# NCERT Solutions for Class 11 Physics Chapter 6

With easy-to-understand examples, the Class 11 Physics Chapter 6 describes the terms ‘Energy,’ ‘Power,’ and ‘Work.’ An everyday understanding of work such as a farmer ploughing a field or a construction worker having to carry bricks is somewhere related to Physics. This chapter teaches students that the term “work” in physics has a specific and defined meaning.

## NCERT Solutions for Class 11 Physics Chapter 6 – Work, Energy and Power

Class 12 is a critical time, and the year goes by so quickly that you will have little time to prepare thoroughly. As a result, it is best to begin the flow in Class 11. Also, if you want to have a successful career in the future, you must have a thorough understanding of the subjects you study in Class 11. To begin with, NCERT Solutions Class 11 is a fantastic start to the entire preparation process.

Chapter 6 of Physics for Class 11 will teach students that energy is defined as the capacity to perform work. The term ‘energy’ used in physics corresponds to work. In daily life, the word ‘power’ has a variety of connotations. This chapter’s goal is to help students understand the three physical quantities of work, energy, and power. Students can access NCERT Solutions Class 11 Physics Chapter 6 to grasp these concepts with ease.

Access NCERT Solutions for Class 11 Physics Chapter 6 – Work Energy and Power

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### NCERT Solutions of Class 11 Physics Chapter Work Energy and Power

Before starting your CBSE Class 11 Physics prep, familiarise yourself with the topics and subtopics of NCERT Class 11 Physics Chapter 6. Knowing the topics and sub-topics in this chapter will aid you in preparing for the exam properly.

The first two subtopics define the scalar product as well as work and kinetic energy. The first section of Work, Energy, and Power also includes the work-energy theorem, the concept of Kinetic Energy, Potential Energy, Mechanical Energy Conservation, and much more. Find the detailed sub-topic description below:

 Section Number Section Title 6.1 Introduction 6.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem 6.3 Work 6.4 Kinetic Energy 6.5 Work Done by a Variable Force 6.6 The Work-Energy Theorem for a Variable Force 6.7 The Concept of Potential Energy 6.8 The Conservation of Mechanical Energy 6.9 The Potential Energy of a Spring 6.10 Various Forms of Energy: The Law of Conservation of Energy 6.11 Power 6.12 Collisions

6.1 Introduction

The term “work” is used in Physics to describe the process of transferring energy from one object or system to another, or of converting energy from one form to another.

6.2 Notions of Work and Kinetic Energy: The Work-Energy Theorem

The term “work” is frequently used in everyday life, and we understand that it refers to the act of doing something. For example, by reading this article, you are improving your understanding of Physics right now! Physics, on the other hand, may disagree. The Work-energy Theorem explains why this Physics of No Work exists! The concepts of kinetic energy, work, and kinetic energy are discussed in this topic. The topic also covers the work-energy theorem and how to prove it.

6.3 Work

We use terms like overworked and hard worker to describe a person’s effort. But what does work mean, and how do we measure it? In many ways, the scientific definition of work differs from its everyday meaning. In physics, the definition of work reveals its connection to energy. Students will learn more about this in this section

6.4 Kinetic Energy

We’ve all heard tales of great explorers in their sailboats sailing the unknown seas. That was back when there was no such thing as an engine. They could only move their massive ships with the help of the wind. What causes such objects to move when the wind blows? The answer can be found in the principles of kinetic energy.

6.5 Work Done by a Variable Force

It’s interesting to note that most of the forces we encounter on a daily basis are variable in nature, as defined by the term variable force. If there is displacement in the system when a force is applied in the direction of the force, the force is said to perform work on the system. Integration is required to calculate the work done in the case of a variable force.

6.6 The Work-Energy Theorem for a Variable Force

The ideal scenario is a constant force, but the forces that occur all around us are all variable. For all forces, the work-energy theorem holds true. Thus, even with a variable force, the work-energy theorem can be proved.

6.7 The Concept of Potential Energy

Potential energy is the amount of energy contained in an object as a result of its position in relation to a zero position.

6.8 The Conservation of Mechanical Energy

One of the most fundamental principles of classical physics, the law of mechanical energy conservation, is at the heart of the complex motion behind a mechanical automatic watch. The conservation of potential energy and its proof are discussed in this topic.

6.9 Potential Energy of a Spring

Have you ever wondered how a spring returns to its original shape no matter how much force you apply to compress or stretch it?  Why is it necessary to apply additional force to change the position of a spring? The key is the potential energy of the spring that has been stored. The topic explains a spring’s potential energy.

6.10: Various Forms of Energy: The Law of Conservation of Energy

Energy is simply an object’s ability to perform work. Most events and processes in the Universe involve a change in energy. Conservation is practised in all of these types of energies. The amount of energy in our universe is fixed. Energy can neither be created nor destroyed, but it can take on various forms. The law of conservation of motion, as well as different types of energy, are covered in this topic.

6.11 Power

Power is a scalar quantity, just like work. The symbol for it is P, and the SI unit is watt (W). The steam engine was invented by Scottish scientist James Watt, who is credited with its invention. Furthermore, this topic emphasises power in terms of force and velocity, as well as power, horsepower, and kilowatt-hour.

6.12 Collisions

Collision is defined as two objects colliding for a very brief period of time. Collisions and collision types are discussed in this topic.

• Elastic Collision

If both the systems’ linear momentum and kinetic energy are conserved, a collision between two particles or bodies is said to be elastic. Collisions between atomic particles, atoms, marble balls, and billiard balls are just a few examples.

• Inelastic Collision

The collision is said to be inelastic when the system’s linear momentum is conserved but not its kinetic energy. When we drop a ball of wet putty on the floor, the impact is inelastic.

On Extramarks, you will find all of the questions and answers from NCERT Book of Class 11 Science Physics Chapter 6 for free. Experts prepare all NCERT Solutions Class 11 Physics Chapter 6, and they are 100 per cent accurate.

### Benefits of NCERT Solutions for Class 11 Physics Chapter 6

• Physics NCERT Solutions are provided in a step-by-step format. Work, Energy, and Power in Class 11 provides step-by-step solutions to complex problems that are required for class tests and final exams. It includes all of the exercises in your book.
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• Our team of experts has prepared NCERT Solutions for Class 11 Physics Chapter 6 in accordance with all CBSE instructions. All of the answers are custom-made to cover the entire topic, explaining all of the key concepts and including illustrations as needed, delivering quality answers for all questions.
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Q.1 The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Ans.

$\begin{array}{l}\text{Work\hspace{0.17em}done\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}relation\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos\theta }\text{\hspace{0.17em}}\\ \text{Here,\hspace{0.17em}}\mathrm{\theta }\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}angle\hspace{0.17em}between\hspace{0.17em}force\hspace{0.17em}}\stackrel{\to }{\mathrm{F}}\text{\hspace{0.17em}and\hspace{0.17em}displacement\hspace{0.17em}}\stackrel{\to }{\mathrm{S}}.\\ \text{(a)\hspace{0.17em}In\hspace{0.17em}this\hspace{0.17em}case,\hspace{0.17em}}\\ \text{force\hspace{0.17em}and\hspace{0.17em}displacement\hspace{0.17em}are\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}same\hspace{0.17em}direction}\\ \text{i.e.,}\\ \mathrm{\theta }=0\mathrm{°}.\text{\hspace{0.17em}}\\ \therefore \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos}0\mathrm{°}\\ =\mathrm{FS}.\\ \text{Hence,\hspace{0.17em}the\hspace{0.17em}sign\hspace{0.17em}of\hspace{0.17em}work\hspace{0.17em}done\hspace{0.17em}is\hspace{0.17em}positive\hspace{0.17em}in\hspace{0.17em}this\hspace{0.17em}case.\hspace{0.17em}}\\ \left(\mathrm{b}\right)\text{\hspace{0.17em}}\mathrm{In}\text{\hspace{0.17em}}\mathrm{this}\text{\hspace{0.17em}}\mathrm{case},\text{\hspace{0.17em}}\mathrm{force}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\mathrm{displacement}\text{\hspace{0.17em}}\mathrm{are}\text{\hspace{0.17em}}\mathrm{in}\text{\hspace{0.17em}}\mathrm{the}\text{\hspace{0.17em}}\mathrm{opposite}\text{\hspace{0.17em}}\mathrm{direction}\text{\hspace{0.17em}}\\ \text{i.e.,}\\ \mathrm{\theta }=180\mathrm{°}\text{\hspace{0.17em}}\\ \therefore \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos}180\mathrm{°}\\ =-\mathrm{FS}.\\ \text{Hence,}\\ \text{the\hspace{0.17em}sign\hspace{0.17em}of\hspace{0.17em}work\hspace{0.17em}done\hspace{0.17em}is\hspace{0.17em}negative\hspace{0.17em}in\hspace{0.17em}this\hspace{0.17em}case.}\\ \text{(c)\hspace{0.17em}As the direction of\hspace{0.17em}the force\hspace{0.17em}of\hspace{0.17em}friction is always\hspace{0.17em}opposite to the direction of motion,}\\ \text{i.e.,\hspace{0.17em}}\mathrm{\theta }=180\mathrm{°}\\ \therefore \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos}180\mathrm{°}\\ =-\mathrm{FS}.\\ \therefore \text{The work\hspace{0.17em}done by\hspace{0.17em}the force of\hspace{0.17em}friction\hspace{0.17em}is negative in this case.}\\ \text{(d) As the applied force acts along\hspace{0.17em}the direction of\hspace{0.17em}motion of the body,\hspace{0.17em}}\\ \text{i.e.,\hspace{0.17em}}\mathrm{\theta }=\text{0}\mathrm{°}\\ \therefore \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos}0\mathrm{°}\\ =\mathrm{FS}\\ \text{The sign\hspace{0.17em}of\hspace{0.17em}work done is positive\hspace{0.17em}in\hspace{0.17em}this\hspace{0.17em}case.}\\ \text{(e)\hspace{0.17em}As\hspace{0.17em}the resistive force of air acts opposite to the direction of motion of\hspace{0.17em}the pendulum\hspace{0.17em}}\\ \text{i.e.,\hspace{0.17em}}\mathrm{\theta }=\text{0}\mathrm{°}\text{\hspace{0.17em}}\\ \therefore \mathrm{W}=\stackrel{\to }{\mathrm{F}}.\stackrel{\to }{\mathrm{S}}\\ =\mathrm{FScos}180\mathrm{°}\\ =-\mathrm{FS}.\\ \therefore \text{The sign\hspace{0.17em}of\hspace{0.17em}work done is negative in this case.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.2 A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) Work done by the applied force in 10 s,
(b) Work done by friction in 10 s,
(c) Work done by the net force on the body in 10 s,
(d) Change in kinetic energy of the body in 10 s, and interpret your results.

Ans.

$\begin{array}{l}\text{According to Newton’s\hspace{0.17em}second law of motion,the acceleration produced in the body by the applied force is given by,}\\ {\text{a}}_{\text{1}}=\frac{\mathrm{F}}{\mathrm{m}}\\ =\frac{7}{2}\\ =3.5\text{\hspace{0.17em}}{\mathrm{ms}}^{-2}\\ \text{Force of\hspace{0.17em}Friction\hspace{0.17em}is given as:}\\ \text{f =}\mathrm{\mu }\text{mg}\\ \text{= 0.1 × 2 × 9.8}\\ \text{= 1.96 N}\\ \text{The retardation produced by the frictional force\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by,\hspace{0.17em}}\\ {\text{a}}_{\text{2}}=\frac{-\mathrm{f}}{\mathrm{m}}\\ =-\frac{1.96}{2}\\ =-0.98\text{\hspace{0.17em}}{\mathrm{ms}}^{-2}\\ \text{Total acceleration of the body,}\\ \text{a}={\text{a}}_{\text{1}}+{\text{a}}_{\text{2}}\\ =3.5-0.98\\ =2.52\text{\hspace{0.17em}}{\mathrm{ms}}^{-2}\\ \text{The distance covered by the body in 10 s is given by the\hspace{0.17em}equation\hspace{0.17em}of motion:\hspace{0.17em}s}=\text{ut}+\frac{1}{2}{\mathrm{at}}^{2}\\ \therefore \mathrm{s}=0+\frac{1}{2}×2.52×{\left(10\right)}^{2}\\ =126\text{\hspace{0.17em}m.}\\ \text{(a) Work done by the applied force\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ {\text{W}}_{\text{a}}\text{= F × s}\\ \text{= 7 × 126}\\ \text{= 882 J}\\ \text{(b) Work done by the force\hspace{0.17em}of\hspace{0.17em}friction\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:}\\ {\text{W}}_{\text{f}}\text{= -F × s}\\ \text{= -1.96 × 126}\\ \text{= -247 J}\\ \text{(c) Total force = F – f}\\ =\text{7 -1.96}\\ \text{= 5.04 N}\\ \therefore {\text{Work done by the total force, W}}_{\text{total}}\text{= 5.04 ×126}\\ \text{= 635 J}\\ \text{(d) According\hspace{0.17em}to the first equation of motion, final velocity\hspace{0.17em}is given as:}\\ \text{v = u + at}\\ \text{v= 0 + 2.52 × 10}\\ {\text{= 25.2 ms}}^{\text{-1}}\\ \therefore \text{Final\hspace{0.17em}kinetic\hspace{0.17em}energy}=\frac{1}{2}{\mathrm{mv}}^{2}\\ =\frac{1}{2}×2×{\left(25.2\right)}^{2}\\ =635\text{\hspace{0.17em}}\mathrm{J}.\text{\hspace{0.17em}\hspace{0.17em}}\\ \text{Initial\hspace{0.17em}kinetic\hspace{0.17em}energy}=\frac{1}{2}{\mathrm{mu}}^{2}=0\\ \text{Change in kinetic energy}\\ =\text{Final\hspace{0.17em}kinetic\hspace{0.17em}energy – Initial\hspace{0.17em}kinetic\hspace{0.17em}energy}\\ =635-0\\ =635\text{\hspace{0.17em}J}\\ \text{It\hspace{0.17em}shows\hspace{0.17em}that\hspace{0.17em}change\hspace{0.17em}in\hspace{0.17em}kinetic\hspace{0.17em}energy\hspace{0.17em}of\hspace{0.17em}a\hspace{0.17em}body\hspace{0.17em}is\hspace{0.17em}equal\hspace{0.17em}to\hspace{0.17em}work\hspace{0.17em}done\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}total\hspace{0.17em}force\hspace{0.17em}on\hspace{0.17em}the\hspace{0.17em}body.}\end{array}$

Q.3 Given in Fig. 6.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions,
if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

Ans.

As total energy, E = K.E. + P.E.
Or K.E. = E – P.E.
As K.E. of an object is a positive physical quantity, it can never be negative. Therefore, the particle cannot exist in a region, where its K.E. would become negative.
(a) For x > a; P.E. (V0) > E
Therefore, K.E. becomes negative. But, Kinetic energy of a body is a positive physical quantity and it can never be negative. Therefore, the particle cannot exist in the region x > a.
(b) In this case, the potential energy (V0) is greater than total energy (E) in all the regions. Therefore, the particle cannot exist in this region.
(c) x > a and x < b; – V1
In this case, the condition of positivity of K.E. is satisfied in the region between x > a and x < b only.
In this case, the minimum potential energy is –Vm. Hence, K.E. = E – (–Vm) = E + Vm. Therefore, for the positive value of the kinetic energy, the total energy of the particle should be greater than –Vm. Thus, the minimum total energy the particle should have is –Vm.

$\begin{array}{l}\left(\text{d}\right)\text{In this case, the value\hspace{0.17em}of\hspace{0.17em}potential energy}\left({\text{V}}_{\text{0}}\right)\text{of the particle becomes greater than the total energy}\left(\text{E}\right)\\ \text{for}\frac{-\mathrm{b}}{2}<\mathrm{x}<\frac{-\mathrm{a}}{2}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}\frac{\mathrm{a}}{2}<\mathrm{x}<\frac{\mathrm{b}}{2}.\text{\hspace{0.17em}Hence,\hspace{0.17em}the\hspace{0.17em}}\\ \text{particle\hspace{0.17em}cannot\hspace{0.17em}exist\hspace{0.17em}in\hspace{0.17em}these\hspace{0.17em}regions.\hspace{0.17em}In\hspace{0.17em}this\hspace{0.17em}case,\hspace{0.17em}the\hspace{0.17em}minimum\hspace{0.17em}potential\hspace{0.17em}energy\hspace{0.17em}is\hspace{0.17em}}-{\mathrm{V}}_{\mathrm{m}}.\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{K}.\mathrm{E}.=\mathrm{E}-\left(-{\mathrm{V}}_{\mathrm{m}}\right)\\ =\mathrm{E}+{\mathrm{V}}_{\mathrm{m}}.\\ \text{Therefore,\hspace{0.17em}for\hspace{0.17em}the\hspace{0.17em}positive\hspace{0.17em}value\hspace{0.17em}of\hspace{0.17em}K.E.,\hspace{0.17em}the\hspace{0.17em}total\hspace{0.17em}energy\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}particle\hspace{0.17em}must\hspace{0.17em}be\hspace{0.17em}greater\hspace{0.17em}than\hspace{0.17em}}-{\mathrm{V}}_{\mathrm{m}}.\\ \text{Hence,\hspace{0.17em}the\hspace{0.17em}minimum\hspace{0.17em}total\hspace{0.17em}energy\hspace{0.17em}the\hspace{0.17em}particle\hspace{0.17em}should have\hspace{0.17em}is\hspace{0.17em}}-{\mathrm{V}}_{\mathrm{m}}.\end{array}$

Q.4 The potential energy function for a particle executing linear simple harmonic motion is given by V(x) = kx2/2, where k is the force constant of the oscillator. For k = 0.5 Nm–1, the graph of V(x) versus x is shown in Fig. 6.12.
Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ± 2 m.

Ans.

$\begin{array}{l}\text{Given,\hspace{0.17em}total energy of the particle, E}=\text{1 J}\\ \text{Potential\hspace{0.17em}energy\hspace{0.17em}of the particle,\hspace{0.17em}V=\hspace{0.17em}}\frac{1}{2}{\mathrm{kx}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Force constant, k}=0.{\text{5 N m}}^{–\text{1}}\\ \text{Let\hspace{0.17em}m\hspace{0.17em}and\hspace{0.17em}v\hspace{0.17em}be\hspace{0.17em}the\hspace{0.17em}mass\hspace{0.17em}and\hspace{0.17em}velocity\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}particle\hspace{0.17em}respectively.}\\ \text{Kinetic energy\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as: K}=\frac{1}{2}{\mathrm{mv}}^{2}\\ \text{According to the law\hspace{0.17em}of\hspace{0.17em}conservation\hspace{0.17em}of\hspace{0.17em}energy:}\\ \text{E}=\text{V}+\text{K}\\ \therefore 1=\frac{1}{2}{\mathrm{kx}}^{2}+\frac{1}{2}{\mathrm{mv}}^{2}\\ \text{At the time of ‘turn back’, velocity (and\hspace{0.17em}therefore\hspace{0.17em}K)\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}particle\hspace{0.17em}becomes zero.}\\ \therefore 1=\frac{1}{2}{\mathrm{kx}}^{2}\\ ⇒\text{\hspace{0.17em}}\frac{1}{2}×0.\text{5}{\mathrm{x}}^{2}=1\\ \therefore {\mathrm{x}}^{2}=4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{x}=±2\\ \therefore \text{The\hspace{0.17em}particle must\hspace{0.17em}turn back when it reaches}\\ \text{x}=±\text{2 m.}\end{array}$

(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In Fig. 6.13(i) the man walks 2 m carrying a mass of 15 kg on his hands. In Fig.6.13 (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Ans.

(a) When the casing burns up, the mass of the rocket decreases. From the law of conservation of energy:

$\begin{array}{l}\text{T}.\text{E}.=\text{P}.\text{E}.\text{}+\text{K}.\text{E}.\\ =\mathrm{mgh}+\frac{1}{2}{\mathrm{mv}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\end{array}$

Because of the reduction in the mass of rocket, its total energy decreases. Therefore, heat energy required for the burning is obtained from the rocket itself and not from the atmosphere.
(b) Gravitational force is conservative in nature. As the work done by a conservative force over a closed path is equal to zero, work done by the gravitational force over every complete orbit of a comet is equal to zero.
(c) When an artificial satellite, orbiting around earth, moves closer and closer to earth, its potential energy decreases due to the reduction in the height. As the total energy of a system remains conserved, the reduction in P.E. results in an increase in K.E. Therefore, the velocity of the satellite increases. But, because of atmospheric friction, the total energy of the satellite decreases by a small amount.

$\begin{array}{l}\left(\text{d}\right)\text{\hspace{0.17em}Case}\left(\text{i}\right)\text{Given,\hspace{0.17em}Mass\hspace{0.17em}carried\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}man,}\\ \text{m}=\text{15 kg}\\ \text{Displacement,}\\ \text{s}=\text{2 m}\\ \text{In\hspace{0.17em}this\hspace{0.17em}case, the direction of the applied\hspace{0.17em}force and\hspace{0.17em}the direction of the displacement are perpendicular\hspace{0.17em}to\hspace{0.17em}each other.}\\ \therefore \mathrm{\theta }=\text{\hspace{0.17em}9}{0}^{0}\\ \text{Work done\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as: W}=\text{Fs cos}\mathrm{\theta }\\ =\text{mgs cos9}{0}^{0}\\ =\text{15}×\text{9}.\text{8}×\text{2}×\text{}0\mathrm{}\text{\hspace{0.17em}}\mathrm{}\left(\because \text{cos9}{0}^{0}=0\right)\\ =0\\ \text{Case}\left(\text{ii}\right)\\ \text{Here,\hspace{0.17em}mass\hspace{0.17em}carried\hspace{0.17em}by\hspace{0.17em}man, m}=\text{15 kg}\\ \text{Displacement, s}=\text{2 m}\\ \text{In\hspace{0.17em}this\hspace{0.17em}case, the direction of the applied\hspace{0.17em}force and\hspace{0.17em}the direction of the displacement of the rope are same.}\\ \therefore \mathrm{\theta }={0}^{0}\\ {\text{As, cos0}}^{\text{0}}=\text{1}\\ \text{Work done, W}=\text{Fscos}\mathrm{\theta }\text{}\\ =\text{mgscos}{0}^{0}\\ =\text{15}×\text{9}.\text{8}×\text{2}×\text{1}\\ =\text{294 J}\\ \text{Thus, work done in the second case is greater.}\end{array}$

Q.6 Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.

Ans.

(a) The work done by a conservative force on a body is positive when it displaces the body in the direction of force. Because of it, the body approaches the centre of force, decreasing x. Therefore, the potential energy decreases.
(b) The work done against the direction of friction decreases the velocity of a body. Therefore, kinetic energy of the body decreases.
(c) Internal forces cannot produce any change in the total momentum of a body. Therefore, the rate of change of total momentum of a many – particle system is proportional to the external forces acting on it.
(d) The total linear momentum of the system always remains conserved whether it is an elastic collision or an inelastic collision. However, in case of an inelastic collision, the total kinetic energy varies as some energy appears in other forms.

Q.7 State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.

Ans.

(a) The given statement is false, because in an elastic collision, the total energy and momentum of the system are conserved, and not of each body.
(b) The given statement is false. This is because, even if internal forces are balanced, they do not cause any work to be done on a body. Only external forces have the ability to do work. Therefore, external forces are capable of changing the total energy of a system.
(c) The given statement is false. This is because; the work done in the motion of a body over a closed loop is zero only when it is moving under the influence of a conservation force only.
(d) The given statement is true. In case of an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in an inelastic collision, there is always a loss of energy in the form of heat, sound, etc.

(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i.e. when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).

Ans.

(a) No, the total initial kinetic energy of the balls will be equal to the total final kinetic energy of the balls, in an elastic collision. This kinetic energy is not conserved at the instant of time the two balls are in contact with each other. In fact, during collision, the kinetic energy of the balls will get converted into potential energy.
(b) Yes, the total linear momentum of the system always remains conserved in an elastic collision.
(c) In an inelastic collision, the loss of kinetic energy always takes place, i.e., the total kinetic energy of the billiard balls before collision is always greater than that after collision.
The total linear momentum of the system of billiards balls remains conserved even in an inelastic collision.
(d) In this case, the forces involved are conservative, as they depend on the distance of separation between the centres of the billiard balls. Therefore, it is an elastic collision.

Q.9 A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) t1/2
(ii) t
(iii) t3/2
(iv) t2

Ans.

$\begin{array}{l}\text{Here, mass of the body}=\text{m}\\ \text{Initial velocity of the body, u}=0\\ \text{Acceleration of the body}=\text{a}\\ \text{According to the first equation of motion:}\\ \text{v}=\text{u}+\text{at}\\ \therefore \text{v}=0+\text{at}\\ \therefore \text{v}=\text{at}.\dots \text{(i)}\\ \text{According to Newton’s second law of motion:}\\ \text{F\hspace{0.17em}=\hspace{0.17em}ma}.\dots \text{(ii)}\\ \text{Power is given as:}\\ \text{P}=\text{F}×\text{v}.\dots \text{(iii)}\\ \text{Substituting equation (i) and (ii) in equation (iii), we get}\\ \text{P}=\left(\mathrm{ma}\right)×\text{at}\\ \mathrm{P}={\mathrm{ma}}^{2}\mathrm{t}\\ \therefore \mathrm{P}\propto \mathrm{t}\\ \therefore \text{Power is directly proportional to time and the correct option is (ii).}\end{array}$

Q.10 A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) t1/2
(ii) t
(iii) t3/2
(iv) t2

Ans.

$\begin{array}{l}\text{Power\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as}:\text{\hspace{0.17em}}\mathrm{P}=\mathrm{Fv}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \therefore \mathrm{P}=\mathrm{mav}\\ =\mathrm{mv}\frac{\mathrm{dv}}{\mathrm{dt}}\\ =\text{\hspace{0.17em}Constant}\left(\text{say,\hspace{0.17em}k}\right)\\ \therefore \mathrm{vdv}=\frac{\mathrm{k}}{\mathrm{m}}\mathrm{dt}\\ \text{Integrating\hspace{0.17em}both\hspace{0.17em}sides,\hspace{0.17em}we\hspace{0.17em}get}\\ \frac{{\mathrm{v}}^{2}}{2}=\frac{\mathrm{k}}{\mathrm{m}}\mathrm{t}\\ \therefore \mathrm{v}=\sqrt{\frac{2\mathrm{kt}}{\mathrm{m}}}\\ \text{Let\hspace{0.17em}displacement\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}body}=\mathrm{x}\\ \therefore \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\\ =\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}{\mathrm{t}}^{\frac{1}{2}}\\ ⇒\mathrm{dx}=\mathrm{k}‘{\mathrm{t}}^{\frac{1}{2}}\mathrm{dt}\\ \text{Here,\hspace{0.17em}k}‘=\sqrt{\frac{2\mathrm{k}}{\mathrm{m}}}\\ =\text{New\hspace{0.17em}constant}\\ \text{Integrating\hspace{0.17em}both\hspace{0.17em}sides,\hspace{0.17em}we\hspace{0.17em}get\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \mathrm{x}=\frac{1}{2}\mathrm{k}‘{\mathrm{t}}^{\frac{3}{2}}\\ \therefore \mathrm{x}\propto {\mathrm{t}}^{\frac{3}{2}}\\ \text{Therefore,\hspace{0.17em}the\hspace{0.17em}correct\hspace{0.17em}option\hspace{0.17em}is\hspace{0.17em}(iii).}\end{array}$

Q.11

$\begin{array}{l}\text{A body constrained to move along the z-axis of the coordinate}\\ \text{system is subject to a constant force given by F =-}\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\text{ }\mathrm{N}\\ \text{Where}\stackrel{^}{\mathrm{i}}\text{, }\stackrel{^}{\mathrm{j}}\text{ and }\stackrel{^}{\mathrm{k}}\text{ are unit vectors along the x-, y-\hspace{0.17em}and z – axis of}\\ \text{the system respectively. What is the work done by this force}\\ \text{in moving the body a distance of 4 m along the z-axis?}\end{array}$

Ans.

$\begin{array}{l}\text{Here, Force applied on the body, }\\ \stackrel{\to }{\mathrm{F}}=\left(–\stackrel{^}{\mathrm{i}}\text{}+\text{}2\stackrel{^}{\mathrm{j}}\text{}+\text{}3\stackrel{^}{\mathrm{k}}\right)\text{ N}\\ \text{Displacement of the body, s = 4 m, along z-axis = 4}\stackrel{^}{\mathrm{k}}\\ \text{Work done is given by the relation, W =}\stackrel{\to }{\mathrm{F}}\cdot \stackrel{\to }{\mathrm{s}}\\ \therefore \text{W =}\left(–\stackrel{^}{\mathrm{i}}+2\stackrel{^}{\mathrm{j}}+3\stackrel{^}{\mathrm{k}}\right)\cdot \left(4\stackrel{^}{\mathrm{k}}\right)\text{= 12}\stackrel{^}{\mathrm{k}}\cdot \stackrel{^}{\mathrm{k}}\text{= 12 J}\\ \therefore \text{12 J of work is done by the force on the body.}\end{array}$

Q.12 An electron and a proton are detected in a cosmic ray experiment, the irst with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass = 9.11 × 10–31 kg, proton mass = 1.67 × 10–27 kg, 1 eV = 1.60 × 10–19 J).

Ans.

$\begin{array}{l}{\text{Here,\hspace{0.17em}mass of electron, m}}_{\text{e}}=\text{9}.\text{11}×\text{1}{0}^{–\text{31}}\text{kg\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ {\text{Mass of proton, m}}_{\text{p}}=\text{1}.\text{67}×\text{1}{0}^{–\text{27}}\text{kg}\\ {\text{Kinetic energy of electron, E}}_{\text{e}}\text{=}\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{v}}_{\mathrm{e}}}^{2}\\ =\text{1}0\text{keV}\\ =\text{1}{0}^{\text{4}}\text{eV}\\ =\text{1}{0}^{\text{4}}×\text{1}.\text{6}0×\text{1}{0}^{–\text{19}}\text{\hspace{0.17em}}\mathrm{J}\\ \therefore {\text{E}}_{\text{e}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{e}}{{\mathrm{v}}_{\mathrm{e}}}^{2}\\ =\text{1}.\text{6}0×\text{1}{0}^{–\text{15}}\text{J}\to \text{(i)}\\ {\text{Kinetic energy of the proton, E}}_{\text{p}}\text{=}\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}{{\mathrm{v}}_{\mathrm{p}}}^{2}\\ =\text{1}00\text{keV}\\ =\text{1}{0}^{\text{5}}\text{eV}\\ =\text{1}{0}^{\text{5}}×\text{1}.\text{6}0×\text{1}{0}^{–\text{19}}\text{\hspace{0.17em}}\mathrm{J}\\ \therefore {\text{E}}_{\text{p}}\text{=}\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}{{\mathrm{v}}_{\mathrm{p}}}^{2}\\ =\text{1}.\text{6}0×\text{1}{0}^{–\text{14}}\text{J}\to \text{(ii)}\\ {\text{As\hspace{0.17em}we\hspace{0.17em}observe\hspace{0.17em}that,\hspace{0.17em}E}}_{\text{e}}>{\text{E}}_{\text{p}}\\ \therefore \text{The\hspace{0.17em}electron\hspace{0.17em}is\hspace{0.17em}traveling\hspace{0.17em}faster\hspace{0.17em}than\hspace{0.17em}the\hspace{0.17em}proton.}\\ \text{Dividing\hspace{0.17em}equation\hspace{0.17em}(i)\hspace{0.17em}by\hspace{0.17em}equation\hspace{0.17em}(ii),\hspace{0.17em}we\hspace{0.17em}obtain:}\\ \frac{{\mathrm{m}}_{\mathrm{e}}{{\mathrm{v}}_{\mathrm{e}}}^{2}}{{\mathrm{m}}_{\mathrm{p}}{{\mathrm{v}}_{\mathrm{p}}}^{2}}=\frac{\text{1}.\text{6}0×\text{1}{0}^{–\text{15}}\text{J}}{\text{1}.\text{6}0×\text{1}{0}^{–\text{14}}\text{J}}\\ =\frac{1}{10}\\ \frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{v}}_{\mathrm{p}}}=\sqrt{\frac{1}{10}\frac{{\mathrm{m}}_{\mathrm{p}}}{{\mathrm{m}}_{\mathrm{e}}}}\\ =\sqrt{\frac{1.67×{10}^{-27}}{10×9.11×\text{1}{0}^{–3\text{1}}}}\\ =13.53\\ \therefore \text{Ratio\hspace{0.17em}of\hspace{0.17em}speed\hspace{0.17em}of\hspace{0.17em}electron\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}speed\hspace{0.17em}of\hspace{0.17em}proton}={\mathrm{v}}_{\mathrm{e}}:{\mathrm{v}}_{\mathrm{p}}\\ =13.53:1\end{array}$

Q.13 A rain drop of radius 2 mm falls from a height of 500 m above the round. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height; it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done by the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms–1?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}radius of the drop,}\\ \text{r}=\text{2 mm}\\ =\text{2}×\text{1}{0}^{–\text{3}}\text{m\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Volume of the drop\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:\hspace{0.17em}}\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^{3}\\ =\frac{4}{3}×3.14×{\left(\text{2}×\text{1}{0}^{–\text{3}}\right)}^{3}{\mathrm{m}}^{3}\\ \text{Density of water,}\\ \mathrm{\rho }=\text{1}{0}^{\text{3}}{\text{kg m}}^{–\text{3}}\\ \text{Mass of drop\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as: m}=\mathrm{\rho }\text{V}\\ \therefore \text{m\hspace{0.17em}=\hspace{0.17em}1}{0}^{\text{3}}×\frac{4}{3}×3.14×{\left(\text{2}×\text{1}{0}^{–\text{3}}\right)}^{3}\\ =3.35×{10}^{-5}\text{\hspace{0.17em}}\mathrm{kg}\\ \text{Gravitational force\hspace{0.17em}acting\hspace{0.17em}on\hspace{0.17em}rain\hspace{0.17em}drop,}\\ \text{F}=\text{mg}\\ =3.35×{10}^{-5}×9.8\text{\hspace{0.17em}N}\\ \text{Work done by the gravitational force on the drop during the first half of its journey:}\\ {\text{W}}_{\text{1}}=\text{Fs}\\ =3.35×{10}^{-5}×9.8×\text{25}0\\ =0.0\text{82 J}\\ {\text{W}}_{\text{1}}\text{=Work done by the gravitational force\hspace{0.17em}on the drop\hspace{0.17em}in the second half of its\hspace{0.17em}journey,}\\ \text{i}.\text{e}.,{\text{W}}_{\text{2}}=0.0\text{82 J}\\ \text{According\hspace{0.17em}to the law of conservation of energy, in\hspace{0.17em}the absence\hspace{0.17em}of any\hspace{0.17em}resistive force,\hspace{0.17em}energy\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}drop\hspace{0.17em}on}\\ \text{reaching\hspace{0.17em}the\hspace{0.17em}ground\hspace{0.17em}will\hspace{0.17em}remain\hspace{0.17em}the\hspace{0.17em}same.}\\ \therefore {\text{E}}_{1}=\text{mgh}\\ =3.35×\text{1}{0}^{–\text{5}}×\text{9.8}×\text{5}00\\ =0.\text{164 J}\\ \text{In the presence of a resistive force,\hspace{0.17em}}\\ \text{velocity\hspace{0.17em}of\hspace{0.17em}the drop on\hspace{0.17em}reaching the ground,}\\ \text{v=\hspace{0.17em}1}0{\text{ms}}^{\text{-1}}\\ \therefore \text{Actual energy of\hspace{0.17em}drop\hspace{0.17em}on\hspace{0.17em}reaching the ground,}\\ {\text{E}}_{\text{2}}=\frac{1}{2}{\mathrm{mv}}^{2}\\ =\frac{1}{2}×3.35×\text{1}{0}^{–\text{5}}×{\left(10\right)}^{2}\\ =1.675×{10}^{-3}\text{\hspace{0.17em}}\mathrm{J}\\ \therefore \text{Work\hspace{0.17em}done\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}resistive force}={\text{E}}_{\text{2}}–{\text{E}}_{\text{1}}\\ =1.675×{10}^{-3}\text{\hspace{0.17em}}\mathrm{J}-0.\text{164 J}\\ =–0.\text{162 J\hspace{0.17em}\hspace{0.17em}}\end{array}$

Q.14 A molecule in a gas container hits a horizontal wall with speed 200 m s up>–1 and angle 300 with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?

Ans.

$\begin{array}{l}\text{The momentum of the gas molecule remains conserved in elastic as well as inelastic collisions.\hspace{0.17em}Now,\hspace{0.17em}let\hspace{0.17em}us\hspace{0.17em}verify\hspace{0.17em}if\hspace{0.17em}K.E.\hspace{0.17em}is\hspace{0.17em}conserved\hspace{0.17em}or\hspace{0.17em}not.}\\ \text{Here,\hspace{0.17em}}\\ \text{speed\hspace{0.17em}of\hspace{0.17em}gas\hspace{0.17em}molecule}=200\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\\ \text{K.E.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}gas\hspace{0.17em}molecule\hspace{0.17em}before\hspace{0.17em}collision,\hspace{0.17em}}\\ {\mathrm{E}}_{1}=\frac{1}{2}\mathrm{m}{\left(200\right)}^{2}\\ =2×{10}^{4}\mathrm{m}\text{\hspace{0.17em}J}\\ \text{As\hspace{0.17em}the\hspace{0.17em}wall\hspace{0.17em}is\hspace{0.17em}very\hspace{0.17em}heavy,\hspace{0.17em}the\hspace{0.17em}recoiling\hspace{0.17em}molecule\hspace{0.17em}does\hspace{0.17em}not produce\hspace{0.17em}velocity\hspace{0.17em}in\hspace{0.17em}the\hspace{0.17em}wall.}\\ \text{Let\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}gas\hspace{0.17em}molecule\hspace{0.17em}}=\mathrm{m}\text{\hspace{0.17em}}\\ \text{Let\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}wall}=\mathrm{M}\text{\hspace{0.17em}}\\ \text{Total\hspace{0.17em}K.E.\hspace{0.17em}after\hspace{0.17em}collision,}\\ {\mathrm{E}}_{2}=\frac{1}{2}\mathrm{m}{\left(200\right)}^{2}+\frac{1}{2}\mathrm{M}{\left(0\right)}^{2}\\ =2×{10}^{4}\mathrm{m}\text{\hspace{0.17em}J}\\ \text{As,\hspace{0.17em}}\\ \text{total\hspace{0.17em}K.E.\hspace{0.17em}after\hspace{0.17em}collision = Total\hspace{0.17em}K.E.\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}gas\hspace{0.17em}}\\ \text{molecule\hspace{0.17em}before\hspace{0.17em}collision}\\ \therefore \text{It\hspace{0.17em}is\hspace{0.17em}an\hspace{0.17em}elastic\hspace{0.17em}collision.}\end{array}$

Q.15 A pump on the ground floor of a building can pump up water to fill a ank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}volume of tank,}\\ \text{V}=\text{3}0{\text{m}}^{\text{3}}\\ \text{Time of operation},\text{}\\ \text{t}=\text{15 min}\\ =\text{15}×\text{6}0\\ =\text{9}00\text{s\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Height of tank,}\\ \text{h}=\text{4}0\text{m}\\ \text{Efficiency of the pump,}\\ \mathrm{\eta }=\text{3}0\mathrm{%}\\ \text{Density of water,}\\ \mathrm{\rho }=\text{1}{0}^{\text{3}}{\text{\hspace{0.17em}kgm}}^{\text{-3}}\\ \therefore \text{Mass of water\hspace{0.17em}pumped,}\\ \text{m}=\mathrm{\rho }\text{V}\\ =\text{3}0×\text{1}{0}^{\text{3}}\text{\hspace{0.17em}kg}\\ \text{Output power is\hspace{0.17em}given\hspace{0.17em}as}:{\text{\hspace{0.17em}P}}_{\text{0}}=\frac{\text{Work\hspace{0.17em}done}}{\text{Time\hspace{0.17em}}}\\ =\frac{\mathrm{mgh}}{\mathrm{t}}\\ =\frac{\text{3}0×\text{1}{0}^{\text{3}}×9.8×40}{900}\\ =13.067×{10}^{3}\text{\hspace{0.17em}}\mathrm{W}\\ {\text{If\hspace{0.17em}P}}_{\text{i}}\text{\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}input\hspace{0.17em}power,then efficiency of\hspace{0.17em}the\hspace{0.17em}pump\hspace{0.17em}is}\\ \text{given\hspace{0.17em}by the relation,\hspace{0.17em}}\mathrm{\eta }\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{P}}_{\text{0}}}{{\text{P}}_{\text{i}}}\\ =30\mathrm{%}\\ \therefore {\text{P}}_{\text{i}}=\frac{{\text{P}}_{\text{0}}}{\mathrm{\eta }}\\ =\frac{13.067×{10}^{3}}{\frac{30}{100}}\\ =43567\text{\hspace{0.17em}}\mathrm{W}\\ =43.567\text{\hspace{0.17em}}\mathrm{kW}\end{array}$

Q.16 Two identical ball bearings in contact with each other and resting on a rictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Ans.

$\begin{array}{l}\text{Total momentum of\hspace{0.17em}a\hspace{0.17em}system\hspace{0.17em}remains conserved elastic as well as inelastic\hspace{0.17em}collisions.\hspace{0.17em}In\hspace{0.17em}case\hspace{0.17em}of an}\\ \text{elastic collision,\hspace{0.17em}the total kinetic energy of a\hspace{0.17em}system remains conserved before\hspace{0.17em}and after collision.\hspace{0.17em}}\\ \text{Let mass of each ball\hspace{0.17em}bearing= m}\\ \text{Total K.E. of the system before collision}\\ \text{\hspace{0.17em}}=\frac{1}{2}{\mathrm{mV}}^{2}+\frac{1}{2}\left(2\mathrm{m}\right)×0\\ =\frac{1}{2}{\mathrm{mV}}^{2}\\ \text{Case}\left(\text{i}\right)\text{Total kinetic energy of the system after}\\ \text{collision}=\frac{1}{2}\mathrm{m}×0=\frac{1}{2}\left(2\mathrm{m}\right){\left(\frac{\mathrm{V}}{2}\right)}^{2}\\ =\frac{1}{4}{\mathrm{mV}}^{2}\\ \therefore \text{Kinetic energy of the system is not conserved in this\hspace{0.17em}case.}\\ \text{Case}\left(\text{ii}\right)\text{Total kinetic energy\hspace{0.17em}of the system after}\\ \text{collision}=\frac{1}{2}\left(2\mathrm{m}\right)×0+\frac{1}{2}{\mathrm{mV}}^{2}\\ =\frac{1}{2}{\mathrm{mV}}^{2}\\ \therefore \text{\hspace{0.17em}Kinetic energy of the system is conserved\hspace{0.17em}in this\hspace{0.17em}case.}\\ \text{Case}\left(\text{iii}\right)\text{Total kinetic energy of the system after \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{collision\hspace{0.17em}}=\frac{1}{2}\left(3\mathrm{m}\right){\left(\frac{\mathrm{V}}{3}\right)}^{2}\\ =\frac{1}{6}{\mathrm{mV}}^{2}\\ \therefore \text{The kinetic energy of the system is not conserved in this\hspace{0.17em}case.}\\ \text{As\hspace{0.17em}kinetic\hspace{0.17em}energy\hspace{0.17em}is\hspace{0.17em}conserved\hspace{0.17em}only\hspace{0.17em}in\hspace{0.17em}case\hspace{0.17em}(ii),\hspace{0.17em}}\\ \therefore \text{\hspace{0.17em}case\hspace{0.17em}(ii)\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}only\hspace{0.17em}possibility.}\end{array}$

Q.17 The bob A of a pendulum released from 30o to the vertical its another bob B of the same mass at rest on a table as shown in Fig. 6.15.
How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Ans.

The bob A will not rise. This is because in an elastic collision between two bodies of equal masses, the velocities of the bodies are interchanged. In such a collision, a complete transfer of momentum takes place from the moving body to the stationary body. Therefore, after colliding with bob B of equal mass, bob A of mass m, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Q.18 The bob of a pendulum is released from a horizontal position. If the ength of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost point, given that it dissipated 5% of its initial energy against air resistance?

Ans.

Here, length of pendulum, l = 1.5 m
Mass of the bob of the pendulum = m
Energy dissipated = 5%
According to the law of conservation of energy, the total energy of the system remains conserved.
At the horizontal position:
Potential energy of the bob of pendulum, EP = mgl
Kinetic energy of the bob of pendulum, EK = 0
Total energy =EP + EK
=mgl … (i)
At the lowermost point (mean position):
Potential energy of the bob of pendulum, EP = 0

$\begin{array}{l}\text{Kinetic energy of the bob\hspace{0.17em}of\hspace{0.17em}pendulum,\hspace{0.17em}}{\mathrm{E}}_{\mathrm{K}}=\frac{1}{2}{\mathrm{mv}}^{2}\\ \therefore {\text{Total energy, E}}_{\text{x}}={\mathrm{E}}_{\mathrm{P}}+{\mathrm{E}}_{\mathrm{K}}\\ =\frac{1}{2}{\mathrm{mv}}^{2}\text{\hspace{0.17em}\hspace{0.17em}}\dots \text{}\left(\text{ii}\right)\\ \text{As the bob travels from the horizontal position to the lowermost point,\hspace{0.17em}energy dissipated= 5}\mathrm{%}\text{}\\ \text{Total energy at the lowermost point = 95%\hspace{0.17em}of the total energy at the\hspace{0.17em}horizontal position, i.e.,}\\ \frac{1}{2}{\mathrm{mv}}^{2}=\frac{95}{100}×\mathrm{mgl}\\ \therefore \mathrm{v}=\sqrt{\frac{2×95×1.5×9.8}{100}}\\ =5.28\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\end{array}$

Q.19 A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly ith a speed of 27 kmh-1 on a frictionless track. After a while, sand starts leaking out of a hole on the floor of the trolley at the rate of 0.05 kg s–1. What is the speed of the trolley after the entire sand bag is empty?

Ans.

Since, the sand bag placed on a trolley is moving with a uniform speed, therefore, the external force acting on the system of the sandbag and the trolley is zero.
When the sand starts leaking from the bag, velocity of the trolley will not change at all. This is due to the fact that according to Newton’s first law of motion, the leaking action does not produce any external force on the system. Therefore, the speed of the trolley will not change.

Q.20

$\begin{array}{l}\text{A body of mass 0.5 kg travels in a straight line with}\\ \text{velocity}\mathrm{v}={\mathrm{ax}}^{\frac{3}{2}},\text{}\mathrm{where}\text{}\mathrm{a}=5\text{}{\mathrm{m}}^{-\frac{1}{2}}{\mathrm{s}}^{-1}.\text{What is the work}\\ \text{done by the net force during its displacement from x}=0\text{to x}=\text{2 m?}\end{array}$

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the body, m}=0.\text{5 kg\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Velocity of the body can\hspace{0.17em}be\hspace{0.17em}calculated\hspace{0.17em}by\hspace{0.17em}using the equation,}\\ \mathrm{v}={\mathrm{ax}}^{\frac{3}{2}}{\text{\hspace{0.17em}where\hspace{0.17em}a\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}5\hspace{0.17em}m}}^{\frac{-1}{2}}{\mathrm{s}}^{-1}\\ \text{At\hspace{0.17em}x=0,\hspace{0.17em}initial velocity},\text{u}=\mathrm{a}×0\\ =0\\ \text{At x}=\text{2 m,\hspace{0.17em}final velocity, v}=10\sqrt{2}\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\\ \text{Work done\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}net\hspace{0.17em}force, W= Change in kinetic}\\ \text{energy}\\ \therefore \text{W=}\frac{1}{2}\mathrm{m}\left({\mathrm{v}}_{}^{2}-{\mathrm{u}}_{}^{2}\right)\\ =\frac{1}{2}×0.\text{5}\left[{\left(10\sqrt{2}\right)}^{2}-0\right]\\ =50\text{\hspace{0.17em}J}\end{array}$

Q.21 The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m2, v = 36kmh-1 and the density of air is 1.2 kg m–3. What is the electrical power produced?

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}area of the circle swept by the windmill}=\text{A}\\ \text{Velocity of the wind}=\text{v}\\ \text{Density of air}=\mathrm{\rho }\\ \left(\text{a}\right)\text{Here,volume of the\hspace{0.17em}wind flowing per sec}=\text{Av}\\ \text{Mass of the\hspace{0.17em}wind flowing per sec}=\mathrm{\rho }\text{Av}\\ \text{Mass}\left(\text{m}\right)\text{of the air passing in time\hspace{0.17em}t}=\mathrm{\rho }\text{Avt}\\ \left(\text{b}\right)\text{Kinetic energy of air\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}as:\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}{\mathrm{mv}}^{2}=\frac{1}{2}\left(\mathrm{\rho Avt}\right){\mathrm{v}}^{2}\\ =\frac{1}{2}{\mathrm{\rho Av}}^{3}\mathrm{t}\\ \left(\text{c}\right)\text{Here,\hspace{0.17em}area of the circle swept by the windmill}=\text{A}\\ =\text{3}0{\text{m}}^{\text{2}}\\ \text{Velocity of air,}\\ \text{v}={\text{36 kmh}}^{\text{-1}}\\ \text{Density of air,}\\ \mathrm{\rho }=\text{1}.{\text{2 kg m}}^{–\text{3}}\\ \text{Electrical energy produced = 25% of the K.E.\hspace{0.17em}of\hspace{0.17em}air}\\ =\frac{25}{100}×\text{K.E.\hspace{0.17em}of\hspace{0.17em}air}\\ =\frac{1}{8}{\mathrm{\rho Av}}^{3}\mathrm{t}\\ =\frac{1}{8}×\text{3}0×{\left(10\right)}^{3}×\text{1}.\text{2}\\ =4500\text{\hspace{0.17em}}\mathrm{W}\\ =4.5\text{\hspace{0.17em}}\mathrm{kW}\end{array}$

Q.22 A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated.
(a) How much work does she do against the gravitational force?
(b) Fat supplies 3.8 × 107 J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?

Ans.

(a) Here, mass of the weight, m = 10 kg
Height to which the weight is lifted, h = 0.5 m
Number of times the person lifts weight, n = 1000
∴Work done against gravitational force, W=n (mgh)
=1000× (10×9.8×0.5)
=49000 J.

$\begin{array}{l}\left(\text{b}\right)\text{Here,}\text{\hspace{0.17em}}\text{energy equivalent of 1 kg of fat}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ =\text{3}.\text{8}×\text{1}{0}^{\text{7}}\text{J}\\ \text{Efficiency rate}=\text{2}0\text{\hspace{0.17em}}%\\ \text{Mechanical energy supplied by the body}\text{\hspace{0.17em}}\text{of}\text{\hspace{0.17em}}\text{dieter}\\ \text{=}\frac{20}{100}×3.8×\text{1}{0}^{\text{7}}\text{J}\\ =\frac{1}{5}×3.8×\text{1}{0}^{\text{7}}\text{J}\\ \text{Equivalent mass of fat lost by dieter}\\ \text{=}\frac{1}{\frac{1}{5}×3.8×\text{1}{0}^{\text{7}}}×49×\text{1}{0}^{3}\text{\hspace{0.17em}}\\ =\frac{245}{3.8}×\text{1}{0}^{-4}\\ =6.45×\text{1}{0}^{-3}\text{\hspace{0.17em}}kg\end{array}$

Q.23 A family uses 8 kW of power. (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{Here,\hspace{0.17em}power consumed by the family,\hspace{0.17em}P}=\text{8 kW}\\ =\text{8}×\text{1}{0}^{\text{3}}\text{\hspace{0.17em}W}\\ \text{Solar energy collected per square metre}=\text{2}00\text{W}\\ \text{Efficiency of conversion from solar energy\hspace{0.17em}to electrical energy = 20 %}\\ \text{Let\hspace{0.17em}area required to generate the desired electricity = A According the information given in the question,\hspace{0.17em}}\\ 8×{10}^{3}=20\mathrm{%}×\left(\text{A}×\text{2}00\right)\\ =\frac{20}{100}×\text{A}×\text{2}00\\ \therefore \mathrm{A}=\frac{8000}{40}\\ =200\text{\hspace{0.17em}}{\mathrm{m}}^{2}\\ \left(\text{b}\right)\text{The area of a solar plate required to generate 8 kW of electricity is almost equivalent to the area of the roof}\\ {\text{of a large\hspace{0.17em}house\hspace{0.17em}of\hspace{0.17em}250\hspace{0.17em}m}}^{\text{2}}\text{.}\end{array}$

Q.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of bullet,}\\ \text{m}=0.0\text{12 kg}\\ \text{Initial speed of bullet,}\\ {\text{u}}_{\text{b}}=\text{7}0{\text{ms}}^{\text{-1}}\\ \text{Mass of wooden block,}\\ \text{M}=0.\text{4 kg}\\ \text{Initial speed of wooden block,}\\ {\text{u}}_{\text{B}}=0\\ \text{Since\hspace{0.17em}the\hspace{0.17em}bullet\hspace{0.17em}comes\hspace{0.17em}to\hspace{0.17em}rest\hspace{0.17em}w.r.t.\hspace{0.17em}the\hspace{0.17em}block,\hspace{0.17em}the\hspace{0.17em}two\hspace{0.17em}behave\hspace{0.17em}as\hspace{0.17em}one\hspace{0.17em}body.}\\ \text{Let\hspace{0.17em}final speed of the system of\hspace{0.17em}the bullet and the\hspace{0.17em}block}=\mathrm{v}\\ \text{As\hspace{0.17em}per\hspace{0.17em}the law of conservation of momentum:}\\ {\text{mu}}_{\text{b}}+\mathrm{M}{\text{u}}_{\text{B}}=\left(\mathrm{m}+\mathrm{M}\right)\mathrm{v}\\ =\mathrm{mv}+\mathrm{Mv}\\ =0.012×\text{7}0+0.\text{4}×0\\ =\left(0.012+0.\text{4}\right)\mathrm{v}\\ \therefore \mathrm{v}=\frac{0.84}{0.412}\\ =2.04\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\\ \text{In\hspace{0.17em}case\hspace{0.17em}of the system of the bullet and the block:\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Mass of system,}\\ \text{m}‘=\mathrm{m}+\mathrm{M}\\ =0.012+0.\text{4}\\ =0.\text{412 kg}\\ \text{Velocity of system}=\text{2}.0{\text{4 ms}}^{\text{-1}}\\ \text{Let\hspace{0.17em}height acquired\hspace{0.17em}by the system}=\text{h}\\ \text{According\hspace{0.17em}to the law of conservation of energy:}\\ \text{P.E.\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}highest\hspace{0.17em}point=K.E.\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}lowest\hspace{0.17em}point}\\ \mathrm{m}‘\mathrm{gh}=\frac{1}{2}\mathrm{m}‘{\mathrm{v}}^{2}\\ \therefore \mathrm{h}=\frac{1}{2}\left(\frac{{\mathrm{v}}^{2}}{\mathrm{g}}\right)\\ =\frac{1}{2}×\frac{{\left(2.04\right)}^{2}}{9.8}\\ =\text{}0.\text{2123 m}\\ \text{The block will reach to a height of 0.2123 m.}\\ \therefore \text{Heat produced = K.E. of the bullet – K.E. of the system}\\ \text{=}\frac{1}{2}{\mathrm{mu}}^{2}-\frac{1}{2}\mathrm{m}‘{\mathrm{v}}^{2}\\ =\frac{1}{2}×0.012×{\left(70\right)}^{2}-\frac{1}{2}×0.412×{\left(2.04\right)}^{2}\\ =\text{29}.\text{4}–\text{}0.\text{857}\\ =\text{28}.\text{54 J}\end{array}$

Q.25 Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16).Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and h = 10 m, what are the speeds and times taken by the two stones?

Ans.

$\begin{array}{l}\text{The\hspace{0.17em}situation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}problem\hspace{0.17em}can\hspace{0.17em}be\hspace{0.17em}shown\hspace{0.17em}by\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}figure.\hspace{0.17em}Here,\hspace{0.17em}AB\hspace{0.17em}and\hspace{0.17em}AC\hspace{0.17em}are\hspace{0.17em}two\hspace{0.17em}frictionless\hspace{0.17em}planes\hspace{0.17em}inclined\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}horizontal\hspace{0.17em}at\hspace{0.17em}}\angle {\mathrm{\theta }}_{1}\text{\hspace{0.17em}and\hspace{0.17em}}\angle {\mathrm{\theta }}_{2}\text{\hspace{0.17em}respectively.}\\ \text{Let\hspace{0.17em}final\hspace{0.17em}velocities\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}two\hspace{0.17em}stones\hspace{0.17em}at\hspace{0.17em}points\hspace{0.17em}B\hspace{0.17em}and\hspace{0.17em}C\hspace{0.17em}be\hspace{0.17em}}{\mathrm{v}}_{1}\text{\hspace{0.17em}and\hspace{0.17em}}{\mathrm{v}}_{2}\text{\hspace{0.17em}respectively.}\\ \text{As\hspace{0.17em}height\hspace{0.17em}of\hspace{0.17em}both\hspace{0.17em}the\hspace{0.17em}planes\hspace{0.17em}is\hspace{0.17em}same,\hspace{0.17em}therefore,both\hspace{0.17em}the\hspace{0.17em}stones\hspace{0.17em}will\hspace{0.17em}have\hspace{0.17em}same\hspace{0.17em}potential\hspace{0.17em}energy\hspace{0.17em}at\hspace{0.17em}point\hspace{0.17em}A.}\\ \text{According\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}law\hspace{0.17em}of\hspace{0.17em}conservation\hspace{0.17em}of\hspace{0.17em}energy,\hspace{0.17em}the\hspace{0.17em}kinetic\hspace{0.17em}energy\hspace{0.17em}of\hspace{0.17em}both\hspace{0.17em}the\hspace{0.17em}stones\hspace{0.17em}at\hspace{0.17em}points\hspace{0.17em}B\hspace{0.17em}and\hspace{0.17em}C\hspace{0.17em}will\hspace{0.17em}also\hspace{0.17em}be\hspace{0.17em}the\hspace{0.17em}equal.}\\ \therefore \frac{1}{2}{{\mathrm{mv}}_{1}}^{2}=\frac{1}{2}{{\mathrm{mv}}_{2}}^{2}\\ \therefore {\mathrm{v}}_{1}={\mathrm{v}}_{2}\\ =\mathrm{v}\left(\mathrm{say}\right)\\ \text{Here,\hspace{0.17em}}\\ \text{m\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}mass\hspace{0.17em}of\hspace{0.17em}each\hspace{0.17em}stone.}\\ \therefore \text{Both\hspace{0.17em}the\hspace{0.17em}stones\hspace{0.17em}will\hspace{0.17em}come\hspace{0.17em}to\hspace{0.17em}the\hspace{0.17em}bottom\hspace{0.17em}with\hspace{0.17em}the\hspace{0.17em}same speed\hspace{0.17em}v.}\\ \text{From\hspace{0.17em}the\hspace{0.17em}given\hspace{0.17em}figure,\hspace{0.17em}it\hspace{0.17em}is\hspace{0.17em}clear\hspace{0.17em}that\hspace{0.17em}the\hspace{0.17em}acceleration\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}two\hspace{0.17em}blocks\hspace{0.17em}are\hspace{0.17em}}\\ {\mathrm{a}}_{1}={\mathrm{gsin\theta }}_{1}\text{\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}}{\mathrm{a}}_{2}={\mathrm{gsin\theta }}_{2}\\ \mathrm{Since}\text{\hspace{0.17em}}{\mathrm{\theta }}_{2}>{\mathrm{\theta }}_{1}\text{\hspace{0.17em}}\\ \therefore {\mathrm{a}}_{2}>{\mathrm{a}}_{1}\\ \therefore {\mathrm{t}}_{2}<{\mathrm{t}}_{1}.\\ \therefore \text{The\hspace{0.17em}stone\hspace{0.17em}moving\hspace{0.17em}down\hspace{0.17em}the\hspace{0.17em}inclined\hspace{0.17em}plane\hspace{0.17em}AC\hspace{0.17em}will\hspace{0.17em}arrive\hspace{0.17em}at\hspace{0.17em}the\hspace{0.17em}bottom\hspace{0.17em}first.}\end{array}$

Q.26 A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of block,}\\ \text{m}=\text{1 kg}\\ \text{Spring constant,}\\ \text{k}=\text{1}00{\text{N m}}^{–\text{1}}\\ \text{Displacement of the block,}\\ \text{x}=\text{1}0\text{cm}\\ =0.\text{1 m}\\ \text{The given situation can be shown as in the given figure.}\\ \text{At equilibrium\hspace{0.17em}position:}\\ \text{Normal reaction, R}=\text{mg cos 37}\mathrm{°}\\ {\text{Force\hspace{0.17em}of\hspace{0.17em}friction, f}}_{\text{R}}\text{=\hspace{0.17em}}\mathrm{\mu R}\\ =\text{mg sin 37}\mathrm{°}\\ \text{Here,}\mathrm{\mu }\text{= Coefficient of friction}\\ \text{Total force acting on the block down\hspace{0.17em}the\hspace{0.17em}incline}\\ =\text{mg sin 37}\mathrm{°}–{\text{f}}_{\text{R}}\\ =\text{mgsin 37}\mathrm{°}–\text{}\mathrm{\mu }\text{mgcos 37}\mathrm{°}\\ =\text{mg}\left(\text{sin 37}\mathrm{°}–\mathrm{\mu }\text{cos 37}\mathrm{°}\right)\\ \text{At equilibrium\hspace{0.17em}position, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Work done by the block = P.E. of the\hspace{0.17em}stretched spring}\\ \text{mg}\left(\text{sin 37}\mathrm{°}\text{}–\text{}\mathrm{\mu }\text{cos 37}\mathrm{°}\right)\mathrm{x}=\frac{1}{2}{\mathrm{kx}}^{2}\\ 1×9.8\left(\text{sin 37}\mathrm{°}–\mathrm{\mu }\text{cos 37}\mathrm{°}\right)=\frac{1}{2}×100×0.1\\ 0.602-\mathrm{\mu }×0.799=0.510\\ \therefore \mathrm{\mu }=\frac{0.092}{0.799}\\ =0.115\end{array}$

Q.27 A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with a uniform speed of 7 ms–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Ans.

Here, mass of bolt, m = 0.3 kg
Speed of elevator = 7 ms-1
Height of elevator, h = 3 m
As the relative velocity of the bolt w.r.t. the elevator is zero at the time of impact, only potential energy of the bolt gets transformed into heat energy.
Amount of heat produced = Loss of potential energy
= mgh
= 0.3 × 9.8 × 3
= 8.82 J
The amount of heat produced will remain the same even if the elevator is stationary. This is because; the relative velocity of the bolt with respect to the elevator will remain zero.

Q.28 B

Ans.

$\begin{array}{l}\text{Here,\hspace{0.17em}mass of the\hspace{0.17em}trolley, M}=\text{2}00\text{kg\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\\ \text{Speed of the trolley, v}={\text{36 kmh}}^{\text{-1}}\text{}\\ =\text{1}0{\text{ms}}^{\text{-1}}\\ \text{Mass of the child, m}=\text{2}0\text{kg}\\ \text{Initial momentum of the system of the child and}\\ \text{the trolley}=\left(\text{M}+\text{m}\right)\text{v}\\ =\left(\text{2}00+\text{2}0\right)×\text{1}0\\ =\text{22}00{\text{kg ms}}^{\text{-1}}\\ \text{Let the final velocity of the trolley w.r.t. ground = v}‘\\ \text{Final velocity of the boy with respect to the ground = v}‘-4\\ \text{Final momentum\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}system}=\text{2}00\text{v}‘+\text{2}0\left(\text{v}‘–4\right)\\ =\text{2}00\text{v}‘+\text{2}0\text{v}‘-80\\ =\text{22}0\text{v}‘–\text{8}0\\ \text{According\hspace{0.17em}to the law of conservation of momentum, Initial momentum\hspace{0.17em}=Final momentum}\\ \therefore \text{22}00=\text{22}0\text{v}‘–\text{8}0\\ \text{228}0=\text{22}0\text{v}‘\\ \therefore \text{v}‘=\frac{\text{228}0}{\text{22}0}\\ =10.36\text{\hspace{0.17em}}{\mathrm{ms}}^{-1}\\ \text{Length of the trolley, l}=\text{1}0\text{m}\\ \text{Speed of the\hspace{0.17em}child, v}‘‘={\text{4 ms}}^{\text{-1}}\\ \text{Time taken by the child to run\hspace{0.17em}over\hspace{0.17em}the\hspace{0.17em}trolley,\hspace{0.17em}}\mathrm{t}=\frac{10}{4}\\ =2.5\text{\hspace{0.17em}}\mathrm{s}\\ \therefore \text{Distance travelled by the trolley}=\text{v}‘‘×\text{t}\\ =\text{1}0.\text{36}×\text{2}.\text{5}\\ =\text{25}.\text{9 m}\end{array}$

Q.29 Which of the following potential energy curves in Fig. 6.18 cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Ans.

The potential energy of a system of two masses is inversely proportional to the distance between them. In this case, the potential energy of the system of two balls will decrease when they come closer to each other, and it will become zero (i.e., V(r) = 0) when the two balls touch each other, i.e., at r = 2R; V(r) = 0, where R is the radius of each billiard ball. Out of the given potential energy curves, curve (v) only satisfies these two conditions. Hence, all other curves do not describe the elastic collisions of two billiard balls.

Q.30 Consider the decay of a free neutron at rest: n → p+ e
Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (Fig. 6.19).

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of β-decay. This particle is known as neutrino. We now know that it is a particle of intrinsic spin ½ (like e, p or n), but is neutral, and either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p +e+ ν]

Ans.

The decay process of free neutron at rest is given by,
n → p +e
According to Einstein’s mass-energy relation, the energy of electron is equal to Δmc2
Where,
Δm = Mass defect = Mass of neutron – (Mass of proton + Mass of electron)
Here, c is the speed of light
Here, Δm and c are constants. Hence, the given two-body decay cannot explain the continuous energy distribution in the β-decay of a neutron or a nucleus. The presence of neutrino on the LHS of the decay describes the continuous energy distribution correctly.

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