CHAPTER 1 PHYSICAL QUANTITIES AND VECTORS

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CHAPTER 1 PHYSICAL QUANTITIES AND VECTORS

1.1 THE NATURE OF PHYSICS

The word `physics' comes from the Greek word which means nature. Physics was conceived as a study of the natural phenomena around us.

Each theory in physics involves: a. A few concept or physical quantities b. Assumptions in order to obtain a mathematical model c. Procedures to relate mathematical models to actual measurement from experiments d. Relationships between various physical concepts e. Experimental proofs to devise explanations to natural phenomena

1.2 BASIC QUANTITIES AND SI UNITS

Physics is based on quantities known as physical quantities. Example: length, mass and time. A physical quantity is clearly defined with a numerical value and a unit. In this text, we emphasize the system of units known as SI units, which stands for the French phrase"Le Systeme International d'Unites".

1.2.1 Base quantities and SI units

In the International System of Units (SI), six physical quantities are selected as base quantities. Units for these base quantities are known as base units.

Basic quantity Length Mass Time Electric current Thermodynamic temperature

Quantity of matter Luminous intensity

Base unit Meter Kilogram Second Ampere Kelvin

Mole Candela

Symbol m kg s A K

mol cd

Table 1: Base quantities and their SI base units

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SHE1215 1.2.2 Derived Quantities and Derived Units Physical quantities other than the base quantities are known as derived quantities.

Derived quantity Force Pressure Energy Power Charge Voltage Resistance Capacitance Inductance Frequency

Unit and Symbol Newton, N Pascal, Pa Joule, J Watt, W Coulomb, C Volt, V Ohm, Farad, F Henry, H Hertz, Hz

In Terms of Base Units N = kgms-2 Pa = Nm-2 = kgm-1s-2 J = Nm = kgm2s-2 W = Js-1 = kgm2s-2

C = As V = JC-1 = kgm2s-3A-1 = VA-1 s = kgm2s-3A-2 F = CV-1 = kg-1m-2s4A2 H = VA-1s = kgm2s-2A-2 Hz = s-1

Table 2: Derived quantities and their units

1.2.3 Other Systems of Units / The Conversion of Units

Sometimes, it is necessary to convert one system of units to another. This is done using the conversion factors. Some conversion factors between the SI system of units and the C.G.S (cm gram second) system are shown in table 3. In any conversion, if the units do not combine algebraically to give the desired result, the conversion has not been carried out properly.

SI to C.G.S

1m = 100 cm

1kg = 1000g 1m2 = 104 cm2 1 m3 = 106 cm3

C.G.S to SI 1 cm = 10-2 m 1 g = 10-3 kg 1 cm2 = 10-4 m2 1cm3 = 10-6 m3

Table 3: Conversion factors

EXAMPLE 1.1 An acre is defined such that 640 acres=1 mi2. How many square meters are in 1 acre?

SOLUTION:

1 mile=1.609 km=1609 m

1 acre= 1 mi2= 1 (1609m) 2=4.05 x 103 m2

640

640

EXAMPLE 1.2: 2

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A geologist finds that a rock sample has a volume of 2.40 in3. Express this volume in cubic centimeters and in cubic meters.

Solution: 1 in = 2.54 cm, so V = 2.40 in3 = 2.40(2.54cm) 3 1 cm = 10-2m, so V = 39.3 cm3 = (39.3) (10-2m) 3=3.93 x 10-5 m3

EXAMPLE 1.3 The mass of the parasitic wasp Caraphractus cintus can be as small as 5x10-6 kg. What is this mass in (a) grams (g), (b) milligrams (mg), and (c) micrograms (g)

REASONING When converting between units, we write down the units explicitly in the calculations and treat them like any algebraic quantity. We construct the appropriate conversion factor (equal to unity) so that the final result has the desired units.

SOLUTION a. Since 1.0 103 grams = 1.0 kilogram, it follows that the appropriate conversion factor is

(1.0 103 g)/(1.0 kg)= 1. Therefore, 5x10-6 kg = 1.0x103 g x 5x10-6 kg=5x10-3g

1.0kg b. Since 1.0 103 milligrams = 1.0 gram,

5x10-3g= 5x10-3g x 1.0x103 mg =5 mg 1.0g

c. Since 1.0 106 micrograms = 1.0 gram, 5x10-3g=5x10-3g x 1.0x106 g =5x103g 1.0g

EXAMPLE 1.4 An engineering student wants to buy 18 gal of gas, but the gas station has installed new pumps that are measured in liters. How many liters of gas (rounded off to a whole number) should he ask for? SOLUTION 18 gal = (18 gal) 3.785 L = 68 L .

1 gal

EXAMPLE 1.5 An automobile speedometer is shown. (a) What would be the equivalent scale readings (for each empty box) in kilometers per

hour?

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SHE1215 (b) What would be the 70-mi/h speed limit in kilometers per hour?

50

40

60

30

70

20

80

10

90

km/h

0 mi/h

100

km/h 0

Speedometer readings

SOLUTION: (a) 10 mi/h = (10 mi/h) 1.609 km = 16 km/h for each 10 mi/h .

1 mi (b) 70 mi/h = (70 mi/h) 1.609 km = 113 km/h .

1 mi

1.3 DIMENSIONS OF PHYSICAL QUANTITIES / DIMENSIONAL ANALYSIS

In Physics, the term dimension is used to refer to the physical nature of a quantity and the type of unit used to specify it. The dimension of a physical quantity relates the physical quantity to the base quantities such as:

Mass (M) Length (L) Time (T) Electric current (I) Temperature () Quantity of matter (N)

EXAMPLE 1.6:

a)

velocity

displacement time

(L) (T )

LT

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b) [force] = [mass] x [acceleration] =M x LT-2 =MLT-2

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USE OF DIMENSIONS 1. To check the homogeneity of physical equations

Homogeneous-the dimension on both sides of an equation must also be the same

EXAMPLE 1.7: Show that this equation s = ut + 1 at2 is dimensionally homogeneous. 2

Left side: [s] = L Right side: [ut] = L .T = L

T [at2]= L 2.T2 = L

T Since all the terms in the equation have the same dimension, the equation is dimensionally homogeneous.

2. To derive a physical equation

Using dimensions, an equation can be derived to relate a physical quantity to the variables that the quantity is dependent on.

EXAMPLE 1.8: The period T of a simple pendulum depends on its length l and the acceleration due to gravity g.

T lxgy T=k lxgy Since the dimensions on both sides of the equation must be the same;

[T] = [k lxgy] T = Lx(LT-2)y

Equating the indices of T;

-2y = 1

y

=

-

1 2

Equating the indices of L; Hence; T=kl1/2g-1/2

x+y=0 x- 12 =0

x=

1 2

So, the value of the constant k can be determined experimentally

EXAMPLE 1.9

The following are dimensions of various physical parameters. Here [L], [T] and [M] denote, respectively, dimensions of length, time and mass.

Parameter Distance (x) Time (t) Mass (m) Speed (v) Acceleration (a)

Dimension [L] [T] [M] [L]/[T] [L]/[T]2

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Force (F) Energy (E)

M[L]/[T]2 M[L]2/[T]2

Which of the following equations are dimensionally correct? a. F = ma b. x = (1/2)at3 c. E = (1/2)mv d. E = max e. v = (Fx/m)1/2

REASONING AND SOLUTION

a. F = [M][L]/[T]2; ma = [M][L]/[T]2 = [M][L]/[T]2

so F = ma is dimensionally correct.

b. x = [L]; at3 = ([L]/[T]2)[T]3 = [L][T] so x = (1/2)at3 is not dimensionally correct .

c. E = [M][L]2/[T]2; mv = [M][L]/[T]

so E = (1/2)mv is not dimensionally correct .

d. E = [M][L]2/[T]2; max = [M]([L]/[T]2)[L] = [M][L]2/[T]2

so E = max is dimensionally correct.

e. v = [L]/[T]; (Fx/m)1/2 = {([M][L]/[T]2)([L]/[M])}1/2 = {[L]2/[T]2}1/2 = [L]/[T] so v = (Fx/m)1/2 is dimensionally correct.

EXAMPLE 1.10 n The speed, v of an object is given by the equation v = Ar3 Bt, where t refers to time. What are the dimensions of A and B? Solutions:

For the equation v At3 Bt , the units of At3 must be the same as the units of v . So the units of A must be the same as the units of v t3 , which would be distance time4 . Also, the units of Bt must

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be the same as the units of v . So the units of B must be the same as the units of v t , which would be distance time2 .

1.4 MEASUREMENT AND ERRORS

1.4.1 Determination of Uncertainty

In Physics, when we take measurements, there are always uncertainties in the values.

Example: determine the diameter of a pencil: By using a ruler (mm scale) : 7.5mm By using a micrometer screw gauge, the reading is 7.55mm.

The difference between the two values is in their uncertainties. The measurement using the micrometer screw gauge is better because it has a smaller uncertainty. The less uncertainty, the reading will be more accurate.

How to write the measurement??? (7.55 0.01) mm.

The accuracy of the measurement depends on: 1) The types and quality of the instruments 2) The skill of the person taking the reading 3) The number of trials made in the measurement

Instrument Meter rule micrometer screw gauge Stopwatch Thermometer Protractor

Smallest scale division

1 mm or 0.1 cm

0.01mm

0.1s 10C 10

Example

(10.00 0.05)cm

(6.28 0.01)mm

(10.5 0.1)s (39.0 0.5)0C (30.0 0.5)0

1.4.2 Types of Errors The error or uncertainty in a measurement is due to factors such as:

the way the measurement is made the instrument used the physical limitations of the observer

There are two common types of errors: 1. systematic errors 2. random errors

1. Systematic errors These are errors made during an experiment. Sources of the systematic errors are: (1) instruments

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(a) zero error of a micrometer screw gauge (e.g. zero setting) (b) fault in the instrument (e.g.: wrong calibration) (2) personal error of the observer/physical limitations of the observer (e.g. reaction time) (3) errors due to the physical conditions of the surroundings Systematic errors cannot be reduced or eliminated by taking several readings using the same method, same instrument and done by same observer However, it can be reduced by taking measurement carefully and using different instruments.

The experimental result is can then be accepted as correct within the error limits.

2. Random errors These are errors which cannot be determined and therefore cannot be controlled during an experiment or measurement. Mistakes made by the observer when taking measurements. The reading obtained may be larger or smaller than the actual value. Examples: a. error made when reading the scale of an instrument ( eye position) b. a wrong count of the number of oscillations in vibrating

Sources of the random errors are: 1) instruments 2) personal error of the observer/physical limitations of the observer 3) errors due to the physical conditions of the surroundings

Random errors can be reduced / minimized by taking several readings (multiple trials) and calculating the mean.

EXAMPLE 1.10 Table below show the diameter of a long wire using a micrometer screw gauge. Six readings have been obtained from several locations of the wires. Suppose the readings obtained are as follows:

Reading for diameter 1.25 1.24 1.23 1.22 1.24 1.25 of wire, d(mm)

The average value of the readings is determined by taking the sum and divided by the number of trials.

The average value of the readings is 1.25 1.24 1.23 1.22 1.24 1.25 1.24mm 6

The error is then calculated as the sum of the deviations of each reading from the average value divided by the number of trials. Error= {1.25 1.24} {1.24 1.24} ........ {1.25 1.24} 0.01mm

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