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Semester 1.

Lecturer : Catherine Walsh Room 326

email :

Lectures : 3 hours per week

Tutorial : 1 hour per week

Practical : 2 hours per week


1. INTRODUCTION. S.I. units, prefixes, scientific notation. Vectors and scalars.

2. MECHANICS. Motion in one direction, Newton’s laws of motion. Work, Energy and Power. Uniform circular motion. Conservation laws, mechanical energy, total energy momentum.

3. PROPERTIES OF MATTER. Structure of matter. Pressure. Elasticity, viscosity, surface tension.

4. WAVE MOTION AND VIBRATIONS. Wave classification and properties.

Sound waves, characteristics of sound. Doppler effect.

For this module no previous knowledge of physics is assumed.

This module will be delivered by a combination of lectures, tutorials and practical work.

Lectures will be used to present new topics and their related concepts. Problem solving will be used in lectures and tutorials to analyse physical situations and find numerical solutions.

There is also an integrated practical programme designed to run in parallel with the lectures and the develop experimental skills.

Lectures. Your attendance is expected at all lectures. You are responsible for all material covered in lectures as well as any problem assignments given.

Lectures will cover ideas, concepts, and sample problems. The lectures will facilitate the acquisition of factual knowledge the application of the knowledge by problem solving.

You will be required to have a calculator at physics lectures.

Tutorials. The class will be divided into smaller groups for a one hour tutorial every week.

The purpose of the tutorial is to facilitate the learning process by providing more individual attention to science students to help them learn and understand physics concepts and problem-solving skills.

Tutorials offer the time for feedback and discussion. Again you will require a calculator.

Practicals : Each student will have a two hour laboratory session every week. You will be required to have two note books for practical work.

1. A direct log book, this note book should be used as a work book during each practical session for recording data etc. This note book should also be signed by your lab supervisor at the each of each practical.

2. A report book for the formal writing of the practical you have carried out. Your lab supervisor will make arrangement for the collection of these report books.

You are required to attend ALL practicals.

If you should miss more than one session per term you should contact your lab supervisor immediately.

The final mark for this subject will be derived from a combination of assignment, laboratory mark and final exam.

Final Exam 50%

Continuous Assessment 50%

Physics is a fundamental science concerned with the interactions of matter and energy, time and space.

Physics is concerned with how things behave and in discovering the general principles which explain natural phenomena.

This course is designed to impart an understanding of the basic physics principles to enable a student to solve a variety of problems and to develop laboratory skills.


All measurement involves a number and a unit. Scientists use the Systeme International ( S. I. ) units which was recommended by the general conference on Weights and Measures in 1960. In physics there are six basic units

Quantity Symbol Unit Symbol

Length l metre m

Mass m kilogram kg

Time t second s

Electric I Ampere A


Temperature T Kelvin K

Luminous I Candela cd



Metre : 1 metre = 1,650,763.73 wavelengths of

Light emitted by a Krypton-86 Atom.

Kilogram : 1 kilogram = mass of international

Standard platinum-iridium cylinder kept at the

International bureau of weights And measures in Serves, France.

Second : 1 second = 9,192,631,770 periods of vibration of the radiation emitted by a caesium-133 atom.

The other three units we will define as we meet them later in the course.


Derived units can be constructed in terms of the independent base units.

The derived units can be written in terms of the basic units and are derived from the definition of the quantity.




In each case length, width, radius , base etc the basic S.I. unit is meter (m)

Therefore the S.I. unit of area is m2

Quantity Symbol Unit Symbol

Area A metre squared m2

Volume V metre cubed m3

Velocity v metre per second m.s-1

Acceleration a metre per second m.s-2


Density d kilogram per kg.m-3

metre cubed

There are other frequently used quantities which are given individually named units.

In each case the unit name honour famous scientists for their work in the fields which use these units.

Quantity Symbol Unit Symbol

Force F Newton N

Pressure P Pascal Pa

Energy,Work E,W Joule J

Power P Watt W

Each of these units can be written in terms of the base units


Newton = kg m s-2

Pascal = kg m-1 s-2

Question 1. Given the following equations derive the S.I. units for each of the following quantities


Question 2. Given that


Determine the unit of Force, Work and Power in terms of the basic S.I. units.


In Physics we use scientific notation to express the numerical value for a physical quantity. Using the POWERS OF TEN

100 = 1

101 = 10

102 = 101 x 101 = 10 x 10 = 100

103 = 101 x 101 x 101 = 10 x 10 x 10 = 1000


10m x 10n = 10 m + n

100 = 1

10-1 = 1 / 10 = 100 / 101

10-2 = 1 / 100 = 100 / 102

10-3 = 1 / 1000 = 100 / 103


10a / 10b = 10 ( a – b )



190 m = 1.9 x 102 m

Rule 4

For every place the decimal point moves to the right SUBTRACT one from the power of 10

Rule 5

For every place the decimal point moves to the left ADD one from the power of 10

Rule 5

10n = 1 x 10n


When entering a number in scientific notation into a calculator it is important that you follow the correct procedure

Pressing the exponential EXP button on the calculator is the same as

Multiplied by 10 to the power of 10


103 is entered into a calculator as

1 EXP 3

10-3 is entered as

1 EXP +/- 3

5.3 x 10-7

5 . 3 EXP +/- 7

NOTE 1: You do not have to enter the 10 into the calculator the EXP button takes care of that

NOTE 2: 5.3 x 10-7 will be displayed on the calculator as

5.3 –07

Always remember to write this down as

8 x 10-7

as 5.3 –07 written down on a sheet of paper means 5.3 to the power of –7 NOT

5.3 by ten to the power of -7

always remember when transferring this back down to a sheet of paper it is

5.3 x 10-7

Question 1. Calculate each of the following values

5 x 106 (2x 103) (3x 103)


______5 x 10 4

(4 x 106) (5x 10-3)


(8 x 10 -4 )(5x 103)

Question 2. Write the following numbers in scientific notation.

1. 1001 2. 53 3. 6,926,300,000 4. -392 5. 0.00361 6. 0.13592 7. -0.0038 8. 0.00000013

Question 3. Calculate each of the following values.

1. 102 x 105 2. 1015 x 1039 3.1016 x 10-13 4. 10-8 x 102 5. 10-8 x 10-5

For quantities very much larger or smaller than the standard unit multiples or sub-multiples are used

Multiple Prefix Symbol

109 giga- G

106 mega- M

103 kilo- k

10-2 centi- c

10-3 milli- m

10-6 micro- μ

10-9 nano- n

10-12 pico- p

1 cm = 1 x 10-2 m

1 mm = 1 x 10-3 m

1 km = 1 x 103 m

1 ms = 1 x 10-3 s

1 μs = 1 x 10-6 s

Question 1. Use prefixes to express the following

(i) 0.000001 m (ii) 3 x 10-9 m

(iii) 4560000 m (iv)9 x 109 N

Question 2. Express the following quantities using standard notation

i) 2.5 pm (ii) 3.6 μs (iii) 500 GN

(iv) 589 nm (v)1.4 MW


S.I. unit of area is m2 however area is often given in either cm2 or mm2 units. We need to relate these to m2.

1 cm2 = 1 cm x 1 cm

= 1 x 10-2 m x 1 x 10-2 m

= 1 x 1 x 10-2 x 10-2 x m x m

1 cm2 = 1 x 10-4 m 2


1 mm2 = 1 mm x 1 mm

= 1 x 10-3 m x 1 x 10-3 m

= 1 x 1 x 10-3 x 10-3 x m x m

= 1 x 10-6 m2


1 cm3 = 1 cm x 1 cm x 1 cm

= 1 x 10-2 m x 1 x 10-2 m x 1 x 10-2m

= 1 x 1 x 1 x 10-2 x 10-2 x 10-2 x m.m.m

= 1 x 10-6 m3

1 mm3 = 1 mm x 1 mm x 1 mm

= 1 x 10-3 m x 1x 10-3m x 1 x 10-3m

= 1 x 1 x 1 x 10-3 x 10-3 x 10-3 x m.m.m.

1 mm3 = 1 x 10-9 m3

Question 1. Convert each of the following into correct S.I. units

i) an area of 500 cm2

ii) an area of 1600 mm2

iii) the area of a circle of diameter 5cm

iv) the area of a square of length 155mm

v) a volume of 6578 cm3

vi) a volume of 442 mm3

vii) the volume of a sphere of radius 23cm

viii) the volume of a cylinder of diameter 7cm and height 15.54cm.

Orders of Magnitude

The following tables give some feeling of the "typical" numbers that are associated with each measured quantity. You need not worry about the details of each number - just the orders of magnitude.

|Object or Distance |  |Size( m ) |

|Universe |  |1026 |

|Milky Way Galaxy |  |1021 |

|Nearest Star |  |1016 |

| Solar System |  |1012 |

|The Sun |  |109 |

|The Earth |  |106 |

|A Mountain |  |103 |

|Humans |  |100 |

|A Cell |  |10-5 |

|An Atom |  |10-9 |

|The Nucleus |  |10-14 |


|Duration/Age |  |Time |

| | |(s) |

|Age of The Solar System |  |1017 |

|Last Ice Age |  |1012 |

|Human Life Time |  |109 |

|A Day |  |104 |

|A Lecture |  |103 |

|A Moment/A Second |  |100 |


Question 1.

(a)The speed of light is 3 x 10 8 meters/second. If the sun is 1.5x1011 meters from earth, how many seconds does it take light to reach the earth. Express your answer in scientific notation.

(b)If the earth is a sphere of Radius = 6.3×106 m.
 and Mass = 5.9742 × 10 24 kilogram. Calculate an approximate value for the density of the earth in kg/m 3 (Note: this method will only yield an approximate value for the density of the earth).

Question 2. Determine the conversion factor from

i) kmhr-1 to ms-1

ii) mileshr-1 to ms-1

Where 5 miles = 8 km

Question 3 An airplane travels at

750 km/h

i) How long does it take to travel 100km

ii) How long does it take to fly from Waterford to London

Question 4 Estimate each of the following

i) the distance from B07 to B18

ii) the distance from Waterford to Dublin

iii) the floor area of B18

iv) the volume of B18

v) the number of litres of water a human drinks in a lifetime

vi) the volume of your body

vii) cross-sectional area of a 5 cent piece

viii) how long it would take one person to mow a football pitch using a mower which has a speed of 1 km/h and a 0.5m width

ix) how many books can be shelved in a college library with 3500 m2 of floor space. Assume 6 shelves high having books on both sides with aisles 1.5 m wide.


Physical quantities can be divided into two categories

Those which have a direction and

Those which do not.


A quantity which is completely specified by magnitude only is defined as a scalar

Magnitude - Numerical value and Associated Unit

e.g mass, time, length, speed


A quantity which is specified by two values magnitude and direction is called a vector

e.g. displacement, velocity, force

Vector- Numerical value and unit and direction

Typical the difference is shown by the terms speed and velocity

Speed is a scalar while velocity is a vector

Speed 10 m . s-1 SCALAR

Velocity 10 m . s-1 due East VECTOR


The Mathematics of scalar quantities is the ordinary algebra with which we are very familiar.

However when dealing with vectors we must conform to the rules of vector algebra.

Vectors are represented by straight lines where

i) The magnitude of the vector is proportional to the length of the line

ii) The direction of the vector is given by the direction of the line shown using an arrow on the line.


10 m s-1 due east represented by a horizontal line of length 10 cm

10. m s-1 due west may be represented by a horizontal line of length 10 cm in the opposite direction

5 m s –1 due east may be represented by a horizontal line of length 5 cm

The negative of a vector is that vector with its direction reversed + 5 N - 5 N

When we write about vector quantities to distinguish them from scalars, the long-standing custom in Physics is to draw a small arrow over the symbol that represents a vector

e.g. s v

representing displacement and velocity


In geometry

1. A parallelogram is a quadrilateral with two sets of parallel sides. The opposite sides of a parallelogram are of equal length, and the opposite angles of a parallelogram are congruent(equal)

2. A triangle has three sides of lengths a, b and c and angles A, B and C. Notice that the side of length a is opposite the angle A, and similarly for b and c


A + B + C = 1800

If a diagonal is drawn through a parallelogram two triangles are formed.

The total angle therefore is 3600

2 ( A + B ) = 3600

A + B = 1800

Parallelogram Law

If two vectors are represented both in magnitude and direction by the adjacent sides of a parallelogram, their resultant will be represented by the diagonal of the parallelogram drawn from the point.

The point being a common starting point for both vectors.


If two vectors A and B are added to give a resultant vector C then by the reverse process the vector C may be represented by the sum of two components A and B.


A + B = C


C = A + B

C is also a vector quantity i.e. it has both a magnitude and a direction.

This process is called RESOLVING VECTORS and we will return to this later.

Example 1. Add two forces where

F1 = 2 N south


F2 = 4 N east.

To solve this we can use two methods

Method 1. By Construction using a scaled diagram

Decide on a scale

1. Construct a vector diagram to scale, so that the two vectors form two adjacent sides of a parallelogram

2. Complete the parallelogram, using a protractor and ruler.

3. Draw in the diagonal of the parallelogram. Measure the length of the diagonal of the parallelogram.

4. The magnitude of the resultant is calculated by measuring the length of the diagonal of the parallelogram and converting to real units using the scaling value.

5. The direction of the diagonal represents the direction of the resulting vector.

Method 2. By calculation using the COSINE and SINE RULES.

1.Draw a vector diagram to represent the vectors to be added (You do not need to use an accurate scale). The two vectors need to form two adjacent sides of a parallelogram

2. Complete the parallelogram. Draw in the diagonal of the parallelogram. The diagonal divides the parallelogram into two triangles.

NOTE: A + B = θ

3. Usually we then look at the triangle formed by the two vectors and the resultant vector and use the Cosine rule to evaluate the magnitude of the resultant vector

Cosine Rule

Often the triangles occurring in real problems are not right-angled, in which case we can use the "sine rule" and the "cosine rule" to help

Cosine Rule For any triangle


Angles in a triangle A + B + C =180

Therefore C = 180 – (A+B)

Recall A + B = ( (the angle between the two vectors a and b)


C = 180 - θ

The equation can be written as


4. Using the same triangle use the sine rule to calculate the direction of the vector


The equation can be rewritten as


Example 1. Calculating Vector Forces

A force 1 of magnitude 5 kN is acting in a direction 80o from a force 2 of magnitude

8 kN. Assume that the first force acts along the +x direction.

By construction.

1. Decide on a scale

1 kN = 1 cm

2. Draw the vector diagram.

Draw vector 1  using appropriate scale and in the direction of its action

From the starting point of vector 1 draw vector 2 from the same point using the same scale in the direction of its action

3. Complete the parallelogram by using vector 1 and 2 as sides of the parallelogram include the diagonal.

4. Measure the length of the diagonal of the parallelogram and convert to real units using the scale.

5. Measure the angle between the diagonal (resulting vector ) and the +x axis. This is also the direction of the resulting vector.

The resulting force can be calculated as

FR = [ (3(kN))2 + (8(kN))2 - 2 5(kN) 8(kN) cos(180o - (80o)) ]1/2

    = 9 kN

The angle between and the resulting vector and the +axis can be calculated as

B = sin-1[ 3(kN) sin(180o - (80o)) / 9(kN) ]

    = 19.1o


Question 1. Forces of 8 N and 10 N act at the origin and are inclined at 60o to each other. Assuming that the 8 N force acts along the + x direction calculate the resultant force using both methods.

Question 2. Two forces each of magnitude 8 N act at a point O and are inclined at 95o to each other. Calculate the resultant.

Question 3. Find the magnitude and direction of the resultant for a force of 5 N due east and a force of 8 N 45o North of east.


The vector - A means a vector equal in magnitude but opposite in direction to the vector + A.

A - B = A + ( - B )


A - A = 0



THEN - C is called the EQUILIBRANT

Example : A is a force of magnitude 5 N acting along the + x axis, B is a force of magnitude 8 N acting along the + y axis. Calculate

i) Their resultant

ii) Their equilibrant and

iii) A - B


We can also use the parallelogram law to convert a single vector into two components.

When this is done the vector is said to be resolved into its components.

The vectors are resolved into two chosen directions at right angles to each other.

The resolved part of the vector, in a particular direction, tells us the effect of the vector in that direction.

The direction of the components are usually taken along an x axis and y axis i.e. horizontal and vertical directions.

The magnitudes of the components can be found by using a scaled diagram or by calculation.

By construction.

Select a scale and accurately draw the vector to scale in the indicated direction. Include the x and y axis.

To find the component of a vector in a particular direction drop a perpendicular from the vector to the direction of the components i.e. to the x axis and to the y axis.

Measure the length of each of the components and use the scale to determine the magnitude of the components in real units.

NOTE : A vector has NO component at right angles to itself.


A component of a vector in a given direction e.g

Fx Fy F1 F2

Tells us the effect of the vector in a particular direction.

The directions of the components do not necessarily have to be in the x and y directions but they do have to be at 90o to each other.

e.g. Suppose C is a VECTOR where

Cx is the component of C in the x direction


Cy is its component in the y direction

To find Cx and Cy by construction.

Draw a vector diagram to represent C, show x and y axis

Cy C



Drop a perpendicular from C to the x axis Then the vector from the origin to the point along the x axis represents the x component Cx

Drop a perpendicular from C to the y axis for Cy

By calculation. In a right angle triangle



Where C = Magnitude of the vector and

θ = the angle between the vector C and the + x or horizontal direction.

Question 1 . A force of 100 N makes an angle of 40o to the + x axis calculate its x and y components.

Question 2. A force acting at an angle of 30o to the + x direction has a y component of 55 N calculate (i) the value of the force and (ii) the x component.

Kinematics describes motions of objects as a function of time but does not consider the causes of the motion.

The study of the causes of motion is called dynamics.

The branch of physics which deals with the kinematics and dynamics of macroscopic

( large scale ) objects is called MECHANICS.

Physical Quantities.

To fully specify a physical quantity we will use the following procedure, using acceleration as an example.

1. Think about what you already know about the quantity. What concept or physical principle is associated with the quantity.

e.g acceleration has to do with

• motion,

• moving ,

• speeding up

• increasing speed, decreasing speed,

• changing velocity

• Force ( what causes the acceleration?)

2. Define the physical quantity using one or more sentences.


Example acceleration is defined as the rate of change of velocity

3. Derive a mathematical expression from the definition. Simplify the expression using appropriate symbols




4. Determine the basic S.I. units for the quantity from the mathematical expression. Rename the basic units if appropriate.


5. Decide whether the quantity is a vector or a scalar quantity. This will determine whether you need to use vector or scalar mathematics to calculate a value for the quantity.


A vector quantity is defined by both a magnitude (the numerical value ) and a direction

A scalar is defined totally by the magnitude.

DISPLACEMENT :When a body moves from one location to another it undergoes a DISPLACEMENT ( Distance with an associated direction ). A change from one position x1 to another position x2 is called a displacement Δx

where Δx = x2 -x1

SPEED : Defined as the rate of change of distance.

Where Rate means per unit time

CONSTANT SPEED: Constant speed is when a body travels equal distances in equal periods of time.

AVERAGE SPEED : Average speed is defined as the total distance travelled during a particular time divided by that time interval.

INSTANTANEOUS SPEED. This is the speed of an object at a given instant.



VELOCITY : Velocity is defined as the rate of change of displacement.


Speed in a given direction.


Distance travelled in unit time in a given direction.

S.I. Units ms-1 [VECTOR]

AVERAGE VELOCITY : is the ratio of the displacement Δx that occurs during a particular time interval Δt to that interval

Average velocity = Δx = x1 - x2

Δt t1 - t2

S. I. Units ms-1 [VECTOR]


This is the rate at which the objects position x is changing with time at any given instant. The instantaneous velocity is obtained from the average velocity my making the time interval Δt closer to zero.


Instantaneous velocity


S. I. Units ms-1 VECTOR

Example 1. A train travels a distance of 270km in 2 hours 35 minutes due East. Calculate (i) total distance travelled (ii) total time taken (iii) average speed and

(iv) average velocity

ACCELERATION : is defined as the rate of change of velocity.


Instantaneous acceleration is the derivative of the velocity with respect to time.


The acceleration of an object at any instant is the rate at which its velocity is changing at that instant.

Example. A car is uniformly accelerating from rest to a velocity of 30 m s-1 due east in 15 seconds. Calculate its acceleration.

Acceleration due to gravity

An object falling freely under the gravitational attraction of the earth is moving with constant acceleration of

9.81 m s-2 in the vertical downward direction.

Acceleration due to gravity is denoted by the symbol g and it is positive in the downward vertical direction.

g = 9.81 m s-2 Vertically down

Deceleration : Negative acceleration .

If an object is thrown vertically upward it slows down i.e the object experiences a deceleration.

Acceleration = - g = - 9.81 m s-2

Since the direction is opposite the sign is opposite i.e negative

Example 1 : Calculate the velocity with which an object hits the ground if it is dropping for 2.26 s.

Example 2. How long will it take an object to reach its maximum height if it is thrown vertically upward with an initial velocity of 16ms-1


Question 1. A car travels along a straight road in time t as shown

s ( m) 0 8 32 72 128

t ( s) 0 2 4 6 8

v ( ms-1) 0 8 16 24 32

Plot a graph of distance s versus time t

Plot a graph of velocity v versus time t and use it to find

i) The total distance travelled and

ii) The acceleration of the car.


1. Read the entire question carefully from start to finish and then return to the start of the question.

2. Write down and name all the quantities given in the question, one at a time, as they are given

3. Check for S.I. Units and correct into S.I. where necessary, then assign the appropriate label or symbol

4. Find a mathematical relationship between the known and unknown quantities

5. Solve the equation by substituting values both numerical and units

6. Consider whether or not the answer is reasonable.


For all equations we will use the following symbols

u = initial velocity in m s-1

v = final velocity after time t in m s-1

s = distance travelled in time t in a given direction i.e displacement in metres

t = time taken in seconds.

Note :

u, v, s, a are all VECTOR quantities and measured in S.I. units

Usually all vectors in a given application are in the same direction.

To use the equations of motion the value of the acceleration, a , must be constant for a given application.

If an application has an acceleration value that changes it must be divided into a number of stages which has a constant acceleration value for each stage, and therefore each stage can be solved using equations of motion.

acceleration = Rate of change of velocity

From definition


From definition


From equation 1



Question 1. An object starts from rest and moves with constant acceleration of 8m.s-2 along a straight line. Calculate (i) the velocity after 5 seconds (ii) distance travelled in 5 seconds and (iii) average speed for 5 seconds.

Question 2. A truck travelling east increases speed uniformly from 15 km hr-1 to 60 in 20 seconds. Calculate (i)acceleration (ii)total distance travelled (iii)average speed(iv) average velocity

Question 3. A bus moving at a velocity of 20 m.s-1 west decelerates at a rate of 3ms-2. How far does it travel before stopping.

Question 4. An object starts from rest and moves at constant acceleration of 2 m.s-2 for 10 seconds, it then travels at constant speed for a further 1 minute before decelerating to rest in a distance of 200 m.

Calculate (i) total distance travelled

(ii) total time taken and (iii) average speed for the journey.

Question 5. A stone is thrown vertically upward and reaches a height of 20 metres. Calculate

i) with what speed was it thrown up

ii) how long would it take for the stone to drop back down to ground from 20 m.

Question 6. A stone is thrown vertically upward with a speed of 30 m.s-1 from the top of a tower 80m high. Calculate

(i) The total distance travelled by the stone before it hits the ground and

ii) the velocity with which it hits the ground.

Question 7. A cyclist starts from rest and accelerates uniformly at a rate of 1 m.s-2 for 8 s, then continues his journey at constant velocity for 1 minute before decelerating to rest at double the acceleration rate. Calculate (i) the total distance travelled (ii) the total time taken (iii) the average speed and (iv) plot a velocity versus time graph to and use it to calculate the total distance travelled.

























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