1 DIMENSIONS AND VECTOR ANALYSIS

1.1 DfTRODUCTIOR All we are aware that the basic purpoee of all sciences is to understand the natural phenomena that occur around us. Amongst all the branches of IlCience, physics is one of the most fundamental. It is the foundation on which the other physical sciences such as chemistry, geology, geophysics, astronomy etc. are based. Physics also plays a very impurtant role in the deve10pment ofbkllogical sciences.

In p~cs we perf'drm experiments, make measurements and then propose theories which predict the results of measurements. In this lesson you will first learn about the units of measurements. Every unit of measurement can be expressed in terms of the basic units. This wi11lead us to the concept of dimensions and their applications in other areas of physics.

We will a1IIo catagorise the physical quantities in two groups namely (i) sea-lanIlind (nl vectors depending upon their nature. Finally, we will learn the simple mathematical operations associated with scalars and vectors. You will find application of vectors in difl'erent fields of physics which you will learn in other leUons during your COW'IIeof study.

After studyjDa this lesson, you should be able to, diStinguiah between theJUndl.unental and derived quantities and thfJir S1

unitIt;. write tI!e dirrumaions 0/ di,fffJl'81l1 physical quantitiea; apply the dirrumaional equations; tli/ffri'htUJte between scaler and !leCtor quantities with eJCl1mplea; find thIJ resultant o/two vectors, andresolve a vector into its components;

and compute the produd o/two vectors.

Physics

1.3 UNITS OF MEASUREMENTS The laws of physics are defined in terms of physical quantities like distance speed, time force, area, volume etc. These quantities in tum are defined in terms of more basic qurntities like mass, length and time and some others which we will study later. If a person measures the quantity of milk, she should express the volume of milk in some accepted units of volume. Like-wise if an engineer measures the length of a road connecting two cities, he should express the di.tance in an accepted unit of length. Such a procedure makes the life more comfort-able. When we travel we have an estimate of distance and time" which helps us proper planning of the journey. If there were no units accepted by all, the life would be miserable. Such units are much more essential in scien-tific measurements to facilitate communication of information at interna-tionallevel.

1.3.1 The SII Units Keeping this point of view in mind, there have been attempts over centuries in SC'Jeral developed civilizations to. suggest standard units of measurements at regional or national level. Without !.~':"'lg into the long histoty of the vari-ous stages of development in the system of units of measurements, we come to the year 1967 when the XIII General Conference on Weights and Mea-sures, rationalised the MKSA (Metre, Kilogram, Second, Ampere) system of units and adopted a system based on six basic units. It was called the 'System International units' known as SI system of units in all languages. In 1"971 the General Conference added another b~c unit 'mole' for the amount of sub"tance to the Sl The present SI system of units bas _n base or fundamental Units. These are listed in Table 1.1: Table 1.1: Ba .. Units of the SI Syiltem

Quantity rJRft Symbol Length Metre m Mass Kilogram kg Time Second s Electric Current Amp.ere A Temperature Kelvin K Lumino1.1.s Intensity Candela Cd Amount of Substance Mole mol

The yard. and mile as units of length are still in use in USA. These are given in Table 1.2. Table 1.2: Units of leli.gth.till in use in claiIy life (USA).

1 mile = 8 furlongs 1 furlong = 220 yards 1 yard m 3 feet 1 foot = 12 inch 1 yard = 0.9144 meter (exactly) 1 inch = 2.54 em (exactly) 1 mile * 1.61km

2

Dimensions and \Lector Analysis

The guiding principle in choosing a unit of measurement is to relate it to ~on man's life as far as possible. As an elllllD.ple, take th.\lnit of mass as lei,.", am or the unit of length as metre. In ou!" day to day business we buy food articles in kg or tens of kg. We buy cloth in metres or tens of metres. If gram had been chosen as the unit of mass:or millimeter as unit of length, we would be unnecessarily using big numbers in our daily life. It is for this Fe8SOn that the basic units of measurements are very closely related to our daily life. The SI system is basically a metric system. The smaller and larger units of the basic units are multiples of ten only. They follow strictly the decimal system. These multiples or submultiples are given special names. Theile are listed in Table 1.3.

Table 1.3: Prefbre. lor Po_n ofTen

Po_oJ PreJfx. AbbntPfatfon bample ten BJImboI

10-' atto a 100's femto f femto metre fin 10-" pica p picofarad pF 10-" nano n nanometre nm lQ-6 micro 1.1 micron !.1M 10""" milli m milligram mg 10-' centi c centimetre em 10-' deci d decimetre dm 10' deca da decagram dag 10' hecto h hectometre hm U)3 kilo k kilogram kg 10" mega M megawatt MW 10" giga G giga hertz GHz 10'" tera T tera hertz THz 10's peta P 10' em E

1.3.2 Standard. of , Length and Time Once we have chosen the fundamental units of the SI, we must decide on the set of sta ndards forthe fundamental quantities. (I) .... : The SI unit of mass is ktlogrtUrL It is the mass of a particular cylinder made of Platinum - Iridium alloy, kept at the International Bureau of Weights and Measures in France. This standard was established in 1887 and there has been no change be-cause this is unusually stable alloy. Prototype kilograms have been made of this alloy and distributed to member states. The national prototype of India is the Kilogram no f,7. This is preselVed at the National Physical Laboratory, New Delhi.

3

Physics

(U, Lellath: The metric system was established in France in 1792. The metre (also written as meter) was defined to be 1/107 times the distance from the Equator to the North Pole through Paris. This standard Ws aban-doned for practical reasons. In 1872 an International Commission was eet up in Paris to decide on more suitable metre standard. In 1875 the new metre was defined. It was defmed as the distance between two lines on a Platinum-Iridium Bar stored under controlled condition. Such standards had to be kept under severe controlled conditions. Even then their safety against natural dis~ters is not gauranteed, and their accuracy is also lim-ited for the present requirements of science and technology. In 1983 the metre was redefmed as follows;

One metre is the distance travelled by light in vacuum in a time in-terval of 1/299792458 second. This definition establishes that the speed of light in vacuum is 299792458 metres per second.

Following this definition a new prototype of one metre can always be pre-pared even if all the existing standards are destroyed in a natural disaster. This is the greatest advantage of this definition. (iii' Time: The time interval second was originally defined in tern1s of the time of rotation of earth about its own axis. This time of rotation is divided in 24 parts, each part is called an hour. An hour is divided into 60 lIiinutes -and each minute is subdivided into 60 seconds. Thus one secondis equal to 1 186400 part of the solar day. To be more specific, the mean solar -rut, the basic unit of time, was defined as toX -t;xf,; part of the mean _far.", for the year 1900. This defmitionwas accepted upto 1960. It is known that the rotation of the earth varies substantially with time and tnerefore the length of a day is a variable quantity, may be very slowly varying. The XIII General Conference on weights and measures in 1967 gave the following defmition of the time interval 'second.

ane second is the time required for Cesium - 133 atom to undergo 9192631770 vibmtions.

This definition has its roots in a device which can be named as atomic clock. The frequency of certain atomic transitions can be measured with an accuracy of 1 part in 101~. Theee frequencies (transitioni are exllem.ely stable and are least affected by the environment,

1.3.3 Derived Units We have defined the three basic units in mechanics. When these basic units, interact, they give rise to quantities which are melllUred in derived units .. Thus, the units which are obtained by the combination of the funda-mental units, are called derftIed WIlD. For example when distance and time interact, they give rise to speed acceleration etc. The speed is mea-sured in metre per second (ml s). Similarly wheI).'length interacts with length. new quantities like area (m2) and volume (m3) etc result. The following tables give some of the derived units commonly used in mechanics and some derived units with special names.

4

Dimensions and Vector An@!ysis

Tabl. 1.4: Bzampl of deri ... d SI Unite

Quantity BIUnit I Symbol area square meter m' wlume cubic meter m 3 speed or velocity meter per second m/s acceleration meter per square se.c m/s" density kilogram per cubic meter kg/m"

Table 1.5: Bzamplee ofden ... d SI Unitnrith Special.ame.

QuanttCJI .MmIe .. Symbol Unit Symbol

force newton N kg mis' pressure pascal Pa N/m' energy, work joule J Nm Power watt W Jls

The SI system of units form a coherent set in the sense that the product of any two unit quantities leads directly to the unit" of the resulting quantity. When unit mass (kg) is divided by unit volume (m3) we stra,ight way get the unito(density kg/m3. We should be careful in writing the units of certain quantities in proper order. Let us take the example of work. The unit of work is Newton - meter which has been givena special name Joule. It should be written as Nm and not as mN. If written as mN it would mean 'milli Newton". Now. it is the time to check your progress. Solve the following intext ques-tions and incase you have any problem. check answers given at the end of this lesson.

IRTEXT QUESTIORS 1.1 __________ _ 1. '" car ;" moving with a _eel of 80 km/hr. What i. th .. speed in mI.?

2. Diatinguillh between the fundamental and the derived unit .

3. The radius of an atom i. 10-'0 m. What will be thiJo value in terms of micro mette? ................................................................... , .................................................. " .. -..

4. Th. total c:owred area of a h"" .. ;" 4500 IqUIU8 fe.t. Expnaa thiJo ..... in jUare metres.

1.4 DIMElfSIORS OF PHYSICAL QUANTITIES It is useful to assign dimension to physical quantities. The three basic di-mensions of the three fundamental units. Lerigth. M8.S$ and Time are sym-bolized respectively as L, M, T. The dimensions of other physical quantities

5

Physics

arc expressed in terms of these symbols. See the following exaJllples.

distance m L ., ilJ speed_ time' =; =T = LT

mass kg M -.3 (:l! df"nsitv = -----=-, = - = ML . volume m" L3

m (3) force = massx acceleration =kgx,:: MLT-2 s

The dim~siona1 analysis is a very useful tool in checking the correctness of expr~ssions or equations relating various physical quantities. Let us exaJIl-ine a few cases as examples. Exemple 1.1: The mechanical energy of a particle can be written in two dif-jtmmt forms as fa) l-'ll mrr and (b) mgh. Are both dimensionally same? Solution: la) >,:, 1m"'" '," M (LIT)' = Y. ML2 'I'"2 fbi mgh = 1M) (LIT') IL) = ML' 'J'"2 We therefore fmd that dimensionally both expressions for energy are equiva-lent. They differ only by a dimensionless multiplier, (Factor 1/2 in this rase). Example 1.2: Suppose a car starts from rest. The car COuenl a distanoe x in time t while moving with uniform acceleration a. Find an expression.jOr x in terms of t and a. Solation: Suppose the expression for x is of the form,

x ~ an t", (~ . is sign of proportionality) This formula will be correct only if the dimensions on both sides are the same. Left Hand Side (LHS) Right Hand Side (RHS)

x= L' - L' MDT" d" t!' - (LIT')" m"

- L"! 'J'"2'" T" - L" '1"'" ..... - Lm MO T""""

If the two sides must have the same dimensions then by comparing the dimensions of L, M and T seprately we get

m=l, n-2m - 0 n- 2 = 0 n - 2

Hence x~ at'. We know (may come to know later) that this IS not the proper form of the expression. The correct expression is x y. at'. The sign of p.opor1ionality is replaced by th, SlgIl of equality with the help of a dimensionless multiplier (1/2 in this case). Example 1.:J: In an experiment with a simple pendulum we come to a,quali-tative conclusion thatthe time period Tofthe pendulum depends on the kngth

6

Dimensions and Vector Analysis

of the pendulum l and theiu:iceleration du8 to grallityg. But we do not know the exact dependance. Find the eKQct expression for Tin terms of land 9 SolatioD : Let us assume that

T~ IT' if Dimensionally L.H.S. - T - LO M' 7'

RHS = "'fI' .. L"'(L/1")O - L- AI' ~ By comparing the powers of L. M and T on both sides

m+ n- 0 and -2n-1 m.--TJ n--2 ... m - +Y. and n - -Yo

... Tal".r or T=2k~ You should bear in mind that the numerical constant (2 It ) cannot be deter-I!lined from dimensional analysis. ~pIe 1.4: It is known that.a particle moving in a circuWr orbit has an acceleration which depends on the orbital speed II and radius r 'afthe orbit. Determine the powers of II and r in the expression acceleration a - krI"r' where k as usual is a dimensionless quantity. SollltioD: LHS a- L/1" - LM' T-2 RHS ""r" - (L/7)"'Lo - L-M'T"". Comparing the powers of L M and T OJl'both sides, we get

m+ n-l and m-2 Wegetn--l

II~ a - kIP:--' - k-

r

Age;" the numerical value of kcannot be obtained from dimensinnalailalvsis. Now, take a pause and do the following questions.

IRTEXT QUESTIORS 1.2 _________ _ 1. A atone ia dropped &om the ......... r .. boll buiIdiIIg. The velocity "with which the alOne

hila the ground cIepeod. on the 'lei&ht h or the ~ and the """"Ioraticn due '.0 ~ g. Obtain the "*' rior eo.. ... ................................................. _ ................................................................ -........ .

2. The diaplarement oCa m6'riu& perticleia Jlivenby the _-on Y- A Sin (X, t+ K,x) lib_ 'y. A and "' ..... in met .... and tin aeCoRd. Obtain the.dim ... siona or K, and K, .................................................................... '-_._ ..... ~ ......................... u .................... .

The accel""",tion or a,I"0vmg objeq.ito

Pb,ysics

L5 ORDER OF MAGNITUDE It is quite often useful to know/ calculate the approximate value of a particu-1ar quantity. It helps us in checking the results of any lengthy calculation or to know the approximate magnitude of a quantity. The order of magni-tude of a quantity is the power of ten of the number that describes the quantity. We know that the speed of light in vacuum is 3 x 10" m/s. We, therefore, say that the order of magnitude is 8 .. If a quantity increases by four orders of magnitude we mean to convey that the quantity has iIi-creased by a factor 104 Ezample 1.S: The size of an average room is 6 x5x 4m:'. The room is to ~ completelyfilled with cricket balls without crushing them. If the diameter of a ball is 5 em, estimate the number of balls that will fill the room. Solution: The volume of the room The approximate volume of a ball

= 6 x 5 x 4 = 120 m3 = 5 x5x 5 x 10'" m3 = 125 x lQ-6 m3

(You know that the actual volume of a ball is given by 41fr' where T is the 3

radius of the ball.)

Hence the total number of balls 120 6

= 125xl0-

Oimensions and Vector Anal}'llia

1.6 VECTORS AIID SCALARS / 1.6.1 Scalar Quantities

We know that the physical quantities are always described by some num-bers with proper units attached with them. We describe the density of material, the volume of ajar, the distance between two cities, the Speed of a moving car, the forcethat pulls all objects towards-the earth and the torque that ~ens or closes a door. Some of these quantities are ~re8l1ed in numbers with units and it is a complete and correct d.:scription of that quantity. The examples are (i) density of copper - 8.9 x loa kg/m3 (ii) mean radius of the earth - 6.4xlo6m (iii) one day - 8.6 x 10" s; Such quantities do not require any direction to be specified for theif descritpion. These are knoWn as __ quantities. '

A _lar CJifAfdlCg has only magnitude and no direction.

1.6.2 Vector Quantities A vector quantity is a physical quantitY that is described by both the magni-tude and the direction. ~is a vector quantity. lfwe apply a fOrce of lOON Oil an object, we must also SPeci1Y the dfrwctfonin which the force is being applied. Veloc:ftgis an other vector quantity. lfwe describe the motion of an object we mwrt specify how fast it is moving and the directiOn ofits motion.

A IMCCDr .-atuwhas magnitude and direction both.

VectorIIcan always be represented graphically. Let us say that a force of '=iiri 501;)' N is applied on a body in the direction west to east. This vector quantity ~ C'lll be represented graphically all shown in Fig 1.1. below

A B

ftc1.1: Gnq1Irk:Dl"""""-'af"- 'The line AB represents this vector. The length of the line AB, say 5 em, represents the magnitude SOON. The direction of the vector is from A to B (West to 'east). The pOint A is called the ~r and the point B with an arrow Diark is called the tip (head) of the vector. y ADDthervector CD ofmagnitude 300N force is1epresented by CD (3 em length) pointing in a different direction. Any two vectors A. and B are said to he .. _ ifthey'bave the same magnit~l(le and point in the same direction. Graphically, therefore, all such vectors which are of the tiame length and parallel to ea,ch other are said to be equal all Shown herein Fig 1.2. The three vectors. A.,. B and c:': are of equal

o '-_______ x

PI&- 103: EqIAal- "'PI> .en(ed grap}.b~i.!l.

length (magnitude) and are parallel to each other (po1nt in the same direction).

We therefore say that A = B =C.

1.6.3 Addition orVectors When two or more vectors are added together, like scalar quantities, they must have the same units. Graphically, it is very easy tC' "dd two v~ctors and find the resultant sum. L,,[ liS have 1"\) \~":llIr, A un.i 1\ :;1'01 ',w have 10 rind their veclor sum R = A + B ...

) A A

Ca) Cb)

nil 1.3. Addition of two ""ctors, graphlcaUy. The two vectors are shown in Fig. 1.3. (a). T~ find the sum of the two vectors we adopt the following procedure. First, draw the vector A graphically. Then draw the vector B in such a manner that the tail of the vector B starts from the tip of Vector A as shown in fig. 1.3.(b). We know that a parallel movement does not change a vector. A. vector drawn from the tail of A to the tip oCB is the resultant vector R ~ A + B. Following the same procedure. We can find the sum of more than two vectors also. Let us take three vectors A, B and C and we have to find the vector R-A+B+C

A.

Flg 1.4: Addition o/three vt."Cfors in. a graphical manner

DrAwtb, \Tctor A. nr"wthevectorB such that the tail of B s1:8 tsfrom the

10

DiIlletlSlOnS and V"rtor Ana1~ s".

tip of A. Then the vector which starts from the tail of A and ends at the tip of vector B is the sum (A+B) .. Following the same rule we can add the vector C to the Vector (A + B) and get to vector R = A t B + C. You can now appreciate that several vectors can be added together in the same manner. Now, following the graphical procedure illustrated above, you can easily sho., that (i) A + B ~ B + A (commutative oj properly vector addition' (1.1) (ii) A + (B + C) = (A + B) + C (Associative property of vector addition' (1.2)

1.6.4 Subtraction of Vectors We define the negattueofvectoras a vector which has the same magnitude but points in .the opposite direction. If A is a vector of magnibloe '300 unit s pointing towards east, then -A is a vector of magnitude 300 units poil1ting towards west. Followmg this defmition of ... negati,'c vector we can evaluate the difference of two vectors in the same way as we evaluated the sum of two vectors. Let us evaluate R = A - B ' Thi.." can also be written as R - A + (-Bl The graphical construction of finding A - B is shown in Fig 1.5. First we draw the vector A. We thm draw the Vector -B. To find A + I-B), we now draw the vector -'8 such that the tail of -B coincides with the tip of A. A vector wbich It joins th .. tail of A to the tip of -B is the vector (A - B).

A

1.6.5 Law of Para11.1o .... 111. ofV.ctor. We can find sum of two vectors anylitically without graphic'al construction

,also. Let us evaluate It A + B

p A Q F

Let A and B be two v"ctors and e the angie bl't\veen them. We construct a parallelogram. PQED where PQ - D& z A

and PD-QE B Let EF be perpendicular to the line PQ. In A PFE

--------_.------------------------------11

Physics '"

~ - (PF)" + (FE)" (1.3) ..; (PQ + QF)" + (FE)' - (PQ)" + (QF)" +2(PQ) (QFl + (J'E)' - (PO)' + [(QF)" + (FE)") + 2 PQ.QF -(PQ)" + (QE)" + 2 1'9. QF -(PQ)' + (QE)",+ 2 PQ.QE,(QF /QE)

R" -.A2 + B' + 2AB cos e (1 .. 4)

(1.5)

This expression for R 4i equation (1.5) gives us .the magnitude of the sw;n of two vectors A and inclined at an RDgle e between them . .... Ic.-

ij Wbm. e "0 the twO ~ are parallel to each.other. thenR -A+B

ii) WbeDe-n/2, R- .JA'+B' (1.68) (1.6 bl

iii) WhflI1e.. n the two vectors are anflparallel to each other. They are directed oppOllite to each other.

R .JA?+B'-2AB, R-A-B (1.6 c) or R-B-A

The direction of the vectorR with respect toa reference direction eM.u.o .. be determined analytically. Let the reference direction be the direct:i.oA Oi one.of the vectors, 1lIIY Vector A and R makes an 8DIle a with ..

EP" EP' B sin 8 tan or. - .. . --:-........... --:-

. PF PQ+QF A+B cos e .

Baine tana=A+Bcose

Since A, and e' are lcnl)WD. the direction of. R' (angle a) can. be eaaib' calculated .,.alal __ :Wecandiscu .. the same speCiaJ. caseaforwbichwe obtained . the magnitude of R .

when e -0, tan a- 0, . a- 0 The direction of R coincide_ with A

iiI' e .. n/2. tan a - B/A , iii) 8- ft. tan a -0." will be ~ the vector A:or. Again. it i. time to check your understanding.

i.

12

11.881

(Ub) . ( 1.SC)

;

I I 1,

Dimension$ and Vector AnalJo:Sis INTEXT QUESTION'S 1.4" ____ ~ __ ~ __ _ 1. There are three vectors of magnitudes 8. 16 and 20 units oriented arbitrarily in the

same plane. What are the minimum and m...mnum values of the;,. resultant?

2. A 'man ~ 3.0 Ian toward. east and then 4 km towarda north. (a) How far away he is from the atartmg point? (b) What i. the direction of hi. final position? ........................................................................ ' .................................................... .

3. There are two vectors A and B of equal magnitude with an angle Df 60' between them .. Determine the following graphica1J,y (a) A + Ii " (b) A - B (c) B -A (d) A + 2B (e) A - 2Il ............................................................................................................ ,...: ............... .

1.7 PRODUCT OF VECTORS 1.7.1 Scalar Product The scalar product of two vectors.A and B is defined 11$ a sca1ar quantity which is B. product of the magnitudes of the two vectors multiplied by the cosine of the lingle between the two vectors A.B - AB cose. (1.9) Where e is the angle between the two vec-tors .A and Bi Fig 1.7. A dot 'sYJ1,lbol between A and. B is put to indicate the scalar product. For this reason the _far product is also called the dot product. The quantities TepTej!lented by the . Vectors A imd B need not have the same . units. The product AB cos e contairis the magnitudes oCthe veCtors onJ.y and is"a sca-lar quantity. It is therefore easy to appreci-ate the following relations. A.B - B:A - AB cos e A.(B +C)" - A.B + A.C

A

(1.10) (1.11)

We Will find a p~ctical Ulle of~ relations in later Chapter. on mechanic-. ___ plel.S: 7he two vectors are o/magnitudes IA I- 30 and IB I-40 and the angle 8 -55' ,calculate the magnitude and direction ofths Vector II - A + B. IIoIutJoll I"

~per equation (1.5) IRI -.J A~' + B~ +:lAB co~~ . - ~(30)~ +(40)~ +2x30x40xcos 55 - (900 + 1600 + 2400 x 0.5736)"

IRJ -. 62.2 (magnitude) Bain9 40xQ.81923 ::::2::;.;...7:.,;:6-=..8 ~a.. =-

A+ Beas9. 30+40xO.57'36 52.944

. 13

Physics

- 0.6189, - 31.80

The l:1!suJl:antR~angleof31.8withvector A. Byemple 1.7 I A carl is being puUed I N by man A towards e.ast with a force of B I 70N. Thesomecarl.isbeingpulledby .~ manBtowardsNorth-westwithaforoe 30N ~. 135" A of30N. "'_..J... _____ ~, . (a) CalmlatethBresultantforoeacting I 70 N

on the cart. (hI Find the direction of this resultant Is '

forc#l, Eblutloll : In this problem .

A-70N, B - 30 N,e - 135

. R- J(70)~+{30)2+2x70x30c~a135 - ~4900+900+4200xcoa(90+4S)" - ~5800-4200ain45-.J2821 ... R,- 53.1 N. . BlliD.9 30x sin (90 +45) _~30.;;.,C;..;08;.;;.4.;;.,S ... ~.;... _

una- - -A+Bcoa9. 70+30coa(90+45) .70+30(41145)

30)(0.7071 21.213 4348 - 70-3D

I I i

Di1r.ensions and Vector Analysis

shown in the figure 1.8 then A x B - C will point vertically upwards and B x A - -C will point vertica1ly downwards. Because of a cross ~ between A.and B, the ~r product is also called as cross product. We will make use of these relations later.

1.8 RESOLUTION. OF VECTORS A vector can be resl'\ved into components with respect to a particular co-ordinate sys-tem. Fig 1.9 shows a vector A in a rectangular coordinate system. The tail of the vector coincides with the origin 0 of the co-oTdi-nate system. Let us draw perpendiculars from the tip of the vector on the coordinate axes. We get quantities A. and A.,. These are, called components of the vector A. The process is called ruolutfon o/~r into com,porwnta. The components A. and A., are

A. A cos e

y

Ay

L-~ __ ~ _________ X

Ax

1'1& 1.!I'"","""'P""0n,..!fhII_ A aIDng 11M! _gular 00I>I'd/naI8 ...... " -,I .

~ - A sin e (1.15) whflre e .1S the ansle which the vector A makes with + X axis. If we know the vector A i.e. we know the magnitude A and the direction with respect to a reference axis. we can obtain the components A. and A.,. Conversely, ifwe know the components, we can obtain the magnitude and direction of the vector.

A. Acose A., - Asine

A~ +A; = A2(cos29+1Ii.n28) = A2 -+ A. ~A; + A; (magnitude) (1.16)

tane A.,/A. (direction) (1.17) The sign of tan e detennines tlle quadrant of the co-ordinate axis.

. A more convenient wa,y to handle the resolution of more than one vectors in a three dimensional space is to introduce the concept of unit vectors. '!be vector A can be written as .

A=ialAl (1.18) Where 1.0 is 'a unit,vector in the direction of A. (A cap over the letter represents a unit vector). In the rectangular co-ordinate system we choose t.l and k as unit vectors along the x, y and z directions respectively. Let us do some simple mathematics for a two dimensional system. This can be easily extended to three dimensions. Following the definition of unit vector.

A. A,.i + Ayj B. B.i+Byl (1.19)

15

This is shoWn in the Fig 1.10.

y +y

n

9 ~""""" __ ~ ___ :It

I

e ..... .)'-.l.,;.~_,....----- x -y

ill I A..i.

..... ~.lOfIl} N ..... 1.10Cb) "","l.lO'~~~""',io_",,'kr ___ , _ A !II .. ~ 9" wll 1_",. __ ... "'" """'""" No

Suppose ~ attempt to find the sum of the two vectors A and B. R=A.+B R,.- A,,+Bz

, thl\ - -VB,. make ..... nl e th X . ,Let e veCtor an _e WI -8XIS.

IRI-J{R~+R~) ~J tanB=-R"

We solve the following problem to illustrate the efficacy or thiil analyticaJ method of addition of vectors. . .... pte I.': 7\uo IIIJCtora are gillen below

.1.- 2i+31 and B .. 3i - 4) Find ths awn fJ/ths IIIJCtora A + s- .If. 101"': Accordina to our nOtation

.

A" - 2, A.. - 3; Bz - 3, B, " ,-4 R,.. 2 + ~- 5 ~-3-4--1 . R_.JS3 +13 -.;J26*5.1 ' tg'e - RJR. - -1/+5 -, -9.2 e' .willlie 1n the lVquadrant.

, ,

- .... 1 ' 7\uoJHIirUAG7IdBinthsXYpItme,hall8~(-.3, ~)m and~. 4) m ,....,ci1l8l1l. . (a) WWte.., .ssionaj'or posiCion 1IeCtcQ qf A. cUad B tA ~-,; (bl BvaluGM (JIA ,+ -,; IUId tA ,- -,; ' ..

5 16

1Io1atloa:

(a) The poai~on vector RA - 1-3 i +2)) W. JIg .; (2 i;':L4 hin

(h) R,.+",-(-31 +-2J)+{2i +4J) - (-2i+6Jt RA - .... -13i +2])-,.(21 +4])

--51 -2J~

....... 1.10: f{avec:tOrCiatldtkdtouedDr. theraultia-9i '-8j. 1/

.ia subtractedfrom C the ~ ia 5i + 4 j _ . What ia the direction of . lIoIattoa: Given,

+ C -'-91 -sl 0-."51+4)

By adding we .e.:; 20--41. 4j; By aubtractlnsWellet, 2.--141 -12ja

, '

. B,. -7 Therefore, tan' e - S- -6 -1.167

" ,

0--214) --71 -61

AcCOfdinl to the tablea IP.ven here e is in the m quadrant. ' , , '8'- 221. .

)r. .., ... 1.111 GiCllm (i) A +.- 61 + 1 (ii) A-.- ..... I+1J What ia the (oJ ~ and (II) ditlill:JCiDia of A

, ."tloal (al By addiq Ii) and(ii) we pt 2A. - 21 + 8]

:. &"- i +4) A.~ 1 and A,. - ... , It. -,~It.:+It.~ -.Ji7. 1t.-4

It., 4 (b) tan e -' A -I' e. tan-I (4).

"

.. TIXT QUB&TIOR 1.1. ____________ _ 1. TMre ill a ..... d'lll block CIa the 1Ioor. '--t A ill JNlIinI; it,,1Iitb. "'"'" .. .uS peI'IIOq

B ill ~ it ~ the _ ,.... ... The cIInaIfaa. of 'the .... with ....,.. to the IIoor ill 30' In' both _ , Wbo WiD lie ... to ....... the blOClk em the IIoor man --uaat\)- .uS wb:T? '

17

Physics

2. A vector A of magnitude 50 m is in the XY plane and makes an angle of 300 with the X - axis. What are the X i!lnd Y components of this veclor? ............................................................................................................................

3. A force of 200 N is applied on a.body at an angle of 120 with the positive X - axi . What are the directions and magnitudes of the rectangular components of this force? ............. -.............................................................................................................. .

1.9 WHAT YOU HAVE LEARNT _______ _ Every physical quantity is measured in some system of units. We in physics adopt

the SI., In mechanics, kg, ~ and "" are the base units of the SI foJ' measuring ma.is,lengtl

, ~d t~mp. r"'!,;Dectl,,-ely. The other 'units are cl~lled der;Vt::d UTl1t:" A standard mass ot an alloy material has been acceptp.d as stanaard kilograrn.

1 One meter i. the diatance travelled by light in Vl\CUum in secona.

, 299192458 One second i. the time required for c. -133 atom to undergo 9192631170 vibratiOn. ,. EvelY physical quantity h .. dimension.. The dimensional lll'a'Iyai. i. a u.eful tool UI

check the correctn of mathematic expre.lion . There ere vector and .calar quantitie . Addition and .ubtrection of vectora - Law of perelle1"11l'&lX'of ""ctora.

(i]Scalar product R A AS CO.' .9. (ii) Vector product R A , I R I AS oi.'l 8. The direction of R i. e1.o,derlllsd.

A vector eon be re.olved inte it. component

1.10 TJtRMUfAJ, QUltSTIOIfS. _______ _ ,

1, Th. aver .. e .peed of the fute.t train in India i. 120 km/h" What Will be the .p.e~ in metre per aecond?

2. Show that the expreuion ut + y, at', where a iii di.tenee, u i. speed lind a i. the aoce1oration of a -inbvina; particle, i. dimen.ional1y co.,..,ct. '

3. If X At + Elf (where x io diotan.e lind t i. time) i. dimenaionll1ly correct, then ah"", that (A + 3 BI'J h .. ~e dimenaion. of velocity.

4. The diameter of lin atom i . about 10" m. E.timate the number of atom. in 1 m' of a .olid. '

S. A his boat i. relltin& on .tiIl water. There ere two rope. tied at the .ome point on the boat. One man pull. the boat 1110111 BOUth with a force 12 N mel another mail. pull. it al0lll eut with a force of lIS N. (aj What i. tho re.ultes)t force with which'the boat i. belni pulled? (h) What is the direction of its displacement?

IS. A river i. flowina: toward. eut with ... peed of S Ie nIh . A owi2nmer io IWimmina from' the BOUth bank with a .peed 4 km/h IIIw~ heoding toward. north. (aj Will he be able'to reach the north bank exactly at the oppoaite point? (h) What i. the direction of hi. effeive .peed in water? (e) What i. hi. eective ape.d in the river?

7. A boy i. pu1lina table toward. south with .. force S N lind m. aioter i. pu1lina the _e table toward, e .. lt with a force 4N on the horizontal floor of !I:1e heu"'! . Th .... i. a force of friction of IN betw,en the floor and the tablit in .wry direction. 'ii1t.at iii the ' moanitude of the reoultes)t force which i. movin& the I":1>le?

8. A man walkin& aero .. hill field. in a cIarIc nicht 10 ... hi. WIlY. He wll1ka 100 .tep. toward. weat and the"; 60 .tepa, toward. north and then .ame .tep. in 'another direction .0 that he apin com back to the .tarting point. How IIW\Y atop. and in

18

Dimensions and Vector Analysis

what di,~etion he walked in his third movemert? Sol"" the problem graphica1ly. 9. A person walks along the paths shown

in Fig 1.11, starting from the point A and ending at the point B. Find the magni tude of his displacement.

10. I). particle moves with a velocity v along east for 5 seconds and then towards north with veloCity V for 5 seconds. (a) How much is the change in velocity

per second. (b) Change in velocity per second is de

fined as acceleration. What is the direction of the' acceleration?

11. A block of mass 5 kg is sliding on an in clined plane making an angle of 30 with horizontal. What is -,;he m~tude of the' force driving the block along the plane?

12. Two vectors-are A"" 8 units and B "" 5 units. - The angle between them is 60". Find

(oj A.B

A 100m

.. ~g 1.11 300m

Fig 1.12

(b) A x B and its direction, for the following orientation (A and B are in horizontal plane). .

A Fill 1.13

13. Two vectors ...... given by A -' 3 i -2 j and B-2 i -5 j , Calculate (A i' B) and (A - B) ANSWERS TO THE INTEXT QUESTIONS latex!: Q1aMtlozuI 1.1

80lan 80 x 1000 200 . 1. --= =-m/a

h 60,,60 9' - - 22.'2 mi.

2, The- _...lndamental unite are indepen-dent units and they are only seven in number. Whereas the derived unit. are obtained by the combination of the fun-damental units and there can be any number of derived units.

3. 1O.,"m - 1 micro metre. 4. 1 feet < .1 feet - 1 sq, feet

1)( 12 )( 12 x 2,54 x '2.54 104

929 - --4 =.O~,

10 2

4500 aq feet -45< 9.29 - 418- m' Ia_ Quatl.oa 1.2 1. vocgmhn

2,

19

RHS - rI' Ii' - ( \.2 r .1," -I, .... T''''' Compare MO L' 1-1 - Lm+n T"":1m. (LHS) m+n-l, n-l-n;, n - 'f,o

(RHS) -2m --1

. Y'hl< .3L :. v oe g. or .,-- a 9A y~ A sin (K,t + K,';

3.

""ee dieplacem""t ia ..... in meter. LHS- Y-L'M"T" RHS - A .... (K,t + K,,>;I- L' . Quanti~ in ( ) mU8t be djrnenoiOll-Ie ... K,f- L"M" T", DimenojOll of K, - 1':' K,x- )("LT", Dim.,oi"" of K. - L-'

.cceJeration -Applied f'on:e

m

F Cl--,

m Fon:e .- _ - M L 'f

i ! I ! I i

!

2 MOTION IN A STRAIGHT LINE

2.1 INTRODUCTION YO)! see a number of things moving around you. People, animals; vehicles can be seen moving on land, fish, frogs and other aquatic animals are seen moving in water. The birds and aeroplane move in air. The sun and the moon appear to-move in the sky. Though we do not see but the earth ( .... 1. which we live also moves. It is, therefore, quite apparent that we live in 8 world that is very much in motion. To understand and describe the phySlc. world around you, the study of motion "is very mucb, essential. Motion car. be in a straight line, in a plane or in the space. If the motin'~ -lfthe object is

- in only one direction, it is said to be the "motion in a straight line'.. For example, motion of a car on a straight road, motion of a train along its straight rdils, motion of a freely falling body, motion of a lift, and motion o(an athlete running on a straight track.

In this lesson we shall confine our attention to we Illotion in a straight line -leading to the description of motion in general. In the next and the following lessons you will study about the laws of motion, motion in a plane and other various types of motions.

2.2 OdJECTlVES

~

Mter studying this lesson, you should be able to,

recall distance, ~placement, speed anq velocity; explain relative velocity and average velocity; define acceleration and instantaneous acceleration; draw and interpret position - time graph;

draw and interpret velocity - time graph jor uniform and nOn-uniform motio"-,; explain Instantaneous velocity; ,,:.mve 0" ~ apply equaUons oj motion with constant acceleration; and dRscrib~ :flotion ul'lc/f.,- gravity.

Physics

2.3 VELoCITY AND ACCELERATION In your earlier classes you would have studied that the total length of the path travelled by a body is distance whereas the difference between the initial and final position of the body is called its dfsplac4lment. Basically the displacement is the Sh~rtest distance between the two position and has a certain direction. Thus ~e displacement is a vector quantity' but the dis-tance is a scalar. You would also h!!.ve learnt that the ra~ of change of distance with time is called speed whereas the rate of change of displace-ment of a body is lmown as its uelocUy_ Unlike speed velpci13T is a vector quanti13T. 2."3.1 Average Velocity When an object travels with different velocities, its rate of motion is mea-sured by its average veloci13T or average speed. The averageveloci13T of an object is defmed as the rate of change of displacement, whereas the average speed of an object is obtained by dividing the total distance travelled by the total time taken. Let XI and X, are the positions of the object at tl and t, times respectively. Hen"e, average veloci13T

displacement v=

time taken

v=X2 -X, Ax i.e. t2 _ tl or v = At

total distance travelled and, average speed = total time taken

(2.1)

For, understanding the average speed and average veloci13T clearly, read the following examples. ,.

Example 2.1: An object is moving along the X - axis whose coordinate is x K 20f' ms-2, where t is the time variable. Calculate the average velocity o/the object over the time interval from 3 s to 3.2 s. Solution ~ Given,

X. 20 fI m!l'2 We lmow, the average velocity is given by the relation

x2 -XI Vav =

t2 -tl Ast, -3s

XI --20)( (3 S)2 ms'" 20mx9s2

= 180m

As t, ~ 3.2 s X, - 20 x (3.2 sl" mr

- 20 )( 10.;2 m - 204 m 204-180

3.2-3 24 -. msl = 120ms1 U.2

22

I

I , ! , I i ,

Motion in a Straight Line

Hence, average velOcity= 120 ms-' Example 2.2: A man nms on a 300m circular track and comes back on the starting point in 2008, What is the average speed and average velocity afthe man? Sohtiou : .Given, Total length of the track. - 300 m, Time taken to cover this length K 200 s The man comes back to th~ same point :, The displacement - 0

. Hence, average speed

and, average velocity

2.03.2 RelativeVeloclty

Total distance travelled = Time taken

300 -1 -1 =--ms =1.5ms

200 displacl!fuent

..

Time taken

o 1 =--=Oms- .

200

Whim we say that a bullock cart is moving at 10 km h-' due south, it means

the velocity of the cart with respect to the earth is 10 km h-1 , Infact, the velocity of a body is always specified with respect to some other body, Thus, the velocity is relative in nature. SupP,OSe a girl is walking in the compartment of a moving train in the direc-tion of the motion of the train. It means the girl is moving with respect to the train. Letthe train is going due west relative to the surface of the earth. The earth in tum is moving due east relative to the sun. The sun itself is moving around our galaxy. Our galaxy is moving relative to other galaxies. Thus if you want to find the absolute velocity of the girl, itttiay be a tedious job. All velocities are thus relative. The relative velocity of an object A with respect to another object -B is the rate at which A changes its position relative to B. For example, ttvA and' lie are the velocities of the two point objects along a straight line, the relative velocjty of B with respect to A will be VB - VA'

The rate of change of the .relative position of an object with respect to the other object is knOwn as the ,..latfw -locCt.rI of that object with respect to the other.

....... 2.3: A car A is moving on roadfrom North to South with a speed of 60 km/ 11. Another car B is movingfrom South to North onthe same road with Q speed of 70 km/ 11. What is the velocity of car B relative to the car'A? .... ~,; ~nsidering the direction from South to North as positive, The velocity (ve) of car..l3 - + 70 km/h And, theveIocity (v~) of car A = -60 kIn/I-.

23

Physics

Hence, the velocity of car B relative to car A . - v - v .~ 70 - ("{;O) = 130 lan/h.

2.3.3 Acceleration Whiletravelling in bus or car, you would have noticed that some titne it speeds up and some time slows down, thus changing its velocity. This change occurs with time: Just as the velocity is defined as the time rate of change of position, the acceleration can b.e dilfVwd as ttrne rt:d:e oJ change oJ velocity. The average acceleration of an object is given by, .

Final velocity - Initial velocity Average acceleration (aa.) = Til;o.e taken for change in .velocity

U2 -VI flv aa. = t2 -tl = M (2.2)

1 he' fnstantaneotIs ciccelenztion a is dejiried as the limiting value of the average acceleration, as we let t.t approach zero;

lim flv dv a = M ~ 0 At = dt (2.3)

.lU::celeration is a vector quantity and its Sf unit is m/s' .

. \cceleration is a vector and therefore has a direction. In gerteral. for horiz!lfJ-tal motion, when the aC,celeration is in the same direction as the motion or velocity (notmally taken: to be in the positive direction), the direction is posi-tive. However, an acceleration.may be in th~ apposite direction ofth~ mo-' tion. Then the acceleration is taken as negative which is often ca1led a deceleration or ~ > . hample 2.4: A cyclist starting from rest attains a veloctty of lS'fanI hiP. 3 min, .compl!1e the accelerQ.t!on of the cyclist. ' f;ollltion : Given,

VI = 0

. (25) 15 lanlh - '"6 m/s t = 3 miri - 180 s.

V -1\1 :. AcCeleration a = 2. I

t a = 0.023 m/s'.

25 1 -x--6 180

, Hence, acceleration of cyclist = 0.023m/ s". Now, it is time to check your progress. Solve the fonowing questions. INTEXT QUESTIONS 2.1 ____________ ----'

. I, Is it pooaible to have some aver.., speed but avera,ge velocity to be -.. for a moW1g object? . ............................................................................................................................

, .

. 2. A womlUl went to the market at a speed of 8 lan/h. Finding market closed she returned back to her home with a speed of 10 lan/h. If the market i. 2 km ~ from her home calculate her average velocity and average speed. ................................................................ H ................................ ; ............ ~ ............ .

24

j

Motion in.a Straigbthme

3. Can a moviDg body have relali'lle velocity ....... with .........,t to ""other body? Clive !Ill example.

4. Two can A and B atart. from the same point to JDOVe in perpendicular direebDllll with uniform _de of 30 km/h and 40 km/h reepectiYely. Calculate the

Physics

the _locft:g of the mOltIng object fa constant fa known as wtfform mo-tion.

In other words we can say that when the moVIng object covers equal distances in equal intervals of time, it i's called unVorm motion.

For uniform motion the position - time graph is a straight line inclined to the time axis.

2.4.2 Position-Time Graph for.l'fon-UniformMotion Let us take an example of a train wh,ic:h starts from one station speeds up, then m()Ves with uniform velocity for certain duration and before stopping at other ,station Slows down. In this case you will find that the d'lstcinca r:ouered In equal fnteroala oftVrur __ not equaL Such Cl motion eanbe called as lion-unifOrm motion. If the distances covered in successive in-tervals are increasing, the Dlotion is said to be accelerated motion. The position-time graph for such an object is as shown in Fig. 2.3, Fromthis graph you will notice that tjJ.e OA, Be and CD portions represent uniform motions but with different velocities. The' portion AB shows the stationaiy position of the body. HoweVer, th.e wholejoumey shown by the graph represents non-uriiform motion.

A A B

i I 1

0

time (I) D

1'I&3.8,_graph/ar a~_

i j ----------I I

B

z. - - .. - . - - - -. -I C

I A I t, It, I

TIme(&)

l'I&a.4, ___ " .. " ~auw.

See the Fig. 2.4, the position-time graph can be a coritinuous curve also. It means the djstances covered in different intervals of time are di!ferent. Hence, the velocity of the body is changing continuouSly. In such a sitUation in any interval of time the average speed ,of the body can, be determined. The fnstantcureous ~ can also be determined which will be equal to the slope of the cUrve at that'instant.

2.4.3 Interpretation of Positlol1 - Time Gr.aph As you have seen, the position - time graph of different moving objects can be different. lfit is.straight line parallel to the time axis:; you cansav that . the body is at rest. (Fig' 2.3) But the straight line having inclination with time axis shows that the motion is unifonn.

26

Motion in a Straight Line

(al Velocity from position - time graph: The slope of the straight line of position - time graph gives the velocity of the object in motion.- For deter-mining the slope, choose any two points (say A and B) on the straight line (Fig. 2.3) and form a triangle b'y drawing lines parallel to y axis and x-axis. Thus. from Fig. 2.3. the velocity ofthe object,

X 2 -X1 v-

IU BC t 2 -t, At AC

Hence, velocity of object = slope of line AB. It shows that more the slope (lUI At) of the straight line of position - time graph, more will be the velocity. Notice that the slope is also equal to" the tangent of the angle that the straight line makes with a horizonta1line, i.e., tan 6 = lUI At. Any two corresponding IU and At intervals can be used to determine the slope and thus the velocity. Example 2.5: The position - time graph pf two bodies A and B is as shown in the figure. Which of them has larger velocity? Solution: The body A has larger ve-locity. Because the slope of the x - t graph for body A is greater. (bl Instantaneous velocity: Asyou have learnt that the velocity of the uniform motion in a straight line is same at every instant. But in the case of the non-uniform motion the posi-tion - time graph comes to be a curved line as shown in Fig.2.4. As a result the slope or the average velocity var-ies, depending on the size of the time intervals selected.' In such a case the velocity of the particle at some one in-stant of time or at some one point of its path, is called its instantaneous velocity.

ii 1 ! B

o """':;'-+--r---r-r--time

..

2 3 Time (seconds)

FIg 2.5 : Displacement-time grophfar a non~unifonn motion.

Taking the limit At~ 0, the slope (lUI At) of a line tangeni: to the curve at that point gives the instantaneous velocity. However, for uniform motion the average and instantaneous velocities are the same. Example 2.6: The position -,time graphfor the motion of a point object is as shown in figure. What distances . 12 with what speeds are travelled by ob- Ii! ject in tiem (i) 0 s to 5 s, (ii) 5 s to 10 s, 'ii 8 (iii)10s to 15 s (iv) 15 s to 20s? , 4 What is the average speedfo" this total

c

journey in time 20 s? A E F OL-+-+--ii---i---!=::";~ 2.5 5

27

10 time (s)

15 17.5 20

Physic;s

Sobattoa: iJ . During 0 s to 5 s distance travened 4 ~

Distance 4 4 The speed - .... - ---,.,-=-=0.8 m Is . Time 5-0 5 .

D) During 5 to 10 s distance travelled - 12 - 4 -8 ~ 12-4 8

'The speed - _-=-=1.6 ~/s 10-5 5

iii) During 10 to 15 s~taI.1ce traveiIed .. 12 -12-6 'The speed- 0

iv) During 15 s to 20 s distance travelled -12~. Distance 12 12 .

The speed - - -. =2.4 ~ /s Time 20-15 5 Stop, and solve the following questions to check your progress.

IlV'tEXT QUBSTIOBS. 2.2,-,-_,-_-,-",--",--_~,--~ 1. Draw the pooition~. time _h r .... amotio;; with -.. lOIXieIendion .

.............................................................................................................................. . .

2. nUl rolb!win& fi&w'e ahowa the m.pi"""",ant time _h for twa _dent. A ";d B who _from th.;.- ochooilOl1clreed>e. their . home . See;t, airefull:y and_' the fo11cnrin&. ... .

~. Who reachea home firat? B .....................................................................

! ....................................... ! - ~

....................... u ! ~ ..................................... .

.. (iVI Who mowa fat? H ....................... ,; ....................................... .

(vi ~;... the apeeda of.A and B when they . ""' .. each other? ..

1.2 3 4 II 6 .1'-_1bI

............................ ~., ............ :! .................. ; ......................................... : ... ~ ............ ~ -

(vil Who.takea m;;.;.;,,~and maximUm time-to reacI1.hqti,.e?' ................................................... , ........ ' .............. ' ....................... ~ ............................. .

. ............... ~ .......................... ~-.................. ~-. ...... ~ .. ~ ... : ............ , ................... ~ ..... ~ ....... :, ... . 4. WhIch of the follcnriD&_h;'nOt ~ ... ,.? aive~ of your .....,.;".-,

- , . -. . .'

.. .' ..... ; .................. : .............. '~ ........... ~ .................. ,.~ ................... ~~~ ......... ~.~ ................. .

c

(1)) (e)

28

Motion in a Straight, Line

2.5 VELOCITY - TIME GRAPH Just like the position - time graph, we can plot a graph between the veloci-ties of the moving body and the corresponding time. Such a graph can be termed as velocity - time graph. For plotting a velocity - time graph, gener-ally the time is taken on x-axis and the velocity on y-aXis.

2.5.1 Velocity - Time Graph for Uniform Motion As you know, in the uniform motion in a straight line the velocity of the body remain constant i.e. there is no change in the velocity with the change in time. The velocity-time graph for such a uniform motion is a straight line parallel to the time axis, as shown in the figure 2.6.

.;- ! i' u-'J 0 m/s l~

10

I 2 3 4 IiIDc(II

n .. 3.6, Velo

Physics

2.05.3 Interpretation ofVeloo1ty - fimeGraph As you ha~ seen the. velocity - time graph 9f a body using v -.tBRlpn - can determ.ine the disUulce travel1edby the body and also the accd.eration of the body at different instants. Let us study qpw can we do so. Ia! Detel'lllliDatloa or the dlst_c.. tta_Ue. b,.. the body: Consider,-it velocity - time graph as shown in the .FiaIu;e 2.7, the porti!in ~.ahows the motion with Constant acCeleration, whel'eas the portion CD. shows the con-stantly retarded motion. The portion BC represents .unifom:a motiOll. The distance travelled by the body from time t, to t.. is given bythe.area under the cmve between t, and t.. . " Thus

distance - area oftrapaziumKLMN - Y.)( (KL + MN) )( KN - Yo)( (v, + v.l )( (t.. - t,l

(It) DeteI'iIIl_tioa orthe _"ratioa or the bod,.: As you haVe relld earlier in this

, lesson:, the acceleration of a body is defined as the ratio of the change in its velocity l;9 the time taken. tfyou look at the velocity -time graph giVen in the Fig 2.9. The aver-age acceleration is represented by the slope of the chord AB, which is giv~ by

, 4V average acceleratlon ((1..) -, 4t

~-- - - - - - B f ".~ "' .....

If the point B is taken ClOser and closer to the point)\, and then the average acceleration be computed over shorter and shorter intervals of :time. ,As you have seen earlier, the ~ _r.ratfDn at the point A is defined as the ,limiting value of the average acceleration when the seccmd point is

, taken closer and closer to the first. ThUll, the instantaneous a.ceeb:ation. ,limit Av dv

a - At -+ 0 At = dt Thus, u..slo'p oJ the ~t at CI point oft u..'.,.lodqf -__ .,...,. , gfNsu..' _z..atfon at that fnstant or at that point. ' 'bemple 2.7: 7JteiJelocity-timegraphsfor

, three different bodies.A, B, and C are given in the folJm.ujng.firJun., (i) Which one htJs the maximum iJccel ation

and how tnuch? (ii) Calculate the distances travelled by

theSe bodies injirst 3 s. . (iii) Which of theSe three bodies cover the

maximum di8tcince at the endof,their journey?

(iv) What are the vef:!ties at the instant of 2 s?

30

6 :!:s .!!.

f: 1

A B

c

Motion in a Straight Lin~

Solution: (i) Since the slope of the v - t graph gives the acceleration.

As the slope of the v - t graph for A body is maximum ... its acceleration is maximum.

fl.v 6-0 6 2 ... a = fl.t = 3 _ 0 = 3 = 2 m / s .

(ii) The distance travelled by a body is equal to the area of the v - t graph ... In first 3 s, " the distance travelled by A = Area OAL

- V.x6x3-9m. the distance travelled by B = Area OBL

- II. x 3 x 3 = 4.5 m. the distance travelled by C .. II. x 1 x 3 = 1.5 m.

(iii) At the end of the journey, the maximum distance is travelled by the bodyB,

(iv) At 2 s, the velocity of A the velocity of B the yelocity of C

- II. )( 6 x 6 - 18 m. 4m/s 2m/s - 0.80 mls (approx.)

2.6 EQUATIONS OF MOTION WITH CONSTANT ACCEL-ERATION

As we have studied earlier, for describing the motion ofan object, the physi-cal quantities like distance, velocity and acceleration are used. For the cases of constant acceleration; the velocitY acquired and the distance travelled in a given time can be calculated by using one or mOl"e of three equations. These equations, generally known as equations oJrrwtionJorconstant acceleration or kinematical equations, are easy to use and will frod many applications in this course. In order to derive these equations, let us take initial time t, to be zero i.e, t, = O. We can then assume t, - tbe the elapsed time. The initial position (x,) and initial velocity (v,) of an object will now be represented by Xu and va and at time tthey will be called x and v (rather than X, and v,). According to the equation 2.1 the average velocity during the time twill be

x-x V= 0 t

(2.4)

2.7.1 Jl'irst Equation of Motion The first equation of motion helps in determining the velOCity of an object after a certain time when the acceleration is given. As yo:u know from the defmitiQIl of accelerat:ion,

....'hange in velocity Acceleration (a) - T' tak

" une en

... a=v-Vo t

31

(2.5)

Physics

This equation gives

I v= va + at I This is known as the first equation of motion.

(2.6)

Example 2.8: If a car startingfrom rest has an acceleTation of 1 Q m/ 8'. Haw fast will it be going after 5 s? sOlution: Given, Initial vel~ity va .. 0 Acceleration a = 10 m/s' Time t = 5 s Using first equation of motion

v = vo+at Mter, time t = 5 s, the velocity

v = 0 + 10 x 5 v - 50m/s

2.6.2 Second Equation of Motion Second equation of motion is used to calculate the position of an object after a time t when it is undergoing constant acceleration a. F'lmn the definition of average velocity ... you know

V .. ,, x-xc

t So, x -!Ii, + v t (2.7)

Since, the velocity increases at a uniform nite . the aver:age velocit;y, v ... will be midway between the ~tial and final veloci~;

:. v .. - (v + vo)/2 (2.8) Combining the equations (2.7) and (2.8), we get,

x=xo +(v+2vo}t

Putting v Vo + at,

(vo+at+Vo) x=xo+ t 2 (2.9)

This is known as the second eqllation of mOUon.

Bample 2.9: A car A is travelling on a straight road with a Wliforrn speed. of 60 /ani h. ,Another car B foUowing it is moving with Wlifarm velocity. of 70 /ani h. When the distance betuJeen them is 2.5 Ian, the car B is given (J deacceleration of20 /anlh". At what distcznoe and time wiU the car Bcatch up with A?

SOlution: Suppose the car B catches up car A, at a distance x after t time. For car A, the distance travelled in t time X 60 )( t For car B, the distance travelled in t time is given by

I I I

I i ! }

Motic.n ill a Straight Line

x = ~ + vot + Y. at' = 0 + 70 x t + y. (-20)x t'

x = 70t- 10 t' . But. the ~ betweenbvo cars is

-- x-x=2.5 :. (70 t- 10 f') - (60 1) = 2.5 or 10 t' - 10 t+ 2.5 = 0

It gives t = .. hour :. x ~ 70 t - 10 t'

= 70 x.- 10 X (Y. . = 35 - 2.5 = 32.15 kill.

2.6.3 Third Equation of Motion. The third equation is used in a situation when the acceleration, position and initial velocity are known, and the final velocity is desired but the time tis not known. From equation (2.7). you have.

x- ~ + v . t

or x _~ + (11+2110 )t But from equation (2.5), you have

V-Vo t - --

a Substituting this into the above equation, we get

x - ( V+vo IV-vo ) x+--~ o 2 a

x - xo+ (v' -v~'

2a ) On aolving this for Ii', we obtain,

11I'=v~+2a(X-Xo) I This is known as third equation of motion. Thus, the three equations for constant acceleration are,

v = vo+at x - ~ + vot + Y2 at' and v2 = II~ +2a (x-xol

(2.10)

Bxample 3.10: A motorcyc1.ist moVes along a srraight road with a constant accelerution 4 ms-2 If initially she was at a position of 5 m and had a velocity of 3 m.s-'. find (i) the position and velocity at time t - 2 s. (ii) the position of the motorcyclist when its velocity is 5 m.s-'. Solutio. : We are given,

~ - 5 m, Vo - 3ms-', a- 4 ms-2 (i) Using equation

33

Physics

x = x., + vot + II. at' = 5 + 3 x 2 + II. x 4 X (2)'

Position, x ~ 19 m. From equation

v = vo:" at = 3+4x:;1

Velocity, v = 11 m .... (Ii) Using equation

v" = vo' + 2a (x-x.,) (5)' (3)' + 2 x 4 x (x- 5)

We get; x = 7m. Hence position of the motorcyclist (x) = 7 m. 2.6.4 Motion under gravity You must 'have noticed that when we throw a body in the upward direction or drop a stone from a cectain height, in both the cases they come down to the earth. Do you' knoY\' why they come to the earth and what type of path they follow? It is because of the gravitational force acting on them. Such type of motions under the influence of gravity only, are also one dimensional or motion in a straight line. Thefreefall ofa bodytowarcb the earth-1$-one of the most common uample of motion with (nearly) constant ac-celeration. In the absence of air resistance it is found that all bodies, irre-spective of their size or weight, fall with the same acceleration at the same point on the earth's surface. Though'the acceleratl,on due to gravity varies with altitude, but for the small distances compared toJhe earth's radius, it remains constant throughout the fall. For our practical use the effect of air resistance and the variation in acceleration with altitude is taken negligible .

. The acceleration of a freely falling body due to gravity is denoted fly d'. At or near the earth's surface its magnitude is approximately 9.8 m/s'-r-More precise values, and its variation with height and. latitude will be discussed in detail in the lesson 5 of this booklet. Example 2.11 : A stone is droppedfrom a height of 10 m and itfalls freety Calculate the following, (i) distance travelled in 2 s. (ii) velocity o/the stone when it reaches the ground. (iii) velocity at 3 s. SolutioD : Given, Height(h) - 10 m. Initial velocity (vo) = 0 Considering, initial position (Yo) to be zero and the origin 0 at the starting point. Thus, the y-axis (vertical axis) below it will be negative. Since, the acceleration is downward .in the negative. y - direction, the value of a--g- -9.8m/s .. (i) Using the equation (2.9),

Y - Yo + .vot + II. affl We get,

Ii - 0 + 0 - II. gff' - -Iia x 9.S )( (2)' --19.6m.

.34

Motion in a Straight Line

The negative sign shows that the distance is below the starting point ill downward direction. (Ii) At the gro~nd Y = -10 m, Using equation (2.10),

v' v; + 2 a(y- Yo) = 0 + 2 (-9 8) (-10 - 0)

v = 14 m./ . (iii) Using v = Vo of at

at t = 3s v - -29.4 m.l.

This shows that the velocity of the stone atf m 3 sis 29.4 mls and it is in downward direction. Take a pause and solve the following questions.

INTEXT QUESTIONS 2.3, ___________ _ 1. A body atarting &:om reat covera a diatance of 40 m in 4 a wttll eonatant IIcceleration

along a straight line. Compute itlllinal veloci1;Y .nd the time i1!quired to cover half of the total distance . ......................................................... 10 .................... ~ .............................. ' ............ ,.

2. A car moves along a straight road with constant acceleration 5 mt r. Initially at 5 m its velocity was 3 m/a. Compute ita position and velocity at 1-2 s . .......... ; ................................................................................................................. .

3. With what velocity should a body be thrown vertically upward 80 that it reaches a height of 25 m? And how long will it be in air? ..............................................................................................................................

4. A ball is thrown upward in the air. 'Is its acCeleration greater while it is heine thrown or after it is thrown? .............................................................................................................................

2.7 WHAT YOU HAVELEARNT _______ _ The ratio of the displacement of an object to the time interval is known as ~

velocity.

The total distance travelled dovided by the' time taken ia a--. speed. The rate of change of the relative polition of an objeCt with roap8Ot to the other object i. known as the relauve velocity of that object -:nth respect to the Clther. The change in the velocity in unit time io called acceleration. The position-time graph for a body at reat, i8 a atr.t line parallel to the time axio.

The position - time graph for'" ~opn motion i ... straight line inclined to the time axis. '

A body covering equal distance in equal intervals of tim. is aaid to be in unitorm motion.

The veloCIty of a particle at Borne one instant of time or at aom.e one point of ita path is called its instantaneous velocity.

The slope of .the position - time graph gives the velocity.

The velocity - time graph for a body moving: with constant acceleration io a straight line inclined to the time axis. .

35

.ote: It is important to mention here that In the above example sign of the acceleration. velocity and position will be positive if you take origin at the ground Ieve.I.

,a

Physics

The area under the velocity - time waph gives the diatance travell4d.

The acceleration of the body can be computed by the slope of velocity- time waph. For explaining the motion of a body J fonowing three equation are used,

~) v - Vo + a.t (ii) X - "b + Vo t +Y.o at' (iii) v - v~ + 2 a.(x- "iJ

2.8 TERJIIlNAL QUESTIONS. ________ _ 1. Distinguish between average speed and average velocity.

2. A car A moWigwith a speed of 65 ian/h on a straight rooul, is ahe~ef motoeach ... the reference (zero) level, upward as the

.p"..,ttive direction, draw the motion graphs. ce. the graph. between (i) distance -' time; (ii) velocity - time, (iii) displace!nent. time, (iv) speed time grapha.

11. 'A body ia thrown vertica1ly upwa..-d, with a velocity of 10 m/a. What will be the value of the velocity and acceleration 'of the bolly, at the highest point?

12. Two objectJo 'Of different in_., one of 10 g and other of 100 g are dropped from the same height. Will thll][ reach the ground at the same time? Explain your answer.

13. What happens to the unifO("m motion of a body when it i. given an acceleralion at right angle to ita motion?

14. What does the slop of velOcity-time graph at any instant represent?

2.9 ABSWERS TO THE INTEXT QUESTIONS ....... QIIptl 2.1 1. Yea

2+2 4 2: Average speed -~ -- X 20

8" + 10. 9 -11.89 km/h, average ""Iocity - 0

3. Yes, two car. movine with same veloc-ity in the same directioo.

4. After 1 hr, the relative distance be-tween A and B will be

- 4(30)2 + (40)2 - 50 km.

36

I I

Hence, re1atiye speed of A with respec..1: to B is 50 km/h. 1a_~2.2 l. See Fig 2.2 2. (i) A, (n) B covera more distance,

(iii) B, (iv) A, (v) SP"'!d 'It.{- - 1.5 km/h, speed of B. - 0.75 (m;.f!!\ (vii) A takes minimum time. .',

3. In the unifonn motion. 4. . (a) is not poaaible. because the distance

covered cannot be less or zero. _Q_2.3 1. Using x - "i. + vot + Y, at'

Q- Sm.a-:I

37

2.

3.

4.

Motion in a Straight Line

Next using ,; - vo' + 2" (x - "tJ fl lOJ2 ma-1 Using eq (2.9), x. 21 m, and using eq. (2.6), ,,- 13 ms-I. At maximum height v . 0, using eq,

(2.10), "0 -7.JW ms- I - 22.6 ms-I The body will be in air for the twice of the time it takes to reach the maxi-mum height - 4.5/ . The acceleration of the ball is greater while it is thrown.

3 LAWS OF MOTION

3.1 INTRODUCTION In the previous lesson you have studied that it is possible to desCribe the motion oCai'!. object in terms of its displacement, velocity and acceleration. But have yo\!. ever wondered what causes an object to move? What causes a ball rolling along the ground to come to a stop, apparently on its own? From our everyday experience we mow that we need to push or pull 'an almirah if we wish to change its position in a room. Sin:l:arly, a football has to be kicked in. order to Send it over a large distance. A I;:ricket ball has to be hit hard with a bat to send it across the boundary for a six. Some kind of muscular activity is involved iri these situations and the action and its effect is quite visible. There are, however, many situations where the action is not visible. .For example, what makes rain drops fali on ground? What makes-the earth go round the sun? In this lesson you will discover a close relation between force and motion in the form of Newton's laws of motion. The concept of foree developed in this lesson will be useful in different branches of phys-ics. Let us study about Newton's laws of motion which enable us to predict the behaviour of iI. particle or a sySt~ of particles under the influence of different forces.

3.2 OBJECTIVES After studying this lesSon, you should be able to, e1qJlain the meaning of inertia and relate it to force; Mate Newton's laws of motion and illustrate them with examples; dlJjfne momentum, impulse and calculate them in a given situation; e1qJlain tluf law of conservation of momentUm and illustrate it with

examples; define coefficient of .friction and distinguish between static.fridioJ1, kinetic

fri.cti.on and rolling friction; sUggest different methods of reducing .friction and highlight the role of

friction in every-thly life; and analyse a given situation and apply Newton's laws of motion using.free

body diagrams.

Laws of Motion

3.3 CORCEPT 0(1' JI'ORCE AND nmRTlA A large number of objects around us are known to remaiI). wherever they are placed. These objects cannot move on their own from one pJe.c;; to another place. These objects have to be forced to change th~.r state of rest. Similarly an object which is moving at constant velocity has to be forc~J to change its state of uniform motion. na. Wn4ent:tI of an oI#ject to nmudn In Its state of rest or In Its Stat. of unfform lfnecrr motion is CCln.d burrtfa. The state of rest or the state of motion of an object are not absolute. In the . prevwus lesson you have already studied that an object at rest with re-

, spect to one observer may appear to be in motion with respect to some other observer. Observations show that u.. cIumtIe In -lodtfI of an oI#ject can on,., 1M brov"lat ff __ net fore. acts on u.. 1HHIg. You are veIY well familiar with the term force. We use it in so m~y situa-tions in our everyday life. We are exerting force when we are pulling pushing, kicking, hitting etc. Though a force is not visib~e but its effect car. be seen and experienced. Forces are known to have two kinds of effect.'! : (a) 2'IIeJI ..... change thII .,... and thII afire of an oI#ject.. For ex-

ample, a balloon changes shape according to the forces acting on it. ill) ~ cdao frr/IuInu:e thII naotfon of an object. A force can set an

object into motion or it can bring a moving object to rest. A force can also change the direction of motion.

In lesson seven you will study about another effect of force - the rotational or turning effect of force.

3.3.1 Force and Change i.D Motion Motion of any body is characterised by its velocity. We come across many situations where the velocity of an object is gradually increasing or de-creasing. For example, in the case of a body falling freely, the velocity of the object increases continuously. Similarly, in the case of a ball rolled on some horizontal surface, the velocity of the ball is seen to decrease gradu-ally to zero. Observations show that some force is responsible for change in velociq. of a body. Dfrectfon ofIMlodtfI of a bodJf .. alWCIJIII In u.. dfrectfon of its motion. na. ___ of thII bocCIt rDfII change dqtmtltng tq.IOn 'the "" .. CIon offlw.torr- adfntI on it. Ifsome force acts on a body in the direction of its motion, the velocity of the body will increase in magnitude. If the direction of force on the body is opposite to the direction of motion the magnitude of velocity will decrease. In both these cases the object moves in a st:raistlt line. However, if some force acts on a body in a direc-tion perpendicular to its velocity, the magnitude of Velocity of the body remains constant. SUch a force is able to change only the direction of veJ.ocityofthe body. It is'important to note that_locftg of a bodJf changes _ ."., _ -fort:e .. aedng on it. Force is a vector quantity, forthis reason when several forces act on a body simultaneously, their single equivalent foree can be found by vec~or acdi-tion about which you have alreadr read in lesson 1. Every fOl' " has a

39

Physics

magnitude and a direc~n. The effect, a force acting on a system can pr(Jdu~e, depends on, (a) the mgnitude and direction of the force; (b) the poi'nt of application offorce; and (c) the duration for which it acts.

3.3.2 Newton's First Law of Motion We see that in order to move a trolley at constant velocity it has to. be continuously pushed or pulled. Similarly, a horse is seen pulling the cart moving at constant velocity. Is there any net force acting on the trolley or the cart in the situations mentioned here?

Galileo was the first to state that in the absence of any-external influence a botly .. annot only be at rest but also moves'.miformly.in a straight line .. He, therefore, regarded uniform rectilinear motion also to be natural siate of borlic

i

I

I !

I

Laws of Motion

3. Can a force change only the direction of velocity of an object keeping ita m8f!)litude constant?

4. What are the different effe.cta a force is capable of producing?

3.4 CONCEPT OF MOMENTUM Study of collision of bodies has revealed that there is a quantity mass multiplied by velocity whose net value for the colliding bodies remains unchanged in a collision. TIle prod.w:t oJ _ moJ a body and fta wlocftB 11 U called its linear momentwlt or sfm.ply momentum p. So

P = mv (3.1) In the SI system of units momentum is measured in kgm/s. Momentum is a vector quantity. The direction of momentum vector is the same as the direction of velocity vector. Momentum of an object, therefore, can change on accoUnt of change in its magnitude alone or direction alone or both. The following examples illustrate this point.

Bxamp1e 3.1: A 2 kg object is allowed to faU freely at t = 0 s. What will be its momentum at fa) t = 0 s, (b) t = 1 sand fe) t = 2 s thping its free-fall ? Solution: (a) As velocity of the object at t - 0 s is zero, the momentum t'f the object will also be zero. (hI At t = 1 s, the Velocity of the object will be 9.8 m/s [use v = Vo +atJ pointing downward. So the momentum of the object will be ~

P, = (2 kg)x (9.8 ml s) = 19.6 kg, ml s pointing downward. (c) At t = 2 s, the velocity of the object will be 19.6 ml s pointing down ward. So the momentum of the object will now be

Po = (2 kg) x (19.6 m/s) = 39.2 kg m/s pointing downWard. Thus ,e see that the momentum of a freelyfalling body increases in magni-tude only and points in the same direction. Now you think what causes the momentum of Ii freely-falling body to change in magnitude?

I. _pie 30:1 : A rubber ball ofmass 0.2 kg strikes a rigid vertical waU with .a speed of 10 m/ s and rebounds along the original path with the same speed. Find the change in momentum of the ball. Bolutloa : Here the momentum of the ball has same magnitude before and after the impact but there is a reversal in its direction. The magnitude of momentum is (0.2 kg) (10 m/s) i.e. 2 kg ml s. If we consider initial momentum vector to be along +x axis, the fmal mo-mentum vector wiD be along-x axis. So if 1', - 2 kg mis, PI = -2 kg m/s and cha"&'l in momentum .0Cthe ball- P,- PI - (-2 kgm/s) - (2 kg m/s)--4kgm/a. Here negative sign shows that the momentum of the ball changes by 4 kg m/s in the direction of -x axis. What causes this change in mome -tum of the ball?

41

Phy~ics

In actual praotice a rubber ball rebounds from a rigid wall with a speed which is less than its speed before the impact. In such a case the magni-tude of the momentum also change ..

3.4.1 Newton's Second Law of Motion It is now known that a body moving at constant velocity will have constatit momentum. It is already known from Newton's fIrst law of motion that no net external force acts on such a body. In example 3.1 we have seen that the momentum of a ball falling freely under gravity increases with time. It is well known that such a body falls under the action of gravitational force acting on it. So there appears to be a connection between change in momentum of an object, net force. acting on it and the time for which it is acting. Newton's second law Of motion gives a quantitative relation between these. According to it tIuI rat:. of change of momentum of a body is dfrectly proportional to tIuI net force acting on tIuI body. Change in momentum of the body takes place in tIuI dfrection of net extern4IJorce acting on tIuI body. This means that if IJ.p is the change in momentum of a body in time IJ.t due to some net external force F

not on it, then I1p F ~

net At

orF =k~ net -At

Here value of k, the constant of proportionality, .depends on the choice of units for .F, p and t. In the SI system of units p is measured in kg mis, tis in s and Fis mea-sured in kg ml s'. That gives k = 1. So

F = I1p net I1t

In the limit !J.t .... 0, we can write dp F =-

net dt

(3.1)

(3.2) From equation 3.1, it follows that change in momentum of a body is in the direction of net external force on it. Change in momentum of a body 4ur-ing any time is taken as fInal momentum of the body minus initial momen tum of the body. From Newton's tirst law of motion it follows that a body at rest,j:a[l not mOve on .its own. Similarly if itis already moving, it can not change its velocity on its own. You may recall that if a body has constant velocity, its acceleration must be zero. You may also recall that by change in veloci.ty we mean change in magnitude of velocity or change in its direction or both. Newton's fIrst law also tells us that some net force must act on an object to change its state of motion. So net force acting on an object is the cause of acceleration of the object. Newton's second law of motion gives a quantita- . tive relation between force and acceleratio

I t i

Then the equation 3.2 gives, dv

F =m-net dt

Laws of Motion

(3.3)

Example 3.3: A ball of mass 0.4 kg which starts rolling on the ground at 2O-m! s when pushed comes to a stop after 10 seconds. Calculate the force which stops the ball, assuming it to be constant in magnitude throughout. Solution :Given,

m=O.4kg u =20m/s v = 0 mls t = 10 s

m(v - u) 0.4 kg (Om Is - 20m Is) SoF= t = lOs

= -D.8 kg m/s' = -0.8 N Here negative sign shows that force on the ball is opposite to its direction of motion. It is evident from the fact that there is a decrease in the momen-tum of the ball. You are already aware that the ball comes to rest because' of the action of force of friction on it.

1m pulse: From equation 3.1 it follows that it is the product p.; At that determines change in momentum of a body. So a desired change in mo-mentum can be brought about by a large force acting for a small duration , or by a small force acting for a long duration of time. The quantity F M is called the impulse. Thus, the cha.nge in momentum of a. body is equal to the fmpulse. Like momentum, impulse is also a vector quantity and is in the direction of the change in momentum. Its SI unit of measurement is kg mls or Ns. In deriving equation 3.1 it was assumed that force Fremains constant over time M of its action. There are many situation where the time of action of force is very small, for example, during the impact of a ball and a bat. In such cases F is considered as the average force.

Inertial Mass: The mass of a body defmed by Newton's second law of motion is called inertial mass of the body. From Newton's second law

F m=-

a The known value offorce and the experimental determination of the accel-eration of a body as a result of the force can help us to measure inertial mass of the body. Now stop and by solving the following questions check how much you have learnt.

INTEXT QUESTIONS 3.2 __________ _ L Two objects of different m4.Sses have same momentum. Which cf them, is moving

faster?

.................................................................................................................................

43

Physics

2. A ball i. thrown up at a speed of 20 mls by a thrower. If the ball return. to the thrower with the aame speed of 20 mls only, will there be any change in (a) momentum of the ball? ................................................................................. (bl magnitude of the momentum of the ball? ...................................... , ........................ .

3. When a ball falls from a height, its momentum increases? What caueee increase in ita momentum? .............................................................................................................................. ~ ...

4. In which case will there be larger change in momentum of the object? (al A 150 N force acta for 0.1 s on a 2 kg object initiatiy at rest. (bl A 100 N force act. for 0.25 on a 4 kg. object initially at reat. .......................................................................................................................... ". .. -

5. I. it con-ect to say ~t a fast moving object poaaessea more force than a slow m~ object? Why? .

6. An ob~~d i. moving at a constant speed in l\ circular path. Does the object have c?natant momentum? Why?

................................................................................................................................ ,

Bampl_ &4 : A ronstant force of 50 N is applied to a body "f 1 0 kg moving initially with a speed of 1 0 m/ s. Haw long will it take the body 10 stop if the force acts Qn it in a direction opposite to its velocity. Solutioa : Given, Mass of the body, m

Tofind :

Fnct "0 " t

-lOkg --SON -lOm/s =0 =?

Formula(s) Foet = m(" -/,o ) or -50 N = 10 kg( O-l~m IS)

-lOOkgm/s lOOkgm/s or t- -SON SOkgm /s2' t.:I .. _ad.

3.S FORCES IN PAIRS It is the gravitational pull of the earth which allows an object to acc:e1era;.c; toward earth. Does the object also pull the earth? Similarly when wi push an almirah, does the almirah also push us? If so, why don't ?til move in the direction of that force? These situations compel us to asJt whether a single force such as a push or a pull exists? It has been obc served that actions of two bodies on each other are always mutual. Here, by 'action' we mean 'force of interaction'. So, whenever two bodies inter-act, they exert force on each other. The force of interaction could be ac-tion-at-a-distance type or a contact force type. Thus forces have been found to emerge in Pairs always. But generally we are concerned with one of the forces in a pair which is acting on the object of interest. 3.5.1 "ewtoll'. Third Law of 1I0Uoll On the basis of his study of interaction between bodies, Newton formu-lated third law oJ motion which states that to -,." cu:tfon tIwN ..

44

I f 1

Laws of Motion

equal and opposite reactl.on. The action and req:ction act on differ-entbodfes. Here by 'action' and 'reaction' we mean force. It follows from here that a single isolated force does not exist. Thus, When we exert some force on a table by pressing our finger against it, the table also exerts a force of equal magnitude on our finger in the up-ward direction as shown in Fig. 3.1. Do the forces f, and J; shown here cancel out? It is important to note that f, and J; are acting on different bodies .

. The action and reaction in a given situation appear as a pair of forces. Anyone of them cannot exist without the other.

Fig 3.1 : Force f, exerted by the finger on table and force J.,. exerted . by the table on the finger.

If one goes by the literal meaning of words, reaction always follows an action. Whereas action and reaction introduced in Newton's third law ex-ist simultaneously. For this reason it is better to state Newton's third law as when two objects fnteract, the force exerted by the first object on the second (the action) is equal fn magnitude and opposite fn direc-tion to the force by the second object on the first (reaction). Vectorially, if .F'2 is the force which object 1 experiences due to object 2 and F2I is the force which object 2 experiences due to object I, then

F'2 = -F2, (3.4) If m, and m, are the masses of objects 1 and 2 respectively then

m, a, =-m,~

3.5.2 -Internal and External Forces As you have studied that some net exter-nal force on an object causes the object to accelerate. If a car .breaks down it ~ ei-ther be towed by some other vehi~le or pushed from behind. Here the force ex-erted on the car by the tOwing vehicle or the push froin behind acts as the external force on the car. Can the car be moved by a person who is sitting in the car and push-ing it from inside? Such a push is called fntemalforce and has no role to playas far asinotion of car is concerned. To make the distinction betWeen in.ternal forces and external forces clear. let us look at some specific examples.

A a--_

A t=

Fig 3.2: Two blocks on a fridtionless horizontal sUrface.

a) Consider two blocks A and B. placed in contact with each other on a f=tionless horizontal suril'ce. When some external force f is applied to

45

F

Physics

them in the direction shown in Fig. 3.2, both the blocks move to the left with the SI!Jlle acceleration. But, what causes the block A on the left to accelerate? It is the force / which the block B on the right exerts on it. In accordance with Newton's third law, block on the left also exerts force/on the block on the. right. However, the forces of magnitude f each which these blocks exert on each other are internal to the system of blocks. b) Whenever bodies collide, the forces involved are 'internal'to the collid-ing bodies taken as a system. So the absence of any external force on the system dem~ds that total momentum'of the system should be consenred. 3.5.3 Forces in Equilibrium Nuntber oJ Jorr:es fICtlng on cui object or a .".,.". _ said to lHI In equilibrium if the 1I8ctor _ oJ all the Jorca is aero. In accordance with Newton's second law, acceleration of such an object or system will be zero. Let us take an example. Consider a block of mass m resting on some horizontal surface as shown in Fig 3.3(a). In this situation we can talk about a number offorces. Such as, .

(i) gravitational pull of the earth on the block equal in magnitude to mg. (u) force of magnitude mg which the block exerts on earlh.

r I "''''.> 77 >;;"')P >" 7>

Pia. 3.31_) PI&- a.31b) (ill) force of magnitude mg which the block exerts on the surface normally

on which it is resting. (iv) force of normal reaction N exerted by the surface on the block in accor-

dance with Newton's third law. Fig. 3.3(b) shows the forces acting on the block. Since the block is in equilibrium N & mg. In the situation described above, is there any fric-tional force acting on the block? In the next section you will study problems where more than two forces are in equilibrium. Now, it is time for you to check how mUch you have learnt.

INTEXT QUESTlO:NS 3.3 __________ _ 1, When a high jumper le_ the ground, where doe. the force which accelerate. the

jumper upward. come from ?

2, Identify action - reaction force. in each of the following situations:

(a) A man kicks a football. , ........................................ , ............................................ .

46

Laws of Motion

(b) EIorth pun. the moon. .. ..................................................................................... . (e) A ball hit. a wall ................................................................................................ .

3. What is the magnitude and direction of the IP"vitationai force which a 60 kg woman exert. on earth? .................................................................................................................................

4. Name the forces which are in equilibrium in each of the following aituationsi' (a) A book reating on a table .................................................................................. .. (b) A cork Boating in water ..................................................................................... . (e) A pendulum bob suspended from the ceiling with the help of a string . .................................................................................................................................

5. Three blocks of JDa8II m each are pieced on top of a table as ebown. Name & force each which i8 internal and external to the system of blocks.

1 2 3

.................................................................................................................................

6. A woman __ a large force on an almirah to pueb it forward. The """""" i. not puebed backward be

Physics

3.6.1 Conservation of Momentum asa Consequence of Newton's Laws According to Newton's second law of motion, equation 3.1, the change in momentum !J.p of a body wheh a force F acts on it for time Mis