Law of Motion Chapter 5 Class 11 Physics NCERT Textbook PDF

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NCERT Class 11 Physics Textbook Chapter 5 With Answer PDF Free Download

Law of Motion

Chapter 5: Law of Motion

5.1 Introduction

In the preceding Chapter, our concern was to describe the motion of a particle in space quantitatively.

We saw that uniform motion needs the concept of velocity alone whereas non-uniform motion requires the concept of acceleration in addition.

So far, we have not asked the question as to what governs the motion of bodies. In this chapter, we turn to this basic question.

Let us first guess the answer based on our common experience. To move a football at rest, someone must kick it.

To throw a stone upwards, one has to give it an upward push. A breeze causes the branches of a tree to swing; a strong wind can even move heavy objects.

A boat moves in a flowing river without anyone rowing it. Clearly, some external agency is needed to provide force to move a body from rest.

Likewise, an external force is needed also to retard or stop motion. You can stop a ball from rolling down an inclined plane by applying a force against the direction of its motion.

In these examples, the external agency of force (hands, wind, stream, etc) is in contact with the object. This is not always necessary.

A stone released from the top of a building accelerates downward due to the gravitational pull of the earth. A bar magnet can attract an iron nail from a distance.

This shows that external agencies (e.g. gravitational and magnetic forces ) can exert force on a body even from a distance.

In short, a force is required to put a stationary body in motion or stop a moving body, and some external agency is needed to provide this force. The external agency may or may not be in contact with the body.

5.2 Aristotle’s Fallacy

The question posed above appears to be simple. However, it took ages to answer it.

Indeed, the correct answer to this question given by Galileo in the seventeenth century was the foundation of Newtonian mechanics, which signaled the birth of modern science.

The Greek thinker, Aristotle (384 B.C– 322 B.C.), held the view that if a body is moving, something external is required to keep it moving.

According to this view, for example, an arrow shot from a bow keeps flying since the air behind the arrow keeps pushing it.

The view was part of an elaborate framework of ideas developed by Aristotle on the motion of bodies in the universe.

Most of the Aristotelian ideas on motion are now known to be wrong and need not concern us. For our purpose here, the Aristotelian law of motion may be phrased thus: An external force is required to keep a body in motion.

Language English
No. of Pages25
PDF Size908 KB

NCERT Solutions Class 11 Physics Chapter 5 Law of Motion

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.


(a)The raindrop is falling at a constant speed. Therefore, acceleration will become zero. When acceleration is zero, the force acting on the drop will become zero since F = ma.

(b) The cork is floating on water, which means the weight of the cork is balanced by the upthrust. Therefore, the net force on the cork will be zero

(c) Since the car moves with a constant velocity, the acceleration becomes zero. Therefore, the force will be zero.

(d) The net force acting on the high-speed electron will be zero since the electron is far from the material objects and free of electric and magnetic fields.

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 Solutions change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance


(a) During the upward motion of the pebble, the acceleration due to gravity acts downwards, so the magnitude of the force on the pebble is

F = mg = 0.05 kg x 10 ms-2 = 0.5 N

The direction of the force is downwards

(b) During the downward motion also the magnitude of the force will be equal to 0.5 N and the force acts downwards

(c) If the pebble is thrown at an angle of 45° with the horizontal direction, it will have both horizontal and vertical components of the velocity.

At the highest point, the vertical component of velocity will be zero but the horizontal component of velocity will remain throughout the motion of the pebble.

This component will not have any effect on the force acting on the pebble. The direction of the force acting on the pebble will be downwards and the magnitude will be 0.5 N because no other force other than acceleration acts on the pebble.

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 with1 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.


(a) Mass of stone = 0.1 kg

Acceleration = 10 ms-2

Net force, F = mg = 0.1 x 10 = 1.0 N

The force acts vertically downwards

(b) The train moves at a constant velocity. Therefore the acceleration will be equal to zero. So there is no force acting on the stone due to the motion of the train. Therefore, the force acting on the stone will remain the same (1.0 N)

(c) When the train accelerates at 1m/s2, the stone experiences an additional force of F’ = ma = 0.1 x 1 = 0.1 N. The force acts in the horizontal direction.

But as the stone is dropped, the force F’ no longer acts and the net force acting on the stone F = mg = 0.1 x 10 = 1.0 N. (vertically downwards).

(d) As the stone is lying on the train floor, its acceleration will be the same as that of the train.

Therefore, the magnitude of the force acting on the stone, F = ma = 0.1 x 1 = 0.1 N.

It acts along the direction of motion of the train.

Law of Motion Textbook With Solutions PDF Free Download

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