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Case Study Questions for Class 11 Physics Chapter 5 Laws of Motion

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Here we are providing case study questions for class 11 physics chapter 5 laws of motion. Students can practice these questions for better understanding of case study type questions.

Case Study Questions:

Question 1:

Conservation of Momentum This principle is a consequence of Newton’s second and third laws of motion. In an isolated system (i.e., a system having no external force), mutual forces (called internal forces) between pairs of particles in the system causes momentum change in individual particles. Let a bomb be at rest, then its momentum will be zero. If the bomb explodes into two equal parts, then the parts fly off in exactly opposite directions with same speed, so that the total momentum is still zero. Here, no external force is applied on the system of particles (bomb).

(i) A shell of mass 10 kg is moving with a velocity of 10 ms -1 when it blasts and forms two parts of mass 9 kg and 1 kg respectively. If the first mass is stationary, the velocity of the second is (a) 1 m/s (b) 10 m/s (c) 100 m/s (d) 1000 m/s

(ii) A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun 5 m/s. The muzzle velocity will be (a) 30 km/min (b) 60 km/min (c) 30 m/s (d) 500 m/s

(iii) A bullet of mass 0.1 kg is fired with a speed of 100 m/s. The mass of gun being 50 kg, then the velocity of recoil becomes (a) 0.05 m/s (b) 0.5 m/s (c) 0.1 m/s (d) 0.2 m/s

(iv) Two masses of M and 4M are moving with equal kinetic energy. The ratio of their linear momenta is (a) 1:8 (b) 1:4 (c) 1:2 (d) 4:1

(v) A unidirectional force F varying with time T as shown in the figure acts on a body initially at rest for a short duration 2T. Then, the velocity acquired by the body is

case study questions class 11 physics chapter 5

(a) πF 0 T/4m (b) πF 0 T/2m (c) F 0 T/4m (d) zero

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CBSE Case Study Questions Class 11 Physics PDF Download

Are you a Class 11 Physics student looking to enhance your understanding and prepare effectively for your exams? Look no further! In this comprehensive guide, we present a curated collection of CBSE Case Study Questions Class 11 Physics that will help you grasp the core concepts of Physics while reinforcing your problem-solving skills.

CBSE 11th Standard CBSE Physics question papers, important notes, study materials, Previous Year Questions, Syllabus, and exam patterns. Free 11th Standard CBSE Physics books and syllabus online. Important keywords, Case Study Questions, and Solutions.

Class 11 Physics Case Study Questions

CBSE Class 11 Physics question paper will have case study questions too. These case-based questions will be objective type in nature. So, Class 11 Physics students must prepare themselves for such questions. First of all, you should study NCERT Textbooks line by line, and then you should practice as many questions as possible.

Chapter-wise Solved Case Study Questions for Class 11 Physics

Chapter 1: Physical World
Chapter 2: Units and Measurements
Chapter 3: Motion in a Straight Line
Chapter 4: Motion in a Plane
Chapter 5: Laws of Motion
Chapter 6: Work, Energy, and Power
Chapter 7: System of Particles and Rotational Motion
Chapter 8: Gravitation
Chapter 9: Mechanical Properties of Solids
Chapter 10: Mechanical Properties of Fluids
Chapter 11: Thermal Properties of Matter
Chapter 12: Thermodynamics
Chapter 13: Kinetic Theory
Chapter 14: Oscillations
Chapter 15: Waves

Class 11 students should go through important Case Study problems for Physics before the exams. This will help them to understand the type of Case Study questions that can be asked in Grade 11 Physics examinations. Our expert faculty for standard 11 Physics have designed these questions based on the trend of questions that have been asked in last year’s exams. The solutions have been designed in a manner to help the grade 11 students understand the concepts and also easy-to-learn solutions.

Class 11 Books for Boards

Why Case Study Questions Matter

Case study questions are an invaluable resource for Class 11 Physics students. Unlike traditional textbook exercises, these questions simulate real-life scenarios, challenging students to apply theoretical knowledge to practical situations. This approach fosters critical thinking and helps students build a deep understanding of the subject matter.

Let’s delve into the different topics covered in this collection of case study questions:

1. Motion and Gravitation

In this section, we explore questions related to motion, velocity, acceleration, and the force of gravity. These questions are designed to test your grasp of the fundamental principles governing motion and gravitation.

2. Work, Energy, and Power

This set of questions delves into the concepts of work, energy, and power. You will encounter scenarios that require you to calculate work done, potential and kinetic energy, and power in various contexts.

3. Mechanical Properties of Solids and Fluids

This section presents case study questions about the mechanical properties of solids and fluids. From stress and strain calculations to understanding the behavior of fluids in different situations, these questions cover a wide range of applications.

4. Thermodynamics

Thermodynamics can be a challenging topic, but fear not! This part of the guide offers case study questions that will clarify the laws of thermodynamics, heat transfer, and thermal expansion, among other concepts.

5. Oscillations and Waves

Get ready to explore questions related to oscillations, simple harmonic motion, and wave characteristics. These questions will deepen your understanding of wave propagation and the behavior of oscillatory systems.

6. Kinetic Theory and Laws of Motion

Kinetic theory and the laws of motion can be complex, but with our case study questions, you’ll find yourself mastering these topics effortlessly.

Discover a wide array of questions dealing with light, lenses, and mirrors. This section will improve your problem-solving skills in optics and enhance your ability to analyze optical phenomena.

8. Electrical Effects of Current

Electricity and circuits are fundamental to physics. The case study questions in this section will challenge you to apply Ohm’s law, Kirchhoff’s laws, and other principles in various electrical circuits.

9. Magnetic Effects of Current

Delve into the fascinating world of magnets and magnetic fields. This set of questions will strengthen your understanding of magnetic effects and their applications.

10. Electromagnetic Induction

The final section covers electromagnetic induction, Faraday’s law, and Lenz’s law. You’ll be presented with scenarios that test your ability to predict induced electromotive forces and analyze electromagnetic phenomena.

In conclusion, mastering Class 11 Physics requires a thorough understanding of fundamental concepts and their practical applications. The case study questions provided in this guide will undoubtedly assist you in achieving a deeper comprehension of the subject.

Remember, practice is key! Regularly attempt these case study questions to strengthen your problem-solving abilities and boost your confidence for the exams. Happy studying, and may you excel in your Physics journey!

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Class 11 Physics Case...

Class 11 Physics Case Study Questions

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Looking for complete and comprehensive case study questions for class 11 Physics? myCBSEguide is just a click away! With extensive study materials, sample papers, case study questions and mock tests, myCBSEguide is your one-stop solution for class 11 Physics exam preparation needs. So, what are you waiting for? Log on to myCBSEguide and get started today!

What is the purpose of physics?

Physics is the study of the fundamental principles governing the natural world. It is a vital part of the scientific enterprise, providing the foundation on which other sciences are built. Physics is essential for understanding how the world works, from the smallest particles to the largest structures in the Universe. In class 11 Physics, students are introduced to the basic concepts of physics and learn about the fundamental principles governing the natural world. Class 11 Physics concepts are essential for understanding the world around us and for further study in physics and other sciences.

What are case study questions in physics?

In physics, case study questions are intended to evaluate a student’s ability to apply theoretical principles to real-life situations. These questions usually ask the student to assess data from a specific experiment or setting in order to discover what physical principles are at play. Problem-solving and critical-thinking skills are developed through case study questions, which are an important aspect of physics education.

CBSE Case Study Questions in Class 11 Physics

CBSE Class 11 Physics question paper pattern includes case study questions. Class 11 Physics case study questions assess a student’s ability to apply physics principles to real-world environments. The questions are usually focused on a situation provided in the Class 11 Physics question paper, and they demand the student to answer the problem using their physics knowledge. Class 11 Physics case study questions are an important aspect of the CBSE physics curriculum. Class 11 Physics case study questions are a useful way to assess a student’s expertise in the subject.

Sample Class 11 Physics Case Study Questions

Expert educators at myCBSEguide have created a collection of Class 11 physics case study questions. The samples of Class 11 physics case study questions are given below. Class 11 physics case study questions are designed to test your understanding of the concepts and principles of physics. They are not meant to be easy, but they should be done if you have a good grasp of the subject. So, take a look at the questions and see how you fare. Good luck!

Class 11 Physics Case Study Question 1

Read the case study given below and answer any four subparts: Potential energy is the energy stored within an object, due to the object’s position, arrangement or state. Potential energy is one of the two main forms of energy, along with kinetic energy. Potential energy depends on the force acting on the two objects.

kinetic energy
potential energy
mechanical energy
none of these
potential energy decreases
potential energy increases
kinetic energy decreases
kinetic energy increases
only when spring is stretched
only when spring is compressed
both a and b
5 × 10 4 J
5 × 10 5 J

Answer Key:

Class 11 Physics Case Study Question 2

distance between body
source of heat
all of the above
convection and radiation
(b) convection
(d) all of the above
(a) convection
(a) increase
(c) radiation

Class 11 Physics Case Study Question 3

internal energy.
1 +(T 2 /T 1 )
(T 1 /T 2 )+1
(T 1 /T 2 )- 1
1 – (T 2 / T 1 )
increase or decrease depending upon temperature ratio
first increase and then decrease
(d) 1- (T 2 / T 1 )
(b) increase
(c) constant

Class 11 Physics Case Study Question 4

It is far away from the surface of the earth
Its surface temperature is 10°C
The r.m.s. velocity of all the gas molecules is more than the escape velocity of the moon’s surface
The escape velocity of the moon’s surface is more than the r.m.s velocity of all molecules
T(H 2 ) = T(N 2 )
T(H 2 ) < T(N 2 )
T(H 2 ) > T(N 2 )

The given samples of Class 11 Physics case study questions will help Class 11 Physics students to get an idea on how to solve it. These Class 11 Physics case study questions are based on the topics covered in the Class 11 Physics syllabus and are designed to test the student’s conceptual understanding. The questions are of varying difficulty levels and cover a wide range of topics. By solving these Class 11 Physics case study questions, students will be able to develop their problem-solving skills and improve their understanding of the concepts.

Examining Class 11 Physics syllabus

Senior Secondary school education is a transitional step from general education to a discipline-based curriculum concentration. The current curriculum of Class 11 Physics takes into account the rigour and complexity of the disciplinary approach, as well as the learners’ comprehension level. Class 11 Physics syllabus has also been carefully crafted to be similar to international norms.

The following are some of the Class 11 Physics syllabus’s most notable features:

Emphasis is placed on gaining a fundamental conceptual knowledge of the material.
Use of SI units, symbols, naming of physical quantities, and formulations in accordance with international standards are emphasised.
For enhanced learning, provide logical sequencing of subject matter units and suitable placement of concepts with their links.
Eliminating overlapping concepts/content within the field and between disciplines to reduce the curricular load.
Process skills, problem-solving ability, and the application of Physics principles are all encouraged.

CBSE Class 11 Physics (Code No. 042)

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Case Study Questions Class 11 Physics Laws Of Motion

Case study questions class 11 physics chapter 5 laws of motion.

CBSE Class 11 Case Study Questions Physics Laws Of Motion. Important Case Study Questions for Class 11 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Laws Of Motion.

At Case Study Questions there will given a Paragraph. In where some Important Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks, 4 marks.

CBSE Case Study Questions Class 11 Physics Laws Of Motion

Case study – 1.

Newton’s first law of motion states that If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body. To summaries, if the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia. Inertia means ‘resistance to change’. A body does not change its state of rest or uniform motion, unless an external force compels it to change that state. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. Consider a book at rest on a horizontal surface. It is subject to two external forces: the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. R is a self-adjusting force. This is an example of the kind of situation mentioned above. The forces are not quite known fully but the state of motion is known. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals that of W. A statement often encountered is : Since W = R, forces cancel and, therefore, the book is at rest”. This is incorrect reasoning. The correct statement is: “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the normal force R must be equal and opposite to the weight W ” .

1) The book on table is at rest. The force of gravity here is balanced by

a) Force of friction

b) Normal reaction by table on book

c) Weight of table

d) none of these

2) If no external force acts on object which is at rest. it will

a) remain at rest

b) start to move

c) both a and b can possible

3) Define inertia.

4) State Newton’s first law of motion.

5) Explain why book on table remains at rest.

Answer Key –

3) the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia.

4) Newton’s first law of motion states that If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body

5) Consider a book at rest on a horizontal surface. It is subject to two external forces: the force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R. This is an example of the kind of situation mentioned above. Magnitude of R equals that of W. This implies that the normal force R must be equal and opposite to the weight W” .

Case Study – 2

Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p:

Momentum is clearly a vector quantity. SI unit is kg m/s. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) is parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.

If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.
Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.

1) SI unit of momentum is

b) Kg m/s 2

d) None of these

2) Momentum is

a) Scalar quantity

b) Vector quantity

3) Define momentum. Give its SI unit.

4) Explain with example how mass of body is important for determining effect of force on its motion?

5) Explain with example how speed is important for determining effect of force on its motion?

Answer key-2

3) Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted By p:

Momentum is clearly a vector quantity. SI unit is kg m/s.

4) If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.

5) Speed is important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time.

Case Study – 3

The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general situation when there is net external force acting on the body. It relates the net external force to the acceleration of the body.

These qualitative observations lead to the second law of motion expressed by Newton as follow:

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval Δt, the velocity of a body of mass m changes from v to v + Δv i.e. its initial momentum

p = m v changes by Δp = mΔv . According to the Second Law

Where k is a constant of proportionality. Mathematically,

F = ma, the unit of force is kg-m/s 2 or Newton, which has the symbol N. Let us note at this stage some important points about the second law:

In the second law, F = 0 implies a = 0. The second law is obviously consistent with the first law.
The second law of motion is a vector law.
The second law of motion given by is applicable to a single point particle as well as to the rigid body but internal forces is not considered in F.
The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant. Answer the following questions.

1) SI unit of force is

d) None of the above

2) According to second law of motion The rate of change of momentum of a body is directly proportional to

a) Velocity of body

b) Applied force

c) Only mass of body

d) None of the above.

3) The second law of motion is

a) Vector law

b) Scalar law

4) State second law of motion.

5) Write a note on 2 nd law of motion. Enlist some deductions from 2 nd law

Answer key -3

4) The second law of motion is quantitative expression of force and it states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force. Mathematically, F = ma, the unit of force is kg-m/s 2 or Newton.

5) The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts. Thus, if under the action of a force F for time interval Δt, the velocity of a body of mass m changes from v to v + Δv i.e. its initial momentum

p = m v changes by Δp = mΔv. According to the Second Law

The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant.

Case Study – 4

4.) The product of force and time which is the change in momentum of the body remains a measurable quantity. This product is called impulse

Impulse = Force × time duration

= Change in momentum

Large force acting for a short time to produce a finite change in momentum is called an impulsive force.

The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object. The two opposing forces are also known as action and reaction forces.

Answer the following questions. The second and third laws of motion lead to an important consequence: the law of conservation of momentum. Take a familiar example. A bullet is fired from a gun. If the force on the bullet by the gun is F, the force on the gun by the bullet is – F, according to the third law. The two forces act for a common interval of time Δt. According to the second law, FΔt is the change in momentum of the bullet and – F Δt is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each. Thus if pb is the momentum of the bullet after firing and pg is the recoil momentum of the gun,

pg = – pb i.e. pb + pg = 0 That is, the total momentum of the (bullet + gun) system is conserved. Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged. This fact is known as the law of conservation of momentum. The total momentum of an isolated system of interacting particles is conserved.

1) Action reaction forces acts on bodies in order that

a) Action acts first then reaction force comes

b) reaction acts first then action force comes

c) both action reaction act at same time

2) Which of the following is correct about action reaction forces?

a) They act on different objects

b) They are equal in magnitude and opposite in direction

c) Both forces acted on different object simultaneously

d) All the above

3) State Newton’s third law of motion

4) Define impulse. Give its formula.

5) State law of conservation of momentum

Answer key – 4

3) The third law of motion states that when one object exerts a force on another object, the second object instantaneously exerts a force back on the first. These two forces are always equal in magnitude but opposite in direction. These forces act on different objects and never on the same object.

4) The product of force and time which is the change in momentum of the body remains a measurable quantity. This product is called impulse

5) The total momentum of isolated system particles is conserved. Isolated system means no external force.

Case Study – 5

Friction: Let us return to the example of a body of mass m at rest on a horizontal table. The force of gravity (mg) is cancelled by the normal reaction force (N) of the table. Now suppose a force F is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. But if the applied force F were the only external force on the body, it must move with acceleration F/m, however small. Clearly, the body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force F, resulting in zero net force on the body. This force f s parallel to the surface of the body in contact with the table is known as frictional force, or simply friction. When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force F increases, fs also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, it is called static friction. Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place) under the applied force, if friction were absent. It is found experimentally that the limiting value of static friction (f s )max f is independent of the area of contact and varies with the normal force(N) approximately as :

(f s )max = μ N

where μs is a constant of proportionality depending only on the nature of the surfaces in contact. The constant μs is called the coefficient of static friction. The law of static friction may thus be written as

f s ≤ μsN

Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by f k . Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction:

f k = μ k N

1) Force of static friction is directly proportional to

a) Normal reaction

b) Force by gravity

c) Velocity of body

2) Coefficient of kinetic friction is independent of area of contact. True or false?

3) Give formula for law of static friction

4) Explain law of static friction

5) Explain kinetic friction.

Answer key-5

3) The law of static friction can be written as

Where μs is coefficient of static friction and N is normal reaction.

4) It is found experimentally that the limiting value of static friction (f s ) max f is independent of the area of contact and varies with the normal force(N) approximately as : (f s )max = μ N . where μs is a constant of proportionality depending only on the nature of the surfaces in contact. The constant μs is called the coefficient of static friction. The law of static friction may thus be written as f s ≤ μsN

5) Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by f k and given by f k = μ k N

Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity.

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Class 11 Physics Case Study Questions PDF Download

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Class 11 Physics Case Study Questions are available here. You can read these Case Study questions by chapter for your final physics exam. Subject matter specialists and seasoned teachers created these quizzes. You can verify the right response to each question by referring to the answer key, which is also provided. To achieve high marks on your Board exams, practice these questions.

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We are providing Case Study questions for Class 11 Physics based on the Latest syllabus. There is a total of 14 chapters included in the CBSE Class 11 physics exams. Students can practice these questions for concept clarity and score better marks in their exams.

Table of Contents

Class 11th PHYSICS: Chapterwise Case Study Question & Solution

Case study questions play a crucial role in the Class 11 Physics curriculum. They are designed to assess your understanding of various concepts and principles in real-life scenarios. These questions help you apply theoretical knowledge to practical situations, enhancing your problem-solving skills.

Case Study-Based Questions for Class 11 Physics

Case Study Based Questions on Class 11 Physics Chapter 2 Units and Measurements
Case Study Based Questions on Class 11 Physics Chapter 3 Motion in a Straight Line
Case Study Based Questions on Class 11 Physics Chapter 4 Motion in a Plane
Case Study Based Questions on Class 11 Physics Chapter 5 Laws of Motion
Case Study Based Questions on Class 11 Physics Chapter 6 Work, Energy, and Power
Case Study Based Questions on Class 11 Physics Chapter 7 System of Particles and Rotational Motion
Case Study Based Questions on Class 11 Physics Chapter 8 Gravitation
Case Study Based Questions on Class 11 Physics Chapter 9 Mechanical Properties of Solids
Case Study Based Questions on Class 11 Physics Chapter 10 Mechanical Properties of Fluids
Case Study Based Questions on Class 11 Physics Chapter 11 Thermal Properties of Matter
Case Study Based Questions on Class 11 Physics Chapter 12 Thermodynamics
Case Study Based Questions on Class 11 Physics Chapter 13 Kinetic Theory
Case Study Based Questions on Class 11 Physics Chapter 14 Waves
Case Study Based Questions on Class 11 Physics Chapter 15 Oscillations

Class 11 Physics MCQ Questions

Before the exams, students in class 11 should review crucial Physics Case Study issues. They will gain a better understanding of the kinds of Case Study questions that may be offered in Physics exams for Grade 11. These questions were created by our highly qualified faculty for standard 11 Physics based on the questions that appeared most frequently in last year’s exams. The solutions have been written in a way that will make them simple to grasp and will aid students in grade 11 in understanding the topics.

Class 11 Books for Boards

Class 11 Physics Syllabus 2024

Unit I: Physical World and Measurement 08 Periods

Chapter–2: Units and Measurements

Need for measurement: Units of measurement; systems of units; SI units, fundamental and derived units. significant figures. Dimensions of physical quantities, dimensional analysis and its applications.

Unit II: Kinematics 24 Periods

Chapter–3: Motion in a Straight Line

The frame of reference, Motion in a straight line, Elementary concepts of differentiation and integration for describing motion, uniform and non-uniform motion, and instantaneous velocity, uniformly accelerated motion, velocity-time and position-time graphs. Relations for uniformly accelerated motion (graphical treatment).

Chapter–4: Motion in a Plane

Scalar and vector quantities; position and displacement vectors, general vectors and their notations; equality of vectors, multiplication of vectors by a real number; addition and subtraction of vectors, Unit vector; resolution of a vector in a plane, rectangular components, Scalar and Vector product of vectors. Motion in a plane, cases of uniform velocity and uniform acceleration projectile motion, uniform circular motion.

Unit III: Laws of Motion 14 Periods

Chapter–5: Laws of Motion

Intuitive concept of force, Inertia, Newton’s first law of motion; momentum and Newton’s second law of motion; impulse; Newton’s third law of motion. Law of conservation of linear momentum and its applications. Equilibrium of concurrent forces, Static and kinetic friction, laws of friction, rolling friction, lubrication.

Dynamics of uniform circular motion: Centripetal force, examples of circular motion (vehicle on a level circular road, vehicle on a banked road).

Unit IV: Work, Energy and Power 14 Periods

Chapter–6: Work, Energy and Power

Work done by a constant force and a variable force; kinetic energy, workenergy theorem, power. Notion of potential energy, potential energy of a spring, conservative forces: non- conservative forces, motion in a vertical circle; elastic and inelastic collisions in one and two dimensions.

Unit V: Motion of System of Particles and Rigid Body 18 Periods

Chapter–7: System of Particles and Rotational Motion

Centre of mass of a two-particle system, momentum conservation and Centre of mass motion. Centre of mass of a rigid body; centre of mass of a uniform rod. Moment of a force, torque, angular momentum, law of conservation of angular momentum and its applications. Equilibrium of rigid bodies, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions. Moment of inertia, radius of gyration, values of moments of inertia for simple geometrical objects (no derivation).

Unit VI: Gravitation 12 Periods

Chapter–8: Gravitation

Kepler’s laws of planetary motion, universal law of gravitation. Acceleration due to gravity and its variation with altitude and depth. Gravitational potential energy and gravitational potential, escape velocity, orbital velocity of a satellite.

Unit VII: Properties of Bulk Matter 24 Periods

Chapter–9: Mechanical Properties of Solids

Elasticity, Stress-strain relationship, Hooke’s law, Young’s modulus, bulk modulus, shear modulus of rigidity (qualitative idea only), Poisson’s ratio; elastic energy.

Chapter–10: Mechanical Properties of Fluids

Pressure due to a fluid column; Pascal’s law and its applications (hydraulic lift and hydraulic brakes), effect of gravity on fluid pressure. Viscosity, Stokes’ law, terminal velocity, streamline and turbulent flow, critical velocity, Bernoulli’s theorem and its simple applications. Surface energy and surface tension, angle of contact, excess of pressure across a curved surface, application of surface tension ideas to drops, bubbles and capillary rise.

Chapter–11: Thermal Properties of Matter

Heat, temperature, thermal expansion; thermal expansion of solids, liquids and gases, anomalous expansion of water; specific heat capacity; Cp, Cv – calorimetry; change of state – latent heat capacity. Heat transfer-conduction, convection and radiation, thermal conductivity, qualitative ideas of Blackbody radiation, Wein’s displacement Law, Stefan’s law .

Unit VIII: Thermodynamics 12 Periods

Chapter–12: Thermodynamics

Thermal equilibrium and definition of temperature zeroth law of thermodynamics, heat, work and internal energy. First law of thermodynamics, Second law of thermodynamics: gaseous state of matter, change of condition of gaseous state -isothermal, adiabatic, reversible, irreversible, and cyclic processes.

Unit IX: Behavior of Perfect Gases and Kinetic Theory of Gases 08 Periods

Chapter–13: Kinetic Theory

Equation of state of a perfect gas, work done in compressing a gas. Kinetic theory of gases – assumptions, concept of pressure. Kinetic interpretation of temperature; rms speed of gas molecules; degrees of freedom, law of equi-partition of energy (statement only) and application to specific heat capacities of gases; concept of mean free path, Avogadro’s number.

Unit X: Oscillations and Waves 26 Periods

Chapter–14: Oscillations

Periodic motion – time period, frequency, displacement as a function of time, periodic functions and their application. Simple harmonic motion (S.H.M) and its equations of motion; phase; oscillations of a loaded spring- restoring force and force constant; energy in S.H.M. Kinetic and potential energies; simple pendulum derivation of expression for its time period.

Chapter–15: Waves

Wave motion: Transverse and longitudinal waves, speed of traveling wave, displacement relation for a progressive wave, principle of superposition of waves, reflection of waves, standing waves in strings and organ pipes, fundamental mode and harmonics, Beats.

FAQs about Class 11 Physics Case Studies

What is the best website for a case study of physics class 11 .

studyrate.in is the best website for Class 11 Physics Case Study Questions for Board Exams. Here you can find various types of Study Materials, Ebooks, Notes, and much more free of cost.

How do you write a case study question for Class 11?

The CBSE will ask two Case Study Questions in the CBSE Class 11th Maths Question Paper. Question numbers 15 and 16 will be case-based questions where 5 MCQs will be asked based on a paragraph.

Are the case study questions based on the latest syllabus?

Yes, the case study questions are curated to align with the latest Class 11 Physics syllabus.

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Important Questions for CBSE Class 11 Physics Chapter 5 – Law of Motion

Important questions for cbse class 11 physics chapter 5 - law of motion, cbse class 11 physics chapter-5 important questions - free pdf download.

Free PDF download of Important Questions with solutions for CBSE Class 11 Physics Chapter 5 - Law of Motion prepared by expert Physics teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

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?

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.

13. 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) When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T , i.e.,

Where F is the net force acting on the particle.

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,

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.

2.Two blocks of masses m 1 , m 2 are connected by light spring on a smooth horizontal surface. The two masses are pulled apart and then released. Prove that the ratio of their acceleration is inversely proportional to their masses.

Ans: The forces F 1 and F 2 due to masses m 1 and m 2 acts in opposite directions

Thus F 1 + F 2 = 0

m 1 a 1 + m 2 a 2 = 0

m 1 a 1 = -m 2 a 2

Hence proved.

3.A shell of mass 0.020kg is fired by a gun of mass 100kg. If the muzzle speed of the shell is 80m/s, what is the recoil speed of the gun?

Ans: Momentum before firing = 0

Momentum after firing = momentum of (bullet+gun)

According to law of conservation of linear momentum

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

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.

11.A monkey of mass 40 kg climbs on a rope which can stand a maximum tension 600 N. In which of the following cases will the rope break? The monkey (a) climbs up with an acceleration of 6m/s 2 (b) climbs down with an acceleration of 4m/s 2 (c) climbs up with a uniform seed of 5m/s (d) falls down the rope freely under gravity. Take g = 10m/s 2 and ignore the mass of the rope.

Ans: m = 40kg, T = 600N (max tension rope can hold)

Rope will break if reaction (R) exceeds Tension (T)

(a) a = 6m/s 2

R = m (g + a) = 40 (10 + 6) = 640 N (Rope will break)

(b) a = 4m/s 2

R = m (g – a) = 40 (10 – 6) = 240 N (Rope will not break)

(d) a = g; R = m (g – a) = m (g – g)

R = zero (Rope will not break)

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.

14.A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must be emitted in opposite directions.

Ans: According to the principle of conservation of linear momentum, total momentum remains constant.

Before disintegration linear momentum = zero

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.

Ans. Mass of the rocket, m = 20,000 kg

Initial acceleration, a = 5 m/s2

Using Newton"s second law of motion, the net force (thrust) acting on the rocket is given by the relation:

F - mg = ma

F = m ( g + a )

Ans. (a) Vertically downward

(b) Parabolic path

(a) At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground.

(b) At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

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

Ans. Mass of each ball = 0.05 kg

Initial velocity of each ball = 6 m/s

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

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

= -0.3 - 0.3 = -0.6 kg m/s

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

Ans. Radius of the circular track, r = 30 m

Speed of the train, v = 54 km/h = 15 m/s

The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton"s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail

The angle of banking θ , is related to the radius ( r ) and speed ( v ) by the relation:

Therefore, the angle of banking is about 36.87°.

Mass of the body, m = 20 kg

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

Final velocity of the body, v = 0

Using Newton"s second law of motion, the acceleration ( a ) produced in the body can be calculated as:

Using the first equation of motion, the time ( t ) taken by the body to come to rest can be calculated as:

21. 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.

Initial momentum of the system (parent nucleus) = 0

According to the law of conservation of momentum:

Total initial momentum = Total final momentum

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

Ans. Mass of the gun, M = 100 kg

Mass of the shell, m = 0.020 kg

Muzzle speed of the shell, v = 80 m/s

Recoil speed of the gun = V

Both the gun and the shell are at rest initially.

Initial momentum of the system = 0

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

Final momentum = Initial momentum

3 Marks Questions

1.A train runs along an unbanked circular bend of radius 30m at a speed of 54km/hr. The mass of the train is 106kg. What provides the necessary centripetal force required for this purpose? The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Ans: (1) The centripetal force is provided by the lateral force acting due to rails on the wheels of the train.

(2) Outer rails

(a) Mass of the block, m = 15 kg

As per Newton"s second law of motion, the force ( F ) on the block caused by the motion of the trolley is given by the relation:

This force is acted in the direction of motion of the trolley.

Force of static friction between the block and the trolley:

= 0.18 x 15 x 10 = 27 N

The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest.

When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation.

(b) An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

3.What is the acceleration of the blocks? What is the net force on the block P? What force does P apply on Q. What force does Q apply on R?

Ans: If a is the acceleration

Then F = (3m)a

(1) Net force on P

(2) Force applied on Q

F 2 = (m + m)a

(3) Force applied on R by Q

4.How is centripetal force provided in case of the following?

(i) Motion of planet around the sun,

(ii) Motion of moon around the earth.

(iii) Motion of an electron around the nucleus in an atom.

Ans:(i) Gravitational force acting on the planet and the sun provides the necessary centripetal force.

(ii) Force of gravity due to earth on the moon provides centripetal force.

(iii) Electrostatic force attraction between the electron and the proton provides the necessary centripetal force.

5.State Newton’s second, law of motion. Express it mathematically and hence obtain a relation between force and acceleration.

Ans: According to Newton’s second law the rate of change of momentum is directly proportional to the force.

6.A railway car of mass 20 tonnes moves with an initial speed of 54km/hr. On applying brakes, a constant negative acceleration of 0.3m/s 2 is produced.

(i) What is the breaking force acting on the car?

(ii) In what time it will stop?

(iii) What distance will be covered by the car before if finally stops?

7.What is meant by coefficient of friction and angel of friction? Establish the relation between the two? OR

A block of mass 10kg is sliding on a surface inclined at a angle of 30 o with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is 0.5

Ans: Angle of friction is the contact between the resultant of limiting friction and normal reaction

Coefficient of static friction

From (1) & (2)

A block of mass 10kg is sliding on a surface inclined at a angle of 30o with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is 0.5

a = 0.657m/s 2

8.State and prove the principle of law of conservation of linear momentum?

Ans: The law of conservation of linear momentum states that if no external force acts on the system. The total momentum of the system remains unchanged.

According to Newton’s third law

Thus momentum gained by one ball is lost by the other ball. Hence linear momentum remains conserved.

9.A particle of mass 0.40 kg moving initially with constant speed of 10m/s to the north is subject to a constant force of 8.0 N directed towards south for 30s. Take at that instant, the force is applied to be t = 0, and the position of the particle at that time to be x = 0, predict its position at t = -5s, 25s, 30s?

Ans. m = 0.40kg

u = l0m/s due North

(1) At t = -5s

(2) At t = 25s

(3) At t = 30s

(4) At t = 30s

x = -50000m

10. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

750 N and 250 N in the respective cases; Method (b)

Mass of the block, m = 25 kg

Mass of the man, M = 50 kg

Force applied on the block, F = 25 x 10 = 250 N

Weight of the man, W = 50 x 10 = 500 N

Case (a): When the man lifts the block directly

In this case, the man applies a force in the upward direction. This increases his apparent weight.

∴Action on the floor by the man = 250 + 500 = 750 N

Case (b): When the man lifts the block using a pulley

In this case, the man applies a force in the downward direction. This decreases his apparent weight.

∴Action on the floor by the man = 500 - 250 = 250 N

If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

11.(a) State impulse – momentum theorem?

(b) A ball of mass 0.1kg is thrown against a wall. It strikes the wall normally with a velocity of 30m/s and rebounds with a velocity of 20m/s. calculate the impulse of the force exerted by the ball on the wall.

Ans: (a) It states that impulse is measured by the total change in linear momentum is

Impulse = m (-20 – 30) = -5Ns

12.Ten one rupee coins are put on top of one another on a table. Each coin has a mass m kg. Give the magnitude and direction of

(a) The force on the 7 th coin (counted from the bottom) due to all coins above it.

(b) The force on the 7 th coin by the eighth coin and

Ans.(a) The force on 7 th coin is due to weight of the three coins lying above it.

Therefore, F = (3 m) kgf = (3 mg) N

Where g is acceleration due to gravity. This force acts vertically downwards.

(b) The eighth coin is already under the weight of two coins above it and it has its own weight too. Hence force on 7 th coin due to 8 th coin is sum of the two forces i.e.

F = 2m + m = (3m) kg f = (3 mg) N

The force acts vertically downwards.

Reaction, R = -F = -4 m (kgf) = - (4 mg) N

-ve sign indicates that reaction acts vertically upwards.

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

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g floating on water,

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

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

Ans (a) Zero net force

The rain drop is falling with a constant speed. Hence, it acceleration is zero. As per Newton"s second law of motion, the net force acting on the rain drop is zero.

(b) Zero net force

The weight of the cork is acting downward. It is balanced by the buoyant force exerted by the water in the upward direction. Hence, no net force is acting on the floating cork.

The kite is stationary in the sky, i.e., it is not moving at all. Hence, as per Newton"s first law of motion, no net force is acting on the kite.

(d) Zero net force

The car is moving on a rough road with a constant velocity. Hence, its acceleration is zero. As per Newton"s second law of motion, no net force is acting on the car.

(e) Zero net force

The high speed electron is free from the influence of all fields. Hence, no net force is acting on the electron.

14. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) during its upward motion,

(b) during its downward motion,

(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction?

Ignore air resistance.

Ans. 0.5 N, in vertically downward direction, in all cases

Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton"s second law of motion as:

F = Net force

m = Mass of the pebble = 0.05 kg

The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction.

If the pebble is thrown at an angle of 45° 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.

Ans. 0.18 N; in the direction of motion of the body

Mass of the body, m = 3 kg

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

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

Time, t = 25 s

Using the first equation of motion, the acceleration ( a ) produced in the body can be calculated as:

As per Newton"s second law of motion, force is given as:

Since 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.

16. 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.

Mass of the body, m = 5 kg

The given situation can be represented as follows:

The resultant of two forces is given as:

As per Newton"s second law of motion, the acceleration ( a ) of the body is given as:

17. 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?

Mass of the stone, m = 0.25 kg

Radius of the circle, r = 1.5 m

The centripetal force for the stone is provided by the tension T , in the string, i.e.,

Therefore, the maximum speed of the stone is 34.64 m/s.

Figure 5.18

Ans. Mass of the man, m = 65 kg

The net force F , acting on the man is given by Newton"s second law of motion as:

20. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

4 Marks Questions

1. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,

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

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

(d) lying on the floor of a train which is accelerating with 1 m s-2, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans. (a) 1 N; vertically downward

Mass of the stone, m = 0.1 kg

As per Newton"s second law of motion, the net force acting on the stone,

F = ma = m g

= 0.1 x 10 = 1 N

Acceleration due to gravity always acts in the downward direction.

(b) 1 N; vertically downward

The train is moving with a constant velocity. Hence, its acceleration is zero in the direction of its motion, i.e., in the horizontal direction. Hence, 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. The magnitude of this force is 1 N.

Therefore, the net force acting on the stone, F " = ma = 0.1 x 1 = 0.1 N

This force is acting in the horizontal direction. Now, when the stone is dropped, the horizontal force F ," 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.

F = mg = 1 N

This force acts vertically downward.

(d) 0.1 N; in the direction of motion of 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.

The net force acting on the stone will be in the direction of motion of the train. Its magnitude is given by:

= 0.1 x 1 = 0.1 N

2. 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. Initial speed of the three-wheeler, u = 36 km/h = 10 m/s

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

Time, t = 4 s

Mass of the three-wheeler, m = 400 kg

Mass of the driver, m " = 65 kg

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

Using the first law of motion, the acceleration ( a ) of the three-wheeler can be calculated as:

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

Using Newton"s second law of motion, the net force acting on the three-wheeler can be calculated as:

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

3. A body of mass 0.40 kg moving initially with a constant speed of 10 ms-1to 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. Mass of the body, m = 0.40 kg

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

(i) At t = -5 s

Acceleration, a " = 0 and u = 10 m/s

(ii) At t = 25 s

(iii) At t = 100 s

As per the first equation of motion, for t = 30 s, final velocity is given as:

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

Ans. (a) 22.36 m/s, at an angle of 26.57° with the motion of the truck

(a) Initial velocity of the truck, u = 0

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

= 0 + 2 × 10 = 20 m/s

The final velocity of the truck and hence, of the stone is 20 m/s.

The resultant velocity ( v ) of the stone is given as:

5. Figure 5.16 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).

Figure 5.16

(a) For t < 0

It can be observed from the given graph 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. Hence, the force acting on the particle is zero.

For t > 4 s

It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle.

For 0 < t < 4

It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero.

(b) At t = 0

Impulse = Change in momentum

Mass of the particle, m = 4 kg

Initial velocity of the particle, u = 0

Final velocity of the particle, v = 0

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

Horizontal force, F = 600 N

Using Newton"s second law of motion, the acceleration ( a ) produced in the system can be calculated as:

When force F is applied on body A:

The equation of motion can be written as:

When force F is applied on body B:

7. 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. The given situation can be represented as shown in the following figure.

AO = Incident path of the ball

OB = Path followed by the ball after deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = θ

Initial and final velocities of the ball = v

Horizontal component of the final velocity = v cos θ along OS

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

∴Impulse imparted to the ball = Change in the linear momentum of the ball

Mass of the ball, m = 0.15 kg

Velocity of the ball, v = 54 km/h = 15 m/s

8. Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Figure 5.17

The given graph shows that a body changes its direction of motion after every 2 s. Physically, this situation can be visualized as a ball rebounding to and fro between two stationary walls situated between positions x = 0 and x = 2 cm. Since the slope of the x - t graph reverses after every 2 s, the ball collides with a wall after every 2 s. Therefore, ball receives an impulse after every 2 s.

Mass of the ball, m = 0.04 kg

The slope of the graph gives the velocity of the ball. Using the graph, we can calculate initial velocity ( u ) as:

(Here, the negative sign arises as the ball reverses its direction of motion.)

Magnitude of impulse = Change in momentum

9. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R . The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

Ans. (a) The free body diagram of the stone at the lowest point is shown in the following figure.

According to Newton"s second law of motion, the net force acting on the stone at this point is equal to the centripetal force, i.e.,

The free body diagram of the stone at the highest point is shown in the following figure.

Using Newton"s second law of motion, we have:

In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc.

Coin placed at 4 cm:

11. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

5 Marks Questions

1.(a) Define impulse. State its S.I. unit?

(b) State and prove impulse momentum theorem?

Ans:(a) Force which are exerted over a short time intervals are called impulsive forces.

(b) Impulse of a force is equal to the change in momentum of the body.

According to Newton’s second law

2. A man of mass 70 kg stands on a weighing scale in a lift which is moving

What would be the readings on the scale in each case?

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

Ans. (a) Mass of the man, m = 70 kg

Acceleration, a = 0

Using Newton"s second law of motion, we can write the equation of motion as:

Where, ma is the net force acting on the man.

As the lift is moving at a uniform speed, acceleration a = 0

= 70 × 10 = 700 N

(b) Mass of the man, m = 70 kg

R + m g = ma

R = m (g + a )

= 70 (10 + 5) = 70 × 15

(d) When the lift moves freely under gravity, acceleration a = g

The man will be in a state of weightlessness.

3. 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.

The given system of two masses and a pulley can be represented as shown in the following figure:

Tension in the string = T

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

Adding equations ( i ) and ( ii ), we get:

Substituting the value of a in equation ( ii ), we get:

Therefore, the tension in the string is 96 N.

4. Explain why

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

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

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

Ans.(a) 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. Therefore, a horse cannot pull a cart and run in empty space.

(b) When 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). As a result, the passenger"s upper body is thrown forward in the direction in which the bus was moving.

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

On the other hand, while pushing a lawn mower, a force at an angle θ is applied on it, as shown in the following figure.

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.

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

(d) According to Newton"s second law of motion, we have the equation of motion:

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

m = Mass of the ball

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

It can be inferred from equation ( i ) that the impact force is inversely proportional to the impact time, i.e.,

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

While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (Δ t ). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

Total mass of the system, m = 1300 kg

Using Newton"s second law of motion, the reaction force R , on the system by the floor can be calculated as:

= 300 (10 + 15) = 300 × 25

Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton"s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward.

(b) Using Newton"s second law of motion, the reaction force R" , experienced by the helicopter can be calculated as:

= m (g + a )

= 1300 (10 + 15) = 1300 × 25

The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton"s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward.

6. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m . Give the magnitude and direction of

(a) Force on the seventh coin is exerted by the weight of the three coins on its top.

Weight of one coin = m g

Weight of three coins = 3 m g

(b) Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin = m g

Weight of the ninth coin = m g

Weight of the tenth coin = m g

Total weight of these three coins = 3 m g

7. A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Mass of the monkey, m = 40 kg

Acceleration due to gravity, g = 10 m/s

T - m g = ma

∴ T = m (g + a )

= 40 (10 + 6)

m g - T = ma

∴ T = m (g - a )

= 40(10 - 4)

The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0.

T - m g = 0

When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g

m g - T = m g

∴ T = m (g - g) = 0

Applied force, F = 200 N

The force of friction is given by the relation:

= 0.15 (5 + 10) × 10

= 1.5 × 15 = 22.5 N leftward

As per Newton"s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force.

Hence, the reaction of the partition will be 177.5 N, in the leftward direction.

(b) Force of friction on mass A:

= 0.15 × 5 × 10 = 7.5 N leftward

As per Newton"s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward.

When the wall is removed, the two bodies will move in the direction of the applied force.

Net force acting on the moving system = 177.5 N

The equation of motion for the system of acceleration a , can be written as:

Net force causing mass A to move:

This force will act in the direction of motion. As per Newton"s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

9. The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Ans. Mass of the box, m = 40 kg

Initial velocity, u = 0

Distance of the box from the end of the truck, s " = 5 m

As per Newton"s second law of motion, the force on the box caused by the accelerated motion of the truck is given by:

= 40 × 2 = 80 N

As per Newton"s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f , acting between the box and the floor of the truck. This force is given by:

∴Net force acting on the block:

The backward acceleration produced in the box is given by:

Using the second equation of motion, time t can be calculated as:

10. You may have seen in a circus a motorcyclist driving in vertical loops inside a "death-well" (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

The equation of motion for the centripetal acceleration ac , can be written as:

Ans. Let the radius vector joining the bead with the centre make an angle θ , with the vertical downward direction.

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Laws of Motion Class 11 Notes CBSE Physics Chapter 5 (Free PDF Download)
Revision Notes

Revision Notes for CBSE Class 11 Physics Chapter 5 (Laws of Motion) - Free PDF Download

If students need reference by their side for exams, they must prefer Chapter 5 physics Class 11 notes. This chapter deals with Newton’s laws of motion and its real-time application. Students are provided with proper formulas that they can directly apply to solve their queries during exams. According to the CBSE board examination pattern, students should prepare well from notes readily available on Vedantu. These Newton’s laws of motion are thoroughly understood with notes of Physics Class 11 Chapter 5. These are highly helpful if it is about solving complex problems at different levels.

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Laws of Motion Class 11 Notes Physics - Basic Subjective Questions

Section-A (1 Mark Questions)

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

Ans. The coefficient of friction will mainly depend upon two factors, they are as following:

1. The materials of the surfaces in contact.

2. The characteristics of the surfaces.

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

Ans. The frictional force between the tyres and the road provides centripetal force.

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

Ans. As the gun recoils after shooting, it must be held softly on the shoulder. Here the gun and the shoulder are one mass system, due to this the back kick will be reduced. A gunman must keep his weapon securely against his shoulder when shooting.

4. Why does a swimmer push the water backwards?

Ans. From the Newton's 3rd laws of motion, we know that "when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted". As a result, in order to swim ahead, the swimmer pushes water backward with his hands.

5. Friction is a self-adjusting force. Justify.

Ans. Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

6. 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. The thief is in free fall during the jump. Both he/she and the box will be weightless during that time. So, the weight of the box experience by the thief during the jump will be zero. So, mathematically it can be written as: Weight of the box, W=m(g-a)=m(g-g)=0.

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

Ans. Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

8. 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. When a force acts in a perpendicular direction on a moving body, the work done by the force is zero. Since W = F.S cosθ, where S = 90° and cos90° = 0, therefore W = 0.

As a result, the magnitude of the body's velocity (or speed) will remain unchanged. The direction of motion of the body, however, will be altered.

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

Ans. As the spring balancing is based on the tension in the spring, it reads weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg , and the reading will be 10kg .

10. 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. There will be an increase in perceived weight. The apparent weight will stay the same as the true weight, if the lift moves at a constant pace.

Section-B (2 Marks Questions)

11. 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.

(a) As the raindrop is falling with a constant speed, so its acceleration a will be 0. As the force acting on a particle is given by F = ma, so the net force acting on the rain drop will be 0.

(b) As the kite is held stationary, so by Newton's first laws of motion, the algebraic sum of forces acting on the kite is zero.

12. Two blocks of masses m 1 , m 2 are connected by light spring on a smooth horizontal surface. The two masses are pulled apart and then released. Prove that the ratio of their acceleration is inversely proportional to their masses.

Ans. Due to inertia, the mass of the two bodies tries to expand, and the acceleration will act in the opposite direction as it shrinks. So let us assume that the F 1 and F 2 be the forces acting in opposite directions due to masses m 1 and m 2 .

Thus $F_{1}+F_{2}=0$

$m_{1}+a_{2}=m_{2}a_{2}=0$

$m_{1}+a_{1}=m_{2}a_{2}=0$

$m_{1}+a_{1}=m_{2}a_{2}$

$\dfrac{a_{1}}{a_{2}}=-\dfrac{m_{1}}{m_{2}}$

Hence the above is proved.

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

Ans. In the given question, we have the force of 16 N and 12 N given and they are acting on a mass of 200 kg in mutually perpendicular directions. We need to find the magnitude of the acceleration produced.

$F=\sqrt{F_{1}^{2}+F_{2}^{2}+2F_{1}F_{2}\;cos\theta }$

Since, the forces are in mutually perpendicular directions. Therefore, (θ = 90°). Hence, the force will become:

$F=\sqrt{F_{1}^{2}+F_{2}^{2}}$

Now substituting the values, we get

$F=\left ( \sqrt{(16)^{2}+(12)^{2}} \right )$

=20 N

Hence, the magnitude of the acceleration will be:

$a=\dfrac{F}{m}$

$=\dfrac{20}{200}$

$=0\cdot 1ms^{-2}$

14. An elevator weighs 3000 kg. What is its acceleration when the tension supporting cable is 33000 N. Given that g = 9.8 ms −2 .

Ans. From the question, we have an elevator having the weighs given as 3000 kg. We need to find the acceleration, if the tension in the supporting cable is given as 33000 N.

Net upward force on the Elevator F is equal to $=T-ms(\vec{}F=ma)$

$\Rightarrow ma=T-mg$

$\Rightarrow T=m(a+g)$

$\Rightarrow T=33000N=3000(a+9\cdot 8)$

$a=\dfrac{33000-3000\times 9\cdot 8}{3000}$

$=1\cdot 2ms^{-2}$

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

Ans. When a curving road is unbanked, the centripetal force is provided by friction between the tyres and the road. Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground's natural response supplies the necessary centripetal force, reducing tyre wear and tear.

16. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6 ms-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Ans. As from the question, we have the mass of each ball given as 0.05 kg

Initial velocity of each ball will be = 6 ms -1

Magnitude of the initial momentum of each ball given as, p i = 0.3 kg ms -1

The balls alter their directions of motion after colliding, but their velocity magnitudes do not change. Final momentum of each ball given as, $p_{f}=-0\cdot 3kg\;ms^{-1}$

Each ball's impulse equals a change in the system's momentum.

$=p_{f}-p_{i}$

$=0\cdot 3-0\cdot 3=-0\cdot 6\;kg\;ms^{-1}$

The negative indication implies that the balls are receiving opposite-direction shocks.

PDF Summary - Class 11 Physics Laws of Motion Notes (Chapter 5)

A force is something that causes a body's rest or motion to alter. An interaction between two bodies is also referred to as force. Two bodies exert forces on each other even if they are not physically in contact, e.g., the electrostatic force between two charges or gravitational force between any two bodies.

It is a vector quantity having SI unit Newton (N) and dimension \[\left[ {ML{T^{--2}}} \right]\] .

Superposition of Force: When many forces are acting on a body then their resultant is obtained by the laws of vector addition:

$\overrightarrow {{F_{res.}}} = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} + \overrightarrow {{F_3}} .............. + \overrightarrow {{F_n}}$

$\overrightarrow {{F_{res.}}} = \sqrt {{{\overrightarrow {{F_1}} }^2} + {{\overrightarrow {{F_2}} }^2} + 2{{\overrightarrow {{F_1}} }^{}}\overrightarrow {{F_2}} \cos {\theta ^{}}} $

$\tan \alpha = \dfrac{{{F_2}\sin \theta }}{{{F_1} + {F_2}\cos \theta }}$

The resultant of the two forces \[\overrightarrow {{F_1}} \] and \[\overrightarrow {{F_2}} \] acting at angle \[\theta \] is given by: \[\overrightarrow {{F_{res.}}} = \sqrt {{{\overrightarrow {{F_1}} }^2} + {{\overrightarrow {{F_2}} }^2} + 2{{\overrightarrow {{F_1}} }^{}}\overrightarrow {{F_2}} \cos {\theta ^{}}} \]

The resultant force is directed at an angle α with respect to force F1 where:

\[\tan \alpha = \dfrac{{{F_2}\sin \theta }}{{{F_1} + {F_2}\cos \theta }}\]

Lami’s Theorem:

If three forces F 1 , F 2 and F 3 are acting simultaneously on a body and the body is in equilibrium, then according to Lami’s theorem:

\[\dfrac{{{{\text{F}}_1}}}{{\sin (\pi - \alpha )}} = \dfrac{{{{\text{F}}_2}}}{{\sin (\pi - \beta )}} = \dfrac{{{{\text{F}}_3}}}{{\sin (\pi - \gamma )}}\] where α , β & γ are the angles opposite to the forces F 1 , F 2 & F 3 respectively.

(Image will be uploaded soon)

Basic Forces:

There are basically four types of forces: Weight, Contact force, Tension and Spring Force.

Weight: It is the force with which earth attracts a body toward itself. It is also called the gravitational force.

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

Normal Force (N) : This is the contact force component that is normal to the surface. It determines how tightly two surfaces are forced together.

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 in contact surfaces.

Tension: It refers to the force exerted by the end of a taut string, rope, or chain. The direction of strain is pulling the body, whereas the natural reaction is pushing it.

Spring Force: It is the resists to change its length; the more you alter its length the harder it resists. The force exerted by a spring is given by \[F{\text{ }} = {\text{ }}--k.x\] , where x is the change in length and k is the spring constant (unit Nm -1 )

Newton's Laws of Motion:

Newton's first laws:.

If a body is at rest or in motion in a straight line, it will remain at rest or in motion unless it is acted by any external force. This is known as Law of Inertia or First laws of Newton

Inertia is the property of the inability of a body to change its position of rest or uniform motion in a straight line unless some external force acts on it.

Newton’s first laws is valid only in a frame of reference of the inertial frame, i.e., if a frame of reference is at rest or in uniform motion it is called inertial, otherwise non-inertial.

Newton's Second Laws:

According to second laws, rate of change of momentum of a body is proportional to the resultant force acting on the body, i.e.,

$F \propto \dfrac{{dp}}{{dt}}$ or $F \propto ma$

$\Rightarrow \dfrac{{dp}}{{dt}} = m\dfrac{{dv}}{{dt}}$ .

The applied resultant force causes a change in momentum in the direction of the applied resultant force is a measure of the total amount of motion in the body.

\[\vec F=\dfrac {d{\vec v}}{dt}\]

\[\Rightarrow m {\vec a}=\dfrac {{\vec P_2}-{\vec P_1}}{t}\]

The external force acting on a body can accelerate it, either by changing the magnitude of velocity or direction of velocity or both.

Special Cases:

Case 1:

If the force is parallel or antiparallel to the motion of the body, then it changes only the magnitude of $\overrightarrow v $ but not the direction. Therefore, the path will be a straight line.

If the force is acting perpendicular to the motion of body, it changes only the direction but not the magnitude of $\overrightarrow v $ . Therefore, the path will be Circular.

If the force acts at an angle θ to the motion of a body, it changes both the magnitude and direction of $\overrightarrow v $ . In this case the path of the body may be elliptical, non-uniform circular, parabolic or hyperbolic.

Newtons Third Laws:

According to this laws, for every action, there is an equal and opposite reaction. E.g. when two bodies A and B exerts a force on each other i.e. ${F_A}$ & ${F_B}$ . Then the force exerted by any of the bodies will be the same as the force exerted by another body but in opposite direction. \[{{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}}\]

The two forces involved in any interaction between two bodies are called action and reaction.

Linear Momentum:

It is defined as the product of the mass of the body and its velocity i.e.

Linear momentum = \[mass\:\times\:velocity\]

If a body of mass $m$ is moving with a velocity $v$ , its linear momentum $\overrightarrow P $ is given by:

\[\overrightarrow P = m\overrightarrow v \]

It is a vector quantity and its direction is the same as the direction of the velocity of the body.

SI unit of linear momentum is \[kg{\text{ }}m{s^{--1}}\] and the cgs unit of linear momentum is \[g{\text{ }}cm{\text{ }}{s^{--1}}\] .

The entire change in linear momentum is used to calculate the force's impulse, which is the product of the average force during impact and the duration of the impact created during the collision.

The force which acts on bodies for short time are called impulsive forces. E.g. hitting a ball with a bat, firing a bullet with a gun etc.

An impulsive force does not remain constant instead, it varies from zero to maximum and then back to zero. Therefore, it is not possible to measure easily the value of impulsive force because it changes with time.

\[{\vec I} = {\vec F_{av}} \times {\text{t}} = {\vec p_2} -{\vec p_1}\]

Apparent Weight on a Body in a Lift

(a) When the lift is at rest, i.e. a=0:

\[mg - R = 0\]

\[\Rightarrow mg = R\]

\[R = mg(1 - \dfrac{a}{g})\]

\[{W_{app.}} = {W_o}(1 - \dfrac{a}{g})\]

\[R + mg - mg = 0\]

\[{F_{AB}}\times \Delta t = {f_{ms}} = {\mu _s}R\]

\[\angle AOC = \theta\]

\[{W_{app.}} = {W_o}\]

(b) When the lift moves upwards with an acceleration a:

\[R - mg - ma = 0\]

\[\Rightarrow R = m(g + a)\]

\[\Rightarrow R = mg(1 + \dfrac{a}{g})\]

\[{\therefore W_{app}} = {W_o}{\left ( 1 + \dfrac{a}{g} \right )}\]

\[R + ma - mg = 0\]

\[\Rightarrow R = mg(1 - \dfrac{a}{g})\]

\[\therefore {W_{app}} = {W_o}(1 - \dfrac{a}{g})\]

(d) When the lift falls freely, i.e., a = g :

\[\Rightarrow R\;=\;0\]

\[\therefore {W_{app}}\;=\;0\]

Principle of Conservation of Linear Momentum-

According to this principle, in an isolated system, the vector sum of all the system's linear momenta is conserved and is unaffected by their interactions reciprocal action and response.

Mutual forces between pairs of particles in an isolated system (i.e., a system with no external force) can thus produce changes in the linear momentum of individual particles. The linear momentum changes cancel in pairs, and the overall linear momentum remains unaltered because the mutual forces for each pair are equal and opposing. As a result, an isolated system of interacting particles' total linear momentum is conserved. This principle is a direct result of Newton's second and third laws of motion.

Consider an isolated system that consists of two bodies A and B with initial linear momenta \[{P_A}\] and \[{P_B}\] . Allow them to collide for a short time t before separating with linear momenta \[{P_{A}}'\] and \[{P_{B}}'\] , respectively.

If \[{F_AB}\] is force on A exerted by B, and \[{F_{BA}}\] is force on B exerted by A, then according to second law of newton:

\[{F_{AB}}\times \Delta t = \] Change in linear momentum of A = \[{\vec P_{A}}' - {\vec P_A}\]

\[{F_{BA}} \times \Delta t = \] Change in linear momentum of B= \[{\vec P_{B}}' - {\vec P_B}\]

Now, according to third law of newton:

\[\Rightarrow {{\text{F}}_{{\text{AB}}}} = - {{\text{F}}_{{\text{BA}}}}\]

\[\therefore\:from\:equations\:we\:get\:\]

\[\Rightarrow {{\vec P_{A}}'-{\vec P_A}}\:=\:{{\vec P_{B}}'-{\vec P_B}}\]

\[\Rightarrow {{\vec P_{A}}'+{{\vec P_B}}'}={{\vec P_{A}}+{\vec P_{B}}}\]

which shows that the total final linear momentum of the isolated system is equal to its total initial linear momentum. This proves the principle of conservation of linear momentum.

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.

It is the force that comes into play when two surfaces come into contact with each other and oppose their relative motion.

I Frictional force is unaffected by the contact area. Because with an increase in the area of contact, the force of adhesion also increases.

When the surfaces in contact are extra smooth, the distance between the molecules of the surfaces in contact decreases, increasing the adhesive force between them. Therefore, the adhesive pressure increases, and so does the force of friction.

Types of Friction:

There are 3 types of friction: Static, Limiting and Kinetic Friction.

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

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

Kinetic Friction - 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.

Laws of Limiting Friction:

(i) Friction always works in the opposite direction of relative motion, making it a perverse force.

(ii) The maximum static friction force, fms (also known as limiting friction), is proportional to the normal reaction (R) between the two in contact surfaces i.e.

\[{f_{ms}} \propto R\]

(iii) The limiting friction force is tangential to the interface between the two surfaces and is determined by the kind and state of polish of the two surfaces in contact.

(iv) As long as the normal reaction stays constant, the limiting friction force is independent of the area of the surfaces in contact.

Coefficient of Static Friction:

We know that,

\[\Rightarrow {f_{ms}} = {\mu _s}R\]

Here, \[{\mu _s}\] is a constant of proportionality and is called the coefficient of static friction and its value depends upon the nature of the surfaces in contact and is usually less than unity i.e. 1 but never 0.

Since the force of static friction (f S ) can have any value from zero to maximum ( \[{f_{ms}}\] ), i.e. \[{f_s} \leqslant {f_{ms}}\] ,

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:

\[{f_k} \propto R\]

\[\Rightarrow {f_k} = {\mu _k}R\]

Rolling Friction:

The opposing force that comes into play when a body rolls over the surface of another body is called rolling friction.

Friction caused by rolling. Consider a wheel that is rolling down a street. As the wheel travels down the road, it presses against the road's surface and is compressed slightly as shown in Fig. :

Angle of Friction:

The angle of friction between any two surfaces in contact is defined as the angle formed by the resultant of the limiting friction force F and the normal reaction direction R. It is represented by θ.

In the given fig. OA represents the normal reaction R that balances the weight mg of the body. OB represent F, the limiting force of sliding friction, when the body tends to move to the right. Complete the parallelogram OACB. Join OC. This represents the resultant of R and F. By definition, \[\angle AOC = \theta \] is the angle of friction between the two bodies in contact.

The angle of friction is determined by the nature of the materials used on the surfaces in contact as well as the nature of the surfaces themselves.

Relation Between 𝛍 and 𝛉:

In $\Delta AOC$ , $\tan \theta = \dfrac{{AC}}{{OA}} = \dfrac{{OB}}{{OA}} = \mu $$\tan \theta = \dfrac{{AC}}{{OA}} = \dfrac{{OB}}{{OA}} = \mu $

Hence, $\mu = \tan \theta $ ... (6)

(i.e.\[\mu\] is the coefficient of limiting friction )

Angle of Repose or Angle of Sliding:

The minimum angle of inclination of a plane with the horizontal at which a body placed on the plane begins to slide down is known as the angle of repose or angle of sliding.

Represented by α. Its value depends on the material and nature of the surfaces in contact.

In fig., AB is an inclined plane such that a body placed on it just begins to slide down. \[\angle BAC\:\alpha\] = angle of repose.

The various forces involved are :

weight, mg of the body,

normal reaction, R,

Force of friction F,

Now, mg can be resolved into two rectangular components: mg cosα opposite to R and mg sinα opposite to F:

$F = mg\sin \alpha $ ... (7)

$R = mg\cos \alpha $ ... (8)

Diving these two eq. we get:

$\tan \alpha = \dfrac{F}{R}$

$\Rightarrow \tan \alpha = \mu $ ... (9)

Hence coefficient of limiting friction between any two surfaces in contact is equal to the tangent of the angle of repose between them.

From (6) and (9): $\mu = \tan \alpha = \tan \theta $

Therefore $\alpha = \theta $

i.e. (Angle of friction = Angle of repose)

Methods of Changing Friction:

Some of the ways of reducing friction are:

By polishing.

By lubrication.

By proper selection of materials.

By Streamlining.

By using ball bearings.

Dynamics of Uniform Circular Motion Concept of Centripetal Force:

The force required to move a body uniformly in a circle is known as centripetal force. This force acts along the circle's radius and towards the centre.

When a body moves in a circle, the direction of motion at any given time is along the tangent to the circle. According to Newton’s first law of motion, a body cannot change its direction of motion by itself an external force is needed. This external force is called the centripetal force.

An expression for centripetal force is:

\[F = m{v^2}/r = m{\omega ^2}r\]

\[R{\text{ }}--{\text{ }}mg{\text{ }} = {\text{ }}0{\text{ }}or{\text{ }}R{\text{ }} = {\text{ }}mg \]

\[F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R{\text{ }} = {\text{ }}{\mu _s}{\text{ }}mg{\text{ }} \]

\[R{\text{ }}cos\theta = {\text{ }}mg{\text{ }} + {\text{ }}F{\text{ }}sin\theta \]

\[From\:{\text{ }}\left( 3 \right),{\text{ }}R{\text{ }}\left( {cos\theta --{\text{ }}{\mu _s}{\text{ }}sin\theta } \right){\text{ }} = {\text{ }}mg \]

On account of a continuous change in the direction of motion of the body, there is a change in velocity of the body, and hence it undergoes an acceleration, called centripetal acceleration or radial acceleration.

Centrifugal Force:

Centrifugal force is a force that arises when a body is moving actually along a circular path, by virtue of the tendency of the body to regain its natural straight-line path.

When a body is moving in a straight line centripetal force is applied on the body, it is forced to move along a circle. The body has a natural inclination to return to its normal straight-line course while moving in a circle. This tendency gives rise to a force called centrifugal force.

The magnitude of centrifugal force = $m{v^2}/r$ , which is the same as that of centripetal force, but opposite in direction i.e. The centrifugal force acts along the circle's radius, away from the centre.

Note : Centripetal and Centrifugal forces, being the forces of action and reaction act always on different bodies. E.g. when a piece of stone tied to one end of a string is rotated in a circle, centripetal force F1 is applied on the stone by the hand. Due to the stone's desire to revert to its natural straight line route, centrifugal force F2 acts on it, pulling the hand outwards.. The centripetal and centrifugal forces are shown in Fig. :

Rounding A-Level Curved Road:

When a vehicle goes around a curved road, it requires some centripetal force. The vehicle's wheels have a tendency to depart the curved path and return to the straight-line path as it rounds the curve. Force of friction between the wheels and the road opposes this tendency of the wheels. This force (friction) therefore, acts, towards the centre of the circular track and provides the necessary centripetal force.

Three forces are acting on the car, fig.

The weight of the car, mg, acting vertically downwards,

Normal reaction R of the road on the car, acting vertically upwards,

Frictional Force F, along the surface of the road, towards the centre of the turn.

As there is no acceleration in the vertical direction,

$R{\text{ }}--{\text{ }}mg{\text{ }} = {\text{ }}0{\text{ }}or{\text{ }}R{\text{ }} = {\text{ }}mg$ ...(1)

The centripetal force required for circular motion is applied along the road's surface, toward the turn's centre. As previously stated, static friction is what supplies the required centripetal force. Clearly $F \leqslant m{v^2}/r$ …(2)

where v is velocity of car and r is the radius of the curved path as,

$F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R{\text{ }} = {\text{ }}{\mu _s}{\text{ }}mg{\text{ }}$ ( ${\mu _s}$ is coefficient of static friction between the tyres and the road)

Therefore from (2),

\[\dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}} \leqslant {{\mu_s}mg}\]

\[v\:\leqslant\:\sqrt{{\mu_s}rg}\]

\[\therefore {\vec v_{max}}= \sqrt{{\mu_s}rg}\] ……(3)

Therefore, the maximum velocity with which car can move without slipping is:

\[{\text{v}} = \sqrt {{\mu _{\text{s}}}{\text{rg}}} \]

Banking of Roads:

The phenomenon of raising the outer edge of the curved road above the inner edge is called banking of roads.

The maximum permissible velocity with which a vehicle can go round a level curved road without skidding depends on μ (Coefficient of friction between the tyre and the road). As a result, the frictional force is not a reliable source of the required centripetal force for the vehicle.

In Fig., OX is a horizontal line. OA is the level of the banked curved road whose outer edge has been raised.

$\angle XOA{\text{ }} = \;\theta $ = angle of banking.

Forces are acting on the vehicle as shown in Fig:

Weight mg of the vehicle acting vertically downwards.

Normal reaction R of the banked road acting upwards in a direction perpendicular to OA.

Force of friction F between the banked road and the tyres, acting along with AO. R can be resolved into two rectangular components-

R cosθ, along a vertically upward direction

R sinθ, along the horizontal, towards the centre of the curved road. F can also be resolved into two rectangular components:

(i) F cosθ, along the horizontal, towards the centre of curved road

(ii) F sinθ, along the vertically downward direction.

As there is no acceleration along the vertical direction, the net force along this direction must be zero. Therefore,

$R{\text{ }}cos\theta = {\text{ }}mg{\text{ }} + {\text{ }}F{\text{ }}sin\theta $ ……(1) If v is the vehicle's velocity on a banked circular road with radius r, then centripetal force is = $m{v^2}/r$ . This is provided by the horizontal components of R and F as shown in Fig.

Therefore, \[R\sin \theta + F\cos \theta = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}\] ...(2)

But F < ${\mu _s}$ R, where ${\mu _s}$ is coefficient of static friction between the banked road and the tyres. To obtain ${v_{\max }}$ , we put $F{\text{ }} = {\text{ }}{\mu _s}{\text{ }}R$ in (1) and (2)

\[{\text{R}}\cos \theta = {\text{mg}} + {\mu _{\text{s}}}{\text{R}}\sin \theta \] … (3)

And \[{\text{R}}\sin \theta + {\mu _{\text{s}}}{\text{R}}\cos \theta = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}\] …(4)

$From{\text{ }}\left( 3 \right),{\text{ }}R{\text{ }}\left( {cos\theta --{\text{ }}{\mu _s}{\text{ }}sin\theta } \right){\text{ }} = {\text{ }}mg$

\[{\text{R}} = \dfrac{{{\text{mg}}}}{{\cos \theta - {\mu _{\text{s}}}\sin \theta }}\] …(5)

From (4), $\quad {\text{R}}\left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right) = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$

$\operatorname{Using} (5),\dfrac{{\operatorname{mg} \left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right)}}{{\left( {\cos \theta - {\mu _{\text{s}}}\sin \theta } \right)}} = \dfrac{{{\text{m}}{{\text{v}}^2}}}{{\text{r}}}$

$\therefore \quad {v^2} = \dfrac{{\operatorname{rg} \left( {\sin \theta + {\mu _{\text{s}}}\cos \theta } \right)}}{{\left( {\cos \theta - {\mu _{\text{s}}}\sin \theta } \right)}} = \dfrac{{\operatorname{rg} \cos \theta \left( {\tan \theta + {\mu _{\text{s}}}} \right)}}{{\cos \theta \left( {1 - {\mu _{\text{s}}}\tan \theta } \right)}}$

\[{\text{v}} = {\left[ {\dfrac{{rg\left( {{\mu _{\text{s}}} + \tan \theta } \right)}}{{\left( {1 - {\mu _{\text{s}}}\tan \theta } \right)}}} \right]^{1/2}}\] …(6)

Discussion:

If \[{\mu _s}\] = 0, i.e., if banked road is perfectly smooth, then from eqn. (51), \[{v_o} = {(rg\tan \theta )^{\dfrac{1}{2}}}\]

Even when there is no friction, this is the speed at which a banked road may be rounded. On a banked road, driving at this speed causes essentially little wear and tear tyres .

\[{v_o}^2 = {\text{ }}rg{\text{ }}\tan \theta \]

\[\Rightarrow {\overrightarrow F _{real}} + {\overrightarrow F _{pseudo}} = {m_p}{a_{P,O}}\]

\[{\overrightarrow F _{real}} - {m_p}{\overrightarrow a _o} = {m_p}{\overrightarrow a _{P,O}}\]

If speed of vehicle is less than \[{v_o}\] , frictional force will be up the slope. Therefore, the vehicle can be parked only if \[\tan \theta \leqslant {\mu _s}\] .

The average speed of vehicles passing through a road is usually banked. However, if a vehicle's speed is slightly less or more than this, the self-adjusting static friction between the tyres and the road will operate, and the vehicle will not slide.

Note that curved railway tracks are also banked for the same reason. The level of the outer rail is raised a little above the level of the inner rail while laying a curved railway track.

Bending of a Cyclist:

When a cyclist turns, he must also exert centripetal force. His weight is balanced by the usual reaction of the ground if he keeps himself vertical while spinning. In that case, he must rely on the friction between the tyres and the road to generate the required centripetal force. Because the force of friction is modest and unpredictable, relying on it is risky.

In order to avoid relying on friction to generate centripetal power, the cyclist must bend slightly inwards from his vertical position while turning. A component of normal reaction in the horizontal direction produces the required centripetal force in this way.

To calculate the angle of bending with vertical, suppose

m = mass of the cyclist,

v = velocity of the cyclist while turning,

r = radius of the circular path,

\[\theta\] = angle of bending with the vertical

In the fig. shown below the weight of the cyclist(mg) acts vertically downwards at the centre of gravity C. R is the normal reaction on the cyclist. It acts at an angle \[\theta\] with the vertical.

R can be resolved into two rectangular components: \[R\:\cos\theta\] , along the vertically upward direction,\[R\:\sin\theta\], along the horizontal, towards the centre of the circular track.

In equilibrium , \[R{\text{ }}cos\theta = {\text{ }}mg\]\[R{\text{ }}cos\theta = {\text{ }}mg\] ….(1)

\[R{\text{ }}\sin \theta = {\text{ }}m{v^2}/r\] ….(2)

Dividing (2) by (1), we get, \[\tan \theta = {\text{ }}\dfrac{{{v_{}}^2}}{{rg}}\]

For a safe turn, θ should be small, for which v should be small and r should be large i.e. Turning should be done slowly and on a greater radius track. This means, a safe turn should neither be fast nor sharp.

Pseudo Force:

\[{\overrightarrow F _{pseudo}} = - {m_p}{\overrightarrow a _0}\] If observer O is non-inertial and still wants to apply Newton's Second Law on particle P, the observer must add a "Pseudo force" to the real forces on particle P.

\[{\overrightarrow F _{real}} + {\overrightarrow F _{pseudo}} = {m_p}{a_{P,O}}\]

i.e \[{\overrightarrow F _{real}} - {m_p}{\overrightarrow a _o} = {m_p}{\overrightarrow a _{P,O}}\]

Where \[{\overrightarrow a _{P,O}}\] is acceleration of P with respect to observer O.

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Physics Class 11 Chapter 5 notes are easy to read that covers all questions asked in the exams. Thus it helps students to prepare well before their final board class. All the answers are described in a precise manner such that students need not refer anywhere else for the same. According to marks assigned, the concepts are quickly clarified, and students become aware of answering a particular question.

What Topics are Covered With Ch 5 Physics Class 11 Notes?

NCERT covers the entire syllabus as preferred by the CBSE board. Thus notes of Laws of Motion Class 11 are well prepared according to the question paper pattern designed by CBSE. These notes aid students while preparation and hence reduce their stress that comes during a hectic study schedule. These revision notes are best for self-study. Experienced teachers have prepared them according to the requirement. Below are the topics covered in physics Chapter 5 Class 11 notes:

The chapter begins with stating Dynamics, which is the study of forces and motions.

Then the first definition is covered: how to define forces?

The next concept is regarding Inertia along with its three different states.

Finally, it covers the definition of three different laws of motion along with the subtopics to prove these laws in detail. Numerical problems are also covered under this concept.

Now let us study each concept that is covered in Class 11 physics ch 5 notes.

Dynamics is defined as the term in physics, which deals with the study and knowledge about motion forces and laws.

What is Inertia? Also, Describe Its Three Different States.

If an object moves in a straight line or is at rest or in a uniform motion, it cannot change its state to another until compelled. This property is known as inertia. It is measured according to the mass of a body.

There Are Three Different Types of Inertia

Inertia of motion

Inertia of rest

Inertia of direction

Class 11 physics laws of motion notes also cover a basic definition of force and its interaction with different objects. Force is the basic pull and pushes applied to an object, whether the object is at rest or moving. We can also say that it is an interaction of one object with another, due to which an object changes its state.

On general terms, there are two different types of forces:

Constant Force

Action Force

Newton’s Laws of Motion

Newton has given three different laws of motion along with the proof to each law.

Newton’s First Law of Motion: Until you apply any external force to an object, the object will not change its state. It is basically if an object is at rest or moving uniformly in a straight line.

Also, this definition is called the Law of Inertia.

Newton’s Second Law of Motion: The net force is equivalent to the magnitude of the net force along with the direction of the applied net force.

Also, many define the second law of motion as the force directly proportional to acceleration produced while moving and the mass of the body.

\[\vec{F} = K \frac{d\bar{p}}{dt} = Km\vec{a}\]

Suppose F is the net force and m is the mass of the body, and a is the acceleration then Force is defined by the expression given above. Here, k is the constant of proportionality.

Newton’s Third Law of Motion: It is defined that if we apply a force to an object, we will experience an equal force in the opposite direction applied by the object on us.

This Law Deals With Some Other Basic Subtopics Like

Linear Momentum

Law of Conservation of Momentum

Concurrent forces and Equilibrium

Notes of Chapter 5 physics Class 11 will also cover the concept of the simple pulley, friction, actual and apparent weight, laws of friction, angle of friction, coefficient of friction, motion in a circle, and many other numerical problems associated with it.

Why Should You Choose Vedantu for Class 11 Chapter 5 Physics Notes?

Students need a definitive tool to prepare well for exams. These physics ch 5 Class 11 notes are the best way to prepare well for the exams and avoid the last-minute rush. The most difficult chapter needs basic clarification for each topic and its subtopic if you want to score good marks.

Also, students get the least time to prepare their notes. Hence, these Vedantu notes will aid them at every step to clarify their concepts. According to CBSE format, the entire content of notes of ch 5 physics Class 11 is prepared by experienced teachers. Thus students need not scroll any other website for their help.

Benefits for Taking Notes of Laws of Motions Class 11 CBSE Physics Chapter 5

The advantages of taking Laws of Motion Class 11 Notes CBSE Physics Chapter 5 in the form of a free PDF download aid students in effective learning, revision, and exam preparation, leading to improved performance in their academic journey.

Taking Laws of Motion Class 11 Notes CBSE Physics Chapter 5 in the form of a free PDF download offers several advantages to students:

Accessibility: The free PDF download of class 11 physics notes allows students to access the study material anytime, anywhere, using their preferred devices.

Comprehensive Coverage: The notes provide a concise yet comprehensive overview of the chapter, covering all the key concepts related to laws of motion.

Simplified Explanations: The notes offer simplified explanations and examples, making complex concepts easier to understand and retain.

Quick Revision: These notes serve as handy reference material for quick revision before exams, helping students recall important points and formulas.

Time-Saving: The PDF format allows students to save time in note-taking, enabling them to focus on understanding the concepts better.

Supplementary Practice : The notes often include practice questions and solutions, facilitating self-assessment and reinforcing learning.

Exam-Oriented: The notes are designed to align with the CBSE exam pattern, ensuring that students focus on topics that are relevant for the board exams.

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The Laws of Motion Class 11 Notes CBSE Physics Chapter 5 in the form of a free PDF download offer a valuable resource to students pursuing physics. These comprehensive notes provide simplified explanations, examples, and practice questions, enhancing students' understanding and problem-solving skills related to motion. The convenience of accessing these notes as a PDF download allows for easy revision and quick reference during exam preparation. By utilizing these notes, students can efficiently cover the entire chapter, reinforcing their conceptual knowledge and preparing effectively for assessments. The free availability of these notes ensures equitable access to quality study material, promoting a holistic learning experience for all students.

FAQs on Laws of Motion Class 11 Notes CBSE Physics Chapter 5 (Free PDF Download)

1. A 5 kg iron sphere feels an upsurge of velocity from 2 m/s to 4 m/s. The velocity rises within the 20 s because a constant force acts on it by unchanging the direction. Find the magnitude & direction of this force.

As per the question

Mass (m) = 5 kg

Time, t = 20 s

Initial velocity, u = 2 m/s

Final velocity, v = 4 m/s

Also, F = m * a …. [1]

As, v = u + at (equation of motion)

a = (v - u) / t

Or, m [(v - u) / t] = Force

Since, a = (v – u) / t

Force = 5 [(4 – 2) / 20]

= 0.5 N (along the direction of motion)

2. What are the chief roles played by Vedantu in CBSE board exams?

Vedantu is involved in so many educational activities. It is paying attention to the quality studies for Physics also. The awareness, along with thorough solved question and answer practice papers, is very helpful to our students for the CBSE board exams.

3. Explain Newton’s law of gravitation?

If two particles possessing different masses (M and m) are mutually attracted with identical and opposite forces such as F and - F.

The relation is F = G(Mm/r 2 )

Here, r = distance between the two particles

G = the universal constant of gravitation

4. How to prepare for Chapter 5 of Class 11 Physics?

The student should make a routine allotting equal hours to every subject and following this regularly to achieve better results. Apart from this, the student should study the given chapter line by line and highlight the important points and topics to retain so that they help during the examination. After this is done, the student should refer to the NCERT solutions and solve the given exercises to attain a better practice at the type of questions that can be asked in the question paper. Consistent practice and hard work will help the student in passing with flying colours.

5. Where can I get the NCERT Solutions for Chapter 5 of Class 11 Physics?

The NCERT solutions are planned and devised to meet the demands and needs of the students. These are planned by experts, containing various exercises considering the syllabus present in the curriculum. These exercises help the student to get a strong idea and understanding of the concept and topics so that the questions asked in the question papers don't perplex them. It is advised that the students have a copy of the NCERT solutions for every subject. These can be availed easily online and practised in the comforts of the home.

6. What are the important topics covered in Chapter 5 of Physics Class 11?

The important topics that are covered in Class 11 Chapter 5 Physics are:

Laws of motion

The law of inertia

Aristotle's fallacy

Newton’s first, second and third laws of motion

Conservation of momentum

Equilibrium of a particle

Common forces in mechanics

Circular motion

Solving problems in mechanics.

All these topics are important for the student to be well versed in. To achieve this they can refer to the NCERT Solutions offered by Vedantu and these solutions can be downloaded online. The exercises that these solutions offer will help the student to grasp a stronghold of the concepts present in Chapter 5 Physics.

7. Why is it important to refer to the NCERT Solutions for Chapter 5 of Class 11 Physics?

The NCERT solutions are devised by student matter experts in order to fulfil the needs and demands of the students. These solutions have numerous exercises that the student needs to practice to retain the important concepts better. Apart from this all these exercises have detailed and explained answers that will help the student in every step of the understanding process, clearing all their doubts and confusions. These solutions also have in the previous year’s question papers that the student can practice getting an idea of how the question might be asked. Thus, practising and acing the NCERT solutions will inevitably lead the student in grabbing good grades in the exam.

8. 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.

Let m 1 and m 2 be the mass of the product respectively

Let v 1 and v 2 be the velocity respectively

Thus, the total linear momentum after disintegration becomes m 1 v 1 +m 2 v 2

But it is to be noted that before disintegration the nucleus is at rest and therefore the linear momentum before disintegration becomes zero

Thus, following the principle of conservation of linear momentum;

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CBSE Study Materials for Class 11

NCERT Solutions for Class 11 Physics Chapter 5 Free PDF Download

Ncert solutions for class 11 physics chapter 5: laws of motion.

Physics is a crucial part of science which aims to explain the behavior of objects. Moreover, it helps a lot to those students that want to be a physicist. In this NCERT Solutions for Class 11 Physics Chapter 5, we are discussing the Laws of Motion and the things related to it. Above all, the topics are explained in a detailed manner and in easy language.

Our team of proficient teachers has prepared these NCERT Solutions for Class 11 Physics Chapter 5. Also, you can learn the topics of the solutions at your convenience. Besides, they will help to improve your grade.

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Download NCERT Solutions for Class 11 Physics.

Download NCERT Solutions for Class 11 here.

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CBSE Class 11 Physics Chapter 5 Laws of Motion NCERT Solutions

Laws of Motion refers to the three physical laws given by Newton that laid the basis for classical mechanics. Also, they define the relationship of a body with the force acting upon it. This NCERT Solutions for Class 11 Physics for Chapter 5 has all the details that you require to realize the concept of the Chapter.

Subtopics covered under NCERT Solutions for Class 11 Physics Chapter 5

5.1 introduction.

This topic describes what we have learned about the laws of motion earlier and what we are going to learn in this chapter.

5.2 Aristotle’s Fallacy

This refers to an argument that does not sound logical. In this topic, the Fallacy theory of Aristotle is described. According to this theory, a body requires an external force to keep it moving.

5.3 The Law of Inertia

This law state that a body of an object does not want to change its state of rest or motion.

5.4 Newton’s First Law of Motion

This law state that an object remains in the state of rest or uniform motion in a straight line unless it is compelled with an external force.

In simple words, objects in motion or rest repel the change in their state of motion.

5.5 Newton’s Second Law of Motion

This law state that the acceleration produced by an object is proportional to the unbalanced force applied to it.

Momentum- It refers to the thrust a body produces which is the product of its mass and velocity.
Impulse- It refers to the impact force which generates for a short period of time. But it is powerful enough to change the momentum of the object.

5.6 Newton’s Third Law of Motion

According to this law, every action we perform has an opposite and equal reaction. Also, both the action and reaction are equal in magnitude. Moreover, forces always occur in pairs.

5.7 Conservation of Momentum

This law states that the momentum of two objects before and after the happening of the event remains the same. But the objects will move in the direction in which the heavier object was moving. Also, this force is equal to 0.

5.8 Equilibrium of Particle

This refers to the state in which the particle or object is not moving or rotating. Moreover, this requires the sum of movements usually called torque.

5.9 Common Forces in Mechanics

This topic describes the different forces that act on the objects of mechanics.

5.9.1 Friction- It refers to the force that resists the relative motion of an object.

Static Friction- the friction that keeps the body at rest.
Impending Motion- It means the motion which would take place but does not do anything when force is applied to it.
Rolling friction- It states that a rolling round object will not face any friction.

5.10 Circular Motion

This topic describes the movement of an object along with the circumference of a circle. Also, this topic explains the circular motion with the help of an example of a car.

5.11 Solving Problems in Mechanics

These topics define the various ways by which the mechanic’s problems could be solved.

You can download NCERT Solutions for Class 11 Physics Chapter 5 PDF by clicking on the button below.

NCERT Solutions for Class 11 Physics Chapter 5 Free PDF Download

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NCERT Solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT Solutions for Class 11 Physics Chapter 5 Work, Energy and Power in Hindi and English Medium updated for CBSE and State board students for 2024-25 with MCQ answers. If you come across any concepts that are unclear, seek clarification from your teacher, classmates, or online resources at Tiwari Academy. It’s crucial to have a strong conceptual understanding before moving forward.

Class 11 Physics Chapter 5 Work, Energy and Power Question Answers

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A ball moves in a friction less inclined table without slipping. What is the work done by the table surface on the ball is

Motion without slipping implies pure rollings. During pure rolling work done by friction force is zero.

View Answer

Which one has higher kinetic energy? Both light and heavy bodies have equal momenta.

The momenta of the heavy body and light body are same, thus the velocity of the light body is greater than that of heavy body. Also kinetic energy is proportional to the square of velocity. Hence the lighter body has greater kinetic energy.

According to the Equivalence of Mass and Energy, it states that mass and energy are not inter convertible

According to Equivalence of mass and energy put forth by Einstein, it states that mass and energy are inter convertible.

The energy passed by the body by virtue of its motion is known as

The energy possessed by a body by virtue of its motion is known as its kinetic energy. Its given by the formula KE = 1/2mv²

Potential energy is the energy stored by any object or system due to its position or the arrangement of its components. However, it is not affected by the environment outside the object or system, such as air or altitude. Potential energy is not transferable, it depends on the height or distance and mass of the object. K.E. can be transferred from one object to another via vibration or rotation. On the other hand, the kinetic energy of a moving object or particle system remains same. Unlike potential energy, the kinetic energy of an object is relative to other fixed and moving objects in its environment.

For example, if an object is placed at a higher height, the kinetic energy of the object will be higher. P.E. is converted into kinetic energy, kinetic energy is converted into potential energy, and then back again. It’s an endless cycle. Kinetic energy is created when potential energy is released, propelled by gravitational or elastic forces and other catalysts.

Kinetic energy is the energy of motion. When work is done on an object and it is accelerated, it increases the kinetic energy of the object. The most important factors in determining kinetic energy are the motion and mass of the object in question. Although mass is a universal measure, the motion of an object can occur in many different ways, including rotation around an axis, vibration, translation, or any combination of these and other motions.

Collision between marble balls is which type of collision

Collision between marble balls is an example of elastic collision.

Which is the type of collision in which both the linear momentum and the kinetic energy of the system remain conserved

An elastic collision is a collision in which there is no net loss in kinetic energy in the system as a result of the collision. Both momentum and kinetic energy are conserved quantities in elastic collisions.

The rate of doing work is called

Power is the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the SI system, the unit of power is the joule per second(J/s).

An electric heater of rating 1000W is used for 5 hrs per day for 20 days. What is the electrical energy utilized?

The power of the Electric heater is 1000 W, and the time period is 20*5 = 100hr Electrical energy = Power * Time, Electrical energy = 1000*100 = 100000 Wh, Electrical energy = 100 kWh

The law of conservation of energy states that energy cannot be created or destroyed – it can only be converted from one energy form to another. This means that a system always has the same amount of energy and there is no need to add it from the outside. This is especially confusing in the case of non-conservative forces, where the energy is converted from mechanical to thermal, but the total energy remains the same. The only way to use energy is to convert it from one to another.

The change in the internal energy of the system can also be determined by mathematical equations given in chapter 6 of 11th Physics NCERT . Although these equations are very powerful, they can make it difficult to see the power of statements. The take home message is that energy cannot be created from scratch. Society has to get its energy from somewhere, although there are many sneaky places to get energy from certain sources, primary fuels, and primary energy streams.

Collision, in physics, is the sudden and violent collision of two objects when they are in direct contact, such as two billiard balls, a golf club and a ball, a hammer and a nail head, or falling objects on the ground. In addition to material properties, two other factors can affect the outcome of an impact: the force with which objects come into contact and time.

A common rule of thumb is that a hard steel ball dropped on a steel plate will almost bounce where it fell, whereas with putty or lead it will not. The collision of a steel ball with a steel plate is called elasticity, and the collision of a putty or lead ball with a steel plate is called in-elasticity or plasticity.

Between these extremes there are varying degrees of elasticity and corresponding responses to impact. Perfect elastic shocks can only be achieved at the atomic level. The kinetic energy of the reacting body is not lost; in a fully plastic shock, kinetic energy is lost the most. An elastic collision is a collision in which there is no net loss of kinetic energy in the system due to the collision.

How many questions are there in the NCERT book of 11th Physics Chapter 5?

Earlier there were 30 questions in Chapter 5 of the NCERT Book Class 11 Physics Exercises but as per the new syllabus there are only 23 questions in 2024-25 curriculum. Moreover, the answers to all these 23 questions are available in Tiwari Academy’s NCERT solutions section. These answers are accumulated in a precise and logical way by expert teachers on India’s leading educational website.

How can the triple point of water be explained through Chapter 5 of Class 11 Physics?

The triple point of any substance is the combination of temperature and pressure at which a substance can exist in all three states – solid, liquid, and gas. Students of class 11 Physics can understand using the examples given in chapter 5. The triple point of water is 273.16 K and the vapor pressure is 611.66 Pascal’s. At this point, water can exist in three states through small changes in pressure and temperature: vapor, liquid, and ice.

How to download NCERT Solutions for Class 11 Physics Chapter 5?

NCERT Solutions for Class 11 Physics Chapter 5 Working Energy and Power is given on various educational websites free to download. Students can download NCERT Solutions for 11th Physics Chapter 5 from Tiwari Academy website also. You have to visit the page NCERT Solutions Class 11 Physics Chapter 5 and download it using the PDF file link. There are two separate file – one is for exercises solutions and other one is additional exercises question answers. These NCERT solutions can help students to prepare for exams and get good grades.

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Important Questions Class 11 Physics Chapter 5

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Important Questions for CBSE Class 11 Physics Chapter 5 – Law of Motion

You can find the best revision questions for Laws of Motion Class 11 Physics Chapter 5 Important Questions here at Extramarks. The questions contain step-by-step solutions that are chosen by subject matter experts. The following Class 11 Physics Chapter 5 Important Questions will help students score well in your final exams.

All of the important topics of the Chapter Law of Motion are covered by the Important Questions and Answers. This set of questions is prepared in such a way that they will aid students in studying the chapter while keeping the important questions and answers in mind. The study material has been created with the types of questions asked in the CBSE Class 11 examination in mind.

This article also includes many important concepts and formulas for Class 11 Physics Chapter 5 that are explained in detail for better understanding.

CBSE Class 11 Physics Chapter 5 Important Questions

These are examples of Chapter 5 Class 11 Physics Important Questions, click here to get the full set of Important Questions for Class 11 Physics Chapter 5.

1 Mark Answers and Questions

Q1. Name the factor on which the coefficient of friction depends.

Ans: The coefficient of friction will mainly depend on two factors, which are as follows:

The materials of the surfaces in contact.
The characteristics of the surfaces.

Q2. What provides the centripetal force to a car taking a turn on a level road?

Ans: Centripetal force is provided by the frictional contact between the tyres and the road.

Q3. Why does a swimmer push the water backwards?

Ans: From Newton’s 3rd Law of Motion, we know that “when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted”. As a result, the swimmer pushes water backward with his hands in order to swim ahead.

Q4. Action and reaction forces do not balance each other. Why?

Ans: Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

Q5. The two ends of a spring-balance are pulled by a force of 10 kg each. What will be the reading of the balance?

Ans: As the spring balancing is based on the tension in the spring, it gauges weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg, and the reading will be 10kg.

Q6. A lift is an acceleration upward. Will the apparent weight of a person inside the lift increase, decrease, or remain the same relative to its real weight? What happens if the lift is going at a uniform speed?

Ans: There will be an increase in perceived weight. The apparent weight will stay the same as the true weight if the lift moves at a constant pace.

Q7. Why is it desirable to hold a gun tight to one’s shoulder when it is being fired?

Ans. Because the gun recoils after being fired, it must be held softly on the shoulder. Because the gun and the shoulder are combined into one mass system, the back kick is reduced. When shooting, a gunman must keep his weapon securely against his shoulder.

Q8. Justify that friction is a self-adjusting force.

Ans: Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

2 Marks Answers and Questions

Q1. 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: (a) As the raindrop is falling at a constant speed, its acceleration will be 0. The net force acting on the raindrop will be 00 because the force acting on a particle is given by.

(b) As the kite is held stationary, by Newton’s first law of motion, the algebraic sum of forces acting on the kite is zero.

Q2. Write two consequences of Newton’s second law of motion.

Ans: The two consequences of Newton’s Second Law of Motion are as follows.

1.It demonstrates that the motion is only accelerated when force is applied to it.

2. It introduces the notion of a body’s inertial mass.

Q3. 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 begins to fly within the cage, the weight of the bird is no longer felt since the air inside is in direct touch with ambient air, making the cage look lighter.

Q4. Why does a cyclist lean to one side, while going along a curve? In what direction does he lean?

Ans: A cyclist leans while riding along a curve because a component of the ground’s natural response supplies him with the centripetal force he needs to turn.

He must lean inward from his vertical posture, towards the circular path’s centre.

Q5. A soda water bottle is falling freely. Will the gas bubbles rise to the surface of the water in the bottle?

Ans: As the water in a freely falling bottle is in a state of weightlessness, As a result, there is no upthrust force on the bubbles, and the bubbles do not ascend in the water.

Q6. Explain why passengers are thrown forward from their seats when a speeding bus stops suddenly.

Ans: When a fast bus comes to a complete stop, the bottom half of the body in touch with the seat comes to a complete halt, while the upper section of the passengers’ bodies prefer to retain their uniform motion. As a result, the passengers are pushed forward.

Q7. How does road banking reduce tyre wear and tear?

Ans: When a curving road is not banked, friction between the tyres and the road provides centripetal force.

Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground’s natural response supplies the necessary centripetal force, reducing tyre wear and tear.

Q8. A force is being applied to a body, but it causes no acceleration. What possibilities might be considered to explain the observation?

Ans: (1) If the force is a deforming force, no acceleration is produced.

(2) Internal force is incapable of causing acceleration.

3 Marks Answers and Questions

Q1. In which of the following cases is centripetal force provided?

(i) Motion of planet around the sun

(ii) Motion of moon around the Earth

(iii) Motion of an electron around the nucleus in an atom

Ans: (i) The centripetal force is provided by the gravitational force acting on the Earth and the sun.

(ii) The centripetal force is provided by the Earth’s gravitational attraction to the moon.

(iii) The centripetal force is provided by the electrostatic attraction between the electron and the proton.

Q2. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass 10 g10 g floating on water,

(d) a car moving with a constant velocity of 30 km/h−130 km/h−1 on a rough road, and

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

Ans: (a) Zero net force

The raindrops are falling at a steady rate. As a result, the acceleration is zero. The net force acting on the raindrop is zero, according to Newton’s second law of motion.

(b) Zero net force

The cork’s weight is acting downward. The buoyant force exerted by the water in an upward direction balances it. As a result, there is no net force acting on the floating cork.

In the sky, the kite is stationary, i.e. it is not moving at all. As a result, according to Newton’s first law of motion, there is no net force acting on the kite.

(d) Zero net force

The car is going at a constant speed down a bumpy route. As a result, it has no acceleration. There is no net force operating on the car, according to Newton’s second law of motion.

(e) Zero net force

All fields have no effect on the high-speed electron. As a result, there is no net force acting on the electron.

Q3. A train runs along an unbanked circular bend of radius 30 m at a speed of 54 km/hr. The mass of the train is 106 kg. What provides the necessary centripetal force required for this purpose, the engine or the rails? What is the angle of banking required to prevent the rail from wearing out? Ans: From the question, we have the radius of a circular bend given as, r=30 m.

Speed of train = v = 54 km h−1 = 54×518 = 15 ms−1

Mass of train given as, m = 106 kg

Then we need to find the angle of banking θ.

(1) The centripetal force is generated by the lateral force exerted by rails on the train’s wheels.

(2) The centripetal force is provided by the lateral thrust of the outer rail.

(3) According to Newton’s third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail, causing its wear and tear.

Therefore, the angle of baking:

tanθ = v 2 rg = 15 2 13×19.8

4 Marks Answers and Questions

Q1. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg

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

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

(d) lying on the floor of a train which is accelerating with 1 m s−2, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans: (a) 1 N; vertically downward

From the question, we have the mass of the stone given as, m=0.1 kg

The acceleration of the stone is given as, a = g = 10 m/s2

The net force exerted on the stone, according to Newton’s second law of motion, is

F = ma = m g

= 0.1×10 = 1 N

Gravitational acceleration always works in the downward direction.

(b) 1 N; vertically downward

The train is travelling at a constant speed. As a result, its acceleration in the horizontal direction, where it is moving, is zero. As a result, there is no horizontal force acting on the stone.

The net force acting on the stone is due to gravity’s acceleration, and it is always vertically downward. This force has a magnitude of 1 N.

It is given that the train is accelerating at the rate of 1 m/s2.

Hence, the net force acting on the stone will be equal to, F′ = ma = 0.1×1 = 0.1 N

This force has a horizontal component to it. The horizontal force F′, no longer acts on the stone when it is dropped. This is due to the fact that the force acting on a body at any one time is determined by the current circumstances rather than previous ones.

As a result, the net force acting on the stone is determined only by gravity’s acceleration.

F = mg = 1 N

This force acts vertically downward.

(d) 0.1 N; in the direction of motion of the train

The typical reaction of the floor balances the weight of the stone. The train’s horizontal motion is the only source of acceleration.

Acceleration of the train, a = 0.1 m/s2

The net force acting on the stone will be directed in the train’s direction of travel. Its magnitude is given by:

= 0.1×1 = 0.1 N

5 Marks Answers and Questions

Q1. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 ms−2. The crew and the passengers each weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

Ans: (a) Mass of the helicopter is given as, mh= 1000 kg

Mass of the crew and passengers is given as, mp= 300 kg

Therefore, the total mass of the system, m = 1300 kg

As the acceleration of the helicopter is given as, a = 15 m/s2

The reaction force R exerted on the system by the floor may be computed using Newton’s second equation of motion.

= 300(10+15) = 300×25

The response force will likewise be directed upward because the helicopter is accelerating vertically.As a result, the force exerted on the floor by the crew and passengers is 7500 N, directed downward, according to Newton’s third law of motion.

(b) The reaction force R experienced by the helicopter may be computed using Newton’s second equation of motion as follows = 32500 N

The helicopter is being pushed higher by the reaction force of the surrounding air. As a result, the rotor’s action on the surrounding air will be 32500N, directed downward, according to Newton’s third law of motion.

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Q.1 Read the assertion and reason carefully to mark the correct option out of the options given below. Assertion: Motion of the rocket is not projectile motion. Reason: An object is in projectile motion only under the force of gravity. However, a rocket is acted upon by a propulsive force which acts against gravity to provide an upward lift to the rocket.

Assertion is true but reason is false., assertion and reason both are false., both assertion and reason are true and the reason is the correct explanation of the assertion., both assertion and reason are true but reason is not the correct explanation of the assertion..

When a body is thrown, shot or launched through the air with some initial velocity we call it as a projectile. A projectile moves due to a force which is applied to it initially without any propulsion. During motion the only force acted upon projectile is the force of gravity.

A rocket is not a projectile. In rocket propulsion, the pressure inside the rocket pushes the gas or liquid downward, produces equal and opposite reaction pushes the rocket upward.

Q.2 A lift of mass 1000 kg, which is moving with an acceleration of 1 m/s 2 in upward direction, has tension developed in its string equal to

T = m (g + a)

= 1000 (10 + 1)

Q.3 A box is lying on an inclined plane. If the box starts sliding when the angle of inclination is 60°, then the coefficient of static friction of the box and plane is

Q.4 A stone tied at the end of a string is whirled in a horizontal circle .When the string breaks, the stone flies away tangentially. Explain why.

When the stone moves in a circular path, its velocity is always tangential to the point of circle. When the string breaks, the force (centripetal) ceases to act on the stone. Hence, according to Newtons first law of motion, the stone flies away in the direction of its motion.

Q.5 A body of mass 10 kg initially at rest explodes and breaks into three fragments of masses in the ratio 1:1:3. The two equal pieces fly away in the direction perpendicular to each other. What will be the velocity of heaviest fragment?

m 1 + m 2 + m 3 = 10 kg m 1 : m 2 : m 3 = 1 : 1 : 3. m 1 = m 2 = 2 kg m 3 = 6 kg v 1 = v 2 = 30 m/s, v 3 = ?

According to law of conservation of momentum.

m 3 v 3 = m 1 v 1 cos 45 0 + m 2 v 2 cos 45 0

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Faqs (frequently asked questions), 1. who discovered the three laws of motion.

Isaac Newton, who was a brilliant mathematician and physicist, discovered the three laws of motion. The following are the three laws:

Objects at rest stay at rest, and objects in motion stay in motion unless acted upon by an external force.

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

For every action, there is an equal and opposite reaction.

2. Why are the laws of motion important?

Newton’s Laws of Motion form a very important part of physics. They explain almost all the types of movements that we experience in our daily lives. The laws pose an explanation for simple things like why we do not fall out of chairs when sitting, how a car moves, what role friction plays in a moving bus. All of these movements can be easily explained using Newton’s laws of motion.

3. How is centripetal force provided for an electron moving around the nucleus in an atom?

Centripetal force is the force required to make an object move in a circular path. In the case of an electron moving around the nucleus in an atom, the required centripetal force is provided by the electrostatic force of attraction between the electron and proton. Extramarks’ Class 11 Physics Chapter 5 Important Questions discusses many such important questions that are carefully selected by subject matter experts.

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NCERT Solutions for Class 11 Physics chapter-5 Laws of Motion

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NCERT Solutions for class-11 Physics Chapter 5 Laws of Motion is prepared by our senior and renowned teachers of Physics Wallah primary focus while solving these questions of class-11 in NCERT textbook, also do read theory of this Chapter 5 Laws of Motion while going before solving the NCERT questions. You can download NCERT solution of all chapters from Physics Wallah in PDF.

Chapter 5 Laws of Motion

Answer The Following Question Answer

Question 1. Give the magnitude and direction of the net force acting on a. a drop of rain falling down with a constant speed, b. a cork of mass 10 g floating on water, c. a kite skillfully held stationary in the sky, d. a car moving with a constant velocity of 30 km/h on a rough road, e. a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Solution : a. As the rain drop is flling with a constant speed, its accleration, a = 0. Hence net force F= ma = 0. b. As the cork is floating on water, its weight is balanced by the upthrust due to water. Therefore, the net force on the cork is 0. c. As the kite is held stationery, in accordance with the first law of motion, the net force on the kite is 0. d. Force is being applied to overcome the force of friction. But as velocity of the car is constant, its accleleration, a = 0. Hence net force on the car F = ma = 0. e. As the high speed electron in space is far away from all gravitating objects and free of electric and magnetic fields, the net force on electron is 0.

Question 2. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, a. during its upward motion, b. during its downward motion, c. at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.

Solution : 0.5 N, in vertically downward direction, in all cases Acceleration due to gravity, irrespective of the direction of motion of an object, always acts downward. The gravitational force is the only force that acts on the pebble in all three cases. Its magnitude is given by Newton’s second law of motion as: F = m × a Where, F = Net force m = Mass of the pebble = 0.05 kg a = g = 10 m/s 2 ∴F = 0.05 × 10 = 0.5 N The net force on the pebble in all three cases is 0.5 N and this force acts in the downward direction. If the pebble is thrown at an angle of 45° with the horizontal direction, 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.

Question 3. Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg, a. just after it is dropped from the window of a stationary train, b. just after it is dropped from the window of a train running at a constant velocity of 36 km/h, c. just after it is dropped from the window of a train accelerating with 1 m s -2 , d. lying on the floor of a train which is accelerating with 1 m s -2 , the stone being at rest relative to the train. Neglect air resistance throughout.

Solution : a. Here, m = 0.1 Kg, a = + g = 10 m/s 2 Net force, F = ma = 0.1 × 10 = 1.0 N This forcer acts vertically downwards. b. When the train is running at a constant velocity, its acceleration = 0, No force acts on the stone due to this motion. Therefore, force on the stone F = weight of stone = mg = 0.1 × 10 = 1.0 N This force also acts vertically downwards. c. When the train is accelerating with 1 m s -2 , an additional force F' = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction. But once the stone is dropped from the train, F' becomes zero and the net force on the stone is F = mg = 0.1 × 10 = 1.0 N, acting vertically downwards. d. As the stone is lying on the floor of the trin, its acceleration is same as that of the train. ∴ force acting on stone, F = ma = 0.1 × 1 = 0.1 NThis force is along the horizontal direction of motion of the train. Note that in each case, the weight of the stone is being balanced by the normal reaction.

Question 4. One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T - mv 2 / l, (iii) T + mv 2 / l, (iv) 0 T is the tension in the string. [Choose the correct alternative].

Solution : (i) T When a particle connected to a string revolves in a circular path around a centre, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e., F = T = mv 2 / l Where F is the net force acting on the particle.

Question 5. A constant retarding force of 50 N is applied to a body of mass 20 kg moving initially with a speed of 15 ms –1 . How long does the body take to stop?

Solution : Retarding force, F = –50 N Mass of the body, m = 20 kg Initial velocity of the body, u = 15 m/s Final velocity of the body, v = 0 Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as: F = ma –50 = 20 × a ∴ a = -50/20 = -2.5 ms -2 Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as: v = u + at ∴ t = -u / a = -15 / -2.5 = 6 s

Question 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?

Solution : Mass of the body, m = 3 kg Initial speed of the body, u = 2 m/s Final speed of the body, v = 3.5 m/s Time, t = 25 s Using the first equation of motion, the acceleration (a) produced in the body can be calculated as: v = u + at ∴ a = (v - u) / t = (3.5 - 2) / 25 = 0.06 ms -2 As per Newton’s second law of motion, force is given as: F = ma = 3 × 0.06 = 0.18 N Since 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.

Question 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.

Solution : Mass of the body, m = 5 kg The given situation can be represented as follows

The resultant of two forces is given as:

Question 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.

Solution : Initial speed of the three-wheeler, u = 36 km/h = 10 m/s Final speed of the three-wheeler, v = 0 m/s Time, t = 4 s Mass of the three-wheeler, m = 400 kg Mass of the driver, m' = 65 kg Total mass of the system, M = 400 + 65 = 465 kg Using the first law of motion, the acceleration (a) of the three-wheeler can be calculated as: v = u + at ∴ a = (v - u) / t = (0 - 10) / 4 = -2.5 ms -2 The negative sign indicates that the velocity of the three-wheeler is decreasing with time. Using Newton’s second law of motion, the net force acting on the three-wheeler can be calculated as: F = Ma = 465 × (–2.5) = –1162.5 N The negative sign indicates that the force is acting against the direction of motion of the three-wheeler.

Question 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.

Solution : Mass of the rocket, m = 20,000 kg Initial acceleration, a = 5 m/s 2 Acceleration due to gravity, g = 10 m/s 2 Using Newton’s second law of motion, the net force (thrust) acting on the rocket is given by the relation: F – mg = ma F = m (g + a) = 20000 × (10 + 5) = 20000 × 15 = 3 × 10 5 N

Question 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

Solution : Mass of the body, m = 0.40 kg Initial speed of the body, u = 10 m/s due north Force acting on the body, F = –8.0 N Acceleration produced in the body, a = F / m = -8.0 / 0.40 = -20 ms -2 (i) At t = –5 s Acceleration, a' = 0 and u = 10 m/s s = ut + (1/2) a' t 2 = 10 × (–5) = –50 m (ii) At t = 25 s Acceleration, a'' = –20 m/s 2 and u = 10 m/s s' = ut' + (1/2) a" t 2 = 10 × 25 + (1/2) × (-20) × (25) 2 = 250 - 6250 = -6000 m (iii) At t = 100 s For 0 ≤ t ≤ 30 s a = -20 ms -2 u = 10 m/s s 1 = ut + (1/2)a"t 2 = 10 × 30 + (1/2) × (-20) × (30) 2 = 300 - 9000 = -8700 m For 30 < t ≤ 100 s As per the first equation of motion, for t = 30 s, final velocity is given as: v = u + at = 10 + (–20) × 30 = –590 m/s Velocity of the body after 30 s = –590 m/s For motion between 30 s to 100 s, i.e., in 70 s: s 2 = vt + (1/2) a" t 2 = -590 × 70 = -41300 m ∴ Total distance, s" = s 1 + s 2 = -8700 -41300 = -50000 m = -50 km.

Question 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 = 11 s? (Neglect air resistance.)

v = (v x 2 + v y 2 ) 1/2 = (20 2 + 10 2 ) 1/2 = 22.36 m/s Let θ be the angle made by the resultant velocity with the horizontal component of velocity, v x ∴ tan θ = (v y / v x ) θ = tan -1 (10 / 20) = 26.57 0 (b) 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. Hence, the acceleration of the stone is 10 m/s 2 and it acts vertically downward.

Question 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.

Solution : (a) Vertically downward At the extreme position, the velocity of the bob becomes zero. If the string is cut at this moment, then the bob will fall vertically on the ground. (b) Parabolic path At the mean position, the velocity of the bob is 1 m/s. The direction of this velocity is tangential to the arc formed by the oscillating bob. If the bob is cut at the mean position, then it will trace a projectile path having the horizontal component of velocity only. Hence, it will follow a parabolic path.

Question 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 , (b) downwards with a uniform acceleration of 5 m s –2 , (c) upwards with a uniform acceleration of 5 m s –2 . What would be the readings on the scale in each case? (d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Solution : (a) Mass of the man, m = 70 kg Acceleration, a = 0 Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma Where, ma is the net force acting on the man. As the lift is moving at a uniform speed, acceleration a = 0 ∴ R = mg = 70 × 10 = 700 N ∴ Reading on the weighing scale = 700 / g = 700 / 10 = 70 kg (b) Mass of the man, m = 70 kg Acceleration, a = 5 m/s 2 downward Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = 70 (10 – 5) = 70 × 5 = 350 N ∴ Reading on the weighing scale = 350 g = 350 / 10 = 35 kg (c) Mass of the man, m = 70 kg Acceleration, a = 5 m/s 2 upward Using Newton’s second law of motion, we can write the equation of motion as: R – mg = ma R = m(g + a) = 70 (10 + 5) = 70 × 15 = 1050 N ∴ Reading on the weighing scale = 1050 / g = 1050 / 10 = 105 kg (d) When the lift moves freely under gravity, acceleration a = g Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = m(g – g) = 0 ∴ Reading on the weighing scale = 0 / g = 0 kg The man will be in a state of weightlessness.

Question 14. Figure 5.16 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).

Solution : (a) For t < 0 It can be observed from the given graph 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. Hence, the force acting on the particle is zero. For t > 4 s It can be observed from the given graph that the position of the particle is parallel to the time axis. It indicates that the particle is at rest at a distance of 3 m from the origin. Hence, no force is acting on the particle. For 0 < t < 4 It can be observed that the given position-time graph has a constant slope. Hence, the acceleration produced in the particle is zero. Therefore, the force acting on the particle is zero. (b) At t = 0 Impulse = Change in momentum = mv – mu Mass of the particle, m = 4 kg Initial velocity of the particle, u = 0 Final velocity of the particle, v = 3 / 4 m/s ∴ Impulse = 4 ( 3/4 - 0) = 3 kg m/s At t = 4 s Initial velocity of the particle, u = 3 / 4 m/s Final velocity o9f the particle, v = 0 ∴ Impulse = 4 (0 - 3/4) = -3 kg m/s

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

The equation of motion can be written as: F – T = m 2 a T = F – m 2 a ∴T = 600 – 20 × 20 = 200 N … (ii) which is different from value of T in case (i). Hence our answer depends on which mass end, the force is applied.

Question 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.

Smaller mass, m 1 = 8 kg Larger mass, m 2 = 12 kg Tension in the string = T Mass m 2 , owing to its weight, moves downward with acceleration a,and mass m 1 moves 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 … (i) For mass m 2 : The equation of motion can be written as: m 2 g – T = m 2 a … (ii) Adding equations (i) and (ii), we get: (m 2 - m 1 )g = (m 1 + m 2 )a ∴ a = ( (m 2 - m 1 ) / (m 1 + m 2 ) )g ....(iii) = (12 - 8) / (12 + 8) × 10 = 4 × 10 / 20 = 2 ms -2 Therefore, the acceleration of the masses is 2 m/s 2 . Substituting the value of a in equation (ii), we get: m 2 g - T = m 2 (m 2 - m 1 )g / (m 1 + m 2 ) T = (m 2 - (m 2 2 - m 1 m 2 ) / (m 1 + m 2 ) )g = 2m 1 m 2 g / (m 1 + m 2 ) = 2 × 12 × 8 × 10 / (12 + 8) = 96 N Therefore, the tension in the string is 96 N.

Question 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.

Solution : Let m, m 1 , and m 2 be 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 According to the law of conservation of momentum: Total initial momentum = Total final momentum 0 = m 1 v 1 + m 2 v 2 v 1 = -m 2 v 2 / m 1 Here, the negative sign indicates that the fragments of the parent nucleus move in directions opposite to each other.

Question 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?

Solution : Mass of each ball = 0.05 kg Initial velocity of each ball = 6 m/s Magnitude of the initial momentum of each ball, p i = 0.3 kg m/s After collision, the balls change their directions of motion without changing the magnitudes of their velocity. Final momentum of each ball, p f = –0.3 kg m/s Impulse imparted to each ball = Change in the momentum of the system = p f – p i = –0.3 – 0.3 = –0.6 kg m/s The negative sign indicates that the impulses imparted to the balls are opposite in direction.

Question 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?

Solution : Mass of the gun, M = 100 kg Mass of the shell, m = 0.020 kg Muzzle speed of the shell, v = 80 m/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. According to the law of conservation of momentum: Final momentum = Initial momentum mv – MV = 0 ∴ V = mv / M = 0.020 × 80 / (100 × 1000) = 0.016 m/s

Question 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.)

Where, AO = Incident path of the ball OB = Path followed by the ball after deflection ∠AOB = Angle between the incident and deflected paths of the ball = 45° ∠AOP = ∠BOP = 22.5° = θ Initial and final velocities of the ball = v Horizontal component of the initial velocity = vcos θ along RO Vertical component of the initial velocity = vsin θ along PO Horizontal component of the final velocity = vcos θ along OS Vertical component of the final velocity = vsin θ along OP The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions. ∴ Impulse imparted to the ball = Change in the linear momentum of the ball = mvCosθ - (-mvCosθ) = 2mvCosθ Mass of the ball, m = 0.15 kg Velocity of the ball, v = 54 km/h = 15 m/s ∴ Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s

Question 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?

Solution : Mass of the stone, m = 0.25 kg Radius of the circle, r = 1.5 m Number of revolution per second, n = 40 / 60 = 2 / 3 rps Angular velocity, ω = v / r = 2πn The centripetal force for the stone is provided by the tension T, in the string, i.e., T = F Centripetal = mv 2 / r = mrω = mr(2πn) 2 = 0.25 × 1.5 × (2 × 3.14 × (2/3) ) 2 = 6.57 N Maximum tension in the string, T max = 200 N T max = mv 2 max / r ∴ v max = (T max × r / m) 1/2 = (200 × 1.5 / 0.25) 1/2 = (1200) 1/2 = 34.64 m/s Therefore, the maximum speed of the stone is 34.64 m/s.

Question 22. 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 ?

Solution : b. the stone flies off tangentially from the instant the string breaks, 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.

Question 23. Explain why a. a horse cannot pull a cart and run in empty space, b. passengers are thrown forward from their seats when a speeding bus stops suddenly, c. it is easier to pull a lawn mower than to push it, d. a cricketer moves his hands backwards while holding a catch.

Solution : a. While trying to pull a cart, ahorse 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. Therefore, a horse cannot pull a cart and run in empty space. b. This is due to inertia of motion. When a speeding bus stops suddenly, the lower part of a passenger's body, which is in contact with the seat, suddenly comes to rest. However, the upper part tends to remain in motion (as per the first law of motion). As a result, the passenger's upper body is thrown forward in the direction in which the bus was moving. c. When the lawnmower is pulled, the vertical component of the applied force acts upwards. This reduces the effective weight of the lawnmower. When the lawn mower is pushed, the vertical component acts in the direction of the weight of the mower. Therefore, there is an increase in the weight of the mower. So, it is easier to pull a lawnmower than to push it. d. According to Newton’s second law of motion, we have the equation of motion: F = ma = m ∆v /∆t ...(i) Where, F = Stopping force experienced by the cricketer as he catches the ball m = Mass of the ball Δt = Time of impact of the ball with the hand It can be inferred from equation (i) that the impact force is inversely proportional to the impact time, i.e., F ∝1/ Δt ....(ii) Equation (ii) shows that the force experienced by the cricketer decreases if the time of impact increases and vice versa. While taking a catch, a cricketer moves his hand backward so as to increase the time of impact (Δt). This is turn results in the decrease in the stopping force, thereby preventing the hands of the cricketer from getting hurt.

Additional Excercises

Question 24. Figure 5.17 shows the position-time graph of a body of mass 0.04 kg. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Question 25.Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with 1 m s –2 . What is the net force on the man? If the coefficient of static friction between the man’s shoes and the belt is 0.2, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (Mass of the man = 65 kg.)

NCERT Solutions for Class 11 Physics Chapter 5 - Laws Of Motion

Solution : Mass of the man, m = 65 kg Acceleration of the belt, a = 1 m/s 2 Coefficient of static friction, μ = 0.2 The net force F, acting on the man is given by Newton’s second law of motion as: F net = ma = 65 × 1 = 65 N The man will continue to be stationary with respect to the conveyor belt until the net force on the man is less than or equal to the frictional force f s , exerted by the belt, i.e., F' net = f s ma' = μmg ∴ a' = 0.2 × 10 = 2 m/s 2 Therefore, the maximum acceleration of the belt up to which the man can stand stationary is 2 m/s 2 .

Question 26. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: [Choose the correct alternative]

T 1 and V 1 denote the tension and speed at the lowest point. T 2 and v 2 denote corresponding values at the highest point.

Using Newton’s second law of motion, we have: T + mg = mv 2 2 / R ...(ii) Where, v 2 = Velocity at the highest point It is clear from equations (i) and (ii) that the net force acting at the lowest and the highest points are respectively (T – mg) and (T + mg).

Question 27. A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s –2 . The crew and the passengers weigh 300 kg. Give the magnitude and direction of the a. force on the floor by the crew and passengers, b. action of the rotor of the helicopter on the surrounding air, c. force on the helicopter due to the surrounding air.

Solution : a. Mass of the helicopter, m h = 1000 kg Mass of the crew and passengers, m p = 300 kg Total mass of the system, m = 1300 kg Acceleration of the helicopter, a = 15 m/s 2 Using Newton’s second law of motion, the reaction force R, on the system by the floor can be calculated as: R – m p g = ma = m p (g + a) = 300 (10 + 15) = 300 × 25 = 7500 N Since the helicopter is accelerating vertically upward, the reaction force will also be directed upward. Therefore, as per Newton’s third law of motion, the force on the floor by the crew and passengers is 7500 N, directed downward. b. Using Newton’s second law of motion, the reaction force R’, experienced by the helicopter can be calculated as: R' - mg = ma = m(g + a) = 1300 (10 + 15) = 1300 × 25 = 32500 N The reaction force experienced by the helicopter from the surrounding air is acting upward. Hence, as per Newton’s third law of motion, the action of the rotor on the surrounding air will be 32500 N, directed downward. c. The force on the helicopter due to the surrounding air is 32500 N, directed upward.

Question 28. A stream of water flowing horizontally with a speed of 15 m s –1 gushes out of a tube of cross-sectional area 10 –2 m 2 , and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Solution : Speed of the water stream, v = 15 m/s Cross-sectional area of the tube, A = 10 –2 m 2 Volume of water coming out from the pipe per second, V = Av = 15 × 10 –2 m 3 /s Density of water, ρ = 10 3 kg/m 3 Mass of water flowing out through the pipe per second = ρ × V = 150 kg/s The water strikes the wall and does not rebound. Therefore, the force exerted by the water on the wall is given by Newton’s second law of motion as: F = Rate of change of momentum = ∆P / ∆t = mv / t = 150 × 15 = 2250 N

Question 29. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of a. the force on the 7 th coin (counted from the bottom) due to all the coins on its top, b. the force on the 7 th coin by the eighth coin, c. the reaction of the 6th coin on the 7 th coin.

Solution : a. Force on the seventh coin is exerted by the weight of the three coins on its top. Weight of one coin = mg Weight of three coins = 3mg Hence, the force exerted on the 7 th coin by the three coins on its top is 3mg. This force acts vertically downward. b. Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top. Weight of the eighth coin = mg Weight of the ninth coin = mg Weight of the tenth coin = mg Total weight of these three coins = 3mg Hence, the force exerted on the 7 th coin by the eighth coin is 3mg. This force acts vertically downward. c. The 6 th coin experiences a downward force because of the weight of the four coins (7 th , 8 th , 9 th , and 10 th ) on its top. Therefore, the total downward force experienced by the 6 th coin is 4mg. As per Newton’s third law of motion, the 6 th coin will produce an equal reaction force on the 7 th coin, but in the opposite direction. Hence, the reaction force of the 6 th coin on the 7 th coin is of magnitude 4mg. This force acts in the upward direction.

Question 30. An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?

Solution : Speed of the aircraft, v = 720 km/h = 720 × 5 / 18 = 200 m/s Acceleration due to gravity, g = 10 m/s 2 Angle of banking, θ = 15° For radius r, of the loop, we have the relation: tan θ = v 2 / rg r = v 2 / g tan θ = 200 2 / (10 × tan 15) = 4000 / 0.26 = 14925.37 m = 14.92 km

Question 31. A train runs along an unbanked circular track of radius 30 m at a speed of 54 km/h. The mass of the train is 10 6 kg. What provides the centripetal force required for this purpose – The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Solution : Radius of the circular track, r = 30 m Speed of the train, v = 54 km/h = 15 m/s Mass of the train, m = 10 6 kg The centripetal force is provided by the lateral thrust of the rail on the wheel. As per Newton’s third law of motion, the wheel exerts an equal and opposite force on the rail. This reaction force is responsible for the wear and rear of the rail The angle of banking θ, is related to the radius (r) and speed (v) by the relation: tan θ = v 2 / rg = 15 2 / (30 × 10) = 225 / 300 θ = tan -1 (0.75) = 36.87 0 Therefore, the angle of banking is about 36.87°.

Question 32. A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding?

750N and 250 N in the respective cases; Method (b) Mass of the block, m = 25 kg Mass of the man, M = 50 kg Acceleration due to gravity, g = 10 m/s 2 Force applied on the block, F = 25 × 10 = 250 N Weight of the man, W = 50 × 10 = 500 N Case (a): When the man lifts the block directly In this case, the man applies a force in the upward direction. This increases his apparent weight. ∴Action on the floor by the man = 250 + 500 = 750 N Case (b): When the man lifts the block using a pulley In this case, the man applies a force in the downward direction. This decreases his apparent weight. ∴Action on the floor by the man = 500 – 250 = 250 N If the floor can yield to a normal force of 700 N, then the man should adopt the second method to easily lift the block by applying lesser force.

Question 33. A monkey of mass 40 kg climbs on a rope (Fig. 5.20) which can stand a maximum tension of 600 N. In which of the following cases will the rope break: the monkey a. climbs up with an acceleration of 6 m s –2 b. climbs down with an acceleration of 4 m s –2 c. climbs up with a uniform speed of 5 m s –1 d. falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

Solution : Case (a) Mass of the monkey, m = 40 kg Acceleration due to gravity, g = 10 m/s Maximum tension that the rope can bear, T max = 600 N Acceleration of the monkey, a = 6 m/s 2 upward Using Newton’s second law of motion, we can write the equation of motion as: T – mg = ma ∴T = m(g + a) = 40 (10 + 6) = 640 N Since T > T max , the rope will break in this case. Case (b) Acceleration of the monkey, a = 4 m/s 2 downward Using Newton’s second law of motion, we can write the equation of motion as: mg – T = ma ∴T = m (g – a) = 40(10 – 4) = 240 N Since T < T max , the rope will not break in this case. Case (c) The monkey is climbing with a uniform speed of 5 m/s. Therefore, its acceleration is zero, i.e., a = 0. Using Newton’s second law of motion, we can write the equation of motion as: T – mg = ma T – mg = 0 ∴T = mg = 40 × 10 = 400 N Since T < T max , the rope will not break in this case. Case (d) When the monkey falls freely under gravity, its will acceleration become equal to the acceleration due to gravity, i.e., a = g Using Newton’s second law of motion, we can write the equation of motion as: mg – T = mg ∴T = m(g – g) = 0 Since T < T max , the rope will not break in this case.

Question 34. Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are

a. the reaction of the partition

b. the action-reaction forces between A and B? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ s and μ k .

Solution : a. Mass of body A, m A = 5 kg Mass of body B, m B = 10 kg Applied force, F = 200 N Coefficient of friction, μ s = 0.15 The force of friction is given by the relation: f s = μ (m A + m B )g = 0.15 (5 + 10) × 10 = 1.5 × 15 = 22.5 N leftward Net force acting on the partition = 200 – 22.5 = 177.5 N rightward As per Newton’s third law of motion, the reaction force of the partition will be in the direction opposite to the net applied force. Hence, the reaction of the partition will be 177.5 N, in the leftward direction. b. Force of friction on mass A: f A = μm A g = 0.15 × 5 × 10 = 7.5 N leftward Net force exerted by mass A on mass B = 200 – 7.5 = 192.5 N rightward As per Newton’s third law of motion, an equal amount of reaction force will be exerted by mass B on mass A, i.e., 192.5 N acting leftward. When the wall is removed, the two bodies will move in the direction of the applied force. Net force acting on the moving system = 177.5 N The equation of motion for the system of acceleration a,can be written as: Net force = (m A + m B ) a ∴ a = Net force / (m A + m B ) = 177.5 / (5 + 10) = 177.5 / 15 = 11.83 ms -2 Net force causing mass A to move: F A = m A a = 5 × 11.83 = 59.15 N Net force exerted by mass A on mass B = 192.5 – 59.15 = 133.35 N This force will act in the direction of motion. As per Newton’s third law of motion, an equal amount of force will be exerted by mass B on mass A, i.e., 133.3 N, acting opposite to the direction of motion.

Question 35. A block of mass 15 kg is placed on a long trolley. The coefficient of static friction between the block and the trolley is 0.18. The trolley accelerates from rest with 0.5 m s –2 for 20 s and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

Solution : a. Mass of the block, m = 15 kg Coefficient of static friction, μ = 0.18 Acceleration of the trolley, a = 0.5 m/s 2 As per Newton’s second law of motion, the force (F) on the block caused by the motion of the trolley is given by the relation: F = ma = 15 × 0.5 = 7.5 N This force is acted in the direction of motion of the trolley. Force of static friction between the block and the trolley: f = μmg = 0.18 × 15 × 10 = 27 N The force of static friction between the block and the trolley is greater than the applied external force. Hence, for an observer on the ground, the block will appear to be at rest. When the trolley moves with uniform velocity there will be no applied external force. Only the force of friction will act on the block in this situation. b. An observer, moving with the trolley, has some acceleration. This is the case of non-inertial frame of reference. The frictional force, acting on the trolley backward, is opposed by a pseudo force of the same magnitude. However, this force acts in the opposite direction. Thus, the trolley will appear to be at rest for the observer moving with the trolley.

Question 36. The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 m s –2 . At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Solution : Mass of the box, m = 40 kg Coefficient of friction, μ = 0.15 Initial velocity, u = 0 Acceleration, a = 2 m/s 2 Distance of the box from the end of the truck, s' = 5 m As per Newton’s second law of motion, the force on the box caused by the accelerated motion of the truck is given by: F = ma = 40 × 2 = 80 N As per Newton’s third law of motion, a reaction force of 80 N is acting on the box in the backward direction. The backward motion of the box is opposed by the force of friction f, acting between the box and the floor of the truck. This force is given by: f = μmg = 0.15 × 40 × 10 = 60 N ∴Net force acting on the block: F net = 80 – 60 = 20 N backward The backward acceleration produced in the box is given by: a back = F net / m = 20 / 40 = 0.5 ms -2 Using the second equation of motion, time t can be calculated as: s' = ut + (1/2)a back t 2 5 = 0 + (1/2) × 0.5 × t 2 ∴ t = √20 s Hence, the box will fall from the truck after √ 20 s from start. The distance s, travelled by the truck in √ 20 s is given by the relation: s = ut + (1/2)at 2 = 0 + (1/2) × 2 × (√20) 2 = 20 m

Question 37. A disc revolves with a speed of 100 / 3 rev / min, and has a radius of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If the co-efficient of friction between the coins and the record is 0.15, which of the coins will revolve with the record?

Solution : Coin placed at 4 cm from the centre Mass of each coin = m Radius of the disc, r = 15 cm = 0.15 m Frequency of revolution, ν = 100 / 3 rev/min = 100 / (3 × 60) = 5 / 9 rev/s Coefficient of friction, μ = 0.15 In the given situation, the coin having a force of friction greater than or equal to the centripetal force provided by the rotation of the disc will revolve with the disc. If this is not the case, then the coin will slip from the disc. Coin placed at 4 cm: Radius of revolution, r' = 4 cm = 0.04 m Angular frequency, ω = 2πν = 2 × (22/7) × (5/9) = 3.49 s -1 Frictional force, f = μmg = 0.15 × m × 10 = 1.5m N Centripetal force on the coin: F cent. = mr'ω 2 = m × 0.04 × (3.49) 2 = 0.49m N Since f > F cent , the coin will revolve along with the record. Coin placed at 14 cm: Radius, r" = 14 cm = 0.14 m Angular frequency, ω = 3.49 s –1 Frictional force, f' = 1.5m N Centripetal force is given as: F cent. = mr"ω 2 = m × 0.14 × (3.49) 2 = 1.7m N Since f < F cent. , the coin will slip from the surface of the record.

Question 38. You may have seen in a circus a motorcyclist driving in vertical loops inside a ‘death-well’ (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is 25 m?

Solution : In a death-well, a motorcyclist does not fall at the top point of a vertical loop because both the force of normal reaction and the weight of the motorcyclist act downward and are balanced by the centripetal force. This situation is shown in the following figure.

Question 39. A 70 kg man stands in contact against the inner wall of a hollow cylindrical drum of radius 3 m rotating about its vertical axis with 200 rev/min. The coefficient of friction between the wall and his clothing is 0.15. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Solution : Mass of the man, m = 70 kg Radius of the drum, r = 3 m Coefficient of friction, μ = 0.15 Frequency of rotation, ν = 200 rev/min = 200 / 60 = 10 / 3 rev/s The necessary centripetal force required for the rotation of the man is provided by the normal force (F N ). When the floor revolves, the man sticks to the wall of the drum. Hence, the weight of the man (mg) acting downward is balanced by the frictional force (f = μF N ) acting upward. Hence, the man will not fall until: mg < f mg < μF N = μmrω 2 g < μrω 2 ω > (g / μr) 1/2 The minimum angular speed is given as: ω min = (g / μr) 1/2 = ( 10 / (0.15 X 3) ) 1/2 = 4.71 rad s -1

Question 40. A thin circular loop of radius R rotates about its vertical diameter with an angular frequency ω. Show that a small bead on the wire loop remains at its lowermost point for ω ≤ √g/R. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for ω = √2g/R? Neglect friction.

Solution : Let the radius vector joining the bead with the centre make an angle θ, with the vertical downward direction.

OP = R = Radius of the circle N = Normal reaction The respective vertical and horizontal equations of forces can be written as: mg = N Cosθ ... (i) mlω 2 = N Sinθ … (ii) In ΔOPQ, we have: Sin θ = l / R l = R Sinθ … (iii) Substituting equation (iii) in equation (ii), we get: m(R Sinθ) ω 2 = N Sinθ mR ω 2 = N ... (iv) Substituting equation (iv) in equation (i), we get: mg = mR ω 2 Cosθ Cosθ = g / Rω 2 ...(v) Since cosθ ≤ 1, the bead will remain at its lowermost point for g / Rω 2 ≤ 1, i.e., for ω ≤ (g / R) 1/2 For ω = (2g / R) 1/2 or ω 2 = 2g / R .....(vi) On equating equations (v) and (vi), we get: 2g / R = g / RCos θ Cos θ = 1 / 2 ∴ θ = Cos -1 (0.5) = 60 0

NCERT Solutions For Class-11 Physics Chapter Wise

Chapter 1 Physical World

Chapter 2 Units and Measurements

Chapter 3 Motion In A Straight Line

Chapter 4 Motion In A Plane

Chapter 6 Work, Energy and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solid

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

Related Chapters

chapter-1 Physical World
chapter-2 Units And Measurements
chapter-3 Motion In A Straight Line
chapter-4 Motion In A Plane
chapter-5 Laws of Motion
chapter-6 Work, Energy And Power
chapter-7 System of Particles and Rotational Motion
chapter-8 Gravitation
chapter-9 Mechanical Properties Of Solids
chapter-10 Mechanical Properties of Fluids
chapter-11 Thermal Properties Of Matter
chapter-12 Thermodynamics
chapter-13 Kinetic Theory
chapter-14 Oscillations
chapter-15 Waves

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Class 11 Physics Chapter 5 Important Questions Laws of Motion

It is important for the students that all the concepts should be very clear for better marks in future. Here, we are providing important conceptual questions and answers for class 11 physics chapter 5 Laws of Motion. In this lesson, students will learn about Laws of Motion . This will not only help the students to know the important questions but will also help them during revision.

Conceptual Questions for Class 11 Physics Chapter 5 Laws of Motion

Q.1. Is force needed to keep a body moving with uniform velocity? Ans. No.

Q.2. What is inertia? What gives the measure of inertia? Ans. The inherent property of the bodies that they do not change their state unless acted upon by an external force is called inertia. Mass of the body give the measure of its inertia.

Q.3. How is inertia related to mass of a body? Ans. More is the mass of a body, more is its inertia.

Q.4. If you jerk a piece of paper under a book quick enough the book will not move. Why? Ans. The book remains there as much due to the inertia of rest.

Q.5. Define the term momentum. Give its SI unit. Ans. Momentum of a body is defined as the total quantity of motion contained in a body and is measured as the product of the mass of body and its velocity. The SI unit of momentum is kg ms -1 .

Q.6. A bullet fired from a rifle is more dangerous than an air molecule hitting a person, though both of them have almost the same speed. Explain. Ans. It is because mass of the bullet is about 1025 times that of the air molecule. As such the momentum of bullet is very large.

Q.7. From which Newton's law of motion the definition of force comes? Ans. First law of motion.

Q.8. Is Newton's second law (F = m.a) always valid. Give an example in support of your answer. Ans. It is valid in an inertial frame of reference. In a non inertial frame of reference (such as a car moving along a circular path) Newton's second law does not hold apparently.

Q.9. the rate of change of momentum of a body is 5 kg m s -1 . What is the force acting on the body? Ans. 5 N.

Read also: Laws of Motion Class 11 Physics Notes Chapter 5

Q.10. Give and state SI unit of force. Ans. The SI unit of force is Newton(N). Force is said to be 1 N, if it produces an acceleration of 1m s -2 in a body of mass 1 kg.

Q.11. What is the difference between mN and nm? Ans. The symbol mN stands for millinewton (10 -3 N) and nm stands for nanometre (10 -9 m).

Q.12. Action and reaction forces do not balance each other. Why? Ans. When one body exerts a force on another body, it gets an equal and opposite reaction. The action and reaction forces do not cancel each other, as they do not act on the same body.

Q.13. What is an impulsive force? Ans. A force which acts for a small time and also varies with time is called an impulsive force.

Q.14. What is the net force on a cork floating on water? Ans. Zero.

Q.15. Which is greater -- the attraction of 10 kg mass for earth or the earth's attraction for 10 kg mass?

Ans. Both are equal.

Q.16. A body is moving along a circular path such that its speed always remains constant. Should there be a force acting on the body? Ans. Yes, the force is needed to change the direction of motion of the body i.e. to deflect the body from its straight path to circular path.

Q.17. State principle of conservation of momentum. Ans. The principle of conservation of momentum states that if no external force acts on a system the momentum of the system remains constant.

Read also: Class 11 Physics Chapter 5 MCQs with Answer Laws of Motion

Q.18. State Newton's first law of motion. Ans. Newton's first law of motion states that every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled by some external force to change that state.

Q.19. State Newton's second law of motion. Ans. Newton's second law of motion states that the time rate of change of momentum of a body is directly proportional to the external force applied on it and the change in momentum takes place in the direction of force.

Q.20. State Newton's third law of motion. Ans. Newton's third law of motion states that to every action there is an equal and opposite reaction.

Q.21. What is an impulse? Ans. Impulse received during an impact is defined as the product of the average force and the time for which the force acts.

Q.22. Can a body remain in state of rest when external forces acting on it. Explain your answer. Ans. Yes, if the external forces acting on the body can be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

Q.23. If the net force acting on a body be zero, will the body remain necessarily in rest position? Explain your answer. Ans. In case the body is in uniform motion along a straight line, it will continue to do so, if the net force acting on a body be zero.

Q.24. Name physical situation, where the mass of a body changes with time. Ans. When the body moves with a speed comparable to the speed of light. For example, the mass of a charged particle increases, when it is accelerated in a cyclotron.

Q.25. When are the two bodies said to possess equal masses? Ans. If the same force acting on the two bodies produces the same acceleration, then the two bodies must be of equal masses.

Q.26. Is a bus moving along a circular track, an inertial frame of reference? Ans. No. It is a non-inertial frame of reference.

Q.27. What can be said about the motion of the vehicle, if a plumb line hanging from its roof drops vertically? Ans. Either the vehicle is at rest or it is moving with constant velocity.

Q.28. The assertion made by the Newton's first law of motion that everybody continues in a state of uniform motion in the absence of external force appears to be contradicted in everyday experience. Why? Ans. When we roll a ball on the floor, it ultimately stops because of the force of friction of the ground. Thus, the state of uniform motion of the object changes due to the external force (friction). On the earth, every change in uniform motion of a body can be connected with an external force acting on it. However, in free space, when no external force acts, the state of motion described by the Newton's first law of motion can be obtained and experienced.

Q.29. According to Newton's first law of motion, a body moving with a uniform speed along a straight line should continue moving. In practice a body in motion stops after some time. Explain the reason. Ans. The force of friction acts as the external retarding force. It brings the body to rest.

Q.30. An astronaut accidentally gets thrown out of his small space ship accelerating in interstellar space at a constant rate of 100 ms -2 . What is the acceleration of the astronaut, the instant after he is outside the spaceship? Assume that there are no nearby stars to exert gravitational force on him. Ans. In the interstellar space, moment the astronaut is out of the spaceship, no external force acts on him. It is because, there are no nearby stars to exert gravitational force on him. The spaceship too can not exert any appreciable gravitational force on him due to its small size. Therefore, once the astronaut is out of the spaceship, his acceleration will be zero.

Q.31. According to Newton's third law, every force is accompanied by an equal and opposite force. How can a movement ever take place? Ans. Since action and reaction do not act on the same body, they do not cancel each other. Therefore, a body may move either under the effect of the action of force or the other body may move under the effect of reaction on it.

Q.32. It is easier to pull than to push a body. Explain. Ans. When we pull a body, the vertical component of the applied pull acts opposite to the weight of the body and it reduces its effective weight. On the other hand, when a body is pushed, the vertical component of the applied push adds to the weight of the body and hence its effective weight increases. Thus, the effective weight is lesser, when the body is pulled. Hence, it is easier to pull than to push a body.

Q.33. A stone when thrown on a glass window smashes the window pane into pieces, but a bullet from the gun passes through making a clean hole. Why? Ans. The velocity of the stone is much less than that of the bullet fired from a gun. Due to its low speed, the stone remains in contact contact with the window pane for a longer time and its motion is shared by whole of the window pane. As a result, it smashes into pieces. On the other hand, the bullet fired from the gun remains in contact with the window pane for such a small time that it can share its motion only with the portion of the window pane, it comes in contact with. As such, it makes a clean hole in the window pane.

Q.34. What is friction? Ans. The opposing force that is set up between the surfaces of contact, when one body slides or rolls or tends to do so on the surface of another body is called friction.

Q.35. Define limiting friction. Ans. The maximum value of the force of friction which comes into play before a body just begins to slide over the surface of another body is called limiting value of static friction.

Q.36. Why do we call friction a self adjusting force? Ans. When applied force is zero, friction is zero. As the applied force is increased, friction also increases and becomes equal to the applied force. It happens so, till the body does not start moving. That is why, friction is called a self adjusting force.

Q.37. What is the unit of coefficient of limiting friction? Ans. It has no unit.

Q.38. What are the factors on which the coefficient of friction between two surfaces depend? Ans. The coefficient of friction between two surfaces depends upon the nature of the two surfaces and their state of roughness.

Q.39. What is the relation between angle of repose and angle of friction? Ans. Angle of friction and angle of repose are equal.

Q.40. Why is friction a non-conservative force? Ans. It is because work done against friction along a closed path is non zero.

Q.41. What happens to limiting friction, when a wooden block is moved with increasing speed on a horizontal surface? Ans. The limiting friction decreases as the wooden block is moved with increasing speed on the horizontal surface.

Q.42. Why are tyres made of rubber and not iron? Ans. It is because, the coefficient of friction between rubber and concrete (material of the road) is less than that between iron and the road.

Q.43. Give any three advantages of friction. Ans.

Friction helps us to walk.
Brakes make use of friction to stop the vehicles.
Friction helps to transmit power from the motors and engines to other machines by making use of belts and clutches.

Q.44. Why are wheels made circular? Ans. The rolling friction is less than the sliding friction. The wheels are made circular so as to convert the sliding friction into the rolling friction.

Q.45. It is easier to roll a barrel than to slide it along the road. Why? Ans. The rolling friction is lesser as compared to the sliding friction.

Q.46. What happens to the fluid friction, as speed of the object moving through it is increased? Ans. The fluid friction increases, as the speed of the object moving through it, is increased.

Q.47. smoother the surface lesser is friction. Comment. Ans. When the surfaces are made smoother, the size of the irregularities in the surfaces decreases. As a result, the area of actual contact decreases. As the number of atoms in contact will also decrease due to decrease in area of contact, the force of molecular attraction and hence the force of friction decreases.

Q.48. Polishing a surface beyond a certain limit may increase friction. Why? Ans. When the surfaces are polished beyond a certain limit, the area at each point of contact becomes very small. However, the actual area of contact between the two surfaces increases appreciably. It is because, the number of points of contact becomes very large on making the surface is highly polished. Since number of atoms (or molecules) of the two surfaces in contact is proportional to their area in contact, the force of friction increases due to the greater force of molecular attraction between the two surfaces.

Q.49. A horse has to apply more force to start a cart than to keep it moving. Why? Ans. When the cart is at rest, the friction between the wheels of the cart and the road is static in nature and once the cart starts moving, it is kinetic in nature. The kinetic friction is always less than the static friction. Due to this reason, a horse has to apply more force to start a cart than to keep it moving.

Q.50. Why do we slip on a rainy day? Ans. On a rainy day, the wet ground becomes very smooth. As a result, the coefficient of friction between our feet and the wet ground gets much reduced. Consequently, the friction between the feet and the ground becomes very small. It may cause us to slip.

Q.51. Explain how friction helps in walking? Ans. When we walk, with push the ground in backward direction. Our foot will get equal and opposite reaction from the ground only, if the food does not slip i.e. there is adequate friction between the food and the ground. The forward horizontal component of the reaction helps us to move forward.

Q.52. How can proper inflation of tyres save fuel? Ans. Under the weight of the vehicle, the road surface gets depressed and a small hump is created just ahead of the wheel. In case the tyres are not properly inflated, the tyres require a large effort to climb the hump. In case the tires are properly inflated, the compression of the tyres as much smaller and hence they experience less resistance (rolling friction). In other words, it will make the vehicle to cover greater distance for the same quantity of fuel spent.

Q.53. What furnishes centripetal force for earth to go round the sun? Ans. The gravitational pull of the sun on the earth.

Q.54. What provides the centripetal force to satellite revolving round the earth? Ans. Gravitational force of attraction on the satellite due to the earth.

Q.55. What furnishes the centripetal force for the electrons to go round the nucleus? Ans. The electrostatic force of attraction on the electrons due to the nucleus.

Q.56. For uniform circular motion, does the direction of the centripetal force depends upon the sense of rotation? Ans. The direction of the centripetal force does not depend, whether the body is moving in clockwise or anticlockwise direction. It is always directed along the radius and towards the centre of the circle.

Q.57. Can centripetal force produce rotation? Ans. No. A centripetal force can move a body along a circular path but it can not produce rotational motion.

Q.58. What provides the centripetal force to a car taking a turn on a level road? Ans. The force of friction between the tyres and the road provides the necessary centripetal force to the car to take a turn on a level road.

Q.59. The speed of the stone is increased beyond the maximum permissible value (the value of speed at which the tension in the string becomes equal to the maximum tension the string can stand) the string breaks suddenly. What will be the trajectory of the stone after the string breaks? Ans. At the instant, the string breaks, the stone will fly off tangentially from its position at that instant. It is because, the speed of the stone at any instant is directed along tangent to the circular path at that point.

Q.60. A train moves on an unbanked circular band of rails. Which rail will wear out faster? Ans. Inner rail. It is because, the inward pressure on the inner rail is more than that on the outer rail.

Q.61. What are the advantages of banking? Ans. The vehicle can be moved along a circular track at a reasonable speed without the fear of skidding. Further, while taking a turn, one may not need to decrease the speed of the vehicle.

Q.62. A stone is moved along a vertical circle so as to just loop the loop. What is the tension in the string, when the stone is at the highest point? Ans. The tension is equal to six times the weight of the stone.

Laws of Motion Class 11 Physics Notes Chapter 5
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CBSE Important Questions for Class 11 Physics

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Case Study Questions for Class 11 Physics Chapter 5 Laws of Motion
Check Answer. Answer: (c) (ii) A bullet of mass 10 g is fired from a gun of mass 1 kg with recoil velocity of gun 5 m/s. The muzzle velocity will be. Check Answer. Answer: (d) (iii) A bullet of mass 0.1 kg is fired with a speed of 100 m/s. The mass of gun being 50 kg, then the velocity of recoil becomes. Check Answer.
Class 11 Physics Case Study Questions Chapter 5 Laws of Motion
Laws of Motion Case Study Questions With Answers. Here, we have provided case-based/passage-based questions for Class 11 Physics Chapter 5 Laws of Motion. Case Study/Passage-Based Questions. Case Study 1: The first law refers to the simple case when the net external force on a body is zero. The second law of motion refers to the general ...
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Chapter-wise Solved Case Study Questions for Class 11 Physics. Chapter 1: Physical World. Chapter 2: Units and Measurements. Chapter 3: Motion in a Straight Line. Chapter 4: Motion in a Plane. Chapter 5: Laws of Motion. Chapter 6: Work, Energy, and Power. Chapter 7: System of Particles and Rotational Motion.
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5) Explain why book on table remains at rest. Answer Key -. 1) b. 2) a. 3) the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity is called inertia. 4) Newton's first law of motion states that If the net external force on a body is zero, its acceleration is zero.
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Answer: (a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms -2 = 0.5 N. (b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards). (c) The pebble is not at rest at highest point but has horizontal component of velocity.
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Calculate the initial thrust (force) of the blast. Solution : Mass of the rocket, m = 20,000 kg. Initial acceleration, a = 5 m/s 2. Acceleration due to gravity, g = 10 m/s 2. Using Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation: F - mg = ma.
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Read also: Class 11 Physics Chapter 5 MCQs with Answer Laws of Motion. Q.18. State Newton's first law of motion. Ans. Newton's first law of motion states that every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled by some external force to change that state. Q.19.
CBSE Important Questions for Class 11 Physics
Students can click the link below to access the chapter-wise important questions of CBSE Class 11 Physics. Important Questions Chapter 1 - Units and Measurements. Important Questions Chapter 2 - Motion in a Straight Line. Important Questions Chapter 3 - Motion in a Plane. Important Questions Chapter 4 - Laws of Motion.
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Laws of Motion Class 11 Notes Physics - Basic Subjective Questions

Section-B (2 Marks Questions)

Lami’s Theorem:

Basic Forces:

Newton's Laws of Motion:

Newton's Second Laws:

Newtons Third Laws:

Linear Momentum:

Apparent Weight on a Body in a Lift

Principle of Conservation of Linear Momentum-

Types of Friction:

Laws of Limiting Friction:

Coefficient of Static Friction:

Angle of Friction:

Angle of Repose or Angle of Sliding:

Methods of Changing Friction:

Dynamics of Uniform Circular Motion Concept of Centripetal Force:

Centrifugal Force:

Rounding A-Level Curved Road:

Banking of Roads:

Discussion:

Bending of a Cyclist:

Pseudo Force:

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