → F_net = 0

Q2. 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) at the highest point (c) during its downward motion (d) just before hitting the ground. Neglect air resistance.
Answer

In all cases, only gravity acts (no air resistance): F = mg = 0.05 × 9.8 = 0.49 N downward in every case (a), (b), (c), (d)

Q3. 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 train running at a constant velocity of 36 km h⁻¹ (b) just after it is dropped from the window of a train accelerating at 1 m s⁻².
Answer

(a) Constant velocity train → no pseudo force. Just after release, only gravity acts. F = mg = 0.1 × 9.8 = 0.98 N downward

(b) Accelerating train → at the instant of release, the stone has the train's horizontal velocity but no horizontal force acts on it (only gravity). Just after release, the net force on the stone is: F = mg = 0.1 × 9.8 = 0.98 N downward (horizontal force is zero; the acceleration of the train doesn't affect the stone once released)

Page 94

Q4. 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

F = −50 N, m = 20 kg, u = 15 m s⁻¹, v = 0

a = F/m = −50/20 = −2.5 m s⁻²

v = u + at → 0 = 15 + (−2.5)t → t = 15/2.5 = 6 s

Q5. A constant force acting on a body of mass 5 kg changes its speed from 2 m s⁻¹ to 3.5 m s⁻¹ in 3 s. The direction of motion remains unchanged. What is the magnitude and direction of the force?
Answer

m = 5 kg, u = 2 m s⁻¹, v = 3.5 m s⁻¹, t = 3 s

a = (v − u)/t = (3.5 − 2)/3 = 0.5 m s⁻²

F = ma = 5 × 0.5 = 2.5 N in the direction of motion

Page 98

Q6. A rocket with a lift-off mass of 20,000 kg is blasted upwards with an initial acceleration of 5.0 m s⁻². Calculate the initial thrust of the blast.
Answer

m = 20000 kg, a = 5.0 m s⁻² (upward)

Net upward force = ma = 20000 × 5 = 100000 N Weight = mg = 20000 × 9.8 = 196000 N

Thrust − Weight = ma Thrust = ma + mg = 100000 + 196000 = 296000 N = 2.96 × 10⁵ N

Q7. A body of mass 5 kg is acted upon by two perpendicular forces of 8 N and 6 N. Find the magnitude and direction of the acceleration.
Answer

F_net = √(8² + 6²) = √(64 + 36) = √100 = 10 N

a = F_net/m = 10/5 = 2 m s⁻²

Direction: tan θ = 6/8 = 0.75 → θ = tan⁻¹(0.75) = 36.87° from the 8 N force (toward the 6 N force)

Page 103

Q8. The driver of a three-wheeler moving with a speed of 36 km h⁻¹ sees a child standing in the middle of the road and brings his vehicle to rest in 4 s just in time to save the child. What is the average retarding force on the vehicle? Mass of the three-wheeler = 400 kg.
Answer

u = 36 km h⁻¹ = 10 m s⁻¹, v = 0, t = 4 s, m = 400 kg

a = (v − u)/t = (0 − 10)/4 = −2.5 m s⁻²

F = ma = 400 × (−2.5) = −1000 N (retarding force = 1000 N)

Q9. A 70 kg man stands on a weighing scale in a lift that is moving (a) upward with uniform speed 10 m s⁻¹ (b) downward with uniform acceleration 5 m s⁻² (c) upward with uniform acceleration 5 m s⁻². What is the reading on the scale in each case?
Answer

(a) Uniform speed → a = 0 → N = mg = 70 × 9.8 = 686 N (scale reads 70 kg)

(b) Downward acceleration a = 5 m s⁻²: N = m(g − a) = 70(9.8 − 5) = 70 × 4.8 = 336 N (scale reads 336/9.8 = 34.3 kg)

(c) Upward acceleration a = 5 m s⁻²: N = m(g + a) = 70(9.8 + 5) = 70 × 14.8 = 1036 N (scale reads 1036/9.8 = 105.7 kg)

Chapter-End Exercise

Q5.1. 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 skilfully held stationary in the sky (d) a car moving with a constant velocity of 30 km h⁻¹ 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 all five cases, the net force is zero. (a) Constant speed means zero acceleration → zero net force. (b) Floating means buoyancy equals weight. (c) Stationary means all forces balance. (d) Constant velocity means zero acceleration. (e) No forces act on the electron.

Q5.2. 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) at the highest point (c) during its downward motion. Neglect air resistance.
Answer

In all cases, the only force is gravity: F = mg = 0.05 × 9.8 = 0.49 N, directed downward (always, regardless of direction of motion)

Q5.3. 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 reading on the scale in each case? What would happen if the lift mechanism failed and it hurtled down freely under gravity?
Answer

(See intext Q9 above for the detailed calculation.)

(a) 70 kg (686 N) — uniform speed, no extra force (b) 34.3 kg (336 N) — downward acceleration reduces apparent weight (c) 105.7 kg (1036 N) — upward acceleration increases apparent weight

If the lift falls freely: a = g downward → N = m(g − g) = 0. The scale reads zero (weightlessness).

Q5.4. 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) the 10 kg body (ii) the 20 kg body. What is the tension in the string in each case?
Answer

Total mass = 30 kg, a = F/m = 600/30 = 20 m s⁻²

(i) Force on 10 kg body: F = 600 N applied on 10 kg, tension T pulls back on 10 kg For 10 kg: F − T = 10a → 600 − T = 10 × 20 → T = 600 − 200 = 400 N

(ii) Force on 20 kg body: F = 600 N applied on 20 kg, tension T pulls back on 20 kg For 20 kg: F − T = 20a → 600 − T = 20 × 20 → T = 600 − 400 = 200 N

Q5.5. 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. (g = 10 m s⁻²)
Answer

m₁ = 8 kg, m₂ = 12 kg, m₂ > m₁ so m₂ goes down, m₁ goes up

a = (m₂ − m₁)g/(m₁ + m₂) = (12 − 8)(10)/(8 + 12) = 40/20 = 2 m s⁻²

T = 2m₁m₂g/(m₁ + m₂) = 2(8)(12)(10)/(20) = 1920/20 = 96 N

Q5.6. 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

By conservation of momentum: initial momentum = 0 (nucleus at rest)

After disintegration: m₁v₁ + m₂v₂ = 0 → v₂ = −(m₁/m₂)v₁

The velocities v₁ and v₂ have opposite signs, meaning the two nuclei move in opposite directions. This is a consequence of momentum conservation.

Q5.7. 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 by the other?
Answer

Before collision: v₁ = +6 m s⁻¹, v₂ = −6 m s⁻¹ After collision: v₁' = −6 m s⁻¹, v₂' = +6 m s⁻¹ (rebound with same speed)

Impulse on ball 1 = m(v₁' − v₁) = 0.05(−6 − 6) = 0.05 × (−12) = −0.6 N s (in the direction of v₂) Impulse on ball 2 = m(v₂' − v₂) = 0.05(6 − (−6)) = 0.05 × 12 = 0.6 N s (in the direction of v₁)

Magnitude of impulse on each ball = 0.6 N s

Q5.8. 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

By conservation of momentum: 0 = m_shell × v_shell + m_gun × v_gun

v_gun = −(m_shell × v_shell)/m_gun = −(0.020 × 80)/100 = −1.6/100 = −0.016 m s⁻¹

The gun recoils at 0.016 m s⁻¹ in the opposite direction to the shell.

Q5.9. 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 if the string can withstand a maximum tension of 200 N?
Answer

m = 0.25 kg, r = 1.5 m, f = 40 rev/min = 40/60 = 2/3 rev s⁻¹ ω = 2πf = 2π(2/3) = 4π/3 rad s⁻¹

T = mω²r = 0.25 × (4π/3)² × 1.5 = 0.25 × (16π²/9) × 1.5 = 0.25 × 16π² × 1.5/9 = 0.25 × 26.32 = 6.58 N (approximately)

Let me recalculate: ω² = (4π/3)² = 16π²/9 = 16(9.87)/9 = 157.9/9 = 17.55 T = 0.25 × 17.55 × 1.5 = 0.25 × 26.32 = 6.58 N

For maximum tension T_max = 200 N: T_max = mv²/r → v² = T_max × r/m = 200 × 1.5/0.25 = 1200 v_max = √1200 = 34.6 m s⁻¹

Q5.10. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed to prevent him from slipping down? (g = 10 m s⁻²)
Answer

For the man to not slip: friction ≥ weight f_s = μ_s N ≥ mg

Where N = centripetal force = mω²r

μ_s × mω²r ≥ mg ω² ≥ g/(μ_s r) = 10/(0.15 × 3) = 10/0.45 = 22.22 ω_min = √22.22 = 4.71 rad s⁻¹

In rev/min: ω = 4.71 × 60/(2π) = 282.6/6.28 = 45 rev/min

Q5.11. Find the maximum speed at which a car can turn round a curve of radius 30 m on a level road if the coefficient of friction between the tyres and the road is 0.4. (g = 10 m s⁻²)
Answer

Centripetal force provided by friction: f = μ_s mg = mv²/r

v² = μ_s gr = 0.4 × 10 × 30 = 120 v = √120 = 10.95 m s⁻¹ ≈ 11 m s⁻¹

Q5.12. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km h⁻¹. The mass of the train is 5 × 10⁵ kg. What provides the centripetal force? What is the role of the flanges? What is the centripetal force?
Answer

v = 54 km h⁻¹ = 15 m s⁻¹, r = 30 m, m = 5 × 10⁵ kg

The centripetal force is provided by the friction between the train wheels and the rails (lateral friction). The flanges on the wheels prevent the train from slipping outward by providing additional lateral support when friction is insufficient.

F_c = mv²/r = 5 × 10⁵ × (15)²/30 = 5 × 10⁵ × 225/30 = 5 × 10⁵ × 7.5 = 3.75 × 10⁶ N

Q5.13. A circular racetrack of radius 300 m is banked at an angle of 15°. If the coefficient of friction between the wheels of a race-car and the road is 0.2, what is the (a) optimum speed to avoid wear and tear (b) maximum permissible speed?
Answer

r = 300 m, θ = 15°, μ = 0.2

(a) Optimum speed (no friction needed): v₀ = √(rg tan θ) = √(300 × 10 × tan 15°) tan 15° = 0.268 v₀ = √(300 × 10 × 0.268) = √804 = 28.4 m s⁻¹

(b) Maximum speed: v_max = √(rg(μ + tan θ)/(1 − μ tan θ)) = √(300 × 10 × (0.2 + 0.268)/(1 − 0.2 × 0.268)) = √(3000 × 0.468/(1 − 0.0536)) = √(3000 × 0.468/0.9464) = √(3000 × 0.4946) = √1483.8 = 38.5 m s⁻¹

Q5.14. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley surface is 0.18. The trolley accelerates from rest with 0.5 m s⁻² for 20 s and then moves with uniform velocity. What is the maximum acceleration of the trolley so that the block does not slip?
Answer

Maximum friction = μ_s × mg = 0.18 × 15 × 9.8 = 26.46 N

For no slipping: ma ≤ μ_s mg → a ≤ μ_s g = 0.18 × 9.8 = 1.76 m s⁻²

Since the trolley's acceleration (0.5 m s⁻²) < 1.76 m s⁻², the block does not slip.

Maximum acceleration of trolley for no slipping = 1.76 m s⁻²

Q5.15. A block of mass 2 kg is placed on the floor (μ_s = 0.4). A force of 7 N is applied. Does the block move? If not, what is the frictional force?
Answer

Maximum static friction = μ_s × N = 0.4 × 2 × 9.8 = 7.84 N

Applied force = 7 N < 7.84 N (maximum static friction)

The block does not move. The frictional force equals the applied force = 7 N (static friction adjusts to match the applied force up to its maximum).

Internal Links

  • [Chapter 5 Guide — Laws of Motion](/school/cbse/class/11/physics/cbse-laws-of-motion)
  • [All Class 11 Physics Solutions](/school/cbse/class/11/physics/solutions)
  • [Class 11 Physics Subject Guide](/school/cbse/class/11/physics)

Frequently Asked Questions

Newton's first law (law of inertia) states that a body at rest stays at rest, and a body in uniform motion stays in uniform motion, unless acted upon by an external unbalanced force. Inertia is the tendency of a body to resist change in its state of motion.