Cengage MECHANICS 1

I-,;;;crm• ng•5 Series™ for IIT-JEE 2012-13

Contents Chapter 1 functions 1.1 Chapter 2 Vectors 2.1 Elementary Algehra 1.2 Scalars 2.2 Common Formulae 1.2 Vectors 2.2 Polynomial, Linear, and Quadratic Equations 1.2 Representation of Vector 2.2 Binomial Expression and Theorem 1.2 Notation of Vector 2.2 Elementary Trigonometry 1.3 Introduction to Different Types of Vectors 2.2 System of Measurement of an Angle 1.3 Collinear Vectors 2.2 Four Quadrants and Sign Conventions 1.3 Equal Vcctors 2.2 The Graphs of sin and cos Functions 1.4 Negative of a Vector 2.2, Trigonomctrical Ratios,of Allied Angles 1.4 Coplanar Vectors 2.3 Inverse Trigonometric f<'tmctions 1.5 Unit Vector 2.3 Basic Coordinate Geometry 15 Position Vector and Displacement Vector 2.4 Origin 1.6 Resultant Vect.or 2.5 Axis or Axes 1.6 Addition of Vectors 25 Position of a Point 1.6 How to Add Two Vectors Graphically (Tip to Tail Method) 2.5 Straight Line Equations 1.7 Triangle Law of Vector Addition 2.6 Parabola: The Quadratic Equations 1.9 Parallelogram Law of Vector Addition 2.6 Differentiation LlO Addition of More Than Two Vectors 2.6 Differential Coefficient or Derivative of a Function 1.10 Veetor Addition by Analytical Method 2.6 Geometrical Interpretation of the Derivative of a Function 1.11 Condition for Zero Resultant Vectors 2.8 Properties of Derivatives 1.11 Lami'sTheorcm 2.8 Derivatives of Some Important F'ul1ctions 1.12 Rectangular Components of a Vector In Two Dimensions 2.9 Maximum and Minimum Values of a Function 1.12 Rectangular Components of a Vector In Three Dimensions 2.9 Use of Maxima and Minima in Physics 1.12 Product of Two Vcctors 210 Integral of a Function 1.I3 Scalar or Dot Product 2.10 Properties of Indcfinite Integral 1.13 Vcctor or Cross Produet 2.11 Standard Formulae for Integration 1.I3 Cross Product Method 1: Using Component Form 2.12 Definite Integral of a Function 1.15 Cross Product Method 2: Determinant Method 2.12 Algebraic Method to Evaluate Definitc Integral 1.15 Properties of Cross Product 2.13 Propelties of Dcfinite Integral 1.15 Solved Examples 2.13 Geometrical Significance of a Definite Integrate 1.15 Exercises 2.16 Geometrical Method to Evaluate Definite Integral 1.15 Subjeclive Type 2.16 Application in Physics 1.16 Objective Type 2.17 Derivation of Linear Kinematical Equations Multiple Correct Answers Type 2.21 using Calculus 1.17 Answers and Solutions 2.22 Subjective Type 2.22

6 Contents 2.25 Translatory Motion 4.2 2.29 Rotatory Motion 4.3 Objective Type Multiple Correct Answers Type 3.1 Oscillatory Motion 4.3 Chapter 3 Units and· Dimensions 3.2 Position Vector and Displacement Vector 4.3 3.2 Systems of Units 3.3 Position Vector 4.3 Dimensions of a Physical Quantity 3.3 Displacement Vector 4.3 Dimensional FOlTImlae 3.5 Displaccmentand Distance 4.3 Uses of Dimensional Analysis 3.5 Significant Figures 3.6 Velocity (Instantaneous Velocity) 4.4 3.6 Rules for Counting Significant Figures 3.7 Speed (Instantaneous Speed) 4.4 Rules for Rounding off the Uncertain Digits 3.7 Significant Figures in Calculations 3.7 Unifonn'Motion 4.5 Errors in Measurements 3.7 Features of Uniform Motion Systematic Enors 3.7 45 Random Errors 3.8 Gross Errors 3.8 Accelerated Motion 4.6 Absolute Errors 3.8 Propagation of Combination ofErrors 3.8 Average Acceleration 4.6 Errol' in Summation 3.8 Error in Difference 3.8 Acceleration (Instantaneous Acceleration) 4.6 Error in Product 3.9 En·or in Division 3.11 Uniform Acceleration 4.7 Error in Power of a Quantity 3.15 Accuracy ilOd Precision 3.15 Variable Acceleration 4.7 Solved Examples 3.14 Exercises 3.21 Formulae for Uniformly Accelerated Motion in a Straight Line Suhjective Type 3.22 4.7 Ol,iective Type 3.23 Multiple Correct Answers Type 3.23 Use of Differentiation and Integration in One-dimensional ~ Assertion-ReasOf~ing Type 3.24 M~oo Comprehensive Type 3.26 Matching-Column Type 3.26 Derivations of Equations of Motions by Calculus Method Archives 3.27 4.10 Answers and Solutions 3.32 Su/Jiective Type 3.33 One-dimensional Motion in a Vertical Line (Motion Under Objective Type 3.33 Multiple Correct Answers Type 3.33 Gravity) 4.11 Assertion-Reasoning Type 3.34 Comprehei'l.'live Type Some Formulae 4.11 Matching Column Type 4.1 Archives Graphs in Motion in one Dimension 4.14 4.2 Chapter 4 Motion in One Dimension 4.2 How to Analyse the Graphs and How to Draw the Graphs 4.14 4.2 Frame of-Reference 4.2 Position-Time Graph of Various Types of Motions of a Motion in One Dimension 4.2 Motion in Two Dimensions Prntic1e 4.14 Motion in Three Dimensions Velocity-Time Graph of Various Types of Motions of 4.15 Rest and Motion a Particle Acceleration-Time Graph of Various Types of Motions of a Pmtic1e 4.16 Derivation of Equations of Uniformly Accelerated Motion from Ve1ocity-Time Graph 4.16 Solved Examples 4.18 ~~u ~U Subjective Type 4.26 O~iective Type 4.30 Graphical Concepts 4.35 Multiple Correct Answers Type 4.39 Assertion-Reasoning Type 4.41 Comprehensive Type 4.41 Matching Column Type 4.45 Answers and Exercises 4.46 Subjective Type 4.46 Ol,iective Type 4.52 Graphical Concepts 4.58 Multiple Correct Answers Type 4.59 Assertion-Reasoning Type 4.61

.. CENGAGE __Learning'- -_.- ........ ......,_._.. Physics for IIT-JEE 2012·13: © 2012, 2011, 2010 Cengage Learning India Pvt. Ltd. Mechanics I ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be B.M. Sharma reproduced, transmitted, stored, Of used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, Of information storage and retrieval systems, without the prior written perf!1ission of the publisher. For permission to use material from this text Of product, submit all requests online at www.cengage.com/permissions further permission questions can be emailed to [email protected] ISBN·13: 978-81-315-1490-0 ISBN·I0: 81-315-1490-0 Cengage learning India Pvt. ltd 418, FLF., Patparganj Delhi 1'1'0092 Cengage Learning is a !eading provider of customized learning solutions with office locations around the g!o~e, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. locate your local office at: www.cengage.com/global (engage Learning products are represented in Canada by Nelson Education, ltd, For product information, visit www.cengage.co.in Printed in India First Impression 2012

Contents 7 Comprehensive 7}pe 4.61 Archives 6.12 Matching Column Type 4.63 Answers and Solutions 6.14 Ans}vers and Solutions 4.22 Objective Type 6./4 Su/\"jective Tvpe 4.22 Multiple Correct Ansl-vers Type 6.19 Objective Tvpe 4.25 Comprehensive Type 6.21 Multiple Correct Answers Type 4.29 Assertion-Reasoning Type 6.24 Chapter 5 Motion in Two Dimensions Matching Colurnn Type 6.24 6.25 5.1 Archives Relative Velocity 5.2 7.1 Graphical Method to Find Relative Velocity 5.2 Chapter 7 Newton's laws of Motion When One Body Moves on the Surface of Other Body 5.4 Introduction 7.2 The Concept of Force 7.2 Problem Solving Tips for Relative Velocity 5.6 Classification of Forces 7.2 Motion with Uniform Acceleration in a Plane 5.7 Displacement 5.7 Newton's Laws of Motion 7.2 7.2 Proiectile Motion 5.8 Newton's First Law of Motion 7.3 7.4 Projectile Given llorizontal Projection 5.11 Newton's Second Law of Motion 7.4 Kinematics of Circular Motion 5.13 . orNewton's Third Law Motion Uniform Circular Motion and Non-uniform Circular-Motion5.] 3 Impulse Angular Position and Angular Displacement 5.13 Some Examples of Impulse: Force Exerted by Liquid Jet ou Angular Velocity 5.13 Wall 7.6 Non-uniform Circular Motion 5.14 Analysis of Uniform Circular Motion 5.14 Free Body Diagrams 7.7 . Centripetal Acceleration 5.15 Relative Angular Velucity 5.16 Weight 7.7 Solved Examples 5.17 Exercises 5.30 Normal Force 7.8 Suf~jecf;ve Type 5.30 Objective 7)1'e 5.33 Tcn:sion 7.8 Friction 7.9 Elastic Spring Forces 7.10 Non-inertial Frame of Reference and Pseudo (Fictitious) Force 7.10 A1ulfiple Correct Answers T.}})e 5.41 Equilibrium of a Particle 7.11 Assertion-Reasoning T.}1J(! 5.42 Concurrent Forces 7.11 Comprehensive Type 5.43 Lamy's Theorem 7.11 Matching Co/ullin 7)I]JC 5.46 Constraint Relation 7.20 Answers and Erercises 5.47 General Constratints 7.21 Sul~iect;ve lSlpe 5.47 Writing Down ConstraintS-Pulley 7.22 O/~jec{ive Type 5.53 Wedge Constraint 7.26 Mulaple Correct Annvers Type 5.62 Pulley and Wedge Coustraint 7.27 Assertion-Reasoning Type 5.63 Spring Force and Combinations of Springs 7.31 Comprehensive 7~'vpe 5.64 Force Constant of Composite Springs 7.32 Matchin.g Co/ullin Type 5.69 Analysis of Friction Force 7.33 Chapter 6 Miscellaneous Assignments and Laws of Limiting Friction 7.34 Angle of Friction 7.34 Archives on Chapters 1-5 6.1 Dynamics of Circular Motion 7.45 Exercises 6.2 Concept of Centripetal Force 7.45 Ol~jective Type 6.2 Centrifugal Force 7.45 Multiple Correct Answers Type 6.4 Solved Examples 7.52 Comprehensive Type 6.8 Exercises 7.62 Assertion-Reasoning Type 6.11 Subjective Type 7.62 i'v1atching Column 7)'JJe 7.65 6./5 Objective Type 7.92 Multiple Correct Answers Type

Brief Contents Chapter 1 Basic Mathematics Chapter 2 Vectors Chapter 3 Units and Dimensions Chapter 4 Motion in One Dimension Chapter 5 Motion in Two Dimension Chapter 6 Miscellaneous Assignments and Archives on Chapters 1-5 Chapter 7 Newton's laws of Motion

8 Contents 7.97 Objective Type 7.125 7.99 Multiple Correct Answers Type 7.150 Assertion-Reasoning Type 7.111 Assertion-Reasoning Type 7.155 Comprehensive Type 7.115 Comprehensive Type 7.156 Matching Column Type 7.ll8 Matching Column Type 7.165 Archives 7.118 Archives 7.169 Answers and Solutions Subjective Type

Preface Since the time the IIT-JEE (Indian Institute of Technology Joint Entrance Examination) started, the examination scheme and the methodology have witnessed many a change. From the lengthy subjective problems of 1950s to the matching column type questions of the present day, the paper-setting pattern and the approach have changed. A variety of questions have been framed to test an aspirant's calibre, aptitude, and attitude for engineering field and profession. Across all these years, however, there is one thing that has not changed about the lIT-JEE, i.e., its objective of testing an aspirant's grasp and understanding of the concepts of the subjects of study and their applicability at the grass-root level. No subject can be mastered overnight; nor can a subject be mastered just by formulae-based practice. Mastering a subject is an expedition that starts with the basics, goes through the illustrations that go on the lines of a concept, leads finally to thc application domain (which aims at using the learnt concept(s) in problcm- solving with accuracy) in a highly structurcd manncr. This series of books is an attempt at coming face-to-face with the latest IIT-JEE pattern in its own format, which is going to be highly advantageous to an aspirant for securing a good rank. A thorough knowledge of' the contemporary pattern of the liT-lEE is a must. This series of books features all types of prohlems asked in the examinat.ion-he it MCQs (one or more than one correct), assertion reason t.ypt;, matrix match type, or paragraph-based, thought-type questions. Not discounting to need for skilled and guided practice, the material in the book has been enriched with a large number of fully solved cnncept-application exercises so that every step in learning is ensured for the understanding and application of the subject. This whole series of books adopts a multi-facetted approach to mastering concepts by including a variety of exercises asked in the examination. A mix of questions helps stimulate and strengthen multi-dimensional problem-solving skills in an aspirant. Each book in the series has a sizeable portion devoted to questions and problems from previous years' IIT-JEE papers, which will help students get a feel and pattern of the questions asked in the examination. The hest part about this series of books is that almost all the exercises and problem have been provided with not just answers but also solutions. Overall the whole content of the book is an amalgamation of the theme of physics with ahead-of-time problems, which an aspirant must follow to accomplish success in IIT-JEE. B. M. SHARMA



Basic Mathematics 1.1

1.2 Physics for IIT·JEE: Mechanics I Roots of a quadratic equation are generally represented hy Q' Mathematics is the supporting tool of physics. The elemen- °and ,8. tary knowledge of basic maths is useful in problem solving in Let (lX 2 + hx + c = be a quadratic equation. Then: physics. Basic knowledge of elementary algebra, trigonometry, coordinate geometry and basic calculus is must before going -b + ~b2 - 4ac -b - ~cb~2---4-a-c into the depth of physics. 1. Its roots arc a = _.\",--,- -~;,8 = 2a 2a , ELEMENTARY ALGEBRA -b ± ~b2 - 4ac Common Formulae 1. (a + h)2 = a2 + Ii + 2ab 2. Hence, its solution is given by x = .-'~~----. 2. (a - b)2 = a 2 + b2 - 2ab 2a 3. (a + b + e)2 = a 2 + b2 + e2 + 2ab + 2hc + 2ea 3. Sum of its roots is given hy a + f3 = .~b. 4. (a +h)(a - b) = a2 - I i a a'5. (II + b)' = + I,' + 3ah(a + b) 4. Product of its roots is given by af3 = .:. 6. (a - b)' = a' - b' - }ab(a - b) 7. (a +b)2 - (a - b)2 = 4ab (/ 8. (a + b)2 + (a - b)2 = 2(a2 + b') . ·)b2 - 4ae S. Difference of its roots is given by a - j3 = ~-~--. Polynomial, Linear, and Quadratic Equations a Real Polynomial Binomial Expression and Theorem Let an, (lj, (i.2 ,\"', all be real numbers and x a real variable, then An algebraic expression containing two terms is ca11ed a bino~ f(x) = ao + ([IX + alx 2 + ,.. + aI/xII is called a real'po/yno- mial expression. mial. 3y), + ~) , +~). etc., Degree or Index of a Polynomial b). (2x - (x (xFor example, «(I + The highest power appearing in a polynomial is called its degree, are binomial expressions. For example, J(x) = x' + 8x + 3 is a polynomial of degree 3. Binomial Theorem for Positive Integral Index Students must note here that it is not necessary that the The general form of a binomial expression is (x + ay, where highest power must be of a single variable only. For example, 1l is any positive integer (caBed index) and x and a are real f(x) = 3x2y + y2 + 2 is a polynomial of degree 3 because of numbers. variable y in the term x 2y. We add powers of the variables in a term to find degree of a polynomial irrespective of the miture Binomial theorem states: of variables. Thus, in the present case, xly is having power of 2 + I :.:: 3 and hence the degree of thc given polynomial is 3. + + + + '\" +(x at = IICoxll nC;!XIf - 1 • (l1 IlC2Xu ..-2 • a 2 Linear Equations + .,. +/I Crxl!·-r . ar Il C • all Il Equations havili.g polynomials of unit degree arc called linear where \"e,= --.n-! ~- andn! = n(11 -1)(n - 2)-··,3 x 2x I , equations, e.g., x + .V ::: 2 or 2x ,+ 3 ::: 5. Such equations always r!(n _1')1 represent a straight line on a graph. the product of first n natural numbers [e,g., 5 ! ::: 5 x 4 x 3 x 2 Quadratic Equations x 1=1201. Equations of second degree arc called quadratic equations. The r---'J---f,mpartantpoints \". general form of a quadratic equation is as given below: • Total number of terms in the expansion = (n + I). ax 2 + hx + c = 0, where a \"# 0. • In every successive term in the expansion, the power of Roots of a Quadratic Equation x goes on decreasing by I and that of a increasing by I, Solutions of a quadratic equation are 'called its roots. Roots are so that the sum of the powers of x and a in each term is those values ofa variable such as x for which the given quadratic equation collapses to zero. As a rule, a quadratic equation always always equal to n. has two roots which mayor may not be equal. • nco, ncIrC!, 2, ... , nCIl are called binomial coefficients. • The expression If! is read as \"factorial n\". So, \". nl n(n--J)(n-2) .. ·3x2xl C1=-----= . -\" =!l.. I!(n-I)! l(n-J)(n-2) .. ·3x2x I J ,Su\"mIary, \"C', = 2!(n11_1 2)! = .n.2(Xn-lI.)....... ,\"C.\" = I. Binomial Theorem for Any Index It'll is positive, negative or fraction and x is any real number such that -1 < x < l, i.e., x lies between -1 and +1, then according to the binomial theorem +I) 2 n(n - I)(n - 2) , (l + x)\" = I + nx + n(n - x --\"~-\"'---X' 3' 2! + ... 00 terms,

Note: Basic Mathematics 1.3 .. Ifn is a positive integer, thentiteexpansioll will have 2. Centesimal system: In this system, (n +1) terms. I right angle = 100 grades (100 g) 1 grade = 100 minutes (100') • lill is anegative ilitegerorajractioll, then the nu/!lber 1 minute = 100 seconds (100\") afterms in the expansion will he infinite, i,e, there will 3. Circular systern: In this system, angle is measured in radian.. rr radians = 180' be no last term. Consider a particle moves from a position P to position Q e Iflxl «1, then ()1!ly.firsttwo termsofthe expansion along a circle of radius r with centre at 0 (sec Fig. 1.1). Then, atesignificant.llis so because the values ofthe secolld Angle () = ~A-rc..1.-en-gt-hP-Q- = .s.. =} s = rO andlligher order terms become verysmall and call Radius of circle r ben.gleeted. Thu\" ill this case, the.binomial expan- p siOn reduces. to the jollowingsimplifie4jorllls When Ixl «1: .. (l+x)n =.1 +nx, (1 +x)-n '\" 1- nx, (1- x)u=l_ nx, and (1- x)-n =1+ nx. Calculate (1001)'/3. Sol. We can write 100 I as: IOC)] = 1000 (I + jolOO), so that Fig. 1.1 ff the length of arc PQ::::::: radius of circle r, thcn 0-::::::: 1 radian. we have ' I ) ] I ](lOOI)'!3 = 1000 1 + --- 1/3 = 10 [ 1+·_·- 1/3 [ ( 1000 1000 Radian = 10(1 + 0.001)'/3 = 10(1 + ~ x 0.(01) eWhen a body completes one rcvolution, then = 2n rad. 3 = H),003333 2rr rad = 36()\" or 2 x3, 14 rad = 36(Y' 360\" Expand (1 + x)-'. 1 rad = - . - - = 57.3·' 2 x 3.14 Sol. _ 1+ (-3)x + ~(--3)-(-:32-!l\")x-2 Four Quadrants and Sign Conventions (1 +x) 3 = Consider two mutually perpendicular lines intersecting at O. + (-3)(-3-1)(-3-2) 3 +.,' These two mutually perpendicular lines divide the plane into four parts called quadrants (sec Fig. 1.2). 3! x· 12 - --60-x-3 + ... =1-3x+ 2 3x2 y p\" = I - 3x + 6x 2 - 10x3 + ' , ' ,,,,,, e ..------1 Concept Application Exercise 1.1 11-----, x , '8 p N X x' N x 0 ,, 0 xr1. Expand (1 + 2 , 2. Using binomial expansion, simplify the following cxprcsM y' (1 ~::) 1],sion: Q [ + (a) (b) 3 ._ assuming 6x to be small in y y comparison to x. , X' _ +N- o '0- - - - + X X'_--+-er..OL..-N+--+X +t,: , ,, ElEMENTARY TRIGONOMETRY , \"p System of Measurement of an Angle p y' Y' There are eli fferent types of measurement of an angle. (e) (d) 1. Sexagesimal system: In this system, 1 right angle = 90\" (degree): 1 degree = 60' (minutes) Fig. 1.2 1 minute = 60\" (seconds)

1.4 Physics for IIT-JEE: Mechanics I Pointsto Remember 2. Addition and subtraction formulae: a. sin(A ± B) = sin A cos B ± cos A sin B I 1. To determine the sign of a trigonometrical ratio in any h. costA ± B) = cos A cos B 'f sin A sin B quadrant, OP will be taken as positive in all [our quadrants, tanA±tanB 2. In first. quadran,t, all trigonolnctrical ratios are positive (see c. tan(A ± B) = :-~----:---;c Fig. 1.3). 'f tan A tan B y 3. Multiple formulae: a. sin2A = 2sinAcosA h. cos 2A = cos' A - sin' A c: cos2A = I - 2sin' A = 2eos' A-I Oi) (i) 2tanA sin and cosec AI! +vc d. tan2A = I- 2 I'Ve tan A X' X (iii) (iv) Trigonometrical Ratios of Allied Angles tan and cot cos and sec eThe angles whose sum or difference with angle is zero or a +vc +ve multiple of 90\" are called angles allied to O. Commonly used }\" T -ratios of some of the allied angles arc given below, Fig. 1.3 1. sine -0) = -sin 0 3. In second quadrant, only sin 0 and cosec 0 arc positive, ecosec(-0) = -cosec e e4. In third quadrant, only tan and cot are positive. COSt-0) = cos (J e5. In fourth quadrant, only cos 8 and sec are positive. sect-0) = sec 0 I 6. The evalue of sin and cos e arc such that -1 :s sin e etan(-0) = -tan lJ· :<: 1 and-l :<:cosO:<: 1. But tan 0 and cot 0 can take a_~2' real value. cot(-8) = -cote Note: '. The Graphs of sin and cos Functions .. Astingle-()lie~in Ihefoutth qu.lldrqnt, only cosO allds~clJ are positive, e.g\"sin(-30~}= --'siu30' and The function y = sin x, where x is any dimensionless quantity, cos(~30').=+cos300 . is called a sine function. The argument x is usually measured in radian. The function y = sin x is plotted in Fig. 1.4(a). The 2. sin(90° - 0) = cos 0 cosee(90\" - 0) = sec 0 maximum positive and negative values of a sine function are +1 eos(90\" - 0) = sin 0 see(90\" - 0) = cosec 0 and 1, respectively, Between x = 0 and x = J{, the fUllction is tan(9(F - 0) = cot f) :2rr\"' , cot(90° - 0) = tan 0 positive, the peak value of +1 occurs at x = Similarly, for thc interval x = J{ to x :::: 2n, the function is negative and the Note: negative peak occurs at x = 23:rr ' The sine function is a periodic =.. Asallgl~ (90? - 0) lies injirst'quadrant,all T-ratios are positive,e.g., si.u30' sin(90~- 60') cos 60', function, with a period of 2:rr. That is, the pattern of the graph e3. sin(90\" + 0) = cos repeats itself after an interval of 2n. Mathematically, it may be eosoe(90° + 0) = sec 0 stated as y = sinx = sin(2:rr + x) = sin(4:rr + x) = ... and so eos(90° + 0) = -sin 0 on. y esee(90\" + 0) = -cosec )' tan(90° + 0)= -cot 0 cot(90\" + 0) = -tan 0 y=sinx ),\"\"cosx (a) (b) Note: .. AStin~le{90'+ 0) lies in 2nd quadrant,. therejoreonly Fig. 1.4 =.sin (J.imd cosec (jarepositive,e.g\" Ifthe h'Taphofthe sine lunction is displaced to the left through siunO'\"\" sin(90'+30'j cos 30'. \":r2r ' we get the graph of cosine function: y :::: cos x as shown in 4. sin(180\" - IJ) = sinO Fig. l.4(b): The cosine function is also a periodic function with cosec(lSO° - e) = cosec 8 a period of2:rr. cos(l80\" - 0) = -cosO see( 1800 - 0) = -sec 8 Some Important Trigonometric Formuloe etan(l80\" - 8) = -tan 1. a. sin' e+ cos' e = 1 eh. 1 + tan' = sec' (J cot(180° - 0) = -cot 0

. Basic Mathematics 1.5 Note: , 2. Similarly, we can find out the value of sin 120\", This angle lies in second quadrant. In second quadrant, sin is positive. .. As angle (180' - 0) lies in 2nd quadrallt, therefore, ollly Therefore, sin 12(f ::::: some p()sitive value, sin 0 and cosec 0 are positive, e~g., ~=} sin 120\" = sin(90\" + 30''') = cos 30\" = sin 150'= sin(1800 - 30') = sin30'. 3. sin 240'\" lies in third quadrant So, it should be negative. 5. sine 180\" + 0) = - sin Ii =} sin 240\" = sin(270' - 30\") = - cos 30\" = - .../3 ecos(180\" + 0) = - cos 2 4. sin 30(P lies in fourth quadrant where sin is negative. tan(180\" + 0) = tan 0 =} sin 300\" = sin(270' + 30') = - cos 30\" = - .../3 Note: 2 5. sin( -3()\") lies in fourth quadrant. so it should be negative, .. As angle (180' + 8) lies in 3rd quadrant, therefore only tan 8 is positive, e.g.\" tan 210' = lall(180' + 30') I =} sin(-30') = -,sin 30\" =-- = t3n30'. 2 6. sin(270\" + 0) = - cos 0 cos(270\" + I)) = sin Ii Inverse Trigonometric Functions tan(270\" + IJ) = -cot 0 Inverse trigonometric functions arc also called anti- Note: trigonometric fUllctions. These are represented hy putting a superscript' -J' on the corresponding trigonometric function .. As angle (270' + 0) lies in the 4'h quadrant, there' whose inverse are to be obtained, c,g., =fore only cos 8 ispositive, e.g., sin 300' sin(270 + 30') inverse of sin () ::::: x, means () = sin --I x It is read as \"sine inverse x\", Just as trigonometric operation = - cos 30°., on any angle gives a particular value, inverse trigonometric op- eration on any value (or number) will return its corresponding 7. sin(300\" - 0) = - sin Ii angle. cos(360\" - 0) = cos 0 Properties of Inverse Trigonometric .Functions: tan(360\" - 0) = -tan 0 1. sin- 1(sin 0) = 0 and sin(sin-- 1 x) = x; provided No!e: .. As angle (360' - 0) lies in the 4'\" quadrant, therefore e-rr12 5 5 n:12 and -I 5 x 5 L only cos 0 is<posi#ve. 2. cos~-I(cose) = Ii and cos(cos- 1x) = x; provided e8. sil1(360\" + 0) = sin o :::: () :::: J\"{ and - 1 :::: x ::: 1, ecos(360\" + 0) = cos e3. tan-1(tan 0) = and tan(tan- J ,r) = x; provided tan(360\" + 0) = tan Ii -~rr/2 5 Ii 5 n:12 and -00 5 x 500, Note: Find the value of sin-I l. +.. As angle (360' 0) lies in IJte first quadrant, ther~rore Sol. Let y = sin-II = sin -'(sin Jf/2) = Jf/2 allther-ratios drepositive, e.g., sin 4000 =sin(360' I'.' sin rr/2 = I and sin -'(sin 0) = () for -Jf/2 5 0 5 n:12] + 40') = sin 40'. Given, that sin 30' = 1/2 and cos 30° = determine the values of sin 600, sin 1200 , Find the value of eos-I(-1/2). sin 240'. sin300'. and sine-30'). -2n-:) -2-n-: 3 3 Sol. Sol. Let y = cos- 1(-1/2) = COs'~\" ( cos = 1. sin 6()o = ? e eI'.' cos(2rr 13) = -1/2 and cos-' (cosO) = for 05 5 Jf] First, we should determine the quadrant in which 60° lies. It is obviollsly first quadrant. Then, we should recall whether sin BASIC COORDINATE GEOMETRY in first quadrant is positive or negative. \"All Silver Tea Cups\" tells us that all the TRs are positive in the first quadrant, If you have to specify the position of a point in space, how will you do it? This is the easiest application of coordinate geometry. therefore, sin 60° must be positive. We can give, assign or find out exact numerical values of the po- Now. we should write 60\" in such a way that it is ±30G with sition of points, lines, curves, slopes, etc, All this is done with the any of the two axes (the horizontal X 0 X' and the vertical ~elp of coordinate systems. There arc many types of coordinate YO yl). So, we can write sin 60\" = SiT~(90G ~ 30(). systems such as rectangular, polar, spherical, cylindrical, etc. It is generally the right handed rectangular axes coordinate system Now, we can recall from the TRs on the previous page which you will be using in physics. This system consists of: that sin(90\" - 0) = cos () =} sin 60\" = sin(90\" - 30\") = cos 30\" = .../3/2,

1.6 Physics for llT-JEE: Mechanics I 1. Origin 2. Axis or Axes 4. y-axis is any fixed direction passing through the origin If the point is known to be on a given line or in a particular perpendicular to the x-axis, convenient to you. Perpen- direction, only one coordinate is necessary to specify its position; dicular means making an angle of +900 with the pos- if it is in a plane, two coordinates arc required; if it is in space, itive direction of x-axis. Students may feel that once three coordinates are needed. the origin and .x-axis have been fixed, the position of y-axis also gets fixed accordingly. But, it is not the Origin case. y-axis can be any fixed direction which is in the This is any fixed point which is convenient to you. Say, in a plane passing through the origin and the x-axis and room, you can consider any corner of the room as the origin, On a sheet of paper, you can mark any point on it and consider it as perpendicular to x-axis, the origin. All measurements arc taken basically with respect to this point called origin. Thus, x and y-axes can be any direction as shown in Fig, 1.6, y Axis or Axes YWX Any fixed direction passing through the origin and convenient A to you can be taken as an axis. If the position of a point or positions of all the- points under consideration always happen to (a) (b) be in a particular direction, then only one axis is required, This is x generally called the x-axis. If the pC'')itions of all the points under consideration are always in a plane, two axes are needed. These x +---.t---- y+---'-+--- are generally caned x and y axes. If the points are distributed in a space, three axes are taken which are called x, y, and z-axes, If y x, y, and z-axes are mutually perpendicular, the system is called (el (d) rectangular axes coordinate system, Fig. 1.6 Important Points 5. Once origin, x- and y-axes are Hxed, z-axis becomes 1. Origin can be any fixed point convenient to you. It is denoted by 0, l m,ltOl,n,atically fixed. convenience, of the observer goes away. z-axis is the fixed direction passing through the 2. x-axis is any fixed direction passing through the origin and ~~~~~ and p~rpendicular ~o both ~- and );-axes. _ ...1 convenient to you (Fig. 1.5). Thus, it is not at all necessary that the (so called) horizontal line passing through the origin is x-axis, ~~)x'-- Position of a Point .o x---... As you already know it well, in ease of plane coordinate geome- try, Le., when the position of a point always remains contained in (,) (b) a plane (called x-y plane), the position of a point is specified by its distances from the origIn along (or parallel to) x and y-axes, ...........-x 0 as shown in Fig. 1.7. (e) (d) (e) You can easily observe that the coordinates (XI, YI), (X2 • .l?). Fig.l.S (x\" n). and (X4. Y4) in Fig, 1.7 are (4, 2), (-4, 3). (-5, -4), 3. Unless otherwise explicitly mentioned, all angles are al- and (2, -2). respectively, ways measured from the direction of x-axis (called the positive direction ofx-axis). Positive angles are measured Distance Formulae in anticlockwise direction and negative in clockwise di- rection, 1. The distance· between two points (x!, YI) and (X2, Y2) = ,1(X2 - XI? + (Y2 - YI)2 2. The distance of the point (Xt, YI) from the origin = j(x? +Yi) The coordinates of the mid-point of the line joining A(xi ,Yt) and B(X2, )12) arc ( -X,-2+X-2-' -YI 2+ Y-2) ,

y Basic Mathematics 1.7 (2, 3) 4 • .•(3,4) o x--+ :··\"e·· ~-. 3 . ; ..; Fig. 1.10 . (x\"y,) . 2 (1,1) ; (x,;y) ·.1 ..•. ... eanticloekwise direction; and m = tan is called its slope or -3 gradient. • (4,-3) • 7. y = rnx, is any line through the origin and having slope m . -4 a, When c.= 0, y = mx The graph between x and y will be a straight line as x Fig, 1.7 bears the direct dependence on y (Fig. L II). SLope of a Line Fig.l.ll The slope of a line joining two points A(XI, y,) and B(x\" Y2) Here, m represents the slope of line. is denoted by m and is given by m = tan e = y, - y, , where e dy ,·,-=m=tan8 X2 - Xl dx is the angle which the line makes with the positive direction of x-axis (Fig. 1.8). b. When c of 0, y = mx + c y Graph for this equation is also a straight line but with a +ve intercept on y-axis. As when x goes to zero, Y accordingly --£--------:.Y2 B(X2,Y2) takes value c. So, the straight line will start from y = c instead of origin (Fig. Ll2). e: YI -- - -------1 ,,,: A(Xl,YJ) : -1-\"'x,------Ox:-,-+ x Fig, 1.8 Straight line Equations Fig. 1.12 1. Ax + By + C = 0, is the general form of the equation of a m = tan e is the slope of the straight line here also. straight line. e, When c = 0, m < 0, y = mx 2. Equation of x-axis is y = o. 3. Equation of y-axis is x = o. For m < 0, e > 90\" (Fig. J.l3). 4. Equation of a straight line parallel to y-axis and at a distance \\~, a from it is given by x = Q. ---'!o,-L-----... x 5. Equation of a straight line parallel to x-axis and at a distance b trom it is given by y = h. Fig, 1.13 a. Constant function, x = a (Fig. 1.9). d. When c of 0, m < 0, y = rnx + c --+-:0-+---\" x eFor m < 0, > 90\" (Fig. Ll4). Fig. 1.9 b, Constant function, y = h (Fig. LlO). 8. ~ + ::: = I, is a line in intercept form where a and b are the 6. y;;;::; mx + c is a line which cuts off an intercept c on y-axis ab and makes an angle e with the +vc direct.ion of x-axis in intercepts on the axes of x and y, respectively (Fig. LIS). 9, y - y, = m(x - x,), is the equation of a line through a given point (x\" yil and having slope m.

1.8 Physics for IIT-JEE: Mechanics I y y 4 4 e_-Li--_-=~4x 3 2 Fig. 1.14 _ _~-+-4_r-+-~-r-2 +~3 ~4+-5~~6 --'x y -1 f -2 be .. 3 Fig. 1.15 Fig. 1.17 Slope 2 If y = O. then -x = 4 =} The slope In of the line Ax + By + C = 0 is given by 3 Coefficient of x -A Method 2: . x +-y = 1 Usmg b -a m= = Coefficient of y B !4 1=} ~ + = lIl!1Wilii.lllll Consider two points PI(2, 7) and IlilWWtlUI =Plot the line -3x - 5y 15. P2(6, 15). Write the equation and draw the straight line Sol. Method I: Given 5y = -3x - 15 through these points. .3 Sol. .. usmgy=mx+c'Y=-Sx-3 Step 1. Obtain the gradient which is m. Here, the slope is negative. i.e., the line makes an angle greater than 90° with x-axis. As intercept is negative, it indicates that Step 2. c can be found by using the (x, y) values of any given the line will cut y-axis at negative side of it (Fig. 1.18). point. When x = O. Y = -3. When y = 0, x = -5. Step 1. . = ~Y2 -- -y, = IS - 7 Height 8 GradlCnt X2 - Xl -6 --2' = Distance 4 = 2. So, Y = 2x + c. (1) Step 2. To find c, put (x = 2. Y = 7) or (x = 6, Y = .15) in equation (1). y 7=2x2+c=}c=3 So, m = 2, c = 3. ,'. the equation becomes y = 2x + 3. y ~pr'I5) / P,(2, 7) , _ _~-5_-_4r--3+--42 r--1 +-i--+--r-+-+-~-r-~x .c~3f x \" }<'ig.l.16 -2 IlI!IJjl'm~11I Plot the line 2x - 3y = 12. -3 -4 -5 Sol, Method 1: -2 x-4 Fig. 1.18 . GIven3y =2x -12=}y = 3 x -y -3x 5y -a 1; -1'5- - -15= I ~.. using y = mx + c, m = and c = -4 Method 2: Using + b = Here, the positive slope (m) means that the angle made by the =} _x_ + _Y_ = 1 line with x-axis s~ould be less than 90° and negative c means (-5) (-3) the line will intercept with negative y-axis (Fig. 1.17).

c------1i ConceptApplication ExetC;ise1.l! :1-----, Basic Mathematics 1.9 = =1. Plot the lines: (i) 3x + 2y 0, (ii) x- 3y + 6 0, • For equation with c = 0, y = axl + bx. The graph be- 2. If a particle starts moving with initial velocity u = I mls and tween x and y is an asymmetric parabola, but the orienta- with acceleration a = 2 m/s2, the velocity of the particle at ti'on of the graph varies with the signs of a and b, Let us take the special case when both a and b are positive. any time is given by v = u + at = 1 + 2t = I + 2t; plot y the velocity-time graph of the particle, 3. A particle starts moving with initial velocity u = 25 m/s Fig. 1.19 with retardation a = -2 m/s2, Draw the velocity-time graph. Parabola: The Quadratic Equations When y = 0, then x = 0 or x = -bla. At x = -bI2a, y is min and Ymin = -b2/4a. It is known as vertex (see Let us now discuss graphs of quadratic equations. Fig. 1.19). For equation y = ax 2 + bx + c (where a, b, and c are con- Plotting Quadratic Equations stants), thc graph between x and y will be an asymmetric 1. General quadratic equation is y = ax 2 + bx + c. parabola. As long as a f 0, this equation represents a quadratic 2. The graph o(a quadratic equation is always a parabola. 3. Orientation of graph depends upon sign of a. function. So, what is the simplest quadratic equation? It is y = ax 2 (obtained by putting b = 0, c = 0), which is the equa- a. When a is +ve, the graph will open up. tion of parabola. h. When a is -ve, the graph will open down, 4. The x-coordinate of the vertex is equal to -bI2a, i.e., Conclusion x = -bI2a, 5. Put this value back in given equation and find y. Point (x, y) Equation of parabola is a quadratic equation in its simplest form. This parabola has its vertex at origin (0, 0) because when we put so obtained represents the vertex. x= 0, it gives y = O. 6. Choose two values of x which are to the right or left of the • The graph for y = ax 2 will be a symmetric parabola about x-coordinate of the vertex, 7. Substitution of these values will give values of y. ,y-axis. The orientation of parabola will be decided by the sign of a. Using these values of (x, y), the graph can be plotted suc- cessfully. I-When a is Positive=c_\"-+~W~h::e::n::a:::..:is:..::N::e,,gc:a.::ti:cv.::e,---:;:-:-1 Note:Sinceaparabola issymmetricilhouttheli./1iipassillg Thc'--'equaiion _..... of the The equation of the thriJugh.itspertex, th\",mirror image ojpoints taken with the same villue willgiv!!other $ide ojparabola, parabola will be y = ax2 , parabola will be y = -ax2. Plot the graph for the equation ±y , y + 4x-1. ~' a = -1, b = 4, c = -1. As a is negative, so parabola • Ifwe exchange x and y in this equation, i.e., x = ay2, then should open down. the axis of symmetry changes and becomes x-axis. As we -b know this orientation changes as per the sign of a, so the Vertex: x = 2a = 2. Put this value of x to get y = 3. orientation will be opposite when a i~ negative, Hence, the vertex of the parabola is (2, 3). a-Wilen a is Positiv;;·....__..· When is-i'iegatiVe--- Assume two values of x as follows and find corresponding values of y. The equation of the The equation of the parabola will be x = ay'. parabola will be x = _ay2. yy , O l t : - - - - -.. x x· ...._ _ _ _~ 0 Points obtained: (l, 2), (-1, -6) ~-.--'--- All points obtained: (2, 3) (Vertex), [I, 2], [...: I, -6] Symmetry of parabola: Mirror image points of (l, 2) and (-1, -6) are (3, 2) and (5, -6). Now, sketch the parabola as shown in Fig. 1.20.

1.10 Physics for IIT-JEE: Mechanics I ~---IIConcept Applitation Exercise 1.3 1--- 1 1. Find the vertex of the following quadratic equations and plot the graph: a. y = x 2 - 8x b. y= -2x2 +3 e. y = x 2 - 6x + 4 2. If a particle starts moving along x-axis from origin with initial velocity u = 1 m/s and acceleration a = 2 m/52, the relationship between displacement and time will be x= at + 1 t 2 = 1x / + -1 x 2x /2 = / + /2 -a 22 Draw the displacement (x )-time (t) graph. (5,-6) Fig. 1.20 DIFFERENTIATION Plot the graph for the equation The purpose of differential calculus is to study the natm:e (i.e., increase or decrease) and the amount of variation in a quantity y =x2 -4x. when another quantity (on which first quantity depends) varies independently. In our day-to-day life, we often face such types Sol. a = I, b = -4, As a is positive, so parabola should open of situations, e.g., growth of plants, expansion of solids on heat- up. As c = 0, so parabola will pass through origin, ing, variation in the velocity of a uniformly accelerated object, growth in the population of a country. -b Quantity: Anything which can be measured is called a quantity. Vertex: x = - = 2, so y = -4=> Vertex = (2, -4) 2a Constants and Variables: A quantity whose value remains con- Assuming two value:;; of x, stant throughout the mathematical operation is called a constant, e.g., integers, fractions, JT, e, etc. On the other hand, a quantity All points obtained: (2, -4), (I, -3), (0, 0) which can have any numerical value within certain specific lim- its is called a variable. A variable is usually represented by u, v, Symmetry of parabola: Mirror image points of (1, -3) and (0,0) are (3, -3) and (4, 0). w, x, y, z, etc. Now, sketch the parabola as shown in Fig. 1.21. Dependent and Indepeudent Variables: A variable which can have any arbitrary value within specific limits is called as in- dependent variable whereas one whose value depends upon the numerical values assigned to the independent variable is defined as dependant variable. Differential Coefficient or Derivative of a Function Suppose y be a function of x, i.e., y = f(x). (i) The value of the function or the dependent variable y depends on the value of independent variable x. If we change the value of independent variable x to x + ~x, then value of the function will also change. Let it becomes y + L'.y. Hence y + L'.y = f(x + L'.x) (ii) Subtracting equation (i) from (ii), we get (iii) y + L'.y - Y = f(x + L'.x) - f(x) or L'.y = f(x + L'.x) - f(x) Above equation provides the change in the value of function y, when the value of variable x is changed from x to x + ~x. Dividing both sides of the equation (iii) by L'.x, we get Fig. 1.21 L'.y f(x + L'.x) - f(x) (iv) L'.x L'.x

Basic Mathematics 1.21 Maximum value of v = 6 x 4 8 = 8 16 In the previous problem, if the particle -- 6x- -- - 9 27 3 9 occupies a position x::;:: 7 m at t = .1 s, then obtain an expression for the instantaneous displacement of the particle. -8 ~I -2 I.n v) = (by puttI.ng t = IDS 93 Sol. Again, we can use the idea that displacement is the integra- On a certain planet, the instantaneous ve- tion of velocity w.r.t. time. locity of a ball thrown up is given by v = 21 - 6 (v is in ms-1 So. x = J vdt = J(2t + 5)dt = 221\"2 + 5t + c = ~ c and 1 is in sec). /' + 5t + 1. Find the expression for the displacement of the particle, where c is the constant of integration. Its value can be determined by using the given condition. (As particular details have been =given that the particle started its journey at x 1 m. given about the particle.) 2. What is the value of g on the surface of this planet? Att = I s, x = 7 m =} 7 = 12 + 5 x I + c =} c = 1 m Hence, the expression becomes x = t 2 +' 5t + 1 Sol. 1. To find the displacements, we need to integrate the ve- f flocity. i.e., 2~2(2t - 6)dt = - 6t + c = t 2 - 6t + c x = vdt = where c is the constant of integration. Its value can be calculated 1.61f~ Concept Application Exercise f--~ from the given boundary condition that x ::::: j, t = O. 1. Displacement of a particle is given by y = (6t 2 + 3f + 4) m, =} c = 1m=} x = t' - 61 + I 2. g is acceleration due to gravity. So, to find acceleration we where t is in second. Calculate the instantaneous speed of the need to differentiate velocity. particle. As v = 2t - 6 =} a= -dv = ~? 2. The velocity of a particle is given by v = 12 + 3(1 + 7t'). dl 2 ms ' So, the acceleration due to gravity on the planet under con- i! 3. What. is the acceleration of the particle? A particle starts from origin with uniform acceleration. Its sideration is 2 m/s-2. Let the instantaneous velocity of a rocket, displacement after t seconds is given in meter by the relation x = 2 + Sf + 7f2. Calculate the magnitude of its just after landing, is given by the expression, v = 21 + 312 a. initial velocity, (where v is in ms- 1 and 1 is in seconds). Find out the distance h. velocity at t ;::: 4 s, traveled by the rocket from t = 2 s to t = 3 s. c. uniform acceleration, and Sol. To find distance traveled we need to integrate v. [The limits d. displacement at t ;:;: 5 s. of integration will be from 2 to 3 as we have to tind the distance i 4. The acceleration of a particle is given by a = tJ~3t2+5, traveled between t = 2 sand t = 3 s. where a is in m8-2 and f in sec. At t ::;:: I s, the displacement 11 11+-11212x = 3 IIdt = and velocity are 8.30 m and 6.25 ms- i , respectively. Cal- 12 2 (2t + 3t2 )dt = \"- 3t 1 2 3, culate the displacement and velocity at 1 ;:;: 2 sec. = It2 + t11~ = 24 m S. A particle starts moving along x-axis from t = 0, its position varying with time as x = 2t 3 ~ 3t 2 + 1. a. At which time instants is its velocity zero? A particle moves with a constant accel- h. What is the velocity when it passes through origin? =eration a 2 ms-2 along a straight line. If it moves with an 6. A particle moves along x-axis obeying the equation x = t(1 - 1)(1 - 2), where x is in meter and t is in second initial velocity of 5 ms-1, then obtain an expression for its instantaneous velocity. a. Find the initial velocity of the particle. Sol. We know that acceleration is time rate of change of veloc- h. Find the initial acceleration of the particle, ity, i.e., a = dv and differentiation is the inverse operation of elt c. Find the time when the displacement of the particle is zero. integration. So, by integrating acceleration we can obtain the d. Find the displacement when the velocity of the particle is expression of velocity. zero, f fSo, v = adt = 2 dt = 2t + c e. Find the acceleration of the particle when its velocity is where c is the constant of integration and its value can be ob- zero. tained from the initial conditions. toIi 7. The speed of a car increases uniformly from zero to . s - 1 At t = 0, v = 5 ms- I . We have 5 = 2xO + c m ::::} c = 5 ms- I in 2 s and then remains constant (Fig. 1.29). Therefore, v = 2t + 5 is the required expression for the instan- a. Find the distance traveled by the car in the first two taneous velocity. seconds.

1.22 Physics for IIT-JEE: Mechanics I F(N) 100 ofL-c12-o),-i--'r4:--,5+-/,\",1:6--+ x (m) ____ ~__________ J -- 50 t (s) Fig. 1.31 Fig. 1.29 a. Calculate the velocity acquired by the particle after getting displaced through 6 m. b. Find the distance traveled by the car in the next two seconds, b. What is the maximum speed allained by the particle and at what time is it attained? c. Find the total distance traveled in 4 s. 8. A car accelerates from rest with 2 ms~2 for 2 s and then 10. A body moving along x-axis has, at any instant, its x-co- ordinate lo be x = a + bt + cP. What is the acceleration of decelerates constantly with 4 ms-2 for to second to come to the particle? rest. The graph for the motion is shown in Fig. 1.30. 11. The displacement of a body at any time t after starting is a (ms· 2) given by s = l5t - OAt 2 Find the time when the velocity of the body will be 7 ms- I 2 (2 + '0) 12. A particle moves along a straight line such that its displace- 0:--\"\"2:'---'--;-:-\"----» t (s) mcnt at any time t is givcn by s = t 3 - 6t' + 3t + 4 m . ,' Find the velocity when the acceleration is n. 4 ------ 13. The coordinates of a moving particle at time t are given by x = ct' and y = bt'. Find the speed of the particle at any Fig. 1.30 time t. a. Find the maximum speed attained by the car. 14. The displacement x of a particle movin.g in one dimension b. Find the value of to. 9. A stationary particle of mass m = 1.5 kg is acted upon by under the action of a constant force is related to time t by the a variable force. The variation of force with respect to dis- placement is plotted in the following Fig. 1.31. equation t = .JX + 3, where x is in meter and t is in seconds. Find the displacement of the particle when its velocity is zero.

Vectors 2.1

2.2 Physics for lIT-JEE: Mechanics I In physics, various quantities arc broadly classified into two 2. By bold Jetters. categories: For example: Force can be represented by :F. In this case, 1. Scalars. and 2. Vectors. I Imagnitude is represented in the same way as above such as F or by same symbol, but Italic F. SCALARS INTRODUCTION TO DIFFERENT TYPES OF VECTORS Scalars are those physical quantities which have magnitude only but no direction. For example: mass, length, time, work, etc. Collinear Vectors VECTORS The vectors which either act along the same line or along thc parallcllines arc called collinear vectors. These vectors may act Vectors arc those physical quantities which have both magnitude either in the same direction or in the opposite dircction. and direction. For example: velocity, acceleration, momentum, force, etc. ParalleL Vectors Further, vectors can be of two types: Two collinear vectors having the same direction are called par- allel or like vectors (sec Fig. 2.2). Angle betwecn them is (l\". 1. Polar vectors: These are vectors which have a starting point or a point of application. • .., • F'or exarnple: velocity, force, linear momentum, etc. ,• x• 2. Axial vectors: These arc vectors whose directions arc along y the axis of rotation. Fig. 2.2 For example: angular veiocity, angular momentum, torque, etc. AntiparaLLeL Vectars Note:. Avectofis meauiugfulonlyifwe kuow both magni- Two collinear vectors acting in opposite directions are called tude onddirectiou of a vector. Without knowiugdirectiou, antiparallel or opposite vectors (see Fig. 2.3). The angle between the descr!ptimi ofa vector quantity isincomplete. them is 180(} or rr -radian. Use of vectors: Many laws of physics can. be expressed in ..-) -) -, compact form by the use of vectors. In this way, the complicated ••_.f.Y_~_~ laws arc greatly simplified and then they become easier to apply. x• For example: F = ma, '[ = Iii, F = qv x 8, etc. , REPRESENTATION OF VECTOR y A vector is represented by a directed line segmcnt, with an arrow Fig. 2.3 -~ Equal Vectors head. For cxample,'a vector F h; represented by a directed line Two vectors are said to be equal Vectors if they have equal mag- PQ (see Fig. 2.1). ..--) ). p nitudc and same direction. Two equal vectors A and B are rep- Fig. 2.1 Point P is called tailor origin of the vector. Point Q. the end resented by two equal and parallel lines having arrow heads in point of vector, is called tip, head or terminal point of the vector. the same direction (see Fig. 2.4), 1. The length of the line represents the magnitude of the vector. 2. The atTOW head represen(~ the direction of the vector. , -) --) •A., AB B• Fig. 2.4 NOTATION OF VECTOR Negative of a Vector 1. By single letter with an arrow overhead. A negative vector of a given vcctor is a vector having same magnitude with the direction opposite to that of given vector For example: Force C~ln be represented by F and its magni- A A.(sec Fig. 2.5). The negative vector of is rcpresented by - Fl·Itude is represented as

Vectors 2.3 -, • Expression of a Vector in terms of i, ), and k and A Magnitude of a Vector •~ A A } Fig. 2.5 y Coplanar Vectors ,, ,,,,, B 52 These arc vectors which lie in the same plane. In Fig. 2.6. o , ,, It, B, and C are acting in the same plane. i.e.. XY plane, ~/ ()C /A x so they are coplanar vectors. ,,i----- ------- ,,I C'(,4/t 5) OA\"\" 4 y : ---------------~/ AB\"\" 2 BC=5 y\\( X Fig. 2.8 ~ B Suppose a particle goes from 0 to A, 4 m along .r-axis; A to 13, 2 m along y-axis; and then B to C, 5 m along z-axis (Fig. 2.8). () Finally, the particle reaches at C. z Coordinates of C: (4, 2, 5) m Fig. 2.6 We can write: 0,1 = 41, AB = 2), and BC = 5k. Unit Vector OCo is the initial position and C is the final position. is the net displacement, which is the resultant of GA, AB. and BC. It is a vector whose magnitude is equal to unity (one). A unit SO, we can write: OC = oA + AS + BC -.vector in a given direction can be obtained by dividing a vector i.e., OC = 41 + 2) + 5k in that direction by its magnitude. Unit vector of A is written as (Hence, here we represent OC in terms of 7, \" and k.) A and is read as A cap. lociMagnitude (modulus) of OC: It is written as or ac. i,So, A = where A is the unit vector along the direction of A In !',OAB: OB' = OA' + AB2 [','LOBC=90'1 A. In !',OBC: OC' = OB' + BC' The direction of unit vector will be same as that of the vector => OC2 = OA 2 + AB' + BC' . It.from which it is obtai-n,ed. It means A is parallel to => OC = ~OA2 + AB' + BC' Also, (i'OID above: A = AA or vector = magnitude x direction = ~4' + 2' + 5' = y'45 In every direction, we can obtain a unit vector. In x, y and => OC = lOCI = y'45. This is the magnitude bf OC. z directions, unit vectors are predefined (Fig. 2.7), Note: Magnitude oIa vector or modulus of a vector is hi)' equal to length ofthe vector. x . _ OC 41 + 2) + 3k AA UOlt vector: OC = - = . - - - - - - ,k ' OC y'45 Fig. 2.7 4.2,3. i is the unit vector along x-axis, . = y'45i +j45j + y'45k } is the unit vector along y-axis, and (Hence, we express a unit vector in the direction of 0(;: in terms k is the unit vector along z-axis. of I,) and.) . Now, a unit vector in any other direction can be obtained in Important: Direction of Dc and OC will be same. 1,terms of 1, and k as shown in the next section. Now, in general, if we have a point P in space whose coordi- Note: A unit vector is a dimensionless quantity, so it has nates are x, y and z, then a vector from origin 0 to P is known no units. It represents only direction. . as position vector of P w.r.t. origin 0 (Fig. 2.9). We can write: -; = \"i5P = xi + y) + zk JMagnitude: r = OP = x 2 + y2 + Z2, +. .. _ r xi + y) + zk VOlt vector: r = ~ = /~2 y-2 + Z2

2.4 Physics for IIT-JEE: Mechanics I y P(X,y,z) rA A(XbYl, Zj) O~~--~--~--~X Fig. 2.11 ' ...... , ,/ • A vector can be completely zero, if all of its individual ...... I , / ...... ! / components are zero. For example: let a vector ------------~j/ a,; (/2.!It A= + + (13k is given, If = 0, then it is possible z only if a, =0, a2 =0 and (/3 =0, Fig. 2.9 • If two vectors are equal, then their individual components Position Vector and Displacement Vector are also equal separately, Consider two points A and B in space whose coordinates are For example: Let two vectors It = (Ill + (l2) + a3k and (x\" y\" 2,) and (X2, y\" z,), respectively (Fig, 2, [0), B = ItbJ + b2) + b3k are given, If = B, then Position vector of A: oA = rj = xJ + y,) + 2,k, (/, = bt, (/2 = b, and (/3 = b3 (but if A = 8, then it means y Jar + ai + aj = Jbi +\"\"i;I+ bi· -, ~ IJImnmnB A particle initially at point A(2, 4, 6) m 1'1 d moves finally to the point 8(3, 2, -3) m. Write the initial ~ position vector, final position vector and displacement vector )o~=-----,-1'1.2- - - - _ + x of the particle. z Sol. Initial position vector: '7, = 21 + 4) + 6k Final position vector: -t 2 = 37 + 2.7 - 3k Fig. 2.10 Displacement: d=r2 r, = (3 - 2)7 + (2 - 4)]+(-3-6)k oRPosition vector of B: = 0 = x2i + )'2} + zi< =?- 2) - 9k Position vector provides us some important information: ~ A particle has the following displace- It gives an idea about the direction and the distance of the point ments in succession: (i) 12 m towards east, (ii) S m towards from origin in space. north, and (iii) 6 m vertically upwards. Find the magnitnde Now consider a particle which goes from point A to B. Then, of the resultant displacement. AB d= is displacement vector of the particle, Displac~ment z yN vector is that vector which tells us how much and in which di- rection an object has changed its position in a given interval oj 6m arne. 5111 ri.Here, is initial position vector of the particle and r! is final ~--~12~'-n----~---4>x E position vector of the particle. Fig. 2.12 .'. displacement vector = f1nal position vector - initial position Sol. dFrom Fig, 2,12: = 121 + 5) + 6k, vector Magnitude: d = \"\" 122 + 52 + 62 = \"\"205 m ~ Determine a vector which when added to d=\"0-rt the resultant of A= 21 + sj - k and B = 31 - 5} - Ii gives d = (X2 - x,)1 + (Y2 - y,») + (Z2 - z,)k unit vector along negative y-direction. Note: .. Positioll. vec((Jrgive,<Hs informatioll abolllpositi(m Sol. Let Cbe the vector which we have to find, ofdpoi1itillspace,displacemelltv~qtoru,\"i!lftdisplac~mellt. Then, given: (A + ii) + C= -) Similarly, ~ velocityvector.or aforce~ectot willtell Ifs about =} (21+5] -k)+(3! -5) -k)+C =-7 velocity orf~r~ealld.theirmagllit~def.~1l44ir~ctiolls.So,u,•• =} C' = -5) - ) + 2k Vector gives usinjot/ll(ltion about the.S(!1f!~.physiccilqlfan- . tiIYby which it is known. . . • Position vector is also known as radius vector. • If A and B are two points whose coordinates are (x\" )'\" 2,) and (X2, )'2, 22), respectively, then a vector from A to B is given by (Fig, 2,11): AB = (x, - x,)? + (Y2 - y,») + (22 - zilk

Resultant Vector Vectors 2.5 The resultant vector of two or more vectors is defined as that Two vectors P=2f - hJ + 2k and single vector which produces the same ejfect both in magnitude and direction as produced by individual vectors taken together. + Ii are parallel. Find the valne of h. P PSol. If and Qare parallel, then we have = m Q, where In Q is some number. 5m -, 27 - b) + 2k = m7 + m} + mk R ---+3m B Equating the components of ; and) from both sides: m = 2 and -b = m =} b =-2 A If vectors 21 + 2J - 2k, Sf + yJ + k, oL--'\"'4-'n'-,- 4 ' p are coplanar, then find the value of y. Fig. 2.13 So!.lf Ii, B, and (; are coplanar, then we have: Ii = mB + 11(; Example: Let a person goes from 0 to P traveling 4 m and then P to Q traveling 3 m perpendicular to OP. The person =} 27 + 2) - 2k = m(57 + y) + k) + n(i + 2) + 2k) could also go directly from 0 to Q, traveling 5 m (Fig_ 2.13). =} 27 +2J - 2k = (5m +11)7 + (my +2n)J + (m +2n)k Equating components of 7, ), and k from both sides: In both the cases, displacement of the person would be same, 5m + n = 2, my + 2n = 2, m + 2n = -2_ Here, oQ produces same effect as produced by oP and PQ Solving them, we get: III = 2/3, n = -4/3 and y = 7. taken together, So, oQ is called the resultant of Of and PQ. We can also write oQ = oP + PQ. Note: R~sultant means (uidition. ADDITION OF VECTORS We cannot write OQ = OP + PQ, because OQ = 5 m and How to Add Two Vectors Graphically (Tip to Tail OP+ PQ =4+3 = 7m_ Method) Points.to Remernber 1. Draw the two vectors by arrow head lines using the same suitable scale. 1. If two vectors A and Bare parallcl, then we can write 2, Put the second vector such that its tail coincides with the head B = mA, where m is a number. If vectors are parallel, or tip of the first vector. then In is +ve and if vectors are antiparallel, then In is -ve 3. Now, draw a single vector from the tail of the first vector to the head of the second vector. This single vector represents (Fig. 2.14). -, the resultant of the two vectors. A • Let us discuss some cases: -, 1. When two vectors are acting in the same direction (Fig. 2.15). B Let the two vectors X and y be acting in the same direction. paralic! vectors ~ •--, -,-.} A• x •x Y + • 11 ~ --t • .0 <lntiparal!e! vectors y R=x+y Pm'a1!c! or antiparallc! vectors (a) (b) arc known as collinear vectors Fig. 2,14 (e) Fig. 2.15 2, If two vectors arc parallel, then their unit vectors are equal, 2. When two vectors are acting in opposite directions: To find the resultant, in this case, coincide the head of;( on i.e., It = B and if two vectors are anti parallel, then we ythe tail of and then draw a single vectors R t~'om the tail have!\\ =-k y. R01'\"7 to the head of The vecto!' gives the resultant of ~} ~,} ~)O vectors \"7 and y (see Fig_ 2.16). 3. If three vectors A, B, and C are coplanar, then we can ~} .0- -+ write: A = m B + n C, where m and n are some numbers. Minimum number of unequal vectors whose sum can be zero is three and these three vectors must be coplanar. -, • • •-, 4. If four vectors II, B, C, and jj are in any arbitrary x x (a) y~ It x')+( )!) directions or in different p,lanes, then we can write: (b) (e) ~.,\" -,> ,,} Fig. 2.16 The direction of the resultant vector is the same as that of A = !YlB +nC+ pD, where m, nand p are some num- bigger vector. bers. The resultant of three non-coplanar vectors can never be zero. Tlu minimum number of non-coplanar vectors ~\" w_ho_s_e _SU_I;'_l c\"_a:nccbce\"'-z\"e:r:.oc,'-\"'=::c.c____________-.l

2.6 Physics for JIT-JEE: Mechanics I (iii) Now complete the II gm (iv) Draw the diagonal 0 as shown. which represents resultant 3. When two vectors are act.ing at some angle: ofx andJ. i/ y xFirst join the tail of with the head of and thcn. to find 75C = OA + DB the resultant in this case, draw a vector R from the tail of X l/I//tSo, R=x+y 0'----::x .;;---'A 0 x-\" -A Rto the head of y .This single vector drawn is the resultant ~ ~___(C_)_ _ _ _ _ vector (Fig. 2.17). _(_d)_ _ _ _ c'/ ADDITION OF MORE THAN TWO VECTORS ? For this, we can use following steps: Fig. 2.17 1. Represent these vectors by arrow head lines-using the sarne suitable scale. R x yrepresents the resultant of and both in magnitude xR y.and direction. So, = + 2. Put these vectors in such a way that the head of one coincides with the tail of second and so on to the last vector. TRIANGLE LAW OF VECTOR ADDITION 3. Then. draw a single vector from the tail of first vector to the If two vectors can be represented both in magnitude and direction head of the last vector. by thc two sides of a triangle taken in the same order, then their resultant is represented by the third side of the triangle (both in 4. This single vector represents the resultant of all the vectors. magnitude and direction) taken.in reverse direction. The above mentioned process may be known as polygon law ;R _~ (~l vector addition. It slates that if any number of vectors act- y ing on a particle at the same time arc rcprcscntt.:d in magnitude and direction by the sides of an open polygon taken in order, • L::Jx• x• then their resultant is represented both in magnitude and direc- tion by the closing side of the polygon. Consider four vectors Fig. 2.18 w, x, y. ZasshowninFig.2.19(a). Suppose X and yare two vectors acting on a particle at the -, same time. They arc represented as two sides of a triangle and x R represents the third side (Fig. 2.18). R xThus, represents the resultant of and y both in mag- nitude and direction. So, we can write: R = X + y. PARALLELOGRAM LAW OF VECTOR yo ADDITION (a) . (b) It states that if two vectors can be represented both in magnitude and direction by two adjacent sides of a parallelogram, then Fig. 2.19 the resultant is represented completely both in magnitude and x w yThen, R = + + + Z, as shown in Fig. 2.19(b), direction by the corresponding diagonal of the parallelogram. (i) Consider two vectors 7 (i1) Draw these two vectors x. y,represents the resultant of W. and Z. and y asshown. X and y from a common •., point O. x B VECTOR ADDITION BY ANALYTICAL METHOD (a) vi Here, we will treat both triangle law and the parallelogram law of vector addition analytically to find the resultant of two vectors. O~A x 1. Analytical treatment of triangle Jaw of vector addition: (b) PLet us consider two vectors and Q acting simultaneously I on a partide and inclined at an angle D. Let these vectors be represented both in magnitude and direction by the two sides CiA ACand of L'.OAC taken in same order. Then, the third side

Vectors 2.7 at: represents the resultant (taken in opposite order) (Fig. 2.20). I Special Cases c I ~ase 1. When the given vectors act in the same direction (8 = 0°). JP2So, R = + Q' + 2P Q cos 0\" [From equation (ii)] o L.l-\"--_.,~-¥ = /1\"+ Q2+2pQ = .j(I' + Q)2 = 1'+ Q pA [.: cos 0\" = 11 Fig. 2.20 I0or IR I = IP I + I which represents the magnitude of Draw CN ..L ON (extended). the resultant. From the right angled \"'CNO. Case 2. When the given vectors act in opposite directions (8 = 180°). OC2 = R2 = ON' + NC' = (OA + AN)' + NC2 \"* R2 = (I' + ANf + NC2 So, R = j\"I'2-:;:-Q\"'+ 2I'Q cos -i SO;; = jP2~- Q2 -- 2I'Q (i) In right-angled \"'ANC. \"*NC AN = .j (I' - Q)2 I' :cos 1800 = -11 NC=QsineandeosO=-Q sinli=Q R = ±(I' - Q) = I' - Q or Q - I' \"* AN = QcosO Ror I I = II pI-loll which represents the magnitude From equation (i): R2 = (I' + Q cos 0)' + Q2 sin2 0 \"* R' = I'2 + Q2 cos2 (I + 2I'Q cos (I + Q' sin2 0 of the resultant. \"* R' = 1\" + Q2 (cos'O + sin' 0) + 2I'Q cos Ii \"* R2 = I'2+ Q'+2pQcose r·.· cos'O+sin2 0 = I] P 0Case 3. When the given vectors and act at right angle \"* R = fP2+ Q' + 2I'Q cos 0 (ii) to each other (8 = 90\"). R = /1\" + Q2+ 2I'Q-Z~;90; = II'2 + Q' The equation of the magnitude of the resultant vector can be [.: cos 90\" = 01 written in either of the following two ways, Ifif~~ror IR I = 1~121.1 R 1= Ilpl' + +217Sllolcos; Q sin 9()\" or tanf3 = pQ and tan f! = I' + Q cos90\" r·: sin = 9()\" = IJ 2·lp+ol=/lpl'+lol'+2Ipllolcosli --- Let f3 be the angle whichR makes withP. Then, Note: Suhtractiol! of two vectors can be followed/rolll NC NC = QsmO addition tanf3 = - = ON OA+AN p+Qcosli Suppose we havetosubtract a vector Qjrom A: SO, we IlOve -1 ( QSinO) -1 ( IOlsinO ) to find A'-\"'Q: Itcalldl~obc writreu aii A +f\" Q). milCii, .'. !J=tan I'+QcosiJ =tan IPI+'I'(2lcosli subtractionofa vector Q/rolliA becomes as the addition of vectors Aand - Q (Fig. 2.22). which gives the direct ion of the resultant vector. S =/A2+ B2 + 2ABcos(180 _ 0) 2. Analytical treatment of parallelogram law of vector ad- \"* .S =/A2+B~-2AB cos~ Pdition: Let us consider two vectors and Q acting simulta- _c-B=-·~si=nc:(1:::80==-..::0:...)-:-c BsinO tan /I -A-c--B--e-o=-s-O-, neously at a point and let us further assume that they can be Bcos(180~0) A+: \"* =tan/l•= represented both in magnitude and direction by the two adjacent =Vectonulditum is commutative: A+Q Q+ A, Vector sides of a parallelogram oA and DB inclined at an angle Ii w.r.t. suhtraction isanti-commutative: A- Q = -(Q- A) each other. Draw CN ..LON (extended) (Fig. 2.21). (Fig. 2.23). Remaining procedure is same as above in triangle law. ,1l~~~~_ _.. C .., B o Fig. 2.22 Fig. 2,23 Fig. 2.21

2.8 Physics for IIT-JEE: Mechanics I lulJt!tmwm Two forces of 10 Nand 15 N are acting ~ ., at a point at an angle of 45° with each other. Find ont the p magnitude and the direction of their resultant force. Fig. 2.27 Sol. Given A = 10 N, B = 15 N, e= 45\" and Magnitude of resultant force: R = y'rA\",c-+:-OB'\"'2C-+-'---;;2'-A\"BC;-c-o-s-';e /2R = 2 + l' + (_4)2 = 51 = v'102 + 152 + 2 x 10 x 15 x cos 45\" - 23.76 N PWe see that = Q+ Rand Q2 = p2 + R2 is satisfied. Hence, B ___________ _ they will form a right-angled triangle, with Qas hypotenuse and Pand Rthe other two sides as shown in Fig. 2.27. R ,,,,,,,,,, \",---\"a,,-e_;=_45_0_ _..,~ Fig. 2.24 Lami's Theorem eDirection: tan a = - -B s-in- = 15 sin 45\" 0.5147 It states that if the resultant of three vectors is zero, then mag- A+Bcose 10+ 15 cos 45\" nitude of a vector is directly proportional to the sine of angle =} a = tan- 1(0.5147) = 27\" between the other two vectors (see Fig. 2.28). Or it can be statcd IlttltUt!!-\" Two forces of equal maguitudes are act- as if the resultant (~fthree vectors is zero, then the ratio a/mag- ing at a point. The magnitude of their resultant is equal to nitude of a vector to the sine of angle between the other two magnitude of the either. Fiud the angle between the force vectors is constant, Le., vectors. ABc e.Sol. Given R = A = B. Using R' = A2 + B2 + 2AB cos --=--=-- sino: sin f! sin y A2 = A2 + A2 + 2AA cose IX ~ 1 =} e = 120° ~> \"~~ =} cose = -- 2 C B Y Condition for Zero Resultant Vectors -, C #A Fig. 2.28 1. The resultant of two vectors can only be zero if they are equal Given that A+ B+ C= O. Out of three in magnitude and opposite in direction (Fig. 2.25). vectors are equal iu magnitude aud the magnitnde of ~ B= _\" A the third vector is .(i times that of either of the two having _ _ _ _ _-'A'--+_ equal magnitude. Find the angles hetween the vectors. A~ + 11= 0 .~.-,B,-_ _ _ _ _ Fig. 2.25 Sol. Given A = B, C \"\" .(iA = .(iB = B. From Fig. 2.29: 2. The resultant of three or more vectors can be zero if thcy a=f!anda+f!+y\",ISO\" =}y=ISO\"-2a constitute a close figure when taken in same order (Fig. 2.26). AC Apply Lami's theorem: - . - = --.- sma sm y A .(iA iOB ~ ~~ ~ ~ ~ sina sin(l80 - 2a) 1 .(i ~ .C A+B+C+D+E=O ~ B~ ~ 4 --.=-- sin a: sin 20: C A+B+C=O ~~ .(i I . -- - oc-;----- =} cos a = v12L0 =} a = 45\" AB 2sina cosa sina Fig. 2.26 f! = 45\" and y = 1800 - 2a = 90\" 1!ilJt!tmWJII. Show that the vectors P = 31 - 2} + k, Q = 1- 3} + 5k and R = 21 + } - 4k form a right-angled BCi ~ triangle. P = )32 + (-2f + l' = v'f4, ~> Sol. C Q = /1-;-;( _3)2 + 52 = 55, #y -, A Fig. 2.29

Vectors 2.9 BAngle between Ii and = 180n - Y = 90\", angle be- y twecn Band C= 180 - a = 135', angle between C and N ~ P Ii = 180\" - fl = lW. M ~ A ~---I Cohcept,Applicatioll Exercise 2.1 f--~ Ay e -, 1. A force of (21 + 3] + k) N and another force of (i + ] + k) N arc acting on a body. What is the magni- 0 Ax X tude of total force acting on the body? Fig. 2.31 2. lfa = 31 + 4] and b = 71 + 24], then find the veetorhav- eComponent along x-axis Ax = A cos ing the same magnitude as that of band parallel to a. (Horizontal component) (1) 3. In the vector diagram given below (Fig. 2.30), what is the eComponent along y-axis Ay = A sin A Enangle between and (Given: C = ~) 2 -> +-c> -c> , . (Vertical component) (2) , So, A = Ax A,. = Axl + Ay.l Squaring and adding equations (I) and (2), we get A; + A; = A2(sin2 e + cos' 0) = A' '* J PM A,.A= A'+A' and t a n 8 =O M- - =A-,' xy So, knowledge of components of a vector gives information about angle which the vector makes with, different axes, Fig. 2.30 A force of 10 N is inclined at an angle 4. What is the angle made by 31 +4) with x-axis? horizontal. Find the horizontal and vertical components of the force. -~ \" \" , , - - - - ) 0 \"A \"- Sol. Let R = 10 N. Horizontal component: + + +S. Three forces A = (i j k), B = (2i - j 3k), and (' arc acting on a body which is kept at equilibrium. Find R, = R cos 30\" = 10./3/2 = 5./3 N C. Vertical component: Ry = R sin 30\" = I()/2 = 5 N 6. At what angle should the two force vectors 2F and ~F act so that the resultant force is ,fiOF? 7. Two forces while acting on a particle in opposite direc- AThe x and y components of vector arc tions, have the resultant of ION. If they act at right angles to each other, the resultant is found to be 50 N. Find the 6 m, respectively. The x and y components of vector two forces. A+ Bare 10 and 9 m, respectively. Calculate for the vector 8. Two forces each equal to FI2 act at right angle. Their B the followings: (i) its x and y compouents; (ii) its length, assuming thatA and B lie in x - y plane; and (iii) the angle effect. may be neutralized by a third force acting along their it makes with the x -axis. (I) isbisector in the opposite direction. What the magnitude Sol. Given Ii = 41 + 6] of that third force? and Ii + B= 101 + 9} 9. The resultant of two forces has magnitude 20 N. One of (2) the forces is of magnitude 20 ./3 N and makes an angle Of 30\" with the resultant. What is the magnitude of the other Subtract equation (l) frqm (2), we get B = 61 + 3] force? 1. Hence, x and y components of Bare 6 and 3 m, respectively. 10. The sum of the magnitudes of two vectors is 18. The mag- nitude of their resultant is 12. If the resultant is perpen- 2. Length of B= magnitude of B= .J6' + 32 = 3.J5 m. dicular to one of the vectors, then find the magnitudes of B3. Let makes an angle ex with x-axis, then tan Q' = 3/6. the two vectors. '* a = tan- I (l/2) = 26.6\" RECTANGULAR COMPONENTS OF A VECTOR IN THREE DIMENSIONS RECTANGULAR COMPONENTS OF A VECTOR When a vector is split into mutually perpendicular directions in IN TWO DIMENSIONS 3-D space, the component vectors obtained arc called rectangu- When a vector is split into two mutually perpendicular\" direc- tions in a plane, the component vectors are called rectangular lar components of the given vector in 3-D space. components of the given vector in a plane. oP.ItFig. 2.32 shows vector represented by Fig. 2.31 shows vector It rcpresented by oP. Here, It = It x + It,. + It, = A,l + A,.) +-.:A\"\"-.:k__ jAT\"+ItThe magnitude of is given by, A = A~ + A;

2.10 Physics for IIT-JEE: Mechanics I Scalar or Dot Product The scalar or dot product of two vectors A and If is definecl y as the product of the magnitudes of two vectors and the cosine z of the smaller angle between them (Fig, 2.34), It is given by Fig. 2.32 ii, B= ABcos!!, Direction Cosines L A'\" Let A is a point in space whose coordinates are (x, y, z), then Fig. 2.34 its position vector W.r.t. the origin of coordinate system is given Special Cases ~ OA + +hy: _ = = \" A A r xi yj zk (see Fig, 2,33), r JxAnd = 0 A = 2 +~V2 + Z2 1. Ifli = 0\", ii, R= AB (maximum value) rAngles of with X-, y- and z-axis, respectively, arc given [':cosO\" = 1] by: xV A· B2. He = 180°, Z = -AB (ncgativernaximum value) cosO' = - =1,cosfJ =:...... =m,cosy = - =n rr r [': cos 180\" = -I] y 3. If I! = 90\". ii ' R= 0 (minimum valuc) [': cos 90\" = 01 ,IA(x,y,z) So, iftwo vectors are pellJendiculm; then their do! product is zero. ,:y-r\"' e4. If is acute, then A. Bis +ve. o»4c\"\"a'----+:-~x y ~,: /; [':cosl! is +ve when I! is acute] ----------:-::-,,/.' 15. If 0 is obtuse, then Ii . Bis -vc. x [': cos e is -vc when e is obtuse] z Fig. 2.33 1_ _ _ _ __ The direction cosines /, In, and n of a vector are the cosines Note: The dot produCt of two vectdrs is always a scalar qualltity. . of the angles a, f3 and y which a given vector makes with X-, Dot Product of Unit Vectors Along X-, y-, and },-, and z-axis, respectively. z-directions Now, squaring and adding l, rn, and 11 Dot product of a unit vector with itself is unity and with other + + ++ x2 y2 Z2 cos2 Q' cos2 f3 cos2 Y = - _. }'l',perpendicular unit vectors is zero (Fig, 2,35), ), \" ,X or Ii ' It means the sum ofsquares ofthe direction cosines afa vector is alYvays undy, z Fig. 2.35 Given A = 51 + 2) + 4k. Find: (i) 1AI i ' i =] ,] = k ' k = 1 and and (ii) the direction cosines of vector A. 7}=]1=]k Sol. =k,]=kl=i,k=O 0) As\"A=SI+2]+4k=} IAI=)25+4+16=)45 In component form, the product is expressed as: (I..I) cosa= I =,r-x = =5 , c o s fj= m = 'rv- = =2 ' Let A = A,! + A,) + AJ, If = Bx! + By] + Bzk. y45 y45 Then z4 cosy = 11 = - = -)4-5 -il ' If = (Ax! + Ay} + AJ)(B,l + B,) + B,k) r PRODUCT OF TWO VECTORS There are two ways of vector multiplication. 1. Scalar or dot product 2. Vector or cross product.

Vectors 2.11 = 11)(B,.7 + By] + Bzk) + Ay)(B,I + 13,) + Bzk) 4. We can find addition of two vectors using dot product: +I1,k(B) + 13,] + B) [n Fig. 2.36, k = A+ ii. So, 11 . B = I1 x B,(i.7)+A,B,(i.])+I1,B,(1.k)+A,B,0-1) LJ \"'; + I1,B,(].]) + A,BzO· k) + I1 zBxCk ·1) A + AzB,.(k.) + I1,B,(k. k) = 11.,13, Fig. 2.36 + 11,13, + I1 z B, '* Ikl = IA + iii or 11. B = 11.,13, + A,By' + I1,B, '* Ikl2 = Iii + sI2 Find the dot product of two vectors '* }?2 = (A + ii)(ii + S) = A2 + 8' + 2.4. ii L~.~,~,2*!,}~?~1=rlY/,,4we, :+~mBf'in+d·s2uAbt·riiac=tiovn'\"A~2f t+wo13V2~+~_~2~AE~B:~c.oc2s~B._JI A = 37 + 2) - 4k and S = 27 - 3) - 6k. Sol. AS = 3 x 2+2 x (-3)+(-4) x (-6) = 24 If the sum of two unit vectors is a unit the magnitude of their difference. Dot product of a vector with itself A vector is parallel to itself. So, the angle of a vector with itself iiis zero. .. A. = I1l1eosO° = A' lience, the dot product (~la vector with itse{fis square (~fits magnitude. We can also write: A\".\"A = 1A\"12 ::::} Ip /<rFor example: + <21 = Sol. Let n! and 112 are the two unit vectors, then their slim is + (2). (I' + Q) \" \"+\" e=>fls=nJ n2 n'S 2= n j + n 2 +2n j 121 2 C O S (Taking A= P+ (2) =1+1+2cosO Ih <21' = (P + (2)(h (2) nsSince it is given that is a unit vector, so 118 = 1. Therefore, Ip + <212 = 1'2 + Q2 + 2p· Q = 1\" + Q' + 2PQcosB '*) = ) + I + 2eose I cosO =-2 Ip -Similarly, '* B = 120\" <212 = 1'2 + Q' - 2P. <2 nd nNow, the ditIerence vector is = I - n2 = p2 + Q2 _ 2PQeosB '* n~ =n;+nl·-2n,I1,cosB = 1+ 1-2eos(120\")=3 Important Points '* 11</ =..[3 1. Angle between two vectors can be calculated from: Find the value of In so that the vector All a! b! +. .a2.b2. .+.[l:.lh.3 .. 37. - 2} + may be perpendicular to the vector 27 +6) +mk. cosl)=~= AB + + +V!al2(122 a:'l / h 2 + /)2' b3' Sol. For the vectors to be perpendicular their dot product has to += 1 IA sl IA - sl,2. If then angle between A and S is be zero. 90\". .. (37 - 2) + k) . (21 + 6] + mk) = 0 sl slIAProof: Given + = IA -'- '* '*6-12+m=0 m-6=0,* 111=6 I\" \"I' l\" \"I'Squaring both side~ w: get A + 8 = A -:: 8 Vector or Cross Product '* II' + B' + 2A . B = A2 + 8 2 - 2A . B 11 BCross product of two vectors and is equal to the product '* 4A. ii = 0 11of the magnitudes of and B and sine of the shortest an- 11 B egie between them, i.e., x = 1113 sin h. where Ii is the '* ii . iiS = O. Hence, is perpendicular to B. unit vector which represents the direction of It x B and it is s) . s)3. If (A + (A - = 0, then Aand S arc equal in perpendicular to the plane containing A and ii. It is given by magnitude, i.e., A = 13. right handed screw rule depicted by Fig. 2.37. Note that h is (A (A -Proof: Given + ii). ii) =0 perpendicular to both Aand S. '* '*A.A·-/i·s+ii·A-S.S=O A2 -13'=0 Iii slMagnitude of A x Ii: x = A13sin8 '* '*112 = 13' 11 = B. Hence proved. I BInFrom here, we can write: Ii x ii = Ii x

2.12 Physics for IlT-JEE: Mechanics I I No n n a-l-+to the plane ) x; = -U x I =),7 x k = -],1< X.1 = -7] I _} BSo, we haveA x = A,BxC-k) + A,B,] + A,B,k ~ ~ ,: A <lndE + A,By(-i) + AxB,(-)) + A,B,(7) AxB = (AyB, - A,B,)! - (A,B, - A,B,)] + (A,By - AyBx)\" ,,---- --------------1, ,, ,,,I B...) ,I ,/ e ,f I ~) I CROSS PRODUCT METHOD 2: DETERMINANT ,\" f A , ',c__________ ~----~--_~ METHOD plane of A and B A BCross product of two vectors and can be obtained easily ~·ig. 2.37 by using the following method. _ AX B A x B = (AJ + A,) + A)) x (B,! + By] + B,k) n.=Iii-X-B-I ;]k Unit Vectors and Their Cross Product = Ax Av A z i,], and k are unit vectors along X-, y-, and z-axis, respectively. 13,13, B, .The magnitude of each vector is 1 and the angle between any of ,, , n n,two unit vectors is 90\". So,; X ) = (1)( I) sin 90\" = where -1-rI -----------.-I ------------ n is a unit vector perpendicular to the plane containing vector A,I' ,,,,,,,,: 7and). Ex : To find out the resultant of any two unit vectors in a cross product use the following rules (see Fig. 2.38). Fig. 2.39 1. Multiplication of any two unit vectors in anticlockwise di- Here, we will use 7, ] , k one hy one. When 1 is chosen, its rection gives third unit vector with positive sign. COlTcsponding row and column become bound and remaining 2. Multiplication of any two unit vectors in clockwise direction clements are subtracted after cross multiplication (Fig. 2.39). gives third unit vector with negative sign. So, i(AyBz - BvA z), is the component along i. ,oJ,, Similarly, in t\"he case of.7, the row and column in which it is k, present become bound and remaining clements are subtracted after cross multiplication (Fig. 2AO). Fig. 2.38 So, f(A,.B, - ByA,) - )(AxB, - BxA,) is the component along I and). . From these rules, we obtain the following results. -7------------]r: ------------k- From Rule 1: Ax A;,:,,, A;c 1. ; x) = k 3. kx;=) From Rule 2: 2. k x } = -i 3.lxk=-] By 1. ] x I = -k Fig. 2.40 CROSS PRODUCT METHOD 1: USING Same argument will follow for k as is fori and] (Fig. 2.41). COMPONENT FORM B:. A x = I(A,B, - B,A,) - )(AxBy - B,A,) A x B = (Ax; + Ay] + A,k) x (Bci + By) + B,1<) = AxB,(i x I) + A,Bxc] x;) + A,BxCk x i) + k(A,B,. - B,A,) + A,B,(7 x)) + AyByc] x]) + A,By(k x )) + A,B,(i x k) + A,BJ] x k) + A,B,(k x k) A.,><AY7------------J------------k4:,, Az ,,,,: [As; x ; = 0, ) x ) = 0, k x k = 0 and; x ] = k, B.y By B; : Fig. 2.41

Vectors 2.13 Properties of Cross Product Calculate the area of the triangle deter- 1. Anticommutative property miued by the two vectors A = 31 + 4} and B = -31 + 7]. The vector product of two vectors is anticornmutative. Sol. We know that the half of magnitude of the cross product of A x B = AB sinen and B x A = BAsine(-il) two vectors gives the area of the triangle. = -Ali sine\" = -(A x B) SO, B x A = -(A x B). It means B x A # A x B. i J Ii 2, Distributive property 340 =7(O-O)-](O-O)+k(21+12)=33k Vector product is distributive, Le., -3 7 0 A x (B + C) = A x B + A x C IATaking magnitude Bx 1 = J33i = 33. So, area of 3, Associative property 1-> \"\"Itriangle = 2I A x tf =\"323 sq. unit. (A + B) x (C + D) = A x co, + A x 15 + B x C Calculate the area of the parallelogram +Bx15 when adjacent sides are given by tbe vectors 4. Cross product of two parallel vectors Cross product of the parallel vectors is zero. A =1+ 2} + 3k and B = 21 - 3] + k. eAs = 0\" (for parallel vcctors), so Sol. We know that the area of the parallelogram is equal to the (A xB)=AlisinO\"n=O magnitude of the cross _product of given vectors. Important Points 1 )k Now. A x B = I 2 3 1. ff two vectors represent the two adjace~t sides of a par- allelogram. then the magnitude of thc cross product will 2 -3 I give the area of the parallelogram. Mathematically: =1(2+9)+J(6-1)+;;(-3-4) Two vectors A and B are represented by the two adja- =111+5]-7k ccnt sides PQ and PS, respectively, of thc parallelogram So, area of parallclogram: as shown in Fig. 2.42. IA BIx = /li2+5'+(-7)2 =v'T95sq. unit. ST ,-------j Concept Application Exercise 2.2 1--- \"\" \"\"\" \\ \" \"-- ! h Ap , L L f ; - - - : : - - - : 1. What is the area of para~lelog~am w\"hose adjacen! side~ are NQ given by vectors If = i - 2j + 3k and B = 4i + 5j? Fig. 2.42 J-2. If the vectors 41 + 3k and 2m1 + Gm] + k arc per- Now, from the magnitude of the cross product: pendicular to each other, then find the value of In. 1A x B 1 = A B sin G = Ah = base x height of parallel- 3. The magnitude of the vector product of two vectors is ../3 ogram = area of parallelogram. times their scalar product. What is the angle between the 2. If two vectors represent the two sides of a triangle, then two vectors? half the magnitude of their cross product will give the J4. What is the angle between 1 + + k and !? area of the triangle. 5. The linear velocity of a rotating body is given by V' A BConsider two vectors and represented by the two ur= x 7.Ifur =1 -2) + 2kand 7 =4) -3k,then sides PQ and PS of triangle PQS (Fig. 2.43). what is the magnitude of V'? A BUsing cross product: x = Ali sinGh 6. Find the magnitude of component of 31 - 2) + k along A B e1'lking magnitude, I x 1 = A B sin the vector 121 + 3] - 4k. = A(B sine) = A x h = base x height S ~ , Q A car is moving around a circular track B ,,,lh with a constant speed v of 20 ms-1 (as sbown in Fig. 2.44). At different times, the car is at A, Band C, respectively. Find 8 -, the cbange in velocity: p 1. from A to C, A Fig. 2.43 Multiplying by 1/2 on both sides, 21 A BI I ' x 1 = 2 x base x height = arca of triangle.. I -- ---'\",-_.\"-_...\"... ,, ~

2.14 Physics for lIT-JEE: Mechanics I where IAI = IA~ + Ai +AI 2. from A to B. , = J22 + 32 + 22 = .,14 + 9 + 4 = ..fi7 . A 27+3)+2k :------- B A = -lA-I = --;Jf7~17~ +~ ~ Find the unit vector of (A B), where v,+-~\"'-'~/ v A= 27 -] + 3k and B= 31 - 2] - 2k. c Fig. 2.44 Sol. (,4 + II) = (27 - ) + 3k) + (37 - 2] - 2k) = 57 - 3] Sol. C+k = (say) 1. Change in velocity, as the car moves from A to C, Magnitude of C = c = [52 + (-3)' + 12]1/2 = .J35 L1vc\" = vc - VA = 20 - (-20) = 40 ms l . C 5;-3j-!-j( f'j. UCA = 40 ms~ 1 in the direction of Vc So,C = - = .J35 C The greatest and least resultant of two forces acting at a point arc 10 and 6 N, respectively. If each force is increased by 3 N, find the resultant of new forces when acting at a point at an angle of 900 with each other. Sol. Let A and B he the two forces Fig. 2.45 Greatest resultant = A + B = IO (i) Least resultant = A - B = 6 (ii) 2. Change in velocity, as the car moves from A to B, S01ving equations (i) and (ii), we get, A = 8 N; B = 2 N ,6.V1111 = VB - VA = VB + (-VA) When each force is increased by 3 N, then L1VBA = )202 + 202 = .,1800 ms- I = 28,28 ms- I A' = A+ 3 = 8 + 3 = II N, Ii' = B + 3 =2 + 3 =5 N. e20 eAs the new forces arc acting at an angle of 90° (i.e\" = 9(P), Also. tllne = 20 = I =? = 45' so R' = .,IA'2 + B'2 = jill)' + (5)2 = y'j46 N, This is the required direction of change in velocity. Two forces whose magnitudes are in the Given it = 0.31 + 0.4] + ck. Calculate ratio 3: 5 give a resultant of 28 N. If the angle of their incli- nation is 600 , find the magnitude of each force. A is a unit vector. Sol, Let A and B be the two forces, Sol. If;\\ is a unit vector, thcn its magnitude must be unity, Then A =3x; B =5x; R =28 Nand e =60' +c=? p+~r+';~ = I, ie\" )(0.3)' + (04)2 2 = I Dividing A by II, AlB = 3/5 or 0,09 + 0, 16 + c2 = I or c' = 1 - 0.25 = 0,75 or c = 0,87 We know that R = .j~A~2-+-1~l2'+~2-A-B cos 0 Determine that vector which when added !c3x=? 28 = j2\"+'(5;)2 + 2(3x)(5-;:) cos 60' to the resultant of -A-). = 3\"i - 5\"j + 7k\" and B-+ = 2\"i + 4\"j - 3k... =? 28 = .j9x2 + 25x' + 1Sx' = 7x 28 =?x=-=4 gives a unit vector along y~direction. 7 A BSol. Here, = 37 - 5] + 7k and = 21 + 4] - 3k Hence, the forces are: A =3 x 4 = 12 N, B =5 x 4 =20 N. -~-}-\">\" ~\" \"\" One of the rectangular components of a + + + +Rcsultant R = A B = (3i - 5 j 7k) (2i 4j - 3k) velocity of 100 ms- 1 is 50 ms- 1, Find the other component. = 57 -] Hk Sol. Here, \\I = JOO ms\" I, Let IIx = 50 ms- I, IIx = \\I cos IJ X,Let the vcctor to he added is SO, thc unit vector along =? 50 = 100cosO I y-direction ] = 51 - ] + 4k + Jt JtSo the required vector = ] - (5i - ] + 4ii) =? cosO = - ore = 60\" 2 = -57 +2; -4k. a.\\I)' = \\I sin = 100 sin 60' = 100 x ..2/3 = 50../3 ms- i Find the unit vector of .A An aeroplane takes offat an angle of60° to it =27 +3) +2k. A= IAI' the horizontal. If muzzle velocity of the plane is 200 kmh- 1, calculate its horizontal and vertical components. Sol. it = 27 + 3] + 2k. We know thllt eSol. Here, II = 200 kmh-I, = 6Q0,

e.'. Horizontal component Vx = ,v cos = 200 cos 60° Vectors 2.15 = 200 x :2I = 100 kmh- 1 So, let us first determine it x B. eVertical component Vy = v sin = 200 sin 60° xNow, A B = (31 - 5] + 10k) x (67 + 5] + 21<) 2= 20Q x F3 = 100,.I,'3C kmh- 1 1 7 I< Prove that (A + zin . (ZA - 3B) 3-510 =1(-1O-50)+)(60-6)+k(15+30) 65 2 = ZA' + AB coso - 6B2. Sol. (A + 2ih·(2,1 - 3ii) =2,1. A- 3,1. B+ 4B· A- 6(R. R) = - 607 + 54] + 45k = 2(A. A) - 3ABcosB +4,1Bcos8 - 6(B. in Magnitude: IA x B I=J(-60)2 + (54)2 +(45)' =,)8541 nSo, required umt vector: = -60; +~54] + 45k = 2A' + ABcosB - 68' ,.,8541 A body constrained to move along the z- A man rows a boat with a speed of 18 axis of a coordinate system is subjected to a constant force F kmh-1 in the north-west direction. The shoreline makes an given by il = -7 +2) +3k N, where 7,), and 1. represent angle of 150 south of west. Obtain the components of the velocity of the boat along the shoreline and perpendicular to unit vectors along X-, y-, and z-axis of the system, respec- the shoreline. tively. Calculate the worl' done by this force in displacing the body through a distance of 4 m along the z-axis. Sol. The north-west direction of the boat makes an angle of 60° with the shoreline (Fig. 2.46). Sol. Displacement = 4k. Force: F= -7 + 2} + 3k N Since work W is the scalar product of force and displacement, ,, .. W =(-7 +2]+3k)·4k = -4(7 . k)+ 8(j .k)+ 12(i( .k) s W = 12joule. because i .k = 0 = ] . k and k .k = I Fig. Z.46 1. Prove that the vectors A = 31 - 2) + k, B = I - 3) Component of the velocity of boat along the shoreline + 51., and C = 2i + ) - 41. form a right-angled triangle. = 18 eos 600 kmh- 1 = 9 kmh-· 1 2. Determine the unit vector parallel to the cross product of Component of the boat velocity along a line normal to the the vectors A = 31 - 5) + 10k and B = 6i + 5) + 2k. shoreline Sol. ~= 18 sin 60° kmh- 1 = 18 x kmh- 1 = 15.59kmh-1 1. The given vectors will conslitute a triangle only if one of the A point P lies in the xy plane. Its posi- given vectors is equal to vector sum of the remaining two tion can be specified by its x, y coordinates or by a radially vectors. In the given problem. B + C = A. SO. the given directed vector \"7 = (xi + y) making an angle 0 with the x-axis. Find a vector ir ot'unit magnitude in th'e direction of vectors do form a triangle. This triangle will be right angled vector \"7 and a vector I, of unit magnitude normal to the only if the dot product of two vectors (out of the given three) vector I, and lying in the xy plane (Fig. 2.47). is zero, y p A .B = (37 - 2) + k) .(1 - 3) + 5k) = 3(7 .1) + 6c] . ]) + 5(k . k) = 3 + 6 -1- 5 = 14 ~,., It .C = (l - 3] + 5k) . (27 + ] - 4k) = 2(7 . i) - 30 . ]) o~L.Le:....-_ _ _~x - 20(k . k) = 2 - 3 - 20 = -21 Fig. 2.47 C . A = (21 + j - 4k) . (37 - 2) + k) = 66 . 1) - 2c] . ]) Sol. When a vector is divided by its magnitude, we get fi unit vector (Fig. 2.48). - 4(k . k) = 6 - 2 - 4 = 0 Since the dot product of C andA is zero. therefore it implies that C is perpendicular toA . 2. The unit vector parallel to (A x B) is given by AXB sf11 = ~--c- IA x

2.16 Physics for JIT-JEE: Mechanics I x1+] ,x ,y x Now. let 10 = 101 + 1fJ. where 01 and fJ arc coefficients to be cosO = - or x = reosO i,. = =I--+}-: r de,termincd, r r rr Making use of scalar product: +.7Agam. , S1\"11e r sm. e. = -Y Of Y = 1, .10 = (1 cos8 + j sinO)· (la + jfJ) (i) r So, we get: 7,. = 7coso sinO Since dot product of perpendicular vectors is zero, y-axis 1, .10 = 0 =} a cos e + fJ sin e = 0 ry a =-fsJin-e e,k::....L_-c-_..L-c- x-axis case 0x Again, since 70 is a\" vector of unit magnitude: 0,'2 + f32 = 1. e e.Put the value of a and get: fJ = ± cos and a = 'I' sin Fig. 2.48 And we get 10 = '1'1 sin e ± ] e.cos Since 10 should have -ve x component and +ve y component, so finally we have: 70 = -7 sin e + ] cos 0 EXERCISES Subjective Type ,SoliJtions onpage 2.?2 +3. At what angle two forces (P Q) and (P - Q) act so that their resultant is 1. a. What is the essential condition for the addition of two vectors? 4. Two forces 7 and 3 N simultaneously act on a body. What b. Is addition of any two scalar quantities meaningful? is the value of their (i) maximum resultant, (ii) minimum resultant, and (iii) what will be the resultant if the forces c. Component of a vector can be a scalar. State true or < act at right angle to each other? false. d. Can two vectors of same magnitude add to give zero 5. Find the resultant forcc of the following forces which are acting simultaneously upon a particle. resultant vector? If yes, under what conditions? a. 30 N due East h. 20 N due North e. Can two vectors of different magnitudes add to give zero resultant vect.or? Can three vectors give the zero c. 50 N due West d. 40 N due South resultant vector on addit.ion. If yes, under what con- A B6. For the vectors and in Fig. 2.49, use a scale drawing ditions? f. Can a rectangular component of a vector have mag- to find the magnitude and direction of nit.ude greater than the vector itself? y -, g. Can a vector be zero if one of its component is not B (18.0 m) zero? h. Can scalar product of two vectors be a negative quan- tity? i. State the condition (regarding the value ofdot or cross A x__~(1~2.0 ~Ill) __(~)~_-L3_7.0_°_____ product) for which two vectors are: i. parallel to each other, Fig. 2.49 ii. perpendicular to each other. a. the vector sum A + B. j. Is possession of magnitude and direction sufficient h. Athe vector difference - B. for calling a quantity a vector quant.ity? k. Is it necessary that sum of two unit vectors is also a unit vector? c. From your answers to parts (a) and (b), find the mag- I. What will be the difference in the product of nitude and direction of i. a real number with a vector, and ii. a scalar with a vector. i.-A-B H.B-A iii. Explain the difference between the following a b7. If two vectors = 4 ms- 1 and = 7 ms- 1 be inclined data? • 4 (5 kmh- 1, east) at an angle of 60° to each other, then determine the direc- .4h(5kmh- l ,east) tion and magnitude of their resultant. 2. Two equal forces have a rcsultant cqual to one and a half times the either force. Find the angle between the forces. 8. Find a unit vector along and opposite to the vector 31 -4} + 12k. 9. If a=21-3].b=61+2]-3k and c=l+k, then find

a.3Ct+26+c Vectors 2,17 b.a-66+2c 10. Given two vectors It = 4.001 + 3.00] and 17. Two vectors It and B have magnitudes A = 3.00 and B = 5.001 - 2.00]. BB = 3.00. Their vector product is It x = -5.00k a. Find the magnitude of each vector. B?+ 2.001. What is the angle between It and b. Write an expression for the vector difference 18. Two vectors have magnitudes 5 units and 12 units, respec- Ii - B using unit vectors. tively. Find their cross product if the angle between them is 30\". c. Find the magnitude and direction of the vector differ- 19; Given two vectors. A= 31 + ] + k and il = 1 - ) -Ii. ence It - 8. Find the d. In a vector diagram show It, B, and It - B, and a. area of the triangle whose two sides arc represented also show that your diagram agrees qualitatively with your answer in part (c). by the vectors Aand k 11. a. Is the vector (i + ) + Ii) a unit vector? Justify your b. area of the parallelogram whose two adjacent sides answer. are represented by the vectors Aand k . b. Can a unit vector have any rectangular component c. area of the parallelogram whose diagonals are repre- . with a magnitude greater than unity? Justify your an- sented by the vectors Aand il. swer. 20. On a horizontal fiat ground, a person is standing at a point c. If It = a(31 + 4), where a is a constant, determine A. At this point, he installs a 5 m long pole vertically. Itthe value of a that makes a unit vector. Now. he moves 5 m towards east and then 2 m towards north and reaches at a point B. There he installs another 12. Resolve the vector R = 21 + 3] along the directions of 3 m long vertical pole. A bird flies from the top of first pole to the top of second pole. Find the displacement and 1 + 2) and 1 - ] and write down the resolved compo- magnitude of the displacement of the bird. nents. 21. Find the vector sum of N coplanar forces, each of magni- R13. Find the rectangular component of vector = 21 + 3) along A=1+). tude F, when each force is making an angle of 2lf with N 14. Find the angle between each of the following pairs of vectors. that preceding it. a. It = -2.001 + 6.00) and B = 2.001 - 3.00) 22, Can you find at least one vector perpcndicular to It Bb. = 3.001 + 5.00] and = 10.001 + 6.00] 31 - 4) + 7k? c. It = -4.0Q1 + 2.00)and B = 7.001 + 14.00] 23, Establish the following inequalities: 15. A cube is placed so that one corner is at the origin and three edges are along the x-, y-, and z-axis of a coordinate a, IA + ill::, IAHill system (Fig. 2.50). Use vectors to compute b. IA + ill?: IIAI-lilll c. IA - ill::, IAI + lill -y d. IA - ill?: IIAI-lilll 24. Two forces P and Q acting at a point are such that if P is reversed, the direction of the resultant is turned through 9D\", then prove that magnitudes of the forces are equal. e,25. Unit vectors F and Qare inclined at an angle then prove that IF - QI = 2sin(ej2). Fig. 2.50 Objective Type Sdlulidnsonpage 2:25, a. the angle between the edge along the z-axis (line ab) 1, The sum and difference of two perpendicular vectors of equal lengths are and the diagonal from the origin to the opposite corner a. also perpendicular and of equal length (line ad). b. also perpendicular and of differeut lengths b. the angle between line ac (the diagonal of a face) and c. of equal length and have an obtuse angle between line ad. them It16, You are given vectors = 51 - 6.5] and d, of equal length and have an acute angle between them B = !O1 + 7]. A third vector C lies in the x - y plane. Vector C is perpendicular to vector It and the scalar product of C with B is 15. From this information, 2. The minimum number of vectors having different planes find the components of vector C. which can be added to give zero resultant is L2 ~3 ~4 ~5

2.18 Physics for IIT-JEE: Mechanics I 3. A vector perpendicular to 1+ } + Ii is A9. Vector is 2 cm long and is 60\" above the x-axis in the first quadrant. Vector ii is 2 cm long and is 6()' below the a.1-}+k b.1-}-k Bx-axis in t:1C fourth quadrant. The sum Ii + is a vector c. -1 -} - k d. 31 + 2} -5k of magnitude 4. From Fig, 2.St, the correct relation is a. 2 em along + y-axis b. 2 em along + x-axis N c. 2 em along - x-axis o d. 2 em along - x-axis -, 10. What is the angle between two veetor forces of equal mag- D nitude such that the resultant is one,third as much as either Fig. 2.51 of the original forces? (1):3b. cos-I (-H)a. cos-I a. A+ii+E=O e. 45\" d. 120\" b. C- D=-A c. i3+E-C=-D 11. The angle between A+ Band Ax ii is d. All of the above a. 0 b. 11:/4 C. 11:12 d. 11: 12. The projection of a vector r = 31 + } + 2k on the x-y S. Out of the following set of forces, the resultant of which plane has magnitude cannot be zero a. 3 b.4 c. v'i4 d. v'TO iii liil, ii '13: IAIIf IA + = = then the angle between Aand a. 10,10,10 b. 10,10,20 c. 10,20,20 d. 10,20,40 is 6. The resultant _of two vectors A and ii is perpendicular iia. 120\" c.90\" to the vector A and its magnitude i,s equal to half of the 14. lfvectors A = 1+ 2} + 4k and = 51 represent the two magnitlilk of V(;;t:lor B, The angle between Ii and B is sides of a triangle, then the third side of the triangle can have length equal to --, a. 6 J! b. J56 c. Both of the above Fig. 2.52 rl. None of the above a. 120\" b. 1500 15. Given IA,I = 2, IA21 = 3 and IA' + A21 = 3. Find the c. 135° d. None of these value of (A, +2A2)' (3A ,-4A2) a. 64 b. 60 c. 62 rl. 61 7. The ratio of maximum and minimum magnitudes of the A.B16. Three vectors -j, B, C satisfy the relation =0 resultant of two vectors --7 and ->- is 3 : 1. Now, I ---,)- ! is C Aand',4 . = O. The vector is parallel to a h a -.-, equal to -, ~ -, a. B b. C -, d. 41 b 1 --, -, rl. BXC a. 1 b\" 1 b. 21 b 1 c. 31 b 1 c. B'C 8. 'Two forces, each equal to F, act as shown in Fig, 2.53. Their resultant is +---;. --} --> ~> ~> - } 17. If A = B c, and the magnitudes of A, B, Care 5, 4, -> --, F and 3 units, then angle between A and C is G)b. coS-I d. 11: 2 F 18. -+ = . s\"ei + A sin e.\"i. A vector ---->- which is per- Fig. 2.53 Given: A A co B , a. FI2 b. F c. .j3F d. .;sF Apendicular to is given by a. Bcose1- B sin e}

Vectors 2.19 b. Bsinel- Bcose) b. the minor diagonal of the parallelogram c. B cos /)1 + B sine) c. any of the above d. Bsine7 + Bcose) d. none of the above A J19. The angle which the vector = 27 + 3 makes with y-axis, where I and} are unit vectors along x- and y_ e. lei8 18128. Given that A+ = If IAI=4, =5and = v'6l. axes, respectively, is The angle between Aand 8 is a. cos-'(3/S) a. 30\" b. 60\" c. 90\" d. 120\" • c. tan--'(2/3) b. cos-'(2/3) 29. Given vector A= 21 + 3J, the angle between A and d. sin~'(2/3) y-axis is p20. Given: = 37 - 4]. Which of the «lllowing is perpen- a. tan- 1(3/2) b. tan-' (2/3) -> dicnlar to P '? c. sin-' (2/3) d. eos-' (2/3) a. 31 b.4) a30. If b= 31 + 4]and = 1-), the vector having the same d. 41 - 3] amagnitude as that of band parallcl to is c. 41 +3] 21. In going from one city to another, a car travels 75 km a. S. . b. J52(i•+•j) j) north, 60 km north-west and 20 km east. The magni- J2(i - tude of displacement between the two cities is (Take c. 5(1 ~ ]) d. 5(1 + ]) I/J2 = 0.7) a. 170 km b. 137 km 31. Choose the wrong statement c. 119 km d. 140 km a. Three vectors of different magnitudes may be com- 22. What is the angle between Aand 8, if Aand 8 are the bined to give zero resultant adjacent sides of a parallelogram drawn from a common ' b. Two vectors of different magnitudes can be combined to give a zero resultant.. point and the area of the parallelogram is ABI2? . a. IS' b. 30° c. 4So d. 60° c. The product of a scalar and a vector is a vector quan- tity. _~_ -4 .-)- -'\" _+ ....... +23. b I = I a - b I. Two vectors a and b are such that-,I a ~ d. All of the above are wrong statements. What is the angle between a and b? a. 0° b. 90° 32. What displacement at an angle 600 to the x-axis has an c. 60\" d, 180\" x-component of 5 m? i and] are unit vectors in x and y -+ \" \" ...) \" \" directions, respectively. + +24. Given: A = 4i 6j and B = 2i 3j. Which of the fol- a. s1 b. S1+5] , lowing is correct? -. ~ d. All of the above c. s1 + 5.[3] ._, ~ ~ b. A'B = 24 -, ---) a. AxB=O 33. Mark the correct statement d. A and B are -0 a. Iii + &1 ::: lal + Ihl anti parallel b. la +bl :s lal + Ihl c. IA = 2 c. Ii - hi :: lal + 1&1 I BI d. All of the above .25. Given: A=21+ p ]+qk and B=s1+7]+3k. If --} -) A II B, then the values of p and q are, respectively, . 14 6 14 6 34. Out of the following forces, the resultant of which cannot be 10 N? a'Sand S sb· 3 and 31 61 d. - and - c. -S and - 3 44 a.ISNand20N b.lONandlON 26. If Ais perpendicular to 8, then c. SNand 12N d. 12 N and I N a.Ax8=0 35. Which of the following pairs of forces cannot be added to give a resultant force of 4 N? b.A.[A+8]=A2 c. A. 8 = AB a. 2 Nand 8 N b. 2 Nand 2 N d. A.[A+8]=A2 +AB c. 2 Nand 6 N d.2Nand4N a27. If the angle between vectors and &is an acute angle, 36. In an equilateral triangle ABC, AL, BM, and CN are medians. Forces along BC and BA represented by them then the difference a- b is will have a resultant represented by a. the major diagonal of the parallelogram a. 2AL b. 2BM c. 2CN d. AC

2.20 Physics for IIT-JEE: Mechanics I 37. The vector sum of two forces is perpendicular to their A B44. A vector when added to the vector = 31 + 4] yields vector difference. The forces are a. equal to each other a resultant vector that is in theJositive y-dircction and h~s h. equal to each other in magnitude a magnitude equal to that of B. Find the magnitude of A. a. JTO b. 10 c. 5 d. v'l5 c. not equa1 to each other in magnitude 45. ABCDEF is a regular hexagon with point 0 as centre. The d. cannot be predicted value of All + AC + AD + AE -I- M is 38. If a parallelogram is formed with two sides represented a. 2AG b. 4AO c. nAG d. 0 a aby vectors and b, then + brepresents the 46. In a two dimensional motion of a particle, the particle a. major diagonal when \"the angle between vectors is moves from point A, position vector (1, to point B, po- acute sition vector r2' If the magnitudes of these vectors are, b. minor diagonal when the angle between vectors is obtuse respectively, 1', = 3 and 1'2 = 4 and the angles they make c. hath of the above with the x-axis are 8, = 75\" and 8, = 15\", respectively. d. none of the above then find the magnitude of the displacement vector. --v ~> .-). .-). 39. The resultant C of A and B is perpendicular to A. Also, A B } .\"} --~ ~~ I A I = I C I. The angle between A and B is a. 4n\": rad 3n: b. - rad 0, 4 82 S;r 7rr Fig. 2.55 c. - rad d. -rad 44 -, ~ d. v'l5 = =40. Two forces of F, 500 N due east and F2 250 N due , .-). --) b. ffi c. 17 a. IS north have their common initial point. F2 - PI is a. 2S0v'S N, tan-'(2)W ofN 47. The sum of the magnitudes of two forces aeting at a point is i6 N, The resultant of these forces is perpendicular to b. 2S0 N, tan-'(2)W ofN the.smallerforce and has a magnitude of8 N.lfthesmaller force is of magnitude x, then the value of x is c. zero d. 750 N, tan-' (3/4) N ofW a.2N b. 4 N c. 6N d. 7 N aB,GA, oC41. The resultant of the three veetors 48. The angle between two vectors Aand Bis e. Resultant of and shown these vectors Rmakes an angle el2 with A. Which of the in Fig. 2.S4 is A following is tme? B a. A = 2B b. A = BI2 c. A = B d. AB = 1 Fig. 2.54 49. The resultant of three vectors 1, 2, and 3 units whose a. r b. 21' directions arc those of the sides of an equilateral triangle is at an angle of a. 30n with the first vector c. r(J + J2) d. 1'(J2 - 1) b. 15\" with the first vector c. -lOO° with the first vector b42. Two vectors -({ and are at an angle of 60° with each d. 1500 with the first veetor other. Their resultant makes an angle of 45\" with tI. If 50. A particle moves in thexy plane with only an x-component of acceleration of2 ms-- 2 . The particle starts from the ori- I;; I = 2 units, then I tIl is gin at t= 0 with an initial velocity having an x-component of 8 ms-' and y-eomponent of -15 ms--'. The total ve- a. v\"l b.v\"l-I locity vector at any time t is c.v\"l+l d. v\"l/2 a. [(8 + 2t)1 - 15]J ms-' 43. The resultant of two vectors Pand Qis k If the magni- b. zero Qtude of is doubled, the new resultant vector becomes c. 211 + IS] P. Rperpendicular to Then, the magnitude of is equal to d. directed along z-axis a. P + Q b. P c.P-Q d.Q

51. A unit vector along incident ray of light is 7. The unit vcc- Vectors 2.21 r,tor for the corresponding refracted ray of light is Ii is y a unit vector normal to the boundary of the medium and directed towards the incident medium. If m be the re- fractive index of the medium, then Snell's law (20d ) of -refraction is ~L-____________ x a. i x fi = !1-(h + -, Fig. 2.56 r) b. I . Ft = !1-(f . h) c. The signs of x- and y-components of dl + (h arc pos- c. I x Pi = !1-(f x Pi) itive. d. None of these. d. /LCI x Pi) = ? x fi 52. The simple sum of two co-initial vectors is 16 units. Their 2. Given two vectors A= 37 + 4] and B= 7+ ].I! is the vector sum is 8 units. The resultant of the vectors is per- pendicular to the smaller vector. The magnitudes of the angle between A and B. Which of the following state- two vectors arc a. 2 units and 14 units ments is/arc correct? b. 4 units and 12 units I I C,~i)a. A cos Ii B.is the component of Aalong c. 6 units and 10 units ~ C;1)b. IAIsin Ais the component of perpendic- d. 8 units and 8 units 53. The components of a vector along x- and y-directions are ularto B. (n + 1) and I, respectively. If the coordinate system is rotated by an angle 0 = 60D , then the components change C-;;)c. IAIcos Ii B.Ais the component of along to nand 3. The value of n is I I C~ j)d. A sin Ii is the component of Aperpendic- a. 2 b. cos 60\" c. sin 60- d. 3.5 54. Two point masses 1 and 2 move with uniform velocities ular to B. v; v;,and respectively. Their initial position vectors arc\" 3. If A=21+)+k and B=7+J+k arc two vectors, ;:; and I:;, respectively. Which of the following should. be then the unit vector satisfied for the collision of the point masses? . (-Ja. perpendicular to A is ..+fi k) v; + v; !v;-v;1 -) -} (21 +] + k) V2 -VI b. Ais -'---'=c--'- Il)~-~ I parallel to .j6 c. Ir; +r; I c. perpendicularto B-. is (-J..+fi k) --;. -;. • 7+J+k V2 -- VI d. parallel to A is M v3 IV; +v; I 4. If (vt + 1:0:) is perpendicular to (vt - ~). then 55. What is the resultant of three coplanar forces: 300 N at a. 1ft is perpendicular to V! 00,400 N at 30\", and 400 N at 150°? b. Ivt I= IVi: I a. SOO N b. 700 N vtc. is null vector vtd. the angle between and 1:0: can have (lny value c.I,IOON d. 300 N 5. Two vectors Aand B lie in one plane. Vector (' lies in a MuLtiple Correct Answers Type different plane. Then, A+ B+ C 1. Which of the following statement is/are correct (sec a. cannot be zero Fig. 2.56)7 b. can be zero ~ c. lies in the plane of Aor B a. The signs of x-component of dl is positive and that 'd. lies in a plane different from that of any of the three --;. vectors of d2 is negative. ..-~ -'>- b. The signs of the y-component ofdl and d, arc positive and negative, respectively.

2.22 Physics for IIT-JEE: Mechanics I ANSWERS AND SOLUTIONS SubjectivE! Type b. Here. R = J2(P2 + Q2); A = P + Q, B=P-Q 1. a. The two vectors should be of the same nature, i.e.• Apply R2 = A2 + B2 + 2AB cos 0 a force vector can be added only into another force =} 2(p2 + Q2) = (P + Q)' + (p_Q)2 + 2(P vector, not in a velocity vector, say. + Q)(P - Q)cosll h. No, they should have same nature, i.c., mass can be e=} cosll = 0 =} = 90° added into mass, not in length, say. 4. a. Resultant is maximum when both vectors act in same c. False, component of a vector is also a vector. direction. Rmox = A + B = 7 + 3 = ION d. Yes, if they are equal and opposite. h. Resultant is minimum when both vectors act in op- c. Two vectors of different magnitudes cannot add to posite direction: Rm;o = A - B = 7 - 3 = 4 N give zero resultant. Three vectors of different magni- tude can add to give zero resultant ifthey are coplanar. c. If both vectors aet at right angle, then R = ,/A2 + B2 _ ,/72 + 32 = y'58 N f. No 5. Resultant force: F = 301 + 20) - SOl - 40) 20N - ~SON .-----~__---4>30N t t40N g. No, a vector can be zero if all components are zero, = 201 - 20) F = '/202 + 202 = 20,/2 s - w h. Yes i . i. cross product is zero 20 ii. dot product is zero fo' j. No, it should follow the vector rules of addition mul- -20 tiplication etc. For example, electric current has both magnitude and direction but still it is a vector quan- Fig. 2.57 tity. 6. a. R=A+B k. No I. . nature of vector remains same, only magnitude R = ../\"1-;022\"+-:-:1\"\"82°-:-+-;2:-x---;-12;:-x-;Ic;;S-e-os-c(C;-1~8:0\"\"\"---:;3\"'7::7o) I. =ll.Im may change 12 sin (180 - 37\") ii. magnitude of the vector changes tan (a - 37°) = \" iii.• 4 (5 kmh I, east) = 20 kmh-l, east 18 + 12cos(1S0 - 37') Here, we are multiplying a velocity 5 kmh-l, east =} a-37° =40.6 =} a =77.6\" with x-axis (Fig. 2.58) with a real number 4. The final result obtained is y 20 kmh- l, cast which is also a velocity. So, nature R of vector remains samc if it is mUltiplied with a ._, real number, only magnitude may change. B =.4 h (5 kmh-l, cast) 20 km, east Here, we multiply velocity with scalar quan- tity, time 4 h. The result obtained is 20 km, east which is a displacement vector. Here, nature of vector. changes. Fig. 2.58 e2. R' = A2 + B2 + 2AB cos Given A = B,R = 3AI2 h.R=A-B R = '/\"\"12\"'2;'-+\"'-'I\"\"S\"2-:+\"\"C2O:x-\"\"\"\"'Ic;;2-x-\"'I\"\"S-co-s-'3\"'7;::\" - 28.5 (~Ar=} =A2+A2+2A2cose 18sin3T tan a = .,..,.---:c;---:c-::- =} a '= 22\" e=} case = \"I8 =} = 830 12 + 18cos37\" 3. a.l)=../3p2+Q2, A=P+Q,B=P-Q Angle with x-axis = ISO\" + 22\" = 202\" Apply R2 = A2 + B2 + 2AB cos e c. - A- Bis opposite to A+ Band havil.g same mag- =} 3P2 + Q2 = (P + Q)' + (p_Q)2 nitude as that of Ii + B. +2(p+ Q)(P - Q)cose d.B - Ii is opposite to Ii - B and having same mag- nitude as that of Ii - B. e=} cos = 1/2 =} = 60°

Vectors 2.23 7. R = ,.)42 + 72 + 2 x 4 x 7 cos 60\" = -!93 mls 11. a. ~'\" 7sin 600 7xf3 Leta =i+j+k tana = 4 + 7 cos 600 15 Magnitude: I(; 1= \")J2 + J2 + 12 = ,jJ C~)=> a = tan-I Since magnitude is not 1, so (; is not a unit vector. -, b. No, a rectangular component cannot be greater than b ______~-- a vector itself. Since rectangular component, say, 7 R,,,,,,,,,' Rx = R cos e, cos e never becomes greater than I. ex , So Rx. never becomes greater than R. lU-'-_-+'4 12. R= 21 + 3], Let A= 1+ 2), B=l - ] 4a Then, we can write R= mA + nB, where rnA is the component of Ralong Aand Fig. 2.59 nB is the component of Ralong B 8. LetA = 37 -4] + 12k, then A = ,.)32 +42 + 122 = 13 => 21+3] =m7 +2m] +n7 -n} A, A 37-4]+12k => m+n=2,2m-n=3 Unit vector along A is A = - '= --~'-- A 13 From these m = 5/3, n = 113 • 5, , rnA = 3\" (i +2j), --> \" A AUnit vector opposite to A is - A = - nB• = -1 (i' - j') = _ (31-4] + 12k) 3 . 13 13. R ARectangular component of along is = (R . A) A 9. a. 3(; +2 b+L' = 3(21 - 3)) + 2(67 + 2] - 3k) (R= A) A= (2 x 1+ 3 x 1)(1 + ]) = ~ (i + ]) + 1+ k =191 - 5} - 5k A2 (,.;2)2 2 b. (; -6 b+2 ,0 = 21 - 3] - 6(61 + 2] - 3k) 14. a. A.B=-2x2+6x(-3)=-22, +2(1 + k) = -327 - 15] + 20k A = ,.)2' + 62= AU, B = ,.;2' + 32= v'T3, m10. a. A = ,.)42+ 32= 5, B = ,.)52+ 22 = cos e = A· B -22 -- = -AoUov~'T=3 AB => e = COS-I Cio~) Fig. 2.60 b. eIn the same way as above: = COS-I [ ;;,2~J '140'113 ._} -} '\" \"- c. A. B= -4 x 7+2 x 14=0 b. A-B=-i+5j c, Magnitude: I A - ,; I = ,.) j2 + 52 = v'26 So, angle between A and B is 90\" tana = ~ => a = tan-I (~) 15. a. Let side of the cube is d, then Uh = dk ad = di latlladl~, d. See Fig. 2.61 Uh .ad d2 I + dF+ dk;cosO = dxf3d xf3 (_1_)=> 0 = COS-I xf3 b. at: = d] +dk,ad = d7 +d} +dk Iat:lladlat: .ad -, 2d2 ,.;2 B , 45 cosO = = ,.;2d,jJd =,jJ 5 B ~ 2 --------------- => 0 = COS-I (,.;2) Fig. 2.61 xf3

2.24 Physics for IIT-JEE: Mechanics! -+ -} '\" \"- + + +22. Let xl y) zk is perpendicular to 31 4) 7k, then +16. Since C lies at x-y plane, so let us assume C = xi y j, their dot product should be zero. ~~ (i) ~ 3x -4y+7z =0. Take x = 1,y=2,z=517 Now, C· A = 0 ~ 5x - 6.5y = 0 C' Band = 15 ~ lOx + 7y = 15 (ii) + + +o 0 50 00 0 So, i 2j 'lk is perpendicular to 3i - 4j 7k. A B v'i9From equatl.Olls (.I) and (I\"I), y = -43, x =. -4390 23. a. Sum of any two sides of a triangle is greater than or equal to third side. 17. 1 x 1= ,,/52 + 22 = ~ PR:cPQ+QR v'i9A Bsine = 1 x 1= ~ Iii +BI :c llil +IBI AB 3 x 3 L1~ R [v'i9]e .~ =sm-t -9- ~~ A B e18. 1 x 1= AB sin = 5 x 12 x sin 30° = 30 units . .B 1 -) ~~ 19. a. Area = 21 A x B 1 PAQ --) -~ Fig. 2.64 b. Area = IA x BI b.PR+QR,,-PQ ~ IIi+BI+IBI\"-11i1 ~ Iii +BI \"-Ilil-IBI ~ Iii +BI \"- 1 ---'> -~ Illil-IBII [(c) and (d) parts, do as (a) and (b)) c. Area = 21A x BI 20. AP = 5k, AQ = 51 + 2) + 3k 24. R= P+ Q,S= -P+ 12, GiventhatS is J. R Displacement of the bird= PQ = AQ- AP So,RS=O~ (1)+Q).(-P+Q)=o +2)= 51 -2k ~ _p2+ Q2=0 '+;-;C;2, +cc2 mPQ = ../\"\"52 2 ,\"2 = .,;'33 ~ p2 = Q2 z Bird p Q ~ P=Q 3m N Hence proved. Y Q sine Sm ---------- B Alternatively: f3 = 90 - a, tan a = P + Q cos 8 W 2m Ci) A 5m (ii) E s x Fig. 2.62 tan f3 = _ ...:Q::.....si_n.'-(1_8_0_8...:)_ 21. Consider a regular polygon of N sides. Each side of the P + Q cos (180 - e) polygon is same. C is the centre of polygon (Fig. 2.63). eQsine e e= ~ ~ is the angle subtended at the centre of poly- tan (90 - a) = P - Q cos --gon by anyone side. ------ Q sine ,,/'\" ,,,, c cot a = -,-\"\"--:c---:- P - Qcose Multiply equations (i) and (ii): Simplify to get: P = Q N ,,,,,,,, s e180 - A IB ~ ~ ,,, R ,,\" Q p--, Fig. 2.63 = =We can prove that a 8 as: LABC + LCBD + a 180' (i) -, 8 But LABC = LCBD = 90' - - -p 2 (ii) Fig. 2.65 eFrom equations (i) and (ii), we get a = 25. IFI = 1121 = 1, because they are unit vectors. 2][ L.H.S. = IF - 121 = J(F - 12)' (F - 12) ~ a=- = )p2 + Q2 _ 2F. 12 = )12 + 12 - 2PQcos8 N So, if we have N veCtors of equal magnitude and they are arranged in such a way that each vector makes an angle 2][ (= a) with its preceding vee;or, we find that head of = ../2 - 2 x 1 x 1 cos e - .,)2 (1 - cos e) N J2= [2sin' Cem] = 2sin(8/2) = RH.S. the last vector will coincide with the tail of first vector. Hence, resultant of all these vectors becomes zero.

Vectors 2.25 Objective type (~r10. a. =12+l'+2x 1 x lcose I/q lsi1. a. Given = or A = B. or 91 = 2(1 + eose) or 1 + eose = 118 Sum: k = A+ S oreose = /8 -1 = -:~ ore = cos- I (-:~) Ikl=l> =.JA2 + B' =./2A 11. e. A+ S will be in the plane containing Aand S. whereas SDifference: = A- S A x S will be ..L to that plane. lsi=l> =.JA2+B'=./2A 12. d. Consider only x and y components: .J32 + 12 = .jlO Ici sl13. a. 14. e. Let third side is C, then (Xl = 45°, (X2 = 45° = IA + or Hence, k and S will be perpendicular and also of equal Ici = IA - sl lengths. B~ ---------- ,,,,,,, 15. a. Al =2, A,=3.IAI +A,I=3 ~ 2 =l> IAI +A,1 =9=l> Ai+Ai+2AI' A, =9 R =l> 22 +32 +2A I ·A2 =9 =l>AI·A2=-2 1E'--I-\"'-,1'-_->i' ~ 0;2 ,,,,,,,! A Now, ( A\" l + 2A\",) . (3\"AI + 4A\" 2 ) = 3A2I - 8A 22 4 HAl' A2 = 3(2)' - 8(3)' + 2( -2) = -64 S .-+ ....\". Fig. 2.66 16. d. A· B = 0 (given) ---c> .-;. A . C = 0 (given) 2. c. The minimum number of vectors having different planes -? -? -) -, ~ . which can be added to give zero resultant is 4. A is perpendicular to both Band C. 3. d. We see that dOl product of 1+ ] + k with 31 + 2] - 5k We know, from the definition of cross product, that B x C is is zero. C.perpendicular to both Band -\"\" -;. -)- 4. d,In t.MNO: A+C- D =O=l> C- D =-A C 3 (3) .. .So, A is parallel to B xc. Hence (b) is correct. 17. a. eose = II = '5 or e = cos-I '5' See Flg. 2.67. Int.MNp:A+S+E=O Hence (a) is correct. ~.~ e A~ InOMPNO: -E - S + C- D = 0 C =l>I3+E-C=-D ~ Hence, (c) is correct. B 5. d. For the resultant of some vectors to be zero, they should form a closed figure taken in same order. Fig. 2.67 6. b. eosfl = BR = 21: =l> fl = , '-, 60 18. b. Clearly, B should be either in second quadrant or fourth. Angle between Aand S.= 90' + fJ = 150' quadrant. In non.e of the given options, we have '-1 term'. ~ -0+-b = -3 or 30 - So, second quadrant is ruled out. Also, B should make an 7. b. a- b 1 3b = 0 +b e~ angle of 90° - with x-axis (Fig. 2.68). So, B should be or2a=4bora=2b B cos(90° - e)i - B sin(90' - el] = B sinel - B cos eJ. 8, b. Note that the angle between two forces is 120' and not ® CDA' 60\". . e R2 = F2 + F2 + 2F2 cos 120' 90- e @) -, or R2 = 2F2 + 2F2 ( - ~) = F2 or R = F ® B 9. b. Here, the angle between two vectors of equal magnitude is 120'. So, resultant has the same magnitude as either of Fig. 2.68 the given vectors, Moreover, it is mid-way between the two vectors, i.e., it is along x-axis. 19. c. Sec Fig. 2.69

2.26 Physics for IIT-JEE: Mechanics I 7]/i G)~t\"n,8 = or,8 = tan' 25. a. A x 1J = 2pq Fig. 2.69 573 -> =or i(3p -7q) + ](5q - 6) + k(14 - 51') 0 56 20. c. p is in fourth quadrant. 3I' = 7q, 5q - 6 = 0 or q = 47 + 3.1 is in the first quadrant. 14 p,Clearly, 47 + 3.1 can be perpendicular to For confirmation, 14 - 51' = 0 or 51' = 14 or I' = - let us check whether their dot produCl is zero. 5 (31 - 4.1) . (41 + 3]) = 12 - 12 = 0 ~ - ) . . . . . ; . -)0- ••-l> .....;. This shows that 41 + 3] is perpendicular to 31 - 4]. 26. h. If A is perpendicular to B, then A . B = 0 and A x If # () S:21. c. = 75.1 + [60 cos 45\"] - 60 sin 45\"1] + 201 s = (20 - 60 x 0.7)1 + (60 x 0.7 + 75») +b27. b. PR. = ,; =} major diagonal Sor = -221 + 117) bsQ = -c;- - => minor diagonal s = ~222 + 1172 = ~484 + 13689 = ~14173 s ,,,,,,,,,-, R = 119km kL_ _ _o-'\" p (/ Q Fig. 2.71 --+ -)- _.:> 28.b.A+B=C --~ ~~ - .•~ .-'). ---i> ...~ (A + If) . (A + If) = C . C => A2 + 13' + 2AB cosO = C2 =} 4' + 5' = 2 x 4 x 5 cos 0 = 61 =2I ~~cos Ii =} 0=60\" ~ (~)29. b. tan Ii = e=} = tan'\" y Fig. 2.70 3 ------=;- A Iii iii22. h. Arcaofparallelogram: x == AB/2(givcn) e =} ABsinO = AB/2 =} sine = 1/2 IL.._ _-'--_--l> x e=} = 30\" e23. b. a2 + l,z + 2ab cos 0 = a 2 + b2 - 2ab cos 2 or 4ab cos () = () Fig. 2.72 But 4ab # 0 =} cos 0 = 0 or 0 = 90' -> 30. a. Let that vector is C. Then -l-ba.5\"\" Aliter c = = =._)-\" (c; - ;;)(; + '/;) and CC iJfj =} C = -;; ~(i - j) are the diagonals of a parallelogram b.whose adjacent sides arc ;:; and 31. b. For the resultant of two vectors to be zero, they should be equal and opposite. +-,} -) -}-~ Since I a b I = I a - b I, therefore the two diagonals of 32. c. In first option (a), vector is along x-axis (Fig. 2.73). a parallelogram arc equal. So, think of square. This leads to 0=90\". A B24. a. x = (41 + 6]) x (21 + 3]) = 12(i x ]) + 12c7 x 7) = 12c7 x ]) - 12c7 x ]) = 0 ~ -> (41 + 6/). (21 + 3]) = 8 + 18 = 26 Again, A· B = . IAI ~16+36 # I Fig. 2.73 Again, I BI = ~4 + 9 2' In (b), angle of vector with x-axis 2--} 1 ··-r e5 = 45\" Also, B = A tan 0 = - = I =} -> ._> 5 In (e), angle ofvector with x-axis =} A and If arc parallel and not antiparallel.

Vectors 2.27 --)--}-)- -, tana = 5v5S = vS- a = 60\" 40. a. F2 - FI = 1\"2 +(- 1\"Jl 33. b. See Fig. 2.74 = 250 N due north + 500 N due west AC :s AB + BC =} 500 tane = - = 2 250 f l~;,,;I> C ~ .~ ob IF, - 1\"1 I = )(500)' + (250)' = 250v's N A o~· 8 Fig. 2.74 Q 34. d. The resultant of two forces can lie between A - Band s A+B,i.e., 12-1 = 11 Nand 12+ I =13N. Fig. 2.78 35. a. Find min (A- B) and maximum (A + B) value of each 0.441. c. OC and are equal in magnitude and inclined to each case, then check if 4 N lies between them. other at an angle of 90\". So, their resultant is../ir. It acts ~~ .~ •...,>- mid-way between OC and 0.4, i.e., along OB. Now, both rand Y'2r are along the same line and in the 36. b. BA+BC=BD same direction. A -----------;;-1 D -,',,,,-,,',,-,,',, .'. resultant = r + .Jir = r(l + .Ji) , , N \" 42. b. tan 45\" = 2 sin 60\" = vS , a+2cos60\" a+l vS . BL or 1 = - - ora + 1 = vS ora = vS -] Fig. 2.75 a+l e e43.d. ta. n 90\" -_ ZQ sin ...-.). --)0 -~ =>OO=~_Z.,,:Q,,--si_n BA + BC = 2 BM. Hence, the answer is2BM. P+2Qcose P+2Qeose +.....\". --+ --). -:> =} P+2Qcose = 0 37. b. (A B) . (A - B) = 0 A2_B2=0 =} A2=B2 Now, R' = p' + Q2 + 2PQcose =} A=B =} R'=Q2+P[P+ZQeose] 38. c. =} R2 = Q2 =} R = Q \",,, b~ ,,,,,,,,,,, ~ -., ,,, S '\" ,, ~ 2Q -, ,, b ,,,, ., o R '\\.'\\, ~--,--,--->-' Q 'lLlL._ _ _';.P a Fig. 2.76 .....\".-+ -~-} .Fig.2.79 a + b -> major diagonal, a + b -> minor diagonal .' C Alternate method: 39. b. tan e' = II = 1 R= 7> + Q =} P= R- Q and S = P+ 2Q =} e' = 45 0 = '::. . = R- Q+2Q 4 PNow, Sand are perpendicular (Fig. 2.78), so S. P= 0 =} (R + Q) . (R - Q) = 0 e = , , -\" - =3-\" 44 =} R2 = Q2 =} R = Q IBI}44. a. Given C= C=} = 5} e Lere = A+ B= A+ 31 + 4} \"'---~::;--0\"'-~-- -- -- -- -- -- -- + =}S} =A+31+4} A IAI=} A= -31 +} =} = )32 +]2 ='v'!D Fig. 2.77

2.28 Physics for IIT-JEE: Mechanics I ~ ~) ~= 2x + 3 x ( - =- 45. c. fIB + AF = AO =} liB = AO - AF AC=fIB+AO. AD = 2AO, AE=AO+AF J3 =} 8 = 30' tan 8 = Ry = - 2 = J3 = _1_ ED R, _~ 3 J3 2 + +--+ _;. --> \" A Vx = Ux axt = 8i 2ti =;\"-} ,,---'> --'> -} ---,} \") ~ -1 Vy = -IS}, V = vx+vy , V [(8+2t)t-15J]ms Fig. 2.80 51. c. You have to try all the options. Let us discuss the conect Now. fIB + AC + AD + AE + AF option only, = SAO + fIB + AF = SAO + AO = 6AO 46. b. Displacement = AS. angle between rl and 1'2: 8 = '75' ; x fi = f.'(r x fi) rr ri --15' = 60\" 2'-11'2 cose (1)(1) sin(180' - i) = 1h(1)(I)(l80' - r) sin i = It sin r From Fig. 2.55, AB2 = + = 32 + 42 - 2 x 3 x 4cos60' = 13 52. c. P+ Q= 16 =} AB=~ (i) p2 + Q2 +2PQcose =64 (ii) 47. c. x + y = 16. Also. y2 = 82 + x2 tan 90° = Q sin 0 or CXJ = _-,Q=si:::n..:.O_ , P + Qcose P + Qcose eP + Q cos 0 = 0 or Q Cos =-P y =From equation (ii), p2 + Q2 + 2P( - P) 64 8N or Q,2 _ p2 = 64 or Q - P= 64 = 4 (iii) x -16 Fig. 2.81 Adding equations (i) and (iii), we get ory2 =64+(16- y)' ['.' x = 16 - y] 2Q = 20 or Q = 10 units or l = 64 + 256 + y2 - 32y or 32y = 320 or y = ]() N From equation (i), P + 10 = 16 or P = 6 units c. x+10=16orx=6N 53. d. Thc length of the vector is not changed by the rotation of the coordinate axes. ____= 48. c. Graphically: fo+ 1)2 + J2 = ~n2 + J2 or n2 + 2n + 2 = n2 + 9 LROQ =0/2. LRQO =812 or 2n =7 or n =3.5 R __________ --- Q 54. b. Fo-\"*r collision, --+ --+ --+ --+ V--+\\)t 8/2 .-;> --+ r! +Vlt =rZ+v2torrl-rZ = (V2- OP~A~RQ ~ OR-S V,t J.'.:{~ -, J\",'1.-x xJ\"...~ V,I /', \",~ Fig. 2.82 -, Hence, !10QR is isosceles /'1 Fig. 2.83 =}OR = RQ =} B = A Equating unit vectors. we get (0/2) = Bsine Analytically: tan _ ,+_B-e-o_s_O_ A rz ---+ --+ --)--+ 1'1 sin(O /2) 2B sinCe /2) eos(O /2) 1)2 - VI A + B(2 eos2(8 /2) - I) I--+ --+ =} I --+---) 1/'2 - rl cos(O /2) V2 - VI I, =} A + 2Bcos2(e12) - B = 2Bcos2(812) =} A = B. 55. a. Net force along x-axis (Fig. 2,84) 49. a. Rx = 1 + 2 cos 120' + 3 cos 240' F, = FI + F, cos 30° - F3 cos 30° R\" = 2 sin 120' + 3 sin 240' F,=300N Fy = F2 sin 30° + F3 sin 30°

= 400 x 2I +400 x 2I =400N Vectors 2.29 Net force: F = .)3002 + 4002 = 500 N (1\")).I I eSo. the other resolved component is Ii sin lJk 3. a., b., c. Ii x B= 2 I I II1 = 7(l - 1) - J(2 - I) + k (2 - 1) = - J+ k Unit vector perpendicular to A4 and B4 is (-J.+,fi k) . So, ________3_0~ _\"\"''''--'-''''---.____ .. choices (a) and (c) are correct. FJ~300N Any vector whose magnitude is K (constant) times Fig. 2.84 (21 + J+ k) is parallel to Ii. Multiple Correct So, unit vector 21+v'6)+k is parallel to - Answers Type A. 1. a., c. Both x and y components of d, are positive. So, choice (b) is correct. x component of d2 is negative and y component is positive. Both x and y components of d, + d2are positive. 4. a., d. If two vectors are normal to each other, then their dot product is zero. Iltl e2. a., b. Component of Ii along Bis easeS for being (-V J + -V2) . (-V 1- -V2) = 0 =? vi - vi = 0 1.the angle between the vectors. =? vi = vi =? VI = V2 or l17rl = 11721 Also S = 1 So, choice (a) is correct. 5. a., d. The resultant of three vectors is zero only if they can The vector (1 - J) is perpendicular to the vector (1 + J). foml a triangle. But three vectors lying in different planes cannot form a triangle.