First mechanics final

Mechanics

2 | Essential Physics

MECHANICS:
Chapter 1. Physical Quantities
Chapter 2. Vectors
Chapter 3. Kinematics
Chapter 4. Laws of motion
Chapter 5. Work and Energy
Chapter 6. Circular Motion
Chapter 7. Gravity and Gravitation
Chapter 8. Equilibrium
Chapter 9. Rotation Dynamics
Chapter 10. Elasticity
Chapter 11. Periodic Motion
Chapter 12. Fluid Statics
Chapter 13. Surface Tension
Chapter 14. Fluid Dynamics

Evaluation Scheme

Teaching Long Short Numerical Mark Distribution
Hour Questions Questions Question
Unit Long Short Numerical Total
Mechanics 70 4 7 4 36
attempt attempt attempt Questions Questions Questions
any three any six three
3×4=12 6×2=12 3×4=12

Ch apter Physical
Quantities
1

Teaching Manual Physics Grade –XI, Higher Secondary Education Board
Curriculum and Training Division Sanothimi, Bhaktapur

Syllabus:

Physical Quantities: Need for measurement; System of units; S.I. unit; Precision and
significant figures; Dimensions; Main use of dimensional equations.

Objectives:

The objectives of this unit are to make the students aware of the importance of the
measurements and let them how various physical quantities are measured.

Activities (micro syllabus):

1. Explain the importance of measurement in physics
2. Define physical quantities
3. Explain SI systems of units
4. Define dimensions of a physical quantity
5. Explain main uses of dimensional equations
6. Explain the meaning of precision and significant number
7. Explain how to estimate order of magnitudes
8. Numerical problems: Focused numerical problems given:

(a) As exercise in the University Physics (Ref.1) and
(b) In advanced level Physics (Ref.2) and also in Physics for XI (Ref.3)

4 | Essential Physics

1.1 Introduction

Physics is a branch of science which deals with nature and natural phenomena. Many natural phenomena
are described quantitatively by using the concept of physics are called physical laws. All the laws of
Physics are expressed in terms of every measurable quantity is called physical quantity. For examples,
mass, length, density, area, time, etc. All the physical quantities are measurable and have magnitude and
unit. i.e., physical quantity = magnitude × unit or numerical value × unit.
The quantity used as a standard of measurement is called the unit. The physical quantities can be divided
into two categories, they are:

1. Fundamental Quantities and 2. Derived quantities

Fundamental Quantities

The physical quantities that do not depend on other quantities are called fundamental quantities. There are
seven fundamental quantities viz, mass, length, time, temperature, luminous intensity, electric current.

Derived Quantities

The physical quantities that cannot be derived from fundamental quantities are called derived quantities.
For examples, area, volume, density, speed, electric intensity etc.

Note: all quantities which can be measured directly or indirectly and can be expressed by a number
are called physical quantities.

1.2 Need for Measurement

Physics with nature and natural phenomena. Now a day, physics has become essential to describe many
natural phenomena quantitatively. For example, if a stone is dropped from the tower, the stone fall down;
there arise several questions why stone fall? What is the speed with which stone fall? Is the speed same
at all points? What is the time taken by the stone to fall on earth? The answer of first questions is known
that stone falls due to earth's gravity but the answers of other questions need measurement of distance
fallen at various times. Thus, without measurement, physics is incomplete. Lord Kelvin to say that,
"physics is a science of measurement”.
The method of comparing any amount of physical quantities with standard amount of same kind is called
measurement. Measurement of a physical quantity consists of a numerical value (n) and a standard
amount or unit (u). Thus, measurement of quantity = numerical value × unit i.e., Q = nu.
The product of numerical value (n) and unit (u) always remains constant for a measurement

i.e., n1u1 = n2u2 = constant. In general, nu = constant.

1.3 Units

Different systems of units are used in measurement. There are mainly three kinds of units. They are given
as below:
1. Fundamental units: The units of fundamental quantities that can’t be derived from other units are

called fundamental units. For examples, Kilogram, Meter, Second etc.
2. Supplementary unit: The unit of plane angle and solid angle are called supplementary units.
3. Derived units: The units that can be derived from fundamental units and supplementary units are

called derived units. For examples, Joule, Square meter (m2), Meter per second (m/sec), Newton
(N) etc.

1.4 System of Units

For the measurement of quantities, there are four systems of units. They are given as below.

Physical Quantities | 5

1. CGS system: In this system, units for length, mass and time are centimeter (cm), gram (g) and

second (s).

2. FPS system: In this system, units of length, mass and time are foot (ft), Pound ( P or lb ) and

second (s).

3. MKS system: In this system, units for length, mass and time are metre (m), kilogram (kg) and

Second (s).

4. SI system: In this system, seven fundamental quantities (units) and two supplementary quantities

(units) are taken. They are given in table below:

Table 1

Units of fundamental quantities (units)

Fundamental quantity Unit Symbol

Length Meter M

Mass Kilogram Kg

Time Second s

Temperature Kelvin K

Current Ampere A

Luminous intensity Candela Cd

Amount of substance Mole Mol

Table 2

Units of supplementary quantities (units)

Quantity Unit Symbol

Plane angle Radian rad

Solid angle Steradian Sr.

1.5 Some of the Non S.I. Units in Common Uses

Some of the non SI units in common uses are given as

1. For length/distance:
i. Astronomical unit: 1 Au = 1.496 × 1011m

ii. Light year unit: 1 light year = 9.46 × 1015m

iii. Parallectic second: 1 parsec = 3.084 × 1016m = 3.26 light years

iv. 1 µ m = 10-6m v. 1 nm = 10-9m
vi. 1 A° = 10–10m vii. 1 x-ray unit = 10–13 m
viii. 1 Fermi = 10–15m
ix. 1 yard = 0.9144m

x. 1 foot = 0.3048 m xi. 1 inch = 0.234

xii. 1 mile = 1609.344m xiii. 1 nautical unit = 1852m

2. For mass:

i. Pound: 1lb = 0.4536kg ii. Slug: slug = 14.59kg

iii. Quintal: 1q = 100kg iv. Metric tons: 1t = 1000kg
v. 1 amu = 1.66 × 10–27 kg

3. For time

i. 1 minute : 1 min = 60sec

ii. Hour: 1 hr = 60 × 60sec

iii. 1 day : 1 day = 24h = 24 × 60 × 60 sec
iv. year: 1 year = 365.25 days = 3.156 × 107 sec
v. shake: 1 shake = 10–8sec

4. For other quantities:
i. Pascal (for pressure): 1 pa = 1 Nm–2

ii. Pressure exerted by earth’s atmosphere : 1 atm = 1.01 × 105 Pa

6 | Essential Physics

iii. Barn (for area) : 1 barn = 10–28m2
iv. Litre (for volume) : 1l = 103 cc = 10–3 m3
v. Bar (pressure) : 1 bar = 1 atm = 1.01 × 105 pa

vi. Torr (for pressure) : 1 torr = 1 mm of Hg = 133.3 pa
vii. Erg: 1 erg = 10–7J
ix. Electron volt: 1 e V = 1.6 × 10–19 J

x. Horse power: 1 hp = 746 watt
xi. Dioptre (D) : 1 D = 1m–1 (power of lens)

xii. Degree: 1° = π radian
180

1.6 Prefixes for Power of Ten

The measurements of physical quantity from micro or smaller (atomic) level to macro or layer (universal)
level in physics. These quantities are expressed in power of 10 as given in table below:

Power of ten Prefixes Symbol
1024 Yotta y
1021 Zetta z
1018 Exa E
1015 Peta p
1012 Tera T
109 Giga G
106 Mega M
103 Kilo K
102 Hecto h
101 Deca da
10–1 Deci d
10–2 Centi c
10–3 Milli m
10–6 Micro µ
n
10–9 Nano p
10–12 Pico f
10–15 Femto q
10–18 Atto z
10–21 Zepto y
10–24 Yocto

1.7 Advantages of SI Units

SI units have following advantages:

i. It is a decimal system

ii. It is a rational system of units

iii. It is a coherent system of units.

iv. It is highly useful in practical and theoretical work etc.

Physical Quantities | 7

1.8 Precision and Significant Figures

Precision is the degree of exactness and gives the limitation of the measuring instrument and significant
figures is a measured quantity indicates the number of digits in which we have confidence.

We know that, measurement of any physical quantity is never absolutely correct. The accuracy depends
upon the least count of the measuring instrument. For example, if the length of steel cylinder measure by
a meter scale is 5.2cm, it may be 5.17cm when measured by vernier calipers and further 5.176cm when
measured by a screw gauge. The results written have 2, 3 and 4 figures respectively. It is clearly shows
that measuring instruments have least count 0.1m, 0.01m and 0.001cm respectively. The figure 2, 3 and 4
in above examples are called significant figure. In these figures, the right most figures (digit) is
reasonably correct while others are absolutely correct. Thus, it is clear that, the measurement of length.
5.176 is more precious than 5.17cm and 5.17cm is more precious than 5.2cm.

1.9 Dimension of a Physical Quantity

The dimensions of a derived unit may be defined as the powers to which the fundamental units of mass,

length and time must be raised to represent it.

For example: Velocity = Displacement = [L] = [L] [T–1] . Therefore, Dimension of velocity = [M° LT–1].
Time [T]

Thus, we say that, dimension of velocity are zero in mass 1 in length and –1 in time.

1.10 Dimensional Formula

The dimensional formula is defined as the expression of the physical quantity in terms of its basic unit
with proper dimensions. For example, dimension of force is F = [MLT–2]

1.11 Dimensional Equation

An equation containing physical quantities with dimensional formula is known as dimensional equation.

For example,

1 at2 is
Dimensional equation of S = ut + 2

Dimensional of (S) = [M° LT°]
Dimension of (u) = [M° LT–1]

Dimension of (t) = [M° L° T]
Dimension of (a) = [M° LT–2]
Dimension of (t2) = [M°L°T2]

Therefore, dimension of s = ut + 1 at2 is
2

[M°LT°] = [M°LT–1] [M°L°T] + 1 [M° LT–2 ] [M° L° T2]
2

1
[M°LT°] = [M°LT°] + 2 [M° LT°]

1
[M°LT°] = [M°LT°], where 2 = nill dimension.

8 | Essential Physics

1.12 Dimensional Formulas of Some Physical Quantities

Dimensional formulas of some physical quantities are presented in table below:

Sn. Physical quantity Dimensions SI units
[ML–3] kgm–3
1. Density (ρ = m ) [MLT–2] kgms–2
v [MLT–1] kgms–1
[MLT–1] kgms–1 (Ns)
2. Force (F = ma) [ML2T–2] kgm2s–2
[ML2T–3] kgm2s–3
3. Linear momentum (p = mv) kgm–1s–2(N/m2)
[ML–1T–2]
4. Impulse (=F.t) No unit
[M°L°T°] = dimensional less
5. Work r energy (w = F × d) [ML–1T2] Nm–2

6. Power (p = w/t) [M°L°T°] – dimensionless rad

7. Pressure or stress ( = Force/area) [T–1] rad sec–1
[T–1] Hz
8. Strain ( = ∆l ) [T–2]
l [ML2] rad sec–2
[ML–2] kgm2
9. Elastic modulus or young's modulus or [ML–2] Nm–1
Bulk modulus ( = stress/strain) [ML2T–2] Nm–1
[ML2T–1] Nm
10. Angular displacement (θ = ^l ) [M–1L3T–2] JS
r
[T–1] kg–1m3s–2 or Nm2
11. Angular velocity (w = θ/t) kg–2
[ML–1T–1]
12. Frequency f or n S–1
[M°L2T–2]
13. Angular acceleration α = ω kgm–1s–1 or Ns/m2
t [ML2T–1]
[LT–2] J/kg
14. Moment of inertia (I = mr2)
[ML–1T–2] JS
15. Surface tension (T = F/l)
[AT] N/kg
16. Force constant (k = F/x) [ML2T–3A–1] Nm–2

17. Torque ( = Fr) [ALT] Ampere sec /Coul.
[L–2A] JC–1
18. Angular momentum (J = wI)
Coul meter
19. Gravitational constant G = mF1rm2 2 Am–2

20. Velocity gradient ∆v
∆x

Coefficient of viscosity =

21. Areata×ngveenloticailtyfogrrcaedient

22. Gravitational potential = wmoarsks

23. energy
Planck's constant (h) = frequency

24. Gravitational intensity = Force/mass

stress
25. Coefficient of elasticity = strain

26. Charge = current × time

27. Potential = w/Q

28. Electric dipole ( = q × 2l)

29. Current density = cuarrreeant

Physical Quantities | 9

30. Permittivity (ε0) [ F = q1q2 [M–1L–3T4A2] Farad/m
4πε0r2 [M–1L–2T4A2] Farad
[ML2T–3A–2] ohm
31. Capacitance c = q/v
[M–1L–2T3A2] ohm–1
w/q w
32. Resistance (R = V/I) = I = qI [MLT–3A–1] Newton/coulomb

33. 1 [ML3T–3A–1] Nm2/C
Conductance = Resistance [ML2T–3A–1] Volt
[MT–2 A –1] Tesla
34. F [ML2 T–2A–1] Weber
Intensity of electric field E = q [MLQ–2] or [MLT–2A–2] Henery m–1
Am2
35. F cosθ [AL2] Joule
Electric flux Q = q [ML2T–2] Jkg–1k–1
[M°L2 T–2k–1] Jkg–1
36. emf = potential difference [T2T–2] J/mole kelvin
J/kelvin
37. Magnetic field induction B = IFl [ML2T–2K–1 mol–1]
k/watt
38. Magnetic flux (φ = B × A) [ML2 . T–2T3 K–1]
J/K
39. Permeability µ B = µ0 Idlsinθ [M–1 L–2 T3 k]
4π r2 
[ML2T–2K–1
40. Magnetic moment (M = NIA)

41. Heat energy ( = energy = work)

42. Specific heat capacity (s)
i.e. [Q = ms ∆θ]

43. Latent heat (L) (i.e., Q = mL)

44. Universal gas constant R

45. R
Boltzmann constant K = N

46. Temp × time
Thermal resistance = heat

47. Q
Entropy = θ

1.13 Principle of Homogeneity

Homogeneity principle means that "the dimensions of fundamental quantities on left hand side of an
equation must be equal to the dimensions of the fundamental quantities on the right hand side of that
equation". Let us consider, A, B, and C are any three physical quantities. Then the relation C = A + B.
According to the principle of Homogeneity, the dimensions of C are equal to the dimensions of A and B.
i.e., dimension of C = dimension of (A + B)

1.14 Uses of Dimensional Equations

The uses of dimensional equations are given as below:

1. Conversion of one System of Units into Another

The method of dimensional analysis can be used to obtain the value of a physical quantity in one
system, when its value in another system is given. Let M1, L1, T1 are fundamental units on one
system and M2, L2, T2 are fundamental units on the other system. We take a, b and c are the
dimensions of the quantity in mass, length and time respectively. If n1 is numerical value of the
quantity in one system and n2 is its numerical value in the other system. Therefore

n1[M1a L1b T1c] = n2 [M2a L2b T2c]

10 | Essential Physics

∴ n2 = n1 MM21a LL21b TT21c

This relation can be used to find the value of a physical quantity in the new system, when its value

in the first system is known. For example;

Convert 1 Newton in dyne: We know that, Newton and Dyne are the units of force in MKS and

CGS systems respectively.
So, force = [M1L1T–2], Now, Newton = ( kg)1 (m)1 (s)–2 and Dyne = (gm)1 (cm)1 (s)–2

Newton = gkmg 1 cmm1 ss–2 = 103.102 = 105
Therefore, Dyne

Thus, 1 Newton = 105 Dyne

2. To Check the Correctness of a Physical Relation (checking the
results)

The correctness of a physical relation depends upon the principle of homogeneity of dimensions.

According to this principle, the dimensions of all the terms on the two sides of an equation must be

same.

For example; we have a relation s = ut + 1 at2
2

Where, Dimension of s = [M°LT°]
Dimension of u = [M°LT–1]

Dimension of t = [M°L°T]

1
Dimension of 2 = Nil (a number)

Dimension of a = [M°LT–2]
Dimension of t2 = [M°L°T2]

Therefore, dimension of s = ut + 1 at2 is
2

or, [M° LT°] = [M°LT–1] [M°L°T] + 1 [M°LT–2] [M°L°T2]
2

or, [M°LT°] = [M° LT°]

i.e., Dimension of LHS = dimension of RHS.

Therefore, dimension of each term on right hand side is the same as that of that of the term on left

hand side. Hence given relation is correct.

3. To derive the Relation Between Various Physical Quantities (Deriving
a result]

The relations between various physical quantities are derived by using the principle of

homogeneity of dimensions. For example; If a time period of a simple pendulum depends on mass

of bob, acceleration due to gravity, length and angle of swing of simple pendulum (θ). Find the

correct relation among them if the proportionality constant is 2π.

The solution for this. Let, t = KMalb gc θd. Where K is constant of proportionality. Taking

dimension on both sides, we have
[T] = [M]a [L]b [LT–2]c
T = Ma Lb + c T–2c

Equating powers of M, L and T on both sides,

a = 0, b + c = 0, and –2c = 1

∴ 11
c = – 2 , b = 2 and a = 0

Thus, t = Kl1/2 g1/2

Physical Quantities | 11

l
t=K g

∴ t = 2π l
g

4. To find the dimension of constants in a given relation

Dimension of a constant appearing in a physical equation can be determined. For example,

Newton's law of gravitational gives the force between two masses M1 and M2 separated by a

distance 'r' as F = GM1M2 where G is a universal gravitational constant. Here we determine the
r2

dimension of G by following way. We have, F = [MLT–1], M1 = M2 = [M], r = [L]
Fr2 [MLT–2] [L2]
∴ G = M1M2 = [M] [M] = [M–1] [L3] [T–2] . Thus, dimension of G is [M–1L3T–2]

1.15 Different Types of Variables and Constant

Physical quantities are classified into four classes. They are

1. Dimensional variables: The physical quantities which have dimensions but do not have fixed
value are called the dimensional variables. For examples, Area, volume, force velocity etc.

2. Non-dimensional variables: The physical quantities which have neither dimensions nor fixed
value are called non-dimensional variables. For example; specific gravity, strain etc.

3. Dimensional constant: The physical quantities which have dimensions and fixed value are called
dimensional constant. For example, gravitational constant, Stefan's constant etc.

4. Dimensionless constant: The physical quantities, which do not possess dimension but have fixed
value are called dimensionless constant. For example, pure numbers (counting numbers) 1, 2, 3,... ,
All trigonometrically functions, π etc.

Note: dimensionless quantities: relative density, angle and solid angle, strain poison's ratio, refractive index
mechanical equivalent of heat, emissivity, magnetic susceptibility, relative permeability, coefficient of friction,
loudness etc.

1.16 Limitations of Dimensional Analysis

The limitations of dimensional analysis are given below

1. It gives no information whether a physical quantity is a scalar or vector

2. We cannot derive the formula containing trigonometric function, exponential functions,
logarithmic function etc.

3. It does not give information about the dimensionalless constant.

4. The exact form of relation cannot be developed when there are more than one part in any relation.

5. If a quality depends on more than three factors having dimension, the formula cannot be derived
etc.

Boost for Objectives

• The physical quantities that do not depend on other quantities are called fundamental quantities.
• There are seven fundamental quantities viz. mass, length, time, temperature, luminous, intensity,

electric current and amount of substance which define other quantities and are independent.
• Measurement of a physical quantity consists of:(i) a numerical value (n) and (ii) a standard amount or

unit (u)
• CGS system: In this system, units for length, mass and time are centimeter (cm), gram (g) and second (s).

12 | Essential Physics

• FPS system: Units for length, mass and time are fool (ft), pound (lb), and second (s).
• MKS system: Units for length, mass and time are meter (m), kilogram (kg) and second (s).
• SI system: In this system, there are seven fundamental quantities and fundamental units.
• A physical quantity may be dimensionless but still may have units. For example; plane angle is

dimensionless but has radian as its unit.

• A physical quantity that does not having any unit must be dimensionless.
• Angular velocity, frequency, velocity gradient, radioactive decay constant = [M0L0T–1].
• Intensity of radiation, solar constant have identical dimensions [ML0T–3]
• Gravitational potential, specific latent heat = [M0L2T–2]
• Electric potential, EMF, electric potential difference = [ML2T–3A–1]
• Angular momentum and Planck's constant = [ML2T–1]
• Magnetizing force and intensity of magnetization have identical dimensions [M0L–1T0A].
• Rydberg's constant and wave number have identical dimension [M0L–1T0]

• If C, R and L are capacitance, resistance and inductance, then [CR], [L/R], [ LC ] have dimensions of
time [T].

• In the Vander Waal's equation P + Va2 (V – b) = RT

• A dimensionless quantity may have units (e.g. Angle - radian)
• A unit less quantity always has zero dimensions.
• Size of nucleus is the order of 10–14 m and size of an atom is in the order f 10–10m.
• Distance of sun from the earth is in the order of 1011 m and size of galaxy is in the order of 1021m.
• Surface tension : [M1L0T–2]
• Angular momentum [ML2T–1]
• Gravitational constant (G) : [M-1L3T-2]
• Planck's constant (h) : [ML2T–1]
• Thermal conductivity: [MLT–3θ–1]
• Universal gas constant (R) = [ML2T–2θ–1]
• Permittivity (ε0) = [M–1L–3T4A2]
• Permeability (µ0): [MLT–2A–2]

Short Questions with Answers

1. Check the correctness of the formula t = 2π l/g using dimensional analysis, t is the time period of
Ans: simple pendulum, l is the length of simple pendulum and g is the acceleration due to gravity.

[HSEB 2069]

We have,t = 2π l/g
Dimension of l = [L]
Dimension of g = [LT-2]
Then, in dimension form

L
[T] = constant LT-2

[T] = [T]. Therefore, dimension of left land side is equal to dimension of right hand side. So, the given
formula is dimensionally correct.

2. Check the correctness of the formula V = 2GM/R where V is the escape velocity of a body, G is
universal gravitational constant, M and R are mass and radius of the earth. [HSEB 2062, 2055]

Ans: Here, V = 2GM

R

Dimensional equation of velocity, V = [LT–1]

Dimension equation of G = [M–1L3T–2]

Physical Quantities | 13

Dimension of R = [L]

Dimension of M = [M]
Dimension of left hand side = [M0LT–1]

Dimension of right hand side = M–1LL3T–2M1/2= [M0LT–1]

Since, dimensions of left hand side are equal to the dimensions of right hand side. Thus, the given
relation is dimensionally correct.

3. The force F is given in terms of time’t’ and displacement 'x' by the equation F=Asin Bx + Csin Dt.
Ans:
What is the dimension of D/B? [HSEB 2069]
4.
Ans: Given formula is F = A sin Bx + C sin Dt

5. Since, Bx and Dt are dimensionless and x has dimension of length and t has dimension of second. So,
Ans:
6. dimension of B is [L-1] and dimension of D is [T-1].
Ans: T-1
7. Therefore, dimension of D = L-1 = [LT-1]
Ans: B

8. Taking force, length and time to be fundamental quantities, find the dimensional formula for the
Ans:
density. [HSEB 2057]

Force, length and time are taken as fundamental quantities. Let F, L and T be the dimensional formula
of force, length and time respectively.

Mass Force
Since, density = Volume = Acceleration × Volume

= Force = Force

Velocity Displacement × (length)3
Time × Volume (Time)2

[F] = [FL–4 T2]
Therefore, density = [LT–2 × L3]

Hence, dimensional formula of density of [FL–4 T2]

The diameter of a steel rod is given as 56.47 ± 0.02mm. What does it mean? [HSEB 2072]

The diameter of a steel rod is given as 56.74 ± 0.02mm. This means that 56.47 mm is the actual

diameter of the rod and ± 0.02 mm is instrumental error. The measuring instrument either measures

0.02mm less or more than the actual diameter.

Check the correctness of the formula v2 = u2 + 2as using dimensional analysis. [HSEB 2059]

Given, v2 = u2 + 2as . . . (1)

Dimensional formula of L.H.S = [LT–1]2 = [L2T]–2

Since, L.H.S = R.H.S, therefore, the given formula is dimensionally correct.

Find the dimensions of Planck's constant 'h' from the given equation: λ = h where λ is
p;

wavelength and p is the momentum of photon. [HSEB 2069S]

The formula for Planck' constant is given as λ = h or h = λ × p,where λwavelength and p is is
p

momentum. The dimension of λ = [L] and dimension of P = [MLT-1]. Then dimensions of h is h = [L]
[MLT-1] or h = [ML2T-1]
Therefore, dimensions of h are 1 on M, 2 on L and -1 on T or [MLT2T-1]

Is dimensionally correct equation necessarily be a correct physical relation? What about

dimensionally wrong equation? [HSEB 2063]

No, dimensionally correct equation may not be a correct physical relation. Consider the following

relation, v = u + 10 at . . . (1)

Dimensional formula of v is [LT–1] and that of (u + 10 at) is also [LT–1]. Hence, the relation (1) is

dimensionally correct. But it is established fact that the relation (1) is not correct physical relation.

14 | Essential Physics

Dimensionally wrong equation can't be correct physical relation. For example, v = u + at2 is

dimensionally wrong equation and it is also incorrect physical relation.

OR

A dimensionally correct equation may be an incorrect physical relation. Pure numbers, trigonometric

terms and some constants like π are dimensionless. So, dimension does not depend on magnitude. Thus,
a dimensionally correct equation does not need necessarily be a correct relation, but a dimensionally

wrong equation must be wrong one.

9. If y = a + bt + ct2, where y is the distance and t is the time. What is the dimension and unit of c?

[HSEB 2070]

Ans:
Given equation is, y = a + bt + ct2, where y is the distance and t is time, so according to principle of
homogeneity, the dimensions and unit of c such that, Unit of c × unit of t2 = unit of y
c × sec2 = m
or, c = m/sec2

Therefore, unit of c is m/sec2 and dimensions of c are 0 on M, 1 on L and -2 in T or [M0LT-2].

R
10. A student writes 2GM for escape velocity check the correctness of the formula by using

dimensional analysis. [HSEB 2062]

R . . . (1)
Ans: Given, V = 2GM

Where, V is the escape velocity, R is the radius, G is universal gravitational constant and M is the mass.

In dimensional form, the equation (1) becomes

[V] =  R . . . (2)
 GM

Here, Dimensional formula of [V] = [LT–1]

Dimensional formula of [R] = [L]
Dimensional formula of [G] = [M–1L3T–2]

Dimensional formula of [M] = [M]

Now, the equation (2) becomes,

[LT–1] = [L]
[M–1L3T–2 × M]

or, [LT–1] = [L–2T2]
or, [LT–1] = [L–1T] which is inconsistent.

Hence, the equation (1) is incorrect dimensionally.

11. Check the correctness of the relation h = 2T cos θ, where symbols have usual meaning.
Ans: r ρg

[HSEB 2070S]

Here, given formula is h = 2T cos θ
r ρg

Here, Dimension of h = [L], dimension of r = [L], dimension of ρ = [ML-3], dimension of g = [LT-2],

Dimension of T = [MT-3] and 2 cos θ is dimensionless. Then, dimension of left hand side, h = [L]

Dimension of right hand side, 2T cosθ = [MT-2 = [M1-1 L-1 + 3 -1 T-2 + 2] = [L]
r ρg [L][ML-3] [LT-2]

12. Give the dimensional formula for force, potential difference and specific heat capacity.
[Model Question]

Ans: We have, dimensional formula of force = Mass × Acceleration

Physical Quantities | 15

Therefore, Dimension of force = [M] [LT–2] = [MLT–2]

Work done
Potential difference = charge

Therefore, Dimension of potential difference = [ML2 T–2 = [ML2T–2Q–1]
[Q]

Q
Specific heat capacity = mdt

[force] × [Distance]
= [mass] × [Temperature]

Therefore, Dimension of heat capacity = [MLT–2] × [L] = [M0L2T–2K–1]
[M] × K

13. A student writes an expression of the force causing a body of mass (m) to move in a circular
Ans: motion with a velocity (v) as F = mv2 use the dimensional method to check its correctness.

14. [HSEB 2070]
Ans:
The given relation is F = mv2. Where F is force, m is mass and v is speed of particle.
The dimension of force F = [MLT-2]
The dimension of velocity v = [LT-1]

The dimension of mass m = [M]
The dimension of left hand side F = [MLT-2]
Dimension of right hand side mv2 = [M] [LT-1]2 = [ML2T-2]

Therefore, dimension on left hand side in not equal to dimension on right hand side of equation. So, the

formula is dimensionally wrong.

What is the different between accurate and precise measurement? [NEB 2074]

An exact measurement of physical quantity is called accurate measurement whereas the measurement
is done with the limitations of measuring scales is called precise measurement.

Numerical Examples

1. Obtain the expression for gravitational potential energy and establish its dimension. [HSEB 2052]
Solution:

The gravitational potential at a distance 'r' from the centre 'O' of the earth is given by,
GM

V = – r2 . . . (1) Where, M is the mass of the earth, G is the universal gravitational constant

and r is the distance. The gravitational potential energy is given by,

U=V×m
GM

U = – r × m . . . (2) Where m is the mass of the small body.

The dimension of the gravitational potential energy is given by,

U = NM× 2L2 × M × M  = [MLT–2L]
L

U = [ML2T–2] . . . (3)

GM
Hence, the expression for the gravitational potential energy is U = – r × m and dimension of

gravitational potential energy is U = [ML2T–2].

2. Obtain the dimensions of specific heat capacity and gravitational constant. [HSEB 2054]

Solution: Here, Q = mc ∆θ
Q [ML2T–2]
∴ C = [L2T–2K–1]
C = ∆θ = [M × K]

16 | Essential Physics

GMm
Again, F = r2

∴ F × r2 [MLT–2] × [L2]
G = M × m = [M] × [M]

∴ G = [M–1L3T–2]

Hence, the dimensional formula of specific heat capacity is C = [L2 T–2K–1] dimensional formula of
universal constant, G = [M–1L3T–2]

3. Calculate the dimensional formula of Universal constant "G". [HSEB 2056]

Solution:

We know that, the force of attraction between two bodies of mass M and m is,

GMm
F = r2

∴ G = F × r2 = [MLT–2] × [L2] = [M–1L3T–2]
M × m [M] × [M]

∴ G = [M–1L3T–2]
Hence, the dimensional formula of universal gravitational constant, G = [M–1L3T–2]

4. Convert 10 ergs in joules. [HSEB 2061]

Solution:

We can convert 10 ergs in joules by the method of dimensions. Since ergs and joules are both units of
energy and dimension, formula for energy is [ML2T–2]

∴ n1[M11L2T–2] = n2 [M2L22T22] . . . (1)

Here, n1 = 10, n2 = ?

M1 = 1 gm M2 = 1 kg

L1 = 1cm L2 = 1m

T1 = 1 sec T2 = 1 sec

From (1), we have,

10 × 1 × (1)2 × (1)–2 = n2 × 1 × (1)2 × (1)–2
or, 10 × 1 × 1 = n2 × 1000 × (100)2

or, 10 = n2 × 1000 × 10000
n2 = 10–6J

Hence, 10 ergs = 10–6J

5. Check the correctness of the formula t = 2π m where, t be the time period, m be the mass and k is
k
the force per unit displacement.
Solution: [HSEB 2064]

Here, t = 2π m
k

L.H.S. = t = [T] . . . (1)

R.H.S = 2π m = 2π M M T2 = [T] . . . (2)
k F/x = MLT–2 =

L

∴ L.H.S. = R.H.S. Hence, the formula is correct.

Physical Quantities | 17

Important Numerical Problems

1. The height 'h' through which water of surface tension 'T' and density 'd' rises in a capillary tube of

radius 'r' is given by, 2T , check the correctness of the relation using the method of dimensions.
rdg

Solution:
2T

We know, the height raised due to surface tension of water is, h = rdg .

The dimension of height (h) of L.H.S. (h) = [L] . . . (1)
. . .(2)
2T = L × L × LT–2 = MMLLT––12TL––21 = [L]
But, in R.H.S = rdg M/L3

Hence, L.H.S = R.H.S. The formula is correct

2. A sphere of radius 'r' moving through a fluid of density 'ρ' with high speed 'v' experienced by a
retarding force F given by F = K rxρyvz where, k = non dimensional coefficient. Use the method of

dimension to find the values of x, y, z. [TU 2045]

Solution:

F = K rxρyvz
[MLT–2] = K[L]x [ML–3]y [LT–1]2

= K[MyLx – 3y + z T–z]

Equating the powers of M, L and T separately we get,

1=y . . . (1)

1 = x – 3y + z . . . (2)

–2 = –z . . . (3)

From equation (1), y = 1
From equation (2), 1 = x – 3y + z,

or, 1 = x – 3 × 1 + 2
or, x = 1 + 3 – 2 = 2
From equation (3), – z = – 2

or, z = 2

Hence, the value of x = 2, y = 1, z = 2

3. The mass 'm' of an object depends upon the velocity 'v' of the object its rest mass 'm0' and velocity of

light 'c'. The relation has been given omitting 'c' as m = m0 v2 . Write the correct formula putting
1–

the missing 'c' at its proper place.

Solution:

From the dimension of L.H.S is [M]. So, the dimension of R.H.S must also [M]. In R.H.S., the

dimension of v must vanish. Hence the dimension of v is cancelled out by adding velocity of light c.

Therefore, the correct formula is,

m= m0
v2
1 – c2

4. Convert 10 dyne to Newton.
Solution:

Dyne and Newton are the units of force in CGS and SI systems respectively.

The basic dimensional relation for the unit conversions is,

18 | Essential Physics

n2 = n1 MM12a LL21 b TT12c

The dimensions of force is expressed as,
[F] = [MLT–2] = [M]1 [L]1[T]–2

So, here a = 1, b = 1 and c = –2.

Therefore, n2 = n1 MM211 LL12 1 TT12–2

In CGS system, M1 = 1gm L1 = 1cm

T1 = 1 sec n1 = 10 (Because force = 10 dynes)

In SI system M1 = 1gm L1 = 1cm

T1 = 1 sec n2 = ?

This gives, n2 = n1 11gkmg 1 11cmm 1 11sseecc–2 = 10 × 1010g0mgm1 110c0mcm 1 11sseecc –2

∴ n2 = 10–4
Hence, 10 dyne = 10–4 Newton

5. Convert 1 Newton into dynes.

Solution:

Newton and Dyne are the units of force in SI and CGS system respectively.

The basic dimensional relation for unit conversions is,

n2 = n1 MM12a LL21 b TT21c

The dimensions of force is expressed as,
[F] = [MLT–2] = [M]1 [L]1 [T]–2

So, here a = 1, b = 1, and c = ?

Therefore, n2 = n1 MM211 LL21 1 TT21–2

In SI system,

M1 = 1kg L1 = 1m
T1 = 1sec n1 = 1(Because force = 1 Newton)

In CGS system, (system 2),

M2 = 1gm L2 = 1cm

T2 = 1 sec n2 = ?

This gives,

n2 = 11kgg1 11cmm 1 11sseecc1 = 1 × 1010g0mgm 1 110c0mcm 1 11sseecc 1

or, n2 = 1 × [1000] 1 [100] 1 [1]–2 = 1 × 1000 × 100 × 1
or, m = 105
Hence, 1N = 105 dynes

6. Convert a Joule into erg using dimension.
Solution:

Joule and erg are the units of work in SI and CGS systems respectively. The dimensional formula of
work is [ML2T–2]

Now, from n2 = n1 MM21a LL21b TT12c

∴ n2 = n1MM211 LL122 TT21–2

In SI system, M1 = 1kg = 1000 gm, L1 = 1m = 100cm
T1 = 1 sec n1 = 1 Joule

Physical Quantities | 19

In CGS system, M2 = 1gm L2 = 1 cm

T1 = 1 sec n2 = ?

∴ n2 = 1 11kgg1 11cmm 2 11sseecc –2 = [1000]1 [100]2= 107 ergs.

Practice Short Questions

1. Obtain dimension of specific heat capacity and gravitational constant.
2. Calculate the dimension formula for universal gravitational constant G.
3. What is fundamental quantity? Find the dimensional formula for the density by taking force, length and time

to be fundamental quantities.
4. The density of gold is 19.39 gm/cc. Express its value in SI unit.
5. Is a dimensionally correct equation necessarily be a correct physical relation? What about dimensionally

wrong question?
6. What is the difference between accurate and precise measurement?
7. What are the limitations of dimensional analysis?
8. The force F is given in terms of time’t’ and displacement x by the equation F = AsinBx + CsinDt. What is the

dimension of D/B?
9. If y = a + b + ct2, where y is the distance and t is the time. What is the dimension and unit of c?
10. Name any two physical quantities which have the same dimensions can a quantity have unit but no

dimension? Explain.
11. The diameter of steel rod is given as 56.47±0.02mm. What does it mean?
12. The length of rod is exactly 1cm. An observer records the readings as 1.0cm, 1.00cm and 1.000cm, which is

the most accurate measurement?

Practice Long Questions

1. What do you know about fundamental and derive units? Is a dimensionally correct equation necessarily be a

correct physical relation? Justify your answer. [HSEB 2066]

2. A student unites R for escape velocity and T = 2π M
2GM K where T be the time period, M is the mass and

K is the force per unit displacement. Check the correctness formula by using dimensional analysis.

[2055 HSEB 2062, 2064, 2069]

3. Find the dimensions of Planck's constant 'h' from the given equation λ = h where λ is wave length and p is the
p

momentum of photon also check the correctness of the formula v2 = u2 + 2as using dimensional analysis.

[HSEB 2059, 2069S]

4. A sphere of radius r moving through a fluid of density ρ with high speed v experienced by a retarding force F
given by F = KrxρyVz where k = non-dimensional coefficient use the method of dimension to find the value

of x, y and z.

Practice Numerical Questions

1. The time period of oscillation of the simple pendulum depends on its length (l), mass of bob (m) and
acceleration due to gravity g. derive the expression for its time period using method of dimensions.

[Ans: T = 2π l/g ]

2. The length and breadth of a rectangular lamina are measured to be (2.3 ± 0.2) cm and (1.6 ± 0.1) cm.

Calculate area of the lamina with error limits. [Ans: 3.68 ± 0.55)cm2

3. A force F is applied on a square plate of side L. If the percentage error in the determination of L is 2% and

then in if F is 4%. What is the possible error in pressure? [Ans: 8%]
4. Convert 10 ergs in Joules. [Ans: 10–6J]

20 | Essential Physics

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. The fundamental quantity is:

a. Specific gravity b. Angle c. Temperature d. pressure

2. The number of fundamental units in MKS system

a. 3 b. 4 c. 5 d. 6

3. Which is a supplementary unit?

a. Radian b. Second c. Micron d. Log

4. The dimension of linear momentum is identical to that of

a. Force b. Angular momentum c. Impulse d. Pressure

5. Planck's constant has identical dimensional formula with:

a. Linear momentum b. Inertia c. angular momentum

d. Energy

6. The dimensional formula for universal gravitational constants G is

a. [M° L° T°] b. [M °L2 T–2] c. [M–1 L3 T–2] d. [M–1 L T–2]

7. Dimension of Angstrom (A°), micro meter (µm ), Fermi (F) and nanometer (nm) is the same. Which of the

following represents correct arrangement of this magnitude in decreasing order?

a. µm, A, nm, F b. A°, µm, F, nm c. µm, nm, A°, F d. nm, A°, µm, F,

8. The dimensional formula of co-efficient of viscosity is

a. [ML–1 T –1] b. [M L° T°] c. [M L–2 T-] d. [M L–1 T–1]

9. NC-1 has the same dimension as

a. Volt/meter b. Farad meter c. Voltmeter d. farad/meter

10. The dimensional formula of impulse is

a. [M LT–2] b. [M LT –1] c. M L2 T –1] d. [M2 L T–1]

11. The star nearest to earth is 4 light years away: the distance is?

a. 9.49 × 1015 m b. 37.84 × 1012 m c. 37.84 ×1015 m d. 9.5 × 1012 m

12. The dimensions of angular momentum are.

a. [MLT–1] b. [M L2 T–1] c. [ML–1] d. [ML T–2]

13. The dimension formula for stress is same as that for

a. Force b. pressure c. Torque d. Work

14. "Light year" is the unit of?

a. Length b. velocity c. Time d. Momentum

15. If the momentum of a body increases by 100%, what is the % change in its kinetic energy?

a. 200% b. 300% c. 400% d. 50%

16. one kilowatt hour has the value of?

a. 1.6 × 106 J b. 3.6 × 106 J c. 764 w d. 9.46 × 103 J

1
17. The relation S= ut + 2 a (2t –1) is correct

a. Only numerically b. only dimensionally

c. Both numerically and dimensionally d. neither both numerically nor dimensionally

18. If energy E, velocity V and time T are taken as fundamental units, the dimensional formula for surface

tension becomes b. [ E° v-2 T –2] c.[E2 V2 T –2] d. [E2 v2 T –2]
a. [E v –2 T–2]

19. The value of Newton in dyne is equal to

a. 104 b. 105 c. 106 d. 107

20. The dimensional formula for angular displacement is

a. [L] b. [M° L T-1] c. [L2] d. none

21. If the unit of force and length are doubled then unit of power become

a. Six time b. four time c. three times d. None

Answer Key

1. c 2. a 3. a 4. c 5. c 6. c 7. d 8. a 9. a 10. b
11. c 12. b 13. b 14. a 15. b 16. b 17. d 18. a 19. b 20. d
21. b

Ch apter Vectors

2

Teaching Manual Physics Grade –XI , Higher Secondary Education Board
Curriculum And Training Division Sanothimi, Bhaktapur.

Syllabus:

Vectors-Graphical presentation of vectors; Addition and subtraction of vectors:
Parallelogram, triangle and polygon laws of vectors; Resolution of vectors; Unit
vectors; Scalar and vector products.

Objectives:

The objectives of this sub unit are to make the students to able to identify whether a
physical quantity is a vector or a scalar, to compose and resolved the vectors, and to
multiply the vectors.

Activities (micro syllabus):

1. Define and explain vector and scalar quantities.
2. Illustrate graphical representation of vectors.
3. State and explain parallelogram, triangle and polygon laws of vector addition

and subtraction.
4. Discuss resolutions of vectors.
5. Explain simple, dot and cross products of vectors
6. Numerical problems: Focused numerical problems given:

(a) As exercises in the University Physics (Ref. 1) and
(b) In advanced level Physics (Ref. 2) and also in Physics for XI (Ref.3).

22 | Essential Physics

2.1 Introduction

In one dimensional motion, the idea of direction of motion of a body may be introduced by saying that the
body is moving to the left or to right or up or down, that is in one dimensional motion. There are only two
possible directions. Analytically, they may be expressed by positive and negative signs. However, in two
and three dimensional motions, there are a very large number of (infinite) directions. The motion of a
particle in a plane is two dimensional; while motion of particle in space is three dimensional. To
represents the direction in these situations, we have to develop a new concept is called the concept of
vectors. Therefore, all the measurable physical quantities can be divided into two classes. They are :

1. Scalar quantity and 2. Vector quantity. They are described as below.

1. Scalars Quantity

The physical quantities which have a magnitude but no direction are called scalars. For example ;
Mass, length, distance covered, time, density, work, specific heat, temperature, charges etc. A
scalar can be completely described by a number representing its magnitude. A scalar may be
positive or negative. They can be added, subtracted, multiplied and divided according to the
ordinary rules of algebra.

2. Vectors Quantity

The physical quantities which have (possess) both magnitude and direction are called vectors. For

example; displacement, velocity, acceleration, force,

momentum, gravitational field, electric field etc. A vector can B

be completely represented by an arrow as shown in figure 2.1.

The arrow represents the magnitude of the length (AB) and the

arrow head at point B gives the direction of the vector. In

mathematical expressions, a vector quantity is denoted by F or F

putting an arrow head over the letter representing the vector

→
quantity or by a bold face letter. For example, if vector AB in

→ A
Figure 2.1 represents force vector, it may be represented by F Fig. 2.1

. It may be pointed out that the vectors cannot be added,
subtracted, multiplied or divided as one may do in case of [Fig 2.1, representation of vector]

scalars. (Since, vectors have also direction). They are added,

subtracted and multiplied according to the vector algebra. The division of a vector by another is

not defined.

2.2 Composition of Vectors

"The process of adding two or more vectors is called the composition or finding the resultant of the

vectors." For the reason that vectors possess direction in addition to their magnitudes, they cannot be

added by simple laws or by simple laws of algebra applicable to scalars. Let us explain it by adding two

displacement vectors in the different cases.

1. When two vectors act in the same direction: Suppose, a particle is displaced through 4m and

then through 3m along the same direction (east) as shown in figure 2.2. If the two displacement

→→ →→
vectors are represented by P and Q , then the ( P + Q ) of the two displacement vectors will be a

displacement of 7 m along the direction of going (east).

→P →Q →P + →Q Vectors | 23
N

WE

Fig. 2.2 S

[Fig. 2.2 Particle displaced]

2. When two vectors acts in the opposite direction: In case a particle is displaced through 4m
→→

along east and then through 3m along west as shown in figure 2.3, then sum ( P + Q ) of the two
displacement vectors will a displacement of 1m along east.

N

→P + →Q = →P + →Q W E

Fig. 2.3 S

[Fig. 2.3 Particle displaced]

3. When two vectors are inclined to each other: Let us first consider, the simple case in which the
two displacement vectors are at right angles to each other. Suppose, a particle is first displaced
through a distance of 4m along east and then through a distance of 3m along north. The two
→→
displacements have been represented by AB and BC according to some suitable scale and the
→
measurements tell that the magnitude of displacement AC is 5m as shown in figure 2.4.

→
The angle θ gives the direction of displacement AC with north of east can also be measured.
Therefore, instead of moving a distance of 4 m from A to B and then a distance of 3m from B to C

(a total distance of 4 + 3 = 7m), if the particle moves a distance of 5m from A to C at an angle

θ with AB, it would reach the same point. Thus, the displacement of 5m along AC produces the
same effect as the displacements of 4m along AB and 3m along BC together do. It amounts to

→ C
saying that displacement vector AC is sum (or

→→
resultant) of the displacement vectors AB and BC.

Therefore, N
→→→
AB + BC = AC . →Q W E
→→ →
θ →P S
If vectors AB, BC and AC are represented by vectors A B
→→ →
P , Q and R respectively, then

→→→
P +Q=R

From the discussion in three cases, it follows that the [Fig 2.4 Displacement of vector]
vectors are not added algebraically but are added

geometrically.

24 | Essential Physics

2.3 Geometrical Addition of Two Vectors

→→
We have seen that when two vectors P and Q are inclined to each other as shown in figure 2.4, their sum

→→→ →→
is given by R = P + Q . A careful look tells that the vectors P and Q have been represented by the

→
two sides AB and BC of a triangle taken in the same order and their sum R is given by the closing side

AC but taken in opposite order. In figure 2.4, the two vectors act at right angles to each other but this

conclusion applied to the two vectors inclined at any angle and leads to the various laws of vector

addition such as triangle law of vector, parallelogram law of vector and polygon law of vector. They are

described as below.

1. Triangle law of vector addition

Triangle law of vector addition states that "if two vectors acting simultaneously on a body are
represented both in magnitude and direction by two sides of a triangle taken in an order, the
resultant of the vectors both in magnitude and direction is given by the third side taken in the
opposite order".

Consider two vectors →P and →Q act on a body. Let, the vectors →P and →Q are represented both in
magnitude and direction by the sides OA and AB of a triangle OAB and θ is the angle between
them. According to triangle law of vector addition, the side OB represents the resultant (→R ) of the
vectors both in magnitude and direction.Thus,

→R = →P + →Q

Let φ be the angle made by the resultant (→R ) with vector →P as in the figure 2.5. Also, let us draw a

perpendicular BC from B to produced part of OA.

From right angle triangle OCB, we have
OB2 = OC2 + CB2

or, OB2 = (OA + AC) 2 + CB2 . . . (1)

In ∆ACB, cosθ = AC
AB

or, AC = AB cosθ

or, AC = Q cosθ and sinθ = BC B
AB

or, BC = AB sinθ

or, BC = Q sinθ R
φ
From equation (1) we get, O Q
or, R2 = (P + Q cosθ)2 +(Q sinθ)2
P θ
or, R2 = P2+ 2PQcosθ + Q2cos2θ + Q2sin2θ AC

or, R2 = P2 + 2PQcosθ + Q2 [Fig 2.5 representation of triangle law
of vector]
∴ R = P2 +Q2 + 2PQcosθ . . . (2)

Equation (2) gives the magnitude of resultant vector →R .

Again, In ∆OCB,

Vectors | 25

tanφ = BC = BC = P Qsinθ
OC OA + AC + Qcosθ

∴ φ = tan–1  Qsinθ  . . . (3)
P + Qcosθ

Equation (3) gives the direction of resultant vector →R .

2. Parallelogram law of vector addition

Parallelogram law of vector addition states that "if two vectors, acting simultaneously at a point,
can be represented both in magnitude and direction by the adjacent sides of a parallelogram
drawn from a point, then the resultant vector is represented completely both in magnitude and
direction by the diagonal of the parallelogram passing through that point".

SQ

B
R

θ α θ
A PN
O

[Fig. 2.6 Representation of parallelogram law of vector]

Let, two vectors →A and→B be completely represented by the two sides OP and OS respectively of a
parallelogram OPQS as shown in the figure 2.6. According to parallelogram law of vectors, the
diagonal OQ will give the resultant vector →R such that, →R = →A + →B . Let < SOP = θ. From Q,
draw QN perpendicular to produced part of OP.

In ∆PNQ we have, Sinθ = QN and cosθ = PN
PQ PQ

or, QN = PQ sinθ or, PN = PQ cosθ

or, QN = Bsinθ . . . (1) and

or, PN = B cosθ . . . (2) (since,PQ = OS = B)

In ∆ONQ, We have

OQ2 = ON2 + QN2

or, R2 = (OP + PN)2 + QN2 (since, OQ = R)

= (A + Bcosθ)2 + (Bsinθ)2 [Using (1) and (2)]

= A2 + 2ABcosθ + B2 cos2 θ + B2 sin2 θ

= A2 + 2ABcosθ + B2 (cos2 θ + sin2 θ)

= A2 + B2 + 2ABcos θ (since, sin2 θ + cos2 θ = 1)

∴ R = A2 + B2 + 2 AB cos θ . . . (3)

The equation (3) gives the magnitude of resultant vector →R . Let α be the angle between →R and →A .

26 | Essential Physics

In ∆ONQ We have,

QN QN
or, tan α = ON = OP + PN

or, tan α = A Bsin θ
+ B cosθ

or, α = tan–1  Bsinθ  . . . (4)
A + B cos θ

The equation (4) gives the direction of resultant vector →R .

Some cases

Case I : If θ = 0° then, equation (3) becomes R = A2 + B2 + 2AB = A + B and α = tan–1

A +BsBinc0o°s0° = 0°. Thus, when two vectors act along the same direction (θ = 0°), the resultant

has magnitude equal to the sum of the magnitude of two vectors and acts along the direction of
→→
A and B .

Case II: If θ = 90° then equation (3) becomes R = A2 + B2 + 2AB cos90° = A2 + B2 and

α = tan–1 A +BBsinc9o0s°90° = tan–1 AB . Thus, when two vectors act perpendicular to each other
then the magnitude and direction of resultant vectors are A2 + B2 and tan–1 AB respectively.

Case III: If θ = 180° then equation (3) becomes, R = A2 + B2 + 2AB cos180° = A – B and α = tan–1

A +BsBinc1o8s108°0° = 0 or 180°. Thus, when two vectors act along opposite direction (θ = 180°),

the resultant has magnitude equals the difference of their magnitude and act along the direction
→

of A .

3. Polygon law of vector addition

It states that "if large number of vectors acting at a point is represented in magnitude and
direction by adjacent sides of polygon taken in same order, then the closing side of the polygon
taken in reverse order gives the magnitude and direction of the resultant vector".

W

D

C B Z
D RC

Y

O A OA X B
(a) (b)

[Fig. 2.7 Representation of polygon law of vector addition]

Vectors | 27

→→→ →
Consider A , B , C and D are vectors used to form a polygon as shown in figure 2.7 (a) and 2.7

→→→ →
(b). In the polygon, the four vectors A , B , C and D are represented in magnitude and direction

→ →→ →
by sides. We supposed OX , XY , YZ and ZW respectively in same order. So that, the resultant

→→
R is represented in both magnitude and direction by closing side OW taken in the opposite order.

To find the resultant →R of this polygon, by using triangle law of vector addition. Thus,

→→→
OY = A + B

→ → → → →→
OZ = OY + C = A + B + C

→ → →→ →→ →
OW = OZ + D = A + B + C + D

Hence, resultant vector in polygon is found by taking the resultant vectors of triangles. i.e.

→→ →→ →
R = A + B + C + D + ...

2.4 Subtraction of Two Vectors

→→ →
The subtraction of a vector Q from vector P is defined as the addition of vector – Q (negative of vector

→ → →→→ →
Q ) to vector P . Thus, P – Q = P + (– Q )

→→
Let us consider two vectors P and Q as shown in figure

2.8 (a). To find the resultant vector on subtraction vector

→→ →→ →
Q and P , draw vector AB = P . Now draw vector BC

equal in magnitude of vector but opposite in direction so

→
that its tail coincides with the arrowhead of vector P as

→→ [Fig 2.8 Representation of vector subtraction]
shown in figure 2.8(b). Then, vector BC = – Q . Draw

→→
vector AC from tail of vector P to arrowhead of vector

→ → → → → → → →→
– Q . Then, AC = AB + BC = P + (– Q ) ∴ AC = P – Q , which is subtraction resultant vector.

2.5 Zero Vectors

Zero vectors is that vector which has zero magnitude and an arbitrary direction. A zero vector is
→

represented by O and is also called null vector. A vector when multiplied with O gives zero vectors.
→→

i.e., OA = O

Properties of Zero Vectors

1. The addition or subtraction of zero vector from a given vector does not alter the given vector.
→→→

Thus, A + O = A

28 | Essential Physics

→→ →
A–O=A

2. The multiplication of a non–zero real number n with a zero vector is again vector. Thus,

→→
nO = O

3. If n1 and n2 are two different non–zero real numbers, where n1 = n2, then the relation,

→→ →→
n1 A = n2 B , can hold only if both A and B are zero vectors.

A zero vector has a lot of physical significance. It is useful in describing the physical situation involving

vector quantities.

Example of Zero Vectors

1. Velocity vector of a stationary particle is a zero vector.
2. The acceleration vector of an object moving with a uniform velocity is a zero vector.
3. The position vector of the origin of coordinate axes is a zero vector.

2.6 Unit Vector and Orthogonal Unit Vector

→
A vector, whose magnitude is unity, is called a unit vector. A unit vector in the direction of vector A is

∧ →→ Y
∧AA ^j
written as A . Mathematically, A =→ =A .

|A |

→→
Where | A | = A is the magnitude of vector A .

The mutually perpendicular unit vectors ∧i , ∧j , ∧k which act along ^i
Z ^k X
x, y and z-axis respectively are called orthogonal unit vector as
[Fig 2.9 Representation of unit vector]
shown in figure 2.9. In three dimensions, a vector is expressed in

terms of unit vector as → = Ax∧i + Ay∧j + Az∧k and magnitude of
A

vector is written as |A| = Ax2 + Ay2+ Az2

Properties of unit vector

Some properties of unit vectors are described as below;

1. Scalar product :

(i) ∧i . ∧i = ∧j . ∧j = ∧k . ∧k = 1 (ii) ∧i . ∧j = ∧j . ∧k = ∧k . ∧i = 0

2. Vector product :

(i) ∧i × ∧i = ∧j × ∧j = ∧k × ∧k = 0 (ii) ∧i × ∧j = ∧k, ∧j × ∧k = ∧i , ∧k × ∧i = ∧j

(iii) ∧j × ∧i = –∧k, ∧k × ∧j = –∧i , ∧i × ∧k = – ∧j

2.7 Resolution of a Vector

The resolution of the vector means that "it is the process of splitting of a vector in to its components".
→→

Let, F be a vector which is acting at an angle θ above positive x-axis. Let, the vector F represented by
the line OA. Draw perpendiculars AB and AC from A on x-axis and y-axis respectively. Here, the

→
distance OB is denoted by Fx and is called x-component (horizontal component) of vector F and the

→
distance OC is denoted by Fy and is called y- component of F . Now, from right angle triangle OBA

Vectors | 29

Cosθ = OB Y
OA

or, Cosθ = Fx C θA
F

or, Fx = F cosθ . . . (1)
Again, right angle triangle OCA,
Fy F
Sinθ = OC
OA
θ
or, Sinθ = Fy O Fx B X
F

or, Fy = F sinθ . . . (2)

Squaring and adding (1) and (2) we get

or, Fx2 + Fy2 = F2cos2θ + F2sin2θ [Fig 2.10 Resolution of a vector]
or, F2 = Fx2 + Fy2

or, F = Fx2 + Fy2 . . . (3)

→
To find the direction of F , let us take right angle triangle OBA,

tanθ = AB = OC = Fy . . . (4)
OB OB Fx

2.8 Rectangular Components of a Vector in a Plane

"When a vector is splitted into two component vectors at right angle to each other, the component

vectors are called rectangular components of a vector".

→→
Consider in figure 2.11, a vector A represented by OR, has to be resolved into two rectangular

component vectors along the direction of x–axis and y–axis. From O, draw two axes OX and OY at right

→→
angles to each other. From point R draw RP and RQ perpendicular to OX and OY. Then, OP = A x and

→→ → →
OQ = A y are the rectangular components of A . Here PR is equal and parallel to OQ. Hence, PR also

→
represents A y in magnitude and direction. From triangle law of vector addition we have,

→ →→
OR = OP + PR
→→ → Y

A = Ax+ Ay . . . (1)

→→ QR
Let in figure 2.11, ∠POR = θ, A x = OP, A y = OQ = PR

In right angled triangle ORP, Ay A

OP = cosθ; j
OR

or, OP = OR cosθ Oi θ X

∴ Ax = A cosθ . . . (2) and Ax P

PR = sinθ; [Fig 2.11 Rectangular component
OR of vector]

or, PR = OR sinθ

∴ Ay = A sinθ . . . (3)
Also, OR2 = OP2 + PR2

30 | Essential Physics

or, A2 =Ax2 + Ay2

or, A= Ax2 + Ay2 and tanθ = PR = Ay ∴ θ = tan–1 AAxy
OP Ax

2.9 Product of Two Vectors

The product of two vectors cannot be obtained as the product of two scalars. There are two ways in which
a product of two vectors is obtained.

1. Scalar Product or Dot Product of Two Vectors

"When two vectors are multiplied in such a way that their product is a scalar quantity then it is

→→
called scalar or dot product. So, the scalar or dot product of two vectors A and B represented by

→→ →→ →→
A . B (read A dot B ) is a scalar, which is equal to the product of the magnitudes of A and B and

→→
the cosine of the angle between them. If θ is the angle between A and B then

→→
A . B = AB cosθ

Geometrical interpretation of dot product of two vectors

→→ →→
Let, the two vectors A and B represented by OP and OQ and ∠POQ = θ. Then from the figure

→→
2.12 (a) and 2.12 (b), OP = A , OQ = B .

Q
QS

→Β

θ R θ →Α P
O →Α P O (b)

Bcosθ
(a)

[Fig. 2.12; geometrical interpretation of dot product of two vectors]

→→
Now, A . B = AB cosθ = A(Bcosθ) = A(OR)

→→
= A(magnitude of component vector of B in the direction of A )

→→
= A( projection of component vector of B in the direction of A ) and

→→
A . B = AB cosθ = B(A cosθ)

→→
= B(magnitude of component vector A in the direction of B )

Thus, scalar product or dot product of two vectors is also defined as the product of the magnitude

of one vector with the magnitude of the component of other vector in the direction of first vector.

Some special cases

1. When two vectors are parallel, then θ = 0°, cos 0° = 1
→→

∴ A . B = AB cos0° = AB
2. When two vectors are perpendicular, then θ = 90°, cos 90° = 0

→→
∴ A . B = AB cos90° = 0
It means, the dot product of two perpendicular vectors is zero.

3. When two vectors are antiparallel or opposite in direction then, θ = 180°, cos180° = –1.

Vectors | 31

→→
∴ A . B = AB cos180° = –AB

Properties of dot product of two vectors

Some properties of dot product of two vectors are given below,

→→ →→
1. Dot or scalar product of two vectors is commutative i.e., A . B = B . A

→→ →→
By definition A . B = AB cosθ and B . A = BA cosθ = AB cosθ

→→ →→
∴ A.B = B.A

This proves that the dot product of two vectors is commutative.

2. Dot or scalar product is distributive i.e.,

→ → → →→ →→
A .( B + C ) = A . B + A . C

3. Dot or product of a vector with itself gives square of its magnitude.

i.e., →→ = AA cos0° = A2 or, A = (→A .→A )1/2
A.A

Examples of scalar or dot product of two vectors

Some examples of scalar product of two vectors are as following.
→→

1. Work done W is defined as the dot product of force vector F and displacement vector S .
→→

i.e., W = F . S = FS cosθ
→

2. The instantaneous power is defined as the dot product of force vector F and the
→ →→

instantaneous velocity v i.e., P = F . v = Fv cosθ
3. Magnetic flux linked with a surface is defined as the dot product of magnetic induction

→ → →→
vector B and the area vector A . i.e., φ = B . A = BA cosθ
Here, work done, power and magnetic flux are scalar quantities.

2. Vector or Cross Product of Two Vectors

"If two vectors are multiplied in such a way that their product is a vector quantity then it is

called vector or cross product of two vectors." So, the vector or cross product of two vectors

→→ →→ → →
A and B represented as A × B (read A cross B ) is a vector whose magnitude is equal to the

product of the magnitudes of the two vectors and sine of the angle between them. If θ is the angle

→ → →→ ^
between A and B , the A × B = AB sinθ C

,

^ [Fig. 2.13 Representation of cross product of vectors
where C is a unit a vector in the direction of
→ →^
C . The direction of C or C (i.e.,vector
product of two vectors) is perpendicular to

→→
the plane containing A and B and points in
the direction as given by right handed screw
rule or right hand thumb rule. They are
explained as below:

32 | Essential Physics

Right handed screw rule: It states that "if a right handed screw placed with its axis
→→

perpendicular to the plane containing two vectors is rotated from A to B through smaller
→→ →

angle, then the sense of the advance of the tip of the screw gives the direction of A × B or C ".
→→

If the two vectors A and B are acting in the plane of paper, then the direction of cross product
→→
A × B , according to right handed screw rule will be perpendicular to the plane of paper directed
outward as shown in figure. 2.13(a).

Right hand thumb rule: It states that "if we curl the fingers of right hand, keeping the thumb
→→

erect, in such a way that they point in the direction of rotation of A to B , then the thumb
→→ →

points the direction of A × B or C " as shown in figure 2.13(b).

Geometrical Interpretation of vector product of two vectors

→→ →→
Consider two vectors A and B are represented in magnitude and direction by OP and OQ and

∠POQ = θ in figure 2.14. Complete a parallelogram OPRQ, join P with Q. Here, OP = A and OQ

→→
= B. Draw QN perpendicular on OP. Magnitude of cross product of A and B is given by

→→ = AB sinθ
|A × B|

= (OP) (OQ sinθ)

= (OP) (NQ) [Since, NQ = OQ sinθ]

= Area of parallelogram OPRQ

Also, → × → = 2[(OP) (NQ)] = 2[Area of ∆OQP] [Fig. 2.14, geometrical representation of
|A B| 2 vector product]

Thus, magnitude of vector product of two vectors

1. is equal to the area of the parallelogram whose two sides are represented by two vectors.

2. is equal to twice the area of a triangle whose two sides are represented by two vectors.

Some special cases

1. When vectors are parallel or antiparallel, then θ = 0° or 180° and sinθ = 0

→→ ^
∴ A × B = AB sin 0° C = 0

It means cross product of two parallel vectors is zero.

2. When vectors are perpendicular, θ = 90° and sinθ = 1

→→ ^^
∴ A × B = AB sin 90° C = AB C .

∧

It means, when vectors are perpendicular then cross product is ABC .

Vectors | 33

Properties of cross product

Some properties of cross product are given as below;

→→ →→
1. Cross product of two vectors is anti-commutative i.e., A × B = – B × A :

→→ →→
Proof: Let, the two vectors A and B be represented by OP and OQ in the plane of paper and

∠POQ = θ.

→→→ ^^
Now, A × B = C = AB sinθC , where C in a unit vector, perpendicular to the plane of

paper directed outwards as shown in figure 2.15(a).

Q

[Fig. 2.15. Properties of cross product of vectors]

→ →→ ^^
Now, B × A = D = BA sinθD , where D is a unit vector perpendicular to the plane of

paper directed inwards as shown in figure 2.15 (b).

^^ ^^
Clearly, C = –D or, D = –C

→→ ^
From, A × B = AB sinθ C

→→ ^ ^ →→
A × B = AB sinθ(–D) = –BAsinθD = –( B × A )

→→ →→
Thus, A × B = – B × A

2. Cross product of two vectors is associative i.e.,

→→ →→ →→→→→→→→
(A + B) × (C + D) = A × C + A × D + B × C + B × D

3. Cross product of two vectors is distributive i.e.,

→ →→ →→→→
A × (B + C) = A × B + A × C

Examples of vector or cross product of two vectors

Some examples of cross product of two vectors are given below:
→

1. Torque τ acting on a particle is equal to the vector or cross product of its position vector
→ → →→→
r and force vector F i.e., τ = r × F
→

2. Angular momentum L of a particle is equal to the vector or cross product of its position
→ → →→→

vector r and its linear momentum P i.e. , L = r × p
→

3. Linear velocity v of particle is rotational motion is equal to the vector product of its
→ → →→→

angular velocity ω and its displacement vector r . i.e., v = ω × r

34 | Essential Physics

→
4. Tangential acceleration a of a particle in rotational motion is equal to its angular

→ → →→→
acceleration vector α and its position vector r i.e. , a = α × r

→
5. Centripetal acceleration ac of a particle is equal to the cross product of its angular velocity

→ → →→→
vector ω and its linear velocity vector v i.e., ac = ω × v .

Boost for Objectives

• A quantity having both magnitude and direction may not be a vector like pressure, surface tension, current

etc. To be a vector, it should follow laws of vector addition as well.

• All potentials, all fluxes and intensity of energy are scalars.

• All flux densities, gradient of all quantities, all field strengths/intensities and all dipole moments are vectors.

• Lami's theorem b
It states that if three coplanar vectors →a , →b and →c acting simultaneously on a

particle are in equilibrium, then magnitude of each vector is proportional to the sine

a bc αγ a
of angle between other two vectors as shown in figure. i.e., sinα = sinβ = sinγ β
• →a + →b = →b + →a (vector addition is commutative)
• →a – →b ≠ →b – →a (vector subtraction is not commutative) c

• A vector can have infinite component vectors.

• Rectangular components of vector can never be greater than the vector itself.

• A vector in plane has two equal rectangular components then that vector is inclined at an angle cos-1  12 or

45° with each axis.

• A vector in space has 3 equal rectangular components then that vector is inclined at cos–1  13 or 54.74°

with each axis.

→ → → →→
• If A + B = A – B , B is zero vector.

→ → → →→ →
• If | A + B | = | A – B |, A and B are perpendicular to each other.

→ → →→ →→
• If | A × B | = A . B , then angle between A and B is π/4

→→ →→
• Angle between A + B and A × B is 90°

• If number of vectors acting simultaneously at a point are such that the largest magnitude is greater than the

sum of other magnitudes then their resultant can never be zero.

• If two vectors of equal magnitude have the magnitude of their resultant equal to either of them, then the angle

between the given vectors is 120°.

→ →→ →→ → →→
• If A + B = R and A + B = R then the angle between A and B is 0°. However, if A + B = R and A – B

→→
= R then angle between A and B is 180°.

•

Short Questions with Answers

1. Should a quantity having magnitude and direction be necessarily a vector? Give examples.
Ans:
[HSEB 2071 S]

No, not only magnitude and direction, but also the quantity must obey the laws of vector addition in
order to be a vector quantity. Hence, a quantity having magnitude and direction both but they are
scalars, since they do not obey the laws of vector addition.

Vectors | 35

2. What is difference between scalar and vector product of two vectors? [HSEB 2062]
Ans:

Scalar Product Vector Product

1. →→ 1. →→

Scalar product of two vectors A and B , Vectors product of two vectors A and B ,

→→ →→ →→ →→

denoted by A . B , is given by, A . B = AB denoted by A × B , is given by A × B = AB

cosθ sinθ n^ where n^ is unit vector which is given by

right hand screw rule.

2. It is also called dot product. 2. It is also cross product.

3. It is a scalar quantity. 3. It is a vector quantity.

4. Scalar product of two vectors follows the 4. Vector product of two vectors doesn't follow the

commutative law. commutative law.

→→ →→ → →→→

A.B = B.A A× B≠B×A

5. Scalar product of two equal vector is square 5. Vectors product of two equal vector is zero.

→→ →→

of the magnitude of either vector i.e., A . A i.e., A × A = 0
= A2

6. Scalar product of two perpendicular vector is 6. Magnitude of vector product of two

zero. perpendicular vectors is equal to the product of

magnitude of each vector.

7. Scalar product of two vector is the product 7. Magnitude of the vector product of two vectors

of magnitude to projection of either vector represents the area of the parallelogram

on another vector and magnitude of the bounded by these vectors.

another vector.

3. Can the resultant of three vectors be zero?
Ans:
Yes, if resultant of any two vectors is equal in magnitude but opposite in direction to the third vector,
4.
Ans: resultant of three vectors will be zero. It is possible only if three vectors are coplanar vectors. If they are
5.
Ans: not coplanar vectors, resultant of two vectors will not be directed opposite to the third vector and they
6.
Ans: cannot cancel to each other.

7. If a vector has zero magnitude, in what sense it is a vector? [Model Question]
Ans:
Since, the sum of two vectors is a vector, then the sum of vectors P and –P is also a vector with zero
8.
Ans: magnitude, called a null vector.

What is a magnitude of component of a vector right angle to itself? [HSEB 2061]

A vector cannot have any component in the direction perpendicular to itself i.e., the magnitude of

component of a vector right angle to itself is zero.

If vector has zero magnitude, is it meaningful to call it a vector? [Model Question, HSEB 2059]

Yes, it is meaningful to call a vector if it has zero magnitude because for the two equal and opposite

vectors, its resultant magnitude will be zero. Also when three vectors are considered, if sum of the two

vectors are equal but opposite to third vector then the resultant vector will be equal to zero.

Can an object with constant acceleration reverse its direction? Explain. [Model Question]

Yes, for example, a ball thrown vertically upward after attaining maximum height falls downward. In

this case, acceleration is always constant acts vertically downwards but the objects at the maximum

height reverse its direction.

Is electric current a vector?

No, it is a usual opinion that since electric current has direction, it must be a vector quantity. In order to

qualify as a vector, a quantity should follow its properties; the current does not follow the law of vector

addition. So, it is not a vector.

36 | Essential Physics

9. The angle between two vectors →A and →B is θ. Find the magnitude and direction of
Ans:
→A × →B and →A . →B . [HSEB 2070]

Let, the angle between two vector →A and →B is θ, then, its cross product is

→A × →B = |→A | |→B | sinθ n^ . Then, magnitude of →A × →B is,

|→A × →B | = |→A | |→B | sinθ

Direction of →A × →B is perpendicular to the direction of plane making by →A and →B .

Again, the dot product is,

→A . →B = |→A | |→B | cosθ

It is scalar product, so, it has only magnitude but not direction.

10. Two equal vectors have a resultant equal to either. At what angle are they inclined to each other?
OR

Under what condition will the sum of two vectors of equal magnitude have magnitude equal to

either vector? OR

Can the sum of two equal vectors be equal to either of the vectors? [HSEB 2061, 2067]

Ans: The resultant 'R' of two vectors 'P' and 'Q' acting at an angle of 'θ' is given by,

R = P2 + Q2 + 2PQ cosθ
or, R2 = P2 + Q2 + 2PQ cosθ

According to the given condition,

R=Q=P
So, P2 = P2 + P2 + 2PP cosθ = 2P2 + 2P2 cosθ
or, P2 = 2P2(1 + cosθ)

or, 1 = 2(1 + cosθ)

or, 1 = 1 + cosθ
2

or, cosθ = 1
–2

or, θ = 120°

Hence, the angle between the vectors should be 120° for the required condition.

11. If →A .→B = 0, what is angle between →A and →B .
Ans: The dot product between →A and →B is given by

→A .→B = A B cosθ. Where θ is angle between →A and →B .
Since →A .→B = 0, then AB cosθ = 0
or, cosθ = 0
or, θ = 90°
So, angle between →A and →B is 90°.
i.e. →A and →B are perpendicular.

Vectors | 37

12. Can you find a vector quantity that has a magnitude of zero but components that are

different from zero? Explain. [HSEB 2070 S]

Ans: No, we cannot find a vector quantity that has a magnitude of zero but components that are

different from zero. Let, →A be a vector inclined at an angle θ with x-axis, then component of →A

along x-axis is A cosθ. If A = 0, then, A cosθ = 0,

13. If B is added to A, under what conditions does the resultant vector have a magnitude equal to

A + B ? Under what condition is the resultant vector equal to zero? [HSEB 2054]

Ans: Let the resultant of the two vectors be C, then according to the question, C = A + B

using the relation from Triangle law of forces,

C = A2 + B2 + 2AB cosθ

Where, 'θ' is the angle between A and B.

A + B = A2 + B2 + 2AB cosθ
(A + B)2 = A2 + B2 + 2AB cosθ
or, A2 + B2 + 2AB = A2 + B2 + 2AB cosθ

` or, 2AB = 2AB cosθ

or, cosθ = 1

∴ θ = 0°.

Thus, if the value of θ is zero, the required condition is applied, i.e. when they are facing towards the

same direction.

If θ = 180° i.e., from C = A2 + B2 + 2AB cosθ we get,
C2 = A2 + B2 + 2AB cos 180°

or, C2 = A2 + B2 – 2AB

or, C2 = (A – B) or (B – A)

→ →→
Thus, magnitude of resultant vector C of the two vectors A and B is zero only if A = B and

angle between them θ = 180°.

14. If →A + →B are non zero vectors, is it possible for →A × →B and →A . →B both to be zero?

Explain. [HSEB 2069]

Ans: No, it is not possible. If →A and →B are two non zero vectors, then their cross & dot products are

→A × →B = AB sinθ n^

If, θ = 0, →A × →B = 0

Similarly, →A . →B = AB Cosθ. Ιf θ = 90°, →A . →B = 0

So, it is not possible for →A × →B and →A . →B both to be zero simultaniously.

→ ∧∧ ∧ → ∧∧ ∧
15. Two vectors are given V 1 = 2 i + 3 j+ 4 k and V 2 = 3 i − 2 j− 4 k which one of the two is

larger in magnitude? Justify your answer. [NEB 2074]

Ans:

Consider a vector → Ax∧i + Ay∧j + Az∧k and magnitude of vector is written as |A| =
A=

→ ∧∧ ∧
given two vectors V 1 = 2 i + 3 j+ 4 k and
Ax2 + Ay2+ Az2 . So that, we have the

→ ∧∧ ∧ → →
V 2 = 3 i − 2 j− 4 k . The magnitude of vector V 1 is V1 = 22 + 32 + 42 = 29 and V 2 is

38 | Essential Physics

( ) ( )V2 = 32 + − 2 2 + − 4 2 = 29 . It shows that the magnitude of both given vectors are

equal.

Numerical Examples

1. If the scalar product of two vectors is equal to the magnitude of their vector product, find the angle

between them . [HSEB 2060, 2068]

Solution:

→→

Let, θ be the angle between A and B is given by,
→→

A .B = AB cosθ →→ . . . (1)

The magnitude of the vector product of A and B is given by,

→→

|A × B | = AB sinθ . . . (2)

→→ →→

According to question, we have, A .B = |A × B |

or, AB cosθ = AB sinθ

or, tanθ = 1 ∴ θ = 45°

Hence, required angle = 45°.

2. The magnitudes of two vectors are 3 and 4, and their product is 6. What is the angle between them?

[HSEB 2063, 2066]

→→

Solution: Let , P and Q are two vectors and their magnitudes are P =3 and Q =4 respectively. The

→→ →→
product of P . Q = 6. We know that, P .Q = PQ cosθ . Since θ be the angle between them.

So that, 6 = 3 × 4 cosθ.

Or, 6 = cosθ ⇒θ = cos−1 1  = 600.
12 2

Hence, angle between the two vectors is 600.

3. Two vectors →A and →B are such that →A – →B = C and A – B = C. Find the angle between them
[HSEB 2064]

Solution:

Suppose, →A and →B are two vectors so that →A – →B = →C and A - B = C. If θ be the angle
between them, then

→C = →A – →B = →A + (-→B ) squaring can both sides we get,

or, C2 = A2 + B2 – 2AB cosθ . . . (1)

Also, we have C = A – B . . . (ii)
or, C2 = A2 + B2 – 2AB

Comparing equation (1) and (2), we get

cosθ = 1 ⇒θ = cos–1 (1) = 0°

Thus, the angle between two vectors →A and →B is 0°.

4. →C is the vector sum of →A and →B i.e. →C = →A + →B for C = A + B to be true, what is the angle

between →A and →B ? [HSEB 2067]

Vectors | 39

Solution:

Suppose →A and →B are two vectors so that →A + →B = →C and A + B = C. If θ be the angle
between them, then

→C = →A + →B . . . (1)
⇒ C2 = A2 + B2 + 2AB cosθ

Also, C = A + B . . . (2)
⇒ C2 = A2 + B2 + 2AB

Comparing Equation (1) and (2), we get
cosθ = 1 ⇒ θ = cos–1 (1) = 0°

Thus, the angle between two vectors →A and →B is 0°.

5. If ^i , ^j and ^k are unit vectors along x, y and z-axis respectively find ^i . (^j × ^k ). [HSEB 2070]

Solution:

Let, ^i , ^j and k^ be the three unit vectors along x, y and z-axis respectively then,

^i . (^j × k^ ) = ^i .(|^j |. |k^ | sin 90 x^ )

= ^i . (1.1.1. x^ ) = ^i .x^

= ^i . ^i = |^i |. |^i | cos 0 = 1.1.1 = 1.

Thus, ^i . (^j × k^ ) is 1.

[where, x^ has the direction of perpendicular to both ^j and k^ or plane made by ^j × k^ which is ^i ]

6. What does→A .→A , the scalar product of a vector with itself gives? What about →A × →A the

vector product of a vector with itself? [2070 S]

Solution:

We have,→A .→A = |→A |. |→A | cosθ = A.A. cos0 (since, θ = 0, angle between →A and →A ] = A2

Also, →A × →A = |→A | × |→A | sin θ n^ (Since, n^ = unit vector)

= AA sin 0n^ (Since, θ = 0 angle between →A and →A ) = 0

∴ →A . →A = A2 , i.e. the scalar product of a vector itself give the square of its magnitude and

→A × →A = 0, i.e. the vector product of a vector itself is zero.

7. A force (in Newton) expressed in vector notation as →F = 2^i + ^j - 3^k is applied on a body so

that the displacement produced in meter is given by →D = ^i - 2^j - 3^k . Express the result and

nature of the work done. [HSEB 2071]

Solution:

Here,→F = 2^i + ^j - 3k^ and →D = ^i - 2^j - 3k^

Then, the work done is

W = →F .→D = (2^i + ^j - 3k^ ). (^i - 2^j - 3k^ ) = 2.1 - 2.1 - 3 × (-3)

= 9J. This is work done which is scalar due to its magnitude only.

8. Given two vectors →A = 4.00 ^i + 3.00^j and →B = 5.00 ^i - 2.00^j . Find the magnitude of each

vector. [HSEB 2072]

Solution:

Given, →A = 4.00 ^i + 3.00^j and →B = 5.00 ^i - 2.00^j .

40 | Essential Physics

According to question, |→A | = ?, |→B | = ? |→B | = (Coeff of ^i )2 + (Coeff of ^j )2
Now, we have the relation
|→A | = (Coeff of ^i )2 + (Coeff of ^j )2,

= 42 + 32 = 52 + (-2)2

= 16 + 9 = 25 + 4

= 5 = 29

Therefore, magnitude of →A is 5 and →B is 29 .

9. A vector →F = ^i + 2^j - 3^k is given. What is the magnitude of the y component of the vector?

[HSEB 2072]

Solution:
Given, vector →F = ^i + 2^j - 3k^

Comparing with a vector with x, y and z- components as,

→F = a x^ + by^ + c^z . Where a, b and c are magnitude of x, y and z components. So, b = 2

or, |→F y| = 22 = 2
Therefore, y-component is 2.

Important Numerical Problems

1. A force (in Newton) is expressed in vector notation as →F = 4^i + 7^j – 3^k is applied on a body
and produces a displacement (in meter). →D = 3^i – 2^j – 5^k in 4 seconds. Estimate the power.

Solution:

Given, →F = 4^i + 7^j – 3k^
Displacement (→D ) = 3^i - 2^j - 5k^
Time taken (t) = 4 sec
Work done (W) = →F × →D = (4^i + 7^j - 3k^ )× (3^i -2^j - 5k^ ) = 12 - 14 + 15 = 13

W 13
Power (P) = t = 4 = 3.35 watt.

Hence, required power P = 3.35 watt

^^ ^ ^^ ^

2. Find the angle between the two vectors, A = 2i + 3j + 4k , B = i – 2j + 3k

Solution:

→ ^ ^ ^ → ^^ ^

We have, the vector, A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk

Then, on comparing the given vector with this vector

Ax = 2, Bx = 1

Ay = 3, By = –2
Az = 4, Bz = 3

The scalar product is given by the equation

→→

A . B = AB cosθ

Vectors | 41

→→

A .B →→ ^^ ^ ^^ ^

Then, cosθ = |A| |B| , where A . B = (Axi + Ayj + Azk ) . (Bxi + Byj + Bzk )

= AxBx + AyBy + AzBz
= 2 × 1 + 3 × (–2) + 4 × 3 = 8

→→

Magnitude of A = |A | = 22 + 32 + 42 = 29

→→

Magnitude of B = |B | = 12 + (–2)2 + 32 = 14

Therefore, cosθ = 8 = 0.397
29 14

∴ θ = 66.6°. Hence, angle between the two vectors is 66.6°.
3. A boat is heading east at 4ms–1 in a river flowing south at 1ms–1. What is the velocity of the boat

relative to the earth?

Solution: VB = 4 ms–1
Given, Velocity of boat due east (VB) = 4ms–1 β
Velocity of river due south (VR) = 1ms–1

Let, β be the angle made by VB with resultant velocity (i.e., velocity of

boat relative to earth) R.

θ is the angle between boat and river, θ = 90°.

Now, using parallelogram law of vector addition we get, R

R = VB2 + VR2 + 2VBVR cos 90° = 42 + 12 + 0 = 17 VR = 1 ms–1
∴ R = 4.1 ms–1

or, tanβ = VR
VB

or, β = tan–1 41

∴ β = 14° i.e., 14° south of east.

4. A ship is travelling due east at 30km/hr and a boy runs across the deck in a south west direction at

10km/hr. Find the velocity of the boy relative to sea.

Solution:
Here, A ship is travelling due east with a velocity (Vs) = 30kmhr–1
A boy on the ship runs south west with a velocity (VB) = 10km/hr
Let, R be the resultant velocity i.e., Velocity relative to sea (R) = ?

Using parallelogram law of vector addition,

R = VS2 + VB2 + 2VSVB cos(90 + 45)° = 12 = 23.99 = 24 km hr–1
900 + 100 – 2 × 10 × 30 ×

→→

Let, θ be the angle made by Vs with R

tanθ = 10 sinβ = 30 10 sin 135°
30 + 20cosβ + 10 cos 135°

∴ θ = 17° i.e. 17° south of east.

Practice Short Questions

1. What is the difference between scalar and vector products of two vectors? Explain.
2. Can the sum of two equal vectors be equal to either of the vectors? Explain.
3. If the scalar product of two vectors is equal to the magnitude of their vector product, find the angle between

them.
4. If B is added to A, under what condition does the resultant vector have a magnitude equal to A + B ? Under

what conditions is the resultant vector equal to zero?

42 | Essential Physics

5. The magnitude of two vectors are 3 and 4 and their product is 6. What is the angle between them?
6. If →A .→B = 0, what is angle between →A and →B ?
7. →C is the vector sum of →A and →B i.e., →C = →A + →B for C = A + B to be true, what is the angle between →A and

→B ?
8. If a vector has zero magnitude, is it meaning full to call it vector?
9. If →A and →B are non zero vectors. Is it possible for →A × →B and →A .→B both to be zero? Explain.
10. If ^i,^j and k^ are unit vectors along x, y and z - axis respectively. Find ^i.(^j × k^)
11. What does→A .→A , the scalar product of a vector with itself gives? What about →A × →A , the vector product of a

vector with itself?
12. If a physical quantity having magnitude and direction necessarily a vector quantity? Explain.

Practice Long Questions

1. What is the resolution of a vector? How will you resolve a vector in a plane in two mutually perpendicular

directions?

2. State triangle law of vector addition obtains an expression for the resultant of two vectors P and Q inclined at

an angle θ. [HSEB 2066]

3. State and explain the parallelogram law of vector addition. Derive the expressions for the magnitude and

direction of the resultant of two vectors inclined at an angle θ from each other.

[HSEB 2055, 2069 2070 S]

4. Explain the geometrical interpretation of the scalar product of two vectors. How will you add two vectors and

more than two vectors graphically? What do you understand by resultant of vectors?

Practice Numerical Questions

1. A force (In Newton) expressed in vector notation as →F = 2^i + ^j – 3k^ is applied on a body so that the

displacement produced in meter is given by →D = ^i – 2^j – 3k^. Express the result and nature of work done.

[Ans: 9J]

2. A force (in Newton) expressed in vector notation us →F = 4^i + 7^j – 3k^ is applied on a body and produces a

displacement (in meter), →D = 3^i – 2^j – 5k^ in 4 seconds. Estimate the power. [Ans: 3.25 w]

3. Calculate the area of the parallelogram when adjacent sides are given by the vectors →a = ^i + 2 ^j + 3k^ and

→ 2^i – 3^j + k^. [Ans: 13.96 sq. units]
b=

4. The velocity of 20m/sec has its x-component 12n/sec. What is its y component? Find the angle at which the

velocity is inclined with the x-axis. [Ans: 16m/sec, 53°]

5. Two equal forces act at a point. The square of their resultant is 3 times their product. Find the angle between

them. [Ans: 60°]

MBBS, B.E., B.Sc. Nursing Entrance Preparation Questions

1. Which of the following is a scalar quantity? b. Electrostatic potential
a. Electric field d. Torque
c. Angular momentum
b. Electrostatic potential
2. Which is a vector quantity? d. Electric flux
a. Electric charge
c. Current density b. Magnetic intensity

3. Which is scalar quantity?
a. Intensity of magnetization

c. Intensity of radiation d. Magnetic Induction

4. Two forces, each of magnitude F have a resultant of the same magnitude F. The angle between two forces
is

a. 45° b. 120° c. 150° d. 60°

Vectors | 43

5. The sum of two forces acting at a point is 16 N . If the resultant forces is 8 N and its direction is
perpendiculars to minimum force, then the forces are

a. 6N and 10N b. 8N and 8N c. 4N and 12N d. 2N and 14N

→→ → →→→
6. The magnitude of vectorsA , B and C are respectively 3,4, and 5 units. If A + B = C , the angle between

→→
A and B is

a. π b. cos-1 (0.6. ( )c. tan-1 7 ( )π
2 5
d. 4

→→ →
7. Two vectors P + Q have their sum equal to P and perpendicular to P

(i) Magnitude of Q is

a. P b. 2 P c. 3 P d. 2 P

→→

ii) The angle between P and Q is

a. 120° b. 60° c. 135° d. 145°

8. A vector of length λ is turned through the angle θ about its tail. What is the change in the position vector of
its head?

a. λ cos θ b. 2λsin θ c. 2λcos θ d. λ sin θ
2 2 2 2

9. If the sum of two unit vector is unit vector, then magnitude of difference of these two unit vector is

a. 2 b. 3 c. 1 d. 5
2

→→ →→ →→
10. If | P + Q | = | P - Q |, then the vectors P and Q are

a. parallel to each other b. inclined at 45° to each other

c. Perpendiculars to each other d. inclined at 60° to each other

11. The maximum no. of components at a vectors can be split are

a. 2 b. 3 c. 4 d. infinite

^^ ^ ^^ ^

12. The angle between A = 2i + 3j + 4k and B = i – 2j + 3k is

a. 0° b. 90° c. 66.6° d. 100°

→→

13. Which of the following is perpendicular to give vector i × j ?

→ →→ →

a. j b. i - j c. k d. both (b) and (c)

→→

14. Which of the following is perpendicular to (A cos θ) i +(A sin θ) j

→→ →→

a. (B cos θ) i + (B sin θ) j b. (B cos θ) i + (B cos θ) j

→→ →→

c. (B sinθ) i – (B cos θ) j d. (B cos θ) i - (B cos θ) j

→→→ → →→ →→ →
15. A particle moves from position (3 i +2 j - 6k) to (14 i +13 j +9k) due to uniform force at (4 i + j +3k)N. If

the displacement is in meter then work done will be

a. 100J b. 200J c. 300J d. 250J

→ →→

16. The position vector of a particle is r =( a cost wt) i +( a sin wt) j the velocity of the particle is

a. Parallel to position vector b. Perpendicular to the position vector

c. Directed towards the origin d. Directed away from the origin

→→ →→
17. The vectors a and b act towards east and vertically upward respectively, then a × b acts

a. due south b. due north

c. vertically downward d. due west

44 | Essential Physics

→→ → →→
18. The angle between the vectors A and B is θ. The value of the triple product A . (B × A) is

a. A2B b. zero c. A2B sinθ d. A2B cos θ

→→ →→

19. The value of ( A + B)× ( A – B) is

a. zero b. A2 –B2 →→ →→

c. B × A d. 2 ×( B × A)

→ →→→ →→ →
20. The torque at the force F = (2 i – 3 j + 4k)N acting at the point r = (3 i – 2 j + 3k)m about the origin be

→→ → →→ →

a. (6 i – 6 j + 12k) b. (17 i – 6 j - 13k)

→→ → →→ →

c. (-6 i – 6 j + 12k) d. (17 i – 6 j + 13k)

21. Two vectors have a sum A and a difference B. If A=B, then angle between the two vector is

a. 0° b. 45° c. 120° d. 90°

22. Resultant of two forces F1 and F2 is F1 and the resultant is at right angle to the force F1 then the force F2
is equal to

a. 2F1 b. 0 c. F1 d. 2 F1

→→→ →→ →
23. If A , B, C have magnitude 6, 8 and 10 respectively and ( A + B) = C. The angle between A and B is

a. 0° b. 90° c. 45° d. 180°

24. Three vectors are arranged to form a right angled triangle of sides 5, 12 and 13 units. The sum two vectors

is equal to the third. The angle between those of magnitudes 12 and 13 will be

a. sin-1 (12/13) b. cos–1(13/13) c. cos–1(5/13) d. tan–1(5/13)

25. Magnetic moment is

a. Scalar quantity b. a vector quantity c. universal constant d. tensor

26. The dot product of vector is 3 times the magnitude of the cross product. The angle between these vectors
will be

a. 30° b. 45° c. 60° d. 75°

→→ →→

27. Two diagonals of a parallelograms are (2 i +2 j ) and (2 i –2 j ) on the area at the parallelogram will be

a. 8 cm2 b. 4 2 cm2 c. 4cm2 d. 2cm2

28. Which sets of forces acting on body never produces zero acceleration?

a. 10N, 10N, 20N b. 6N, 8N, 20N c. 7N, 12N, 20N d. 1N, 2N, 3N

29. A boat which has a speed of 5 km/hr crosses a river of width 1km along the shortest path in 154 min.
Velocity of river in km/hr is

a. 1 b. 3 c. 4 d. 4

Answer Key

1. b 2. b 3. c 4. b 5.a 6.a 7.(i)b(ii)c 8. b 9. b 10. c 11. d 12. c
13.d 14.c 15. a 16.b 17. a 18. b 19. d 20. b 21.d 22.d 23.b 24. b
25.b 26. a 27. c 28. c 29. b

Ch apter Kinematics

3

Teaching Manual Physics Grade –XI , Higher Secondary Education Board
Curriculum And Training Division Sanothimi, Bhaktapur.

Syllabus:

Kinematics- Uniform and non-uniform motion; average velocity and acceleration,
Instantaneous velocity and acceleration; Equation of motion (graphical treatment);
Motion of a freely falling body; Relative velocity; Projectile motion

Objectives:

The objectives of this sub unit are to make the student be able to differentiate uniform
and non uniform motions, calculate distance, velocity and acceleration from distance -
time and velocity - time graph, and describe the motion of projectiles.

Activities (micro syllabus):

1. Review the terms displacement, distance, velocity, speed, acceleration and
decelerations.

2. Discuss graphical representation of velocity and acceleration.
3. Write equations of motions for uniform acceleration and freely falling body.
4. Explain the concept of the relative velocity.
5. Define a projectile and give its example.
6. Obtain the mathematical expression for the time of flight, the maximum

height, and the horizontal range of a projectile.
7. Numerical problems: Focused numerical problems given:

(a) An exercises in the University Physics (Ref.1) and
(b) In Advanced Level Physics (Ref. 2) and also in Physics for XI (Ref. 2).

46 | Essential Physics

3.1 Introduction

Mechanics deals with the theory of motion (state of rest or moving) of material objects which is described
in two classes. They are statics and dynamics. The branch of mechanics which deals with study of objects
at rest is called statics but a branch of mechanics which deals with study of objects in motion is called
dynamics. The study of motion also divided in two classes. They are

1. Kinematics: Kinematics is the term which study of the motion of objects using words, equations,
graphs diagrams and numbers. Kinematics is a study aimed at explaining the motion of objects.
So, kinematics deals with the study of the motion of objects without taking into the account of the
cause of the motion in the objects.

2. Dynamics: the study of motion of objects by taking into account their causes of motion is called
dynamics. Various aspects of motion are described with the help of displacement, velocity,
acceleration and time. Both the velocity and the acceleration are intimately connected with the
displacement. The terms related with motion are described as below:

Note: A particle is said to be in motion, if it changes its position w.r.t. its surroundings. On the other hand, if the
particle does not change its position w.r.t. its surroundings, then the particle is said to be at rest. The rest and
motion are relative terms. The rate of change of position of a particle in a particular direction is called its
velocity. A body at rest has zero velocity, whereas all moving particles possess velocity .The rate of change of
velocity of a particle is called acceleration. A moving particle may or may not possess acceleration.

3.2 Displacement

Suppose, a particle is moving from point A to point B. The path
followed by the particle in going from A to B may be of any shape
curved as shown in figure 3.1 (a) or a series of a straight path as
shown in figure 3.1 (b). In fact, the particle can go from point A to B
along infinite number of paths in each case. The length of the actual
path between the initial and final positions of the particle is called the
distance covered by it. In figure 3.1 (a), the distance covered by the
particle is the length of curved path ACB, while in figure 3.1 (b), the

sum of the lengths of series of straight paths between A and B
i.e., (AL + LM + MB) is the distance covered by the particle. [Fig. 3.1, Representation of displacement]
It is a scalar as it has got magnitude only. A vector drawn
from initial position to the final position is called linear displacement or simply displacement.

→
Therefore, in figure 3.1 (a) and 3.1 (b), vector AB is the displacement of the particle in going from A to
B. The arrowhead at B shows that particle is displaced from A to B. If the particle moves from B to A, the

→
magnitude of displacement is same but the direction will be opposite i.e., the displacement will be BA (a
vector drawn from B to A). Therefore, we conclude that
1. Displacement of a particle between two points tells not only the shortest distance between the two

points but also about the direction of motion.
2. Displacement of a particle between two points is a unique path which can take the particle from its

initial to final position. To explain it, let us consider that a man goes from Kathmandu to Pokhara.
The distance between the two stations is 200 km. By travelling a distance of 1500 km from
Kathmandu in any direction, he cannot reach Pokhara. It will be possible for him only if he moves
a distance of 200 km and in a particular direction.
Thus, displacement of a particle is defined as the change in position of the particle in a particular
direction. It is a vector as it possesses both magnitude and direction and is independent of the
actual path that may have been followed by the body, when it moves from one point to another.

Kinematics | 47

Note: Displacement is directed distance between initial and final position and the length is never less than the
magnitude of displacement for the motion of same object in same time. Displacement is vector quantity but
distance is a scalar quantity.

3.3 Velocity and Speed

Velocity: The time rate of change of displacement is called velocity. Since, displacement is a vector and
time is a scalar so that, velocity is a vector. In S.I., its unit is ms–1 and dimensional formula is [M0LT–1].

Speed: The rate of covering the distance by a particle is called its speed. It is a scalar quantity, Its unit
and dimensional formula is same as that of velocity. It is evident that velocity is the speed of a body in a
particular direction.

Note: If a body moves with a variable velocity i.e. body covers unequal distance in equal interval of time, the
motion is said to be non-uniform motion. But if equal distance are covered in equal time intervals during a
motion along a straight line, the motion is called uniform motion.

3.4 Average, Instantaneous, Uniform and Variable Velocity in
One Dimensional Motion

The motion of a particle is called one dimensional,
if it moves along a straight line. It is also called
rectilinear motion. The various kinds of velocities in
one dimensional motion are explained as below.

Average Velocity

Consider, a particle moving along X–axis. Suppose,

the particle is at points A and B at times t1 and t2 [Fig. 3.2, representation of velocity]
such that the distances from the starting point O are

OA = x1 and OB = x2 respectively as shown in figure 3.2. Then, average velocity of the particle in interval

between t1 and t2 (or between points A and B) is given by

vav = x2 – x1 . . . (1)
t2 – t1

The average velocity of a particle does not tell anything about the nature of motion between the two

points during the time interval (t2 – t1). The motion of the particle may be uniform or variable. If a body

covers a distance S in time t, then

S
vav = t

∴ S = vav. t . . . (2)
Therefore, distance covered = average velocity × time

Instantaneous Velocity

The velocity of a particle at a particular point of its path or at a particular instant of time is called
instantaneous velocity and is defined as the limiting value of the average velocity of the particle in a small
time interval around that instant, when the time interval approaches zero. Mathematically, it is

limit dS . . . (3)

∆t → 0 , V = dt

Uniform Velocity

A particle is said to be moving with uniform velocity, if it undergoes equal displacement in equal
intervals of time, however small these intervals may be

distance covered = uniform velocity × time . . . (4)

48 | Essential Physics

If a body moves with uniform velocity, then
i. Its direction of motion as well as magnitude of its velocity (speed) will always remain same.

Obviously, the motion of particle will be along a straight line.
ii. Instantaneous velocity at all the instants of time will be same as the uniform velocity.
iii. Velocity time graph will be a straight line parallel to time axis.

Variable Velocity

A particle is said to be moving with variable velocity, if either its speed or its direction of motion or both
change with time. Therefore, for a particle moving with variable or non–uniform velocity,
i. Either speed or direction of motion or both will change with time. Accordingly, particle may move

along a straight line or curved path i.e. in one, two or three dimensions.
ii. The instantaneous velocity of the particle at different instants of time during a time interval will

not be equal to the average velocity in that time interval.
iii. The velocity time graph may or may not be a straight line but it will not be parallel to time axis.

Note: (1) Direction of velocity is equal to speed (2) Average velocity is equal to instantaneous velocity. (3) The
speed of object can never be less than the magnitude of velocity (4) Speed is a scalar and velocity is vector.

3.5 Acceleration

The time rate of change of velocity of body is called acceleration. Acceleration is a vector quantity. Its
unit in SI is ms–2 and its dimensional formula is [M0LT–2]. If the velocity of a particle increases, the body
is said to be moving with positive acceleration and on the other hand, if velocity decreases it is said to
possess negative acceleration or retardation.

3.6 Average, Instantaneous, Uniform and Variable Acceleration

in One Dimension

Average acceleration

Consider, a particle moving along X–axis as shown in

figure 3.3. Suppose the velocity of the particle at time t1 [Fig. 3.3 Average acceleration]
and t2 becomes v1 and v2 respectively. Then, average
acceleration of the particle is given by

aav = v2 – v1 . . . (1)
t2 – t1

The acceleration of the particle during the time interval (t2 – t1) may be uniform or variable. Like average

velocity, average acceleration also does not tell about the nature of motion between the two points and

hence, we define instantaneous acceleration.

Instantaneous Acceleration

If the particle moves with variable acceleration, one may need to know its acceleration at any instant of

time or any point of its path and it is called instantaneous acceleration. It is defined as the limiting value

of the average acceleration of the particle in a small time interval around that instant, when the time

interval approaches zero. Mathematically, it is denoted by

limit ∆v dv . . . (2)
∆t → 0 a = ∆t = dt

Uniform Acceleration

A body is said to be moving with uniform acceleration, if its velocity changes by equal amounts in equal
intervals of time. However, small the intervals may be, if particle has uniform acceleration, then
i. The magnitude and direction of acceleration will always be the same and hence particle will move

along a straight line (one dimensional motion).
ii. The instantaneous acceleration at all the instants of time will be the same as the uniform

acceleration.

Kinematics | 49

iii. The velocity time graph will be a straight line sloping upwards. In case the acceleration is negative
(retardation), the straight line graph will slope downward. The slope of the graph will be equal to
the acceleration.

Variable Acceleration

A body is said to be moving with variable or non–uniform acceleration, if either the magnitude of the
acceleration or its direction or both change with time.The motion of a particle moving with variable
acceleration
i. May be in one, two or three dimensions.
ii. The velocity time graph is not a straight line.

Note : Velocity changes at constant rate and acceleration is constant.

3.7 Graphical Representation of Uniform Motion and Uniformly
Accelerated Motion

1. When a body moves with uniform velocity, then the velocity time graph will be straight line
parallel to time axis as shown in figure 3.4 (a)

vv v

t tt

[Fig 3.4, Graphical representation of motion with time]

2. When a body moves with positive uniform acceleration, then the velocity time graph will be a
straight line having positive slope i.e., the straight line will make an acute angle with the time axis
as shown in figure 3.4 (b).

3. When a body moves with negative uniform acceleration, then the velocity time graph will be a
straight line having negative slope i.e., the straight line which make an obtuse angle with the time
axis as shown in figure 3.4 (c).

3.8 Equations of Motion in a Straight Line

When a body moves along a straight line with uniform acceleration, the relation between velocity,
acceleration, time and the distance covered can be deduced both analytically and graphically as below:

1. Distance Covered by a Body Moving with Uniform Velocity

Analytical treatment: If a body moves with uniform velocity u for a
time t, then as said earlier distance covered = uniform velocity × time

∴ S = ut . . . (1)

Graphical treatment : The velocity – time graph for a body moving

with uniform velocity u for a time t is a straight line AB parallel to [Fig 3.5, time velocity graph]
time axis such that OA = u and OM = t as shown as figure 3.5.
Therefore, distance covered is equal to area OABM

i.e., S = OA × OM

or, S = ut

2. Velocity of a Uniformly Accelerated Body after time t

Analytical treatment: Let, a body having in initial velocity u moves for a time t with uniform
acceleration a. If v is the final velocity of the particle, then

50 | Essential Physics

change in velocity
acceleration = time taken

v–u
or, a = t

or, v = u + at . . . (1)

Graphical treatment: The velocity–time graph for a body having

initial velocity u and moving with positive uniform acceleration a

for a time t is a straight line AB such that OA = u, BM = v and

OM = t as shown in figure 3.6. Draw AL perpendicular to BM.

The acceleration of the motion of body is equal to slope of the

velocity–time graph v

i.e., acceleration = slope of line AB [Fig. 3.6, time velocity graph]

BL
or, a = AL

Now, BL = BM – LM = BM – OA

or, BL = v – u

Also, AL = OM = t

∴ v–u
a= t

or, v = u + at . . . (2)

3. Distance Covered by a Uniformly Accelerated Body in time t

Analytical treatment: Let, a body having initial velocity 'u' move for a time 't' with uniform

acceleration a. Let v be the final velocity of the body and S be the distance covered by it.

Then, distance covered = average velocity × time

u+v
or, S = 2 × t , but v = u + at

u + u + at
or, S = 2 × t

∴ S = ut + 1 at2 . . . (1)
2

Graphical treatment: Let us again consider velocity–time graph for the uniformly accelerated

motion shown in figure 3.6. Then, distance covered is equal to area of the trapezium OABM

i.e., S = area OALM + area ABL

Now, area OALM = OA × OM = ut and

Area ABL = 1 BL × AL = 1 BL AL2
2 2 AL

BL
But, AL = slope of velocity time graph a = acceleration of the body.

Therefore, area ABL = 1 at2
2

Hence (s) = ut + 1 at2
2

4. Velocity of Uniformly Accelerated Body after Covering a Distance S

Analytical treatment: Consider a body moving with initial velocity u and uniform acceleration 'a'
for time t. Let, v be the final velocity of the body and S is the distance covered by it in time t.
Then, Squaring both sides we get,

v2 = (u + at)2 = u2 + 2u at + a2t2