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) Distance-time graphs;

(b) Displacement – time graphs

(c)  speed – time graphs

(d) Velocity-time graphs

 

a)DISTANCE-TIME GRAPHS

(i) For a body moving with constant speed

 

(ii) For a body moving with variable speed

 

(iii) Stationary body

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(b) DISPLACEMENT-TIME GRAPHS

(i) Uniform velocity

(ii) Variable velocity

(iii) Stationary body

(c) SPEED-TIME GRAPHS

(i) Constant speed

 

(ii) Variable speed

 

AREA UNDER SPEED-TIME GRAPH

This gives the distance covered

Distance covered=Area under the graph=(½ ^x 3 ^x15)+( 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 ten-tick = 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

1. 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

 

2. 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 at²    ……….     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  V² = u² + 2aS  …………….  This is the 3^rd equation of linear motion

The 3equations include(i) v=u+at

(ii) s=ut+¹/₂ as

(iii) v²=u²+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=u-at

s=ut-¹/₂at²

v²=u²-2as

 

QUESTIONS

1. 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)

2. 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)
3. 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+¹/₂gt²

v²=u²+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

1. 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 – ¹/₂gt²

v² = u² – 2gs

 

Time to reach maximum height & time of flight

(a)   Time to reach maximum height

At maximum height   v = 0

Hence from v = u-gt

0 = u-gt

t = ^u/_gwheret 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

1. Set up apparatus as shown below
2. Starting with l = 50cm, set the pendlum bob swinging through an angle of about 10⁰. N/B –  Length of pendlum L = Length of thread + Radius of bob.

see fig. below

 

3. Time 20 oscillations.
4. 4. Repeat the experiment and obtain the average time for 20 oscillations.

Hence complete the table below

6

5.  Determine the periodic time T (time for one oscillation ) and fill in the table.

QUESTIONS

(a)  Plot a graph of T²  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  T² = 4∏²L/_g

 

HORIZONTAL PROJECTION

Consider a body projected horizontally with horizontal velocity V_hfrom 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

1. 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)

 

2. 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 10⁸ 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
20o	13
30o
40o
50o
60o
70o 80o

 

-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 2-6

REFRACTIVE INDEX IN TERMS OF VELOCITY-

Other applicable formula

1. Refractive index for any medium = velocity of light in air

Velocity of light in medium

 

QUESTIONS: From KLB  BK 3. Example 7-10

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 63-65 Example 14- 17.  Example 19 Pg 67-68

 

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 glass-air interface as shown below

QUESTIONS:

1. 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)

 

 

1. A ray of light travels fro a transparent material to perspex as shown below

Calculate the critical angle c     (24.6)

1. Calculate the critical angle of diamond given that its refractive index is 2.42

 

 

 

1. The critical angle for water is 48.6^o. Calculate the refractive index for water.

 

 

 

 

1. A ray of light travels through air into medium as shown in the fig. below

Calculate the critical angle for the medium

 

 

 

1. 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

1. 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.

2. 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

1. 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

1. 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

1. A van of mass 3 tonnes is travelling at 72km/h. Calculate the momentum of the vihecle.
2. A car is moving at 36km/h. What velocity will double its momentum?

 

2. 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

1. 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

 

2. 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

1. 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.

 

2. 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_sR

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_kR

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

1. 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/s²

 

 

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  =  10³J

1MJ = 10⁶J

Questions on work and energy

1. Calculate the work done by a stone mason in lifting a stone of mass 15kg through a height of 2M. ( Take g=10N/kg)
2. 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 MV²

Where   M  =  Mass of the body

V    =  Velocity

But we can have,

Final  K.E  =  ½ MV²_final           Where  V_final is the final velocity,         and

Initial  K.E  =  ½ MU²_initial       Where  U_initial is the initial velocity.

QUESTIONS;

1. 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 at²   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 MV²_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 136-137

 

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;

1. The single fixed pulley
2. The single movable pulley
3. 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    Cross-section 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

πR2 πr2

 

 

V.R  =                     Where R is the radius of the load piston

r is the radius of the effort piston

 

0r       V.R  =  R²/r²

 

Qns; Example 20, 21 from KLB SEC PHY BK 3.

Other qns for general practice – Exercise 4 pg 159-161 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

1. 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 168-169)

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.

Current-Voltage 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. Current-Voltage 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

1. Temperature: Resistance of good conductorse.g metals increase with increase in temperature.
2. Length of the conductor: Resistance of a uniform conductor increases with increase in length (Resistance is directly proportional to length).
3. Cross-sectional area: Resistance of a conductor is inversely proportional to its cross-section area. i.e the larger the cross-section area the lower the resistance and the smaller the cross-section area, the larger the resistance.

 

RESISTIVITY OF A MATERIAL

Is the resistance of the material of unit length and unit cross-section 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 cross-section (A)  x  Resistance (R)

Length of the material  (L)

 

ρ  =

AR

L

 

 

 

 

 

 

 

RESISTIVITY OF SOME MATERIAL AT 20⁰ 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	1015

 

QUESTIONS

KLB PHY BK 3, 3^rded.  Pg 173-174 example 7-9

 

RESISTERS

Are conductors specially designed to offer particular resistance to flow of electric current.

 

TYPES OF RESISTERS

They include:

1. 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.

3. 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:

1. The voltmeter ammeter method
2. The wheatstone bridge method

 

1. 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  =  100-38  =62cm

 

But   R     =     25                      R = 38 x 25   = 15.32Ω

38           62   62

 

RESISTER NETWORKS

1. RESISTERS IN SERIES

 

Here    VT  = V1 +  V2  +  V3

RT  = R1  +  R2  +  R3

 

QnsExample  11  and 12 pg 181 KLB BK 3  PHYSICS

 

2. 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 183-184 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 185-186

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  =  y-intercept  =  ————   ( read from the graph plotted)

 

(ii)  r

Solution

-r  =  gradient

 

Questions;  Example 21, 22, 23 pg 193-194 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 x-ray 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

 

1. AREA OF OVERLAP; Increasing area of overlap increases the capacitance of a parallel plate capacitor while reducing the area of overlap reduces the capacitance.

 

2. 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.

 

3. 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 4cm². Given that the potential difference between the plates is 100V, calculate the charge stored in the capacitor.    (answer = 5.9 x 10^-10C)

CAPACITOR COMBINATIONS

Just like resistors, capacitors can be arranged in series or in parallel.

1. 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  =  I² R

 

Also by substituting  I  with  V/R

P   =    V²

R

 

Substitution of  V as IR in equation   E  =  VIt   gives;

E  =  I²Rt

 

QUESTIONS

1. 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 ultra-violet 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 15⁰C to 45⁰C.

 

SOLUTION

C  =  460 J/K^-1

Temperature change   θ  =  45 –  15   =   30⁰C

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^-1K^-1

 

Q  =mcθ

 

 

From the above equation,   mc  =  Q

θ

=  C

But    Q

θ

Hence   C  =  mc

 

 

 

QUESTIONS

1. A block of metal of mass 1.5kg which is suitably insulated is heated fro 30⁰C to 50⁰C 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  =  20⁰C

 

 

C  =  27000

20

 

= 1350  J/K

 

• The specific heat capacity c

 

C  =  mc

1350  =  1.5  x  c

c  =  900Jkg^-1K^-1

 

 

2. Find the final temperature if a heater rated at 42W heats 50g of water from 20⁰C in five mins. (Specific heat capacity of water is 4200 Jkg^-1K^-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  =  80⁰C

 

 

3. A piece of copper of mass 60g and specific heat capacity 390Jkg^-1K^-1 cools fro 90⁰C to 40⁰C. Find the quantity of heat given out.

 

Solution

Q  =mcθ

=  0.06  x  390  x  (90-40)

=  1170J

 

 

SPECIFIC HEAT CAPACITIES OF SOME MATERIALS

MATERIAL	Specific heat capacity ( x 103 Jkg-1K-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 8⁰C.

 

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 =  22⁰C
• Ammeter reading =   5A
• Voltmeter reading  = 3.0V
• Final temperature of the block =  30⁰C
• Time of heating =  7  mins

Use the data to calculate specific heat capacity c of copper

(Ans = 394 JKg^-1K^-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_sc_sѲ_s=  M_wc_wѲ_w  +  M_c c_cѲ_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  =  100⁰C
• 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 = 22⁰C
• When the hot solid was transferred into the cold water in the container, the final temperature of the mixture was = 25⁰C

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 10⁰C.

 

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 20⁰C to 23⁰C.

 

 

 

 

 

 

 

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 -20⁰C to 104⁰C.

 

 

 

In Region AB, temperature rises steadly from  -20⁰C to 0⁰C.

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)

1. How long does it take for the liquid to change to change to gas.
2. 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

1. Determine the amount of heat required to change 0.5kg of ice at

-10⁰C to liquid water at 20⁰C.  (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 10⁵J/Kg/K).

 

SOLUTION

H for raising temp of ice from -10⁰C to 0⁰C

H  =  M c Ө

H  = 0.5 x 2100 x (10-0)

= 10500J

 

H for melting the ice

H = ML

H = 0.5 x 3.36 x 10⁵

= 1.68 x 10⁵J

 

H for raising temp of water from 0⁰C to 20⁰C

H =  M c Ө

= 0.5 x 4200 x (20-0)

= 42000J

TOTAL HEAT REQUIRED = 10500 + 1.68 x 10⁵ + 42000

=      220500J

 

 

 

2. Calculate the amount of heat required to change 2kg of ice at -20⁰C to liquid water at 100⁰C.(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 10⁵J/Kg/K).

 

 

3. Calculate the amount of heat required to change 2kg of ice at -20⁰C to steam at 100⁰C.(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 10⁵J/Kg/K, Specific Latent heat of vapourization of steam = 2.26 x 10⁶ 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 cross-section 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   P₁V₁  =  P₂V₂

 

THE SHAPE OF AGRAPH OF P PLOTTED AGAINST V IS SHOWN BELOW

 

QUESTIONS

1. The pressure of fixed mass of gas is 760mmHg when its volume is 38cm³. What will be its pressure when the volume increases to 100cm³.

 

2. Complete the table by filling in the missing values.

Pressure (cmHg)	0	_	90	_
Volume (cm3)	36	80	_	40

 

3. The volume V of a gas at a pressure P is reduced to ³/₈V without change in temperature. Determine the new pressure of the gas.

 

4. 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 ⁰C TO K-

 

 

QUESTIONS

1. 02m³ of a gas at 27⁰C is heated at a constant pressure until the volume is 0.03m³. Calculate the final temperature of the gas in ⁰C.
2. A mass of air of volume 750cm³ is heated at a constant pressure from 10⁰C to 100⁰ 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

1. A cylinder contains oxygen at 0⁰C and 1 atmosphere pressure. What will be the pressure in the temperature rises to 100⁰
2. 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

1. A mass of 1200cm³ of oxygen at 27⁰C and pressure 1.2 atmospheres is compressed until its volume is 600cm³ and its pressure is 3.0 atmospheres. What is the new temperature of the gas in ⁰C?
2. The volume of a fixed mass of air at 27⁰C and 75cmHg is 200cm³. Find the volume of the air at -73⁰C and 80cmHg.