Physics lesson note for SS1 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Physics.
Physics lesson note for SS1 First Term has been provided in detail here on schoolgist.ng
For prospective school owners, teachers, and assistant teachers, Physics lesson note is defined as a guideline that defines the contents and structure of Physics as a subject offered at SS level. The lesson note for Physics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.
To further emphasize the importance of this document, the curriculum for Physics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.
Physics Lesson note for SS1 carries the same aims and objectives but might be portrayed differently based on how it is written or based on how you structure your lesson note. Check how to write lesson notes as this would help make yours unique.
The SS1 Physics lesson note provided here is in line with the current scheme of work hence, would go a long way in not just helping the teachers in carefully breaking down the subject, topics, and subtopics but also, devising more practical ways of achieving the aim and objective of the subject.
The sudden increase in the search for SS1 Physics lesson note for First Term is expected because every term, tutors are in need of a robust lesson note that carries all topics in the curriculum as this would go a long way in preparing students for the West African Secondary Examination.
This post is quite a lengthy one as it provides in full detail, the Physics-approved lesson note for all topics and sub-topics in Physics as a subject offered in SS1.
Please note that Physics lesson note for SS1 provided here for First Term is approved by the Ministry of Education based on the scheme of work.
I made it free for tutors, parents, guardians, and students who want to read ahead of what is being taught in class.
SS1 Physics Lesson Note (First Term) 2024
SUBJECT:
PHYSICS
CLASS:
SENIOR SECONDARY SCHOOL 1
TERM:
FIRST
SCHEME OF WORK
WEEK TOPIC
- Introduction to Physics
- Measurement of Mass, Weight, Length and Time.
- Motion in Nature, Force, Circular Motion, Centripetal and Centrifugal Forces
- Friction
- Vector and Scalar Quantity, Distance/Displacement, Speed/Velocity, Acceleration, Distance/Displacement-Time Graph, Speed/Velocity-Time Graph
- Density and Relative Density
MID-TERM PROJECT
- Upthrust, Archimedes Principle, Law of floatation, Pressure
- Work, Energy and power
- Work Done in a Force Field, Types of Energy and Energy Conversion
- Viscosity
- Revision
- Examination
WEEK ONE
INTRODUCTION TO PHYSICS
CONTENT
- Meaning of Physics
- Aspects of Physics
- Importance of physics
- Scholars Achievement in the Field of Physics
MEANING OF PHYSICS
Physics is the scientific study of matter and energy and how they interact with each other. The interests and concerns of physicists have always formed the basis of future technology.
Physics has the capability of playing a major role in finding solutions to many of the problems facing the human race. In a broader sense, physics can be seen as the most fundamental of the natural sciences.
ASPECTS OF PHYSICS
1.Mechanics
2.Electricity
3.Electromagnetism
4.Nuclear and quantum physics
5.Optics and Light
6.Heat and thermodynamics
IMPORTANCE OF PHYSICS
Physics being the bedrock of technology has in no way been the foundation of global technological advancement. Physics contributes to the technological infrastructure and provides trained personnel needed to take advantage of scientific advances and discoveries.
Physics is important to man’s life because it is used in
-Cooking food
-Cleaning clothes
-Watching TV
-Heating your house
-Playing sports
Everything else in your life
- Physics plays an important role in health
- Economic development
- Education
- Energy and
- The environment
SCHOLARS ACHIEVEMENT IN THE FIELD OF PHYSICS
1.William Gilbert (1544-1603) – An English physicist who hypothesized that the Earth is a giant magnet.
2.Galileo Galilei (1564-1642) – An Italian physicist. He performed fundamental observations, experiments, and mathematical analyses in astronomy and physics
3.Willebrod Snell (1580-1626) – A Dutch physicist who discovered law of refraction (Snell’s law)
4.Blaise Pascal (1623-1662) – A French physicist who discovered that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of its container (Pascal’s principle)
5.Christiaan Huygens (1629-1695) – A Dutch physicist. He proposed a simple geometrical wave theory of light, now known as “Huygen’s principle”
6.Robert Hooke (1635-1703) – An English who discovered Hooke’s law of elasticity
7.Sir Isaac Newton (1643-1727) – An English physicist who developed theories of gravitation and mechanics, and invented differential calculus
8.Daniel Bernoulli (1700-1782) – A Swiss physicist. He developed the fundamental relationship of fluid flow now known as Bernoulli’s principle
9.Benjamin Franklin (1706-1790) – An American. He was the first American physicist; characterized two kinds of electric charge, which he named positive'' andnegative”
10.Charles Augustin de Coulomb (1736-1806) – A French physicist who performed experiments on elasticity, electricity, and magnetism; established experimentally nature of the force between two charges
11.André Marie Ampère (1775-1836) – A French physicist and father of electrodynamics
12.Georg Ohm (1789-1854) – A physicist from Germany. He discovered that current flow is proportional to potential difference and inversely proportional to resistance (Ohm’s law)
13.Michael Faraday 1791-1867) – An English physicist who discovered electromagnetic induction and devised first electrical transformer
14.Lord Kelvin (1824-1907) – A British physicist who proposed absolute temperature scale, of essence to development of thermodynamics
CLASSWORK
1.What is physics?
2.State five aspects of physics
ASSIGNMENT
SECTION A
1.Which of the following statements is/are not true of physics (i) it is the building block of technology (ii) it is the basis for all technological invention (iii) it is irrelevant in electricity (a) i (b) ii (c) iii (d) i and ii
2.Which of these is not a physicist (a) Albert Einstein (b) Sir Isaac Newton (c) Charles G. Finney (d) Neil Armstrong
3.The following are invention related to physics except (a) electricity (b) raincoats (c) matches (d) driving skills
4.Physics relies upon mathematics to provide the logical framework in which physical laws may be precisely formulated and predictions quantified (a) true (b) false (c) cannot say (d) all of the above
5.Which of this physicist is known for the law of elasticity (a) Robert Hooke (b) Georg Ohm (c) Lord Kelvin
SECTION B
1.Briefly describe how physics has helped in technological advancement
2.Mention three scientists and their contributions to physics
WEEK TWO
MEASUREMENT OF MASS, WEIGHT, LENGTH & TIME
CONTENT
- Meaning of Measurement
- Measurement of Mass
- Measurement of Weight
- Measurement of Length
- Measurement of Time
MEANING OF MEASUREMENT
Measurement is the process of observing and recording the observations that are collected as part of a research effort. To get the exact measurement of an object we make use of tools used in the field of science, especially in physics called measuring instrument. Examples of such instrument are chemical balance, spring balance, meter rule, caliper, micrometer screw guage, clock, thermometer etc.
MEASUREMENT OF MASS
Mass is the quantity of matter contained in a body. The instrument used in measuring mass is chemical/beam balance. It is a scalar quantity & measured in kilogram (kg). Mass is a fundamental quantity & is constant from place to place.
MEASUREMENT OF WEIGHT
Weight is the earth pull on a body or the downward force produced when a mass is in a gravitational field. The instrument used in measuring weight is spring balance. It is a vector quantity & measured in newton (N). Weight is a derived quantity & varies from place to place.
Relationship Between mass & weight
W=mg
Where: W=weight (N); m=mass (kg) & g=acceleration due to gravity (m/s2)
MEASUREMENT OF LENGTH
Length is simply distance extended. It is a fundamental quantity & measured in meters (m).
Length can be measured using tape rule, meter rule, caliper, vernier caliper and micrometer screw guage
MEASUREMENT OF TIME
Time can be defined as the interval between events. It is a fundamental quantity & measured in seconds(s).Time can be measured using stop watch/clock
FUNDAMENTAL & DERIVED QUANTITY
Fundamental quantities are the basic quantities that are independent of others. They are length (m), mass (kg) and Time (s), electric current (A), temperature (k), amount of substance (moles) and luminous intensity (candela).
Derived quantities are those obtained by simple combination of basic quantities e.g. Area, Volume, Density, Velocity, Acceleration, Force, energy, work, power, momentum, pressure, electric charge, electric potential etc.
S/N QUANTITY DERIVATION DIMENSION UNIT
- Area length x breadth L2
m2
- Volume length x breadth x heights L3 m3
- Density
mass ÷ volume ML-3 Kg/m3
- Speed or velocity distance ÷ time LT-1 m/s
- Acceleration
change in velocity ÷ time LT-2 m/s2
- Force
Mass x acceleration M LT-2 N
- Weight
Mass x acceleration due to gravity MLT-2 N
- Momentum mass x velocity M LT-1 kgm/s
- Impulse
force x time M LT-1 NS
- Pressure Force ÷ Area M L-1T-2
Pa or NM2
- Energy or work force x distance M L2T-2 J or Nm
- Power Work ÷ time M L2T-1 W or Nm/s
CLASSWORK
1.Define these: (i) mass (ii) weight (iii) length
2.Calculate the weight of an object of mass 5000g if g = 10m/s2
3.List five instrument for measuring length
ASSIGNMENT
SECTION A
1.The following are the fundamental quantities except (a) Length (b) weight (c) mass (d) time
2.The reading accuracy of meter rule is (a) 0.01cm (b)0.1cm (c) 10.005cm (d) 0.004cm
3.The best instrument for measuring the diameter of a thin wire is (a) vernier caliper (b) steel rule (c) micrometer screw gauge (d) meter rule
4.The SI unit of weight is (a) N (b) m (c) mls2 (d) kg
5.What is the reading on the micrometer screw gauge below:
0 15
(a)8.511 (b) 8.011 (c) 8.519 (d) 8.151
6.What is the reading the vernier caliper is showing?
7 8 9 10
(a)7.13 (b) 7.21 (c) 7.12 (d) 7.31
SECTION B
1.Derive the dimension for (a) work (b) volume (c) acceleration (d) force
2.Distinguished between fundamental and derived quantity
3.Distinguished between mass and weight
WEEK THREE
MOTION IN NATURE
CONTENT
- Definition of Motion
- Causes of Motion
- Circular Motion
- Centripetal Acceleration and Force
DEFINITION OF MOTION
Motion is the change of position of a body with time. The study of motion without involving the force causing the motion is called kinematics. The study of motion of objects and the forces acting on them is called dynamics.
Types of Motion
1.Translational motion: This type of motion occurs when a body moves in a fixed direction without rotating e.g. A car moving in one direction from one town to another, movement of a man etc. It is also called rectilinear motion
2.Rotational or Circular motion: This is the movement of a body in a circular manner about its axis e.g. the movement of car wheels, electric fan blade, earth about its axis etc.
3.Random Motion: This is a type of motion in which a body moves in a zigzag or disorderly manner with no specific direction e.g. motion to molecules of gasses, butterflies etc.
4.Vibratory or Oscillatory Motion: This is a to and fro or up and down movement of a body about a fixed point e.g. the simple pendulum, vibration of plucked guitar string, etc.
Relative Motion
If two bodies, A and B are moving on a straight line, the velocity of A relative to B is found by adding the Velocity of B revered to the velocity of A. For instance, if a car traveling on a straight road at 100km/hr passes a bus going in the same direction at 60km/hr., the velocity of the car relative to the bus is (-60+100) = 40km/hr. If the car and the bus are traveling in opposite direction with the same velocities of 100km/r and 60km/hr respectively, the velocity of the car relative to the bus is ( -(-60) + 100) = (60 +100) = 160 km/hr.
NB: When the velocities are not in the same straight line, the parallelograms law should be used to add this since velocities are vectors, and their magnitudes and direction must be taken into consideration.
CAUSES OF MOTION
All objects will continue in their state of rest unless acted upon by force. Only the application of a force can cause visible motion. Hence, force causes motion. There are two types of force (a) Contact force (b) Field force
(a)Contact Force: They are forces that are in contact with the body they affect e.g. tension, reaction frictional forces, forces of pull & push, viscous force etc.
(b)Field Force: They are forces whose sources do not require contact but the effect of such forces is felt in a field of the force e.g. electrical force, magnetic, gravitational pull etc.
CIRCULAR MOTION
In physics, circular motion is movement along a circular path or orbit. It can be uniform (i.e. with constant angular rate of rotation) or non-uniform (i.e. with a changing rate of rotation)
Formulae for uniform circular motion
Consider a body of mass (kg), moving in a circle of radius r (m), with an angular velocity of ω (rads-1)
- The angular velocity is w = Ө/t where Ө-angle subtended (rad) & t-time (s)
- The linear speed is v = s/t where s-distance covered (m)
- The linear speed is v = r×ω (m/s)
- The centripetal (inward) acceleration is a = r×ω2 = v2/r (m/s2)
- The centripetal force is F = m×a = r×m×ω2 = mv2/r (N)
- The momentum of the body is p = m×v = r×m×ω(kgm/s)
- The moment of inertia is I = r 2×m (kgm2)
- The angular momentum is L = r×m×v = r 2×m×ω = Iω(kg·m2·s−1).
- The kinetic energy is E = ½(m×v2) = ½(r2×m×ω2) = ·p2 (2·m)−1 = ½(I·ω2) = ½(L2) (J).
- The circumference of the orbit is 2πr (m).
- The period of the motion is T = 2πω−1 (s)
- The frequency is f = T−1 (Hz)
(To convert radian from degree 3600=2∏)
CENTRIFUGAL FORCE
Centrifugal force is a force that acts in opposite direction to the centripetal force. Centrifugal force is an outward force associated with rotation.
CLASSWORK
1.What is kinematics?
2.Define motion
3.List two types of motion and explain briefly each of them giving in each case one example
ASSIGNMENT
SECTION A
1.An airplane from Kano to Lagos is an example of which if these types of motion? (a) Linear (b) rotational (c) random (d) oscillatory.
2.Which of the following correctly gives the relationship between linear speed v & angular velocity w of a body moving uniformly (a) v=w r (b) v=w2r (c) v=wr2(d) v=w/r
3.Which of the following is odd? (a) Random motion (b) rotational motion (c) nuclear motion (d) oscillatory motion.
4.The motion of the prongs of a sounding turning fork is (a) random (b) translational(c) rotational (d) vibratory
5.A body moving in a circle at constant speed has (i) a velocity tangential to the circle (ii) a constant kinetic energy (iii) an acceleration directed towards the circumference of the circle. Which of the statement above are correct (a) I & ii only (b) ii & iii only (c) i & iii only (d) i, ii& iii
SECTION B
1.Explain the causes of motion
2.Define centripetal and centrifugal force
3.A body of mass 5kg moving in a circular path with a velocity of 5m/s for 10 complete revolution within 4s. If the radius of the circular path is 30m, find: (a) the centripetal force (b) the centripetal acceleration (c) angle subtended in radian (d) angular velocity
WEEK FOUR
FRICTION
CONTENT
- Definition of Friction
- Laws Governing Solid Friction
- Advantages & Disadvantages of Friction
- Reducing Friction
DEFINITION OF FRICTION
Friction (Fr) is defined as an opposing force which acts at the surface of two objects or bodies in contact. It is simply force of opposition. We have two types of friction:
(a) Static friction, Fs
(b) Dynamic friction, Fd.
NOTE: Fs is greater than Fd for object at rest while Fd is greater than Fs for object in motion.
LAWS GOVERNING SOLID FRICTION
1.Friction opposes the relative motion of two surfaces in contact
2.It is independent of the area of the surface of contact
3.It depends on the nature of the surface
4.It is proportional to normal reaction (R)
5.It is independent of relative velocity between the surfaces
Fr α R
Fr = μR 1
Where: Fr=frictional force: μ=coefficient of friction and R=normal reaction
R
Fr F P
W=mg
At equilibrium, R = mg, this implies that
Fr = μmg 2
μ = tanӨ 3
Fr = RtanӨ 4
If μ> 0, P< Fr + mgsinӨ
ADVANTAGES OF FRICTION
1.It makes walking and running possible
2.It enables gripping of belt in machines possible
3.It enables nails to stay in the wall when driven
4.It stops tyre from slipping
5.Enable cars to stop when breaks are applied
6.Enables human to use mouse in surfing web
DISADVANTAGES OF FRICTION
1.It causes wear and tear
2.It reduces the efficiency of the machines
3.It causes a lot of energy to be consumed by the machine
4.It causes loss of resources
METHODS OF REDUCING FRICTION
- Lubricating surfaces with grease, oil etc.
- Using ball or roller on wheels
- Smoothing or polishing the surface
- Streamlining bodies
CLASSWORK
1.Define friction
2.Mention four effect of friction
3.State 2 ways of reducing friction
ASSIGNMENT
SECTION A
1.Which of the following statements about the concept of solid friction is NOT true? (a) it always acts in the direction of motion (b) it causes wear and tear in car tyres (c) it depends on the nature of the surfaces in contact (d) it reduces the efficiency of machines
2.A metal block of mass 8kg lies on a rough horizontal platform. If the horizontal resistive force is 10N, find the coefficient of static friction (g=10m/s2) (a) 0.25 (b) 0.125 (c) 0.8 (d) 0.124
3.Which of the statement is correct (a) static friction is less than dynamic friction (b) static friction equals dynamic friction (c) static friction is greater than dynamic friction (d) none of the arrange
4.A metal block of mass 5kg lies on a rough horizontal platform. If a horizontal force of 8N applied to the block on the platform, then the coefficient of limiting friction between the block and the platform is: (a) 0.16 (b) 0.63 (c) 0.80 (d) 1.06
5.If the angle between the incline length and the horizontal platform of an incline plane is 600 calculate the coefficient of friction (a) 0.86 (b) 1.73 (c) 0.50 (d) 0.73
SECTION B
1.State two (a) laws governing solid friction (ii) advantages of friction (iii) disadvantage of friction (iv) methods of reducing friction
2.A body of weight 6N rest on a plane inclined at an angle of 300to the horizontal (a) what force keeps it sliding down the plane? (b) what is the coefficient of friction
3.A body of mass 25kg, moving at 3m/s on a rough horizontal floor is brought to rest after sliding through a distance of 2.5m on the floor. Calculate the coefficient of sliding friction (g=10m/s2)
WEEK 5
VECTOR & SCALAR QUANTITY, DISTANCE/DISPLACEMENT, SPEED/VELOCITY, ACCELERATION, DISTANCE/DISPLACEMENT-TIME GRAPH, SPEED/VELOCITY-TIME GRAPH
CONTENT
- Distance & Displacement
- Speed & Velocity
- Acceleration & Retardation
- Distance/Displacement – Time Graph
- Speed/Velocity – Time Graph
LINEAR MOTION
Terminologies used in linear motion:
1.Distance: This defined as the total length of path traversed. It is also the separation between two points. It is denoted as “s” or “x”. It is a scalar quantity. The SI unit of distance is meters (m)
2.Displacement: this is distance moved in a specified direction. It is denoted as “s” or “x”. It is a vector quantity. The SI unit of displacement is meters (m)
3.Speed: This is the rate of change of distance with time. It is a scalar quantity. Its SI unit is meter per seconds (m/s or ms-1)
4.Uniform speed: This is when the rate of change of distance with time is constant.
5.Velocity: This is the rate of change of displacement with time. It is a vector quantity. Its SI unit is meter per seconds (m/s or ms-1).
6.Uniform velocity: This is when the rate of change of displacement with time is constant.
NOTE: Velocity is often used interchangeably with speed during calculations
7.Acceleration: This is the increasing rate of change of velocity with time. It is a vector quantity. Its SI unit is meter per seconds-square (m/s2 or ms-2)
8.Uniform acceleration: This is when the increasing rate of change of velocity with time is constant
9.Deceleration: This is the decreasing rate of change of velocity with time. It is a vector quantity. It is commonly referred to as negative acceleration or retardation.
10.Uniform deceleration: This is when the decreasing rate of change of velocity with time is constant
Equation of Uniformly Accelerated motion
S = (v+u) t 7
2
v = u + at 8
v2 = u2 + 2 aS 9
S = ut + ½ at2 10
Equations (7) to (10) are called equations of uniformly accelerated motion and could be used to solve problems associated with uniformly accelerated motion where u- initial velocity, v – final velocity, a – acceleration, S – distance covered and t – time
Example – A car moves from rest with an acceleration of 0.2mls2. Find its velocity when it has moved a distance of 50m.
Solution
Given:
a = 0.2mls2, S = 50m, u = 0m/s, v =?
v2 = u2 + 2 as
v2= 02 + (2×0.2×50) = 20
v = √20 m/s
Distance/Displacement- Time Graph
The slope of this time graph gives speed/velocity.
For a uniform speed/velocity, the time graph is given below:
If the velocity is non – uniform, the velocity at a point is the gradient or slope of the tangent at that point.
S (m)
t (s)
Speed/Velocity – Time Graph
The slope of the speed/velocity-time graph gives acceleration.
Example – A car starts from rest and accelerates uniformly until it reaches a velocity of 30mls after 5 seconds. It travels with uniform velocity for 15 seconds and is then brought to rest in 10s with a uniform retardation. Determine (a) the acceleration of the car(b) The retardation (c) The distance covered after 5s(d) The total distance covered (use both graphical and analytical method)
The velocity – time diagram for the journey is shown above, from this diagram
a.the acceleration = slope of OA
= AE / EO
= (30-0) /(5-0)=30/5
= 6mls2
b.the retardation = slope of BC = CB / CD
= (0-30) / (30-20) = -30/10
= -3mls2 (the negative sign indicate that the body is retarding)
c.Distance traveled after 5s = area of A E O
= ½ x b x h
= ½ x 5 x 30
= 75m
d.total distance covered = area of the trapezium OABC
= ½ (AB + OC) AE
= ½ (15 + 30) 30
= 675m.
Using equations of motion:
a)U = O, V = 3, t = 5
V = u + t
a = v-u/t = 30 – 0 / 5
a = 30/5 = 6ms-2
b)a = (v – u) / t
a = (0-30) / 10
a = -3 mls2
- c) S = (u + v) 5
2
S= (30 x 5)/2
S= 75m
- d) To determine the total distance traveled we need to find the various distance for the three stages of the journey and then add them.
For the 1st part S= 75m from (c) above
For the 2nd stage: where it moves with uniform velocity.
S = vt
= 30 x 15
= 450m
For the last stage S = ½ (u + v) t
= ½ (30 + 0) 10
= 150m.
Total distance = 75 + 450 + 100 = 675m.
CLASSWORK
1.Define the following terms as used in linear motion: i. Acceleration ii. Speed iii. Displacement
2.A car moves with a velocity of 72kmhr-1. It is brought to rest in 10s. Find (i) the velocity in ms-1 (ii) the retardation
3.State the differences & similarity between speed & velocity
ASSIGNMENT
SECTION A
1.The area under the curve of a velocity-time graph represents (a) distance covered (b) acceleration (c) instantaneous speed (d) work done
2.Which of the following statements is correct about speed and velocity (a) speed and velocity are both scalar quantities (b) speed and velocity have the same unit (c) velocity relates to translational motion while speed relates to circular motion (d) velocity and speed cannot be represented graphically
3.A car moving with speed 90kmhr-1 was brought uniformly to rest by the application of brakes in 10s. How far did the car travel after brakes were applied (a)120m (b) 150m (c) 125m(d)15km
4.The slope of distant-time graph f0r a uniform rectilinear motion of a body represents (a) its acceleration (b) its total distance travelled (c) its speed (d) the force causing the motion
5.The distance traveled by a particle starting from rest is plotted against the square of the time elapsed from the commencement of motion. The resulting graph is linear. The slope of this graph is a measure of (a) initial displacement (b) initial velocity (c) acceleration (d) half of acceleration
SECION B
1.(a) Explain the terms uniform acceleration and average speed (b) a body at rest is given an initial uniform acceleration of 8.0ms-2 for 30 seconds after which the acceleration is reduced to 5.0ms-2 for the next 20 seconds. The body maintains the speed for 60seconds after which it is brought to rest in 20 seconds. Draw the velocity-time graph of the motion using the information given above (c) using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (iii)
2.A car starts from rest and accelerates uniformly for 10s, until it attains a velocity of 25ms-1; it then travels with uniform velocity for 20s before decelerating uniformly to rest in 5s. (i) calculate the acceleration during the first 10s (ii) calculate the deceleration during the last 5s (iii) sketch a graph of the motion and calculate the total distance covered throughout the motion
WEEK SIX
DENSITY& RELATIVE DENSITY
CONTENT
- Definition of Density
- Determination of Density
- Relative Density
- Determination of Relative Density of Solids & Liquid
DEFINITION OF DENSITY
The density of a substance is the mass per unit volume of the substance.
Density = mass of a given substance
Volume of the substance
Density is scalar quantity& measured in kgm-3 (kilogram per cubic meter)
DETERMINATION OF DENSITY
The determination of density involves the determination of a mass and a volume. The mass can be found by weighing. The density of a substance can be determined using a graduated density bottle.
Relative Density
Relative density is also known as specific gravity. Relative density of a substance is defined as the density of the substance per density of water.
R.D = Density of the substance
Density of water
R.D is also equal to the ratio weight of a substance to weight of an equal volume of water. As weight is proportional to mass
R.D = mass of substance
Mass of equal volume of water
DETERMINATION OF R.D OF SOLID (E.G. SAND)
Mass of empty bottle = m1
Mass of bottle + sand = m2
Mass of bottle + sand + water = m3
Mass of bottle + water only= m4
Mass of sand = m2 – m1
Mass of water added to sand = m3 –m2
Mass of water filling the bottle = m4 – m1
Mass of water having the same volume as sand = (M4-M1) – (M3-M2)
Relative density = Mass of sand
Mass of equal volume of water
R.D = m2 – m1
(m4 – m1) – (m3 -m2)
DETERMINATION OF R.D OF LIQUID
Mass of empty density bottle = m1
Mass of bottle filled with water = m2
Mass of bottle filled with liquid = m3
R.D of liquid = m3 – m1
m2 – m1
Example – A glass block of length 10cm width 8cm and thickness 2cm has a mass of 400g. Calculate the density of the glass.
Solution
l = 10cm = 0.1m, b = 8cm = 0.08cm, h = 2cm = 0.02m, m = 400g = 0.4kg
V = l x b x h = 0.1 x 0.08 x 0.02 = 0.00016m3
Density = Mass (m) = 0.4 = 2500kgm3
Volume (V) 0.00016
Example 1
Calculate the volume in m3 of a piece of wood of mass 500g and density 0.76gcm-3
Mass of the wood = 500g
Density = 0.76gcm-3
Volume =?
Volume = mass / density
= 500
0.76
Volume = 658cm3 = 6.58 x 10-4 m3
Example 2
An empty relative density bottle has a mass of 15.0g. When completely filled with water, its mass is 39.0g. What will be its mass if completely filled with acid of relative density 1.20?
Solution
m1, mass of empty bottle = 15.0g
m2, mass of bottle + water = 39.0g
Mass of acid = n – 15.0g
Mass of water = 39.0 – 15.0g
= 24.0g
R.D = 1.20
R.D = n – 15.0g
39.0 – 15.0g
1.20 = n –15.0
24.0
n- 15.0= 1.20×24.0
n- 15= 28.8
n =28.8+15
n=43.8g
NOTE: The hydrometer is an instrument used to measure the relative density of liquids
CLASSWORK
1.Define density
2.What does it mean by the statement that the density of gold is 19.3gcm-3
3.Differentiate between density & relative density
4.A glass block of length 100cm width 60cm and thickness 20cm has a mass of 4000g. Calculate the density of the glass
ASSIGNMENT
SECTION A
1.The relative densities of zinc, brass, copper, gold and silver are respectively 7.1, 8.5, 8.9, 19.3 and 10.5. A metal ornament which weighs 0.425kg and can displace 50×10-6m3 of water is made of (a) zinc (b) brass (c) copper (d) gold (e) silver
2.Find the density of a substance, if the mass of the substance is 150,000g and the dimension is 20m by 10m by 500cm (a) 0.5kg/m3 (b) 0.24 kg/m3 (c) 1.50 kg/m3 (d) 2.40 kg/m3.
3.What is the height of a cylindrical iron if the density is 7900kg1m? The mass is 700kg and the radius is 0.1m (a) 2.918m (b) 2.819m (c) 3.418m
4.Which is the correct unit of density? (a) m3/kg (b) kg/m (c) kg/m3 (d) m/v
5.What volume of alcohol with density of 8.4x102kgm-3 will have the same mass as 4.2m3 of alcohol whose density is 7.2x102kgm-3? (a) 1.4m3 (b) 6.3m3 (c) 4.9m3 (d) 3.6m3
SECTION B
1.40 m3 of liquid P is mixed with 60m3 of another liquid Q. if the density of P and Q are 1.00kgm-3 and 1.6kgm-3 respectively. What is the density of the mixture?
2.The density of 400cm3 of palm oil was 0.9gcm-3 before frying. If the density of the oil was 0.6cm-3 after frying, assuming no loss of oil due to spilling, its new volume was?
MIDTERM PROJECT
Using a white cardboard draw these instruments, write short note and explain how to take readings from them:
vVernier caliper
vMicrometer screwguage
vMetre rule
vSpring balance
vBeam balance
WEEK SEVEN
PRESSURE, ARCHIMEDES’ PRINCIPLES, UPTHRUST&LAWS OF FLOATATION
CONTENT
- Pressure
- Archimedes’ Principles & Upthrust
- Laws of Floatation
PRESSURE
Pressure is defined as force per unit surface area. It is a scalar quantity & measured in N/m2 or Pascal (pa).
P = F 1
A
Where P-pressure, F- force & A-area
NOTE: 1 bar = 105 N/m2 = 105 pa
Example – A force of 40N acts on an area of 5m2. What is the pressure exerted on the surface?
Solution
F = 40N, A = 5m2, P =?
P = F/A = 40/5 = 8pa
Pressure in Liquid
Pressure in liquid has the following properties
1.Pressure increases with depth
2.Pressure depend on density
3.Pressure at any point in the liquid acts equally in all direction
4.Pressure at all points at the same level within a liquid is the same
5.It is independent of cross-sectional area
P = ρgh 2
Where: p-pressure, ρ-density, h-height & g-acceleration due to gravity.
ARCHIMEDES’ PRINCIPLES AND UPTHRUST
Archimedes’ principle is a law that explains buoyancy or upthrust. It states that when a body is completely or partially immersed in a fluid it experiences an upthrust, or an apparent loss in weight, which is equal to the weight of fluid displaced.
From pressure, p is given by p = hρg, where:
h is the height of the fluid column
ρ is the density of the fluid
g is the acceleration due to gravity
Let us confirm this principle theoretically. On the figure on the left, a solid block is immersed completely in a fluid with density ρ. The difference in the force exerted, d on the top and bottom surfaces with area a is due to the difference in pressure, given by
d = h2aρg – h1aρg = (h2 – h1)aρg
But (h2– h1) is the height of the wooden block. So, (h2 – h1)a is the volume of the solid block, V.
d = Vρg
Upthrust = Vρg
Weight in air – upthrust = weight in fluid
Upthrust = weight in air – weight in fluid
Upthrust = Apparent loss in weight
NB: When an object is wholly immersed, it displaces its volume of fluid.
So;
Upthrust = weight of fluid displaces
Upthrust = Volume of fluid displaced x its density x g
Upthrust= volume of object x density of fluid x g
Determination of Relative Density by Archimedes’ Principle
1.Relative density of solid
The body is weighed in air w1, and then when completely immersed in water w2
Relative density of solid = Weight of solid in air
Weight of equal volume in water
= w1
W1-W2
2.Relative density of liquid
A solid is weighed in air (w1), then in water (w2) and finally in the given liquid (w3)
Relative density of liquid = apparent loss of weight of solid in liquid
apparent loss of weight of solid in water.
= W1 – W3
W1 – W2
Example – The mass of a stone is 15g when completely immersed in water and 10g when completely immersed in liquid of relative density 2.0. What is the mass of the stone in air?
Solution:
Relative density = upthrust in liquid
upthrust in water
Let W represents the mass of the stone in air
2 = w – 10
w – 15
2(w – 15) = w –10
2w – 30 = w – 10
2 w – w = -10 + 30
w = 20g
LAW OF FLOATATION
A floating object displaces its own weight of the fluid in which it floats or an object floats when the upthrust exerted upon it by the fluid is equal to the weight of the body. When an object is floating freely (i.e. neither sinking nor moving vertically upwards), then the upthrust must be fully supporting the object’s weight. We can say
Upthrust on body = Weight of floating body. By Archimedes’ principle,
Upthrust on body = Weight of fluid displaced.
Therefore, Weight of floating body = Weight of fluid displaced.
This result sometimes called the “principle of floatation”, is a special case of Archimedes’ principle
CLASSWORK
1.Define pressure
2.State three characteristics of pressure in liquids
3.State Archimedes’ principle.
ASSIGNMENT
SECTION A
1.A force of 40N acts on an area of 10m2. What is the pressure exerted on the surface? (a) 8pa (b) 4pa (c) 400pa (d) 10pa
2.For which of the following sets are the units fundamental? (a) density, length and pressure (b) impulse, mass and time (c) volume, mass and density (d) length, time and mass
3.What is the height of a cylindrical iron if the density is 7900kglm3 the mass is 700kg and the radius is 0.1m (a) 2.918cm (b) 2.819m (c) 3.418m (d) 4.328m
4.A piece of cork density 0.25x103kgm-3 floats in a liquid of density 1.25x105kgm-3, what fraction of volume of the cork will be immersed? (a) 5 (b) 1/5 (c) 2/5 (d) 1/3
5.The SI unit for pressure, density and upthrust is respectively (a) Nm-2; gm-3; F (b) Nm-2; Kgm-3; N (c) F; Kgm-3; Nm-2 (d) Nm-3; Kgm-2, N
SECTION B
1.Explain the following terms: (i) viscosity (ii) terminal velocity
2.State two (i) effect of viscosity (ii) applications of viscosity
3.What is the pressure due to water at the bottom of a tank which is 20cm deep and is half of water? (Density of water = 103kg/m3& g = 10m/s2)
WEEK EIGHT
WORK, ENERGY AND POWER
CONTENT
- Work
- Energy
- Power
WORK
Work is defined as the product of force and distance in the direction of the force. It is a scalar quantity & measured in Joules.
Mathematically:
W = F X d 1.
W = mgh 2.
If a force is applied on a body at an angle Ø to the horizontal
Work done = FcosØ x d 3.
Work done to raise the body to an appreciable height
= FsinØ x d 4.
Example – A boy of mass 50kg runs up a set of steps of total height 3.0m. Find the work done against gravity
Solution
m = 50kg, h = 3m, g = 10m/s2
Work done = m x g x h
= 50 x 10 x 3
= 1500 Joules
ENERGY
Energy is defined as the ability to do work. It is a scalar quantity &measured in Joules. There are many forms of energy. These include:
i.Mechanical energy
ii.Thermal energy
iii.Chemical energy
iv.Electrical energy
v.Nuclear/Atomic Energy
vi.Solar/Light energy
vii.Sound Energy
Types of Mechanical Energy
Mechanical energy is classified as
1)Potential energy
2)Kinetic energy
Potential Energy – is simply “stored energy” i.e. energy possessed by a body by virtue of its states:
P.E = mgh 5.
Kinetic Energy: is the energy possessed by a body by virtue of its motion. Examples area student running a race, wind or air motion, electrical charges in motion, a moving bullet
K.E = ½ mv2 6.
Example I – An object of mass 5kg is moving at a constant velocity of 15m/s. Calculate its kinetic energy.
Solution
K.E = ½ mv2 = ½ x 5 x 15 x 15= 562.5 J
Example II- Find the potential energy of a boy of mass 10kg standing on a building floor 10m above the ground level. g = 10m/s2
Solution:
P.E =m x g x h= 10 x 10 x10 =1000 J
POWER
Power is defined as the rate of doing work or the rate of transfer of energy. It is a scalar quantity & measured in watt
Power = work done
Time 7
P = (F X d)/t = F X d/t = FV 8
Example -A boy of mass 10kg climbs up 10 steps each of height 0.2m in 20 seconds. Calculate the power of the boy.
Solution
Height climbed = 10 x 0.2 = 2m
Work done = m x g x h = 10 x 10 x 2= 200 Joules
Power = work
Time
= 10 x 10 x 2 = 10watts
20
CLASSWORK
1.Define these terms (i) power (ii) work done (iii) energy
2.A boy of mass 960g climbs up to 12 steps each of height 20cm in 20 seconds. Calculate the power of the boy
ASSIGNMENT
SECTION A
1.Under which of the following conditions is work done (a) a man supports heavy load above his head with his hands (b) a woman holds pot of water (c) a boy climb unto a table (d) a man pushes against stationary petrol tanker
2.An object of mass 0.5kg has kinetic energy of 25J. Calculate the speed of the object (a) 50ms1 (b) 25ms-1 (c) 2.ms-1 (d)10ml-1
3.A man of mass 50kg ascends a flight of stairs 5m high in 5seconds. If acceleration due to gravity is 10m/s2, the power expended is (a) 100W (b) 250W (c) 500W (d) 400W
4.The kinetic energy of a bullet fired from a gun is 40J. If the mass of the bullet is 0.1kg, calculate the initial speed of the bullet (a) 4.0ms-1 (b) 40.0ms-1 (c) 28.28 ms-1 (d) 20.0 ms-1
5.A diver is 5.2m below the surface of water of density 103 kg/m3. If the atmospheric pressure is 1.02 x 105 pa. Calculate the pressure on the diver. (g=10m/s2 ) (a) 6.02 x 104 pa (b) 1.02 x 105 pa (c) 1.54 x 105 pa (d) 5.20 x 105 pa
6.Force (N)
100
A Displacement (m)
0,0 10 20
Using the force-displacement diagram shown above, calculate the work done.
(a)2000J (b) 1000J (c) 20J (d) 5J
SECTION B
1.Explain work done.
2.(a) State the law of conservation of energy. (b) A body is displaced through a certain distance x by a force of 30N. If the work done is 100J and the displacement is in the direction of force, what is the value of x?
3.A motor can converts chemical energy of petrol to mechanical energy at 30% efficiency. Calculate the mechanical energy obtained from 10litres of petrol. (1 litre of petrol contains 2.8kJ of chemical energy)
4.A student eats a dinner container 8.0x106J of energy. He wishes to do an equivalent amount of work in a nearby gym by lifting a 60kg object. How many times must he raise the object to expand this much energy? Assume that he raises it a distance of 2.0m each time
WEEK 9
WORK DONE IN A FORCE FIELD& ENERGY CONVERSION
- Work done in Lifting a Body & Falling Bodies
- Conservation & Transformation of Energy
- World Energy Resources
WORK DONE IN LIFTING A BODY & FALLING BODIES
The magnitude of work done in lifting a body is given by
Work = force x distance = m x g x h = mgh
Also, the work done on falling bodies is given by
Work = force x distance = m x g x h = mgh
CONSERVATION & TRANSFORMATION OF ENERGY
Energy can be converted from one form to another in a closed system. The law of conservation of energy states that energy can neither be created nor destroyed but can be converted from one form to the other.
World Energy Resources
World energy resources can be classified as
- Renewable Energy Resources: They are energy that can be replaced as they are used e.g. solar energy, wind energy, water energy & biomass
- Non-renewable Energy Resources: Energy that cannot be replaced after use e.g. nuclear energy, petroleum & natural gas
CLASSWORK
1.State the law of conservation of energy
2.A loaded sack of total mass 100kg falls down from the floor of a lorry 2m high. Calculate the work done by gravity on the load
3.Differentiate between renewable & non-renewable energy
ASSIGNMENT
SECTION A
1.Which of the following correctly explain energy conversion in a food eaten by a student to the energy he gets to play around? (a) potential energy – kinetic energy (b) chemical energy – kinetic energy (c) heat energy – potential energy (d) mechanical energy – chemical energy
2.Electric cell convert ……… to electrical energy (a) nuclear (b) chemical (c) mechanical (d) heat
3.A body rolls down a slope from a height of 100m. Its velocity at the foot of the slope is 20ms-1. What percentage of its initial potential energy is converted into kinetic energy? (g=10ms-2)
4.The following are examples of renewable energy except (a) biomass (b) solar (c) wind (d) nuclear
5.A boy of mass 50kg runs up a set of steps of total height 3.0m. Find the work done against gravity (a) 1200J (b) 1500J (c) 1000J (d) 1300J
SECTION B
1.A loaded sack of total mass 100kg falls down from the floor of a lorry 2m high. Calculate the work done by gravity on the load
2.A body of mass 0.6kg is thrown vertically upward from the ground with a speed of 20ms-1. Calculate its: (a) potential energy at the maximum height reached; (b) kinetic energy just before it hits the ground
WEEK 10
VISCOSITY
CONTENT:
- Meaning of Viscosity
- Experiment to Determine the Terminal Velocity of a Steel Ball Falling in a Fluid
- Factors Affecting Viscosity
- Effect of Viscosity
- Application of Viscosity
MEANING OF VISCOSITY
Viscosity is the internal friction which exists between layers of the molecules of fluid (liquid or gas) in motion. The viscosity of a fluid can also be defined as the measure of how resistive the fluid is to flow. It is a vector quantity & measured in Pascal-seconds (Pa.s).
It can be defined mathematically as the ratio of the shearing stress to the velocity gradient in a fluid
Viscosity (ŋ) = Force
Area x Velocity gradient 1.
Velocity gradient = velocity
Length 2.
W = U + V
W – U- V = 0 ……………………………………………………………3.
V = W – U (apparent or effective weight)
Where: V-viscous force, W- weight, U- upthrust
NOTE: Substances with low viscosity include water, kerosene, petrol, ethanol etc.
Those with high viscosity are glue, syrup, grease, glycerin etc.
Experiment to Determine the Terminal Velocity of a Steel Ball Falling Through a Fluid
Aim: To determine the terminal velocity of a steel ball falling in through a jar of glycerin
Apparatus: steel ball, cylindrical calibrated jar, glycerine
Diagram:
Procedure: Set-up the apparatus as shown above &gently drop the steel ball in the jar of glycerin
Observation: It will be observed that the ball is accelerating in the liquid. Also the time taken for the ball to move from A-B will be different from B-C and so on. A time will be reached when the ball will be moving at a constant speed or velocity. It is that point that terminal velocity is experience.
Graph:
Conclusion: Terminal velocity is attained when W = V + U. At a point when the ball is moving at a constant speed through the glycerine.
Precaution:
- The steel ball should be dropped gently on the liquid
- Experiment should be done under constant temperature
- Avoid error of measurement when taken the reading.
NB: Terminal velocity is the maximum velocity of an object when the viscous force due to motion of the object equals the apparent (effective) weight of the object in the fluid where there is no longer net force on the object.
Drag force is the force that keeps the object continuously moving after the terminal velocity has been attained.
Stokes’ Law state that at the terminal velocity, the upward frictional force
(F) = 6ΠŋrV
Where F- Frictional/Drag force, ŋ- viscosity, r- radius of sphere, Vt- Terminal velocity
Factors Affecting Viscosity
1.Viscosity varies with material
2.The viscosity of simple liquids (a) decreases with increasing temperature (b) increases under very high pressure
3.The viscosity of gases (a) increases with increasing temperature (b) is independent of pressure & density
Effect of Viscosity
1.Viscosity is responsible for different rate of fluid flow
2.Viscosity affect motion of body in fluid
Application of Viscosity
1.It is use as a lubricant
2.The knowledge of viscous drag/drag force is applied in the design of ship & aircraft
3.Use to estimate the enlarge size of particles
NB: A liquid is said to be VISCOSTATIC if its viscosity does not change (appreciably) with change in temperature.
CLASSWORK
1.What is viscosity?
2.State two application of viscosity
3.Mention three viscous liquid that you know
ASSIGNMENT
SECTION A
1.Viscosity of a liquid does not depend on the (a) nature of liquid (b) normal reaction between the liquid layers (c) area of the surface in contact (d) temperature of the liquid
2.Viscosity opposes motion of an object in (a) solid (b) liquid only(c) gas only (d) liquid & gas
3.Which of the following statements about viscosity are correct? When a ball falls through a viscous liquid (i) viscosity opposes the gravitational force on the ball (ii) viscosity opposes the upthrust on the body (iii) viscosity is in the same direction as the upthrust on the ball (iv) the ball falls faster in a more viscous liquid (a) I and II (b) I and III (c) II and IV (d) III and IV
4.The SI unit of velocity gradient is (a) m/s (b) s-1 (c) m/s2 (d) ms
5.Terminal velocity is attained when (a) w + v = u (b) w = v – u (c) w + u = v (d) w = v + u
SECTION B
1.Derive the dimension of viscosity
2.State two effect of viscosity
3.Describe an experiment to determine the terminal velocity of a steel ball falling in a fluid
WEEK 11
Revision
WEEK 12
Examination
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