SPH 101 MECHANICS LECTURES

SOUTH EASTERN KENYA UNIVERSITY

DEPARTMENT OF PHYSICAL SCIENCES

SPH 101: MECHANICS

FIRST YEAR 1ST SEMESTER JAN, 2013

Credit Hours: 3

Lecturer: Mr. Ngumbi, P.K

Objective

To provide fundamental aspects

of mechanics and illustrate some

of its basic phenomena

Expected Learning Outcomes

By the end of the course, the

learner should be able to:

• Describe unit and dimension

of measurements and error

analysis

• State and prove the Newton’s

laws of motion and state

their applications in real

life systems

• Explain the conservation of

energy and momentum and

static equilibrium

• Describe the plane motion and

projectiles

Course Assessment

Examination - 70%

CATs - 30%

Total - 100% -

References

Halliday and resnik; (1988-).

Fundamentals of physics. (3rd ed.)

N.Y. John Wiley.

Mahendra K Verma; (2005).

Introduction to Mechanics. University

Press.

Daniel Kleppner; (200). An

Introduction to Mechanics; MC Graw

Hill Science Engineering 1973-

London

Nelkon and parker; (1995).

Advanced Level Physics. (7th Ed.).

Oxford: Heineman.

U.S Department of war; (2005).

Applied Physics For Airplane Mechanics.

University Press.

1

COURSE CONTENT

1.0 Introduction

1.1. Units and Dimensions

1.1.1. Definitions

1.1.2. Basic units and

dimensions

1.1.3. Dimensional

analysis

1.1.4. Examples

1.2. Scalar and Vectors

1.2.1. Definition

1.2.2. Vector notation

and representation

1.2.3. Position vectors

1.2.4. Unit vectors

1.2.5. Vector operations

1.3. Composition and

resolution of coplanar

vector

2.0 Part II: Motion in 1-D

and 2-D

2.1 Introduction

2.2 Rectilinear motion

2.2.1 Distance and

displacement

2.2.2 Speed and

velocity

2.2.3 Position-time

graphs

2.2.4 Velocity-time

graphs

2.2.5 Acceleration

2.2.6 Relative velocity

2.3 Kinematic equations

2.4 Free fall

2.5 Projectile motion

3.0 Part III: Dynamics and

Statics

3.1 Force, Impulse.

3.2 Newton’s laws of motion

and their application.

3.3 Linear momentum and its

conservation.

3.4 Free body diagrams and

analysis

3.5 Moments, Couples, Torque

and Applications.

3.6 Center of gravity.

3.7 Work, Energy, Power,

Principle of conservation

of energy.

3.8 Elastic and inelastic

collision

2

3.9 Circular motion: Angular

velocity, angular

acceleration, rotation

with constant angular

acceleration.

3.10 Rotational motion of a

rigid body about a fixed

axis and Moments of

inertia.

3.11 Angular momentums and

its conservation.

Rotational kinetic energy.

3.12 Hydrostatics

3.12.1 Pressure in a

fluid.

3.12.2 Pressure gauges

3.12.3 Archimedes

Principle.

3.13 Hydrodynamics

3.13.1 Equation of

continuity

3.13.2 Bernouille’s

Principle

3.13.3 Applications

(Fluid flow, Pitot

tube, sprays)

3

AN OVERVIEW: INTRODUCTION

Definition of physics

Physics is defined as the study

of the laws that determine the

structure of the universe with

the reference to matter and

energy of which it consists.

Definition of mechanics

It is due study of the

interactions between matter and

the forces acting on matter.

There are three broad classes of

mechanics dealing with solid

bodies.

Kinematics- the study of motion

of bodies without reference to

the forces causing the motion

Dynamics- the study of motion

concerned with the action of

forces resulting in a change in

momentum.

Statics- the study concerned

with actional forces where there

is the change in momentum. i.e.

when a body is in equilibrium.

Physics and Units

It is a systematic study of the

laws. It is very closely

connected to measurements

For accurate reproducible

measurements, a frame of

reference is needed i.e. agreed

upon by every one.

These standard reference points

are called units. International

standard system currently in use

is the SI Units system. It is an

adaptation of the earlier mks

system

m-metre

k-kilogram

s-second

whereby four quantities are

added to the mks system to

produce the standard reference

system.

The table below compares the

first 3 basic quantities in four

stems

4

Quantity

System

SI mks scg Imperial

MASS kg kg g Poud (lb)

Measurements and Quantities

In physics we usually deal with

measurable quantities, for these

reasons unless dealing with

counting of parameters you will

be dealing with quantities.

A QUANTITY = A NUMBER + UNIT

Under the SI system of units,

the following are the seven

basic quantities measures

Derived Quantities

Any other quantities which could

be encountered can be derived

from the basic SI units. They

are hence referred to as derived

quantities and have derived

units.

Basic Units and Dimensions

The term dimension can be used

in two senses;

In the first sense, it refers to

the basic SI quantities; mass,

length, time, ….. etc.

In this sense it is said that

when any derived unit depends on

the rth power of the basic unit

it is said to be of r dimensions

in the basic unit.

Example

Therefore it

is said, area has two dimensions

in length.

Therefore it is said, area has

three dimensions in length.

We have a special type of

notation with units and

dimension involving i.e.

which refers to units

and or its dimensions in terms

5

Quantity SI Unit Abbr

e1 Mass Kilogram

me

kg

2 Length Metre m3 Time Second s4 Electric

current

Ampere A

5 Thermodynamic Kelvin K

of the basic units. This is

denoted in the following way for

the first three basic quantities

which we denote by the three

abbreviations M, L & T.

i.e.

When we express the derived

quantities, we express them

(usually) in terms of M, L & T

all together e.g.

These abbreviations can also be

used to represent the general

unit of the quantity in

question.

Hence

Thus

From this expression we should

see that the magnitude of area

is independent of mass and time

and depends only on the square

of the unit of length used.

Definition of dimensions

The dimensions of a physical

quantity are the powers which

the basic units must be raised

to express the quantity

completely.

Note on parameters ratios

Certain physical quantities are

measured by the ratio of two

similar quantities i.e. the two

similar quantities have the same

units and thus the quantity has

no units or dimensions e.g.

Strain, specific gravity,

Poisson’s ratio etc

The Dimension of Equation

(Dimensional analysis)

This is the relation stating the

units or dimensions of a derived

quantity in terms of the basic

units or dimensions. The

dimension of equation of a

quantity is found by expressing

6

it is in terms of other physical

quantities whose dimensions are

known.e.g.

or

So, the dimensions of ,

or

So, the dimensions of ,

In the SI system , .

So, in terms of basic units

In this way we can determine the

units of ant quantity which can

be expressed by a mathematical

relation.

Example

1. Show that the expression

is dimensionally

correct, where v represents

speed, acceleration, and

t a time interval.

2. Given that the equation

is dimensionally

correct and that and

have the dimensions of

length, determine the

dimensions of .

3. Hooke’s law states that the

force, F, in a spring

extended by a length is

given by . From

Newton’s second law ,

where is the mass and

is the acceleration,

calculate the dimension of

the spring constant .

4. The coefficient of thermal

expansion, of a metal

bar of length whose

length expands by when

its temperature increases

by is given by

7

. What are the

dimensions of ?

5. Suppose we are told that

the acceleration of a

particle moving with

uniform speed v in a circle

of radius r is proportional

to some power of r, say rn,

and some power of v, say vm.

How can we determine the

values of n and m?

Exercise .

Determine whether the following

equations are dimensionally

correct.

(a) The volume of a cylinder of

radius r and length h.,

.

(b) for an object with

initial speed u, (constant)

acceleration a and final speed v

after a time t.

(c) where E is energy, m

is mass and c is the speed of

light.

(d) , where c is the speed

of light, is the wavelength

and is the frequency

NB: Dimensional analysis is a

way of checking that equations

might be true. It does not prove

that they are definitely

correct.

SCALARS AND VECTORS

Definitions

At this level quantities can be

broadly separated into two types

Scalars:- a scalar is a quantity

which has magnitude but without

direction.

(e.g. Mass = 5 kg, Length = 1m

or Time = 2.5 s)

Vectors:-a vector is a quantity

which has both magnitude as well

as direction.

Consider length which is a

scalar (e.g. length of 1 m has

no direction). If we travel or

take this length in a particular

direction, then it becomes a8

vector. Examples of vector

include;

Displacement- distance travelled

in a particular direction e.g.

if you travel 100 metres east or

in any other direction this is a

vector.

Velocity- the rate of change of

displacement

Acceleration- the rate of change of

velocity e.t.c.

Mathematical Notation of Vectors

A vector being a quantity with

magnitude and direction, it can

be denoted in two ways

Graphically – using diagrams

Symbolically – using notation

Since a vector also includes

direction, it is important to

note that direction can only be

given with respect to a system

of reference. This system is

commonly referred to as the

coordinate system.

NB: (i) vector notation

A vector may be denoted by a

variable. There are special ways

of writing these variables. A

vector quantity A is written as

or . For our purpose we

shall denoted vector A as .

(a) Graphic Vectors I

Graphical description of vector

requires a frame of reference

especially for the vector

direction. The frame of

reference is the coordinate

system. There are three common

types:

Cartesian system

Cylindrical polar system

spherical polar system

We will only use the

system and although it is 3-

dimensional, one dimension

corresponds to a given direction

or along a straight line and two

dimensions correspond to a

plane.

In graphical method, a vector is

represented by a line with an

arrow at one end. The line is

9

usually proportional to the

magnitude of the vector and the

arrow points in the direction in

which the vector is acting. E.g.

if we have a displacement of 1 m

due north, then diagrammatically

the can be written as;

There are two types of vectors

Free vector – These are non-

localised vectors which are the

same wherever they are. They can

be moved around in a given

coordinate system.

Bound vectors – (Also called

localised vectors). They are

vectors which are measured from

a particular reference point

e.g. from the origin of the

coordinate system.

(b) Graphic Vectors II

The coordinate system is

as shown below.

y

z

x

O origin 1cmy

It is a orthogonal system i.e.

all the axes are at right angles

to each other.

We can measure length along any

of the axis as long as we know

what units we are using. The

direction of the vector is

specified by the use of certain

free vectors called unit

vectors. It is called a unit

vector because it has a unit

magnitude i.e. they all have

magnitudes of one, 1. These are

vector of unit magnitude along

each of the axes.

10

Length ∝ 1

m

E

S

N

W

Distance between consecutive divisions ≡ 1 unit of length

z

x

y

k

i j

There are three unit vectors in

the system i.e.

- unit vector along the

x-axis direction

- unit vector along the

y-axis direction

- unit vector along the

z-axis direction

Note: The negative unit

vectors denote unit vectors

along direction of the

respective axis.

(c) Position Vector

There is a localised position

vector which is the vector or a

point in space P measured with

respect to the coordinate

system from the origin

. Let P have the

coordinates where

and are all

variables.

In diagrammatic form it’s

represented as;

P

1x

1y

1z

B

O A

C

The position of point P is

denoted by vector OP, . OP

denotes the line segment from O

to P while the arrow gives the

direction. From the diagram we

can go from O to P if we follow

the path

We can also say that in terms of

the unit vectors , and :

11

The position vector is generally

denoted (in most books as )

Symbolic Analysis

Components of a Vector.

Still refering to the same

diagram of OP, we can go from O

to P by travelling along or

parallel to the coordinate axis

i.e. by travelling distance; -

along the x-axis

- along the y-axis

- along the z-axis

The components of a vector are

the parts of the vector

travelled along a particular

coordinate axis. In this way a

three dimensional vector in

coordinate system is said

to have and components.

i.e.

the - components of

the - components of

the - components of

so that we can also describe the

component in the following i.e.

by using a subscript. e.g.

the - components of

the - components of

the - components of

so that it is also possible to

write as

Consider in 2-dimensions where P

has coordinates

But so,

Magnitude of Vectors

We have looked at the magnitude

of the part of a position

vector. These are the and

. We can now define the

magnitude of the total position

vector. The magnitude of OP is

simply the distance travelled

12

from O to P. i.e. It is denoted

by modulus of a vector

As , then or

Using Pythagoras theorem, we

can find the magnitude of in

both 3- and 2-dimensions.

In 3-dimensions

In 2-dimensions

But

and

so,

In 2-dimension, if OP makes an

angle with the x-axis, then

from trigonometry, it should be

apparent that,

or and

or

From which we get if we make the

coordinate variable the subject,

and

So in 2-dimension we can

further describe the position

vector in terms of as;

Unit Vectors

Vector quantities often are

expressed in terms of unit

vectors. A unit vector is a

dimensionless vector having a

magnitude of exactly 1. Unit

vectors are used to specify a

given direction and have no

other physical significance.

They are used solely as a

convenience in describing a

direction in space. It is

13

usually denoted by a hat or

caret. E.g. the unit vector

. If a vector has a unit

i.e , then it is also a

unit vector.

e.g. if , then the

unit vector is defined as

so as

then

Equality of Two Vectors

For many purposes, two vectors A

and B may be defined to be equal

if they have the same magnitude

and point in the same direction.

That is, only if and

if A and B point in the same

direction along parallel lines.

This property allows us to move

a vector to a position parallel

to itself in a diagram without

affecting the vector.

Vector Operations

Let there be vectors

Then the following are defined

Addition:

If

Then,

Grouping similar components and

operating on them accordingly

gives

Subs

tituting into both LHS and RHS

of components in terms of

component gives

For the equality to be true, the

coefficients of , and

must be equal. i.e.

14

Example

A particle undergoes three

consecutive displacements:

,

and

. Find the

components of the resultant

displacement and its magnitude.s

Example

A vector and

. Find the

values of m and n such that

.

Subtraction:

If

Note: subtraction of a vector is

the addition of a negative

vector

So we can rewrite as

The negative of a vector has the

same magnitude as the +ve vector

only that it acts on the

opposite direction. i.e. the

unit vectors are –ve. In

component form,

Grouping similar components on

RHS gives

Substituting into (2) in

component form gives

Which is only true if

Multiplication of a vector by a

scalar.

If

Where is a scalar

In component form we can write

(3) as

Simplifying the RHS gives

15

For this expression to be true,

the components on either side

must be equal to

Dot product (scalar product)

The dot product between two

vectors

and is denoted as and

is defined in the following

ways;

Definition 1

- the scalar and magnitude of

vector and

- the magnitude of vector

- the magnitude of vector

- the angle between and

Let us apply this definition to

the unit vector.

In this case (since unit

vectors have a unit magnitude.

In terms of the angle between

and in this case are parallel

vectors; such that the angle

between them is zero (0). i.e

.

So, or (since

)

so, from the sketch, ,

So, or (since )

It also follows that

It should be apparent that the

dot product of any two

perpendicular unit vector is

zero. Conversely,

Definition 2

16

Examples

17

Motion in 1-D and 2-D

Introduction

A body is said to be in motion

if it is continuously changing

its position w.r.t the

surrounding. E.g. a running

athlete, earth rotating round

the sun.

In our study of motion, we

describe the moving object as a

particle (point mass) regardless

of its size. This just means

that the object in question is

small in size compared to the

distance that it moves for the

times of interest.

The motion of a particle is

completely known if the

particle’s position in space is

known at all times.

A particles position is the

location of the particle w.r.t a

chosen reference point which is

considered to be the origin of

the coordinate system.

We shall describe motion in

terms of space and time while

ignoring the agents that caused

that motion. This portion of

classical mechanics is called

kinematics.

Motion in one dimension (linear

motion) and motion in two

dimensions (projectile motion) are

the subtitles of kinematics they

are also called as 1D and 2D

kinematics. Some mathematical

symbols, equations and graphics

will be used to show the

relations of basic concepts.

Types of motion

• Linear motion: A body is said

to be in linear motion if it

moves in a straight line

(rectilinear motion) or along

a curved path (curvilinear

motion).

• Rotational motion: Occurs when

a body stays in one place and

turns round about a fixed

18

axis. e.g. a fan, earth’s

rotation.

• Oscillatory: This is when a

body moves to and fro about a

mean position e.g. a pendulum.

Distance and Displacement

Distance is a scalar quantity

representing the interval

between two points. Measured in

metre

Displacement is a vector

quantity and can be defined as

distance between the initial

position to a final position

of an object. It must be the

shortest interval connecting the

initial and final points, which

is a straight line.

Changes in position are given by

changes in the value of and

written as .

The change in coordinate is

the displacement of the particle

and occurs over time interval

.

If a particle is moving, we can

easily determine its change in

position as it moves from an

initial position to a final

position .

Its displacement is given by

and written as

From this definition we see that

is positive if is greater

than and negative if is

less than . Displacements

have units of length (meters)

Speed and Velocity

Speed

Speed can be defined as “how

fast something moves” or it can

be explained more scientifically

as “the distance covered in a

unit of time” or rate of change

of distance. It is a scalar

quantity.

SI is

Velocity

19

This is the speed in a given

direction or simply the “rate of

change of displacement”.

Given that the displacement of a

body is and the time taken

is , then;

or SI is

NB: velocity can be:

Positive i.e. when the body

moves away from the reference

point or origin

Negative i.e. when the body

moves towards the reference

point or origin

Average and instantaneous

velocity

Average velocity for a particle

for the ratio of the particle’s

displacement to the time

interval during which that

displacement occurred: i.e

Average velocity tell us about

the average behavior of a body

during the given time interval.

Instantaneous velocity is the

velocity a body at a given

instant of time. It is gotten by

the gradient to the curve at the

point of interest.

It is given by

In the language of calculus,

instantaneous velocity is given

by the limit

.

This is a derivative of w.r.t

Examples

1. A particle moves along the x

axis. Its coordinate varies

with time according to the

expression where

is in meters and is in

seconds.

20

(a) Determine the

displacement of the

particle in the time

intervals and

(b) Calculate the average

velocity during these two

time intervals.

(c) Find the instantaneous

velocity of the particle at

2. A car travels 17 km due east,

then does a U-turn, and

travels 23 km due west.

(a) What total distance has the

car traveled?

(b) What is the total

displacement of the car?

(c) If the entire trip took 2.0

hours, determine the average

speed of the car.

(d) If the entire trip took 2.0

hours, determine the average

velocity of the car

3. If a car moves 150m to the

east and then 70m to the west,

calculate the average speed

and velocity of the car if the

travel takes10 seconds.

Position – Time Graph

Consider the position-time

graphs below;

II

I

III

IV

Time

Posit

ion

The gradient of a position –time

graph gives velocity/speed.

21

From the above graphs, three

cases arise:

Case I

For every increase in time ,

. Therefore the gradient is

zero hence velocity .

This means the body is at rest/

stationary.

Case II

For every increase in time ,

is positive and constant.

Therefore the gradient hence

velocity is constant and

positive.

This means that the body is

moving at a constant speed, away

from the reference point/origin

Case III

For every increase in time ,

is negative and constant.

Therefore the gradient hence

velocity is constant and

negative.

This means that the body is

moving at a constant speed, towards

the reference point/origin.

NB: The values of velocity in

case I, case II and case III is

the average velocity of the body

during the time of study.

Case IV

For every increase in time ,

cannot be clearly defined

for the whole period of motion.

The gradient hence velocity can

only be obtained for a

particular instant of time by

obtaining the tangent to the

curve at that point of interest.

This is the so called

instantaneous velocity

It is given by;

Acceleration

In most occurrences, the

velocity of a body changes while

the body is moving. When this

22

occurs with time, the body is

said to be accelerating.

Consider a particle moving along

the -axis at a velocity at

time and a velocity at

time .

The average acceleration of the

particle is defined as the ratio

of change in velocity to the

time interval during which

the change occurred.

Velocity – Time graph

Consider the velocity-time

graphs below;

II

I

III

IV

Time

Velo

city

The gradient of a velocity –

time graph gives acceleration.

From the above graphs, three

cases arise:

Case I

For every increase in time ,

. Therefore the gradient is

zero hence acceleration, .

This means the body is moving at

a constant speed.

Case II

For every increase in time ,

is positive and constant.

Therefore the gradient hence

23

acceleration is constant and

positive.

This means that the body is

accelerating.

Case III

For every increase in time ,

is negative and constant.

Therefore the gradient hence

acceleration is constant and

negative.

This means that the acceleration

of the body is decreasing.

NB: The values of acceleration

in case I, case II and case III

is the average acceleration of

the body during the time of

study.

Case IV

For every increase in time ,

cannot be clearly defined

for the whole period of motion.

The gradient hence acceleration

can only be obtained for a

particular instant of time by

obtaining the tangent to the

curve at that point of interest.

This is the so called

instantaneous acceleration.

It is given by;

Example

The position of a particle

moving along the x -axis depends

on the time as per the following

equation for four seconds

where x is in meters and t in

seconds. Determine it speed and

acceleration at 2 seconds.

Applications of speed - time

graph

These can be used in deriving

the equations of motion

(kinematic equations)

The equations of motion

In a uniformly accelerated

rectilinear motion, the variable

24

quantities are time, velocity,

displacement and acceleration.

The equations of motion are

simple relations that exist

between these quantities.

The following are the three

equations of motion:

i.

ii.

iii.

Derivation of the First Equation

of Motion

Consider a particle moving along

a straight line with uniform

acceleration 'a'. At t = 0, let

the particle be at A and be

its initial velocity and time t,

v be its final velocity.

…………………… (i)

Second Equation of Motion

…….1

Average velocity can also be

written as

…….2

From equations (1) and (2)

……3

The first equation of motion is

Substituting the value of v in

equation (3), we get

……… (ii)

Third Equation of Motion

The first equation of motion is

.

. .. .. .. ……………………….….

(a)

25

………………………. (b)

………………….

(c)

From equation (b) and equation

(c) we get,

…………………...….. (d)

Multiplying equation (a) and

equation (d) we get,

  …………… (iii)

Derivations of Equations of

Motion (Graphically)

First Equation of Motion

A

t time

velo

city

y B v

u C

D x O

Consider an object moving with a

uniform velocity in a straight

line. Let it be given a uniform

acceleration ‘ ’ at time

when its initial velocity is .

As a result of the acceleration,

its velocity increases to

(final velocity) in time .

Slope of the v - t graph gives the

acceleration of the moving

object.

Thus, acceleration = slope = AB

=

  ………….. (i)

Second Equation of Motion

Let be the initial velocity

of an object and ' ' the

acceleration produced in the

body. The displacement in time

is given by the area enclosed

by the velocity-time graph for

the time interval 0 to t.

26

Distance travelled S = area ABDO

of the trapezium

S = area of rectangle ACDO +

area of DABC

(v = u + at in equation (i) of

motion; v - u = at)

……………………….. (ii)

Third Equation of Motion

Let ' ' be the initial

velocity of an object and ‘a’ is

the acceleration produced in the

body. The displacement' ' in

time‘t’ is given by the area

enclosed by the v - t graph.

S = area of the trapezium OABD.

………… 1

But

or

Substituting the value of t in

equation (1) we get,

…………………… (iii)

Examples

1. A jet lands on an aircraft

carrier at 63m/s.

a) What is its acceleration if

it stops in 2 secs.

b) What is the displacement of

the plane while it is

stopping.

Solutions

27

2. A car traveling at a constant

speed of 45m/s passes a

trooper hidden behind a

billboard. One second after

the speeding car passes the

billboard, the trooper sets

out from the billboard to

catch it, accelerating at a

constant rate of 3m/s2. How

long does it take her to

overtake the car?

Relative Velocity

Relative velocity: Is the velocity of

one object relative to

another’s point of view

Reference Frame: The frame of

reference for the observer’s

point of view - the observer

will always be at rest in

his/her frame of reference.

Relative velocity of an object

moving relative to an observer

is determined by subtracting

vectors. i.e.

Consider two cars, car A moving

at velocities VA to the right

and car B moving at VB to the

left. VB

VA VAVA

Determine the velocity of

a) B relative to A

b) A relative to B

a) Here we find the velocity

that B appears to be moving

at when A is the frame of

reference (i.e to someone in

A)

Car B is in the opposite

direction of A hence its

velocity is negative.

28

b) Here we find the velocity

that A appears to be moving

at when B is the frame of

reference (i.e to someone in

B)

Car B is in the opposite

direction of A hence its

velocity is negative.

Example

Free Falling Bodies

A freely falling object is any

object moving freely under the

influence of gravity alone,

regardless of its initial

motion. Objects thrown upward or

downward and those released from

rest are all falling freely once

they are released. Any freely

falling object experiences

acceleration directed downward,

regardless of its initial

motion.

If we neglect air resistance and

assume that the free-fall

acceleration does not vary with

altitude over short vertical

distances, then the motion of a

freely falling object moving

vertically is equivalent to

motion in one dimension under

constant acceleration. The

equations of motion developed

for objects moving with constant

acceleration can be applied.

Down ward

motion

i.

ii.

iii.

Upward motion

29

i.

ii.

iii.

Examples

1. A stone thrown from the top of

a building is given an initial

velocity of 20.0 m/s straight

upward. The building is 50.0 m

high, and the stone just

misses the edge of the roof on

its way down, as shown in

Figure below. Using tA=0 as the

time the stone leaves the

thrower’s hand at position A,

determine the;

(a) Time at which the

stone reaches its maximum

height,

(b) Maximum height,

(c) Time at which the

stone returns to the height

from which it was thrown,

(d) Velocity of the stone

at this instant

Solution

a) Initial velocity=20 m/s

Final velocity=0

To calculate the time t B at which the

stone reaches maximum height, we

use equation

b)

c) When the stone is back at the height

from which it was thrown the y

coordinate is again zero.

This is a quadratic equation and so

has two solutions One solution is t=0

corresponding to the time the stone

starts its motion. The other solution is

t=4.08s which is the solution we are

after.

d) Velocity at t=4.08s

30

but

displacement at point of throw is

zero. Hence

+20m/s is the initial velocity of the

body moving upwards while -20m/s

is the velocity of the falling object at

the point of throw.

The body hits the point of throw with

a velocity equal in magnitude but

opposite in direction to the initial

velocity.

2. A ball thrown vertically

upward is caught by the

thrower after 20.0 s. Find

(a) the initial velocity

of the ball and,

(b) the maximum height it

reaches.

3. A ball is thrown vertically

upward from the ground with an

initial speed of 15.0 m/s.

(a) How long does it take

the ball to reach its

maximum height?

(b) What is its maximum

height?

(c) Determine the velocity

and acceleration of the

ball at t = 2.00 s.

4. A ball is dropped from rest

from a height h above the

ground. Another ball is thrown

vertically upward from the

ground at the instant the

first ball is released.

Determine the speed of the

second ball if the two balls

are to meet at a height h/2

above the ground.

5. A stone is dropped from the

top of a 95 m building. A

second stone is dropped 0.75 s

later. How far from the ground

is the second stone when the

first stone hits the ground?

31

6. A student through a stone

vertically upwards from the

base of a 51 m building at the

same time a second student on

the top of the building drops

a stone. If the two stones

collide 12 m from the ground,

what was the initial velocity

of the thrown stone?

7. A model rocket is launched

straight upward with an

initial speed of 50.0 m/s. It

accelerates with a constant

up-ward acceleration of 2.00

m/s2 until its engines stop at

an altitude of 150 m.

(a) What can you say about the

motion of the rocket after

its engines stop?

(b) What is the maximum height

reached by the rocket?

(c) How long after lift-off

does the rocket reach its

maximum height?

(d) How long is the rocket in

the air?

8. How long does it take to

travel 3.0 km if you are

traveling at a constant speed of

20 m/s?

9. A stone thrown vertically

downwards from the top of a

building at a velocity of

5.0m/s. this stone is observed

passing by a window that has a

height of 2.0 m. assuming that

the bottom of the window is 25

m from the ground, how long

does it take for the stone to

pass by the window

10. A ball is dropped from a

very high ledge.

a) how far does it travel in

the first 3.0 seconds?

b) how fast is it moving at

that time?

11. A car traveling at 40 m/s

hits the brakes when it sees a

red light 100 meters away.

What acceleration is required

32

in order for the car to stop

before it reaches the light?

12. A ball is thrown upwards at

23 m/s. How far up does it

go?

13. A ball which is thrown

straight upward reaches a

height of 26 meters. At what

speed was it thrown? How much

time elapses before it returns

to the point where it was

thrown?

14. A rocket speeds up

uniformly from 120 m/s to 160

m/s during a 10 second

interval. How far does it

travel during this time?

15. A ball is dropped over the

edge of a building. How fast

is the ball moving 2.0 seconds

after being dropped? v=19.6

m/s

16. A rock is dropped from a

bridge that is 28 meters above

a river. How long does it

take the rock to hit the

water? t=2.39 s

17. A lacrosse ball that is

thrown straight upwards

reaches a maximum height of

4.5 meters. At what speed was

it thrown? vo=9.39 m/s

18. A soccer player heads the

ball and sends it flying

vertically upwards at a speed

of 12.0 m/s. How high above

the players’ head does the

ball travel? d=7.35 m

19. A snow ball is thrown

upward at a speed of 27 m/s.

How high (above the position

from which it was thrown) is

the snow ball 3.5 seconds

later? Also, how fast is it

moving at that time? In what

33

direction? d=34.5m, v=7.3 m/s

down

20. A mis-hit golf ball flies

straight up in the air.

Exactly 4.0 seconds later it

lands right next to the tee.

How high up did the golf ball

go? d=19.6 m

21. A cliff diver on an alien

planet dives off of a 32 m

tall cliff and lands in a sea

of hydrochloric acid 1.20

seconds later. Assuming that

the extra-terrestrial’s

initial speed was zero, what

is the free fall acceleration

of this strange world? g =

44.4 m/s2

22. Two people are standing on

the edge of a building that is

42 meters high (back on

earth!). One person throws a

tennis ball straight downward

at a speed of 16 m/s. At the

same exact time, the other

person throws a tennis ball

straight upward at 16 m/s.

How long after the first

tennis ball lands will the

second tennis ball arrive at

the ground? t2-t1= (4.985 s) –

(1.719 s) = 3.27 s

Motion in two dimensions

The position, velocity and

acceleration vectors

Let describe the position of a

particle by its position vector

r, drawn from the origin of some

coordinate system to the

particle located in the xy

plane, as in Figure below.

At time to the particle is at

point A, and at some later time

tf it is at point B. The path

from A to B is not necessarily a

34

straight line. As the particle

moves from A to B in the time

interval its position vector

changes from ro to rf.

displacement is a vector

quantity and the displacement of

the particle is the difference

between its final position and

its initial position. We now

formally define the displacement

vector for the particle as

being the difference between its

final position vector rf and its

initial position vector ro:

We define the velocity of a

particle during the time

interval as the displacement

of the particle divided by that

time interval:

Multiplying or dividing a vector

quantity by a scalar quantity

changes only the magnitude of

the vector, not its direction.

Because displacement is a vector

quantity and the time interval

is a scalar quantity, we

conclude that the average

velocity is a vector quantity

directed along .

The instantaneous velocity v is

defined as the limit of the

average velocity as

approaches zero:

That is, the instantaneous

velocity equals the derivative

of the position vector with

respect to time. The direction

of the instantaneous velocity

vector at any point in a

particle’s path is along a line

tangent to the path at that

point and in the direction of

motion. The magnitude of the

instantaneous velocity vector is

called the speed.

As a particle moves from one

point to another along some

path, its instantaneous velocity

vector changes from vo at time to

35

to v at time tf . Knowing the

velocity at these points allows

us to determine the average

acceleration of the particle.

The average acceleration of a

particle as it moves from one

position to another is defined

as the change in the

instantaneous velocity vector

divided by the time

during which that change

occurred:

When the average acceleration of

a particle changes during

different time intervals, it is

useful to define its

instantaneous acceleration a.

The instantaneous acceleration

a is defined as the limiting

value of the ratio as

approaches zero:

NB: It is important to recognize

that various changes can occur

when a particle accelerates.

i. The magnitude of the

velocity vector (the speed)

may change with time as in

straight-line (one-

dimensional) motion.

ii. The direction of the

velocity vector may change

with time even if its

magnitude (speed) remains

constant, as in curved-path

(two-dimensional) motion.

iii. Both the magnitude and the

direction of the velocity

vector may change

simultaneously.

Projectile Motion

Projectile is a body thrown with an initial

velocity in the vertical plane and then it

36

moves in two dimensions under the action

of gravity.

Its motion is called projectile

motion. The path of a projectile

is called its trajectory.

Examples: A golf ball in flight, a bullet

fired from a rifle and a jet of water from a

hole near the bottom of a water tank.

Projectile motion is a case of two-

dimensional motion. In dealing

with projectile motion two

assumptions are made

• The free fall acceleration g is constant

over the range of motion and is

directed down ward

• The effect of air resistance is negligible.

With these assumptions, we find

that the path of a projectile is

a parabola. To show that the

trajectory of a projectile is a

parabola, let us choose our

reference frame such that the y

direction is vertical and

positive is upward. Because air

resistance is neglected, we know

that and that

Furthermore, let us assume that

at t = 0, the projectile leaves

the origin with initial velocity

, as shown in Figure 1. The

vector makes an angle with

the horizontal, where is the

angle at which the projectile

leaves the origin.

From the definitions of the

cosine and sine functions we

have and

therefore, the table below gives

a summary of y and x components

of this motion.

X axis Y axis1. Component

of initial

1. Component

of initial

37

velocity along

x-axis.

velocity along

y-axis.

2. Velocity

component along

the x-axis at any

instant t.

This means that

the horizontal

component of

velocity does not

change throughout

the projectile

motion

2. Velocity

component

along the y-

axis at any

instant t.

3. Displacement

along x-axis at

any time t

3. Displacement

along y-axis

at any time t

Net velocity at any instant of

time t

………………………………2

Where is the

angle that the resultant

velocity (v) makes with the

horizontal at any instant?

Time to reach the maximum height

Angular Projectile motion is

symmetrical about the highest

point. The object will reach the

highest point in time . At the

highest point, the vertical

component of velocity becomes

equal to zero.

At

Time of Flight T

38

Angular Projectile motion is

symmetrical about the highest

point. The object will reach the

highest point in time .At

the highest point, the vertical

component of velocity becomes

equal to zero.

At

……………………………..3

Maximum height H

Let consider the figure below

Equation for vertical distance

(y component)

At

Substituting the value of T from

equation 3 we get

Range R

Range is the total horizontal

distance covered during the time

of flight. From equation for

horizontal motion, when

thus;

39

Using the trigonometric identity

The maximum value of R from

Equation above is . This

result follows from the fact

that the maximum value of

is 1, which occurs when .

Therefore, R is a maximum when

. In addition, a point

having Cartesian coordinates (R,

0) can be reached by using

either one of two complementary

values of , such as 75° and 15°

as shown in the figure below. Of

course, the maximum height and

time of flight for one of these

values of are different from

the maximum height and time of

flight for the complementary

value.

Proof that the path of a

projectile is parabolic

If we start from the origin,

This gives

From , making t the

subject,

Substituting into

40

Equation of Trajectory (Path of

projectile)

Take the form

At any instant t

or

Also,

Substituting for t

From trigonometry,

or

……….1

This equation is of the form where 'a' and 'b are

constants. This is the equation

of a parabola. This equation of

motion is valid for launch

angles in the range .

Examples

1. A stone is thrown horizontally

at +15 m/s from the top of a

cliff that is 44 m high.

a) How long does the stone

take to reach the bottom

of the cliff?

b) How far from the base of

the cliff does the stone

strike the ground?

c) Sketch the trajectory of

the stone.

Solution

Known: Horizontal velocity, vx = +15 m/s

Initial height = 44 m

Initial vertical velocity, vy = 0 m/s

Vertical acceleration, g = -

9.8 m/s2

41

Unknowns: Time interval, t

Horizontal displacement,

x

If the initial height is 44 m,

then the vertical displacement

is y = -44 m

x = vx t and y = vy t + ½ gt2

a. y = vyo t + ½ gt2

with vyo = 0, y= ½ gt2

t2 = 2 y /g = 2 (-44 m)/ -9.8 m =

9.0 m/s2

hence t = 3.0 s

b. x = vx t

x = (+15 m/s) (3.0 s) = +45 m

2. A ball is thrown with a

velocity of 4.36 m/s at an

angle of 81 above the

horizontal.

a) How many seconds is the

ball in the air? (time of

flight)

b) How high does the ball go?

(maximum height)

c) How far (horizontally) does

the ball go? (range)

solution

Known:

Initial velocity (4.36 m/s, 81

above horizontal

Unknowns:

Time of flight (t)

Maximum vertical displacement

(y)

Range (x)

Always start by resolving the

components of the original

velocity of the projectile.

(You’ll need these every time!)

4.36 m/s

vvi

81

vh

vxi = 4.36 m/s (cos 81) = 0.682 m/s

vyi = 4.36 m/s (sin 81) = 4.31 m/s

42

a. Calculate the time of flight.

Remember, the ball goes up,

then it comes down.

All we need to calculate is the

time for the ball to go up to

its maximum height and stop.

(The time it takes to come down

is the same!)

Let’s see… the ball starts out

going “up” at

vyi = 4.31 m/s.

It’s vyf = 0 m/s

It’s acceleration is always -9.8

m/s2.

Using g = (vyf – vyi)/t

we can solve for t by rearranging

the equation to

t = (vyf – vyi)/g or t = (0 – 4.31 )/g sec

Then t = 0.44 s

But… That’s only to go to the

maximum height. (Up only!)

For the entire trip:

2 (t) = 2 (.44 s) = 0.88 s

b. Calculate the maximum height

reached by the ball.

Let’s use an equation for

distance and acceleration.

Remember… this is just “up”. So

only use the 0.44 s for the “up”

trip.

y = vyi t + ½ gt2

y = vyi t + ½ gt2 = (4.31 m/s)

(0.44 s) + ½ (-9.8 m/s2) (0.44 s)2

y = + 0.95 m

d. Calculate the range (distance the ball

goes horizontally)

Since the ball has only constant

velocity horizontally, simply

use:

x = vx t

x = (0.682 m/s) (0.88 s)

x = 0.60 m

43

Why 0.88 s this time??? It’s

for the entire trip! Up and

Down!

Exercise

1. A cannon shoots a cannonball

horizontally off a cliff at

100m/s. The cannonball hits

the ground 150m away from the

base of the cliff. How high

was the cliff and with what

velocity did the cannonball

strike the ground?

2. A pelican flying along a

horizontal path drops a fish

from a height of 5.4m. The

fish travels 8 m horizontally

before it hits the water

below. What is the pelican’s

initial speed?

3. A rocket is launched at 100m/s

at an angle of 55˚ above the

ground. Find the total time

in the air, the rocket’s max

height and its range.

4. A ball is thrown with a speed

of 40m/s at an angle of 30°.

When the baseball is thrown

the fielder is 100m from the

plate. It is thrown from 1m

above the ground and caught

2.5m above the ground. With

what velocity should the

fielder have to run to make

the catch? At what distance

from the plate is the catch

made?

v = ?

5. A projectile is launched at an

angle of 30° and lands 60 m

away. What was its initial

speed?

6. A projectile is launched with

a speed of 40 m/s at an angle

44

of 60° from the top of a 30 m

cliff. Find the projectile’s

a. Max height

b. Total time in the air

7. A ball is kicked towards a

fence from a point 32.0 m

away. The velocity of the

ball as it leaves the kicker’s

foot is 20.0 m/s at an angle

of 37.0 with the horizontal.

The top of the fence is 2.50

m. Air resistance is

negligible.

a) Find the time it takes for

the ball to reach the plane

of the fence.

b) Will the ball hit the

fence? If so, how far from

the top of the fence will

it hit? If not, how far

above the fence will it

pass?

8. The missile is launched at

150/s at an angle of 35˚ above

the horizon from 250 high.

a) Find the max height of the

missile.

b) Find the range if it is a

direct hit.

c) Find the velocity with

which it strikes the ship.

9. A missile is launched with a

velocity of 250m/s at an angle

of 45˚ above the ground.

a) Find d for direct hit

b) Show the missile clears the

cliff

c) Max height the missile

reaches

d) Find the projectile’s

velocity at 2sec

45

2 0 .0 m /s3 7 .0

3 2 .0 m

2 .5 0 m

Plane of fence