assignment on laws of motion class 11

CBSE Class 11 Physics Revision Notes

All the CBSE Class 11 Physics Notes are also available in the PDF format on the official website of eSaral. Once you download these notes in PDF, you do not need an internet connection to go through the notes for a quick revision. These 11th class 11 Physics notes pdf can also be printed as a hard copy to have an additional mode of revising all your chapters easily. These notes are designed keeping in mind the understanding level of Class 11th students; so you will get clarification at the concept level so that they do not need to mug up the solutions. The Physics Notes for Class 11 by our expert teacher prepares you better for your exams as you will be able to manage your time well and deal with stressful exam environments more efficiently with these solutions in your hands.

CBSE Class 11 Physics Weightage 2024-25

Overview of Revision Notes CBSE Class 11 Physics

Students of Class 11 have Physics as one of their main subjects for their boards. The subject of Physics introduces students to different concepts such as Thermodynamics, Kinematics, the Physical World & Measurements, Gravitation, Motion of Particles, Oscillation & Waves, Kinetic Theory of Gases and so many. There are important concepts, principles, functions, processes, and equations that are included in the chapters so that you can study the chapters in more detail with the help of 11th Physics Notes. Prepared by experts at esaral with vast experience in the field of Physics, these notes provide a detailed understanding of all the chapters in the Physics syllabus for students.

Solving these questions in the textbook and preparing the Physics topics properly will help you perform well in your board exams. You can download and practice from Physics Notes for Class 11 PDF in order to gain a proper understanding of the topics. These notes contain precise and step-by-step explanations of all the concepts and principles mentioned in the chapters. 

Revision Notes of Physics Class 11

Chapter 1 - Physical World

In this Revision Notes of Physics Class 11 Chapter 1, students will get acquainted with the basics of physics and its origin. The main points covered in this chapter are:

Details on Scientific method of observation and Mathematical modelling.

The two principal types of approaches in Physics (Unification and Reduction).

How Physics has impacted the development of devices and the use of concepts of Physics in different facets.

Different scopes of Physics and its categorization into two types based on the scope (Classical Physics and Modern Physics).

Macroscopic and Microscopic phenomena in Physics.

Chapter 2 - Units and Measurements

Physics Quick Revision Notes for Chapter 2 talks about various units of measurements and their use in measuring distance or time between multiple objects. The main topics covered in this chapter are:

Measurement and units for fundamental quantities like length, time, etc.

How to measure large distances using the parallax method.

Use of an electron microscope to measure very small distances.

Measurement of time using an atomic clock.

Chapter 3 - Motion in a Straight Line

Our revision notes of Chapter 3 cover the important terms related to kinematics like:

Motion in one dimension and important terms related to it.

Rest and motion.

Position, distance, and placement.

Difference between speed and velocity.

Chapter 4 - Motion in a Plane

Our Class 11 Physics revision Notes of Motion in a Plane cover the basics of motion in two dimensions. The main feature points discussed in Chapter 4 are:

Scalar and vector - unit vector, parallel vectors.

Addition, subtraction, and scalar multiplication of vectors

Average and instantaneous velocity.

Position vector and displacement.

Projectile motion.

Chapter 5 - Law of Motion

In our key 11th Class Physics Notes, you will understand what force is and the laws of motion. The chapter covers:

Definition of force and basic forces.

Newton’s law of motion.

Linear momentum.

Principle of conservation of momentum.

Chapter 6 - Work, Energy, and Power

Our revision notes for Chapter 6 will give you brief of formulas and equations on Work, Power, and Energy. Your concept on the following will get cleared:

Work and dimensions of the unit of work.

Conservative and non-conservative forces.

Kinetic energy and its relationship with momentum.

Chapter 7 - Systems of Particles and Rotational Motion

The notes of Chapter 7 contain detailed knowledge of how the system of particles moves in our daily lives. The main focus is on the following key points:

Kinematics of a system of particles- various types of motions like translational motion, rotational motion, and rotational plus translational motion.

Rotational dynamics - This comprises Newtons' laws, the moment of inertia, and theorems related to these concepts.

Angular momentum and impulse.

Work and Energy - The work is done by a torque, kinetic energy.

Rolling - rolling and sliding on an inclined plane.

Chapter 8 - Gravitation

In our revision notes for gravitation, we explain the law of gravitation given by Newton and many examples and problems that will clarify this law to you. It has the following topics in detail:

Newton’s law of gravitation

The universal constant of gravitation (g) and variation in g based on the distance of the object above the earth’s surface

Satellite - How a satellite is projected to form a circular orbit around a planet

Orbital velocity

Period of revolution of a satellite

Chapter 9 - Mechanical Properties of Solids

Our crucial revision notes in this chapter will help you understand how even rigid bodies can be bent and stretched by applying sufficient external force. As part of this chapter, you would learn about:

Deforming force

Perfect elastic body

Types of stress - Longitudinal, Tangential or shearing, Normal

Hooke’s law

Chapter 10 - Mechanical Properties of Fluids

Our revision notes of Physics Chapter 10 explain Pascal’s and Archimedes' principles related to fluid dynamics. Here is a synopsis of what all you would learn in this chapter:

Fluid mechanics - fluid pressure, atmospheric pressure

Pascal’s law

Archimedes Principle

Fluid dynamics - steady flow, line of flow

The velocity of Efflux

Bernoulli’s theorem

Chapter 11 - Thermal Properties of Matter

The important notes of the matter's thermal properties will take you through a detailed course on heat and thermodynamics. You would learn about following rules and theorems in this chapter:

Thermal properties of matter - temperature, heat, absolute temperature scale, and measurement of temperature

Thermal expansion

Heat transfer - Conduction, Convection, and Radiation

Newton’s law of cooling

Chapter 12 - Thermodynamics

In our revision notes of Chapter 12, students would learn about the interrelation between heat and other forms of energy. It covers the following key concepts:

Thermal equilibrium

Heat, work, and internal energy

Important thermodynamic terms like static variables, equation of state, Quasi-static process

Isothermal process

1st law of thermodynamics and its applications

Chapter 13 - Kinetic Theory

These crucial notes on kinetic theory will teach students about the following important topics:

Definition of kinetic theory

Molecular nature of matter - Dalton’s atomic theory, Avogadro’s law, Gay Lussac’s law

Molecular structure of solids, liquids, and gases

The behaviour of gases and the perfect gas equation

Deduction of Boyle’s and Charle’s Laws

Chapter 14 - Oscillations and Waves

Our revision notes of Class 11 Chapter 14 will teach you about what oscillation is and different types of oscillations. You can learn the following in this chapter:

Definition of oscillation and waves

Important terms related to oscillation like amplitude, period, acceleration, etc.

The time of simple harmonic motion

Damped and forced oscillations

Benefits of Class 11 Physics Notes PDF Download

All Class 11 Physics revision notes have been explained in detail in these revision notes. The learned experts at eSaral have used simple language to get the concepts explained. You can rely on these notes to get proper guidance about the chapters and develop their knowledge.

The notes are amazing study resources that can help you build a strong foundation in Physics subjects. You will get familiar with the concepts such as units of fundamental quantities, Newton’s laws of gravitation, Satellites, Angular Momentum, Energy, Work, and much more. The concepts have been clearly explained using different examples to make you understand better.

The notes for class 11 Physics will help students clarify any doubts that they might have about the chapters. Students can easily refer to the revision notes and find out the areas where they need to make improvements. Thus, these class 11 Physics notes pdf  will also help them in rectifying their errors and strengthen their answering skills.

The revision notes have been designed in a very systematic manner for the students. Thus, if you are looking for Class 11 Physics notes, you will be able to find the details right there. This makes it easier for you to access the files and get what they are looking for.

Before the exams start, you might not have the time to go through the whole Physics textbook and learn the details one by one. You can refer to the revision notes and get details about the chapters easily and this will make sure that students are able to revise their concepts before the exams.

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NCERT Class 11 Physics Chapter 4 Laws of Motion

NCERT Class 11 Physics Chapter 4 Laws of Motion   Solutions , NCERT Class 11 Physics Chapter 4 Laws of Motion Notes to each chapter is provided in the list so that you can easily browse throughout different chapters  NCERT Class 11 Physics Chapter 4 Laws of Motion Question Answer and select needs one.

Also, you can read the SCERT book online in these sections NCERT Class 11 Physics Chapter 4 Laws of Motion Solutions by Expert Teachers as per SCERT ( CBSE ) Book guidelines. These solutions are part of SCERT  All Subject Solutions . Here we have given NCERT Class 11 Physics Chapter 4 Laws of Motion Solutions for All Subjects, You can practice these here.

Laws of Motion

(For simplicity in numerical calculations, take g = 10 m s -2 )

1. Give the magnitude and direction of the net force acting on:  

(a) A drop of rain falling down with a constant speed.

Ans: The drop is moving with a constant speed, so according to Newton’s first law, the net force on it is zero.

(b) A cork of mass 10g floating on water.

Ans: As the cork is floating on water, its weight is balanced by the upthrust due to water. Hence net force on the cork is zero.

(c) A kite skillfully held stationary in the sky.

Ans: Kite is held stationary, according to Newton’s first law the force on the kite is zero.

(d) A car moving with a constant velocity of 30 km/h on a rough road.

Ans: The car moving with a constant velocity, the net force on the car is zero.

(e) A high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Ans: The net force on electron is zero.

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.

Ans: Given:

F = net force.

m = mass of the pebble = 0.05kg.

g = 9.8 m/s.

F = 0.05 × 9.8 = 0.49N.

(b) During its downward motion.

F = 0.05 × 9.8 = 0.49N. 

(c) At the highest point where it is momentarily at rest. 

F= net force.

g= 9.8 m/s.

F = 0.05 x 9.8 = 0.49N.

Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?  

Ans: If the pebble was thrown at an angle of 45° the answers remain the same. Regardless of the angle of projection, as long as we ignore air resistance.

3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg. (Neglect air resistance throughout.)

(a) Just after it is dropped from the window of a stationary train.

Ans: The only force acting on the stone is gravity:

Force = mass x acceleration due to gravity (F = mg)

F = 0.1kg × 9.8 m/s -2

= 0.98 N. 

(b) Just after it is dropped from the window of a train running at a constant velocity of 36 km/h.

Ans: The window of a train running at a constant velocity of 36 km/h, no force acts on the stone due to the motion of the train. Thus, Force on stone = Mg – 0.98N.

(c ) Just after it is dropped from the window of a train accelerating with 1 m s-2.

Ans: Force on the stone F = 0.1 × 1 = 0.1 N.

This force also acts vertically downwards.

(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.  

Ans: The net force on the stone is given by: F = 0.1 × 1 = 0.1 N.

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:

T is the tension in the string. [Choose the correct alternative].  

Ans: (i) is correct.

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?

Ans: Given that:

The retarding force F = 50N

The mass m = 20 kg

Acceleration a = F/m 

= 50 N / 20 kg = 2.5 m/s 2

Since this is a retarding force the acceleration will be negative 

a = – 2.5 m/s 2

v is the final velocity

0 = 15 m/s + (- 2.5 m/s 2 ) × t

– 15 m/s = – 2.5 m/s 2 × t

t = 15m/s / 2.5 m/s 2

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?  

Ans: Given that:  

v = 3.5 ms-1 

v = v – u 

= 3.5m/s – 2.0m/s

Calculate of acceleration:

a = 1.5 m/s / 25s

a = 0.06 m/s 2

Newton’s second law to find the force:

F = 3.0 kg x 0.06 m/s 2

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.

Ans: Here F 1 = 8N

F = √(F 1 ) 2 + (F 2 ) 2

= √ (8) 2 + (6) 2 

a = F/M 

= 10/5 = 2ms -2

If F makes angle θ with the direction F 1 , then

cos θ  = F 1 /F 

= 8/10 

= 0.8 

= θ = cos-1 (0.8) = 36.87°.

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.  

Ans: Given u = 36 km/h

36km/h = 36 x 1000m/3600 = 10 m/s.

Initial speed u = 10m/s

a = u – v /t 

= 0 – 10m/s/ 4.0s

= – 2.5 m/s 2

m = 400 kg + 65 kg

F = ma = 465 (-2.5)

= 1162.5 N.

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.

Ans: Here m = 2 × 10 4 kg, 𝚫v/ 𝚫t = 5 ms -2

As, 𝚫v/𝚫t = vr/m × 𝚫m/𝚫t -g

Hence vr = 𝚫m/𝚫t

= m. 𝚫v/𝚫v/𝚫t + mg

= 2 × 10 4 × 5 + 2 × 10 4 × 9.8

= 10 5 + 1.96 × 10 5

= 2.96 × 10 5 N. 

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.

Ans: Given: Here m = 0.40 kg

u = 10 ms -1

F = – 8N (retarding force) 

Time duration of force application t force = 30s

a = F/m 

= – 8.0/0.40 

= – 20m/s 2

Determination of t = -5s, t = 25s and t=100s.

t = – 5s

Before the velocity is applied the body’s constant  velocity of 10m/s

Position at t = -5s

S -5 = 10 x (-5) + ½ x 0 x (-5) 2

(ii) Position at, t = 25s

S 25 = 10 x 25 + ½ x (-20) x (25) 2

= – 6000m 

= – 6km

(iii) Position at t = 30s

= S30 = 10 × 30 + ½ × (-20) × (30) 2

= 300 – 0.5 × 20 v 9000

= 300 – 9000

= – 8700

Velocity at t = 30s

v 30 = 10 + (-20) × (30)

= 10 – 600

= – 590 m/s

Additional time after force stops: t additional  

= 100 – 30 = 70s 

S 100-30 = -590 × 70 + ⅕ x 0 × (70) 2

= – 41300 ms -1

Total distance = S 30 + S 30-100

= – 41300 – 8700

= – 50000m

= – 50km.

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 = 11s? (Neglect air resistance.) 

Ans: (a) As v = u + at, when v = vx, u= 0

a = 2 ms -2 , t = 10s

We get v = 0 + 2 × 10s = 20 ms -1

v y = 0 + 9.8 × 0.1 = 0.98 ms -1

= 10.1 – 10 = 0.1s

v R = √v 2 x + v 2 y

= √(20) 2 + (0.98) 2

= 20.02 ms -1

And tan θ = v y /v x

= 0.98 / 20 

= 0.049 

Or θ = tan-1 (0.049) = 2.8°

(b) Acceleration at 10.1s = g = 9.8 ms -2 .

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.

Ans: (a) As the bob of a simple pendulum has no velocity at the extreme position. If the string is cut, it will fall vertically downwards.

(b) At the mean position, f the string is cut, bob is acted by vertical gravitational force = a = 9.8 𝑚𝑠−2 Hence bob will behave like a projectile and follows a parabolic path.

The bob will have a periodic path as it is having horizontal velocity.

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.

Ans: R = mg

Given m = 70kg

g = 9.8 m/s

= 70kg × 9.8 

(b) Downwards with a uniform acceleration of 5 m s-2.

Ans: R’ = m (g – a) 

= 70 (9.8 – 5)

(c) Upwards with a uniform acceleration of 5 m s-2. What would be the readings on the scale in each case?

Ans: R’’ = m (g + a)

= 70 ( 9.8 + 5)

= 70 × 14.8

= 1036 N. 

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Ans: R’’’ = m (g – g)

= 70 kg ( 9.8 – 9.8)

= 0 N. 

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).

Ans: (a) For t < 0 and t > 4s, the particle is at rest as the position does not change w.r.t time. Clearly no force acts on the particle during these intervals.

Further, for 0 < t < 4s, the position of the particle continuously changes with respect to time. As the position – time graph is a straight line, it represents uniform motion and there is no acceleration. Hence it is also clear that no force acts on the particle during these intervals.  

(b) Because the velocity is uniform from O and A, hence velocity at O = velocity at A = slope of the graph OA = ¾ ms -1

Impulse (at t= 0) = change in momentum. 

= Final momentum-initial momentum

= 0 – mv 

= – 4 ( ¾) 

= – 3 kg ms -1 .

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 the string. What is the tension in the string in each case?

Horizontal force F = 600 N

Mass of body, A, M 1 = 10 kg

Mass of body B, m 2 = 20 kg

Total mass of the system m = m 1 + m 2 = 30 kg

Here, a = 600 / 10 + 20

Force is applied on 10 kg mass,

= 600 – T = 10 × 20 

Force applied on 20 kg mass,

= 600 – T = 20 × 20 

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.  

Ans: Let m 1 , = 8 kg and m 2 = 12

For mass m 1 , m 1 a = T – M1g

8a = T – 8 × 9.8 (i) 

For mass m 2 = mg – T

= 12a = 12 × 9.8 – T (ii) 

Adding equation of (i) and (ii) 

(8 + 12) a = T – 8 × 9.8 + 12 × 9.8 – T

20a = 4 × 9.8 

a = 4 × 9.8 / 20 

= 1.96 ms -2  

Equation (i) 

T = 8 × 1.96 MS -2 + 8 × 9.8 

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. 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?

Ans: Let m = initial mass of the nucleus.

m 1 and m 2 are masses after disintegration and v 1 and v 2 are their respective velocities.

Now, initial monuments of the nucleus 

= m × 0 = 0 

Final momentum of the nucleus

= m 1 v 1 + m 2 v 2

Using laws of conservation of momentum i.e., initial momentum of the system. 

= Final momentum of the systems, 

= 0 = m 1 v 1 + m 2 v 2

= m 2 v 2 = m 1 v 1

Or v 2 = (-) m 1 v 1 / m 2 . 

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?

Ans: Initial momentum of ball 

= mu = 0.05 × 6 

= 0.3 kg ms-1.

Final momentum of ball 

= mv = 0.05 × (-6)

= – 0.3 kg ms-1

Change in momentum 

Impulse J imparted to each ball is the change in momentum

J1 = Pfinal1 – Pinitial

J1 = 0. 3 kg m/s – 0. 3 kg m/s

J1 = – 0.6 kg m/s

J2 = Pfinal2 – Pinitial

J2 = 0.3 kg m/s – 0.3 kg m/s

J2 = 0 kg m/s

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?

Ans: Given: 

m shell = 0.020 kg 

M gun = 100 kg

v shell = 80 m/s

v gun to determine: ?

According to the conservation of momentum:

= m shell x  v shell = – m gun ×  v gun

= 0.020 × 80 = – 100 × v gun

= 1.6 = – 100 × v gun

= v gun = 1.6 / – 100

= v gun = – 0.0016 m/s

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.)

Ans: Suppose the point O as the position of the bat. AO line shows the path along which the ball strikes the bat with velocity v and OB is the path showing defection such that <AOB = 45°

Here, Initial momentum of the ball = mu cos θ 

= 0.15 × 54 × 1000 × 22.5 / 3600°

= 0.15 × 15 × 0. 9239 along NO

Final momentum of the ball = mu cos θ – (- mu cos θ)

= 2 × 0.25 × 15 × 0.9239

= 4.16 kg ms -1 .  

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?

Ans: M = 0.25 kg , r = 1.5

v = 40 rev.min -1 = 40/60 rev s -1

ω = 2rv = 2𝜋 × 40 / 60 

= 1.33𝜋 rad s -1  

Here, tension = centripetal force = Mv2 / r 

= Mrω = 0.25 x 1× (1.33𝜋) 2  

The string can withstand a maximum tension of 200 N. if v max be the maximum speed of the stone, then,

200 = Mv2 max / r 

= v max = 

 = 25.82ms -1 .

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?

Ans: Option b is correct. When the string breaks the stone will move in the direction of the velocity at that instant. It is because the speed of the stone at any instant is directed along a tangent to the circular path at the point. 

23. Explain why:

(a) A horse cannot pull a cart and run in empty space.

Ans: An empty space has no such response force. As a result, a horse cannot draw a cart and run in open space.

(b) Passengers are thrown forward from their seats when a speeding bus stops suddenly.

Ans: It is because of the Inertia of motion.

(c) It is easier to pull a lawn mower than to push it.

Ans: Pulling creates a downward force on the handles, increasing friction between the mower and the ground, which improves traction.

(d) A cricketer moves his hands backwards while holding a catch.

Ans: By increasing time, force is reduced.

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Class 11 Computer Science Complete Notes

These complete Class 11 Computer Science Notes for every unit are tailored to help you easily grasp the essential topics covered in your CBSE Class 11 computer science curriculum. So, whether you are studying for exams, or working on assignments, these notes provide a clear and straightforward guide about the topics.

This guide covers all 3 units of Class 11 Computer Science, like Computer Systems and Organisation where you get to know all the fundamentals of computer systems. Computational Thinking and Programming where you will learn the basics of programming, including key concepts like variables, data types, and basic algorithms, to help you start coding confidently. And in last, you will learn about Society, Law, and Ethics here you will get to know all about the computer society, IT laws, and ethics.

Introduction to Unit I: Computer Systems and Organisation

Welcome to Unit I: Computer Systems and Organisation ! This unit is all about understanding computers’ essential components and workings, helping you grasp how they process, store, and manage information. Here’s a quick tour of what we’ll cover:

• Computer System Basics : We’ll start with the foundation of what a computer system is, including both hardware (the physical parts like the CPU, memory, and input/output devices) and software (the programs and operating systems that run on the hardware).
• Hardware : Dive into the major components such as the Central Processing Unit ( CPU ), which acts as the brain of the computer, and the various types of memory—primary ( RAM ), cache, and secondary storage (hard drives and SSDs). We’ll also look at how these elements work together to perform tasks.
• Software : Learn about the different types of software, from system software (like operating systems and utilities) that manage hardware to application software that helps you perform specific tasks such as word processing or web browsing. We’ll also touch on programming tools that help developers create software.
• Boolean Logic : Understand the basics of Boolean logic, including operators like AND, OR, and NOT, and how they form the foundation of computer operations and digital circuits.
• Number Systems : Explore different number systems (binary, octal, decimal, hexadecimal) and how to convert between them. This is crucial for understanding how computers process and represent data.
• Encoding Schemes : Discover how computer characters and symbols are represented using encoding schemes like ASCII , ISCII, and Unicode.

This unit sets the stage for your journey into computing, giving you the tools to understand how computers work and interact with the digital world. Whether you’re curious about the inner workings of your device or eager to dive deeper into programming and technology, this unit provides the essential knowledge you’ll build on throughout your studies.

Introduction to Unit II: Computational Thinking and Programming

Welcome to Unit II: Computational Thinking and Programming ! This unit is designed to introduce you to the fundamental concepts of problem-solving and programming, essential skills for anyone interested in technology and computing. Here’s a quick overview of what you’ll learn:

• Computational Thinking : We’ll begin by exploring computational thinking, a method of problem-solving that involves breaking down complex problems into smaller, manageable parts. You’ll learn about key strategies such as pattern recognition, abstraction, and algorithm design, which are crucial for developing efficient solutions.
• Programming Basics : Dive into the world of programming with a focus on understanding how to write and interpret code. You’ll be introduced to basic programming concepts such as variables, data types, control structures (like loops and conditionals), and functions. These are the building blocks of creating software and solving computational problems.
• Algorithms and Flowcharts : Learn how to design algorithms—step-by-step procedures for solving problems—and represent them using flowcharts. This helps in visualizing and organizing your approach to solving a problem before writing code.
• Introduction to Programming Languages : Get familiar with different programming languages and their uses. We’ll cover the basics of popular languages and highlight how they can be applied to various tasks, from web development to data analysis.
• Debugging and Testing : Discover techniques for finding and fixing errors in your code. Understanding how to debug and test your programs ensures they run smoothly and perform as expected.
• Project Development : Apply what you’ve learned by working on practical programming projects. These projects will help you understand how to implement computational thinking and programming concepts to build functional software.

This unit provides the foundational skills needed to tackle computational problems and start programming. Whether you’re interested in building your own apps, solving complex problems, or just understanding how software works, Unit II equips you with the essential tools and techniques to get started in the world of programming.

Introduction to Unit III: Society, Law, and Ethics

Welcome to Unit III: Society, Law, and Ethics ! This unit explores the important intersection of technology with societal norms, legal frameworks, and ethical considerations. Here’s a quick guide to what we’ll cover:

• Digital Footprints : Understand the concept of digital footprints—how your online activities leave traces that can impact your privacy and reputation. We’ll discuss ways to manage and protect your personal information in the digital world.
• Digital Society and Netiquette : Explore the norms and etiquette for interacting online. This includes understanding netiquette (internet etiquette), communication etiquette, and social media behavior to ensure respectful and effective online interactions.
• Data Protection : Learn about intellectual property rights like copyrights, patents, and trademarks. We’ll also cover what happens when these rights are violated through plagiarism, copyright infringement, or trademark misuse, and why respecting these rights is crucial.
• Open Source Software and Licensing : Discover what open-source software is and the various licensing models that govern its use, including Creative Commons, GPL, and Apache licenses. This helps in understanding how software can be shared and used legally.
• Cyber Crime : Get to know different types of cyber crimes , such as hacking, eavesdropping , phishing, ransomware , cyber trolls, and cyberbullying . We’ll discuss their impact and how to stay safe from these threats.
• Cyber Safety : Learn essential practices for safe online behavior, including browsing safely, protecting your identity, and maintaining confidentiality. This section emphasizes practical steps to enhance your security in the digital world.
• Malware : Understand various types of malware, including viruses, trojans, and adware. We’ll explore how they infect systems and what you can do to protect your computer from these malicious threats.
• E-Waste Management : Learn the importance of properly disposing of electronic gadgets to minimize environmental impact. We’ll discuss ways to recycle and manage electronic waste responsibly.

This unit equips you with the knowledge to navigate the digital world responsibly, understand legal and ethical implications, and protect yourself from various online risks. By integrating societal and legal perspectives with technology, you’ll be better prepared to handle the challenges and responsibilities of the modern digital landscape.

In conclusion, these Class 11 Computer Science Notes cover all key topics in the CBSE curriculum, from computer systems basics to programming and digital ethics. Designed to be clear and concise, these notes will help you excel in exams and deepen your understanding of computer science. Use them to confidently master the subject and achieve top grades.

Class 11 Computer Science Notes – FAQs

What topics are covered in the cbse class 11 computer science notes.

The notes cover all key topics in the CBSE Class 11 Computer Science syllabus, including Computer Systems and Organization, Computational Thinking and Programming, Boolean Algebra, Number Systems, and Society, Law, and Ethics in technology.

How can these notes help me prepare for my Class 11 Computer Science exams?

These notes provide a clear, concise, and structured overview of all the topics you need to know for your exams. They simplify complex concepts, making it easier for you to understand and retain information, which is crucial for exam preparation.

Are the notes suitable for beginners in computer science?

Yes, the notes are designed to be student-friendly and easy to understand, making them perfect for beginners who are new to computer science concepts and programming.

Can I use these notes for assignments and classwork?

Absolutely! These notes are well-organized and comprehensive, providing detailed explanations and examples that can help you with both assignments and classwork.

Do these notes include practical examples and exercises?

Yes, the notes include practical examples and exercises to help you apply the concepts you learn, enhancing your understanding and skills in computer science.

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CBSE Class 11 Physics Chapter 5 Laws of motion Study Materials

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Law of Motion Class 11 Notes Physics Chapter 5

• Dynamics  is the branch of physics in which we study the motion of a body by taking into consideration the cause i.e., force which produces the motion. • Force Force is an external cause in the form of push or pull, which produces or tries to produce motion in a body at rest, or stops/tries to stop a moving body or changes/tries to change the direction of motion of the body. • The inherent property, with which a body resists any change in its state of motion is called inertia. Heavier the body, the inertia is more and lighter the body, lesser the inertia. • Law of inertia states that a body has the inability to change its state of rest or uniform motion (i.e., a motion with constant velocity) or direction of motion by itself. • Newton’s Laws of Motion Law 1.  A body will remain at rest or continue to move with uniform velocity unless an external force is applied to it. First law of motion is also referred to as the ‘Law of inertia’. It defines inertia, force and inertial frame of reference. I here is always a need of ‘frame of reference’ to describe and understand the motion of particle, lhc simplest ‘frame of reference’ used are known as the inertial frames. A frame of referent, e is known as an inertial frame it, within it, all accelerations of any particle are caused by the action of ‘real forces’ on that particle. When we talk about accelerations produced by ‘fictitious’ or ‘pseudo’ forces, the frame of reference is a non-inertial one. Law 2. When an external force is applied to a body of constant mass the force produces an acceleration, which is directly proportional to the force and inversely proportional to the mass of the body. Law 3.  “To every action there is equal and opposite reaction force”. When a body A exerts a force on another body B, B exerts an equal and opposite force on A. • Linear Momentum The linear momentum of a body is defined as the product of the mass of the body and its velocity. • Impulse Forces acting for short duration are called impulsive forces. Impulse is defined as the product of force and the small time interval for which it acts. It is given by Impulse of a force is a vector quantity and its SI unit is 1 Nm. — If force of an impulse is changing with time, then the impulse is measured by finding the area bound by force-time graph for that force. — Impulse of a force for a given time is equal to the total change in momentum of the body during the given time. Thus, we have • Law of Conservation of Momentum The total momentum of an isolated system of particles is conserved. In other words, when no external force is applied to the system, its total momentum remains constant. • Recoiling of a gun, flight of rockets and jet planes are some simple applications of the law of conservation of linear momentum. • Concurrent Forces and Equilibrium “A group of forces which are acting at one point are called concurrent forces.” Concurrent forces are said to be in equilibrium if there is no change in the position of rest or the state of uniform motion of the body on which these concurrent forces are acting. For concurrent forces to be in equilibrium, their resultant force must be zero. In case of three concurrent forces acting in a plane, the body will be in equilibrium if these three forces may be completely represented by three sides of a triangle taken in order. If number of concurrent forces is more than three, then these forces must be represented by sides of a closed polygon in order for equilibrium. • Commonly Used Forces (i) Weight of a body.  It is the force with which earth attracts a body towards its centre. If M is mass of body and g is acceleration due to gravity, weight of the body is Mg in vertically downward direction. (ii) Normal Force. If two bodies are in contact a contact force arises, if the surface is smooth the direction of force is normal to the plane of contact. We call this force as Normal force. Example.  Let us consider a book resting on the table. It is acted upon by its weight in vertically downward direction and is at rest. It means there is another force acting on the block in opposite direction, which balances its weight. This force is provided by the table and we call it as normal force. (iii) Tension in string. Suppose a block is hanging from a string. Weight of the block is acting vertically downward but it is not moving, hence its weight is balanced by a force due to string. This force is called ‘Tension in string’. Tension is a force in a stretched string. Its direction is taken along the string and away from the body under consideration. • Simple Pulley Consider two bodies of masses m 1  and m 2  tied at the ends of an in extensible string, which passes over a light and friction less pulley. Let m 1  > m 2 . The heavier body will move downwards and the lighter will move upwards. Let a be the common acceleration of the system of two bodies, which is given by • Apparent Weight and Actual Weight — ‘Apparent weight’ of a body is equal to its ‘actual weight’ if the body is either in a state of rest or in a state of uniform motion. — Apparent weight of a body for vertically upward accelerated motion is given as Apparent weight =Actual weight + Ma = M (g + a) — Apparent weight of a body for vertically downward accelerated motion is given as Apparent weight = Actual weight Ma = M (g – a). • Friction The opposition to any relative motion between two surfaces in contact is referred to as friction. It arises because of the ‘inter meshing’ of the surface irregularities of the two surfaces in contact. • Static and Dynamic (Kinetic) Friction The frictional forces between two surfaces in contact (i) before and (ii) after a relative motion between them has started, are referred to as static and dynamic friction respectively. Static friction is always a little more than dynamic friction. The magnitude of kinetic frictional force is also proportional to normal force. • Limiting Frictional Force This frictional force acts when body is about to move. This is the maximum frictional force that can exist at the contact surface. We calculate its value using laws of friction. Laws of Friction: (i) The magnitude of limiting frictional force is proportional to the normal force at the contact surface. (ii) The magnitude of limiting frictional force is independent of area of contact between the surfaces. • Coefficient of Friction The coefficient of friction (μ) between two surfaces is the ratio of their limiting frictional force to the normal force between them, i.e., • Angle of Friction It is the angle which the resultant of the force of limiting friction F and the normal reaction R makes with the direction of the normal reaction. If θ is the angle of friction, we have • Angle of Repose Angle of repose (α) is the angle of an inclined plane with the horizontal at which a body placed over it just begins to slide down without any acceleration. Angle of repose is given by α = tan-1 (μ) • Motion on a Rough Inclined Plane Suppose a motion up the plane takes place under the action of pull P acting parallel to the plane. • Centripetal Force Centripetal force is the force required to move a body uniformly in a circle. This force acts along the radius and towards the centre of the circle. It is given by where, v is the linear velocity, r is the radius of circular path and ω is the angular velocity of the body. • Centrifugal Force Centrifugal force is a force that arises when a body is moving actually along a circular path, by virtue of tendency of the body to regain its natural straight line path. The magnitude of centrifugal force is same as that of centripetal force. • Motion in a Vertical Circle The motion of a particle in a horizontal circle is different from the motion in vertical circle. In horizontal circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the motion is not effected by the acceleration due to gravity (g) whereas in the motion of vertical circle, the value of ‘g’ plays an important role, the motion in this case does not remain uniform. When the particle move up from its lowest position P, its speed continuously decreases till it reaches the highest point of its circular path. This is due to the work done against the force of gravity. When the particle moves down the circle, its speed would keep on increasing. Let us consider a particle moving in a circular vertical path of radius V and centre o tide with a string. L be the instantaneous position of the particle such that Here the following forces act on the particle of mass ‘m’. (i) Its weight = mg (verticaly downwards). (ii) The tension in the string T along LO. We can take the horizontal direction at the lowest point ‘p’ as the position of zero gravitational potential energy. Now as per the principle of conservation of energy, From this relation, we can calculate the tension in the string at the lowest point P, mid-way point and at the highest position of the moving particle. Case (i) : At the lowest point P, θ = 0° When the particle completes its motion along the vertical circle, it is referred to as “Looping the Loop” for this the minimum speed at the lowest position must be √5gr • IMPORTANT TABLES

CBSE Class 11 Physics Chapter-5 Important Questions

1 Marks Questions

1.What is the unit of coefficient of friction?

Ans:  It has no unit.

2.Name the factor on which coefficient of friction depends?

Ans: Coefficient of friction  depends on the nature of surfaces in contact and nature of motion.

3.What provides the centripetal force to a car taking a turn on a level road?

Ans:  Centripetal force is provided by the force of friction between the tyres and the road.

4.Why is it desired to hold a gun tight to one’s shoulder when it is being fired?

Ans:  Since the gun recoils after firing so it must be held lightly against the shoulder because gun and the shoulder constitute one system of greater mass so the back kick will be less.

5.Why does a swimmer push the water backwards?

Ans:  A swimmer pushes the water backwards because due to reaction of water he is able to swim in the forward direction

6.Friction is a self adjusting force. Justify.

Ans:  Friction is a self adjusting force as its value varies from zero to the maximum value to limiting friction.

7.A thief jumps from the roof of a house with a box of weight W on his head. What will be the weight of the box as experienced by the thief during jump?

Ans:  Weight of the box W = m (g – a) = m (g – g) = 0.

8.Which of the following is scalar quantity? Inertia, force and linear momentum.

Ans: Inertia and linear momentum is measured by mass of the body and is a vector quantity and mass is a scalar quantity.

9.Action and reaction forces do not balance each other. Why?

Ans: Action and reaction do not balance each other because a force of action and reaction acts always on two different bodies.

10.If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?

Ans: No change in speed, but there can be change in the direction of motion.

11.The two ends of spring – balance are pulled each by a force of 10kg.wt. What will be the reading of the balance?

Ans: The reading of the balance will be 10kgwt.

12.A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Ans: The apparent weight will increase. If the lift is going with uniform speed, then the apparent weight will remain the same as the real weight.

14.  If, in Exercise 5.21, 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 ?

Ans.(b) When the string breaks, the stone will move in the direction of the velocity at that instant. According to the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. Hence, the stone will fly off tangentially from the instant the string breaks.

2 Marks Questions

1.Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed.

(b) A kite skillfully held stationary in the sky.

Ans: (1)  According to first law of motion F = 0 as a = 0 (particle moves with constant speed)

(2)  Since kite is stationary net force on the kite is also zero.

4.A force is being applied on a body but it causes no acceleration. What possibilities may be considered to explain the observation?

Ans: (1)  If the force is deforming force then it does not produce acceleration.

 (2)  The force is internal force which cannot cause acceleration.

5.Force of 16N and 12N are acting on a mass of 200kg in mutually perpendicular directions. Find the magnitude of the acceleration produced?

6.An elevator weighs 3000kg. What is its acceleration when the in the tension supporting cable is 33000N. Given that g = 9.8m/s 2 .

Ans: Net upward force on the

ElevatorF = T – mg

7.Write two consequences of Newton’s second law of motion?

Ans: (1) It  shows that the motion is accelerated only when force is applied.

(2)  It gives us the concept of inertial mass of a body.

8.A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans: When the bird starts flying inside the cage the weight of bird is no more experienced as air inside is in free contact with atmospheric air hence the cage will appear lighter.

9.Why does a cyclist lean to one side, while going along curve? In what direction does he lean?

Ans: A cyclist leans while going along curve because a component of normal reaction of the ground provides him the centripetal force he requires for turning.

He has to lean inwards from his vertical position i.e. towards the centre of the circular path.

10.How does banking of roads reduce wear and tear of the tyres?

Ans: When a curved road is unbanked force of friction between the tyres and the road provides the necessary centripetal force. Friction has to be increased which will cause wear and tear. But when the curved road is banked, a component of normal reaction of the ground provides the necessary centripetal force which reduces the wear and tear of the tyres.

12.A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle?

Ans: bubbles will not rise in water because water in freely falling bottle is in the state of weight – lessens hence no up thrust force acts on the bubbles.

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

Ans: Initial momentum to the ball A = 0.05(6) = 0.3 kg m/s

As the speed is reversed on collision,

final momentum of ball A = 0.05(-6) = -0.3 kg m/s

Impulse imparted to ball A = change in momentum of ball A = final momentum – initial momentum = -0.3 -0.3 = -0.6 kg m/s.

15.Explain why passengers are thrown forward form their seats when a speeding bus stops suddenly.

Ans: When the speeding bus stops suddenly, lower part of the body in contact with the seat comes to rest but the upper part of the body of the passengers tends to maintain its uniform motion. Hence the passengers are thrown forward.

• Laws of motion
• Active page

It is an agent that changes or tends to change the state of rest or of uniform motion of a body in a straight line. Unit:N (newton)

2. Newton's Laws Of Motion

3. Commonly Used Forces

(a) Normal force: Normal to the surfaces of contact and towards the body under consideration.

(b) Weight of body: Equals to Mg and acts vertically downward. (c) Tension in string: Along the string, away from the body under consideration.

4.Pulley Mass Systems

Unless stated otherwise, pulleys and strings are massless (i) When unequal masses m 1 and m 2 are suspended from a pulley

(m 1 > m 2 )

m 1 g – T = m 1 a, and T – m 2 g = m 2 a

On solving equations, we get

a = ((m 1 – m 2 ) / (m 1 + m 2 )) * g

T = 2m 1 m 2 / (m 1 + m 2 ) * g

(ii) When a body of mass m 2 is placed frictionless horizontal surface, then

Acceleration a = m 1 g / (m 1 + m 2 )

Tension in string T = m 1 m 2 g / (m 1 + m 2 )

(iii) When a body of mass m2 is placed on a rough horizontal surface, then

Acceleration a = ((m 1 – μm 2 ) / (m 1 + m 2 )) * g

Tension in string T = (m 1 m 2 (1 + μ) / (m 1 + m 2 )) * g

(iv) When two masses m 1 and m 2 are connected to a single mass M as shown in figure, then

m 1 g – T 1 = m 1 a …..(i)

T 2 – m 2 g = m 2 a ……(ii)

T 1 – T 2 = Ma …….(iii)

Acceleration a = ((m 1 – m 2 / (m 1 + m 2 + M)) * g

Tension T 1 = (2m 2 + M / (m 1 + m 2 + M) * m 1 g

T 2 = (2m a + M / (m 1 + m 2 + M) * m 2 g

(v) Motion on a smooth inclined plane, then

m 1 g – T = m 1 a …..(i)

T – m 2 g sin θ = m 2 a ……(ii)

Acceleration a = ((m 1 – m 2 sin θ/ (m 1 + m 2 )) * g

Tension T = m 1 m 2 (1 + sin θ) g / (m 1 + m 2 )

(vi) Motion of two bodies placed on two inclined planes having different angle of inclination, then

Acceleration a = (m 1 sin θ 1 – m 2 sin θ 2 ) g / m 1 + m 2

Tension T = (m 1 m 2 / m 1 + m 2 ) * (sin θ 1 – sin θ 2 ) g

5. Stepwise Procedure to Solve Questions Based On Motion Of Connected Bodies

   (a) Identify the unknown forces and accelerations.    (b) Draw FBD of bodies in the system.    (c) Resolve forces in the direction of motion and perpendicular to it.    (d) Apply  F = ma in the direction of motion and a = 0 in the direction of equilibrium.    (e) Write constraint relation if required and possible.    (f) Solve the equations written in above steps to get the results.

6. Pseudo Force

7. Conical Pendulum

It consists of a string whose upper end is fixed and bob is tied at the other free end. The string traces the surface of the cone, the arrangement is called a conical pendulum.

8. Apparent Weight in a Lift

(i) When a lift is at rest or moving with a constant speed, then

The weighing machine will read the actual weight.

(ii) When a lift is accelerating upward, then apparent weight

R 1 = m(g + a)

The weighing machine will read the apparent weight, which is more than the actual weight.

(iii) When a lift is accelerating downward, then apparent weight

R 2 = m (g – a)

The weighing machine will read the apparent weight, which is less than the actual weight.

(iv) When lift is falling freely under gravity, then

R 2 = m(g – g)= 0

The apparent weight of the body becomes zero.

(v) If lift is accelerating downward with an acceleration greater than g, then body will lift from floor to the ceiling of the lift.

9. Frictional Force

Frictional forces are produced due to intermolecular interactions acting between the molecules of the bodies in contact.

10. Types Of Frictional Force

(a) Static frictional force: Self adjusting force having magnitude less than or equal to μs N.

(b) Limiting frictional force: Maximum value of frictional force having value μsN. If (f)m is the maximum value of static friction, then is the coefficient of static friction. In general,  = μs × normal contact force.

μ s = tan θ

(c) Kinetic frictional force: If the body begins to slide on the surface, the magnitude of the frictional force rapidly decreases to a constant value f k kinetic friction.

It has value equal to μ k N.

where μ k = coefficient of kinetic friction and N = normal force.

Kinetic friction is of two types:

(a) Sliding friction

(b) Rolling friction

As, rolling friction < sliding friction, therefore it is easier to roll a body than to slide.

Kinetic friction (f k ) = μ k R

where μ k = coefficient of kinetic friction and R = normal reaction.

Angle of repose or angle of sliding It is the minimum angle of inclination of a plane with the horizontal, such that a body placed on it, just begins to slide down.

If angle of repose is a. and coefficient of limiting friction is μ, then

μ s = tan α (d) For plane inclined at an angle θ with horizontal: If the body just slides (acceleration = 0)          μ = tan θ If θ < tan –1 μ, the body will not slide If θ = tan –1 μ, the body will just slide down If θ > tan –1 μ, the body will slide with acceleration a.                

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Law of Motion Class 11 Notes Physics Chapter 5

January 19, 2024 by Bhagya

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Laws Of Motion Class 11 Notes PDF (Handwritten & Short Notes)

The Laws Of Motion is one of the important chapters of the Class 11th Physics. Students can easily cover the chapter with the help of the Laws Of Motion Class 11 notes. In the notes, all the concepts and topics are explained in a creative and precise manner. 

The Laws Of Motion Class 11 notes are considered to be important study material throughout the preparation. As it helps students to understand the chapter Laws Of Motion in many different ways. One of the important ways is by practising questions given in the chapter Laws Of Motion notes. By practising many questions students can increase their overall score in the examination by performing too well in the examination. 

Laws Of Motion Notes PDF

The Class 11 Physics notes is a written explanation for every passage or concepts included in the chapter Laws Of Motion. These Laws Of Motion notes are available in the Portable Document Format. The PDF of the Class 11 Physics Notes can be easily downloaded without any difficulty. With the help of the downloaded revision notes, Class 11 students can pay attention to the chapter Laws Of Motion to be able to perform well in the board examinations as well as in the competition exam. 

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Features of the Laws Of Motion Class 11 Notes

Features are considered to be an important or noticeable part of the Laws Of Motion Class 11 notes. Those important features are: 

• Explanations are Provided: In the Laws Of Motion notes, explanations are provided to each and every topic as well as definitions. According to the explanations, students can understand the chapter Laws Of Motion in a fine way. 
• Different Types of Questions are Included: Inside the Class 11 Laws Of Motion notes, different types of questions are included. Those questions are: example questions, objective type questions, very short type answers, short type questions, long type questions, numerical problems, etc. 
• Simple Language: These Class 11 Physics notes are created by subject matter experts in a very easy and simple language. It is formed in an easy language so that Class 11 students can understand each topic and question of the chapter Laws Of Motion. 
• Colourful Content are Given: Content included in the Class 11 Laws Of Motion notes are in a colourful manner. This colourful content can attract many Class 11 students to cover the chapter Laws Of Motion in a proper way. 
• Hints and Solutions are Provided: To tackle the questions given in Laws Of Motion Class 11 Notes, students can refer to the Hints and Solutions given in the PDF file of Class 11 Revision Notes of Laws Of Motion. With the help of hints and solutions, students can not only tackle the questions but can develop a good understanding in the overall chapter.

Advantages of the Laws Of Motion Class 11 Notes

Completing the chapter with the help of Laws Of Motion Class 11 notes can provide good results to students. This is one of the important advantage of Class 11 Physics notes, other than this are: 

• Helps to Memorise Topics: With the help of Class 11 Physics notes, students can easily memorise topics discussed in the chapter Laws Of Motion. 
• Summary is Given: After covering all topics of the chapter Laws Of Motion, a brief summary is given. Through the summary, students can get an idea about the important topics and definitions. 
• A Quick Revision Can Be Done: With the help of Class 11 Physics notes, students can have a quick revision of the chapter Laws Of Motion. Through the last minute revision, students can identify the weak points for the chapter Laws Of Motion. 
• All Topics are Covered: In the Laws Of Motion notes, all the topics and concepts are covered in an elaborate manner. Understanding the topics and concepts in a detailed way can help students to increase their comprehensive skills. 
• Increases the Accuracy Level: Routine exercise of questions from the chapter Laws Of Motion can help students to increase their accuracy level. Accuracy level mainly helps Class 11 students to increase their quality in giving answers for the chapter Laws Of Motion.

When Is The Right Time To Go Through Laws Of Motion Class 11 Notes?

After having a brief knowledge about the chapter Laws Of Motion, students can look through the Laws Of Motion Class 11 notes. With the help of notes, students can recall all topics and concepts of Laws Of Motion in a better way. 

Although, generally, it is advised to all the students to use the revision notes on a weekly basis to recall the previous learnings. Along with this, using the Laws Of Motion Class 11 notes at the time of exam preparation helps a lot so, using the Class 11 Physics Notes during exam preparation can be the right time.

Tips for Students to Cover the Chapter Laws Of Motion Using Notes

It is a must that Class 11 students should follow some strategic tips to understand Laws Of Motion. By understanding topics, students can score well in the questions related to the Laws Of Motion. Below, we have mentioned some strategic tips, students can use -

• Look Through the Chapter: First tip is to look through all the topics included in the chapter Laws Of Motion. Students can get a brief idea about all the topics and concepts of the Class 11th Physics chapter using Notes or Syllabus. 
•   Complete the Concepts: After getting some idea about the chapter and its topics, students can complete all the topics and concepts with the help of Laws Of Motion Class 11 notes. However, notes are only ideal to recall the previou learning so, with the help of Laws Of Motion Class 11 notes students can only revise the complete concepts of the Class 11 Physics chapter.
• Practising Questions: To understand topics of the chapter Laws Of Motion, students need to practise questions in a fine way. Different kinds of questions are included in the Laws Of Motion notes. Those questions given to practise are: objective questions, short type questions, very short type questions, short type questions, etc. 
• Note Down the Mistakes: While practising Laws Of Motion questions from the given notes, students are advised to note down all the mistakes. By noting down the mistakes, students can easily improvise their preparation level for the chapter Laws Of Motion. 
• Correction of Mistakes: Correcting the errors are equally important as solving questions from the chapter Laws Of Motion. It can aid students to solve all their doubts regarding the chapter Laws Of Motion. 
• Take Frequent Breaks: It is very necessary for Class 11 students to take frequent breaks while preparing for Laws Of Motion. Frequent breaks can help students to focus and grab more information from the revision notes of Laws Of Motion Class 11. 
• Proper Revision: Try to do proper revision of all the concepts and topics you have studied so far. Irregular revision can put you in devastating situations while recalling all the topics you have studied so far in Laws Of Motion Class 11 notes. 
• Sleep Well and Eat Healthy: Taking care of yourself is very important during study and when you are trying to revise the topics like Laws Of Motion Class 11 using notes then you must pay attention to the quality sleep and healthy diet so that your brain can efficiently manage all the topics you have studied so far.

Why is it Important for Class 11 Students to Refer to the Laws Of Motion Class 11 Notes?

It is necessary for Class 11 students to refer to the Laws Of Motion Class 11 notes as it can aid to progress in the chapter. It is also important to refer as it provides many important contents from concepts to important formulas and derivation. Necessary information included in the Class 11 Physics notes are: definitions, concepts, topics, questions, etc.

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Laws of Motion Class 11 Physics Notes And Questions

Please refer to Laws of Motion Class 11 Physics notes and questions with solutions below. These revision notes and important examination questions have been prepared based on the latest Physics books for Class 11. You can go through the questions and solutions below which will help you to get better marks in your examinations.

Class 11 Physics Laws of Motion Notes and Questions

Newton’ 1st law or Law of Inertia

Every body continues to be in its state of rest or of uniform motion until and unless and until it is compelled by an external force to change its state of rest or of uniform motion.

The property by virtue of which a body opposes any change in its state of rest or of uniform motion is known as inertia. Greater the mass of the body greater is the inertia. That is mass is the measure of the inertia of the body.

Numerical Application

Physical Application 1. When a moving bus suddenly stops, passenger’s head gets jerked in the forward direction. 2. When a stationery bus suddenly starts moving passenger’s head gets jerked in the backward direction. 3. On hitting used mattress by a stick, dust particles come out of it. 4. In order to catch a moving bus safely we must run forward in the direction of motion of bus. 5. Whenever it is required to jump off a moving bus, we must always run for a short distance after jumping on road to prevent us from falling in the forward direction.

Key Concept In the absence of external applied force velocity of body remains unchanged.

Newton’ 2 nd law Rate of change of momentum is directly proportional to the applied force and this change always takes place in the direction of the applied force.

Note :-  Above result is not Newton’s second law rather it is the conditional result obtained from it, under the condition when m = constant.

Physical Application Horizontal Plane i) Case – 1 N Body kept on horizontal plane is at rest.

For vertical direction N = mg(since body is at rest)

ii) Body kept on horizontal plane is accelerating horizontally under single horizontal force.

For vertical direction N = mg (since body is at rest) For horizontal direction F = ma mg

iii) Body kept on horizontal plane is accelerating horizontally towards right under two horizontal forces. (F 1  > F 2 ) 

For vertical direction N = mg (since body is at rest)  For horizontal direction F 1  – F 2  = ma

iv) Body kept on horizontal plane is accelerating horizontally under single inclined force

For vertical direction N + FSinθ = mg (since body is at rest)

For horizontal direction FCosθ = ma

v) Body kept on horizontal plane is accelerating horizontally towards right under an inclined force and a horizontal force.

For vertical direction N + F1Sinθ = mg (since body is at rest)

For horizontal direction F 1 Cosθ – F 2  = ma

vi) Body kept on horizontal plane is accelerating horizontally towards right under two inclined forces acting on opposite sides.

For vertical direction N + F 1 Sinθ = mg + F 2  SinФ (since body is at rest)

For horizontal direction  F 1 Cosθ – F 2 CosФ = ma

Inclined Plane

i) Case – 1  Body sliding freely on inclined plane.

Perpendicular to the plane N = mgCosθ (since body is at rest)

Parallel to the plane  mgSinθ = ma

ii) Case – 2 Body pulled parallel to the inclined plane.

Parallel to the plane  F – mgSinθ = ma

iii) Case – 3 Body pulled parallel to the inclined plane but accelerating downwards.

Parallel to the plane  mgSinθ – F = ma

iv) Case – 4 Body accelerating up the incline under the effect of two forces acting parallel to the incline.

Parallel to the plane  F 1  – F 2  – mgSinθ = ma

v) Case – 5 Body accelerating up the incline under the effect of horizontal force.

Perpendicular to the plane N = mgCosθ + F 1 Sinθ (since body is at rest)

Parallel to the plane  F1Cosθ – mgSinθ = ma

vi) Case – 6 Body accelerating down the incline under the effect of horizontal force and gravity.

Perpendicular to the plane N + FSinθ = mgCosθ (since body is at rest)

Parallel to the plane  FCosθ + mgSinθ = ma

vii) Case – 7 Body accelerating up the incline under the effect of two horizontal forces acting on opposite sides of a body and gravity.

Perpendicular to the plane  N + F 1 Sinθ = mgCosθ + F 2 Sinθ(since body is at rest)

Parallel to the plane  F2Cosθ – F1Cosθ – mgSinθ = ma

Vertical Plane

i) Case – 1 Body pushed against the vertical plane by horizontal force and moving vertically downward.

For horizontal direction mg = ma (since body is at rest)  For vertical direction F = N

ii) Case – 2 Body pushed against the vertical plane by horizontal force and pulled vertically upward. For vertical direction F2 – mg = ma

For horizontal direction (since body is at rest) N = F1

iii) Case – 3 Body pushed against the vertical plane by inclined force and accelerates vertically upward.

For horizontal direction  N = FSinθ (since body is at rest)

For vertical direction FCosθ – mg = ma

iv) Case – 3 Body pushed against the vertical plane by inclined force and accelerates vertically downward.

For vertical direction  FCosθ + mg = ma

Tension In A Light String Force applied by any linear object such as string, rope, chain, rod etc. is known as it’s tension. Since string is a highly flexible object so it can only pull the object and can never push. Hence tension of the string always acts away from the body to which it is attached irrespective of the direction.

Physical Application

i) Flexible wire holding the lamp pulls the lamp in upward direction and pulls the point of suspension in the downward direction. ii) Rope holding the bucket in the well pulls the bucket in the upward direction and the pulley in the downward direction. iii) Rope attached between the cattle and the peg pulls the cattle towards the peg and peg towards the cattle. iv) When a block is pulled by the chain, the chain pulls the block in forward direction and the person holding the chain in reverse direction.

Key Point In case of light string, rope, chain, rod etc. tension is same all along their lengths.

Consider a point P on a light (massless) string. Let tensions on either side of it be T 1  and T 2  respectively and the string be accelerating towards left under these forces. Then for point P T 1  – T 2  = ma Since string is considered to be light mass m of point P is zero or, T 1  – T 2  = 0 or, T 1  = T 2

i) Case – 1 Two bodies connected by a string are placed on a smooth horizontal plane and pulled by a horizontal force.

For vertical equilibrium of m1 and m2 N 1 = m 1 g and N2 = m 2 g For horizontal acceleration of m1 and m2 F – T = m1a and T = m 2 a (Since both the bodies are connected to the same single string they have same acceleration)

ii) Case – 2 Two bodies connected by a horizontal string are placed on a smooth horizontal plane and pulled by a inclined force.

For vertical equilibrium of m1 and m2 N1 + FSinθ = m1g and N2 = m2g For horizontal acceleration of m1 and m2 FCosθ – T = m1a and T = m2a (since both the bodies are connected to the same single string they have same accelerations)

iii) Case – 3 Two bodies connected by a inclined string are placed on a smooth horizontal plane and pulled by a inclined force.

For vertical equilibrium of m1 and m 2 N1 + FSinθ = m1g + TSinθ and N2 + TSinθ = m2g For horizontal acceleration of m1 and m 2 FCosθ – TCosθ = m1a and TCosθ = m 2 a (since both the bodies are connected to the same single string they have same accelerations)

iv) Case – 4 Two bodies connected by a string made to accelerate up the incline by applying force parallel to the incline.

For equilibrium of m 1  and m 2  in the direction perpendicular to the plane N1 = m1gCosθ and N2 = m 2 gCosθ For acceleration of m1 and m 2  up the incline F – T – m 1 gSinθ = m1a and T – m 2 gSinθ = m 2 a

Tension of A light Rigid Rod Force applied by rod is also known as its tension. Since rod is rigid, it cannot bend like string. Hence rod can pull as well as push. Tension of rod can be of pulling as well as pushing nature but one at a time. Tension of a rod attached to the body may be directed towards as well as away from the body.

Physical Application i) Pillars supporting the house pushes the house in the upward direction and pushes the ground in the downward direction. ii) Wooden bars used in the chair pushes the ground in the downward direction and pushes the seating top in the upward direction. iii) Parallel bars attached to the ice-cream trolley pushes the trolley in the forward direction and pushes the ice-cream vendor in the backward direction.(when the trolley is being pushed by the vendor) iv) Rod holding the ceiling fan pulls the fan in the upward direction and pulls the hook attached to the ceiling in the downward direction. v) Parallel rods attached between the cart and the bull pulls the cart in the forward direction and pulls the bull in the backward direction.

Different Cases of Light Rigid Rod i) Case – 1 Rod attached from the ceiling and supporting the block attached to its lower end. Since the block is at rest

ii) Case – 2 Rod is attached between two blocks placed on the horizontal plane and the blocks are accelerated by pushing force.

For vertical equilibrium of m1 and 

N 1 = m1g and N 2 = m 2 g (Since both the bodies connected to the rod will have same acceleration)

iii) Case – 3 Rod is attached between two blocks placed on the horizontal plane and the blocks are accelerated by pulling force.

For vertical equilibrium of m1 and m 2 N 1 = m1g and N2 = m2g

For horizontal acceleration of m1 and m 2 F – T = m1a and T = m 2 a (Since both the bodies are connected to the same rod they have same acceleration)

iv) Case – 4 Rod is attached between two blocks placed on the incline plane and the blocks are accelerated by pushing parallel to the incline.

For vertical equilibrium of m1 and m 2   N 1 = m 1 gCosθ and N 2 = m 2 gCosθ

For acceleration of m1 and m 2 parallel to  the incline m1gCosθ F – m 1 gSinθ – T = m1a, θ m1g T – m 2 gSinθ = m 2 a

Fixed Pulley It is a simple machine in the form of a circular disc or rim supported by spokes having groove at its periphery. It is free to rotate about an axis passing through its center and perpendicular to its plane.

Key Point In case of light pulley, tension in the rope on both the sides of the pulley is same (to be proved in the rotational mechanics)

Anticlockwise Torque – Clockwise Torque = Moment of Inertia x Angular acceleration T1 x r – T2 x r = Iα Since the pulley is light and hence considered to be massless, it’s moment of inertia I = 0 or, T1 x r – T2 x r = 0 or, T1 x r = T2 x r or, T1 = T2

Different Cases of Fixed Pulley

i) Case – 1 Two bodies of different masses (m1 > m2) are attached at two ends of a light string passing over a smooth light pulley

For vertical equilibrium of pulley  T1 = T + T = 2T

For vertical acceleration of m 1 and m 2 m1g – T = m 1 a and T – m 2 g = m 2 a  m1 accelerates downwards and m 2 accelerates upwards(m 1 >m 2 )

ii) Case – 2 Two bodies of different masses are attached at two ends of a light string passing over a light pulley. m1 is placed on a horizontal surface and m2 is hanging freely in air.

For vertical equilibrium m 1 N = m1g

For horizontal acceleration of m 1 T = m 1 a

For vertically downward acceleration of m2  m2g – T = m 2 a

iii) Case – 3 Two bodies of different masses are attached at two ends of a light string passing over a light pulley. m1 is placed on an inclined surface and m2 is hanging freely in air.

For equilibrium of m1 perpendicular to incline plane  N = m1gCosθ

For acceleration of m1 up the incline plane  T – m1gSinθ = m1a

For vertically downward acceleration of m2 m2g – T = m2a

Movable Pulley The pulley which moves in itself is known as movable pulley.

Key Point In case of light movable pulley, acceleration of a body (pulley) goes on decreasing on increasing the number of strings attached to it. That is the body attached with two ropes moves with half the acceleration of the body attached with single rope.

Length of the string is constant  x + 2y + z = L (Constant) Differentiating both sides with respect to t (Time) dx + 2dy + dz = dL dt dt dt dt y or, v1 + 2v 2 + 0 = 0 (z and L are constant)  or, v1 + 2v 2 = 0 Again differentiating both sides with respect to dv1 + 2dv 2 = 0 dt dt  or, a 1 + 2a 2 = 0 or, a 1 = – 2a 2

That is acceleration of m1 (body attached to a single string) is opposite and twice the acceleration of m2 (body attached to a double string)

Different Cases of Light Movable Pulley i) Case – 1 Mass m1 is attached at one end of the string and the other end is fixed to a rigid support. Mass m2 is attached to the light movable pulley.

For vertical acceleration of m1  m1g – T = m 12 a (m1 is connected to a single string)

For vertical acceleration of m 2 T 2  – m 2 g = m 2 a (m1 accelerates downwards and m 2  accelerates upwards since m1>2m2)

For the clamp holding the first pulley T1 = 2T

For the clamp holding the movable pulley  2T – T2 = mpulleya or, 2T – T2 = 0 (light pulley) or, 2T = T2

ii) Case – 2 Mass m1 is attached at one end of the string and placed on a smooth horizontal surface and the other end is fixed to a rigid support after passing through a light movable suspended pulley. Mass m 2 is attached to the light movable pulley.

For vertical equilibrium of m1 N = m1g

For horizontal acceleration of m1 T = m12a

For vertical motion of m2  m2g – 2T = m2a

iii) Case – 3 Mass m 1  is attached to the movable pulley and placed on a smooth horizontal surface. One end of the string is attached to the clamp holding the pulley fixed to the horizontal surface and from its other end mass m 2  suspended.

For vertical equilibrium of m 1 N = m1g

For horizontal motion of m 1 2T = m1a

For vertical motion of m2  m 2 g – T = m 22 a

iv) Case – 4 Mass m1 is attached to a movable pulley and placed on a smooth inclined surface. Mass m2 is is suspended freely from a fixed light pulley.

For acceleration of m1 up the incline plane  2T – m1gSinθ = m 1 a

For vertically downward acceleration of m2  m2g – T = m22a

Newton’ 3 rd law or Law of Action and Reaction Every action is opposed by an equal and opposite reaction. or For every action there is an equal and opposite reaction.

F 12  is the force on the first body (m1) due to second body (m 2 ) F 21  is the force on the second body (m2) due to first body (m 1 )

Force on the first body due to second body (F 12 ) is equal and opposite to the force on the second body due to first body (F 21 ).

i) When we push any block in the forward direction then block pushes us in the backward direction with an equal and opposite force.

ii) Horse pulls the rod attached to the cart in the forward direction and the tension of the rod pulls the cart in the backward direction.

iii) Earth pulls the body on its surface in vertically downward direction and the body pulls the earth with the same force in vertically upward direction.

iv) While walking we push the ground in the backward direction using static frictional force and the ground pushes us in the forward direction using static frictional force.

v) When a person sitting on the horse whips the horse and horse suddenly accelerates, the saddle on the back of the horse pushes the person in the forward direction using static frictional force and the person pushes the saddle in the backward direction using static frictional force.

Note –  Normal reaction of the horizontal surface on the body is not the reaction of the weight of the body because weight of the body is the force with which earth attracts the body towards its center, hence its reaction must be the force with which body attracts earth towards it.

Linear Momentum It is defined as the quantity of motion contained in the body. Mathematically it is given by the product of mass and velocity. It is a vector quantity represented by p.

Principle Of Conservation Of Linear Momentum It states that in the absence of any external applied force total momentum of a system remains conserved. Proof- We know that,

Physical Application i) Recoil of gun – when bullet is fired in the forward direction gun recoils in the backward direction. ii) When a person jumps on the boat from the shore of river, boat along with the person on it moves in the forward direction. iii) When a person on the boat jumps forward on the shore of river, boat starts moving in the backward direction. iv) In rocket propulsion fuel is ejected out in the downward direction due to which rocket is propelled up in vertically upward direction.

Different Cases of Conservation of Linear Momentum

Recoil of gun Let mass of gun be mg and that of bullet be mb. Initially both are at rest, hence their initial momentum is zero. pi = mgug + mbub = 0 Finally when bullet rushes out with velocity vg, gun recoils with velocity vb, hence their final momentum is pf = mgvg + mbvb Since there is no external applied force, from the principal of conservation of linear momentum pf = pf or, mgvg + mbvb = 0 or, mgvg = -mbvb or, vg = – mbvb/ mg From above expression it must be clear that 1. Gun recoils opposite to the direction of motion of bullet. 2. Greater is the mass of mullet mb or velocity of bullet vb greater is the recoil of the gun. 3. Greater is the mass of gun mg, smaller is the recoil of gun.

Impulse and Impulsive Force Impulsive Force The force which acts on a body for very short duration of time but is still capable of changing the position, velocity and direction of motion of the body up to large extent is known as impulsive force. Example – 1. Force applied by foot on hitting a football. 2. Force applied by boxer on a punching bag. 3. Force applied by bat on a ball in hitting it to the boundary. 4. Force applied by a moving truck on a drum. Note- Although impulsive force acts on a body for a very short duration of time yet its magnitude varies rapidly during that small duration. Impulse Impulse received by the body during an impact is defined as the product of average impulsive force and the short time duration for which it acts. I = Favg x t

Relation Between Impulse and Linear Momentum Consider a body being acted upon by an impulsive force, this force changes its magnitude rapidly with the time. At any instant if impulsive force is F then elementary impulse imparted to the body in the elementary time dt is given by dI = F x dt Hence total impulse imparted to the body from time t 1 to t 2 is

But from Newton’s second law we know that F = dp dt or, Fdt = dp

or, I = p2 – p1 Hence impulse imparted to the body is equal to the change in its momentum.

Graph Between Impulsive Force and Time With the time on x axis and impulsive force on y axis the graph of the following nature is obtained

Area enclosed under the impulsive force and time graph from t 1  to t 2  gives the impulse imparted to the body from time t 1  to t 2 .

Physical Application i) While catching a ball a player lowers his hand to save himself from getting hurt. ii) Vehicles are provided with the shock absorbers to avoid jerks. iii) Buffers are provided between the bogies of the train to avoid jerks. iv) A person falling on a cemented floor receive more jerk as compared to that falling on a sandy floor. v) Glass wares are wrapped in a straw or paper before packing.

Equilibrium of Concurrent Forces If the number of forces act at the same point, they are called concurrent forces. The condition or the given body to be in equilibrium under the number of forces acting on the body is that these forces should produce zero resultant. The resultant of the concurrent forces acting on a body will be zero if they can be represented completely by the sides of a closed polygon taken in order.

Lami’s Theorem –  It states that the three forces acting at a point are in equilibrium if each force is proportional the sine of the angle between the other two forces.

Inertial and Non-inertial Frame of Reference Frame of reference is any frame with respect to which the body is analyzed. All the frames which are at rest or moving with a constant velocity are said to be inertial frame of reference. In such frame of reference all the three laws of Newton are applicable. Any accelerated frame of reference is said to be non-inertial frame of reference. In such frames all the three laws of Newton are not applicable as such. In order to apply Newton’s laws of motion in a non-inertial frame, along with all other forces a pseudo force F = ma must also be applied on the body opposite to the direction of acceleration of the frame.

Reading of Spring Balance Reading of a spring balance is equal to the tension in the spring of the balance but measured in kilogram. Reading = T kgf/ g

Reading of Weighing Machine Reading of a weighing machine is equal to the normal reaction applied by the machine but measured in kilogram. Reading = N kgf/ g

Observer Outside the Lift

Observer Inside the Lift (Body is at rest according to the observer inside the lift)

Friction –  The property by virtue of which the relative motion between two surfaces in contact is opposed is known as friction. Frictional Forces –  Tangential forces developed between the two surfaces in contact, so as to oppose their relative motion are known as frictional forces or commonly friction. Types of Frictional Forces – Frictional forces are of three types :- 1. Static frictional force 2. Kinetic frictional force 3. Rolling frictional force

Static Frictional Force –  Frictional force acting between the two surfaces in contact which are relatively at rest, so as to oppose their relative motion, when they tend to move relatively under the effect of any external force is known as static frictional force. Static frictional force is a self adjusting force and its value lies between its minimum value up to its maximum value. Minimum value of static frictional force – Minimum value of static frictional force is zero in the condition when the bodies are relatively at rest and no external force is acting to move them relatively. fs(min) = 0 Maximum value of static frictional force – Maximum value of static frictional force is μsN (where μs is the coefficient of static friction for the given pair of surface and N is the normal reaction acting between the two surfaces in contact) in the condition when the bodies are just about to move relatively under the effect of external applied force.

fs(max) = μsN Therefore, fs(min) ≤ fs ≤ fs(max) or, 0 ≤ fs ≤ μsN

Kinetic Frictional Force – Frictional force acting between the two surfaces in contact which are moving relatively, so as to oppose their relative motion, is known as kinetic frictional force. It’s magnitude is almost constant and is equal to μkN where μk is the coefficient of kinetic friction for the given pair of surface and N is the normal reaction acting between the two surfaces in contact. It is always less than maximum value of static frictional force. fk = μkN Since, fk < fs(max) = μsN Therefore, μkN < μsN or, μk < μs

Limiting Frictional Force –  The maximum value of static frictional force is the maximum frictional force which can act between the two surfaces in contact and hence it is also known as limiting frictional force. Laws of Limiting Frictional Force – 1. Static friction depends upon the nature of the surfaces in contact. 2. It comes into action only when any external force is applied to move the two bodies relatively, with their surfaces in contact. 3. Static friction opposes the impending motion. 4. It is a self adjusting force. 5. The limiting frictional force is independent of the area of contact between the two surfaces.

Modern View –  According to modern theory the cause of friction is the atomic and molecular forces of attraction between the two surfaces at their actual point of contact. When any body comes in contact with any other body then due to their roughness at the microscopic level they come in actual contact at several points. At these points the atoms and molecules come very close to each other and intermolecular force of attraction start acting between them which opposes their relative motion.

Contact Force –  The forces acting between the two bodies due to the mutual contact of their surfaces are known as contact forces. The resultant of all the contact forces acting between the bodies is known as resultant contact force. Example

friction (f) and normal reaction (N) are contact forces and their resultant (Fc) is the resultant is the resultant contact force.

Fc = √ f2 + N2 Since maximum value of frictional force is Limiting frictional force (μsN) Therefore maximum value of contact force is

Fc(max) = √ (μsN) 2 + N2 or, Fc(max) = N√ μs 2 + 12 or, Fc(max) = N√ μs 2 + 1

Angle of Friction –  The angle between the resultant contact force (of normal reaction and friction) and the normal reaction is known as the angle of friction.

Tan λ = f / N or, λ = Tan -1  f/ N  or, λ max = Tan -1  f max/N or, λ max = Tan-1 μsN/N or, λ max = Tan-1 μs Angle of Repose – The angle of the inclined plane at which a body placed on it just begins to slide is known as angle of repose.

Parallel to the plane when body is at rest  mgSinθ = fs

When body is just about to slide

mgSinθ = fs(max) = μsN = μsmgCosθ or, Tanθ = μs or, θ = Tan-1/μs

Note –  Angle of repose is equal to the maximum value of angle of friction Rolling Frictional Force – Frictional force which opposes the rolling of bodies (like cylinder, sphere, ring etc.) over any surface is called rolling frictional force. Rolling frictional force acting between any rolling body and the surface is almost constant and is given by μrN. Where μr is coefficient of rolling friction and N is the normal reaction between the rolling body and the surface. fr = μrN Note – Rolling frictional force is much smaller than maximum value of static and kinetic frictional force. fr << fk < fs(max) or, μrN << μkN < μsN or, μr << μk < μs

Cause of Rolling Friction –  When any body rolls over any surface it causes a little depression and a small hump is created just ahead of it. The hump offers resistance to the motion of the rolling body, this resistance is rolling frictional force. Due to this reason only, hard surfaces like cemented floor offers less resistance as compared to soft sandy floor because hump created on a hard floor is much smaller as compared to the soft floor.

Need to Convert Kinetic Friction into Rolling Friction –  Of all the frictional forces rolling frictional force is minimum. Hence in order to avoid the wear and tear of machinery it is required to convert kinetic frictional force into rolling frictional force and for this reason we make the use of ball-bearings.

Friction: A Necessary Evil –  Although frictional force is a non-conservative force and causes lots of wastage of energy in the form of heat yet it is very useful to us in many ways. That is why it is considered as a necessary evil.

Advantages of Friction – i) Friction is necessary in walking. Without friction it would have been impossible for us to walk. ii) Friction is necessary for the movement of vehicles on the road. It is the static frictional force which makes the acceleration and retardation of vehicles possible on the road. iii) Friction is helpful in tying knots in the ropes and strings. iv) We are able to hold anything with our hands by the help of friction only.

Disadvantages of Friction – i) Friction causes wear and tear in the machinery parts. ii) Kinetic friction wastes energy in the form of heat, light and sound. iii) A part of fuel energy is consumed in overcoming the friction operating within the various parts of machinery.

Methods to Reduce Friction – i) By polishing –  Polishing makes the surface smooth by filling the space between the depressions and projections present in the surface of the bodies at microscopic level and there by reduces friction. ii) By proper selection of material –  Since friction depends upon the nature of material used hence it can be largely reduced by proper selection of materials. iii) By lubricating –  When oil or grease is placed between the two surfaces in contact, it prevents the surface from coming in actual contact with each other. This converts solid friction into liquid friction which is very small.

Physical Application Horizontal Plane i) Body kept on horizontal plane is at rest and no force is applied. For vertical equilibrium N = mg

ffriction = 0 (friction is a opposing force and there is no external applied force) ii) Body kept on horizontal plane is at rest under single horizontal force.

For vertical equilibrium  N = mg (since body is at rest)

For horizontal equilibrium (since body is at rest)  F = fs

iii) Body kept on horizontal plane is just about to move. For vertical direction  N = mg (since body is at rest)

For horizontal direction (since body is just about to move)  F = fs = fs(max) = μsN

iv) Body kept on horizontal plane is accelerating horizontally.

For vertical direction  N = mg (since body is at rest)

For horizontal direction  F – fk = ma or, F – μkN = ma

v) Body kept on horizontal plane is accelerating horizontally towards right under single upward inclined force.

For horizontal direction  FCosθ – fk = ma or, FCosθ – μkN = ma

vi) Body kept on horizontal plane is accelerating horizontally towards right under single downward inclined force.

For vertical direction N = FSinθ + mg (since body is at rest)

For horizontal direction  FCosθ – fk = ma  or, FCosθ – μkN = ma

vii) Body kept on horizontal plane is accelerating horizontally towards right under an inclined force and an opposing horizontally applied force.

For horizontal direction FCosθ – F1 – fk = ma  or, FCosθ – F1 – μkN = ma

For vertical direction(since body is at rest) N + F1Sinθ = mg + F2 SinФ

For horizontal direction F1Cosθ – F2CosФ – μkN = ma

Inclined Plane i) Case – 1 Body is at rest on inclined plane.

Parallel to the plane (since body is at rest) mgSinθ = fs

ii) Case – 2 Body is just about to move on inclined plane.

Parallel to the plane (since body is at rest) mgSinθ = f s  = f s(max)  = μ s N

iii) Case – 3 Body is accelerating downwards on inclined plane.

Parallel to the plane mgSinθ – f k  = ma or, mgSinθ – μ k N = ma

iv) Case – 4 Body is accelerating up the incline under the effect of force acting parallel to the incline.

Parallel to the plane F – fk – mgSinθ = ma or, F – μkN – mgSinθ = ma

Perpendicular to the plane N = mgCosθ + FSinθ (since body is at rest)

Parallel to the plane FCosθ – mgSinθ – fk = ma  or, FCosθ – mgSinθ – μkN ma

Vertical Plane i) Case – 1 Body pushed against the vertical plane by horizontal force and is at rest. For horizontal direction (since body is at rest) F = N

For vertical direction  mg = fs

ii) Case – 2 Body pushed against the vertical plane by horizontal force and pulled vertically upward

For horizontal direction (since body is at rest) F = N

For vertical direction F1 – mg – fs = ma

For vertical direction FCosθ – mg – fs = ma

Circular Motion –  When a body moves such that it always remains at a fixed distance from a fixed point then its motion is said to be circular motion. The fixed distance is called the radius of the circular path and the fixed point is called the center of the circular path.

Uniform Circular Motion –  Circular motion performed with a constant speed is known as uniform circular motion.

Angular Displacement –  Angle swept by the radius vector of a particle moving on a circular path is known as angular displacement of the particle. Example :– angular displacement of the particle from P1 to P2 is θ.

Relation Between Angular Displacement and Linear Displacement – Since, Angle = arc radius Anglular Displacement = arc P 1 P 2  / radius θ = s/ r

Angular Velocity –  Rate of change of angular displacement of a body with respect to time is known as angular displacement. It is represented by ω. Average Angular Velocity –  It is defined as the ratio of total angular displacement to total time taken. ωavg = Total Angular Displacement/Total Time Taken ωavg = Δθ/Δ t

Instantaneous Angular Velocity –  Angular velocity of a body at some particular instant of time is known as instantaneous angular velocity. Or Average angular velocity evaluated for very short duration of time is known as instantaneous angular velocity. ω = Lim ωavg = Δθ Δ t→0 Δt ω = dθ/dt

Relation Between Angular Velocity and Linear Velocity We know that angular velocity

ω = dθ/dt Putting, θ = s/r ω = d (s/r)/dt or, ω = 1 ds/r dt or, ω = v/r or, v = rω

Time Period of Uniform Circular Motion –  Total time taken by the particle performing uniform circular motion to complete one full circular path is known as time period. In one time period total angle rotated by the particle is 2π and time period is T. Hence angular velocity

ω = 2π /T or, T = 2π/ω

Frequency – Number of revolutions made by the particle moving on circular path in one second is known as frequency. f = 1/T = ω/ 2π

Centripetal Acceleration –  When a body performs uniform circular motion its speed remains constant but velocity continuously changes due to change of direction. Hence a body is continuously accelerated and the acceleration experienced by the body is known as centripetal acceleration (that is the acceleration directed towards the center).

Consider a particle performing uniform circular motion with speed v. When the particle changes its position from P1 to P2 its velocity changes from  → v 1  to  → v 2  due to change of direction. The change in velocity from P 1  to P 2  is  → Δv which is directed towards the center of the circular path according to triangle law of subtraction of vectors. From figure ΔOP 1 P 2  and ΔABC are similar, hence applying the condition of similarity BC/AB = P1/O P 1 /P 1 or, Δv/v = Δs/r or, Δv = vΔs/r Dividing both sides by Δv/Δt , = vΔs/rΔt Taking limit Δt  →  0 both sides,

Since the change of velocity is directed towards the center of the circular path, the acceleration responsible for the change in velocity is also directed towards center of circular path and hence it is known as centripetal acceleration. Centripetal Force –  Force responsible for producing centripetal acceleration is known as centripetal force. Since centripetal acceleration is directed towards the center of the circular path the centripetal force is also directed towards the center of the circular path.

If a body is performing uniform circular motion with speed v and angular velocity ω on a circular path of radius r, then centripetal acceleration is given by Net Acceleration of a Body Performing Non-Uniform Circular Motion When a body performs non-uniform circular motion its speed i.e. magnitude of instantaneous velocity varies with time due to which it experiences tangential acceleration aT along with the centripetal acceleration aC. Since both the accelerations act simultaneously on a body and are mutually perpendicular to each other, the resultant acceleration aR is given by their vector sum.

Physical Application of Centripetal Force i) Case – 1 Circular motion of a stone tied to a string.

Centripetal force is provided by the tension of the string Fc = mv 2 /r = T

ii) Case – 2 Circular motion of electron around the nucleus. Centripetal force is provided by the electrostatic force of attraction between the positively charged nucleus and negatively charged electron Fc = mv 2 /r = F E iii) Case – 3 Circular motion of planets around sun or satellites around planet.

Centripetal force is provided by the gravitational force of attraction between the planet and sun Fc = mv 2 /r = F g

iv) Case – 4 Circular motion of vehicles on a horizontal road. Centripetal force is provided by the static frictional force between the road and the tyre of the vehicle. Fc = mv2/r = fs

v) Case – 5 Circular motion of a block on rotating platform. Centripetal force is provided by the static frictional force between the block and the platform. Fc = mv2/r = fs

vi) Case – 6 Circular motion of mud particles sticking to the wheels of the vehicle. Centripetal force is provided by the adhesive force of attraction between the mud particles and the tyres of the vehicle. Fc = mv2/r = Fadhesive

At very high speed when adhesive force is unable to provide necessary centripetal force, the mud particles fly off tangentially. In order to prevent the particles from staining our clothes, mud-guards are provided over the wheels of vehicle.

vii) Case – 7 Circular motion of a train on a horizontal track. Centripetal force is provided by the horizontal component of the reaction force applied by the outer track on the inner projection of the outer wheels  Fc = mv2/r = NHorizontal

viii) Case – 8 Circular motion of a toy hanging from ceiling of vehicle.

Whenever car takes a turn, string holding the toy gets tilted outward such that the vertical component of the tension of string balances the weight of the body and the horizontal component of tension provides the necessary centripetal force. TSinθ = mv2/r TCosθ = mg Therefore, Tanθ = v2/rg

ix) Case – 9 Conical pendulum.

Whenever bob of a pendulum moves on a horizontal circle it’s string generates a cone. Such a pendulum is known as conical pendulum. The vertical component of the tension of the string balances the weight of the body and the horizontal component of tension provides the necessary centripetal force.

TSinθ = mv2/r TCosθ = mg Therefore, Tanθ = v2/rg

x) Case – 10 Well of death.

In the well of death, the rider tries to push the wall due to its tangential velocity in the outward direction due to which wall applies normal reaction in the inward direction. The vertical component of the normal reaction balances the weight of the body and its horizontal component provides the necessary centripetal force. NSinθ = mv2/r NCosθ = mg Therefore, Tanθ = v2/rg

xi) Case – 11 Turning of aero plane.

While taking a turn aero-plane tilts slightly inwards due to which it’s pressure force also gets tilted inwards due to which it’s pressure force also gets tilted inwards such that it’s vertical component balances the weight of the body and the horizontal component provides the necessary centripetal force.

FPSinθ = mv2/r FPCosθ = mg Therefore, Tanθ = v2/rg

xi) Case – 11 Banking of Roads In case of horizontal road necessary centripetal force mv2/r is provided by static frictional force. When heavy vehicles move with high speed on a sharp turn (small radius) then all the factors contribute to huge centripetal force which if provided by the static frictional force may result in the fatal accident. To prevent this roads are banked by lifting their outer edge. Due to this, normal reaction of road on the vehicle gets tilted inwards such that it’s vertical component balances the weight of the body and the horizontal component provides the necessary centripetal force.

nSinθ = mv2/r nCosθ = mg Therefore, Tanθ = v2/rg

xii) Case – 12 Bending of Cyclist In case of a cyclist moving on a horizontal circular track necessary centripetal force is provided by static frictional force acting parallel along the base. As this frictional force is not passing from the center of mass of the system it tends to rotate the cycle along with the cyclist and make it fall outward of the center of the circular path. To prevent himself from falling, the cyclist leans the cycle inwards towards the center of the circle due to which the normal reaction of the surface of road on the cycle also leans inward such that that its vertical component balances the weight of the body and the horizontal component provides the necessary centripetal force.

NSinθ = mv2/r NCosθ = mg Therefore, Tanθ = v2/rg

xiii) Case – 13 Motion of a Ball in a Bowl

When the bowl rotates with some angular velocity ω. The vertical component of the normal reaction of the bowl on the ball balances the weight of the body and its horizontal component provides the necessary centripetal force. NSinθ = mv2/r NCosθ = mg Therefore, Tanθ = v2/rg

xiv) Case – 14 Motion of a train on the banked tracks. At the turns tracks are banked by slightly elevating the outer tracks with respect to the inner ones. This slightly tilts the train inwards towards the center of the circular path due to which the normal reaction of the tracks on the train also gets slightly tilted inwards such that the vertical component of the normal reaction balances the weight of the train and it’s horizontal component provides the necessary centripetal force.

Vertical Circular Motion Whenever the plane of circular path of body is vertical its motion is said to be vertical circular motion.

Vertical Circular Motion of a Body Tied to a String

Consider a body of mass m tied to a string and performing vertical circular motion on a circular path of radius r. At the topmost point A of the body weight of the body mg and tension TA both are acting in the vertically downward direction towards the center of the circular path and they together provide centripetal force. TA + mg = mvA2/r

Critical velocity at the top most point As we go on decreasing the v A  , tension T A  also goes on decreasing and in the critical condition when vA is minimum tension T A  = 0. The minimum value of vA in this case is known as critical velocity TA(Critical) at the point A. From above 0 + mg = mvA(Critical) 2 /r or, vA(Critical) 2  = rg or, vA(Critical)= √rg

If the velocity at point A is less than this critical velocity then the string will slag and the body in spite of moving on a circular path will tend to fall under gravity.

Critical velocity at the lower most point

Taking B as reference level of gravitational potential energy and applying energy conservation

This is the minimum possible velocity at the lower most point for vertical circular motion known as critical velocity at point B.

Tension at lowermost point in critical condition For lowermost point B net force towards the center is centripetal force. Tension TB acts towards the center of the circular path whereas weight mg acts away from it. Hence,

TB – mg = m5gr/r or, TB = 6mg Hence in critical condition of vertical circular motion of a body tied to a string velocities at topmost and lowermost be √(rg) and √(5rg) respectively and tensions in the strings be 0 and 6mg respectively.

General Condition for Slagging of String in Vertical Circular Motion For the body performing vertical circular motion tied to a string, slagging of string occurs in the upper half of the vertical circle. If at any instant string makes angle θ with vertical then applying net force towards center is equal to centripetal force, we have

T + mgCos θ = mv 2 /r For slagging T = 0, 0 + mgCos θ = mv 2 /r or, v = √rgCos θ Case-1 At Topmost point θ = 0, therefore v = √rg Case-2 At θ = 60o, therefore v = √rgCos60 = √rg/2

Case-3 When string becomes horizontal that is at θ = 90o, v = √rgCos90 = 0

Velocity Relationship at different Points of Vertical Circular Motion Let initial and final velocities of the body performing vertical circular motion be v1 and v2 and the angle made by string with the vertical be θ1 and θ2. Taking lowermost point of vertical circular path as reference level and applying energy conservation,

Vertical Circular Motion of a Body Attached to a Rod Since rod can never slag hence in the critical situation a body attached to the rod may reach the topmost position A of the vertical circular path with almost zero velocity. In this case its weight mg acts in vertically downward direction and tension of rod acts on the body in the vertically upward direction. Applying net force towards center is equal to centripetal force,

Putting vA = 0 (for critical condition) mg – TA = 0 or, TA = mg

Critical velocity and Tension at the lower most point

This is the minimum possible velocity at the lower most point for vertical circular motion known as critical velocity at point B. vB(Critical) = √4rg

Tension at lowermost point in critical condition For lowermost point B applying net force towards center is equal to centripetal force. Tension TB acts towards the center of the circular path whereas weight mg acts away from it in vertically downward direction. Hence,

Hence in critical condition of vertical circular motion of a body attached to the rod velocities at topmost and lowermost be 0 and √4rg respectively and tensions in the rod be mg (pushing nature) and 5mg (pulling nature) respectively.

Motion of A Body Over Spherical Surface

A body of mass m is moving over the surface of the smooth sphere of radius r. At any instant when the radius of sphere passing through the body makes angle θ with the vertical the tangential velocity of the body is v. Since net force towards the center is centripetal force we have

mgCosθ – N = mv2/r or, N = mgCosθ – mv2/r

if v increases N decreases and when the body just loses contact with the sphere N = 0. Putting N = 0, 0 = mgCosθ – mv 2 /r or, mv 2 = mgCosθ/r or, v = √rg Cosθ

This is the minimum velocity at which the body loses contact and it is the maximum velocity at which the body remains in contact with the surface.

CENTRIFUGAL FORCE

It is a pseudo force experienced by a body which is a part of the circular motion. It is a non-realistic force and comes into action only when the body is in a circular motion. Once the circular motion of the body stops, this force ceases to act. Its magnitude is exactly same as that of centripetal force but it acts opposite to the direction of the centripetal force that is in the radially outward direction. Frame of reference attached to a body moving on a circular path is a non-inertial frame since it an accelerated frame. So when ever any body is observed from this frame a pseudo force F = ma = mv2/r = mrω2 must be applied on the body opposite to the direction of acceleration along with the other forces. Since the acceleration of the frame in circular motion is centripetal acceleration a = v2/r directed towards the center of the circular path, the pseudo force applied on the bodies observed from this frame is F = mv2/r directed away from the center of the circular path. This pseudo force is termed as a centrifugal force.

It is an apparatus used to separate cream from milk. It works on the principal of centrifugal force. It is a cylindrical vessel rotating with high angular velocity about its central axis. When this vessel contains milk and rotates with high angular velocity all the particles of milk start moving with the same angular velocity and start experiencing centrifugal force FCentrifugal = mrω 2  in radially outward direction. Since centrifugal force is directly proportional to the mass of the particles, massive particles of milk on experiencing greater centrifugal force starts depositing on the outer edge of the vessel and lighter cream particles on experiencing smaller centrifugal force are collected near the axis from where they are separated apart.

Critical Condition of Vertical Circular MOtion

Very Short Answer

Question. Action and reaction forces do not balance each other. Why? Answer.  Since they are acting on different bodies.

Question. The string is holding the maximum possible weight that it could withstand. What will happen to the string if the body suspended by it starts moving on a horizontal circular path and the string starts generating a cone? Answer.  It will break because tension in the string increases as soon as the body starts moving.

Question. Can a body remain in state of rest if more than one force is acting upon it? Answer.  Yes, if all the forces acting on it are in equilibrium.

Question. Is the centripetal force acting on a body performing uniform circular motion always constant? Answer.  No, only its magnitude remains constant but its direction continuously goes on changing.

Question. Is net force needed to keep a body moving with uniform velocity? Answer.  No.

Question. Is Newton’s 2nd law (F = ma) always valid. Give an example in support of your answer? Answer.  It is valid in an inertial frame of reference. In non-inertial frame of reference (such as a car moving along a circular path), Newton’s 2 nd  law doesn’t hold apparently.

Question. What is the maximum acceleration of a vehicle on the horizontal road? Given that coefficient of static friction between the road and the tyres of the vehicle is μ. Answer.  a max  = fs(max)/m = μN/m = μmg/m = μg.

Question. Why guns are provided with the shoulder support? Answer.  So that the recoil of gun may be reduced by providing support to the gun by the shoulders.

Question. While paddling a bicycle what are the types of friction acting on rear wheels and in which direction? Answer.  Static friction in forward direction and rolling friction in backward direction.

Question. What is the reaction force of the weight of a book placed on the table? Answer.  The force with which the book attracts the earth towards it.

We hope the above Laws of Motion Class 11 Physics are useful for you. If you have any questions then post them in the comments section below. Our teachers will provide you an answer. Also refer to MCQ Questions for Class 11 Physics

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• Introduction To Motion
• Laws Of Motion

Newton's Laws of Motion

This article will go through Sir Isaac Newton’s Laws of Motion, which revolutionised our understanding of the physical world centuries ago. This article explores Newton’s three laws and provides a deep understanding of their implications. Starting with Newton’s First Law of Motion, also known as the Law of Inertia, we delve into how objects behave when at rest or in uniform motion. Moving on to Newton’s Second Law of Motion, we unravel the relationship between mass, acceleration and external forces. Next, we explore Newton’s Third Law of Motion, shedding light on the concept of action and reaction. A concise summary of Newton’s laws offers a recap of the key concepts, while numerical examples in the Laws of Motion Numericals section demonstrate practical applications. Finally, our Frequently Asked Questions (FAQs) section covers additional queries, ensuring a comprehensive understanding of Newton’s Laws of Motion.

Newton’s First Law of Motion

Newton’s First Law of Motion, also known as the Law of Inertia, is a fundamental principle that describes the behaviour of objects in the absence of external influences. The term “Law of Inertia” emphasizes the concept of inertia, which refers to the property of massive objects to resist changes in their state of motion. This idea stems from the observation that objects naturally maintain their current state of rest or motion, resisting any changes unless acted upon by an external force.

By naming the first law of motion the “Law of Inertia,” Newton highlighted this inherent property of objects and laid the groundwork for understanding how forces can cause changes in motion. Newton’s first law of motion states that objects persist in their current state of motion unless compelled to do otherwise by an external force. Whether an object is at rest or in uniform motion, it will continue in that state unless a net external force acts upon it.

One crucial insight provided by Newton’s First Law is that the object will maintain a constant velocity in the absence of a net force resulting from unbalanced forces acting on an object. If the object is already in motion, it will continue moving at the same speed and direction. Likewise, if the object is at rest, it will remain stationary. However, introducing an additional external force will cause the object’s velocity to change, responding to the magnitude and direction of the force applied.

Understanding Newton’s First Law of Motion sets the stage for a deeper exploration of the subsequent laws that govern the complexities of motion. By comprehending this fundamental principle, we gain crucial insights into how objects behave independently and how external forces influence their motion. The first law of motion provides a strong foundation for further understanding the dynamics and behaviour of objects in the physical world.

Newton’s Second Law of Motion

This section will explore Newton’s Second Law of Motion, which provides a deeper understanding of how bodies respond to external forces.

The second law of motion describes the relationship between the force acting on a body and the resulting acceleration. According to Newton’s second law, the force acting on an object is equal to the product of its mass and acceleration.

Mathematically, we express Newton’s Second Law as follows:

Here, F represents the force, m is the object’s mass and a is the acceleration produced. This equation reveals that the acceleration of an object is directly proportional to the magnitude of the net force applied in the same direction as the force and inversely proportional to the object’s mass.

By understanding Newton’s Second Law, we can determine how much an object will accelerate when subjected to a specific net force. The equation highlights the intricate relationship between force, mass, and acceleration, providing a quantitative framework for analysing the dynamics of objects in motion.

In the second law equation, a proportionality constant is represented by the letter “k.” When using the SI unit system, this constant is equal to 1. Therefore, the final expression simplifies to:

The concise and powerful expression of Newton’s Second Law showcases the fundamental principle that governs the relationship between force and acceleration in physics. With this law, we gain a quantitative understanding of how external forces impact the motion of objects based on their mass and the resulting acceleration they experience.

By exploring Newton’s Second Law of Motion , we deepen our insights into the mechanics of motion, setting the stage for further exploration of the principles that govern the complexities of physical phenomena.

Newton’s Third Law of Motion

This section will discuss Newton’s Third Law of Motion, revealing a fascinating relationship between forces exerted by interacting bodies.

Newton’s Third Law of Motion states that for every action, there is an equal and opposite reaction. When two bodies interact, they apply forces on each other that are equal in magnitude and opposite in direction. This law highlights the concept that forces always occur in pairs.

To illustrate this principle, consider the example of a book resting on a table. As the book applies a downward force equal to its weight on the table, the table, in turn, exerts an equal and opposite force on the book. This occurs because the book slightly deforms the table’s surface, causing the table to push back on the book, much like a compressed spring releasing its energy.

This third law of motion has profound implications, including conserving momentum. Momentum is a property of moving objects determined by an object’s mass and velocity. According to Newton’s third law, the total momentum of an isolated system remains constant. This means that in any interaction, the total momentum before and after the interaction remains the same, regardless of the forces involved.

Understanding Newton’s third law of motion deepens our comprehension of the interconnectedness and equilibrium within the physical world. It provides a framework for analysing and predicting the effects of forces in various scenarios, from everyday interactions to complex mechanical systems.

As we delve further into the subsequent sections on the laws of motion, we will continue building upon the foundational principles of inertia, force, and action-reaction relationships.

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Explaining Everyday Phenomena with Newton’s Laws of Motion

Jack wonders why his wallet falls from the passenger seat to the floor while driving to work. How can we explain this phenomenon using physics?

We can explain this to Jack using Newton’s first law of motion. Due to its inertia, the wallet tends to maintain its state of motion. As the car accelerates or decelerates, the wallet continues moving forward with the same velocity before the car’s motion changes. However, when the car suddenly stops or changes direction, an external force (in this case, the force exerted by the car floor) acts on the wallet, causing it to slide off the seat and onto the floor. This is because the wallet resists changes in its state of motion, as Newton’s first law of motion described.

Using Newton’s laws, how can we explain the magician’s ability to pull a tablecloth from underneath dishes?

Newton’s first law of motion best explains the magician’s trick of pulling a tablecloth from underneath dishes. The magician carefully applies a negligible horizontal force to the tablecloth while quickly pulling it. According to Newton’s first law, objects at rest (the dishes and glasses) tend to remain in their state of motion or rest unless acted upon by an external force. In this case, the sudden pull of the tablecloth applies a minimal frictional force on the dishes and glasses. Since the tablecloth is made extremely slippery, it reduces the friction between the tablecloth and the dishes, allowing them to remain undisturbed and stay in their original state of motion or rest.

We gain insights into various phenomena by understanding and applying Newton’s laws of motion. We can explain seemingly perplexing situations like the movement of objects in a moving car or the magician’s illusionary tricks involving objects on a table. These laws provide a solid foundation for comprehending the principles governing motion in our everyday lives.

Laws of Motion Summary

This section presents a visual summary of Newton’s Laws of Motion in the form of a flowchart. The flowchart provides an easy and digestible format to remember the key principles underlying the three laws of motion.

The flowchart highlights the three laws of motion established by Sir Isaac Newton:

• Newton’s First Law of Motion: The law of inertia states that an object at rest will remain at rest, and an object in motion will continue moving with a constant velocity, unless acted upon by an external force.
• Newton’s Second Law of Motion: This law relates the force acting on an object to its mass and acceleration. The force is equal to the product of mass and acceleration, where acceleration is the rate of change of velocity.
• Newton’s Third Law of Motion: The law of action and reaction states that for every action, there is an equal and opposite reaction. When one body exerts a force on another body, the second body simultaneously exerts a force of the same magnitude but in the opposite direction on the first body.

By referring to this flowchart, you can quickly grasp the fundamental principles of Newton’s Laws of Motion and understand how they govern the behaviour of objects in various scenarios. It serves as a useful tool for remembering Newton’s three laws of motion.

Laws of Motion Numericals

1. suppose a bike with a rider on it having a total mass of 63 kg brakes and reduces its velocity from 8.5 m/s to 0 m/s in 3.0 seconds. what is the magnitude of the braking force.

The combined mass of the rider and the bike = 63 kg Initial Velocity = 8.5 m/s Final Velocity = 0 m/s The time in which the bike stops = 3 s

The net force acting on the body equals the rate of change of an object’s momentum.

The momentum of a body with mass m and velocity v is given by  p = mv

Hence, the change in momentum of the bike is given by

Hence, the net force acting on the bike is given by

Substituting the value, we get

The magnitude of the braking force is -178.5 N.

2. Calculate the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s 2 .

We calculate the force using the following formula.

Substituting the values in the equation, we get

Frequently Asked Questions – FAQs

Who discovered the three laws of motion, why are the laws of motion important, what are newton’s laws of motion all about, what is the difference between newton’s laws of motion and kepler’s laws of motion, what are some daily life examples of newton’s 1st, 2nd and 3rd laws of motion.

• The motion of a ball falling through the atmosphere or a model rocket being launched up into the atmosphere are both excellent examples of Newton’s 1st law.
• Riding a bicycle is an excellent example of Newton’s 2nd law. In this example, the bicycle is the mass. The leg muscles pushing on the pedals of the bicycle is the force.
• You hit a wall with a certain amount of force, and the wall returns that same amount of force. This is an example of Newton’s 3rd law.

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NCERT Solutions Class 11 Physics Chapter 4 Laws of Motion

NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion - FREE PDF Download

NCERT for Chapter 4 Laws of Motion Class 11 Solutions by Vedantu, explores the principles governing the motion of objects. This chapter builds on the concepts of force and motion that explains the laws formulated by Sir Isaac Newton.

The chapter covers the practical applications of these laws through free-body diagrams, the equilibrium of forces, motion on inclined planes, and circular motion. These concepts are fundamental for solving real-world problems in engineering, design, and everyday activities, making the chapter essential for students pursuing further studies in physics and engineering. Also, understanding these laws is crucial for analysing and predicting the motion of objects in various physical situations. With Vedantu's Class 11 Physics NCERT Solutions , you'll find step-by-step explanations of all the exercises in your textbook, ensuring that you understand the concepts thoroughly. 

Glance on Physics Chapter 4 Class 11 - Laws of Motion

Chapter 4 of Class 11 Physics likely covers Newton's Laws of Motion, which are the foundation of classical mechanics.

The chapter begins with Newton's First Law, which states that an object will remain at rest or continue to move at a constant velocity unless acted upon by an external force. This law introduces the concept of inertia, which is the tendency of objects to resist changes in their state of motion.

The chapter emphasizes the use of free-body diagrams to visually represent the forces acting on an object. These diagrams are essential tools for analysing problems involving multiple forces and predicting the resulting motion.

The chapter includes a detailed analysis of forces acting on objects on slopes, considering both gravitational components and frictional forces. It addresses the forces involved in circular motion, particularly centripetal force, which keeps an object moving in a curved path.

This article contains chapter notes, important questions, exemplar solutions, exercises, and video links for Chapter 4 - Laws of Motion, which you can download as PDFs.

There are 23 fully solved questions in the exercise of class 11th Physics chapter 4 Laws of Motion.

Access NCERT Solutions for Class 11 Physics Chapter 4 – Laws of Motion

1. Give the magnitude and direction of the net force acting on 

a drop of rain falling down with a constant speed, 

Ans: The net force is zero.

As the speed of the rain drop falling down is constant, its acceleration is zero.

Therefore, from Newton’s second law of motion, the net force acting on the rain drop is zero.

a cork of mass $10g$ floating on water, 

It is known that the weight of a cork floating on water acts downward. 

The weight of the cork is balanced by buoyant force exerted by the water in the upward direction.

Therefore, no net force acts on the floating cork.

a kite skilfully held stationary in the sky, 

The kite is stationary in the sky indicates that it is not moving.

Therefore, from Newton’s first law of motion, the net force acting on the kite is zero.

a car moving with a constant velocity of $30{km}/{h}\;$ on a rough road, 

As the car is moving with constant velocity, its acceleration is zero.

Therefore, from Newton’s second law of motion, net force acting on the car is equal to zero.

a high-speed electron in space far from all material objects, and free of electric and magnetic fields. 

As the high speed electron is free from the influence of all the fields, no net force acts on the electron.

2. A pebble of mass \[\mathbf{0}.\mathbf{05kg}\] is thrown vertically upwards. Ignore air resistance and give the direction and magnitude of the net force on the pebble,

during its upward motion, 

Ans: It is known that,

Acceleration due to gravity always acts downward irrespective of the direction of motion of an object. The only force that acts on the pebble thrown vertically upward during its upward motion is the gravitational force.

From Newton’s second law of motion: $F=m\times a$ 

$F$ is the net force 

$m$ is the mass of the pebble, $m=0.05kg$ 

a is the acceleration due to gravity, $a=g=10{m}/{{{s}^{-2}}}\;$ 

$\Rightarrow F=0.05\times 10=0.5N$ 

Therefore, the net force on the pebble is $0.5N$ and this force acts in the downward direction. 

during its downward motion, 

Ans: The only force that acts on the pebble during its downward motion is the gravitational force.

Therefore, the net force on the pebble in its downward direction is same as in upward direction i.e., $0.5N$ and this force acts in the downward direction. 

at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of ${{45}^{\circ }}$ with the horizontal direction? 

Ans: When the pebble is thrown at an angle of ${{45}^{\circ }}$with the horizontal, it will have both the horizontal and vertical components of velocity. 

At the highest point, only the vertical component of velocity becomes zero. However, the pebble will have the horizontal component of velocity throughout its motion. This component of velocity produces no effect on the net force acting on the pebble. 

Therefore, the net force on the pebble is $0.5N$.

3. Neglect air resistance throughout and give the magnitude and direction of the net force acting on a stone of mass $0.1kg$ ,

just after it is dropped from the window of a stationary train, 

Ans: It is given that,

Mass of the stone, \[\mathbf{m}=\mathbf{0}.\mathbf{1kg}\] 

Acceleration of the stone, \[\mathbf{a}=g=10{m}/{{{s}^{2}}}\;\] 

From Newton’s second law of motion, 

The net force acting on the stone is $F=ma=mg$ 

$\Rightarrow F=0.1\times 10=1N$ 

It is known that acceleration due to gravity always acts in the downward direction. 

Therefore, the magnitude of force is $1N$ and its direction is vertically downward.

just after it is dropped from the window of a train running at a constant velocity of $36{km}/{h}\;$, 

The train is moving with a constant velocity. 

Therefore, its acceleration is zero in the direction of its motion, i.e. in the horizontal direction. 

Thus, no force is acting on the stone in the horizontal direction. 

The net force acting on the stone is because of acceleration due to gravity and it always acts vertically downward. 

just after it is dropped from the window of a train accelerating with $1m{{s}^{-2}}$ 

Ans: It is given that, 

The train is accelerating at the rate of $1m{{s}^{-2}}$.

Therefore, the net force acting on the stone is $F'=ma=0.1\times 1=1N$ 

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force stops acting on the stone. This is because of the fact that the force acting on a body at an instant depends on the situation at that instant and not on earlier situations. 

Therefore, the net force acting on the stone is given only by acceleration due to gravity i.e., $F=mg=1N$.

lying on the floor of a train which is accelerating with $1m{{s}^{-2}}$, the stone being at rest relative to the train. 

The weight of the stone is balanced by the normal reaction of the floor. The only acceleration is provided by the horizontal motion of the train. 

Acceleration of the train, $a=1m{{s}^{-2}}$

The net force acting on the stone will be in the direction of motion of the train. 

Magnitude: $F=ma=0.1\times 1=0.1N$ 

Therefore, the magnitude of force is $0.1N$ and its direction is in the direction of motion of the train.

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: 

ii. $T-\frac{m{{v}^{2}}}{l}$ ,

iii. $T+\frac{m{{v}^{2}}}{l}$,

iv. $0$ 

$T$  is the tension in the string. (Choose the correct alternative). 

Ans: (i) $T$ 

The centripetal force of a particle connected to a string revolving in a circular path around a centre is provided by the tension produced in the string.

Therefore, the net force on the particle is the tension $T$ , i.e., 

$F=T=\frac{m{{v}^{2}}}{l}$ 

Where $F$ is the net force acting on the particle. 

5. A constant retarding force of $50N$ is applied to a body of mass $20kg$ moving initially with a speed of $15m{{s}^{-1}}$ . How long does the body take to stop?

Retarding force, \[F=50N\] 

Mass of the body, $m=20kg$ 

Initial velocity of the body, \[u=15m/s\] 

Final velocity of the body, $v=0$ 

From Newton’s second law of motion,

The acceleration $(a)$ produced in the body: $F=ma$ 

$\Rightarrow -50=20\times a$ 

$\Rightarrow a=\frac{-50}{20}=-2.5m{{s}^{-2}}$ 

From the first equation of motion, 

The time $(t)$ taken by the body to come to rest: $v=u+at$ 

$\Rightarrow 0=15+(-2.5)t$ 

$\Rightarrow t=\frac{-15}{-2.5}=6s$ 

Therefore, the time taken by the body to stop is $6s$.

6. A constant force is acting on a body of mass $3.0kg$ changes its speed from $2.0m{{s}^{-1}}$ to $3.0m{{s}^{-1}}$  in $25s$ . The direction of the motion of the body remains unchanged.  What is the magnitude and direction of the force? 

Mass of the body, $m=3kg$ 

Initial speed of the body, \[u=2m/s\] 

Final speed of the body, \[v=3.5m/s\]

Time, $t=25s$ 

The acceleration $(a)$ produced in the body: $v=u+at$

$\Rightarrow a=\frac{v-u}{t}$ 

$\Rightarrow a=\frac{3.5-2}{25}=\frac{1.5}{25}=0.6m{{s}^{-2}}$ 

Force, $F=ma$ 

\[\Rightarrow F=3\times 0.06=0.18N\] 

As the application of force does not change the direction of the body, the net force acting on the body is in the direction of its motion. 

Therefore, the magnitude of force is $0.18N$ and direction is along the direction of motion.

7. A body of mass $5kg$ is acted upon by two perpendicular forces $8N$ and $6N$. Give the magnitude and direction of the acceleration of the body.

Mass of the body, $m=5kg$ 

Representation of given data:

Law of Vectors

Resultant of two forces $8N$ and $6N$, $R=\sqrt{{{\left( 8 \right)}^{2}}+{{\left( -6 \right)}^{2}}}$ 

$\Rightarrow R=\sqrt{64+36}$ 

$\Rightarrow R=10N$ 

Angle made by $R$ with the force of $8N$ 

$\theta ={{\tan }^{-1}}\left( \frac{-6}{8} \right)=-{{36.87}^{\circ }}$ 

The negative sign indicates that $\theta $ is in the clockwise direction with respect to the force of magnitude $8N$. 

$a=\frac{F}{m}=\frac{10}{5}=2m{{s}^{-2}}$ 

Therefore, the magnitude of acceleration is $2m{{s}^{-2}}$ and direction is ${{37}^{\circ }}$ with a force of $8N$.

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 $4.0s$  just in time to save the child. What is the average retarding force on the vehicle? The mass of the three-wheeler is $400kg$ and the mass of the driver is $65kg$.

Initial speed of the three-wheeler, \[u=36\text{ }km/h\] 

Final speed of the three-wheeler, \[v=0m/s\] 

Time, $t=4s$ 

Mass of the three-wheeler, $m=400kg$ 

Mass of the driver, $m'=65kg$ 

Total mass of the system, \[M=400+65=465kg\] 

From the first law of motion, 

The acceleration $(a)$ of the three-wheeler can be calculated from: $v=u+at$ 

$\Rightarrow a=\frac{v-u}{t}=\frac{0-10}{4}$ 

$\Rightarrow a=-2.5{m}/{{{s}^{2}}}\;$ 

The negative sign indicates that the velocity of the three-wheeler is decreasing with time. 

The net force acting on the three-wheeler can be calculated as: $F=ma$ 

$\Rightarrow F=465\times (-2.5)$ 

$\Rightarrow F=-1162.5N$ 

The negative sign indicates that the force is acting against the direction of motion of the three-wheeler. 

Therefore, the average retarding force on the vehicle is $-1162.5N$.

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

Mass of the rocket, $m=20,000kg$ 

Initial acceleration, $a=5m{{s}^{-2}}$ 

Acceleration due to gravity, $g=10m{{s}^{-2}}$

By using Newton’s second law of motion, 

The net force (thrust) acting on the rocket can be written as:

$F-mg=ma$ 

$\Rightarrow F=m(g+a)$ 

$\Rightarrow F=20000(10+5)=20000\times 15$ 

$\Rightarrow F=3\times {{10}^{5}}N$  

Therefore, the initial thrust(force) of the blast is $3\times {{10}^{5}}N$.

10. A body of mass \[\mathbf{0}.\mathbf{40kg}\] moving initially with a constant speed of $10m{{s}^{-1}}$ subject to a constant force of $8.0N$ directed towards the south for $30s$. Take the instant the force is applied to be $t=0$ , the position of the body at that time to be predict its position at \[\mathbf{t}=-\mathbf{5}\text{ }\mathbf{s},\text{ }\mathbf{25}\text{ }\mathbf{s},\text{ }\mathbf{100}\text{ }\mathbf{s}\].

Mass of the body, $m=0.40kg$ 

Initial speed of the body, \[u=10m/s\] due north 

Force acting on the body, $F=-8.0N$ 

Acceleration produced in the body, $a=\frac{F}{m}$

At $t=0$ 

$\Rightarrow a=\frac{-8}{0.4}=-20m{{s}^{-2}}$ 

At $t=-5s$ 

Acceleration, $a'=0$ and $u=10m/s$ 

$s=ut+\frac{1}{2}a'{{t}^{2}}$ 

$\Rightarrow s=10\times (-5)=-50m$ 

Acceleration, $a=-20m{{s}^{-2}}$ and $u=10m/s$ 

$s'=ut'+\frac{1}{2}a{{(t')}^{2}}$ 

$\Rightarrow s'=10\times (25)+\frac{1}{2}\times (-20){{(25)}^{2}}$ 

$\Rightarrow s'=250+(-6250)$ 

$\Rightarrow s'=-6000m$ 

At $t=100s$

For $0\le t\le 30s$, $a=-20m{{s}^{-2}}$ and $u=10m/s$

${{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}$

$\Rightarrow {{s}_{1}}=10\times 30+\frac{1}{2}\times (-20)\times {{(30)}^{2}}$

$\Rightarrow {{s}_{1}}=300-9000$

$\Rightarrow {{s}_{1}}=8700m$

For $30s\le t\le 100s$

First equation of motion: $v=u+at$ 

$\Rightarrow v=10+(-20)\times 30$ 

$\Rightarrow v=-590m{{s}^{-1}}$ 

Velocity of body after $30s=-590m/s$ 

For motion between $30s$ to $100s$ i.e., in $70s$ 

${{s}_{2}}=vt+\frac{1}{2}a'{{t}^{2}}$

${{s}_{2}}=-590\times 70=-41300m$

Total distance, $s''={{s}_{1}}+{{s}_{2}}$ 

$\Rightarrow s''=-8700+(-41300)=-50000m$ 

Therefore, the position of the body at $t=-5s$ is $-50m$ at $t=25s$ is  $-6000m$ and at $t=100s$ is$-50,000m$.

11. A truck starts from rest and accelerates uniformly at $2.0m{{s}^{-2}}$. At $t=10s$, a stone is dropped by a person standing on the top of the truck ($6m$ high from the ground). Neglect air resistance. What are the

velocity, and 

Initial velocity of the truck, $u=0$ (Initially at rest)

Acceleration, $a=2m{{s}^{-2}}$ 

Time, $t=10s$ 

From first equation of motion: $v=u+at$ 

$\Rightarrow v=0+2\times 10=20{m}/{s}\;$ 

Therefore, the final velocity of the truck and the stone is$20{m}/{s}\;$.

At $t=11s$:

The horizontal component $({{v}_{x}})$  of velocity, in the absence of air resistance, remains unchanged, i.e. \[{{v}_{x}}=20m/s\].

The vertical component of velocity $({{v}_{y}})$ of the stone is given by the first equation of motion as: 

${{v}_{y}}=u+{{a}_{y}}\delta t$ 

Where, 

$\delta t=11-10=1s$ 

$a=g=10m/{{s}^{2}}$ 

$\Rightarrow {{v}_{y}}=0+10\times 1=10{m}/{{{s}^{2}}}\;$ 

The resultant velocity $(v)$  of the stone is:

Two Velocities of Different Magnitude

$\Rightarrow v=\sqrt{v_{x}^{2}+v_{y}^{2}}$ 

$\Rightarrow v=\sqrt{{{20}^{2}}+{{10}^{2}}}=\sqrt{400+100}$

$\Rightarrow v=\sqrt{500}=22.36m/s$

Consider $\theta $  as the angle made by the resultant velocity with the horizontal component of velocity, ${{v}_{x}}$.

$\Rightarrow \tan \theta =\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{{{v}_{y}}}{{{v}_{x}}} \right)$ 

$\Rightarrow \theta ={{\tan }^{-1}}\left( \frac{10}{20} \right)$

$\Rightarrow \theta ={{\tan }^{-1}}\left( 0.5 \right)$

$\Rightarrow \theta ={{26.57}^{\circ }}$

Therefore, the magnitude of resultant velocity is $22.36{m}/{s}\;$ making an angle of ${{26.57}^{\circ }}$ with the horizontal component of velocity.

acceleration of the stone at $t=11s$?

Ans: When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. 

Therefore, the acceleration of the stone is $10m{{s}^{-2}}$  and it acts vertically downward. 

12. A bob of mass \[\mathbf{0}.\mathbf{1kg}\] hung from the ceiling of a room by a string $2m$ long is set into oscillation. The speed of the bob at its mean position is $1m{{s}^{-2}}$. What is the trajectory of the bob if the string is cut when the bob is

at one of its extreme positions, 

Ans: If the string is cut when the bob is at one of its extremes then the bob will fall vertically on the ground.

Therefore, at the extreme position, the velocity of the bob becomes zero.

at its mean position. 

Ans: If the string is cut when the bob is at its mean position then the bob will trace a projectile path having the horizontal components of velocity only.

The direction of this velocity is tangential to the arc formed by the oscillating bob. At the mean position, the velocity of the bob is $1m/s$. 

Therefore, it will follow a parabolic path. 

13. What would be the readings on the scale of a man of mass \[\mathbf{70kg}\] stands on a weighing scale in a lift which is moving

upwards with a uniform speed of $10m{{s}^{-1}}$,

Mass of the man, $m=70kg$ 

Acceleration, $a=0$(uniform speed)

From Newton’s second law: $R-mg=ma$ 

$ma$ is the net force acting on the man.

$\Rightarrow R-70\times 10=0$ 

$\Rightarrow R=700N$ 

Reading on the weighing scale$=\frac{700}{g}=\frac{700}{10}=70kg$ 

Therefore, the mass of the man, $m=70kg$ 

downwards with a uniform acceleration of $5m{{s}^{-2}}$,

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ downward

From Newton’s second law: $R=m(g-a)$

$\Rightarrow R=70\left( 10-5 \right)=70\times 5$ 

$\Rightarrow R=350N$ 

Reading on the weighing scale$=\frac{350}{g}=\frac{350}{10}=35kg$ 

Therefore, the mass of the man, $m=35kg$ 

upwards with a uniform acceleration of $5m{{s}^{-2}}$.

Ans: Acceleration,$a=5{m}/{{{s}^{2}}}\;$ upward

From Newton’s second law: $R=m(g+a)$

$\Rightarrow R=70\left( 10+5 \right)=70\times 15$ 

$\Rightarrow R=1050N$ 

Reading on the weighing scale$=\frac{1050}{g}=\frac{1050}{10}=105kg$ 

Therefore, the mass of the man, $m=105kg$ 

What would be the reading if the lift mechanism failed and it hurtled down freely under gravity? 

Ans: When the lift moves freely under gravity,

Acceleration, $a=g=10m{{s}^{-2}}$ 

$\Rightarrow R=70\left( 10-10 \right)=0$ 

Reading on the weighing scale$=\frac{0}{g}=\frac{0}{10}=0kg$ 

Therefore, the man will be in a state of weightlessness.

14. Figure shows the position-time graph of a particle of mass $4kg$ . Consider one-dimensional motion only and find the 

Position- Time Graph for a Particle

force on the particle for $t<0,t>4s$,$0<t<4s$ ?

Ans:  For $t<0$ :

From the given graph, it is observed that the position of the particle is coincident with the time axis. It indicates that the displacement of the particle in this time interval is zero. 

Therefore, the force acting on the particle is zero. 

For $t>4s$ :

From the given graph, it is observed that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of $3m$ from the origin.

Therefore, no force acts on the particle. 

For $0<t<4$ :

From the given position-time graph, it is observed that it has a constant slope. Thus, the acceleration produced in the particle is zero. 

impulse at $t=0$ and $t=4s$ ? 

Ans: At $t=0$:

\[Impulse=Change\text{ }in\text{ }momentum=mv-mu\] 

Mass of the particle, $m=4kg$

Initial velocity of the particle, $u=0$ 

Final velocity of the particle, $v=\frac{3}{4}m/s$ 

Impulse $=4\left( \frac{3}{4}-0 \right)=3kgm{{s}^{-1}}$ 

Initial velocity of the particle, $u=\frac{3}{4}m/s$ 

Final velocity of the particle, $v=0$ 

Impulse\[=4\left( 0-\frac{3}{4} \right)=-3kgm{{s}^{-1}}\]

Therefore, the impulse at $t=0$ is $3kgm{{s}^{-1}}$ and at $t=4s$is $-3kgm{{s}^{-1}}$.

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

Horizontal force, $F=600N$ 

Mass of body A, ${{m}_{1}}=10kg$ 

Mass of body B, ${{m}_{2}}=20kg$

Total mass of the system, $m={{m}_{1}}+{{m}_{2}}=30kg$

The acceleration $(a)$ produced in the system is: $F=ma$ 

$\Rightarrow a=\frac{F}{m}=\frac{600}{30}=20m{{s}^{-2}}$

When force $F$ is applied on body A: 

Two Bodies of Masses 

The equation of motion can be written as: $F-T=ma$

$\Rightarrow T=F-{{m}_{1}}a$ 

\[\Rightarrow T=600-10\times 20=400N\]

Therefore, the tension in the string is $400N$.

B along the direction of the string

Ans: When force $F$  is applied on body B: 

The equation of motion can be written as:  $F-T={{m}_{2}}a$ 

$\Rightarrow T=F-{{m}_{2}}a$ 

\[\Rightarrow T=600-20\times 20=200N\]

Therefore, the tension in the string is $200N$.

16. Two masses $8kg$ and $12kg$ 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.

Ans: The given system of two masses and a pulley are represented in the following figure: 

Smaller mass, ${{m}_{1}}=8kg$ 

Larger mass, ${{m}_{2}}=12kg$

Tension in the string $=T$ 

Mass \[{{m}_{2}}\], owing to its weight, moves downward with acceleration $a$, and mass moves ${{m}_{1}}$upward.

Applying Newton’s second law of motion to the system of each mass: 

For mass ${{m}_{1}}$: 

The equation of motion can be written as: \[T-{{m}_{1}}g=ma\]    .......$(1)$ 

For mass ${{m}_{2}}$:

The equation of motion can be written as: \[{{m}_{2}}g-T={{m}_{2}}a\]   ........$(2)$ 

Adding equations $(1)$ and $(2)$, we get:

$({{m}_{2}}-{{m}_{1}})g=({{m}_{1}}+{{m}_{2}})a$ 

$\Rightarrow a=\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g$        ........$(3)$ 

$\Rightarrow a=\left( \frac{12-8}{12+8} \right)\times 10=\frac{4}{20}\times 10$

$\Rightarrow a=2m{{s}^{-2}}$ 

Thus, the acceleration of the masses is $2m{{s}^{-2}}$ . Substituting the value of  $a$ in equation $(2)$:

\[\Rightarrow {{m}_{2}}g-T={{m}_{2}}\left( \frac{{{m}_{2}}-{{m}_{1}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( {{m}_{2}}-\frac{m_{2}^{2}-{{m}_{1}}{{m}_{2}}}{{{m}_{2}}+{{m}_{1}}} \right)g\]

\[\Rightarrow T=\left( \frac{2{{m}_{1}}{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)g\]

\[\Rightarrow T=\left( \frac{2\times 12\times 8}{12+8} \right)\times 10\]

\[\Rightarrow T=96N\]

Thus, the tension in the string is $96N$.

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. 

Ans: Consider $m$, ${{m}_{1}}$ and ${{m}_{2}}$ as the respective masses of the parent nucleus and the two daughter nuclei. The parent nucleus is at rest. 

Initial momentum of the system (parent nucleus) $=0$ 

Let ${{v}_{1}}$ and ${{v}_{2}}$  be the respective velocities of the daughter nuclei having masses ${{m}_{1}}$ and ${{m}_{2}}$.

Total linear momentum of the system after disintegration$={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

From the law of conservation of momentum: 

\[\mathbf{Total}\text{ }\mathbf{initial}\text{ }\mathbf{momentum}=\mathbf{Total}\text{ }\mathbf{final}\text{ }\mathbf{momentum}\] 

 $\Rightarrow 0={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}$ 

$\Rightarrow {{m}_{1}}{{v}_{1}}=-{{m}_{2}}{{v}_{2}}$

$\Rightarrow {{v}_{1}}=\frac{-{{m}_{2}}{{v}_{2}}}{{{m}_{1}}}$

The negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

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

Mass of each ball$=0.05kg$ 

Initial velocity of each ball$=6m/s$ 

Magnitude of the initial momentum of each ball, ${{p}_{i}}=0.3kgm{{s}^{-1}}$ 

After collision, the balls change their directions of motion without changing the 

magnitudes of their velocity. 

Final momentum of each ball, ${{p}_{f}}=-0.3kgm{{s}^{-1}}$

Impulse imparted to each ball$=$ Change in the momentum of the system 

$\Rightarrow \operatorname{Im}pulse={{p}_{f}}-{{p}_{i}}$ 

$\Rightarrow \operatorname{Im}pulse=-0.3-0.3=-0.6kgm{{s}^{-1}}$

The negative sign indicates that the impulses imparted to the balls are opposite in direction. 

19. A shell of mass \[\mathbf{0}.\mathbf{020kg}\] is fired by a gun of mass \[\mathbf{100kg}\]. If the muzzle speed of the shell is $80m{{s}^{-1}}$ what is the recoil speed of the gun?

Mass of the gun, $M=100kg$ 

Mass of the shell, $m=0.020kg$ 

Muzzle speed of the shell, $v=80m/s$ 

Recoil speed of the gun  $=V$.

Both the gun and the shell are at rest initially. 

Initial momentum of the system$=0$

Final momentum of the system$=mv-MV$ 

Here, the negative sign appears because the directions of the shell and the gun are opposite to each other. 

From the law of conservation of momentum:

\[Final\text{ }momentum=Initial\text{ }momentum\] 

$\Rightarrow V=\frac{mv}{M}$ 

$\Rightarrow V=\frac{0.020\times 80}{100\times 1000}=0.016m/s$ 

Therefore, the recoil speed of the gun is $0.016m/s$.

20. A batsman deflects a ball by an angle of ${{45}^{\circ }}$ 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 $0.15kg$) 

Ans: The given situation can be represented as:

Deflection of a Ball Hit by a Batsman

$AO=$ Incident path of the ball 

$OB=$ Path followed by the ball after a deflection 

$\angle AOB=$ Angle between the incident and deflected paths of the ball$={{45}^{\circ }}$ 

$\angle AOB=\angle BOP={{22.5}^{\circ }}=\theta $

Initial and final velocities of the ball$=v$ 

The horizontal component of the initial velocity$=v\cos \theta $ along $RO$ 

Vertical component of the initial velocity$=v\sin \theta $ along $PO$ 

The horizontal component of the final velocity$=v\cos \theta $ along $OS$ 

The vertical component of the final velocity $=v\sin \theta $ along $OP$ 

The horizontal components of velocities suffer no change. The vertical components of velocities are in opposite directions. 

It is known that Impulse imparted to the ball$=$ Change in the linear momentum of the ball.

$\operatorname{Im}pulse=mv\cos \theta -(-mv\cos \theta )=2mv\cos \theta $ 

It is given that,

Mass of the ball, $m=0.15kg$ 

Velocity of the ball, $v=54km/h=54\times \frac{5}{18}=15m/s$ 

Impulse$=2\times 0.15\times 15\cos {{22.5}^{\circ }}=4.16kgm{{s}^{-1}}$ 

Therefore, impulse imparted to the ball is $4.16kgm{{s}^{-1}}$.

21. A stone of mass $0.25kg$  tied to the end of a string is whirled round in a circle of radius $1.5m$ with a speed of $40rev./\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 $200N$?

Mass of the stone, $m=0.25kg$ 

Radius of the circle, $r=1.5m$ 

Number of revolution per second, $n=\frac{40}{60}=\frac{2}{3}rps$ 

Angular velocity, $\omega =\frac{v}{r}=2\pi n$     ........$(1)$ 

The centripetal force for the stone is provided by the tension $T$ , in the string, i.e., $T={{F}_{Centripetal}}$

$\Rightarrow \frac{m{{v}^{2}}}{r}=mr{{\omega }^{2}}=mr{{\left( 2\pi n \right)}^{2}}$ 

$\Rightarrow {{F}_{Centripetal}}=0.25\times 1.5\times {{\left( 2\times 3.14\times \frac{2}{3} \right)}^{2}}$

$\Rightarrow {{F}_{Centripetal}}=6.57N$

Maximum tension in the string, ${{T}_{\max }}=200N$ 

${{T}_{\max }}=\frac{mv_{\max }^{2}}{r}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{{{T}_{\max }}\times r}{m}}$

$\Rightarrow {{v}_{\max }}=\sqrt{\frac{200\times 1.5}{0.25}}$

\[\Rightarrow {{v}_{\max }}=\sqrt{1200}=34.64m/s\]

Thus, the maximum speed of the stone is \[34.64m/s\].

22. If, in Exercise 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: 

the stone moves radially out wards, 

the stone flies off tangentially from the instant the string breaks, 

the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle? 

Ans: $(ii)$ 

From the first law of motion, the direction of velocity vector is tangential to the path of the stone at that instant. So, if the string breaks, the stone will move in the direction of the velocity at that instant. 

Therefore, the stone will fly off tangentially from the instant the string breaks. 

23. Explain why

a horse cannot pull a cart and run in empty space, 

Ans: In order to pull a cart, a horse pushes the ground backward with some force. The ground in turn exerts an equal and opposite reaction force upon the feet of the horse. 

This reaction force causes the horse to move forward. An empty space is devoid of any such reaction force. 

Hence, a horse cannot pull a cart and run in empty space.

passengers are thrown forward from their seats when a speeding bus stops suddenly, 

Ans: If a speeding bus stops suddenly, the lower portion of a passenger’s body, which is in contact with the seat, suddenly comes to rest. However, the upper portion tends to remain in motion (as per the first law of motion).

So, the passenger’s upper body is thrown forward in the direction in which the bus was moving. 

it is easier to pull a lawn mower than to push it, 

Ans: While pulling a lawn mower, a force at an angle $\theta $  is applied on it, as shown in the following figure:

Resolution of Force while Pulling a Lawn Mower

The vertical component of this applied force acts upward. This reduces the effective weight of the mower. 

While pushing a lawn mower, a force at an angle $\theta $ is applied on it, as shown in the following figure:

Resolution of Force while pushing a Lawn Mower

In this case, the vertical component of the applied force acts in the direction of the weight of the mower. This increases the effective weight of the mower. 

As the effective weight of the lawn mower is lesser in the first case, pulling the lawn mower is easier than pushing it. 

From Newton’s second law of motion: $F=ma=m\frac{\Delta v}{\Delta t}$  ........$(1)$ 

$F=$ Stopping force experienced by the cricketer as he catches the ball 

$m=$ Mass of the ball 

$t=$ Time of impact of the ball with the hand 

From equation $(1)$it can be observed that the impact force is inversely proportional to the impact time, i.e., $F<\frac{1}{\Delta t}$       ........$(2)$ 

Equation $(2)$  shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. 

Therefore, it is easier to pull a lawn mower than to push it.

a cricketer moves his hands backwards while holding a catch.

Ans: While taking a catch, a cricketer moves his hand backward so as to increase the time of impact $\Delta t$. This in turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

Therefore, a cricketer moves his hands backwards while holding a catch.

Laws of Motion Chapter Summary - Class 11 NCERT Solutions

Force .

A force is something which changes or tends to change the state of rest or motion of a body. It causes a body to start moving if it is at rest or stop it, if it is in motion, or deflect it from its initial path of motion.

Force is a vector quantity having SI unit Newton (N) and dimension [MLT –2 ].

Types of Force 

There are, basically, four forces, which are commonly encountered in mechanics.

(a) Weight : Weight of an object is the force with which earth attracts it. It is also called the force of gravity or the gravitational force.

(b) Contact Force : When two bodies come in contact, they exert forces on each other that are called contact forces.

(i) Normal Force (N): It is the component of contact force normal to the surface. It measures how strongly the surfaces in contact are pressed together.

(ii)  Frictional Force (f) : It is the component of contact force parallel to the surface. It opposes the relative motion (or attempted motion) of the two surfaces in contact.

(c) Tension: The force exerted by the ends of a taut string, rope or chain is called the tension. The direction of tension is so as to pull the body while that of normal reaction is to push the body.

(d) Spring Force: Every spring resists any attempt to change its length; the more you alter its length the harder it resists. The force exerted by a spring is given by F = –kx, where x is the change in length and k is the stiffness constant or spring constant (unit Nm –1 ).

Newton’s Laws of Motion 

First Laws of Motion

(a) Everybody continues in its state of rest or of uniform motion in a straight line unless it is compelled by a resultant force to change that state

(b) A frame of reference in which Newton’s first law is valid is called an inertial frame, i.e., if a frame of reference is at rest or in uniform motion it is called inertial, otherwise non-inertial.

Second Laws of Motion

(a) This law gives the magnitude of force.

(b) According to second Laws of Motion, the rate of change of momentum of a body is directly proportional to the resultant force acting on the body, i.e.,

$\vec{F}\alpha \left ( \frac{d\vec{p}}{dt} \right )$

$\vec{F}=K\dfrac{d\vec{p}}{dt}$

Here, the change in momentum takes place in the direction of the applied resultant force. Momentum, $\vec{p}=m\vec{v}$  is a measure of the sum of the motion contained in the body.

Third Laws of Motion

(a) According to this law, for every action, there is an equal and opposite reaction. When two bodies A and B exert force on each other, the force by A on B (i.e., action represented by $\vec{F}_{BA}$ ), is always equal and opposite to the force by B on A (i.e., reaction represented $\vec{F}_{BA}$ ). Thus, $\vec{F}_{BA}=-\vec{F}_{BA}$ .

(b) The two forces involved in any interaction between two bodies are called action and reaction. But we cannot say that a particular force is action and the other one is reaction. Action and Reaction force always acts on different bodies.

Applications of Newton’s Laws of Motion

There are two kinds of problems in classical mechanics :

(a) To find unknown forces acting on a body, given the body’s acceleration.

(b) To predict the future motion of a body, given the body’s initial position and velocity and the forces acting on it. For either kind of problem, we use Newton’s second law . The following general strategy is useful for solving such problems :

(i) Draw a simple, neat diagram of the system.

(ii) Isolate the object of interest whose motion is being analysed. Draw a free body diagram for this object, that is, a diagram showing all external forces acting on the object. For systems containing more than one object, draw separate diagrams for each object. Do not include forces that the object exerts on its surroundings.

(iii) Establish convenient coordinate axes for each body and find the components of the forces along these axes. Now, apply Newton’s second law, $\sum \vec{F}=m\vec{a}$ , in component form. Check your dimensions to make sure that all terms have units of force.

(iv) Solve the component equations for the unknowns. Remember that you must have as many independent equations as you have unknowns in order to obtain a complete solution.

(v) It is a good idea to check the predictions of your solutions for extreme values of the variables. You can often detect errors in your results by doing so.

Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

Thus, the force of friction is the force that develops at the surfaces of contact of two bodies and impedes (opposes) their relative motion.

Static Friction, Limiting Friction and Kinetic Friction

The opposing force that comes into play when one body tends to move over the surface of another, but the actual relative motion has yet not started is called Static friction.

Limiting friction is the maximum opposing force that comes into play, when one body is just at the verge of moving over the surface of the other body.

Kinetic friction or dynamic friction is the opposing force that comes into play when one body is actually moving over the surface of another body.

Coefficient of Static Friction

We know that, $f_{ms}\alpha N$ or $f_{ms}=\mu _{s}N$ or $\mu _{s}=\frac{f_{ms}}{N}$  ...(2)

Here, μ s is a constant of proportionality and is called the coefficient of static friction. Thus : Coefficient of static friction for any pair of surfaces in contact is equal to the ratio of the limiting friction and the  normal reaction. μ s , being a pure ratio, has got no units and its value depends upon the nature of the surfaces in contact. Further, μ s , is usually less than unity and is never equal to zero.

Since the force of static friction (f s ) can have any value from zero to maximum (f ms ), i.e. f s < f ms , eqn. (2) is generalised to f s < μ s N

Kinetic Friction

The laws of kinetic friction are exactly the same as those for static friction. Accordingly, the force of kinetic friction is also directly proportional to the normal reaction, i.e.,

$f_{k}\alpha N$ or $f_{k}=\mu _{k}N$ ...(4)

μ k is coefficient of kinetic friction.  μ k < μ s .

Circular Motion

It is the movement of particles along the circumference of a circle.

The uniform circular motion is that in which the particle is moving at a constant speed on a circular path.

The non-uniform circular motion is that in which the particles move with variable speed on its circular path.

Variables in Circular Motion

Angular Displacement: It is the angle subtended by the position vector at the centre of the circular path. Angular displacement, $\Delta \theta =\Delta s/r$ where, $\Delta s$ is the arc length and r is the radius

Angular Velocity: The time rate of change of angular displacement $(\Delta \theta )$ is called angular velocity.

Angular velocity, $\omega =\Delta \theta /\Delta t$

Angular velocity is a vector quantity

Relation between linear velocity (v) and angular velocity $(\omega )$ is given by

$v=r\omega $

Angular Acceleration: The rate of change of angular velocity is called angular acceleration.

Angular acceleration,

$\alpha =\lim_{\Delta t\rightarrow 0}\dfrac{\Delta \omega }{\Delta t}=\dfrac{d\omega }{dt}=\dfrac{d^{2}\theta }{dt^{2}}$

Its SI unit is rad/s 2 and dimensional formula is [T -2 ]

Acceleration in a circular motion has two components, as given below:

(a) Tangential acceleration is the change in magnitude of linear velocity and act along tangent to the circular path. It is given by:

$\alpha _{T}=r\alpha $

(b) Radial Acceleration is the change in direction of linear velocity and acts along the radius towards the centre of circle. It is given by $\alpha _{r}=\dfrac{v^{2}}{r}=\omega ^{2}r$

It is also called centripetal acceleration.

Relation between linear acceleration (a) and angular acceleration $(\alpha )$ , 

$a=r\alpha$ , where r = radius

Relation between angular acceleration $(\alpha )$ and linear velocity (v)

$\alpha =\dfrac{v^{2}}{r}$

Centripetal and Centrifugal Force

Centripetal Force: In uniform circular motion, the force acting on the particle along the radius and towards the centre keeps the body moving along the circular path. This force is called centripetal force.

Centrifugal Force: The pseudo force experienced by a particle performing uniform circular motion due to an accelerated frame of reference which is along the radius and directed way from the centre is called centrifugal force.

Motion of a Car on a Plane  Circular Road

For motion without skidding

$\dfrac{Mv^{2}max}{r}=\mu M_{g}$

$\Rightarrow v_{max}\sqrt{\mu rg}$

Motion on a Banked Road

Angle of banking $=\theta$

$tan\theta =\dfrac{h}{b}$ 

Maximum safe speed at the bend $v_{max}=\left [ \dfrac{rg(\mu +tan\theta )}{1-(\mu tan\theta )} \right ]^{1/2}$

If friction is negligible $v_{max}=\sqrt{rg \tan\theta }=\sqrt{\frac{rhg}{b}}$ and $tan\theta =\dfrac{v^{2}_{max}}{rg}$

Motion of Cyclist on a Curve

In equilibrium angle with vertical is $\theta $ , then $tan\theta =\dfrac{v^{2}}{rg}$

Maximum safe speed $=v_{max}=\sqrt{\mu rg}$

Overview of Deleted Syllabus for CBSE Class 11 Physics Laws of Motion

NCERT Class 11 Physics Chapter 4 Solutions on Laws of Motion provided by Vedantu provide a comprehensive understanding of the principles that govern the motion of objects. The solution cover exercise in the NCERT textbook offers clear explanations and step-by-step methods for solving problems. Important points to focus on include Newton's First Law or the Law of Inertia, Newton's Second Law, Newton's Third Law, the importance of free-body diagrams for visualizing forces, conditions for equilibrium of forces, and the analysis of motion on inclined planes and circular paths. From previous year's question papers, typically around 6–7 questions are asked from this chapter. These questions test students' understanding of theoretical concepts as well as their problem-solving skills.

Other Study Material for CBSE Class 11 Physics Chapter 4

Chapter-Specific NCERT Solutions for Class 11 Physics

Given below are the chapter-wise NCERT Solutions for Class 11 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

FAQs on NCERT Solutions Class 11 Physics Chapter 4 Laws of Motion

1. Why should I refer to the NCERT Solutions for Chapter 5 Laws of Motion of Class 11 Physics?

NCERT Solutions for Class 11 Physics Chapter 5- Laws of Motion will help students understand and answer correctly the questions provided at the end of the chapter by NCERT. These questions and exercises are important since many of them can be asked in the exams. NCERT Solutions are very useful in case you face difficulty in grasping the concept behind any questions. Referring to the solutions will provide you with an idea as to how to solve any given question.

2. What are the concepts covered in Chapter 5 of NCERT Solutions for Class 11 Physics?

Chapter 5 covers the fundamental concepts based on Newton’s laws. These concepts can be correlated to nearly all activities in our daily lives. This chapter talks about one of the most basic and important concepts that is taught in Class 11 Physics. The NCERT Solutions of chapter 5 in Class 11 Physics will help students answer questions that are based on the laws of motion and other related concepts in the chapter.

3. What are the most important topics in Class 11 Physics Chapter 5?

Chapter 5 - Laws of Motion is one of the most important chapters in Mechanics. It talks about the three of Newton’s Laws which are certainly the most important topic that has been covered in this chapter. However, there are more topics that hold high importance and need focus when preparing for your Class 11 Physics exam. These topics are Conservation of Momentum, Problem-Solving in Mechanics, and Circular Motion.

4. Where can I get the NCERT Solutions for Class 11 Physics Chapter 5?

You can find NCERT Solutions for Class 11 Physics Chapter 5 - Laws of Motion on  Vedantu  app or website. Experts at Vedantu have carefully designed these solutions to help students achieve in-depth knowledge while answering the questions provided in the NCERT for Class 11 Physics. These solutions have been framed based on the latest syllabus provided by CBSE. Access and download the NCERT Solutions for Class 11 Physics Chapter 5 free of cost.

5. How do I score well in Class 11 Physics Chapter 5?

Scoring well in Class 11 Physics Chapter 5 simply means having proper knowledge of all concepts that have been taught throughout the chapter and being able to use this knowledge in solving all the questions based on this chapter. Students must regularly practice and revise everything they study in school. Making notes of important details and formulas can be very helpful during your preparation for the Class 11 Physics exams.

6. What are the important questions in Laws of Motion class 11 NCERT Solutions?

The important questions in the laws of motion class 11 ncert solutions often revolve around the fundamental principles and applications of Newton's laws. Key questions include:

Deriving and explaining Newton's three laws of motion.

Problems involving free-body diagrams and calculating net forces.

Questions on equilibrium of forces, including both static and dynamic situations.

Analysing motion on inclined planes and circular motion.

Real-life applications and conceptual questions about action-reaction pairs.

7. What are the two marks questions in Class 11 Physics Laws of Motion NCERT Solutions?

Two-mark questions in the Laws of Motion Class 11 Solutions typically involve concise explanations or simple problem-solving exercises. Examples include:

Stating and explaining each of Newton's laws.

Drawing and interpreting free-body diagrams for simple systems.

Short problems involving the calculation of force, mass, or acceleration using $F=ma$.

Explain the concept of inertia and give examples.

Briefly discussing applications of Newton’s third law in everyday scenarios.

8. How many types of Newton's law are there in Class 11 Physics Ch 4 NCERT Solutions?

There are three types of Newton's laws of motion mentioned in laws of motion class 11 ncert solutions:

Newton's First Law (Law of Inertia): An object will remain at rest or in uniform motion unless acted upon by an external force.

Newton's Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass $F=ma$.

Newton's Third Law: For every action, there is an equal and opposite reaction.

9. How is the concept of inertia explained in Class 11 Physics Chapter 4 Exercise Solutions?

According to class 11 physics laws of motion ncert solutions, Inertia is the tendency of an object to resist changes in its state of motion. It is directly related to the mass of an object; the greater the mass, the greater the inertia. The concept is explained through examples such as why passengers lurch forward in a sudden stop or why a heavy object is harder to move than a lighter one.

10. What are some real-life applications of Newton’s Third Law discussed in Class 11 Physics Chapter Laws Of Motion NCERT Solutions?

Real-life applications of Newton’s Third Law in laws of motion class 11 solutions include:

The propulsion of rockets and jet engines, where exhaust gases are expelled backward, results in a forward thrust.

The recoil experienced when firing a gun.

Swimming, where the swimmer pushes water backward, propelling themselves forward.

11. How do the Laws of Motion NCERT Solutions for Class 11 Physics explain the use of free-body diagrams?

Free-body diagrams are visual tools used to represent the forces acting on an object. The Ch 4 Physics Class 11 NCERT Solutions explains how to draw these diagrams, identify forces such as tension, normal force, gravitational force, and friction, and use them to solve problems involving equilibrium and motion.

12. What is the importance of equilibrium in the Laws of Motion Class 11 Physics Chapter 4 NCERT Solutions?

Equilibrium is a state where the net force acting on an object is zero. The class 11 physics ch 4 ncert solutions discusses two types of equilibrium:

Static Equilibrium: Where an object is at rest and remains at rest.

Dynamic Equilibrium: Where an object is moving with constant velocity. Understanding these concepts is essential for analysing situations where forces balance each other.

NCERT Solutions for Class 11 Physics