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.
4.5 Newton's second law of motion
Newton's Second Law of Motion generalizes the relationship between force and motion when a net external force acts on a body. It states that the rate of change of momentum of a body is directly proportional to the applied force and occurs in the direction of the force. Momentum (p) is defined as the product of mass (m) and velocity (v), p = m v, and is a vector quantity. The law can be expressed mathematically as F = d p/d t, where F is the net external force and d p/d t is the time derivative of momentum. For constant mass, this reduces to F = m a, where a is acceleration. The section discusses the importance of momentum in understanding force effects, including examples such as heavier vehicles requiring more force to accelerate, and the direction of force affecting velocity components. It also introduces impulse as the product of force and time interval, equal to the change in momentum, useful for analyzing forces acting over short durations. The vector nature of force and momentum is emphasized, and the law's applicability to systems of particles is noted.
📊 Diagram: See figure_5: Fig. 4.3 Force not only depends on the change in momentum but also on how fast the change is brought about. A seasoned cricketer draws in his hands during a catch, allowing greater time for the ball to stop and hence requires a smaller force.; See figure_6: Fig. 4.4 Force is necessary for changing the direction of momentum, even if its magnitude is constant. We can feel this while rotating a stone in a horizontal circle with uniform speed by means of a string.; See figure_7: Fig. 4.5 Acceleration at an instant is determined by the force at that instant. The moment after a stone is dropped out of an accelerated train, it has no horizontal acceleration or force, if air resistance is neglected. The stone carries no memory of its acceleration with the train a moment ago.
🔗 Connection: This section naturally leads to Newton's third law of motion, which explains the mutual nature of forces.
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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