Rahul sardana Mechanics 2 JEE

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text overview, goals & focus

In the past few years, the IIT-JEE has evolved itself as an examination designed to check out true scientific skills. The examination pattern wants us to see those little details which, others fail to see. Tlrose details which tell ushow much in depth we should know to describe asmuch aspossible. Keeping the present day scenario in mind, this book is written for students, to allow them not only to learn the tools that "Mechanics" provides but also to see

why they work so nicely inexplaining the beauty of ideas behind the subject. The central goal ofthis text isto help .the students develop a thorough understanding of the principles of"Mechanics". This book stresses on building a rock solid technical knowledge based on firm foundation of the fundamental principles than on a largecollection offormulae. The primary philosophy of this book is to act as a guide who creates a careful, detailed groundwork for strong conceptual understanding and development of problem solving skills like mature and experienced physicists. features of book

theory with illustrations

"Mechanics" is an important topic, and in thisbook I havetried to make thistopic lively, clear and precise to

the greatest levels. I have generally seen students not stressing on the theoretical details. They always feel that doing more numerical problems will solve their purpose. But letme tell you here, that numerical problems are just the special cases of the theoretical concepts. The' entire Physics is based on a simple program "IF THEN -> ELSE". Tryto follow this and seehow you get to your ultimate goal i.e., IIT-JEE. So^ keeping this in mind, the entire theory part of all the chapters has been kept elaborative, simple to understand with supportive Illustrations at all the places. DO NOT TRY TO ATTEMPT ILLUSTRATIONS WITHOUT GOING THROUGH THE THEORY.

conceptual notes, remarks, words of advice, misconception removal '

Throughout the text, the Conceptual Notes andRemarks are liighlighted which focus onthe principal ideas and concepts that a student must take care of. Places where students commonly develop a misconception have been supported by Misconception Removal, highlighted in grey' and supported by Words of Advice. Throughout my teaching career of18 years I have always foimd my students getting benefitted from these Conceptual Notes, Remarks, Words of Advice, Misconception Removals. All these are actually used to provide warnings to the students about common errors and ways to avoid them. problem solving techniques

..

These techniques, highlighted in grey, always ensure thatlhe students become capable of solving a variety of problems in an easy way. Wherever necessary, the text is supplemented with them for having a thorough understanding to the application processes. in chapter exercises (Ice); topic wise

After youstudy the theory andapply it to the Illustrations, its time you practice something onyour own and that too topic wise. For this purpose I have created In Chapter Exercises (ICE) (except for Mathematical Physics). Each ICE has the name of the topic(s) clearly mentioned on it. Please note that ICE are based on simple, single

concept classification technique. They arefully solved, so thatif youcome across only problem, then youjustrefer to the solutions.

solved problems

After you have gone through the entire Theory (with Illustrations) and all the supplements (ICE, Conceptual Notes, Remarks, Words of Advice, Misconception Removal, Problem Solving Techniques), its high time to do problems that are a true mix of concepts studied. This section has problems that involve multiple concept usage so that your brain is exposed to the ultimate throttle required to extract the best from you at IIT-JEE.

practice exercise sets (fully solved) Now comes the time when you are very much ready to do the practice as per the IIT-JEE pattern. This section contains all the variety of questions that have been asked in the IIT-JEE. In this section you will come across the following variety of questions. single correct choice type (SCCT)

Each question, in this section, has four choices (A), (B), (C) and (D), out of wliich ONLY ONE is correct. multiple correct choice type (MCCT)

Each question, in this section, has four choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

reasoning based questions / assertion-reason type (ART)

This section contains Reasoning type questions, also called Assertion-Reason type question, each having four choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Each question contains STATEMENT1 (Assertion) and STATEMENT 2 (Reason). You have to mark your answer as Bubble (A) If both statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1.

Bubble (B) If both statements are TRUE but STATEMENT 2 is not the correct explanation of STATEMENT 1. Bubble (C) If STATEMENT 1 Is TRUE and STATEMENT 2 is FALSE. Bubble (D) If STATEMENT 1 is FALSE but STATEMENT 2 is TRUE. linked comprehension type (LCT) / paragraph type

>

This section contains Linked Comprehension Type Questions or Paragraph based Questions. Each set consists of a Paragraph followed by questions. Each question has four choices (A), (B), (C) and (D), out of which only one is correct. (For. the sake of competitiveness there may be a few questions that may have more than one correct options). matrix match type (MiVlT) / column matching

Each question in this section contains statements given in two columns, which have to be matched. The statements in COLUMN-I are labelled A, B, C and D, while the statements in COLUMN-II are. labelled p, q, r,

s (and t). Any given statement in COLUMN-I can have correct matching with ONE OR MORE statement(s) in COLUMN-H. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following examples: If the correct matches are A ^ p, s and t; B darkening of bubbles will look like the following :

q and r; C P

A

® ® © ®

(9) © ® ®

p and q; and D -> s and t; then the correct

r

s

t

© © © ©

© © © ©

® ® ® ®

r

\

integer answer type questions (lATQ) / numerical type questions

In this section the answer to each of the question is a four digit integer, ranging from 0 to 9999. The

appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answer to question number X (say) is 6092, then the correct darkening of bubbles will look like the following: X.

CS) •

(S> (H)

CD CD (D CD

(D (D d) • (3) (2) (3) CD

CD 0

(D (3)

CD ® ® CD B) CD CD

% (D CD CD CD answers & soiutions

Each chapter contains answers followed by solutions to the problems. The solutions are exhaustive with complete methods and reasons which will help you a lot to understand a particular concept. Short cuts are also included (wherever necessary) forenhancing youproblem solving skills.

This book, I hope, will nourish you with the concepts involved such that you get a great rank atIIT-JEE. To conclude, 1apologise in advance for the errors (if any) that may have inadvertently crept in the text. 1would be grateful to the readers who bring errors of any kind to my attention. 1truly welcome all comments, critiques and suggestions [email protected]. PRAYING TO GOD FOR YOUR SUCCESS AT llT-JEE, GOD BLESS YOU!

The Author RAHUL SARDANA M.Sc.(Hons.) Physics

vO

\

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I'-.'l-

•

• • •; •

CHAPTER 7

Work, Energy,Power& LawofConservation ofEnergy Work, Energy, Power &Lawof Conservation ofEnergy. Solved Practice Problems.

7.1

Answersto in Chapter Exercises (ICE) &PracticeExercise Sets

9.722

Solutions to InChapterExercises (ICE)

9.728

Solutions to Practice Exercise Sets

9.749

7.27 CHAPTER 10

Practice Exercise Sets

Gravitation 8eSatellites

o

Single Correct Choice Type Questions

7.34

0

Multiple Correct Choice Type Questions.

7.48

Gravitation & Satellites

70.7

9

Reasoning BasedQuestions (Assertion Reason Type)

7.52

Solved Practice Problems

o

Linked Comprehension Type Questions (ParagraphType)

7.55

Practice Exercise Sets

Z

Matrix MatchType Questions (Column Matching Type)

7.62

Z

SingleCorrectChoice Type Questions

70.58

7.65

Z

MultipleCorrectChoiceType Questions

70.49

7.68

z

Reasoning BasedQuesf/ons'Wssert/on Reason Type)

70.55

Solutions to InChapterExercises (ICE)

7.75

Z

Linked ComprehensionType Questions(Paragraph Type)

70.55

Solutions to Practice ExerciseSets

7.85

Z

Matrix MatchType Questions (Column Matching Type)

70.50

Z

IntegerAnswerType Questions

10.63

z

Integer Answer Type Questions

Answers to InChapterExercises (ICE) &Practice Exercise Sets

CHAPTERS

Centre ofMass, Conservation ofLinear Momentum & Collisions Centre ofMass

S.I

• Conservation ofLinear Momentum & Collisions Solved Practice Problems

8.15

-

8.59

Z Single Correct Choice Type Questions.

8.47

Practice Exercise Sets

Z

Multiple Correct Choice Type Questions

8.60

z

ReasoningBasedQuestions(Assertion ReasonType)

8.64

Z ' Linked ComprehensionType Questions(Paragraph Type)

8.66

z

Matrix MatchType Questions (Column Matching Type)

8.72

z

IntegerAnswer Type Questions

8.76

Answersto In Chapter Exercises (ICE) &PracticeExercise Sets

8.78

Solutionsto In Chapter Exercises (ICE)

8.82

-

Solutions to Practice Exercise Sets

8.95

CHAPTER 9

Rotational Dynamics RotationalDynamics Solved Practice Problems

•'••••9.1 9.55

Practice Exercise Sets

z • o

SingleCorrectChoice Type Questions

9.72

Multiple Correct ChoiceType Questions

9.95

Z

ReasoningBasedQuestions(Assertion ReasonType)

9.701

Z

Linked Comprehension Type Questions (Paragraph Type)

9.104

Z

MatrixMatch Type Questions(Column Matching Type)

9.775

Z

Integer Answer TypeQuestions

9.179

70.52

Answersto in Chapter Exercises (ICE) &PracticeExercise Sets

70.55

Solutionsto In Chapter Exercises (ICE)

10.69

Solutions to Practice Exercise Sets

-

70.85

Work, Energy, Power iP

& Law of Conservation

r**

of Energy

I

Contents

WORK. ENERGY, POWER &LAW OF CONSERVATION OF ENERGY Solved Practice Problems

1

7.1 7.27

Practice Exercise Sets

C Single Correct Choice Type Questions

7.34

O Multiple Correct Choice Type Questions

7.48

S Reasoning Based Questions (Assertion ReasonType)

7.52

C Linked Comprehension Type Questions (ParagraphType)

7.55

O Matrix Match Type Questions (Column Matching Type)

7.62

S

IntegerAnswer Type Questions

7.65

Answers to InChapter Exercises (ICE) &Practice Exercise Sets

7.68

Solutions to in Chapter Exercises (ICE)

7.73

Solutions to Practice Exercise Sets

7.85

no

Work, Energy/ Power

displacement, the calculated work ^so depends on the

INTRODUCTION

reference frame.

WORK DONE BY A CONSTANT i=ORCE

T

he work W done by a constant force F when the

Illustration 1

point of application

A box is moved over a horizontal'path by applying force F = 80 N at an angle 0 = 60° to the horizontal. What is the work done during the displacement of the box over a

of force

displacement Ar is defined as '

undergoes

a

'

W = FArcos0

.••(!)

distance of 0.5 km. Solution

By definition, W = Fscos0 Here F = 80 N, s = 0.5 km = 500 m, 0 = 60°

where 0 is the angle between F and Ar as indicated in figure. Only the component of F along Ar , that is, Fcos0> contributes to the work done. Strictly speaking, the work is

=>

W = (80)C500)cos(60°) = 20kJ

.

Illustration 2

done by the source or agent that applies the force. Work is a scalar quantity and its SI imit is the joule (J). From

an acceleration a = 2 ms"^.Find the work done during the

equation (1),we see that

first one and a half seconds &om the beginning of.motion.

lJ =lNm

'

Solution

From (1) we can also conclude that, work done is also defined as the dot product of force and its displacement as given by the following equation (2) IV = F-Ar

.

.

...(2)

In terms of rectangular components, die two vectors are

F- FJ +FJ +F^k and Ar - Axi +Ay; +Azk Hence, equation (2) may be written as

W=F-Ar =F,Ax +Fj,Ay+F,Az ^

' , •

A load of mass m = 3000 kg is lifted by a winch with

The height to which the body is lifted during the first t /

second is h = —at 2

,...(3)

The work done by a given force on a body depends only on the force, the displacement, and the angle between .them. It does not depend on the velocity or the acceleration of die body, or on the presence of other forces. • . Since the work is a scalar, its value also does^ not depend on ,the orientation of the coordinate axes. Since the,magnitude of

•

The tension in the rope is giyen by T = mg + ma

Work done is given by

'

T

I

w

IV =TTicos(0°) =m'( g+fl)^ af^

mg

Here m= 3000kg, fl = 2ms'"^, g = lp.ms'^,, f=1.5s W = (3000)(lO + 2) lV = 81kJ

a displacement in a given time, interval depends on the velocity of the frame of reference used to measure the 7.1

Advanced JEE Physics

Mechanics -11

Using ^F =mfl ,we get

Illustration 3

A block of mass w = 4kg is pulled by a force F = 20 N upwards through a height h = 2m. Find the work done on the block by the applied force F and its weight. Take g = 10 ms"^.

T-20 = 2fl => 'm

a-.

PVj =(r)(S)cos(0'*) =>

Solution

Weight of the block is

r = 24N

Work done by string (tension) on 2 kg block in 2 s is

W, =(24)(4)(1) = 96J

Similarly, work done by string on 3 kg block in 2 s will be Wj=(T)(S)(cosl80°)

= (4) (10) = 40 N

Work done by the applied force Wp =F?icos(0°)

W,=(24)(4)(-1) = -96J

The angle between force and displacement is 0°, so

Wp=C20)(2)(l) = 40J Similarly, work done by the weight of the block is

POSITIVE AND NEGATIVE WORK

Work done by a forcemay be positive or negative depending

^-.i; =('"5)(^)cos(l80°)

on the angle 9 between the force and displacement. If the

PV„^ =C40)(2)(-1) =-80J

angle 0 is acute (090°

.G

Illustration 4

Two

unequal

masses

Positive work done

(b) Negative work done

by a force F

by a force F

(a)

of

3 kg and 2 kg are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in

If the angle 0 is obtuse (0>9O°), then the component of force is antiparallel to the displacement and the work done by force is negative.

figure. If the system is released from rest, find the work done by string on both the blocks in 2 s.

ZERO WORK DONE

Take ^ =10 ms"^.

Since we know that PV = FArcos0, so the work done by a force is zero when

Solution

The acceleration of the blocks in the system is /•

S

m^-m2

(a)

EITHER F = 0

(b)

OR Ar = 0

(c)

OR cos0 = O i.e. 0 = 90° N

3-2 a-

Wf = 0^--

V

10 = 2 ms ^

3+ 2

W, = 0 W''

Displacement of both the blocks in 1 s is

y^

S = -/if^=-(2)(2)^-4m 2

2

Free body diagram of 2 kg block is shown in figure.

(a) (b) When the force and the displacement are perpendicular, the work done by the force is zero. (a) the normal reaction N is perpendicular to displacement, therefore, Wf^=0. (b) the centripetal force is perpendicular to displacement, thusWF=0.

WORK DONE BY FRICTION 3 kg

2 kg

2 kg

There is a naisconception that the force of friction always does negative work. In reality, the work done by friction may be zero, positive or negative depending upon the situation.

2g 7.2

20 N

Work, Energy,Power & Law of Conservation of Energy

In figure (a), when a block is pulled by a force F and the block does not move, the work done by friction is zero.

A

s = 0 —

pr

Let us assume the coordinate axes as shown in the figure, to specify the components of the two vectors - although the value of work will not depend on the orientation of the axes.

Now, the forceof gravity, F^ = -mgj

fS

I-- —

^

f«-«—

W

and the displacement is given by Ar Axi + Ayj+ Azk

(b)

(a)

The work done by gravity is

=F^ •Ar =-mgj •(Axi +Ayj +Azic) => Wg=-mgAy {•.•]-1 =0,] •] =!,]•k =0} Since Ay =

- i/j = -h

W^=-mg{yf-yi) =+mgh (c)

In figure (b), when a block is pulled by a force f on a stationary surface, the work done by the kinetic friction is

If the block moves in the upward direction, then the work

done by gravity is negative i.e., work is done against the gravitational pull and is given by W = -mgh

negative.

In figure (c), block A is placed on the block B. When the block A is pulled with a force F, the friction force does negative work on block A and positive work on block B.

a)

The kinetic friction and displacement are oppositely directed in case of block A while in case of block B they are in the

b)

same direction.

DEPENDENCE OF WORK ON FRAME OF REFERENCE

Work as defined, depends on frame of reference also. When we change from one inertial reference frame to another inertial reference frame, the force does not change, while displacement may change, so the work done by a force will

CONCEPTUAL NOTE(S) The work done by the force of gravity depends only on the Initial and final vertical coordinates, not on the path taken,

The work done by gravity is zero for any path that returns to its initial point. When several forces act on a body one may calculate the work done by each force individually. The net work done on the body is the algebraic sum of individual contributions.

W„«=F,.A^ + F2-a^+

^

=

+ F„-A^,

+W„

be different.

WORK DONE BY A VARIABLE FORCE

FOR EXAMPLE

When the magnitude and direction of a force vary in three

Suppose a person is pushing a box in a moving train (in the

dimensions, it can be expressed as a function of the position

same direction as the movement of train) by applying a force

vector F(r), or in tenfis of the coordinates F(x,y,z). The work done by such a force in ah infinitesimal displacement

F on it. In the reference frame of the train, the work done by the force will be F-f where f is the displacement of the block with respect to the train. But in the reference frame of the

earth the work will be F*(f +fo) where 7^ is the displacement

df is dW^F-dr

of the train with respect to the earth.

WORK DONE BY GRAVITY Consider a block of mass m which slides down a smooth

inclined plane of angle 0 as shown in figure. s

A particle moves along a curved path subject to a

variable force F. The work done by the force In a displacement ds is dW = F; dfi

The total work done in going from point A to point B as shown in the figure, i.e.. The work done by gravity is W = -mg(y, - y,) = +mgh

7.3

Advanced JEE Physics B

B

A

A

Mechanics - II

F=(21 +3/ +4k) N. Calculate the work done by this force

== Jp'rfr =J(Fcos0) dr

in the process.

In terms of rectangular components,

Solution

F=Fj+Fj +F,k

W=jf-d?

and dr = idx+ jdy + kdz therefore, yn

Ze

•

^A^B =jF^^x+.jF^dy+jF^dz

= J [li+3j+4:k)-{dxi +dyj+ dzk)

w

(2.3,4)m

Va

w=(2x+3y..4z)i;;-;;;>-9j

Illustration 5

A force F =(5+2x+3a:^) acts on a particle in X-direction where F is in newton and x in metre. Find

Alternate Solution:

Since, F = constant, we can also use.

the work done by this force during a displacement from A: = lm to A:= 2m.

VS/ = F-Ar

Here, Ar =^-5

Solution

Ar =(i+2j+3k) - (21 +3]+4k) =I-i-i-k.

As the force is variable, we shall find the work done in a

smalldisplacement from x to x+dx and then integrateit to find

the total work.

The work done in

this small

Illustration 8

displacement is

An

2

2

Thus, IA' =JdPV =|(5+2x+3x')dx lV = 5(l) + (4-l) + (8-l)

is

displaced from

position vector

force. Solution

=> W=(5x +x^+x^)|i =>

object

r, =(2t +3/) m to position vector =(4r +6^) m under a force F=(3xh'+2yy) N. Calculate the work done by this

dVV=Fdx = (5 + 2x + 3x=^)dx

IV = 15J

W

=jp-dr ,where dr =idx +jdy+icdz *2

Illustration 6

W=j[3x^i +2yj)-{idx+jdy+kdz)

A force F = —j-Cxi^O) acts on. a particle in xdirection. Find the work done by this force in displacing

'2

IV =j(3xVx+2ydy)

the particle from x = a to x = +2a, where fc is a positive constant.

(4.6)

Solution

W= J. rf(i')+ i rf(y=)=(A;'+y^)|[*; 'la

I PROBLEMS SOLVING TRICK

I Itis important to note that work comes out to be negative which Is quite obvious as the force acting on the particle is In negative x-

direction I

while displacement is along positive x-

direction (from x = a to x = 2a)

(2. 3)

=>

(2.3)-

W = 83]

illustration 9

An object is displaced from a point A(0, 0, O) to

B(1 m, 1m, 1m) under aforce F=(yf +x;) N. Calculate the work done by this force in this process. Solution B

illustration 7

An object is displaced from point 'A(2,3,4) m to a point B(1, 2,3) m under the influence of a constant force 7.4

ZTTT^

W

=jp-dr. where dr = idx + jdy + kdz

Work, Energy, Power &Law of Conservation ofEnergy B

WORK DONE AS AREA UNDER F-x GRAPH

A

Graphically, the work done by the spring force in a displacement from at,- to Xf is the shaded area (as shown in the figure) which is the difference in the areas of two

1^=J(yi +xj)-(Ux+Jdi/+kdz] B

B

W=j(ydx+xdi/} =jd(xy}

triangles.

^ ^=(^^)i(o;o!o) =>

W=1J

WORK DONE BY A SPRING

If .t be the displacement of the free end of the spring from

its equilibrium position then, the restoring force (Fj) in the spring is given by F.^-kx

The

negative

(a) The work done by a non-constant force is approximately

sign

equal to sum ofthe areas ofthe rectangles.

signifies that the restoring

(b) Theareaunderthecurveisgivenbytheintegral W=|F(x)dx

force is always opposite to the extension (x > 0) or the compression

In general, thework done by a variable force F(a:) from an initial point x, to final point x^ is given by the area under

(x 2m^^™=|Kx^

'O

Solution

If I is the stretched length of the spring, then from figure

-"K

(b) At x=^

| =cos(37°)=|

2

^(say)/ then

rC

Then,

- '=f So the extension in the spring is

Decrease in

'' Increase'

gravitational potential

in elastic

x=/-d =^-d=^and

potential

=

^

in kinetic +

energy of

energy

both the

.of spring.

blocks

energy

of block

^ Increase

Now taking point B as reference level and applying Law of Conservation of MechanicalEnergy between A and B,

(U+K),„={U +K)„. . =5>

1 + —kx l,-2 =—my 1 2 mgh

=>

«

Im

f = 2