Please copy and paste this embed script to where you want to embed

UNITS, MEASUREMENTS & MOTION These topics are taken from our Book: ISBN : 9789386320018 Product Name : Units, Measurements & Motion for JEE Main & Advanced (Study Package for Physics) Product Description : Disha's Physics series by North India's popular faculty for IIT-JEE, Er. D. C. Gupta, have achieved a lot of acclaim by the IIT-JEE teachers and students for its quality and indepth coverage. To make it more accessible for the students Disha now re-launches its complete series in 12 books based on chapters/ units/ themes. These books would provide opportunity to students to pick a particular book in a particular topic. Unit, Measurement & Motion for JEE Main & Advanced (Study Package for Physics) is the 1st book of the 12 book set.

• The book provides basic concepts related to Mathematics useful in Physics. • The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems and Solved Examples. • Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than 1 correct, Assertion & Reason, Passage and Matching based Questions. • The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are also incorporated at their appropriate places. • The present format of the book would be useful for the students preparing for Boards and various competitive exams.

Contents

Contents

0.

Mathematics Used in Physics

1-16

1.

Units and Measurements

17-60

3.

Motion in a Straight Line

93-146

3.1

Concept of a point object

94

3.2

Rest and motion are relative terms

94

3.3

Motion

94

3.4

Motion parameters

94

Definitions Explanations and Derivations

18

1.1

Fundamental quantities

18

1.2

Derived quantities

18

1.3

The SI system of units

18

3.5

Equations of motion

97

1.4

Definitions of SI units

19

3.6

Study of motion by graphs

105

1.5

Advantages of SI system

20

3.7

Relative velocity

112

1.6

Dimensions of a physical quantity

20

1.7

Order of magnitude

22

3.8

Motion with variable acceleration

118

1.8

Rules of significant figures

22

3.9

Problems based on maxima and minima 118

1.9

Errors in measurement

26

1.10

Indirect methods of measuring large distances

31

1.11

Indirect method of measuring small distances

33

Exercise 3.3 (Assertion and Reasoning type questions)

Vernier callipers and ccrew gauge

34

Exercise 3.4 (Passage & Matrix)

1.12

Exercise 3.1 Level 1 (Single correct option) Exercise 3.1 Level 2 (Single correct option) Exercise 3.2 (more than one correct options)

Exercise 1.1 Level 1 (Single correct option)

Exercise 3.5 (Past years JEE-(Main and Advance)

Exercise 1.1 Level 2 (Single correct option)

Hints and Solutions (Solution of all exercises)

Exercise 1.2 (more than one correct options) Exercise 1.3 (Assertion and Reasoning type questions) Exercise 1.4 (Passage & Matrix) Exercise 1.5 (Past years JEE-(Main and Advance) Hints and Solutions (Solution of all exercises)

2. Vectors

61-92

Definitions Explanations and Derivations

62

2.1

Scalar quantity or scalar

62

2.2

Vector quantity or vector

62

2.3

Vectors operations

65

2.4

Addition or subtraction of two vectors

65

2.5

Addition or subtraction of more than two vectors

68

2.6

Product of two vectors

73

2.7

Geometrical interpretation of scalar triple product

77

4.

Motion in a Plane 147-202 4.1

Introduction

148

4.2

Position vector and displacement

148

4.3

Average velocity

148

4.4

Average acceleration

149

4.5

Motion in a plane with constant acceleration

150

4.6

Relative velocity in two dimensions

151

4.7

Projectile motion

157

4.8

Projection up on an inclined plane

169

4.9

Projection down the inclined plane

170

4.10

Motion along a curved path

171

4.11 Constraint relations

Exercise 2.1 Level 1 (Single correct option) Exercise 2.1 Level 2 (Single correct option) Exercise 2.2 (more than one correct options) Exercise 2.3 (Assertion and Reasoning type questions)

179

Exercise 4.1 Level 1 (Single correct option) Exercise 4.1 Level 2 (Single correct option) Exercise 4.2 (more than one correct options) Exercise 4.3 (Assertion and Reasoning type questions)

Exercise 2.4 (Passage & Matrix)

Exercise 4.4 (Passage & Matrix)

Exercise 2.5 (Past years JEE-(Main and Advance)

Exercise 4.5 (Past years JEE-(Main and Advance)

Hints and Solutions (Solution of all exercises)

Hints and Solutions (Solution of all exercises)

Chapter 1 Units and Measurements

1.9 ERRORS

IN MEASUREMENT

Every measurement is limited by the reliability of the measuring instrument and skill of the person making the measurement. If we repeat a particular measurement, we usually do not get the same result every time. This imperfection in measurement can be expressed in two ways :

Accuracy and precision Accuracy refers to the closeness of observed values to its true value of the quantity while precision refers to closeness between the different observed values of the same quantity. High precision does not mean high accuracy. The difference between accuracy and precision can be understand by the following example : Suppose three students are asked to find the length of a rod whose length is known to be 2.250 cm. The observations are given in the table. Student

Measurement-1

Measurement-2

Measurement-3

Average length

A.

2.25 cm

2.27 cm

2.26 cm

2.26 cm

B.

2.252 cm

2.250 cm

2.251 cm

2.251 cm

C.

2.250 cm

2.250 cm

2.251 cm

2.250 cm

It is clear from the above table, that the observations taken by student A are neither precise nor accurate. The observations of student B are more precise. The observations of student C are precise as well as accurate. Error : Each instrument has its limitation of measurement. While taking the observation, some uncertainty gets introduced in the observation. As a result, the observed value is somewhat different from true value. Therefore, Error = True value – Observed value Systematic errors : The errors which tend to occur of one sign, either positive or negative, are called systematic errors. Systematic errors are due to some known cause which follow some specified rule. We can eliminate such errors if we know their causes. Systematic errors may occur due to zero error of an instrument, imperfection in experimental techniques, change in weather conditions like temperature, pressure etc. Random errors : The errors which occur randomly and irregularly in magnitude and sign are called random errors. The cause of random errors are not known. If a person repeat the observations number of times, he may get different readings every time. Random errors have almost equal chances for positive and negative sign. Hence the arithmetic mean of large number of observations can be taken to minimize the random error. Mean value of a quantity : Since the probability of occurrence of positive and negative errors are same, so the arithmetic mean of all observations can be taken as the true value of a observed quantity. If a1, a2, ................an are the observed values of a quantity, then its true value a can be given by a

=

amean =

=

1 n

a1 + a2 + ................ + an n

n

å ai i =1

The absolute errors in individual observations are: Da1 = a - a1

The mean absolute error is defined as Da

Da2 = a - a2 ........................... Dan = a - an

=

| Da1 | + | Da2 | +.................+ | Dan | n

Units and Measurements n

= `

1 | Dai | n i= 1

å

Thus the final result of the observed quantity can be expressed as a = a ± Da . It is clear from above that any observed value can by (a - Da ) £ a £ (a + Da ) . Relative or fractional error : The ratio of the mean absolute error to the true value of the quantity is called relative error. Da a Percentage error : If relative error is expressed in percentage is called percentage error.

Thus relative error =

Thus percentage error =

Note:

Da ´ 100 a

Absolute error has the unit of quantity. But relative error has no unit.

Combination of errors 4 3 pr . There involves multiplication of radius three times. 3 The measurement of radius has some error, then what will be error in calculating the volume of sphere? The error in final result depends on the individual measurement as well as the mathematical operations involved in calculating the result. Following rules are used to evaluate maximum possible error in any computed quantity. 1. Error in addition Let Z = X + Y. Suppose ± Dx be absolute errors in X and ± Dy be the absolute error in Y, then we have Z + Dz = (X ± Dx) + (Y ± Dy) = (X + Y) ± (Dx + Dy) Dz = (Z + Dz) – Z \ = ± (Dx + Dy)

Let we want to get the volume of sphere, V =

Note: The maximum value of Dx or Dy can be least count of the instrument used. Example : x = 2.20 cm, Dx will be 0.01 cm.

RULE : The maximum possible error in the addition of quantities is equal to the sum of their absolute error. % error in Z, 2.

Error in subtraction Let

\

Dz × 100 = Z

Z = Z + Dz = = Dz = =

é Dx + Dy ù ±ê ú × 100 ë X +Y û

X–Y (X ± Dx) – (Y ± Dy) (X – Y) ± (Dx m Dy) (Z +Dz) – Z ± Dx m Dy

For maximum possible error Dx and Dy must be of same sign. Dz = ± (Dx + Dy) \ RULE : The maximum possible error in subtraction of quantities is equal to the sum of their absolute errors.

27

28

MECHANICS % error in Z, 3.

é Dx + Dy ù Dz × 100 = ± ê ú. Z ë X -Y û

Error in product Let

Z = XY Z + Dz = (X ± Dx) (Y ± Dy) = XY ± DxY ± XDy ± DxDy \ Dz = (Z + Dz) – Z = ± (DxY + XDy) ± DxDy If Dx and Dy are both small, their product be very small, therefore we can neglect it. Dz = ± (DxY + XDy) \ The maximum fractional error in Z, Dz Z

=

é Dx Dy ù ±ê + Y úû ëX

and maximum percentage error in Z, Dz × 100 = Z

é Dx Dy ù ±ê + × 100 Y úû ëX

RULE : The maximum fractional error in the product is equal to the sum of the fractional errors in the individual quantities.

Note: The product Dx Dy can not be neglected if the errors in x and y are order of 10% or more. The product can be neglected, if the error in x and y are 1% or little more than this (say 2 to 3%). 4.

Error in quotient or division Let Then

or

Z = Z + Dz =

Dz Z

X ± Dx Y ± Dy

=

æ Dx ö X ç1 ± X ÷ø è æ Dy ö Y ç1 ± Y ÷ø è

=

X Y

æ Dx öæ Dy ö ç1 ± ÷ç 1 ± ÷ X øè Y ø è

=

X Y

æ Dx öæ Dy ö ç1 ± X ÷ç1 m Y ÷ è øè ø

Z + Dz =

1+

X Y

-1

Dx öæ Dy ö æ 1m Z ç1 ± ÷ç X øè Y ÷ø è

æ Dx ö æ Dy ö 1m = ç1 ± X ÷ø çè Y ÷ø è

=

1±

Dy Dx Dy Dx ± . m X X Y Y

Units and Measurements Dx Dy and are small, so their product can be neglected. The maximum fractional X Y error is given by

As the term

æ Dx Dy ö Dz + = ±ç ÷ Y ø è X Z And maximum possible percentage error in Z, æ Dx Dy ö Dz + × 100 = ± ç × 100 Y ÷ø è X Z RULE : The maximum fractional error in the quotient is equal to the sum of their individual fractional errors. Error in the power of a quantity Let Z = Xn Z + Dz = (X ± Dx)n

\

5.

=

Xn

æ Dx ö ç1 ± ÷ X ø è

n

Dx ö æ ; Z ç1 ± n X ÷ è ø

or

Z + Dz Z

or

1+

\

Dx ö æ = ç1 ± n X ÷ø è

Dz Z

= 1±n

Dz Z

= ±n

Dx X

Dx X

Maximum percentage error in Z æ Dx ö Dz ´ 100 ÷ × 100 = ± n ç è X ø Z

RULE : The fractional error in the quantity with n powers is n times the fractional error in that quantity.

Note:

Here n may have any value. It may be a whole number, fraction, positive or negative.

General case : If Z =

X aY b

Wc

, the maximum possible fractional error in Z,

Dz =± Z

Dy Dw ù é Dx êa X + b Y + c W ú ë û

The maximum possible percentage error Dz Dy Dw ù é Dx × 100 = ± ê a × 100 +b +c Z X Y W úû ë

The above used algebraic method in many operations become difficult to operate. In such situations we can used differential method to find the error.

Differential method of calculation of errors

29

30

MECHANICS 1.

Let

Z =

k

X aY b

Wc

where k is a constant. Taking logarithms of both sides of equation, we get ln Z = ln k + a ln X + b ln Y – c ln W Now differentiating partially the above expression, we have dx dy dw dz = a +b -c X Y W Z We can write above equation by writing D in place of d;

Dx Dy Dw Dz +b -c = a X Y W Z Errors calculated by above equation, is known as mathematical error. But our interest is in finding the maximum possible error. Dy Dw ù é Dx Dz +b +c × 100 = ± ê a ×100 X Y W úû ë Z

\ 2.

Let

Z =

W (X +Y )

Taking logarithms of both sides of above equation, we have ln Z = ln W – ln (X + Y) Differentiating partially, we get dz dw d( x + y ) = = Z W (X +Y )

d w (d x + d y ) W X +Y

(a) The maximum possible error in Z Dz Z

(b) For

Z = Dz Z

3.

=

Let

=

Z =

é Dw Dx + Dy ù ±ê + ú ë W (X +Y)û

W X -Y é Dw Dx + Dy ù ±ê + X - Y úû ëW

XY U +V

Taking logarithms of both sides of above equation, we have ln Z = ln X + ln Y – ln (U + V) Differentiating partially, we get dz Z

=

dx dy d (u + v ) + X Y U +V

=

dx dy (du + dv ) + X Y U +V

=

é Dx Dy ( Du + Dv ) ù ±ê + + Y U + V úû ëX

The maximum possible error in Z (a)

Dz Z

Units and Measurements (b) For 4.

Z =

XY é Dx Dy Du + Dv ù Dz + , =±ê + Y U - V úû U -V Z ëX

Let Z = sinx Differentiating partially, we get dz = cos x dx or Dz = cos x Dx and

Dz Z

=

or

Dz Z

=

cos x 1 - sin 2 x Dx = Dx sin x sin x 1 - z2 Dx Z

31

Chapter 4 Motion in a Plane 4.7 PROJECTILE

MOTION

When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and vertical directions. Motion of such a particle is called projectile motion. Since there is no force acting in horizontal direction, the velocity remains constant in this direction. In vertical direction gravitational pull of earth produces the acceleration.

Assumptions used in projectile motion (i) (ii) (iii)

Neglecting the effect of air resistance on the projectile. Assuming the acceleration due to gravity is constant at each point of projectile. Neglecting the effect of curvature of earth. We will discuss the following types of projections in details :

Type 1 : Projectile fired at some angle with the horizontal.

Type 2 : Horizontal projection

Type 4 : Projection on an inclined plane

Type 3 : Projectile fired from some height

Type 5 :Projection down the inclined plane Figure. 4.21

ANALYSIS OF PROJECTILE OF TYPE 1 Let us consider a particle is projected with initial velocity u at an angle q with the horizontal (called angle of projection). The velocity u has two rectangular components: (i) The horizontal component ucosq, which remains constant throughout the motion of particle. (ii) The vertical component usinq, which changes with time due to effect of gravity. Thus we have initial velocity ur u = u x ˆi + u y ˆj ur or u = ucosq ˆi + usinq ˆj

Figure. 4.22

158

MECHANICS Velocity at any time t : Using first equation of motion in vertical direction, we have vy = uy – gt = u sinq – gt ur \ Velocity at any time t, v = v ˆi + v ˆj x

y

ur or v = ucosqˆi + (usinq - gt )jˆ. Velocity at any height : At any height h vx = ux = u cosq Figure. 4.23 and vy2 = uy2 – 2gh = (u sinq)2 – 2gh Squaring (i) and adding with equation (ii), we get v

…(i) …(ii)

u 2 - 2 gh .

=

Position at any time t Position of particle at any time t, is given by ur r = xˆi + yˆj where x = u cosq t

…(1) 1 2 gt 2

and

y

= u sinq t –

\

r r

1 2 = u cos q t ˆi + (u sin q t - gt )ˆj 2

or

1 æ ö (u cos q t )2 + ç u sin q t - gt 2 ÷ è ø 2

r =

Figure. 4.24

2

= and

or

tanf =

ut

gt sin q æ gt ö 1+ ç ÷ è 2u ø u

y x

1 2ù é ê u sin qt - 2 gt ú ú f = tan –1 ê ut cos q ê ú ë û

é 2u sin q - gt ù . = tan–1 ê ë 2u cos q úû The angle of elevation f of the highest point of the projectile; t

=

\

tanf =

or

tanf =

u sin q g u sin q g 2u cos q

2u sin q - g ´

tan q 2

Equation of trajectory We have,

x

= u cosq t

or

t

=

æ x ö çè ÷ u cos q ø

2

Motion in a Plane and

y = u sinq t –

159

1 2 gt 2

æ x ö 1 = u sinq çè ÷ – g u cos q ø 2

x tan q -

æ x ö çè ÷ u cos q ø

2

gx 2

. 2u 2 cos2 q On comparing this equation with general equation of parabola, y = ax ± bx2, we find that path of projectile is parabolic in nature. y

=

Time of flight (T) : Total time of motion of particle in air is called time of flight. The displacement in vertical direction (y-axis) becomes zero in whole time of motion. So we have

or

which gives

y = uy t –

1 2 at 2 y

uyT –

1 2 gT 2

0

=

T

=

2u y g

=

2u sin q . g

Maximum height attained (H): The maximum vertical distance achieved by particle is Figure. 4.25 called maximum height. At the highest point of projection vy = 0, so we have, vy2 = uy2 – 2gh 0 = uy2 – 2gH

or

which gives

H

=

u y2 2g

=

u 2 sin 2 q . 2g

Horizontal range (R) : The horizontal distance moved by particle in total time of flight is called horizontal range. Horizontal range,

R = ux × T = ux × =

2u x u y g

=

2u y g

2u cos q u sin q g

For maximum range,

sin2q

u 2 sin 2q . g = 1 or 2q = 90° or q = 45°.

Thus

Rmax

=

or

Corresponding,

R

=

H =

u2 . g

u2 sin2 45° u 2 . = 2g 4g

There are two angles of projection for same range: Replacing q by (90° – q) in the formula of range, we get

160

MECHANICS R'

=

u2 sin 2(90° -q) g

=

u 2 sin(180° - 2q) g

=

u 2 sin 2q = R. g

Thus, for a given velocity of projection, a projectile has the same range for angle of projection q and (90° – q). Time of flight for angle of projection q,

2u sin q g and time of flight for angle of projection (90° – q), T1 =

T2 = =

Figure. 4.26

2u sin(90° - q) g 2u co s q . g

Multiplying T1 and T2, we get \

T1T2 =

2u sin q 2ucosq × g g

or

T1T2 =

2 æ u 2 sin 2 q ö ç ÷ gè g ø

or

T1T2 =

2R . g

More about projectile motion 1.

If t1 is the time taken by projectile to reach a point P at height h and t2 is the time taken from point P to ground level, then t1 + t2 = T =

or

u sinq

=

2u sin q g

g (t1 + t 2 ) . 2

The height of point P, h = u sinq t1 –

or 2.

h

1 gt 2 2 1

=

g (t1 + t 2 ) 1 t1 – gt 2 2 2 1

=

1 g t1 t2. 2

Change in momentum : Change in momentum between two positions of projectile is given by ur ur uur uur uur ΔP = P f - Pi = m(v f - v i ) . (a) Between point of projection and highest point

Motion in a Plane ur vi

and

\

ur vf

= u cosq ˆi + u sinq ˆj = u cosq ˆi

uur ΔP

=

m [(u cosq ˆi ) – (u cosq ˆi + u sinq ˆj )]

or DP = mu sin q (b) For the complete projectile motion r vi

r vf \ or

uur ΔP

= u cosq ˆi + u sinq ˆj

Figure. 4.27

= u cosq ˆi – u sinq ˆj =

m [(u cosq ˆi – u sinq ˆj ) – (u cosq ˆi + u sinq ˆj )]

DP = 2 mu sin q

ANALYSIS OF PROJECTILE OF TYPE 2 : HORIZONTAL PROJECTION Let a particle be projected horizontally with initial velocity u from height h. Velocity at any time t We have, vx = u and vy = uy + gt or = 0 + gt r = u ˆi - gt ˆj \ v and Also

u 2 + ( g t )2

v = tana =

gt . u

Position at any time t Taking point of projection as the origin, the position vector at any time t ur r = x ˆi – y ˆj . where

x = ut

and

y =

1 2 gt 2

\

r r

utˆi -

Displacement

=

s = r=

1 2ˆ gt j 2

æ1 ö (u t ) 2 + ç g t 2 ÷ è2 ø

Equation of trajectory We have,

x = ut or t =

and

y

= –

x u

1 2 gt 2

1 æ xö = – gç ÷ 2 è uø

2

2

Figure. 4.28

161

162

MECHANICS y

=

1 x2 - g 2 2 u

We have,

h

=

uyt +

or

h

= 0+

which gives

T =

or Time of flight (T)

Horizontal range (R)

1 a t2 2 y

1 g T2 2

2h . g

R = ux × T

2h . g The average acceleration in total time of flight is g downward.

= u

ANALYSIS OF PROJECTILE OF TYPE 3 Method - I : Let us consider a particle is projected with initial velocity 100 m/s at an angle 30° with the horizontal. The height of projection is 100 m. Time of flight (T) We have

y

= uy t +

1 2 gt 2

100 = – 100 sin 30° T +

or

1 × 10 × T2 2

T2 – 10 T – 20 = 0

or Figure. 4.29

which gives

T =

10 ± (-10)2 - 4 ´1´ (-20) 2

= 11.71 s (consider only positive value) The horizontal range (R) R = ux × T = 100 cos 30° × 11.71 = 100 ×

3 × 11.71 2

= 1014 m. Method - II : Taking point of projection as the origin, the coordinates of point of strike are (R, – 100 m). We have, y = x tanq – Here

\ or

gx 2 2u 2 cos 2 q

y = – 100 m and – 100 = R tan 30° – R2 – 866 R – 150000

= 0

x=R 10 R 2

2(100)2 cos 2 30°

Figure. 4.30

Motion in a Plane which gives,

R =

866 ± (866)2 + 4 ´150000 2

= 1014 m (consider only positive value) Time of flight,

T =

R = 11.71 s. u cos q

FORMULAE USED y

Projectile Type 1 1.

2.

ux = u cos q, uy = u sin q

u

ax = 0, ay = –g .

q

R

Position after time t x = u cos q t,

3.

O

H

1 2 gt 2

y = u sin q t –

Equation of trajectory y = x tan q –

gx 2 2u 2 cos 2 q

.

2u sin q . g

4.

Time of flight, T =

5.

Maximum height, H =

u 2 sin 2 q . 2g

6.

Horizontal range, R =

u 2 sin 2q g

7.

Maximum range, Rmax =

u2 , for q = 45° g

u

Projectile Type 2 1.

x

Position after time t x = ut, y =

h

1 2 gt 2

gx 2

.

2.

Equation of trajectory, y =

3.

Velocity after time t, v = u 2 + (gt)2

4.

Time to hit the ground, T =

5.

Horizontal range, R = uT = u

2u

2

2h g 2h . g

R

x

163

View more...
• The book provides basic concepts related to Mathematics useful in Physics. • The chapters provide detailed theory which is followed by Important Formulae, Strategy to solve problems and Solved Examples. • Each chapter covers 5 categories of New Pattern practice exercises for JEE - MCQ 1 correct, MCQ more than 1 correct, Assertion & Reason, Passage and Matching based Questions. • The book provides Previous years’ questions of JEE (Main and Advanced). Past years KVPY questions are also incorporated at their appropriate places. • The present format of the book would be useful for the students preparing for Boards and various competitive exams.

Contents

Contents

0.

Mathematics Used in Physics

1-16

1.

Units and Measurements

17-60

3.

Motion in a Straight Line

93-146

3.1

Concept of a point object

94

3.2

Rest and motion are relative terms

94

3.3

Motion

94

3.4

Motion parameters

94

Definitions Explanations and Derivations

18

1.1

Fundamental quantities

18

1.2

Derived quantities

18

1.3

The SI system of units

18

3.5

Equations of motion

97

1.4

Definitions of SI units

19

3.6

Study of motion by graphs

105

1.5

Advantages of SI system

20

3.7

Relative velocity

112

1.6

Dimensions of a physical quantity

20

1.7

Order of magnitude

22

3.8

Motion with variable acceleration

118

1.8

Rules of significant figures

22

3.9

Problems based on maxima and minima 118

1.9

Errors in measurement

26

1.10

Indirect methods of measuring large distances

31

1.11

Indirect method of measuring small distances

33

Exercise 3.3 (Assertion and Reasoning type questions)

Vernier callipers and ccrew gauge

34

Exercise 3.4 (Passage & Matrix)

1.12

Exercise 3.1 Level 1 (Single correct option) Exercise 3.1 Level 2 (Single correct option) Exercise 3.2 (more than one correct options)

Exercise 1.1 Level 1 (Single correct option)

Exercise 3.5 (Past years JEE-(Main and Advance)

Exercise 1.1 Level 2 (Single correct option)

Hints and Solutions (Solution of all exercises)

Exercise 1.2 (more than one correct options) Exercise 1.3 (Assertion and Reasoning type questions) Exercise 1.4 (Passage & Matrix) Exercise 1.5 (Past years JEE-(Main and Advance) Hints and Solutions (Solution of all exercises)

2. Vectors

61-92

Definitions Explanations and Derivations

62

2.1

Scalar quantity or scalar

62

2.2

Vector quantity or vector

62

2.3

Vectors operations

65

2.4

Addition or subtraction of two vectors

65

2.5

Addition or subtraction of more than two vectors

68

2.6

Product of two vectors

73

2.7

Geometrical interpretation of scalar triple product

77

4.

Motion in a Plane 147-202 4.1

Introduction

148

4.2

Position vector and displacement

148

4.3

Average velocity

148

4.4

Average acceleration

149

4.5

Motion in a plane with constant acceleration

150

4.6

Relative velocity in two dimensions

151

4.7

Projectile motion

157

4.8

Projection up on an inclined plane

169

4.9

Projection down the inclined plane

170

4.10

Motion along a curved path

171

4.11 Constraint relations

Exercise 2.1 Level 1 (Single correct option) Exercise 2.1 Level 2 (Single correct option) Exercise 2.2 (more than one correct options) Exercise 2.3 (Assertion and Reasoning type questions)

179

Exercise 4.1 Level 1 (Single correct option) Exercise 4.1 Level 2 (Single correct option) Exercise 4.2 (more than one correct options) Exercise 4.3 (Assertion and Reasoning type questions)

Exercise 2.4 (Passage & Matrix)

Exercise 4.4 (Passage & Matrix)

Exercise 2.5 (Past years JEE-(Main and Advance)

Exercise 4.5 (Past years JEE-(Main and Advance)

Hints and Solutions (Solution of all exercises)

Hints and Solutions (Solution of all exercises)

Chapter 1 Units and Measurements

1.9 ERRORS

IN MEASUREMENT

Every measurement is limited by the reliability of the measuring instrument and skill of the person making the measurement. If we repeat a particular measurement, we usually do not get the same result every time. This imperfection in measurement can be expressed in two ways :

Accuracy and precision Accuracy refers to the closeness of observed values to its true value of the quantity while precision refers to closeness between the different observed values of the same quantity. High precision does not mean high accuracy. The difference between accuracy and precision can be understand by the following example : Suppose three students are asked to find the length of a rod whose length is known to be 2.250 cm. The observations are given in the table. Student

Measurement-1

Measurement-2

Measurement-3

Average length

A.

2.25 cm

2.27 cm

2.26 cm

2.26 cm

B.

2.252 cm

2.250 cm

2.251 cm

2.251 cm

C.

2.250 cm

2.250 cm

2.251 cm

2.250 cm

It is clear from the above table, that the observations taken by student A are neither precise nor accurate. The observations of student B are more precise. The observations of student C are precise as well as accurate. Error : Each instrument has its limitation of measurement. While taking the observation, some uncertainty gets introduced in the observation. As a result, the observed value is somewhat different from true value. Therefore, Error = True value – Observed value Systematic errors : The errors which tend to occur of one sign, either positive or negative, are called systematic errors. Systematic errors are due to some known cause which follow some specified rule. We can eliminate such errors if we know their causes. Systematic errors may occur due to zero error of an instrument, imperfection in experimental techniques, change in weather conditions like temperature, pressure etc. Random errors : The errors which occur randomly and irregularly in magnitude and sign are called random errors. The cause of random errors are not known. If a person repeat the observations number of times, he may get different readings every time. Random errors have almost equal chances for positive and negative sign. Hence the arithmetic mean of large number of observations can be taken to minimize the random error. Mean value of a quantity : Since the probability of occurrence of positive and negative errors are same, so the arithmetic mean of all observations can be taken as the true value of a observed quantity. If a1, a2, ................an are the observed values of a quantity, then its true value a can be given by a

=

amean =

=

1 n

a1 + a2 + ................ + an n

n

å ai i =1

The absolute errors in individual observations are: Da1 = a - a1

The mean absolute error is defined as Da

Da2 = a - a2 ........................... Dan = a - an

=

| Da1 | + | Da2 | +.................+ | Dan | n

Units and Measurements n

= `

1 | Dai | n i= 1

å

Thus the final result of the observed quantity can be expressed as a = a ± Da . It is clear from above that any observed value can by (a - Da ) £ a £ (a + Da ) . Relative or fractional error : The ratio of the mean absolute error to the true value of the quantity is called relative error. Da a Percentage error : If relative error is expressed in percentage is called percentage error.

Thus relative error =

Thus percentage error =

Note:

Da ´ 100 a

Absolute error has the unit of quantity. But relative error has no unit.

Combination of errors 4 3 pr . There involves multiplication of radius three times. 3 The measurement of radius has some error, then what will be error in calculating the volume of sphere? The error in final result depends on the individual measurement as well as the mathematical operations involved in calculating the result. Following rules are used to evaluate maximum possible error in any computed quantity. 1. Error in addition Let Z = X + Y. Suppose ± Dx be absolute errors in X and ± Dy be the absolute error in Y, then we have Z + Dz = (X ± Dx) + (Y ± Dy) = (X + Y) ± (Dx + Dy) Dz = (Z + Dz) – Z \ = ± (Dx + Dy)

Let we want to get the volume of sphere, V =

Note: The maximum value of Dx or Dy can be least count of the instrument used. Example : x = 2.20 cm, Dx will be 0.01 cm.

RULE : The maximum possible error in the addition of quantities is equal to the sum of their absolute error. % error in Z, 2.

Error in subtraction Let

\

Dz × 100 = Z

Z = Z + Dz = = Dz = =

é Dx + Dy ù ±ê ú × 100 ë X +Y û

X–Y (X ± Dx) – (Y ± Dy) (X – Y) ± (Dx m Dy) (Z +Dz) – Z ± Dx m Dy

For maximum possible error Dx and Dy must be of same sign. Dz = ± (Dx + Dy) \ RULE : The maximum possible error in subtraction of quantities is equal to the sum of their absolute errors.

27

28

MECHANICS % error in Z, 3.

é Dx + Dy ù Dz × 100 = ± ê ú. Z ë X -Y û

Error in product Let

Z = XY Z + Dz = (X ± Dx) (Y ± Dy) = XY ± DxY ± XDy ± DxDy \ Dz = (Z + Dz) – Z = ± (DxY + XDy) ± DxDy If Dx and Dy are both small, their product be very small, therefore we can neglect it. Dz = ± (DxY + XDy) \ The maximum fractional error in Z, Dz Z

=

é Dx Dy ù ±ê + Y úû ëX

and maximum percentage error in Z, Dz × 100 = Z

é Dx Dy ù ±ê + × 100 Y úû ëX

RULE : The maximum fractional error in the product is equal to the sum of the fractional errors in the individual quantities.

Note: The product Dx Dy can not be neglected if the errors in x and y are order of 10% or more. The product can be neglected, if the error in x and y are 1% or little more than this (say 2 to 3%). 4.

Error in quotient or division Let Then

or

Z = Z + Dz =

Dz Z

X ± Dx Y ± Dy

=

æ Dx ö X ç1 ± X ÷ø è æ Dy ö Y ç1 ± Y ÷ø è

=

X Y

æ Dx öæ Dy ö ç1 ± ÷ç 1 ± ÷ X øè Y ø è

=

X Y

æ Dx öæ Dy ö ç1 ± X ÷ç1 m Y ÷ è øè ø

Z + Dz =

1+

X Y

-1

Dx öæ Dy ö æ 1m Z ç1 ± ÷ç X øè Y ÷ø è

æ Dx ö æ Dy ö 1m = ç1 ± X ÷ø çè Y ÷ø è

=

1±

Dy Dx Dy Dx ± . m X X Y Y

Units and Measurements Dx Dy and are small, so their product can be neglected. The maximum fractional X Y error is given by

As the term

æ Dx Dy ö Dz + = ±ç ÷ Y ø è X Z And maximum possible percentage error in Z, æ Dx Dy ö Dz + × 100 = ± ç × 100 Y ÷ø è X Z RULE : The maximum fractional error in the quotient is equal to the sum of their individual fractional errors. Error in the power of a quantity Let Z = Xn Z + Dz = (X ± Dx)n

\

5.

=

Xn

æ Dx ö ç1 ± ÷ X ø è

n

Dx ö æ ; Z ç1 ± n X ÷ è ø

or

Z + Dz Z

or

1+

\

Dx ö æ = ç1 ± n X ÷ø è

Dz Z

= 1±n

Dz Z

= ±n

Dx X

Dx X

Maximum percentage error in Z æ Dx ö Dz ´ 100 ÷ × 100 = ± n ç è X ø Z

RULE : The fractional error in the quantity with n powers is n times the fractional error in that quantity.

Note:

Here n may have any value. It may be a whole number, fraction, positive or negative.

General case : If Z =

X aY b

Wc

, the maximum possible fractional error in Z,

Dz =± Z

Dy Dw ù é Dx êa X + b Y + c W ú ë û

The maximum possible percentage error Dz Dy Dw ù é Dx × 100 = ± ê a × 100 +b +c Z X Y W úû ë

The above used algebraic method in many operations become difficult to operate. In such situations we can used differential method to find the error.

Differential method of calculation of errors

29

30

MECHANICS 1.

Let

Z =

k

X aY b

Wc

where k is a constant. Taking logarithms of both sides of equation, we get ln Z = ln k + a ln X + b ln Y – c ln W Now differentiating partially the above expression, we have dx dy dw dz = a +b -c X Y W Z We can write above equation by writing D in place of d;

Dx Dy Dw Dz +b -c = a X Y W Z Errors calculated by above equation, is known as mathematical error. But our interest is in finding the maximum possible error. Dy Dw ù é Dx Dz +b +c × 100 = ± ê a ×100 X Y W úû ë Z

\ 2.

Let

Z =

W (X +Y )

Taking logarithms of both sides of above equation, we have ln Z = ln W – ln (X + Y) Differentiating partially, we get dz dw d( x + y ) = = Z W (X +Y )

d w (d x + d y ) W X +Y

(a) The maximum possible error in Z Dz Z

(b) For

Z = Dz Z

3.

=

Let

=

Z =

é Dw Dx + Dy ù ±ê + ú ë W (X +Y)û

W X -Y é Dw Dx + Dy ù ±ê + X - Y úû ëW

XY U +V

Taking logarithms of both sides of above equation, we have ln Z = ln X + ln Y – ln (U + V) Differentiating partially, we get dz Z

=

dx dy d (u + v ) + X Y U +V

=

dx dy (du + dv ) + X Y U +V

=

é Dx Dy ( Du + Dv ) ù ±ê + + Y U + V úû ëX

The maximum possible error in Z (a)

Dz Z

Units and Measurements (b) For 4.

Z =

XY é Dx Dy Du + Dv ù Dz + , =±ê + Y U - V úû U -V Z ëX

Let Z = sinx Differentiating partially, we get dz = cos x dx or Dz = cos x Dx and

Dz Z

=

or

Dz Z

=

cos x 1 - sin 2 x Dx = Dx sin x sin x 1 - z2 Dx Z

31

Chapter 4 Motion in a Plane 4.7 PROJECTILE

MOTION

When a particle is projected obliquely near the earth surface, it moves simultaneously in horizontal and vertical directions. Motion of such a particle is called projectile motion. Since there is no force acting in horizontal direction, the velocity remains constant in this direction. In vertical direction gravitational pull of earth produces the acceleration.

Assumptions used in projectile motion (i) (ii) (iii)

Neglecting the effect of air resistance on the projectile. Assuming the acceleration due to gravity is constant at each point of projectile. Neglecting the effect of curvature of earth. We will discuss the following types of projections in details :

Type 1 : Projectile fired at some angle with the horizontal.

Type 2 : Horizontal projection

Type 4 : Projection on an inclined plane

Type 3 : Projectile fired from some height

Type 5 :Projection down the inclined plane Figure. 4.21

ANALYSIS OF PROJECTILE OF TYPE 1 Let us consider a particle is projected with initial velocity u at an angle q with the horizontal (called angle of projection). The velocity u has two rectangular components: (i) The horizontal component ucosq, which remains constant throughout the motion of particle. (ii) The vertical component usinq, which changes with time due to effect of gravity. Thus we have initial velocity ur u = u x ˆi + u y ˆj ur or u = ucosq ˆi + usinq ˆj

Figure. 4.22

158

MECHANICS Velocity at any time t : Using first equation of motion in vertical direction, we have vy = uy – gt = u sinq – gt ur \ Velocity at any time t, v = v ˆi + v ˆj x

y

ur or v = ucosqˆi + (usinq - gt )jˆ. Velocity at any height : At any height h vx = ux = u cosq Figure. 4.23 and vy2 = uy2 – 2gh = (u sinq)2 – 2gh Squaring (i) and adding with equation (ii), we get v

…(i) …(ii)

u 2 - 2 gh .

=

Position at any time t Position of particle at any time t, is given by ur r = xˆi + yˆj where x = u cosq t

…(1) 1 2 gt 2

and

y

= u sinq t –

\

r r

1 2 = u cos q t ˆi + (u sin q t - gt )ˆj 2

or

1 æ ö (u cos q t )2 + ç u sin q t - gt 2 ÷ è ø 2

r =

Figure. 4.24

2

= and

or

tanf =

ut

gt sin q æ gt ö 1+ ç ÷ è 2u ø u

y x

1 2ù é ê u sin qt - 2 gt ú ú f = tan –1 ê ut cos q ê ú ë û

é 2u sin q - gt ù . = tan–1 ê ë 2u cos q úû The angle of elevation f of the highest point of the projectile; t

=

\

tanf =

or

tanf =

u sin q g u sin q g 2u cos q

2u sin q - g ´

tan q 2

Equation of trajectory We have,

x

= u cosq t

or

t

=

æ x ö çè ÷ u cos q ø

2

Motion in a Plane and

y = u sinq t –

159

1 2 gt 2

æ x ö 1 = u sinq çè ÷ – g u cos q ø 2

x tan q -

æ x ö çè ÷ u cos q ø

2

gx 2

. 2u 2 cos2 q On comparing this equation with general equation of parabola, y = ax ± bx2, we find that path of projectile is parabolic in nature. y

=

Time of flight (T) : Total time of motion of particle in air is called time of flight. The displacement in vertical direction (y-axis) becomes zero in whole time of motion. So we have

or

which gives

y = uy t –

1 2 at 2 y

uyT –

1 2 gT 2

0

=

T

=

2u y g

=

2u sin q . g

Maximum height attained (H): The maximum vertical distance achieved by particle is Figure. 4.25 called maximum height. At the highest point of projection vy = 0, so we have, vy2 = uy2 – 2gh 0 = uy2 – 2gH

or

which gives

H

=

u y2 2g

=

u 2 sin 2 q . 2g

Horizontal range (R) : The horizontal distance moved by particle in total time of flight is called horizontal range. Horizontal range,

R = ux × T = ux × =

2u x u y g

=

2u y g

2u cos q u sin q g

For maximum range,

sin2q

u 2 sin 2q . g = 1 or 2q = 90° or q = 45°.

Thus

Rmax

=

or

Corresponding,

R

=

H =

u2 . g

u2 sin2 45° u 2 . = 2g 4g

There are two angles of projection for same range: Replacing q by (90° – q) in the formula of range, we get

160

MECHANICS R'

=

u2 sin 2(90° -q) g

=

u 2 sin(180° - 2q) g

=

u 2 sin 2q = R. g

Thus, for a given velocity of projection, a projectile has the same range for angle of projection q and (90° – q). Time of flight for angle of projection q,

2u sin q g and time of flight for angle of projection (90° – q), T1 =

T2 = =

Figure. 4.26

2u sin(90° - q) g 2u co s q . g

Multiplying T1 and T2, we get \

T1T2 =

2u sin q 2ucosq × g g

or

T1T2 =

2 æ u 2 sin 2 q ö ç ÷ gè g ø

or

T1T2 =

2R . g

More about projectile motion 1.

If t1 is the time taken by projectile to reach a point P at height h and t2 is the time taken from point P to ground level, then t1 + t2 = T =

or

u sinq

=

2u sin q g

g (t1 + t 2 ) . 2

The height of point P, h = u sinq t1 –

or 2.

h

1 gt 2 2 1

=

g (t1 + t 2 ) 1 t1 – gt 2 2 2 1

=

1 g t1 t2. 2

Change in momentum : Change in momentum between two positions of projectile is given by ur ur uur uur uur ΔP = P f - Pi = m(v f - v i ) . (a) Between point of projection and highest point

Motion in a Plane ur vi

and

\

ur vf

= u cosq ˆi + u sinq ˆj = u cosq ˆi

uur ΔP

=

m [(u cosq ˆi ) – (u cosq ˆi + u sinq ˆj )]

or DP = mu sin q (b) For the complete projectile motion r vi

r vf \ or

uur ΔP

= u cosq ˆi + u sinq ˆj

Figure. 4.27

= u cosq ˆi – u sinq ˆj =

m [(u cosq ˆi – u sinq ˆj ) – (u cosq ˆi + u sinq ˆj )]

DP = 2 mu sin q

ANALYSIS OF PROJECTILE OF TYPE 2 : HORIZONTAL PROJECTION Let a particle be projected horizontally with initial velocity u from height h. Velocity at any time t We have, vx = u and vy = uy + gt or = 0 + gt r = u ˆi - gt ˆj \ v and Also

u 2 + ( g t )2

v = tana =

gt . u

Position at any time t Taking point of projection as the origin, the position vector at any time t ur r = x ˆi – y ˆj . where

x = ut

and

y =

1 2 gt 2

\

r r

utˆi -

Displacement

=

s = r=

1 2ˆ gt j 2

æ1 ö (u t ) 2 + ç g t 2 ÷ è2 ø

Equation of trajectory We have,

x = ut or t =

and

y

= –

x u

1 2 gt 2

1 æ xö = – gç ÷ 2 è uø

2

2

Figure. 4.28

161

162

MECHANICS y

=

1 x2 - g 2 2 u

We have,

h

=

uyt +

or

h

= 0+

which gives

T =

or Time of flight (T)

Horizontal range (R)

1 a t2 2 y

1 g T2 2

2h . g

R = ux × T

2h . g The average acceleration in total time of flight is g downward.

= u

ANALYSIS OF PROJECTILE OF TYPE 3 Method - I : Let us consider a particle is projected with initial velocity 100 m/s at an angle 30° with the horizontal. The height of projection is 100 m. Time of flight (T) We have

y

= uy t +

1 2 gt 2

100 = – 100 sin 30° T +

or

1 × 10 × T2 2

T2 – 10 T – 20 = 0

or Figure. 4.29

which gives

T =

10 ± (-10)2 - 4 ´1´ (-20) 2

= 11.71 s (consider only positive value) The horizontal range (R) R = ux × T = 100 cos 30° × 11.71 = 100 ×

3 × 11.71 2

= 1014 m. Method - II : Taking point of projection as the origin, the coordinates of point of strike are (R, – 100 m). We have, y = x tanq – Here

\ or

gx 2 2u 2 cos 2 q

y = – 100 m and – 100 = R tan 30° – R2 – 866 R – 150000

= 0

x=R 10 R 2

2(100)2 cos 2 30°

Figure. 4.30

Motion in a Plane which gives,

R =

866 ± (866)2 + 4 ´150000 2

= 1014 m (consider only positive value) Time of flight,

T =

R = 11.71 s. u cos q

FORMULAE USED y

Projectile Type 1 1.

2.

ux = u cos q, uy = u sin q

u

ax = 0, ay = –g .

q

R

Position after time t x = u cos q t,

3.

O

H

1 2 gt 2

y = u sin q t –

Equation of trajectory y = x tan q –

gx 2 2u 2 cos 2 q

.

2u sin q . g

4.

Time of flight, T =

5.

Maximum height, H =

u 2 sin 2 q . 2g

6.

Horizontal range, R =

u 2 sin 2q g

7.

Maximum range, Rmax =

u2 , for q = 45° g

u

Projectile Type 2 1.

x

Position after time t x = ut, y =

h

1 2 gt 2

gx 2

.

2.

Equation of trajectory, y =

3.

Velocity after time t, v = u 2 + (gt)2

4.

Time to hit the ground, T =

5.

Horizontal range, R = uT = u

2u

2

2h g 2h . g

R

x

163

Thank you for interesting in our services. We are a non-profit group that run this website to share documents. We need your help to maintenance this website.

To keep our site running, we need your help to cover our server cost (about $400/m), a small donation will help us a lot.