Laws of Motion | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Laws of Motion – this guide gives you a concise, exam-ready overview of Laws of Motion from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
4.4 Newton's first law of motion
Newton's First Law of Motion restates Galileo's law of inertia precisely: A body remains at rest or moves with uniform velocity in a straight line unless acted upon by a net external force. This implies that if the net external force on a body is zero, its acceleration is zero. The law distinguishes two situations: (1) when the net external force is zero, the acceleration is zero, and (2) when the acceleration is zero, the net external force must be zero. Examples include a spaceship in interstellar space moving with constant velocity due to absence of external forces, and a book resting on a table where the normal force balances the weight, resulting in zero net force. The section also explains the concept of inertia through everyday experiences, such as the jerk felt when a bus starts or stops suddenly, due to the body's resistance to change in motion. The law emphasizes that external forces cause acceleration, and internal forces cannot change the motion of the whole body.
📊 Diagram: See figure_3: (a); See figure_4: (b)
🔗 Connection: This section prepares the ground for Newton's second law of motion, which relates force and acceleration quantitatively.
Table on page 19 (6×5)
| Quantity | Symbol | Units | Dimensions | Remarks |
|---|---|---|---|---|
| Momentum | p | kg m s-1or N s | [MLT-1] | Vector |
| Force | F | N | [MLT-2] | F = m a Second Law |
| Impulse | kg m s-1or N s | [MLT-1] | Impulse = force × time = change in momentum | |
| Static friction | fS | N | [MLT-3] | fs ≤ μsN |
| Kinetic friction | fK | N | [MLT-3] | fs = μsN |
Frequently asked questions
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 10g 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.
Solution: (a) Since the drop of rain is falling with constant speed, acceleration is zero. Hence, net force = 0. (b) Cork floating on water is at rest, so net force = 0. (c) Kite held stationary means velocity is zero and not changing, so net force = 0. (d) Car moving with constant velocity means acceleration is zero, so net force = 0. (e) Electron moving at high speed in free space with no fields or objects experiences no force, so net force = 0.
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.
Solution: Mass, m = 0.05 kg Acceleration due to gravity, g = 10 m/s²
(a) During upward motion, the only force acting is gravity downward. Net force = mg downward = 0.05 × 10 = 0.5 N downward.
(b) During downward motion, gravity acts downward. Net force = mg downward = 0.5 N downward.
(c) At the highest point, velocity is zero but acceleration due to gravity still acts downward. Net force = mg downward = 0.5 N downward.
If thrown at 45°, the vertical component of motion is similar, so net for
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/s², (d) lying on the floor of a train which is accelerating with 1 m/s², the stone being at rest relative to the train. Neglect air resistance throughout.
Solution: Mass, m = 0.1 kg Acceleration due to gravity, g = 10 m/s²
(a) Dropped from stationary train: Net force = mg downward = 0.1 × 10 = 1 N downward.
(b) Dropped from train moving at constant velocity: Velocity constant means no acceleration, so net force = mg downward = 1 N downward.
(c) Dropped from accelerating train: Just after dropping, stone has horizontal velocity equal to train's velocity but no horizontal acceleration. Net force = mg downward = 1 N downward.
(d) Stone lying on a
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 - \frac{mv^2}{l}, (iii) T + \frac{mv^2}{l}, (iv) 0 T is the tension in the string. [Choose the correct alternative].
Answer: (i) T
Explanation: The net force acting on the particle moving in a circle is the centripetal force directed towards the centre. This force is provided by the tension T in the string. Hence, net force = T.
Options (ii) and (iii) incorrectly add or subtract the centripetal force term from T. Option (iv) zero force is incorrect as centripetal force is required for circular motion.
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