PHYSICS FORM III
^{LINEAR MOTION}
Definition of terms.
Distance;Is the length between two points.
Displacement; Is the distance moved by a body but in specified direction.
Hence distance is a scalar quantity while displacement is a vector quantity
Speed;Is the distance covered per unit time
Velocity; Is the change of displacement per unit time.
Speed is therefore a scalar quantity while velocity is a vector quantity.
Motion time graphs
They include; (a) Distancetime graphs;
(b) Displacement – time graphs
(c) speed – time graphs
(d) Velocitytime graphs
a)DISTANCETIME GRAPHS
(i) For a body moving with constant speed
(ii) For a body moving with variable speed
(iii) Stationary body
(b) DISPLACEMENTTIME GRAPHS
(i) Uniform velocity
(ii) Variable velocity
(iii) Stationary body
(c) SPEEDTIME GRAPHS
(i) Constant speed
(ii) Variable speed
AREA UNDER SPEEDTIME GRAPH
This gives the distance covered
Distance covered=Area under the graph=(½ ^{x} 3 ^{x}15)+( ½^{ x} 3 ^{x} 4)
= 45M
(d) VELOCITY TIME GRAPHS
(i) Constant velocity
(ii) Uniform acceleration
THE AREA UNDER A VELOCITY – TIME GRAPH
This gives displacement. See the fig. below
Displacement = Area under graph = ½ ^{x }8^{x} 10
= 40M
(iii) Increasing acceleration
(iv) decreasing acceleration
MEASURING SPEED, VELOCITY AND ACCELERATION
METHOD 1; Finding the speed of a student running around the field.
Using a tape measure, measure the perimeter of the field.
– Record the time it takes the student to run round the field once.
– Calculate the speed by the formula
Perimeter
Time taken 
Average speed =
If the time taken by the student to run 100m from the stating point straight to the finishing line, the average velocity can be determined because here the direction is the same.
METHOD 2; Using a ticker timer
This method is used to determine velocity for a shorter distance.
A ticker timer has an arm which vibrates regularly due to the changing current in the mains supply. As the arm vibrates, it makes dots on the paper tape. Successive dots are marked at the same interval of time.
Most ticker timers operate at frequency of 50 HZ (50 cycles/Sec).
Such a timer makes 50 dots per second.
Here the time interval for 2 consecutive dots will be equivalent to the time for 1 cycle i.e
1×1
50 
50 cycles → 1 sec
1 cycle → = 0.02s
This time interval is called a tick.
The distance between 2 adjacent dots is thus the distance moved by the paper tape in 0.02s. Since this distance is usually very small; it is necessary to measure the distance moved in 10 tick interval and devide this by the time for covering the 10 tick interval.
– This time = 0.02 x 10 = 0.2 for ticker timer operating at 50HZ
N/B The dots on the paper pulled at constant velocity are equally spaced while those on the tape pulled with changing velocity unequally spaced.
See the diagrams below
N/B When the dots are closely packed together, the tape is moving slowly and when the dots are far apart, it is moving fast.
Whenthe spaces between the dots increase uniformly, then the tape is accelerating, and when this distance decreases uniformly, the tape is decelerating.
Consider the following tapes obtained in similar experiments
CASE 1; Frequency = 50HZ
Length of 1 ten tick = 5cm
Time to cover 1 tentick = 0.02 x 10 = 0.2
Speed = D/t = 0.05/0.2 = 0.25M/S
N/B From the diagram, the velocity is constant
CASE 2: FREQUENCY = 50HZ
In this case
V_{BC}= ^{0.04}/_{0.2} = 0.2M/S
V_{CD} = ^{0.08}/_{0.2} = 0.4M/S
V_{AB} = ^{0.03}/_{0.2} = 1.5M/S
Here, the trolley is accelerating
CASE 3 Frequency = 50HZ
V_{AB} = ^{0.1}/_{O.2} = 0.5m/s
V_{BC} = ^{0.12}/_{0.2} = 0.1m/s
Here the body is accelerating
OTHER QUESTIONS
 A tape is pulled through a ticker timer which makes 1 dot every second. If it makes 3 dots, and the distance between the 1^{st} and the 3^{rd} dot is 16cm. Find the velocity of the tape.
Solution
Frequency = 1HZ = 1 cycle per sec.
Hence time between 2 dots (consecutive) = 1 seconds
Distance between the 1^{st} and the 3^{rd} dot = 16cm
Hence velocity = ^{0.16m}/2sec.= 0.08m/s
 A tape is pulled steadly through a ticker timer of frequency 50HZ. Given the outcome shown in the fig. below, calculate the velocity with which the tape is pulled. ( The diagram is drawn to scale)
Frequency = 50HZ
Hence time between 2 consecutive dots = 0.02sec.
Distance between consecutive dots = 5cm = 0.05m
V = D/t = ^{0.05}/_{0.02} = 2.5m/s
Other questions; Exercise 1.3 pg 26 KLB BK 3 qn 1 (a) and 2
LINEAR MOTION
Equations of linear motion
Consider a body moving in a straight line with uniform acceleration a so that its velocity increases from an initial value u to final value V in time t.
Acceleration a = V – u
t
Making V the subject of the formula V = u + at ..……this the 1^{st} equation of linear motion
The displace of the body s is = Average velocity x time
S = u + V x t
2
Hence S = ut + ½ at^{2} ………. This is the 2^{nd} equation of linear motion
Displacement S, can be given by = Average velocity x time
But t = v – u
t
+ 
Hence S = u + v v – u
2 t
Therefore V^{2} = u^{2} + 2aS ……………. This is the 3^{rd} equation of linear motion
The 3equations include(i) v=u+at
(ii) s=ut+^{1}/_{2} as
(iii) v^{2}=u^{2}+2as
Where u is the initial velocity
v is the final velocity
a acceleration
t time
s distance
N/B Retardation is a. The equations for abody undergoing retardation are:
v=uat
s=ut^{1}/_{2}at^{2}
v^{2}=u^{2}2as
QUESTIONS
 A body is uniformly accelerated from rest to a final velocity of 100 m/s in 10s. Calculate the distance covered
(use first equation to get a then 2^{nd} equation to get s.) (s=500m)
 A body whose initial velocity is 30 m/s moves with a constant retardation of 3 m/s. Calculate the time taken for the body to come to rest. (t=10s)
 Abody moving with uniform acceleration of 10 m/s covers a distance of 320m. If its initial velocity was 60 m/s, calculate its final velocity. (v=100 m/s)
MOTION UNDER GRAVITY
FREE FALL
The 3 equations of abody under constant acceleration can be applied in free fall. Herea is replaced by g. So the 3 equations become.
v=u+gt
s=ut+^{1}/_{2}gt^{2}
v^{2}=u^{2}+2gs Where g is the acceleration due to gravity
Under free fall, a body dropped from a certain height to the ground will have initial velocity u=0
A velocity – time graph for a body dropped from a certain height to the ground is a straight line from the origin i.e.
N/B The velocity of the body increases from city with which it hits the 0 to maximum when the body hits the ground.
QUESTIONS
 A body is dropped/released from the top of a cliff 180m high. Taking g=10m/s^{2} , calculate
(a) The time it takes to hit the ground
(b) The velocity with which it hits the ground
The equations still hold here except the value of g which is 10 and not +10.
Hence the equations include
v = u – gt
s = ut – ^{1}/_{2}gt^{2}
v^{2} = u^{2} – 2gs
Time to reach maximum height & time of flight
(a) Time to reach maximum height
At maximum height v = 0
Hence from v = ugt
0 = ugt
t =^{ u}/_{g}wheret is time to reach maximum height
(b) Time of flight
This is the time taken by the body to reach maximum height and fall back to the original point.
Hence it is twice the time to reach the greatest height i.e
Time of flight = ^{2u}/_{g}
(c) Maximum height: This is at apoint where v = 0
N/B On returning / falling back to the ground, a vertically projected body hits the ground with the same speed to one with which it was projected.
QUESTION
A stone is projected vertically upwards with a velocity of 30m/s from the ground. Calculate:
(a) The time it takes to reach maximum height. (t = 3 seconds)
(b) Time of flight. ( t = 6 seconds)
(c) The maximum height reached. (s = 45m)
(d) The velocity with which it hits the ground. ( v = 30m/s)
Determination of acceleration due to gravity (g) using simlependlum
Apparatus:Pendlum bob, thin thread, stand and clamp, metre rule & stop watch.
Procedure
 Set up apparatus as shown below
 Starting with l = 50cm, set the pendlum bob swinging through an angle of about 10^{0}. N/B – Length of pendlum L = Length of thread + Radius of bob.
see fig. below
 Time 20 oscillations.
 4. Repeat the experiment and obtain the average time for 20 oscillations.
Hence complete the table below
6
 Determine the periodic time T (time for one oscillation ) and fill in the table.
QUESTIONS
(a) Plot a graph of T^{2} against L (in metres)
(b)
Calculate the slope of the graph. What does it represent?
(c) Use the graph to calculate g.
SOME NOTES
For a simple pendlum oscillating with small amplitude
Where T is the period, L length of pendlum and g acceleration due to gravity.
Thus T^{2} = 4∏^{2}L/_{g}
HORIZONTAL PROJECTION
Consider a body projected horizontally with horizontal velocity V_{h}from point O as shown
below:
The horizontal velocity remains unchanged throughout the flight
The path followed by the projectile is called the trajectory.
The maximum horizontal distance covered by the projectile R is called Range& iscalculated as
Range = V_{h} x time
N/B
The time the projectile takes to travel from O and land at X is the same time it would take to land at Y if is dropped with zero velocity from O (When it is dropped initial velocity at O = 0 )
QUESTIONS
 A ball is thrown from the top of a cliff 20m high with a horizontal velocity of 10m/s. Calculate
(a) The time taken for it to strike the ground. (Ans = 2 seconds)
(b) The distance from the foot of the cliff to where the ball strikes the ground. (Range = 20m)
 A stone is thrown horizontally from a building that is 45m high above the horizontal ground. The stone hits the ground at a point which is 60m from the foot of the building.p0
Calculate the horizontal velocity of the stone. (Use R = V_{h} x t )
(Other questions from KLB BK 3 Exercise 1pg 37 – 41 )
REFRACTION OF LIGHT
REFRACTION is the bending of light at the interface when it travels from one medium to another at an angle. i.e the figure below shows refraction of a ray as it travels from air to water.
N/B
 When light travels from an optically less dense medium to an optically more dense medium it is refracted towards the normal e.g when light travels from air to water as shown in the diagram above. (Here the angle of incidence is larger than the angle of refraction)
 But when light travels from an optically more dense medium to an optically less dense medium, it is refracted away from the normal.
(Angle of refraction > Angle of incidence) e.g when light travels from water to air.
A ray through the normal is not refracted i.e
N/B Light travels with a velocity of 3.0 x 10^{8} m/s in vacuum. It travels with a velocity slightly lower than this in air. In other optically dense material such as water, glass and Perspex it travels at a much reduced speed.
TO INVESTIGATE THE PATH OF LIGHT THROUGH A RECTANGULAR GLASS BLOCK USING PINS.
APPARATUS: Soft board, White sheet of paper, Drawing pins, Rectangular glass block.
PROCEDURE
 Fix the white plain paper on the soft board using pins.
 Place the glass block on the plain paper, trace its outline and label it ABCD. Remove the glass block
 Draw a normal NON at a point O on side AB.
 Draw a line PO making an angle of 20^{o} with the normal.
 Replace the glass bock to its original position.
 Stick two pins P1 and P2 on the line PO such that they are upright and at least more than 6cm apart.
 View pins P1 and P2 through the opposite side of AB and stick and stick two pins P3 and P4 such that they appear to be on a straight line with P1 and P2. Mark the positions P3 and P4.
 Remove the pins and the block.
 Draw a line joining P3 and P4 and produce it to meet the outline face CD at a point O^{I}.
 Join O to O^{I}
See figure below
– Measure and record angle NOO^{I} in the table below.
Angle of incidence I (Degrees)  Angle of refraction r (Degrees)  Sin i  Sin r  Sin i
Sin r 
20^{o}  13  
30^{o}  
40^{o}  
50^{o}  
60^{o}  
70^{o}
80^{o} 

Repeat the procedure for the other angles in the table and complete the table.
Compare all the values of sin i
Sin r
Results :These values are the same /very close/constant – This constant is the refractive index.
 Plot a graph of sin i against sin r
`
Result: This graph is a straight line through the origin
 Determine the slope of this graph
Explanation: This slope of the graph is the refractive index.
The symbol for refractive index is n
LAWS OF REFRACTION
Law 1 :The incident ray, the refracted ray and the normal at the point of incidence, all lie in the same plane.
Law 2 :The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media. i.e
= 
n= Sin i Refractive index for medium 2
1 2 
Sin r
When light travels from medium 2 to medium 1 along the same
Path
1 
n 
(this is the refractive index for medium 1) 
2 1 
=
`refractive index for medium 2
QUESTIONS pg 51 53, Example 26
REFRACTIVE INDEX IN TERMS OF VELOCITY
Other applicable formula
 Refractive index for any medium = velocity of light in air
Velocity of light in medium
QUESTIONS: From KLB BK 3. Example 710
REFRACTIVE index for glass with respect to water
REFRACTION IN TERMS OF REAL AND APPARENT DEPTH
An object under water/ glass block, when viewed normally, appears to be nearer the surface than it actually is. See the figure below
The actual depth is called the real depth
The false depth is the apparent depth, (as in the figure above)
The distance from the real position to the apparent position of the coin is the Vertical displacement of the coin.
Here
QUESTIONS
From KLB BK III Pg 6365 Example 14 17. Example 19 Pg 6768
The critical angle ( c )
Is the angle of incidence (in the optically more dense medium) for which the angle of refraction (in the optically less dense medium) is 90^{o}.
See the fig below
TOTAL INTERNAL REFLECTION :This occurs when the angle of incidence has exceeded the critical angle.
See the figure below
At this stage i=r and i>C
For total internal reflection to occur;
 Light must be travelling from an optically dense medium to an optically less dense medium
 The angle of incidence must be greater than the critical angle
Relationship between C and n
Consider a ray of light striking a glassair interface as shown below
QUESTIONS:
 The fig. below shows the path of light passing through a rectangular block of perspex placed in air
Calculate the refractive index of Perspex (1.48)
 A ray of light travels fro a transparent material to perspex as shown below
Calculate the critical angle c (24.6)
 Calculate the critical angle of diamond given that its refractive index is 2.42
 The critical angle for water is 48.6^{o}. Calculate the refractive index for water.
 A ray of light travels through air into medium as shown in the fig. below
Calculate the critical angle for the medium
 Calculate the critical angle for glass water interface (refractive indices for water and glass are ^{4}/_{3} and ^{3}/_{2} respectively.
EFFECTS OF TOTAL INTERNAL REFLECTION
 MIRAGE
This happens 0n a hot day when then ground gets heated and in turn heats the air above it. This heated air is optically less dense than the air far above the ground.
Therefore, a ray from the sun passes through the colder (optically more dense) air to the warmer (optically less dense) air and is hence refracted away from the normal.
For some of these rays, total internal reflection results. See figure below
To the observer, the ray seems to come from point I. This appears like a pool of
water. This phenomenon is called mirage.
Two theories have been advanced to explain mirage. One is of total internal reflection as explained above, and the other advocates the direct rays travelling through air of the same temperature to the observer as shown in the diag. above.
Mirages are also seen in very cold regions , but here light curves in the opposite direction as shown below
Here the air next to the ground is colder than the one far away from the ground. The mirage appears above the ground.
 ATMOSPHERIC REFRACTION
This is a phenomenon which enables us to see the sun after it has set. (Wee see the sun in its apparent position). See the fig below
Similarly, the sun is seen before it rises.
TOTAL INTERNAL REFLECTION PRISMS
Using a right angled glass or perspex prism. ( angles are 90, 45, 45).
(a) To turn a ray of light through 90^{o}
Consider a ray of light incident to face AB of a right angled isosceles prism shown below:
The incident ray is unrefracted because it passes through the normal. It meets face AC at a point O, where it makes an angle of 45^{o} with the normal. This angle is greater than the critical angle for glass (42^{o}), hence the ray is totally internally reflected. The reflected ray meets BC normally (through the normal) and passes on unrefracted.
(b) To turn a ray through 180^{o}
(c) Inversion with deviation
(d) Inversion without deviation.
APPLICATIONS OF TOTAL INTERNAL REFLECTION
 In periscopes: A periscope is an instrument used to view objects over obstacles.
Prisms rather than plane mirrors are used in periscopes because plane mirrors have the following disadvantages:
– Mirrors absorb some of the incident light
– The silvering on mirrors can become tarnished and peel off.
– Mirrors, especially if they are thick, produce multiple images. (See fig 2.50 pg 76 KLB BK III)
A prism periscope
Here light is deviated through 90^{o} by the first prism before the second prism deviates it a further 90^{o} in the opposite direction.The image formed is erect, vitual and the same size as the object.
OTHER APPLICATIONS OF TOTAL INTERNAL REFLECTION
 Used in prism binoculars – Instrument used for viewing distant objects.
 In optical fibre. Optical fibre is used in:
 Transmitting sinals in communication.
 In medicine to view the internal parts of human body
DISPERSION OF WHITE LIGHT
 When a beam of white light is directed to an equilateral prism and a white screen placed infront of the prism, a band of 7 colours is formed on the screen as shown below
White light is a mixture of 7colours and the separation is due to their different velocities in the prism. The velocity of red light is the greatest hence it is deviated least while the violet light with least velocity is deviated most.
Light from the sun is an example of white light.
DISPERSION OF WHITE LIGHT IN THE RAINBOW
The rainbow is a bow – shaped colour band of visible spectrum seen in the sky when white light from the sun is refracted, dispersed and totally internally reflected by rain drops.
It can also be seen on spray fountains and water falls when the sun shines on the drops of water.
NEWTON LAWS OF MOTION
The effects of a force on motion of a body are based on 3 laws known as newton’s laws of motion
 NEWTON’S FIRST LAW OF MOTION
It states that a body remains in its state of rest or uniform motion in a straight line unless acted upon by an external force.
(Illustrate the examples on page 87 & 88)
INERTIA: Is the property of bodies to resist change in state of motion. This explains why cars have seat belts. They hold passengers on the seats when a vihecle comes to stop or when it decelerates sharply.
Momentum of a body: Is the product of mass of the body and its velocity.
i.e Momentum = Mass (kg) xVelocity (m/s)
Momentum = MV, SI unit is Kgm/s
Momentum is a vector quantity. The direction of momentum is the same as that of the velocity of the body.
QUESTIONS
 A van of mass 3 tonnes is travelling at 72km/h. Calculate the momentum of the vihecle.
 A car is moving at 36km/h. What velocity will double its momentum?
 NEWTON SECOND LAW OF MOTION
The rate of change of momentum of a body is directly proportional to the resultant external force producing the change, and takes place in the direction of the force.
Relationship between mass, force and acceleration
If the resultant force F
Acts on a body of mass M
For time t
and causes velocity to change from U to V
Then change in momentum = Final momentum – Initial momentum
Change in momentum =MV – MU
t 
Rate of change of momentum = MV – MU
QUESTIONS
 What is the mass of an object which is accelerated at 3m/s by a force of 125N?
Others are examples 4,5& 6 pg 94 KLB BK 3
IMPULSE: When a force acts on a body for a very short time, the force is referred to as an impulsive force. The result produced is called the impulse of the force. Impulsive forces occur when two moving bodies collide.
If a force F acts on a body of mass M for a time t,
Impulse = Force x time
Impulse = Ft
t 
From newton’s second law F = MV – MU , Ft = MV – MU,
Hence impuse is change in momentum
QUESTIONS pg 96 to 97 example 8 to 10
 NEWTON’S THIRD LAW OF MOTION: Action and reaction are equal and opposite
ACTION AND REACTION FORCES ON A STATIONARY BLOCK
QUESTIONS
Example 13 pg 102, No 3 & 4 pg 103 KLB BK III
LAW OF CONSERVATION OF LINEAR MOMENTUM
For a system of colliding bodies, the total linear momentum remains constant, provided no external forces act.
Qn 1. A body of mass 5kg moving with a velocity of 3m/scollides head – on with another body B of mass 4kg moving in the opposite direction at 6m/s. If after collision the bodies move together (coalese), calculate the common velocity V.
2.Pg 107 example 15 KLB BK III
COLLISIONS
There are two types of collisions namely:
 Elastic collision
 Inelastic collision
 Elastic collision: This is one in which both kinetic energy and momentum are conserved.
(b)Inelastic collision: This is one in which momentum is conserved but kinetic energy is not.
QUESTIONS; Example 16 & 17 pg 106 to 108 KLB BK III
APPLICATIONS OF THE LAW OF CONSERVATION OF MOMENTUM
 In the rocket and jet propulsion; The rocket propels itself forward by forcing out its exhaust gases. The hot exhaust gases are pushed out of the exhaust nozzle at high velocity and gain momentum in the one direction. The rocket thus gains momentum in the opposite direction which gives it a forward thrust.
 Garden sprinkler. (see fig. on pg 108 KLB BK III)
FRICTION
Is the force that opposes relative motion between two bodies in contact.
Molecular explanation of friction: Surfaces of bodies are made of tiny bumps and troughs when viewed under powerful microscope.
Hence when two surfaces are in contact the bumps and troughs interlock as shown below
The interlocking opposes relative motion , hence friction.
Factors affecting friction between solid surfaces
Consider a wooden block resting on a wooden surface as shown below;
The block exerts a force F = Mg = weight, and this equals to the normal reaction R.
When the block is pushed e.g in direction A, it experiences a friction force in the opposite direction.
The following factors are true about this friction force
(a) Normal reaction: Friction is directly proportional to the norm al reaction R, i.e friction increases with increase in normal reaction
(b) Nature of surface: Smooth surfaces under relative motion yield low friction, while rough surfaces yield high friction.
(c) Friction force does not depend on the area of surfaces in contact.
N/B The applied force F_{A } is equal to friction when the block just starts to move. The friction at this point called limitingfriction / static friction (friction on a body that is still stationary).
Since friction is directly proportional to the normal reaction, F_{A} is therefore also directly proportional to normal reaction R.
Limiting friction here (F) here α R
F α R
Since at this point the applied force F_{A} = Limiting friction F
F_{A} α R
The constant of proportionality here is µ_{s} (coefficient of static friction)
Hence F = µ_{s}R
Where F is either limiting friction or the applied force.
Similarly, for body in motion, friction force acting on it is directly proportional to the normal reaction. i.e
F α R
But here the constant of proportionality is µ_{k} (coefficient of kinetic friction). Hence
F = µ_{k}R
N/B
 µ_{s} and µ_{k} have no units
 When the applied force moves the body with constant speed, then the applied force = Friction force. i.e force that overcomes friction will give the body uniform speed.
 If a larger force is applied, then this force is called the resultant force.
Resultant force = Applied force – Force needed to overcome friction
 The force needed to start motion is higher than that needed to maintain motion.
Questions
 Awooden box of mass 5kg rests on a rough floor. The coefficient of friction between the floor and the box is 0.6
(a) Calculate the force required to just move the box. (Take g = 10)
R = 5×10=50N
F = 0.6 x 50
= 30
(b) If a force of 200N is applied on the box, with what acceleration will it move? (Take g = 10)
solution
Resultant force (F) = Applied force – friction
Resultant force = 200 – 30 = 170
=
F = Ma
170 = 5 x a
a = 34m/s^{2}
 A block of metal with a mass of 20kg requires a horizontal force of 50N to pull it with uniform velocity along a horizontal surface. Calculate the coefficient of Friction between the surface and the block. (Take g = 10)
F = µ_{k }xR
50 = µ_{k} x 200
50
200 
µ_{k} =
µ_{k} = 0.25
Methods of reducing friction
 Placing rollers between the two rough surfaces
 Lubrication The application of oil or grease between moving parts.
 Use of ball bearings in the rotating axles
 Air cushioning done by blowing air between the rough surfaces to prevent the surfaces fro coming into contact.
Uses of friction
 Walking
 Motor vehicles
 Brakes: Friction between the brake drum and the brake lining halts the vehicle
 Match stick: Friction between the match stick head and the rough surface develops heat, igniting the stick head.
Limitations of friction
 Causes wear and tear between moving parts
 Causes noise
 Causes energy loss since work has to be done against it.
VISCOSITY
This is the frictional resistance to motion in fluids
For example,
 It is more difficult to wade through water than to move the same distance in open air space – water has higher viscosity than air.
 A steel ball dropped in glycerinetakes a longer time to reach the bottom than when dropped into cylinder full of water.
Terminal velocity
Is the constant velocity attained by a body falling in a fluid when the sum of the upward forces equal to the weight of the body
N/B Viscosity decreases with temperature
QUESTIONS : ON pg 119 to 120
WORK, ENERGY, POWER & MACHINES
ENERGY; Is the ability to do work. It is measured in joules (J)
WORK ; Is done when an applied force makes its point of application to move in the direction of the force.i.e work done = Force x Distance moved by the object in t the direction of the force
W = F X D
Units NM
1NM = 1J Hence work is also measured in joules.
POWER;Is the rate of doing work.
i.e power =
MACHINE; Is a device that makes work to be done more easily or conveniently.
SOURCES OF ENERGY
These include
 Wind For driving wind mills, pumping water or generating electricity.
 Fuels Wood and charcoal, petroleum and natural gas.
 Geothermal
 High dams and water falls – used to turn turbines in HEP stations to produce electricity.
 Oceans – Waves and tidal energy
 Nuclear/ Atomic energy.
FORMS OF ENERGY
 Chemical energy
 Mechanical energy
 Heat energy
 Wave energy
 Electric energy
TRANSFORMATIONS OF ENERGY
Any device that facilitates transformation of energy is called a transducer. E.g
ENERGY TRANSFORMATION TRANSDUCER
Chemical to electrical energy _________________ Battery
Electrical to sound energy _________________ Loudspeaker
Heat to electrical energy __________________Thermocouple
Solar to electrical energy __________________ Solar cell
Kinetic energy to electric energy _______________ Dynamo
Electric energy to kinetic energy _______________ Motor
Solar energy to heat energy ________________ Solar panel
Note the following;
1KJ = 10^{3}J
1MJ = 10^{6}J
Questions on work and energy
 Calculate the work done by a stone mason in lifting a stone of mass 15kg through a height of 2M. ( Take g=10N/kg)
 A boy of mass 40kg walks up a flight of 12 steps. If each step is 20M high, calculate the work done by the boy. (g = 10N/kg)
POTENTIAL ENERGY / GRAVITATIONAL POTENTIAL ENERGY
This is the work done to lift an object through a height h. i.e
P.E = Force x height
But force = Weight of the object = Mg
Hence P.E =Mgh
Qn; A student climbs a vertical rope 10M long. If the mass of the student is 50kg, how much work does the student perform?
Solution
P.E =Mgh = 50 x10 10 =5000J
ELASTIC POTENTIAL ENERGY
This is the work done in stretching or compressing a spring. (It is the same as the energy stored I spring).
In stretching spring, the applied force varies from 0 to maximum force F.
Below is a sketch for extension plotted against force for a stretched spring.
Since force has changed 0 to F
But W = ½ Fe = ½ x 12 x 0.08
W = 0.48 J
KINETIC ENERGY (K.E)
This is the energy a body possesses due to motion.
Mathematically, K.E = ½ MV^{2}
Where M = Mass of the body
V = Velocity
But we can have,
Final K.E = ½ MV^{2}_{final} Where V_{final} is the final velocity, and
Initial K.E = ½ MU^{2}_{initial} Where U_{initial} is the initial velocity.
QUESTIONS;
 A trolley of mass 2.0kg is pulled from rest by a horizontal force of 5N for 1.2 seconds. Assuming that there is no friction between the horizontal surface and the wheels of the trolley, calculate;
(a) The distance covered by the trolley
Solution; Use S =ut + ½ at^{2} to get S and F = Ma to find a
Answer = 1.8M
(b)The K.E gained by the trolley
Solution; K.E gained by the trolley is the final K.E ( because initial K.E was zero)
Hence K.E = ½ MV^{2}_{final}
Use V = U + at to find V_{final}
Answer = 9J
Qn 2. Example 6 Pg131 KLB PHY BK 3
THE PENDLUM
The fig below shows a pendulum bob released so that it swings to and from a vertical axis.
At points A and C, the pendulum bob has maximum potential energy and no K.E. At point B, it has maximum K.E and no P.E.
At x K.E = P.E
At y K.E = P.E
QUESTIONS ON POWER
Joules Hence units are J/S
Seconds Alternative = Watt (W) 
Work done
Time taken 
P = =
T 
1W = 1 J/S
Qn. 1. A person weighing 500N takes 4 seconds to climb to climb upstairs to a height of 3m. What is the average power in climbing up the height.
Work
Time 
Solution:
P = W = F x D = 500 x 3 = 1500J
1500
4 
P = = 375N
Qn. 2. An electric motor raises 50kg load at a constant velocity. Calculate the power of the motor if it takes 40 seconds to raise the load through a height of 24M (Take g = 10N/kg)
Solution; W = 500 x 24 = 12000J Time = 40s
300W 
= 
W
t 
= 
1200
40 
P =
Assignment; Example 8 pg 134 K.L.B PHY BK 3 & Exercise 4.1 pg 136137
MACHINES
A machine is a device that makes work to be done more easily or conveniently.
Simple machines include; Levers, pulleys, hydraulic press, gears etc. If a machine, say a pulley, is used to raise a stone, then the weight of the stone is the load and the fore applied is the effort.
TERMS ASSOCIATED WITH MACHINES
EFFORT (E) ; Is the force applied to the machine. Measured in N.
LOAD (L);Is the force exerted to the machine. Measured in N.
Load
Effort 
MECHANICAL ADVANTAGE (MA); Is the ratio of load to the effort.
IeMA = MA has no units

VELOCITY RATIO (V.R);
Distance moved by effortDE
Distance moved by load DL

V.R =
DE
DL 
ie V.R =
EFFICIENCY (ƞ)
100 
x 
Work done on load
Work done by effort 
Is the ratio of work done on the load (work output) to work done by effort (work input), usually expressed as a percentage. i.e efficiency is either,
(i) ƞ =
Work output
Work input 
100 
x 
or (ii) ƞ =
X 100 
Work done on load
Wok done by effort 
RELATIONSHIP BETWEEN M.A, efficiency & V.R
In (i) above, ƞ =
Load x Distance moved by load
Effort x Distance moved by effort 
and Work done = Force x Distance moved by force
ƞ = x 100
1
V.R 
Hence ƞ = M.A xx 100
M.A
V.R 
Ƞ = x 100
Load x Distance moved by load
Effort x Distance moved by effort 
QUESTION; In a machine, the load moves 2M when the effort moves 8M. If an effort of 80N is used to raise a load of 60N, What is the efficiency of the machine?
Ƞ =x 100
60 x 2
20 x 8 
=x 100
= 75^{o}/_{o}
LEVERS
A SIMPLE LEVER
Questions; Answer Example 12 pg 140 KLB secondary PHY BK 3.
THERE ARE 3 CLASSES OF LEVERS NAMELY;
(i) LEVERS WITH THE PIVOT BETWEEN THE LOAD AND THE EFFORT; Ee.g pliers, hammer, beam balance, crow bar, pair of scissors.
(ii) LEVERS WITH THE LOAD BETWEEN THE PIVOT AND THE EFFORT; E.g wheelbarrow, nut crackers, bottle openers etc.
(iii) LEVERS WITH THE EFFORT BETWEEN THE LOAD AND THE PIVOT; e.g sweeping brooms, fishing road, human arm, spade etc.
INCLINED PLANES
QN; A man uses an inclined plane to lift a load of 50kg through a vertical height of 4M. The inclined plane makes an angle of 30^{o} with the horizontal. If the efficiency of the inclined plane is 72%, calculate;
(a) the effort needed to move the load up the inclined plane at a constant velocity.
1
0.5 
1
Sin 30 
1
Sin ϴ 
From expts, V.R = = = = = 2
72
100 
M.A = ƞ x V.R = x 2 = 1.44
50×10
1.44 
L
MA 
E = = = 347.2N
(b) The work done against friction in raising the load through a height of 4M.
Soln; Work against friction = Work input – Work output
Work output = 50 x 10 x 4 = 2000N
Work input = Effort x Distance moved by effort
4
Sin 30 
= E x AC
Work input = 347.2 x= 2777.6
Work done against friction = 2777 – 2000 =777.6 J
The distance between two successive threads is called the pitch. In one revolution, the screw moves through a distance equal to one pitch.
V.R of screw = Circumference of the screw head (handle)
Pitch
V.R = ^{2πR}/_{Pitch} where R is the radius of the screw head/ handle
N/B A screw combined with a lever can be used as a jack for lifting heavy loads such as cars.
GEARS: A gear is a wheel which can rotate about its centre.
Below is an arrangement of two gears
The driver wheel; Is the wheel on which the effort is applied
The load wheel: Is the driven wheel
Assuming that the driver wheel has n teeth and the driven wheel N teeth, then when the driver wheel makes 1 revolution, the driven wheel makes ^{n}/_{N} revolutions.
_{V.R =} Revolutions made by the driven wheel
Revolutions made by the driver wheel
V.R = 1
^{n}/_{N}
V.R =^{ N}/_{n}
_{Therefore the V.R of the gear =}^{No of teeth on the driven wheel}
^{No of teeth in the driver wheel }
PULLEYS
PULLEY: Is also a type of machine.
There are several types of pulley systems. The 3 common ones include;
 The single fixed pulley
 The single movable pulley
 The block and tackle pulley
V.R of a pulley;Is the number of ropes supporting the load.
 THE SINGLE FIXED PULLEY
Here V.R = 1
(b) SINGLE MOVABLE PULLEY
The velocity ratio for the two arrangements above is the same = 2
( The number of ropes supporting the load is 2)
(c) THE BLOCK AND TACKLE
Here the velocity ratio (number of ropes supporting the load) = 4
Questions to be answered here from page 151 example 17 to pg 153 example 153 example 18 KLB sec. PHY BK III
THE HYDRAULIC MACHINE (LIFT)
Here
Distance moved x Crosssection area = Distance moved x Cross
by effort piston of effort piston by load piston section
area of
load piston
HENCE
Distance moved by effort = Cross section area of load piston
Distance moved by load Cross section area of effort piston
πR^{2}
πr^{2} 
V.R = Where R is the radius of the load piston
r is the radius of the effort piston
0r V.R = R^{2}/r^{2}
Qns; Example 20, 21 from KLB SEC PHY BK 3.
Other qns for general practice – Exercise 4 pg 159161 KLB PHY BK 3.
CURRENT ELECTRICITY II
Electric current: Is the rate of flow of charge through a conductor.
Ammeter: Is the instrument used to measure electric current.
SI unit for current is Ampere (A)
Potential difference (p.d) of a cell: Is the voltage across the cell in a closed circuit ( when it is supplying current).
Electromotive force (Emf) of a cell: Is the voltage across the cell in an open circuit (when it is supplying no current).
Voltmeter: Is an instrument used to measure voltage (Emf or P.d) .
N/B Potential difference between two points A and B V_{AB} of a conductor , (see the fig. below)
is the work done in moving a unit charge from point B to A of the conductor.
Hence P.d = Work done (in joules)
Charge moved (in coulombs)
Or V_{AB} =^{W}/_{Q}
 In moving charge of 10 coulombs from point B to A, 120 J of work is done. What is the Pd between A and B?
V_{AB} =^{W}/_{Q} = ^{120}/_{10} = 12V
A voltmeter is usually connected in parallel with the circuit (across the appliance whose voltage is to be determined e.g the bulb). See fig. below.
Reason: Because it has very high resistance.
The ammeter on the other hand is usually connected in series with the circuit because it has very low resistance. See the figs. below
CURRENT IN A PARALLEL CIRCUIT ARRANGEMENT
CURRENT IN SERIES CIRCUIT ARRANGEMENT
VOLTAGE IN PARALLEL CIRCUIT ARRANGEMENT
VOLTAGE IN SERIES
Qn ; Example 2 pg 169
OHM’S LAW; The current flowing through a conductor is directly proportional to the potential difference across it,provided the temperature and other physical conditions are kept constant.
OHM’S LAW; The current flowing through a conductor is directly proportional to the potential difference across it,provided the temperature and other physical conditions are kept constant.
(Carryout expt 5.3 pg 168 KLB phy BK 3 3^{rd} ed. Pg 168169)
TABLE
Current (A)  0.01  0.02  0.03  0.04  0.05  0.06  0.07  0.08  0.10 
Voltage (Volts)  0.30  0.50  0.75  1.05  1.10  1.35  1.50  1.85  2.00 
Here a graph of current against voltage is a straight line through the origin
Sample graph
= Constant 
The gradient of the graph ∆ Voltage
∆ Current
The constant is the resistance (R) of the nichrome wire used in the experiment.
The SI unit of resistance is the Ohm (Symbol Ω)
Ohms law can also be verified using a standard resister is used in place of nichrome wire in the experiment and a graph of current agaist voltage plotted.
The graph will also be a straight line through the origin as shown below
From Ohm’s law, an Ohm is defined as the resistance of a conductor when a current of 1 ampere flowing through it produces a voltage drop of 1 volt across its ends.
The multiples of an Ohm are;
1 Kilo Ohm (1KΩ) = 1000Ω
1 Mega Ohm (1MΩ) = 1000000Ω
Qns on pg 170 Example 4, 5 and 6.
OHMIC AND NON OHMIC CONDUCTORS
Ohmic conductors: Are the conductors which obey Ohm’s law.
CurrentVoltage graphs for Ohmic conductors is a straight line through the origin.
Examples of Ohmic conductors are metals and electrolytes e.g copper II sulphate.
Non Ohmicconductors : Are the conductors which do not obey Ohm’s law. CurrentVoltage graphs for non Ohmic conductors are not straight lines. Examples of non Ohmic conductors are: Torch bulb, Thermister, Semiconductor diode, Thermionic diode. See the figs. below:
Electric resistance: Is the opposition offered by a conductor to flow of electric current. It is measured in Ohms (Ω).
A material with high conductance has very low resistance e.g copper metal. The instrument used to measure resistance is called the Ohm meter.
FACTORS THAT AFFECT RESISTANCE OF A METALLIC CONDUCTOR
 Temperature: Resistance of good conductorse.g metals increase with increase in temperature.
 Length of the conductor: Resistance of a uniform conductor increases with increase in length (Resistance is directly proportional to length).
 Crosssectional area: Resistance of a conductor is inversely proportional to its crosssection area. i.e the larger the crosssection area the lower the resistance and the smaller the crosssection area, the larger the resistance.
RESISTIVITY OF A MATERIAL
Is the resistance of the material of unit length and unit crosssection area at a certain temperature. The SI unit of resistivity is Ohm Meter (ΩM). The symbol for resistivity is ρ. Hence mathematically,
Resistivity ρ = Area of crosssection (A) x Resistance (R)
Length of the material (L)
ρ = 
AR
L
RESISTIVITY OF SOME MATERIAL AT 20^{0} AND THEIR USES
MATERIAL 
RESISTIVITY (ΩM) 
USE 
Silver  1.6 x10^{8}  Contacts on some switches 
Copper  1.7×10^{8}  Connecting wire 
Aluminium  2.8 x10^{8}  Power cables 
Tungsten  5.5 x10^{8}  Lamp filaments 
Constantan  49×10^{8}  Resistance boxes, variable resisters 
Nichrome  100×10^{8}  Heating elements 
Carbon  3000 x10^{8}  Radio resisters 
Glass  10^{8} 10^{14}  
Polystyrene  10^{15} 
QUESTIONS
KLB PHY BK 3, 3^{rd}ed. Pg 173174 example 79
RESISTERS
Are conductors specially designed to offer particular resistance to flow of electric current.
TYPES OF RESISTERS
They include:
 FIXED RESISTERS: They are resisters designed to give fixed resistance. They include: Wire wound resisters and carbon resisters.
2.VARIABLE RESISTERS: These are resisters with varied range of resistance.
They include:
(a) Rheostat (b) Potentiameter
(a) Rheostat: Is a two terminal variable resister. It is represented in electric circuits by the symbols shown below:
N/B Moving the sliding contact along the length of the resister varies the resistance between points A and B. When the contact is nearer A, the resistance of the rheostat is lower.
(b) Potentiameter: Is a variable resister with 3 terminals. See the fig below.
 NON LINEAR RESISTERS: The current flowing through these resisters not changelinearly with the change in the applied voltage. Such resisters include; Thethermister, the light dependent resister (LDR).
MEASUREMENT OF RESISTANCE
There are 2 ways of determining resistance namely:
 The voltmeter ammeter method
 The wheatstone bridge method
 VOLTMETER AMMETER METHOD
Apparatus: Two cells, Switch, Voltmeter, Ammeter, Variable resister, Resister R.
Set up apparatus as shown below
– With the switch open, record the voltmeter reading V and the corresponding ammeter reading I. Here no current is flowing and hence both V and A read 0.
– Close the switch and by adjusting the variable resister to the given lengths, record the other 5 values of V and the corresponding values of I in the table.
Length of resistance wire (cm)  0  20  40  60  80  100 
Voltage V (Volts)  0  
Current I (Amps)  0  
^{V}/_{I} (V/A)  0 
It is observed that, as I increases V also increases.
– Compare all the values of ^{V}/_{I}
All the values of ^{V}/_{I} are found to be the same/ almost the same.
– Plot a graph of V against I.
– Determine the gradient of the graph.
It is a straight line graph whose gradient = resistance R
Question on using a meter bridge to determine resistance
In an expt. to determine resistance a nichrome wire using a meter bridge, the balance point was found to be the 38cm mark. If the value of the resistance in the right hand gap needed to balance the bridge was 25 Ω. Calculate the value of the unknown resister.
Since AB = 100cm and AC = 38cm, CB = 10038 =62cm
But R = 25 R = 38 x 25 = 15.32Ω
38 62 62
RESISTER NETWORKS
 RESISTERS IN SERIES
Here VT = V1 + V2 + V3
RT = R1 + R2 + R3
QnsExample 11 and 12 pg 181 KLB BK 3 PHYSICS
 RESISTERS CONNECTED IN PARALLEL
The figure below shows R1, R2, and R3 connected in parallel
Here IT = I1 + I2 + I3
and 1 = 1 + 1 + 1
RT R1 R2 R3
Qns: Example 13, 14, 15 and 16 pg 183184 KLB BK 3 PHY 3^{rd} Ed
N/B Equivalent resistance, total resistance and effective resistance mean the same thing.
Qns: Example 17, 21 and 19 pg 185186
Other qns : Example 20 pg 188, Example 20 pg189
ELECTROMOTIVE FORCE AND INTERNAL RESISTANCE
Electromotive force of a cell is the p.d across its terminals when it is supplying no current, (Cell in open circuit).
Once a cell supplies current to an external circuit, the p.d across it it drops by a value called ‘ lost voltage’. This loss in voltage is due to the internal resistance (r) of the cell.
Internal resistance: A cell or any source of emf is made up of material that are not perfect conductors of electricity. They therefore offer some resistance to the flow of current that they generate. This resistance is usually low and is called internal resistance (r) of the cell or battery.
Relationship between emf and r
If a resister R is connected in series with a cell (see fig. below),
I = Emf
Total resistance
I = E
R + r
Hence E = I (R+r) or E = IR + Ir
N/B IR is the voltage drop across resister and Ir the voltage drop across r
To determine the internal resistance of a cell
Apparatus: Voltmeter, ammeter, variable resister, cells, connecting wire.
– Connect apparatus as shown below;
Adjust the resister to minimum value of current.
– Increase the current in stages and record the corresponding values of current in the table below.
Current I (Amps)  
Voltage V (Volts) 
Plot a graph of voltage against current
The graph is straight line as shown below
– If the equation of the graph is E = V + Ir
(a) Find the value of :
(i) E
Solution;
V = r I + E
y = m x + c
Hence E = yintercept = ———— ( read from the graph plotted)
(ii) r
Solution
r = gradient
Questions; Example 21, 22, 23 pg 193194 KLB PHY BK3
Other qnsexercise 5qn 7.
WAVES II
A Wave: Is the disturbance that moves through a medium.
In this topic, we shall study characteristics of waves which include:
– Reflection
– Refraction
– Diffraction
– Interference
A ripple tank; Is an apparatus used to demonstrate the properties of wave like reflection, refraction, diffraction and interference.
Below is the diagram of a ripple tank
It consists of a transparent tray containing water having a lamp above and a screen below the tank.
Circular waves shown below are seen on the screen when a finger is dipped into the water.
When a ruler is drugged in the water, straight pulses are formed on the screen as shown below;
A wave front: Is an imaginary line which joins a set of particles which are in phase in wave motion.
See figs. below
CHARACTERISTICS OF WAVES
(a) REFLECTION
(i) Reflection of a plane wave front by a plane reflector
PROCEDURE
– Generate plane wave fronts in a ripple tank e.g by letting the frame of a running motor touch the water surface.
– Observe on the screen how the waves are reflected on the straight walls of the ripple tank, (Plane reflector).
The following will be observed
Diagram to show reflection of plane waves on a plane reflector
It is observed that waves obey the laws of reflection i.eagle of
incidence = angle of reflection.
– The lines drawn from the reflecting surface cutting the incident waves perpendicularly are called incident wave fronts.
– The lines drawn from the reflecting surface cutting the reflected waves perpendicularly are called the reflected wave fronts
(ii) Reflection of plane waves by curved reflectors
(I) CONCAVE REFLECTOR
Just like parallel rays of light are reflected through the principle focus (F) of the concave mirror, (or converge at F and spread out),
The reflected waves too converge at F and spread out as if originating from F as shown below.
N/B The reflected waves here are circular.
(II) CONVEX REFLECTOR
Just like the convex mirror where parallel rays converge to F behind the mirror, where they appear to diverge from on the real side of the mirror,
(see fig. below)
reflected waves too appear to diverge from the virual principle focus of the reflector as shown below;
The direction and velocity of a wave changes during reflection.
The reflected wave is also 180^{o} out of phase with the incident wave, (hence troughs of the incident wave fall below the crests of the reflected wave). See the figs below;
λ Is the distance from crest to crest of incident wave or crest to crest of the reflected wave.
(b) REFRACTION OF WAVES
Is the bending of waves as they travel from one medium to another.
TO DEMONSTRATE REFRACTIONOF WAVES
Here a glass plate is placed inside a ripple tank as shown below:
A plane wave is then introduced from the deeper end of the ripple tank.
OBSERVATION
As the plane wave travels across the water the wavelength in the shallow region λ2 is less than the wavelength in the deeper region λ1.
See the fig. below
Since during refraction the frequency of the wave does not change and V = f λ, the speed of wave reduces in the shallow as the wavelength reduces.
Speed of wave in deep water
Speed of wave in shallow water 
So therefore here , n =
From V =fλ and given that f does not change,
fλ1
fλ2

n =
λ1
λ2 
n =
If the fig. above, λ1 = 1cm and λ2 = 0.15cm
(c) DIFFRACTION OF WAVES
Is the spreading of waves behind an obstacle.
This can be demonstrated by allowing water waves to pass through holes of various sizes as shown below;
(I) Plane (water) wave through a small gap
(II) Plane waves through a wide gap
(III) Diffraction of plane waves around an obstacle
(IV) Diffraction of circular waves through a small gap / Small arpature
N/B (a) A narrow gap diffracts waves more than a wide gap.
(b) Increase in wavelength reduces diffraction.
(d) INTERFENCE OF WAVES
Is the interference between waves to cancel or reinforce each other.
This happens when waves from one source meet others from another source so that they either reinforce or cancel each other depending on whether the meet in phase or out of phase
To show interference using plane waves and 2 small slits S1 and S2.
(I) CONSTRUCTIVE INTERFERENCE (Constructive superposition)
Is the interaction between waves in phase such that they reinforce each other to give a bigger amplitude e.g
During this process, the resultant amplitude doubles i.e
Amplitude of wave 1 + Amplitude of wave 2 = Total amplitude
but wavelength and hence frequency does not change
Coherent sources: Are sources which produce waves in phase and with the same frequency.
DESTRUCTIVE INTERFERENCE (Destructive superposition)
Is the interaction between waves which are out of phase such that they cancel each other.e.g
APPLICATIONS OF DIFFRACTION
 Study of crystal structure e.g in determining the spacing of atoms and their arrangement.
 In xray photography for medical diagonosis.
 Used in electron diffraction microscope
Interference is used
 To cancel unwanted noise in hall (silencer) through destructive interference.
 In stereo radios and T.V systems to produce louder sound where sound from different loudspeakers interfere constructively at most of the points thus producing an enhanced louder sound
STATIONARY WAVES
This is the wave in which the amplitude of oscillation depends on the position of a particle.
Stationary waves are formed when two ends of a tight rope are made fixed and the middle of the rope plucked as shown below:
He displacement increases gradually from the node towards A (antinode) . The A is the position of maximum displacement. (See the fig. above).
The wave above has only 1 loop (A) and hence its wavelength L = ^{λ}/^{2} .
When the frequency of the wave is increased 2 , 3, 4 etc loops can be formed within the same region as shown below:
QUESTION.
The apparatus below show how to set up a stationary wave on stretched string. The stationary wave is produced when the frequency of the vibrator is 40HZ.
Calculate the speed of the wave in the string.
Solution: V = f λ
= 40 x 1 = 40m/s
STATIONARY WAVES IN AN OPEN TUBE
This can be demonstrated using a vibrating tuning fork, a beaker, an open tube and water as shown below;
A vibrating tuning fork is held above the open tube and the tube raised up gradually until there is resonance (loud sound is heard). This will be at the first node, N (Resonance occurs at the node)
Measure the length L1 Say = 0.25 cm
Continue raising the open tube until there is another resonance. This will be at the next node. i.e
Measure the length L2 Say = 0.75 cm
Then ½ λ = L2 – L1 = 0.75 – 0.25 = 0.5cm
½ λ = 0.5cm
λ = 0.5 x 2 = 1cm (This is the λ for this wave)
If the frequency of the tuning fork is known, the velocity of the wave can be calculated using the formula:
V = f λ
N/B The loud sound (at resonance) is also called overtone
– When the end correction (e) is there, then it is added to the length L of
the wave before it is used the calculation.
ELECTROSTATICS II
Electrostatics: Is the study of charge at rest.
The charge is acquired by rubbing.
Like charges repel while unlike charges attract.
The force of attraction/repulsion is stronger when the charged bodies are closer to one another. The force diminishes when they are moved far apart.
The amount of charge on a body can be determined using the gold leaf electroscope.
ELECTRIC FIELD PATTERNS
(a) Positive and negative point charges
(b) Positive and positive point
N/B – The lines of force move from the positive charge to the negative charge.
– The lines of force do not cross one another.
– The lines of force are close where the field is strong and far apart where
the field is weak.
CHARGE DISTRIBUTION THE SURFACE OF CONDUCTORS.
 spherical shaped conductor
(b) Pear shaped body
(c) Cuboid
N/B Sharp points have high charge concentration, See fig. (b) and (c) above.
CHARGE ON SHARP POINTS
Since charge on sharp point is extremely concentrated, when the sharp part of the conductor is brought close to a candle flame, the flame is diverted as if wind is emanating from the sharp point. See the dgm below
EXPLANATION : If the charge on the conductor is positive (as above), the high concentration of positive charges at the sharp point on the conductor causes ionization of the sorrounding air to produce positive ions and electrons. The electrons are attracted towards the positive conductor while heavy ions drift towards the flame forming an electric wind.
If the conductor is brought close to the conductor from above, the flame splits as shown below:
In this case the flame causes ionization of the surrounding air molecules. The positive ions formed move away from the charged rod towards the flame by repulsion, causing the division of the flame as shown in the diagram.
LIGHTINING ARRESTOR
Movement of the clouds in the atmosphere produce large amounts of static charges due to friction with air. The static charges in the cloud induce large opposite charges on the Earth, producing high P.d between the Earth and the cloud. The high P.d makes air to become a charge conductor. The opposite charge strongly attract and neutralize causing lighting and thunder.
Lightning can cause destruction to buildings and other objects on the Earth’s surface. To save the buildings from being struck, a lightning arrestor is used.
It comprises of thick copper wire with sharp spikes at the top. See the fig. below:
The wire is connected to large thick copper plate buried deep into the ground.
When the cloud gets –vely charged, it induces a positive charge on the spikes of the arrestor.
The +ve charges concentrated at the spikes, ionize the air around it.
Positive charges formed go to neutralize the –ve on the cloud as the –ve charges neutralize the +ve on the spikes and the building. Hence the building is protected.
CAPACITOR
A capacitor is a device for storing charge.
It consists of two or more plates separated by either a vacuum or material media called dielectric. A dielectric can be air, glass or plastic.
There are 3 main types of capacitors namely
 Paper capacitors
 Electrolytic capacitors
 Variable air capacitors
CHARGING AND DISCHARNING A CAPACITOR
FACTORS THAT AFFECT CAPACITANCE OF A PARALLEL PLATE CAPACITOR
 AREA OF OVERLAP; Increasing area of overlap increases the capacitance of a parallel plate capacitor while reducing the area of overlap reduces the capacitance.
 THE DISTANCE OF SEPARATION; Increasing the distance of separation of plates of parallel plate capacitor, reduces the capacitance while reducing the distance of separation increases the capacitance.
 THE NATURE OF THE DILECTRIC; Using glass as the dielectric material give a different value of capacitance from when air is used. Plastic will also give a different value when used as the diletric.
From 1 and 2 above, capacitance is directly proportional to the area of overlap and inversely proportional to the distance of separation,
C α
C =
QUESTION
Two plates of a parallel plate capacitor are 0.6mm apart and each has an aea of 4cm^{2}. Given that the potential difference between the plates is 100V, calculate the charge stored in the capacitor. (answer = 5.9 x 10^{10}C)
CAPACITOR COMBINATIONS
Just like resistors, capacitors can be arranged in series or in parallel.
 CAPACITORS IN SERIES
APPLICATIONS OF CAPACITORS
 Rectification: This is the conversion of the a.c to d.c . During this conversion, in order to maintain high voltage, capacitors are included in the circuit.
 A capacitor is included in the primary circuit of the induction coil to eliminate sparking at the contacts
 A variable capacitor is connected in parallel with an inductor in tuning of radio in order to receive the signal
 Capacitors re used in delay circuits designed to give intermittent flow of current in car indicators.
 A capacitor is included in the flash circuit of a camera. The camera flashes during the discharging process of the capacitor.
HEATING EFFECT OF AN ELECTRIC CURRENT
SIMPLE EXPERIMENTS TO SHOW THE HEATING EFFECT OF AN ELECTRIC CURRENT
 Set up apparatus as shown below
Note the initial temperature of water.
Close the switch for about 10 mins and note the new temperature.
Results: The temperature of the water increases
Explanation: The hot coil (heated by the heating effect of an electric current) heats up the water.
here experiments are:
 When an immersion heater is dipped into water and the switch closed, the temperature of the water rises. Here the electric energy is converted to heat energy.
 A bulb feels warm/hot to touch after lighting for some time.
Factors affecting heating by electric current
 The amount of current
 The resistance of the conductor
(ii) The time for which the current flows
Electrical energy formulae
Since Voltage V = Work done W
Charge Q
V = W
Q W = VQ …………………… (i) Where W=E
and Q = I t …………………….(ii)
Combining (i) and (ii) E = W = VQ = V It
Hence electric energy E =VIt The SI unit of electrical energy is the Joule (J)
Electrical Power
Power = Work = Electrical energy = VIt
timetime t
Hence P = VI
From Ohm’s law, V = IR
Hence P = I^{2} R
Also by substituting I with V/R
P = V^{2}
R
Substitution of V as IR in equation E = VIt gives;
E = I^{2}Rt
QUESTIONS
 A torch bulb is lebelled 2.5V, 0.3A. Calculate the power of the bulb.
Answer other questions from KLB and foundation PHY
APPLICATIONS OF HEATING EFFECT OF AN ELECTRIC CURRENT
The heating effect of an electric current is used in the following electric appliances:
 THE FILAMENT LAMPS
 A conductor may be heated to white hot without melting. Under this condition, it emmits light. The electric bulb operatesthis way.
 e, when current flows through the lamp filament, it heats up to a high temperature and becomes white hot . For this reason it is made of tungsten (a metal with high melting point.).
DIAGRAM OF AN ELECTRIC BULB longhorn pg 228
 The filament is enclosed in a glass bulb from which air has been removed to prevent the oxidation of the filament
 Since hot metals evaporate in the vacuum the bulb is filled with inactive gas like nitrogen and argon to slow down the rate of evaporation and hence increase the life of the filament.
 FLUORESCENT LAMPS
 They are more efficient than filament lamps.
DIAGRAM OF A FLUORESCENT LAMP/TUBE KLB PG 256
 When the lamp is switched on, the mecuryvapouremmits ultraviolet radiation which makes the powder on the inside wall of the tube to fluoresce/ glow/ emmit visible light.Different powders emmit different colours.
 THE FUSE
 Is a short length of wire of material of low melting point which melts and breaks the circuit when the current through it exceeds a certain value.
 The breaking of the fuse, saves the wiring from becoming hot and catching fire.
DIAGRAM OF THE FUSE KLB PG 257
 A 15A fuse will blow out if a current of 15A flows through the circuit.
Other appliances in which the heating effect of an electric current is used include:
 Electric iron
 Electric kettle
 A radiation room heater
 Immersion heater
QUANTITY OF HEAT
 Heat is a form of energy that flows from one body to another due to temperature difference.
 The absorption of heat by a body results in the rise in temperature while the loss of heat results in the fall in temperature.
HEAT CAPACITY AND SPECIFIC HEAT CAPACITY
 HEAT CAPACITY
Is the quantity of heat required to raise temperature of a given mass of a material by one Kelvin. It is denoted by C.
C = Heat energy absorbed Q
Temperature change θ
i.e C = Q
θ The SI unit for heat capacity is J/K^{1}.
QUESTION
Calculate the quantity of heat required to raise temperature of metal block with a heat capacity of 460 J/K^{1} from 15^{0}C to 45^{0}C.
SOLUTION
C = 460 J/K^{1}
Temperature change θ = 45 – 15 = 30^{0}C
Q =Cθ
= 460 x 30
= 13800J
 SPECIFIC HEAT CAPACITY
Is the quantity of heat required to raise the temperature of a unit mass of a substance
by 1 K ie heat capacity per unit mass .
c = Heat capacity
Mass
It is denoted by c .
c = Q
θ
m
c = Q
mθ The SI unit for c is J kg^{1}K^{1}
Q =mcθ
From the above equation, mc = Q
θ
= C 
But Q
θ
Hence C = mc
QUESTIONS
 A block of metal of mass 1.5kg which is suitably insulated is heated fro 30^{0}C to 50^{0}C in 8mins and 20 seconds by an electric heater coil rated at 54W. Find:
(a) The quantity of heat supplied by the heater
Solution
Q = P x t
Q = 54 x 500
= 27000 J
(b) The heat capacity of the block
C = Q
θ
But Q = 27000 and θ = 50 – 30 = 20^{0}C
C = 27000
20
= 1350 J/K
 The specific heat capacity c
C = mc
1350 = 1.5 x c
c = 900Jkg^{1}K^{1}
 Find the final temperature if a heater rated at 42W heats 50g of water from 20^{0}C in five mins. (Specific heat capacity of water is 4200 Jkg^{1}K^{1})
SOLUTION
Heat lost by heater = heat gained by water
P x t = mcθ
42 x 5 x 60 = 0.05 x 4200 x θ
θ = 60
The final temp = 20 + 60 = 80^{0}C
 A piece of copper of mass 60g and specific heat capacity 390Jkg^{1}K^{1} cools fro 90^{0}C to 40^{0}C. Find the quantity of heat given out.
Solution
Q =mcθ
= 0.06 x 390 x (9040)
= 1170J
SPECIFIC HEAT CAPACITIES OF SOME MATERIALS
MATERIAL  Specific heat capacity ( x 10^{3} Jkg^{1}K^{1}) 
Water
Alcohol Kerosene Ice Aluminium Glass Iron Copper Mercury Lead 
4.2
2.3 2.2 2.1 0.9 0.83 0.46 0.39 0.14 0.13 
DETERMINATION OF SPECIFIC HEAT CAPACITY
EXPERIMENT TO DETERMINE SPECIFIC HEAT CAPACITY OF A SOLID BY ELECTRICAL METHOD
Apparatus: Cylindrical solid metal block with 2 holes, connecting wires, voltmeter, ammeter, dry cells, variable resistor, stop watch.
Measure the mass m of the metal block and set up apparatus as shown below
Close the switch and start the stop watch. Record the time taken for the temperature to rise by 8^{0}C.
N/B Cotton wool acts as lagging material to prevent heat loss by radiation from the metal block to the outside.
 The silver foil is used to minimize heat loss by radiation
 The wooden container minimizes heat loss by conduction
Electrical energy E spent by the heater in time t is given by E = IVt. This energy is converted into heat energy that is absorbed by the metal block (mcѲ). i.e
Heat lost by the heater = Heat gained by the metal block
IVt =mcѲ
From this specific heat capacity c, of the solid can be calculated as:
c = VIt
mѲ
QUESTION.
In an experiment to determine specific heat capacity of copper, the following data was obtained:
 Mass of copper block = 200g
 Initial temperature of the block = 22^{0}C
 Ammeter reading = 5A
 Voltmeter reading = 3.0V
 Final temperature of the block = 30^{0}C
 Time of heating = 7 mins
Use the data to calculate specific heat capacity c of copper
(Ans = 394 JKg^{1}K^{1})
DETERMINATION OF SPECIFIC HEAT CAPACITY OF WATER BY METHOD OF MIXTURES
This can be done by heating a solid to a certain temperature and transferring to cold water in a beaker as shown below.
Here,
Heat lost by the hot solid = Heat gained by the cold water + Heat gained by the container
M_{s}c_{s}Ѳ_{s}= M_{w}c_{w}Ѳ_{w} + M_{c} c_{c}Ѳ_{c}
QUESTION
In an experiment to determine the specific heat capacity of water, the following data was obtained;
 Mass of solid = 50g
 Specific heat capacity of the solid = 400J/kg/k,
 Initial temperature of the hot solid = 100^{0}C
 Mass of the container = 200g
 Specific heat capacity of the material of the container = 400J/Kg/K
 Mass of water = 100g
 Initial temperature of water and the container = 22^{0}C
 When the hot solid was transferred into the cold water in the container, the final temperature of the mixture was = 25^{0}C
Use the data to determine the specific heat capacity of the water.
Answer = 4200
TO DETERMINE SPECIFIC HEAT CAPACITY OF A LIQUID ELECTRIC METHOD
Apparatus: Lagged Copper container, heating coil, thermometer, connecting wires, ammeter, voltmeter stop watch
Measure the mass m_{c} of the copper container and mass m_{l} of the liquid and set up apparatus as shown below
Close the switch and start the stop watch. Record the time taken for the temperature to rise by say 10^{0}C.
Here,
Heat energy produced by the heater =Heat energy gained + Heat energy gained
by the liquidby the container
QUESTION
In an experiment to determine specific heat capacity of a liquid, the following data was obtained:
 Power of the heater = 30W
 Mass of the container = 200g
 Specific heat capacity = 400J/Kg/K
 Mass of water in the container = 100g
 Specific heat capacity of water = 4200J/Kg/K
Use se the data to calculate the time taken by the heater to raise the temperature of the water and container from 20^{0}C to 23^{0}C.
^{ }
CHANGE OF STATE
Heating a substance is known to cause an increase in temperature of the substance. However there are situations when heating does not cause any increase in temperature. These include:
 When a solid is melting to liquid
 When a liquid is boiling to ga
Below is a temperature – time graph for a solid heated from 20^{0}C to 104^{0}C.
In Region AB, temperature rises steadly from 20^{0}C to 0^{0}C.
In this region heating causes an increase in temperature.
This happens during the first 10 seconds.
Region BC is the melting point of the solid. Here, the heat energy supplied to the solid does not cause a rise in temperature of the solid, it is used to change (melt) the solid to liquid by breaking the forces of attraction between the solid molecules. This heat energy absorbed by a solid during melting is called LATENT HEAT OF FUSION.
DEFINITION OF LATENT HEAT OF FUSION; Is the quantity of heat required to change the state of a material from solid to liquid without temperature change.
SPECIFIC LATENT HEAT OF FUSION: Is the quantity of heat required to change a unit mass (1kg) of substance from solid to liquid without change in temperature.
Its SI unit is J/kg
QUESTION
From the graph above,
 how long does it take for the solid to melt?
 At what temperature does the solid melt?
After the melting process, in region CD, as further heating takes place, temperature increases steadly to the boiling point.
Region DE is the boiling point of this substance. Here the heat energy supplied is not used to increase the temperature of the liquid. It is used to change (boil) the liquid to gas. Such heat energy absorbed by a solid during boiling is called LATENT HEAT OF VAPOURIZATION.
DEFINITION OF LATENT HEAT OF VAPOURIZATION: Is the quantity of heat required to change the state of a material from liquid to gas without change in temperature.
SPECIFIC LATENT HEAT OF VAPOURIZATION: Is the quantity of heat required to change a unit mass of a material from liquid state to gas without change in temperature.
Its SI unit is J/kg
QUESTIONS (From the graph)
 How long does it take for the liquid to change to change to gas.
 At What temperature does the solid boil?
After the boiling process, as heating continues (Region EF), the temperature rises steadily.
N/B During cooling;
 Latent heat of vapourization is lost as the gas changes to liquid.
 Also latent heat of fusion is lost as the liquid changes back to solid.
N/B – When there is temperature change, heat is determined using the formula H =M x c x Ө
Where H is the heat loss/gain
M is the mass of the substance
c is the specific heat capacity
Ө is the temp change
When there is no temperature change, heat is determined using the formula H = ML
Where M = Mass
L = Latent heat of fusion/vapourization
QUESTIONS
 Determine the amount of heat required to change 0.5kg of ice at
10^{0}C to liquid water at 20^{0}C. (Specific heat capacity of ice 2100J/kg/K, specific heat capacity of water is 4200J/kg/K, specific latent heat of fusion of ice is 3.36 x 10^{5}J/Kg/K).
SOLUTION
H for raising temp of ice from 10^{0}C to 0^{0}C
H = M c Ө
H = 0.5 x 2100 x (100)
= 10500J
H for melting the ice
H = ML
H = 0.5 x 3.36 x 10^{5}
= 1.68 x 10^{5}J
H for raising temp of water from 0^{0}C to 20^{0}C
H = M c Ө
= 0.5 x 4200 x (200)
= 42000J
TOTAL HEAT REQUIRED = 10500 + 1.68 x 10^{5} + 42000
= 220500J
 Calculate the amount of heat required to change 2kg of ice at 20^{0}C to liquid water at 100^{0}C.(Specific heat capacity of ice 2100J/kg/K, specific heat capacity of water is 4200J/kg/K, specific latent heat of fusion of ice is 3.36 x 10^{5}J/Kg/K).
 Calculate the amount of heat required to change 2kg of ice at 20^{0}C to steam at 100^{0}C.(Specific heat capacity of ice 2100J/kg/K, specific heat capacity of water is 4200J/kg/K, specific latent heat of fusion of ice is 3.36 x 10^{5}J/Kg/K, Specific Latent heat of vapourization of steam = 2.26 x 10^{6} J/kg).
Extra question; Example 9 &10 pg 277 to 279 KLB BK 3 THIRD ED.
FACTORS AFFECTING MELTING AND meltng POINTS
These include:
 Pressure
 Impurities
EFFECT ON MELTING POINT
 Pressure
Increase in pressure lowers the melting point.
(b)Impurities
Adding impurities to substance lowers its melting point.
EFFECT ON BOILING POINT
 Pressure
Increase in pressure increases boiling point of a liquid.
(b) Impurities
The presence of impurities in a liquid raises its boiling point.
EVAPORATION
Molecules in a liquid are in continuous random motion. A molecule on the surface of the liquid may acquire sufficient K.E to overcome the attraction force from the neighbouring molecules in the liquid.
This process is known as evaporation and takes place at all temperatures.
FACTORS AFFECTING THE RATE OF EVAPORATION
 Temperature
Increasing temperature of liquid makes the molecules on its surface to move faster. This makes it easier for more of them to escape.E.g it takes a shorter time for clothes to dry on hot day than on a cold day.
Hence increase in temperature increases the evaporation rate.
(b)Surface area
Increasing the surface area increases the rate of evaporation e.g a wet bed sheet dries faster when spread out than when folded.
(c )Draught
Passing air over a liquid surface increases the rate of evaporation, this is why wet clothes dry faster on windy day.
(d) Humidity
This is the concentration of water vapour in the atmosphere. High humidity reduces the rate of evaporation, this is why wet clothes take a longer time to dry up on a humid day.
DIFFERENCES BETWEEN BOILING AND EVAPORATION
EVAPORATION  BOILING 
Takes place at all temperatures  Takes place at a fixed temperature 
Takes place on the surface of the liquid  Takes place throughout the liquid 
Decreasing the atmospheric pressure increases the rate of evaporation  Decreasing the atmospheric pressure lowers the boiling point 
APPLICATIONS OF COOLING BY EVAPORATION
 Sweating
 Cooling of water in a porous pot (water pot).
 The refrigerator
THE MAIN PARTS OF A REFRIGERATOR ARE SHOWN BELOW
In the upper coil, the volatile liquid (Freon) takes latent heat from the air around and evaporates causing cooling in the cabinet. The vapour is moved by the pump into the lower coil where it is compressed and changes back to liquid form (Freon). During this process, heat is given out and is conducted away by the copper fins. The liquid (Freon) goes back to the upper part of the coil and the cycle is repeated.
REVISION EXERCISE 9 PG 288 KLB BK 3, 3^{RD} EDITION.
GAS LAWS
These are laws which show the relationship that exists between pressure, temperature and volume of gases.
They include:
 Boyles law
(b) Charles law
(c ) Pressure law
BOYLE’S LAW
It states that ‘pressure of a fixed mass of gas is inversely proportional to its volume provided temperature is kept constant’.
This can be demonstrated using the arrangements shown below:
When the nozzle of the syringe is closed with a finger and the piston slowly pushed inwards as shown above, it is observed that an increase in pressure of the fixed mass of gas results in decrease in volume.
THE APPARATUS BELOW CAN ALSO BE USED TO SHOW THE RELATIONSHIP BETWEEN PRESSURE AND VOLUME OF FIXED MASS OF GAS
Any pressure here recorded by the pressure gauge is shown by a fall in the level of oil in the right arm. This results in the rise of the oil level in the right arm by a certain height hence reducing the height h. The height h is the represents the volume of the air because the glass tube has uniform crosssection area.
Hence in this experiment, as pressure increases, volume decreases.
From the statement of Boyle’s law,
P α 1 Since K is the constant,
V
P = K x 1
V
Hence PV = K
Meaning that P_{1}V_{1} = P_{2}V_{2}
THE SHAPE OF AGRAPH OF P PLOTTED AGAINST V IS SHOWN BELOW
QUESTIONS
 The pressure of fixed mass of gas is 760mmHg when its volume is 38cm^{3}. What will be its pressure when the volume increases to 100cm^{3}.
 Complete the table by filling in the missing values.
Pressure (cmHg)  0  _  90  _ 
Volume (cm^{3})  36  80  _  40 
 The volume V of a gas at a pressure P is reduced to ^{3}/_{8}V without change in temperature. Determine the new pressure of the gas.
 A column of air 26cm long is trapped by mercury thread 5cm long as shown in fig. (a) below. When the tube is inverted as in fig. (b), the air column becomes 30cm long.
What is the value of the atmospheric pressure?
CHARLE’S LAW
It states that ‘ volume of a fixed mass gas is directly proportional to its absolute temperature if the pressure is kept constant’.
The apparatus shown below can be used to illustrate this law
As the temperature rises (shown on the thermometer), the height h (volume) also increases i.e the sulphuric acid index moves up.
This shows that an increase in temperature of the air increases volume.
A graph of volume against temperature is a straight line as shown below
If the graph is extrapolated as shown above, it cuts the temperature axis
at 273C (absolute zero).This is the lowest temperature a gas can fall to. At this temperature, the volume of the gas is assumed to be zero (from the graph).
From the statement of Charle’s law above,
V α T
V = KT Where K is a constant
= K 
V
T
Which implies that
= 
V1 V2
T1 T2
NB The absolute temperature is the temperature on the Kelvin scale. When carrying out calculations use the temperature in K.
ASK AQUESTION ON CONVERSION OF ^{0}C TO K
QUESTIONS
 02m^{3} of a gas at 27^{0}C is heated at a constant pressure until the volume is 0.03m^{3}. Calculate the final temperature of the gas in ^{0}C.
 A mass of air of volume 750cm^{3} is heated at a constant pressure from 10^{0}C to 100^{0} What is the final volume of the air?
PRESSURE LAW
It states that the pressure of fixed mass of gas is directly proportional to its absolute temperature provided the volume is kept constant.
The apparatus below can be used to illustrate the pressure law
In the experiment above, an increase in the thermometer reading results to an increase in the reading of the pressure gauge.
From the statement of Pressure law above, it is true that;
= 
P1 P2
T1 T2
QUESTIONS
 A cylinder contains oxygen at 0^{0}C and 1 atmosphere pressure. What will be the pressure in the temperature rises to 100^{0}
 At 200C, the pressure of a gas is 50cm of mercury. At what temperature would the pressure of the gas fall to 10cm of mercury?
EQUATION OF STATE
Consider a fixed mass of gas being changed from state 1 to state 2 through an intermediate state C as shown in the fig. below:
QUESTIONS
 A mass of 1200cm^{3} of oxygen at 27^{0}C and pressure 1.2 atmospheres is compressed until its volume is 600cm^{3} and its pressure is 3.0 atmospheres. What is the new temperature of the gas in ^{0}C?
 The volume of a fixed mass of air at 27^{0}C and 75cmHg is 200cm^{3}. Find the volume of the air at 73^{0}C and 80cmHg.