FIRST TERM
CLASS: SS2.
SUBJECT: PHYSIC
SCHEME OF WORK
WEEK TOPIC
THEME 1: Scalars and Vectors
Scalars and Vectors, addition of vectors, parallelogram law, components of a vector, resolution
of two or more vectors
THEME 2: Motion
Speed, Velocity, Acceleration
Equations of Uniformly Accelerated Motion
Equation of Motion of bodies falling under Gravity
THEME 3: Projectile
Projectile Motion
Time of Flight
Maximum Height
Range
Application of Projectile
THEME 4: Equilibrium of Forces I
Static and dynamic Equilibrium
Resultant and Equilibrant Forces
Moment of Force
Resultant Moments
Conditions of Equilibrium under the action of Coplanar Forces
Principle of Moments
THEME 5: Equilibrium of Forces II
Couple
Moment of a Couple
Conditions of Equilibrium under the Action of Non-Parallel Coplanar Forces
Centre of Gravity
THEME 6: Stability of a Body
Equilibrium of Bodies in Fluids
Upthrust
Archimedes Principle
Floatation Principle
Density
Relative Density
THEME 8: Simple Harmonic Motion (SHM)
Definition of SHM, Illustration of SHM from String
Simple Pendulum
THEME: SHM
Definition of Terms
Speed and Velocity
Linear and Angular Speed
Linear and Angular Velocity
Energy in SHM
Resonance and Forced Vibration
THEME 10: Linear Momentum
Momentum
Impulse
Newton’s Laws of Motion
Conservation of Linear Momentum
Types of Collision
THEME 11: Machine
Mechanical Energy
Definition of Machine
Mechanical Advantage
Velocity Ratio, Efficiency of Machine, Lever, Classes of Lever
1
WEEK 1
TOPIC: SCALAR AND VECTORS
CONTENT
By the end of the lesson, the students should be able to:







Define Scalar and Vector Quantities with Examples
Add Vectors in
(i)
Same direction
(ii)
Opposite direction
Find their resultants
State parallelogram law of Vectors
Calculate resultants using Cosine rule
Cosine components of a Vector
Resolve two or more vectors acting at a point
Scalars and Vectors
Scalars are physical quantities which have magnitude only or numerical value
but no direction. Examples are mass, length, density, time etc.
Representation of Vector
A vector is always represented by a straight line with an arrow at the one end
pointing in a particular direction. The length of the line represents the
magnitude of the vector e.g.
A
The vector
AB
10N
B
is 10N in eastern direction.
Addition of Vector
If two vectors B and D moved in the same direction and are 10N and 12N
respectively, then, their resultant R is
1.
B
10N
R
22N (10+ 12)
D
12N
2. If B
and D
are in opposite direction
10N
12N
= D-B=R
D
B
R = 2N
The resultant is in the direction of the larger force i.e.
D
B-
2
D=
=
3. When the two vectors are equal, the total resultant is zero.
12N
= R = 0.
12N
D
B
R = 0.
iv. When two vectors are inclined at 900 to each other, their resultant is
obtained using the Pythagoras theorem.
∅
Using Pythagoras,
(i) R2 = B2 + D2
= 102 + 122
(ii) The directionθ is given by
B
10
Tan θ = D or Tan θ = 12
= 100 + 144
= 244
= 0.8333
∴ θ = 39.80
R = √244
= 15.6N
The resultant can also be found by scale drawing.
The resultant vector is a single vector which would have the same effect in
magnitude and direction as the original vectors acting together.
Parallelogram Law of Vectors
This is used when two vectors are inclined at an angle to each other and it
states that:
If two forces acting at a point are represented in magnitude and direction by
the sides of the parallelogram drawn from that point, their resultant is
represented in magnitude and direction by the diagonal of the parallelogram
drawn from that point.
3
If B and D are inclined at an acute angle of 600, the resultant can be found
using scale drawing or using Cosine rule;
R2 = (0B)2 + (0D)2 + 2(0B)(0D) Cos600
R2 = B2 + D2 + 2BD Cos600
And the other angles by Sine rule
B
SinB
=
D
SinD
=
R
SinR
To find the resultant, from the diagram,
R2 = (0B)2 + (0D)2 + 2(0B)(0D) Cos600
R2 = 52 + 42 + 2 x 4 x 5 Cos600
R2 = 16 + 25 + 2 x 20 x Cos 60
R2 = 41+ 40 x 0.5
R2= 60
R = √61
= 7.81N
Components of a Vector
A vector, V, generally has two components, namely the vertical component of a
vector Fy , and horizontal component of a vector, Fx
4
Ɵ
The component of a vector is the effective value of the vector in a particular
direction.
Horizontal component of a vector is the effective value of the vector in
horizontal direction (Fx )
Cosθ =
FR
Fx
Fx = FRCosθ
The effective value of any force or vector in horizontal direction (Fx ) or
component is FCos(Fx )
For the vertical component of a force, is the effective value of a force/vector in
vertical direction (Fy )
Sin θ =
Fy
F
Fy = F Sin θ
The effective value of any vector or force in vertical direction Fy is F Sin θ
R2 = Fy2 + Fx2
5
R = √Fy2 + Fx2
The direction of the force
tan θ =
Fy
Fx
Fy
θ = tan−1 ( F ).
x
Two or More Vectors Acting at a Point
When there are two or more vectors acting at a point, each of the vectors is
resolved into the vertical and horizontal components
F
θ
Vertical Component (Fy )
Horizontal Component (Fx )
F1
θ1
F1 sin θ1
F1 cos θ1
F2
θ2
F2 sin θ2
−F2 cos θ2
F3
θ3
−F3 sin θ3
−F3 cos θ3
F4
θ4
−F4 sin θ4
F4 cos θ4
Fy
Fx
Resultant
direction
R2 = Fx2 + Fy2
R = √Fy2 + Fx2 , tan θ =
Fy
Fx
6
Find the magnitude
and direction of the
resultant forces
F
θ
Vertical Component (Fy )
Horizontal Component (Fx )
10
300 F0 cos 30 = 10x 0.8660 = 8.67N
10 sin 300 = 10 x 0.5 = 5.0N
15
600 −15 cos 60 = -15 x 0.5 = -7.5N
15 sin 600 = 15 x 0.8660 = 12.99N
12
450 −12 cos 45 = -12 x 0.7071 = -8.49N
−12 sin 450 =-12 x 7.7071 = -8.49N
14
500 14 cos 50 = 14 x 0.6428 = 9.00N
−14 sin 500 =-14 x7.7660= -10.73N
Fx = 1.68N
R = √Fy2 + Fx2
Fy = -1.23N
The direction
Fy
R = √(−1.23)2 + (1.68)2
R2 = 1.5129 + 2.8224
R = √4.3353
R = 2.082N
tan θ = F =
x
−1.23
1.68
tan θ = -0.7321
θ = tan−1 (−0.7321)
θ = 36.20
7
WEEK 2
TOPIC: MOTION
CONTENT
By the end of the lesson, the students should be able to:

Define
Speed , Velocity and Acceleration
State their uses



Derive equations of uniformly accelerated motion
State equation of motion under gravity
Solve simple problems on using the equations
Speed
Speed is defined as the distance moved per time taken. It is a scalar quantity
i.e. it has only magnitude no direction.
Speed =
distance moved
𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
kilometre
=
hour
Its unit is Kmh-1 or ms-1
Velocity is the time rate of increase displacement. It is a vector quantity. It is
measured in ms-1.
V=
increase in displacement
time
Acceleration is the time rate of increase velocity. It is a vector quantity and its
unit is ms-2.
a=
increase in velocity
a=
V−U
time
t
V – Final velocity
U – Initial velocity
t – time in seconds
Retardation is the time rate of decrease velocity. It is a vector quantity and
measured in ms-2.
a=
U−V
t
8
Equations of Uniformly Accelerated Motion
Suppose a body is moving with uniform acceleration, a, and increase in
velocity, v, in time, t, in seconds.
acceleration of the body, a.
a=
change in velocity
a=
V−U
time
t
∴ at = v-u
V = u + at
_______________ (1)
V – Final velocity,
u- Initial velocity,
a- acceleration,
t- time in seconds
This is called the first equation of motion.
If the velocity of the body increases as it undergoes uniform acceleration, the average
velocity is equal to half of the initial and final velocities
i.e. average velocity = av =
u+v
2
The distance covered, s, during the journey is
S = average velocity x time
=
u+v
2
xt
*
From equation (1), v = u+at; substituting for v in equation *; then
S=(
u+u+at
S=(
2u+at
)xt
2
2
)xt
at
S = (𝑢 + 2 ) x t
S = 𝑢𝑡 +
at2
2
1
S = 𝑢𝑡 + at 2
2
___________ (2)
This is the second equation of motion.
Using the above equations (1) and (2), and eliminating t,
V = u + at
Squaring both sides,
V2= u2 +2aut+a2t2
9
1
V2 = U2+ 2a(𝑢𝑡 + 2 at 2 )
From equation (2)
1
S = 𝑢𝑡 + 2 at 2
1
Let 𝑢𝑡 + 2 at 2 be replaced by S, then
V2=u2+2aS.
______________ (3)
The four equations of motions are;
(i)
(ii)
(iii)
(iv)
V = u + at
1
S = 𝑢𝑡 + at 2
2
V2=U2+2aS
u+v
S=(2 )xt
Examples
A bike travels from rest and accelerates uniformly to a velocity of 20ms-1 in 5
seconds. Calculate the distance travelled.
Solution:
S=
u+v
S=
0+20
2
2
u = 0.
V= 20ms-1,
t= 5s
xt
x5
= 10 x 5
S = 50m
Motion under Gravity
Scientists carried out various researches on motion of objects falling under
gravity and were able to come up with acceptable concept that all bodies falling
under gravity freely to the earth moved with the same acceleration irrespective
of their masses. This gravitational acceleration is the acceleration on an object
caused by the force of gravitation.
Neglecting friction such as air-resistance, all bodies accelerate
gravitational field at the same rate relative to the centre of mass.
in
a
This acceleration is known as acceleration due to gravity denoted by g.
When an object is falling to the earth, it obeys the law of gravity and it is
positive g (+y) but when it is an object moving vertically upward, it is against
the gravity hence, it is denoted as negative g (-g), therefore, for a freefall of
bodies under gravitational motion, a is replaced by g, hence, the equation of
motion for the uniformly accelerated bodies becomes;
10
V = u ± gt
1
H = ut± 2 gt 2
V2 = u2± 2gt
S=H
At maximum height, H, the velocity = 0.
The initial velocity is zero, when an object is released from a height, h.
When an object strikes the ground, h = 0,
If an object falls from a height, h, u = 0
1
H = ut + 2 gt 2
1
H = 0 + 2 gt 2
t2 =
2H
g
2H
t=√g
Example
An orange falls from a tree of height 40m, calculate the time of fall and velocity
with which it strikes the ground.
Solution: h = 40cm,
g= 10ms-2, u = 0, v= ?
1
H = 2 gt 2
2h
t = √g
=√
2 x 40
10
= = √8
t = 2.83Secs.
Velocity = u + gt
v = 0 + 10 x 2.83
= 28.3ms-1.
11
WEEK 3
TOPIC: PROJECTILES
CONTENT
By the end of the lesson, the students should be able to;






Define projectile and projectile motion
Give examples of projectile in
(i)
Sport
(ii) Warfare
Define (i) Time of Flight (ii) Maximum Height
(iii) Range
State equations, formulae for the time of flight, maximum height and
range
State applications of projectile
Solve simple problems
Projectile
It is common experience that children love to throw stone to a bird on a tree, or
throw a stick to a tree in order to pluck a mango fruit from a mango tree, the
stone thrown into the air or stick is regarded as a projectile.
Any object thrown into the air and allowed to fall freely under the influence of
gravity is called a projectile.
The trajectory or curved motion or path taken or described by a projectile while
in the air is a trajectory.
A projectile motion is the curved motion described by a falling object in the air.
Examples of bodies exhibiting projectile motion are:
In sport:
A
A
A
A
A
thrown javelin
thrown discuss
thrown shotput
bouncing table tennis ball from the wall
fired missile
In Warfare:
A thrown bomb
A thrown spear
A fired bullet from a gun
A shot stone from catapult
Time of Flight
12
If an object is projected with a velocity, u, and at an angle of inclination, θ,
with the horizontal, the vertical component of the velocity is usin θ, while the
horizontal component ux , is ucos θ.
Uy
U
H
ϴ
Ux
R
At the maximum height, the final vertical velocity of the projectile is 0, v = 0,
then if
V = u + gt
0 = usin θ-gt
t=
usinθ
g
This is the time taken for the projectile to reach the maximum height, H, the
projectile also use the same time to return to the plane of projection.
If the time taken to reach maximum height is t1 and t 2 is the time taken to
return to the plane of projection, and t1 = t 2 = t, then the total time of flight for
the projectile to travel and return to the plane of projection is t1 + t2 = t + t = T.
T = 2t
T=
2usinθ
g
Time of flight (T) of a projectile is the time required for a projectile to return to
the plane of projection.
Maximum Height
The maximum height (H) of a projectile is the highest vertical distance attained
by a projectile as measured from the horizontal plane of projection.
To find the maximum height, H.
Using the equation under gravity,
V2 = U2 – 2gH
But U = u sin θ, V = 0 at maximum height
13
then 0 = (u sin θ)2- 2gH
2gH = U2Sin2θ
H=
U2 Sin2 θ
2g
Range
Range of a projectile is the horizontal distance covered by a projectile from the
point of projection to the point where it strikes the projection plane again.
If the time taken to cover horizontal distance is T, and the horizontal
component of the velocity is u cos θ, then
Range = horizontal velocity x Time
R= u cos θ x T
R= u cos θ x
R=
2usinθ
g
2U2 sinθcosθ
g
From trigonometry, 2sinθcosθ = 2θ
Then
R=
U2 sin 2θ
g
When θ =450, Range is maximum = 2 x 45 = 90
Then, R =
Rmax =
U2 sin 90
g
U2
g
Hence, maximum horizontal range can be achieved when θ is 450.
Applications of Projectile


It is used in warfare to hit enemy target
It is used in sports by throwing javelin, discuss, etc. at 450 to obtain
maximum range.
Example
An missile is fired at a speed of 200ms-1 at an angle of 450. Find
(i)
(ii)
(iii)
the height attained by the missile
How long does it take in the air?
The horizontal range
Solution
θ= 450, U = 200ms-1, g = 10ms-2
14
(i)
Height attained =
U2 Sin2 θ
2g
2002 x (0.7071)2
H=
20
2002 x 0.5
H = 20 =
H = 1000m.
(ii)
Time of Flight, T =
=
=
2000
20
2Usinθ
g
2x200x sin450
10
400x 0.6494
10
=
259.779
10
= 25.97Secs
(iii)
Range =
=
U2 x sin2𝜃
g
2002 sin2X45
10
40000 x 1
=
2002 xsin90
10
= 10
= 4000m
Evaluation Questions
1. A body is projected upwards with initial speed of 400ms-1 at an angle of
300. Find (i) Time of Flight
(ii) horizontal distance travelled
2. A stone is fired from a catapult at an angle of 600 and initial speed of
60ms-1. Calculate the time of flight
3. Maximum range is obtained when the angle of projection is
A. 600
B. 900
C. 300
D. 450.
4. What is a projectile?
5. Give two examples of projectile in
(i)
Sport
(ii)
Warfare
15
WEEK 4
TOPIC: EQUILIBRIUM OF FORCES
CONTENT
By the end of the lesson, the students should be able to:






Define Static and dynamic equilibrium
Distinguish between resultant and equilibrant forces
Define moment of force
State conditions of equilibrium under the action of coplanar forces
State principles of moments
Solve simple problems of it
Equilibrium of Forces
When forces act on a body, it may change the position of rest of the body or
cause the body to rotate about its fixed axis.
A body is said to be in equilibrium when it is at rest or in a state of uniform
motion. When the body is at rest, it experiences static equilibrium and when it
undergoes uniform motion, it experiences dynamic equilibrium. A body is said
to be in equilibrium when all the forces acting on the body cancel out that is
the resultant of all the forces acting on equal to zero.
Resultant and Equilibrium Forces
For a body to be in equilibrium, the sum of forces in one direction must be
equal to sum of the forces in the opposite direction.
R
B
A
O
16
From force board experiment, the resultant R of two forces A and B acting at
point 0, is found using parallelogram method. It is found to be equal in
magnitude but opposite in direction to the third force. This third force which is
equal in magnitude but opposite in direction of resultant force is called
equilibrant force, therefore,
A resultant force is that single force that act alone and has the same effect in
magnitude and direction as two or more forces acting together.
An equilibrant force is a single force which will balance all other forces taken
together.
If F1 , F2 , and F3 are in equilibrium, F1 is the equilibrant of F2 and F3 , F3 is the
equilibrant of F1 and F2 , F2 is the equilibrant of F1 and F3 .
Moment of a Force
When taps are opened, a turning effect of force is experienced, likewise when
doors are opened, the applied force brings about a turning effect about a point
or hinges attached to the wall of the door. Turning effect experienced in each
case is called moment of a force.
Moment of a force about a point is the product of the force and the
perpendicular distance from the line of action to the point or pivot.
Moment = Force x Perpendicular distance of pivot to the line of action of
the force
= Newton x Metre
Its unit is Newtonmetre (Nm), hence, it is a vector quantity.
If the force is inclined at an angle θ.
17
o
ϴ
d
Moment = F x dSinØ
= FdSin Ø
F
The magnitude of moments depends on:
(i)
(ii)
Force applied
Perpendicular distance from the pivot to the line of action of the force.
Resultant Moments
When more than two forces act on a body, the resultant moment on the body
about any point can be obtained using algebraic moments using the clockwise
moment and anticlockwise moments about the same point.
It the clockwise moment is taken as positive and anticlockwise moments are
negative.
∴ Clockwise moment about 0 = Anticlockwise moment about 0
F1 x X1 = F2 x X2
At equilibrium,
F1 x X1 - F2 xX2 = 0.
Conditions of Equilibrium under the Action of Coplanar Forces
Coplanar forces are forces that lie in the same plane and parallel forces are
forces whose lines of action are parallel to each other.
The conditions for equilibrium of coplanar parallel forces are;
(i)
The algebraic sum of the forces acting on a body in any given
direction must be zero. In other words, the sum of the upward forces
must be equal to the sum of the downward forces.
18
(ii)
The algebraic sum of the moments of all the forces about any point on
a body must be zero. i.e. the total clockwise moments about a point
must be equal to the total anticlockwise moments about the same
point.
Principle of Moment
The principle of moments states that if a body is in equilibrium, the sum of
clockwise moments about any point must be equal to the sum of the
anticlockwise moments about the same point.
Examples
1. A 100m uniform rod has a load of 40N suspended at 20cm from one end. If
the fulcrum is at the centre of gravity.Calculate the force that must be
applied at its other end to keep it in horizontal equilibrium.
50
20
0
100
30
50
40N
F
Taking moments about 0,
Clockwise moments about 0 = Anticlockwise moment about the same point.
F x 50 = 40 x 30
F=
40 𝑥 30
50
= 24N.
2. A light rod CD, sits on a pivot AB. A load of 30N hangs at 4m from the pivot
support at A. Find the value of the reaction forces at A and B as shown
F
E
CW
ACW
8m
6m
C
2m
A
D
B
4m
30N
19
Solution
From conditions of equilibrium of parallel forces, total upward forces = total
downward forces.
E + F = 30N.
Using principle of moments,
The total clockwise moments = total anticlockwise moments
Taking moments about A,
30 x 6 = F x (8+6)
30 x 6 = F x 14
F=
30 𝑥 6
14
= 12.9N
Evaluation Questions
1.
2.
3.
4.
Distinguish between equilibrant and resultant forces
What is dynamic equilibrium?
State the conditions of equilibrium under the action of coplanar forces.
A metre rule is balanced at 45cm mark. When a body of mass 50g is
suspended at 8cm mark, the balance point is found to be 25cm mark.
Find the mass of the metre rule.
20
WEEK 5
TOPIC: COUPLE
CONTENT
By the end of the lesson, the students should be able to:






Define couple with examples
Define moment of a couple
State conditions of equilibrium under the action of non-parallel coplanar
forces
Define centre of gravity and its position on different objects shape
State types of equilibrium and the effects of centre of gravity on it
Solve simple problems on it
Couple
A couple is a system of two equal and parallel but opposite forces not acting in
a straight line. This system of forces in parallel can cause a body rotating and
not moving in straight line.
COUPLE
F
F
Wheel of a bicyle
Moment of a couple is the product of one of the forces and the perpendicular
distance between the lines of action of the forces.
Moment of a couple = Force x radius of the wheel.
The distance between the two equal forces is called arm of the couple.
Some of applications of torque or moment of couple include;
(i) The easiness to turn a tap on or off
(ii) It is easier to turn a steering wheel of a vehicle by couple using the two
hands
(iii) The wind vane
(iv) Spinning top
Condition of Equilibrium under the Action of Non-Parallel Coplanar Forces
The conditions of equilibrium under the action of non-parallel coplanar forces
are:
Forces: The vector sum of all the forces acting on a body must be zero i.e.
(i)
The algebraic sum of the horizontal components must be zero ∑ Fx = 0
21
(ii)
The algebraic sum of the vertical components of the forces must be
zero ∑ Fy = 0
2. Moments: The algebraic sum of the moments of all the forces about any
axis perpendicular to the plane of the forces must be zero.
Example
A metal frame weighing 40N is hung on a nail using two ropes as shown.Find
the tension in the rope.
Solution
Resolving in vertical direction, ∑ Fy = 0
Tcos600 + Tcos60 – 40N = 0
2Tcos600 = 40N
Tcos600=
20
cos60
= 40N
Centre of Gravity
Centre of gravity of a body is the point at which the total weight of a body
appears to be concentrated
OR
Centre of gravity of a body is the point at which the total weight of a body
appears to pass through.
Centre of mass of a body is the point at which the total mass of a body appears
to pass through or concentrated.
Centre of gravity for objects varies as shown.
22
for a uniform rod, the
c.g is at mid point of the
metre rule
G
the c.g is at the center of the plate
G
uniform circular plate
c.g at the center of the ring
G
uniform circular ring
c.g is at the point of intersection
of its diagonals
G
Uniform rectangular sheet
c.g of a triangular plate is at
its point of intersections of the medians
C
G
For irregular shaped objects,
The C.G. position is obtained by balancing method: The object is balanced on a
knife edge and a line of the edges are drawn on it. A repeat is carried out
turning the object. The C.G is the point of intersection of the lines drawn.
Use of Plumbline
This method is applied by suspending the object and a plumbline from two
different points on the objects. The point of intersection drawn along the line of
the thread of the plumbline is the C.G.
Types of Equilibrium
23
There are three types of equilibrium namely:
(1)
(2)
(3)
Stable
Unstable
Neutral
(1) Stable Equilibrium: A body is said to be in stable equilibrium if when
slightly displaced or tilted and released, it returns to its original position.
For stable equilibrium, the conditions are:
(i)
(ii)
(iii)
A body must have a low centre of gravity
The vertical line through C.G must fall within the base even if slightly
tilted
The height of C.G must increase from base as body is slightly tilted
(2) Unstable Equilibrium: A body is said to be in unstable equilibrium if when
slightly displaced and released, it tends to move further away from its original
position.
Conditions for unstable equilibrium are:
(i)
(ii)
(iii)
High C.G
Narrow base
Low P.E due to decrease in height of C.G when slightly tilted.
(3) Neutral Equilibrium: A body is said to experience neutral equilibrium, if
when slightly displaced or tilted, it tends to come to rest in its new position.
Evaluation Questions
1. What is a couple?
2. State the conditions of equilibrium under the action of parallel coplanar
forces
3. Define (i) Centre of gravity of a body
(ii) Centre of mass of a body
4. State the types of equilibrium you know.
24
WEEK 6
TOPIC: EQUILIBRIUM OF BODIES IN LIQUIDS
CONTENT
By the end of the lesson, the students should be able to:



Explain upthrust and show its equation
State Archimedes principle and show its verification
State principle of floatation
Equilibrium of Bodies in Liquids
It is a common experience that when an object is immersed in water, it
becomes lighter whether it is wholly or partially immersed.
There is some upward force exerted by the liquid on an object when it is
immersed in a fluid or liquid. This upward force experienced is called upthrust.
If the true weight of the object in air is W1 and the apparent weight in liquid is
W2 , then,
Upthrust, U = W1 - W2
Upthrust depends on:
(i)
(ii)
Volume of object immersed
Density of the liquid into which it is immersed
u
The magnitude of the upthrust is
given by Archimedes principal
Archimedes principle states that when a body is wholly or partially immersed
in a fluid, it experiences an upthrust which is equal to the weight of the fluid
displaced.
U= V𝜌 g
= Vol. of object x density of liquid x acceleration due to gravity
If 1⁄3 volume of the object is immersed in the liquid,
Then, U =
V
3
x density of liquid x g
25
=
V
3
𝜌g.
Experimental Verification of Archimedes Principle
Suspend an object (glass Stopper) freely from a spring balance and read its
weight. The object is then lowered gently into water in Eureka can, and the new
reading is noted.
Weight of object in air = W1
Weight of object in water = W2
Weight of empty beaker = W3
Weight of beaker + displaced water = W4
Apparent loss in weight of object = W1 – W2
Weight of displaced water = W4 – W3
If W1 – W2 is compared to W4
Archimedes principle.
–
W3, they will be found to be equal, confirming
Principle of Floatation
When a cork is pushed inside water, the cork is seen to bob up and float on top
of water when it is released due to up thrust of the water on the cork.
Floatation principle states that when a body is wholly or totally immersed in a
fluid, it experience an up thrust which is equal to the weight of the fluid
displaced.
Due to upthrust, a swimmer can float in water, ice cubes float in water, a
balloon floats in air.
If the weight W, of a body is greater than the upthrust, U, the body will sink.
If the weight is less than or equal to U, the body will rise.
26
If the weight is equal to the upthrust, the body will rest.
Density
Density of a substance is the ratio of mass to a volume of a substance. OR
Density is mass per unit volume of a substance.
Density =
=
mass
volume
kilogramme
m3
= Kgm-3
Its unit is kgm-3.
Different materials have different densities as shown
Material
Density gm-3.
Copper
8.9x103
Lead
11.3x103
Iron
7.8x103
Water
1x103
Mercury
13.6x103
Zinc
7.1x103
Aluminium
2.7x103
Determination of Density of a Solid
(i) Regular Shaped Object: The mass of a regular object is measured with a
chemical balance or beam balance and its volume measured by metre rule,
VernierCalliperor micrometer screw gauge depending on the shape of the body.
Applying the formula,
Density =
𝜌=
mass
volume
m
v
(ii) For Irregular Object: The volume of the object is determined by
displacement method using Eureka can.
27
The volume of water displaced by
suspending the object in water is
equal to the volume of the object.
mass
The formula of density = volume is
used to determine the density of the
body.
Relative Density
Relative density of a substance is the ratio of mass of a given volume of a
substance to the mass of an equal volume of water
Relative density =
R.d =
weight of a substance
mass of substance
mass of equal volume of water
(Because mass is directly proportional to the weight of a body)
weight of equal volume of water
R.d =
Density of a substance
Density of equal volume of water
Relative density has no unit.
Examples
1. A solid weighs 1.80N in air and 0.90N in a liquid of density 900kgm-3.
Calculate
(i) theupthrust of liquid in solid (ii) volume of the liquid.
Solution
(i) Upthrust = Weight in air – weight in liquid
= 1.80-0.90
= 0.90N
W
(ii) mass = 0.09kg (1kg = 10N) = g =
0.9
10
= 0.09Kg
= 9.0g
Volume of liquid =
=
mass
density
9
density
W
=p
=
0.09
900
= 0.0001m3=10-4m3
= 10cm3
28
2. A solid material of volume 2x10-5m3 and density 2.5x103kgm3 is suspended
from a spring balance with half of the volume of the solid immersed in
water. What is the reading on the spring balance? (g = 10ms-2).
Solution
Weight = Vρg
= 2.5x103x2.0x10-5x10
= 0.5N
Upthrust in water = weight of water displaced
V
= Vρg = ρg
2
=
2.0x10−5
2
x 1.0x103x10
= 0.1N.
Spring balance reading = Weight in air ________ Upthrust
= 0.5 – 0.1
= 0.4N
Evaluation Questions
1. A solid weights 0.04N in air and 0.024 in a liquid of density 800kgm-3.
Find the volume of the solid. (g = 10ms-3).
2. State Archimedes principle
3. Describe an experiment to verify Archimedes principle
4. State the principle of floatation
5. Briefly explain why a solid metal sinks in water while a ship made of the
same material does not sink.
29
WEEK 7
TOPIC: SIMPLE HARMONIC MOTION
CONTENT
By the end of the lesson, the students should be able to:



Design Simple Harmonic Motion (SHM)
Give examples of bodies exhibiting SHM
Illustrate SHM for (i) a spring (ii) a simple pendulum
Simple Harmonic Motion to
This is the periodic motion of a body or a particle along a straight line such
that the acceleration of the body is directed towards a fixed point and is also
directly proportional to the displacement from equilibrium position.
Examples of bodies that exhibit simple harmonic motion are:
Motion of balance wheel of a wrist watch
Motion of a simple pendulum
Motion of a child’s swing
The heartbeat
Motion of a loaded test tube oscillating vertically in a liquid
Motion of pendulum of a wall clock
Motion of piston in a gasoline engine
Illustration of Simple Harmonic Motion
Mass on a String
Let a mass M be hung from the lower end of a spring and the other end firmly
clamped to a rigid support.
When the mass originally at position B is pulled down to C and then released,
it is observed to move up and down in regular pattern.
30
M
Using Hooke’s law, the restoring force, F, in the spring would be directly
proportional to the extension, e, produced, provided the spring is not deformed
F = Ke
K – the spring constant
With the attached mass, W = mg = F.
W = Ke.
Mg = Ke _____________ *
When attached mass is displaced vertically through distance x and allowed to
oscillate
e
x
Restoring force, F = -K(e+x) _______ 1
The net force towards centre
F = K(e+x) – mg
i.e. ma= [-K(e+x) – mg] _______ 2
Since mg = Ke, then
Ma = -mg-kx+mg
31
ma = -Kx
a=
−Kx
m
__________ 3
Comparing with SHM equation,
W2 =
k
m
∴ The period of motion, T =
2π
w
2π
m
T = √k
For a Simple Pendulum
If a bob of mass, m, of a simple pendulum is slightly displaced through a small
angle θ and allowed to oscillate about 0,
1
The K.e at A would be minimum, but maximum at 0. K.emax= 2 mv 2 .
But from circular motion, V = wr
1
∴ k.emax = 2 m(w 2 r 2 ) _________
4
At B, or A, the p.e is maximum as a result of the height to which the bob is
raised when displaced.
Maximum p.e = mgh
The force that tends to move it towards the centre 0 = -mgsin θ, and F= ma
∴ -mgsin θ = ma
a = gsin θ ________ **
For the period, T, of oscillation,
The angle of displacement is small, such that the arc 0A form an approximate
straight line, if 0A = y.
32
∴equation ** becomes
y
-mg l = ma
a=
−g
l
y, = (w 2 y) __________ 5
angular velocity (w) of oscillating pendulum
g
w = √𝑙
∴T=
2π
w
2π
l
= √g ________ 6
Example
A loaded tube of 2.5kg mass placed in a liquid of density 9.5g/cm-3 was made
to oscillate vertically when slightly pushed down. If it performs 50 oscillations
in 36 seconds, calculate (i) periodic frequency
(ii) angular velocity
Solution
𝜌 = 9.5g/cm-3, m = 2.5kg, number of oscillations = 50, t = 35s.
Period, T =
=
no.of oscillations
time
time
no.of oscillations
,f=
30
35
f = 1.43Secs
Angular velocity, w =
2𝜋
T
= 2𝜋𝑓
=
2 𝑥 3.14 𝑥 1.43
1
= 8.98rads-1.
Evaluation Questions
1. An object performing SHM has an amplitude of 0.02m and frequency of
20Hz. Find the period of oscillation.
2. Define SHM
Give 3 examples of a body performing SHM
3. Calculate the period of oscillation of a body that makes 50 complete
cycles in 2seconds.
33
WEEK 8
TOPIC: SPEED, VELOCITY AND ACCELERATION AND ENERGY OF SIMPLE
HARMONIC MOTION
CONTENT
By the end of the lesson, the students should be able to:






Define (i) amplitude (ii) period (iii) frequency of a body performing SHM
State the formulae for linear and angular velocity
State the equations for linear and angular acceleration
State the energy of bodies exhibiting SHM
Define resonance and forced vibrations
Solve simple problems on it
Definition of Terms
Amplitude A- This is the maximum displacement of a body performing SHM
from its equilibrium position.
Period, T- This is the number of complete revolutions per second made by a
vibrating body. It is measured in Hertz (Hz)
f=
1
T
or f =
𝑛
t
Speed
B
W
A
Ø
Y
C
x
S
P
Q
Z
D
From the diagram, a particle P, moves round the circle once it sweeps through
an angle𝜃 = 3600. Where 360 = 2𝜋 radians, in time, t, in seconds.
The time rate of change of angle 𝜃 with time (t) is angular velocity (w)
angle turned through by the body
w=
time taken
w=
𝜃
t
but θ = wt
34
distance = velocity x time
S=vxt
s
∴ V = t for linear motion
but
2𝜋 = 3600, w =
θ
t
When θ changes with time, the length of the arc pz = s, also changes with time.
θ=
s
r
s = rθ
Let r = A = radius of the circle.
w=
s
t
s
t
.
θ
t
s
=rx
1
t
1
r
=v
w = V.
1
r
V = wr = wA.
Linear speed is the product of the angular speed and the radius or amplitude of
motion.
Example
A stone tied to one end of inelastic string and whirled round in a circular path
of radius 30cm. If the stone makes 9 complete revolutions in 3secs, find its
angular and linear velocities during the period.
Solution
1Rev = 2π
9v = 9 x 2π
= 18π
w=
θ
t
=
18π
3
= 6𝜋rads-1.
V = wr
= 6𝜋 x 30
35
= 180𝜋ms-1.
Acceleration
Angular acceleration is the time rate of change of angular velocity.
∝=
wf−w0
t
v
w = r , v = wr
1 Vt−V0
∝ = 𝑟(
∝=
t
)
a
r
a = ∝r,∝ = angular acceleration.
a = 𝑤 2r
Angular acceleration is the time rate of increase in angular velocity. It is
measured in rads-2.
Linear acceleration is the product of angular acceleration and the radius or the
displacement of the particles from its central position.
Energy of S.H.M
When force and displacement are involved in SHM, energy or workdone is also
involved.
If a massis suspended from the end of a spring and it undergoes SHM, the
force required to stretch the string through distance x is
F = Kx
where K is the force constant of the spring.
The workdone in extending the spring is
1
Work = average force x distance/displacement = 2 k𝑥. 𝑥
1
= 2kx2.
If A= x, then, A- is the amplitude of oscillation
1
w = 2kA2
1
w = 2mw2A2
,V = wA
Resonance and Forced Vibrations
A body undergoing SHM loses energy slowly due to friction and air resistance
and its amplitude decreases gradually until it becomes zero. When the body
36
experiences such type of motion, it is said to be damped, but when the motion
is free it is said to experience free vibrations.
Free vibration is said to occur when an object vibrate on its own after being
displaced and released.
Forced vibration occurs when a body vibrates under continuous influence of an
external force which is periodic in nature e.g.
Vibration of turning fork
-
when subjected to periodic force of sound wave
Vibration of a bridge when soldiers are marching.
Turning of a radio set
Resonance
Resonance is said to occur when a body or system is set into oscillation or
vibration at its own frequency as a result of impulse received from other system
which is vibrating with the same frequency e.g.A body’s own frequency is better
referred to as its natural frequency (f0). Resonance occurs when the forcing
frequency (ff) of a body equates the natural frequency i.e (ff = f0).
A tuning radio set, car bodies rattles at certain speed
Evaluation Questions
1. Define
(i) Linear Velocity
(ii) Angular velocity
(iii) Linear acceleration
(iv) Angular acceleration
2. A body executing SHM has an angular velocity of 48rads-1. If it has a
maximum displacement of 8cm, calculate its linear velocity.
3. A body of mass 8kg is set into vibration with SHM, if the period of
oscillation is 4secs,find the energy of the motion if the amplitude of
vibration is 5cm.
WEEK 9
TOPIC: LINEAR MOMENTUM
CONTENT
By the end of the lesson, the students should be able to:



Define linear momentum and state its unit
Define impulse and state its unit
State Newton’s laws of motion
37



Use the 2nd law to derive equation F = ma
State the law of conservation of linear momentum
Distinguish between elastic and inelastic collisions
Linear Momentum
Momentum of a body is the product of mass of the body and its linear velocity
Momentum = mass x velocity
= kg x ms-1.
Its unit is kgms-1. It is a vector quantity. The momentum of a body depends on
the mass and the velocity with which the body is moving.
Impulse
Impulse of a force is the product of average force and the time during which it
acts on the body. A force which acts for a very short time is called impulsive
force or simply impulse.
Impulse = Force x time
= Newton x seconds
The unit of impulse is Ns. It is a vector quantity.
Example;
i. A body of mass 20kg moves with a velocity of 10ms-1, calculate its linear
momentum
Solution
Momentum = mass x velocity
= 20 x 20
= 200kgms-1.
ii. A force of 40N acts on a body for 5seconds. What is the impulse experienced.
Solution
F = 40N,
t = 5secs.
Impulse = Force x time
= 40 x 5
= 200Ns.
Newton’s Laws of Motion
Sir Isaac Newton studied the behaviour of bodies in motion and was able to
come up with three laws governing motion and they are as follows:
38
Newton’s First Law of Motion
A body will continue to be in a state of rest or of uniform motion, unless it is
acted upon by an external force.
Second Law
The time rate of change of momentum of a body is directly proportional to the
applied force and this takes place in the direction of the force.
Therefore,
a=
F∝
Change in momentum
time
F∝
mv−mu
F∝
m(v−u)
t
t
v−u
t
, from equation of motion
F ∝ ma
Introducing a constant, k, = 1
F = kma,
k=1
F = ma
The unit of force is Newton, and it is a vector quantity.
Newton is a force which gives us a mass of 1kg on acceleration of 1ms-2.
From second law
F=
mv−mu
t
Ft = mv – mu.
Impulse of a force is equal to =the change in momentum of the body.
F = ma
a=
F
m
Third Law of Newton
Actions and reactions are equal and opposite. When a wooden block is resting
on a table, the weight of the block is acting vertically down on the table while
the table is also exerting equal and opposite forces to balance the downward
force so that the block can remain at rest.
39
R
reactional force
W
Weight of
the block
This opposite force that allows the object to balance is called reactional force
(W=R).
Other examples of third law explanation are:
Recoil of a gun, jet propulsion and walking on the ground.
Conservation of Linear Momentum
The law of conservation of linear momentum states that for a system of
colliding bodies the total momentum remainsconstant, provided no external
forces act. OR
The total momentum of a body before collision is equal to the momentum of
abody after collision provided no external forces act.
If masses, m1, and m2, move with initial velocities U1 and U2 respectively
collides with each other and moved in different directions
Before Collision
After Collision
M
M
1U1
2U2
M1U1 + M2U2
M
M
1V1
2V2
M1V1 + M2V2
From conservation of linear momentum, the total momentum before collision is
equal to the total momentum after collision; therefore,
m1U1 +m2U2 = m1V1 +m2V2
When two bodies collide in opposite direction,
40
Before Collision
After Collision
M1
M2
U1
U2
=
M1U1 - M2U2
=
M1
M2
V1
V2
M1V1 + M2V2
When a stationary body collides with a moving body, and move apart.
BeforeCollision
After Collision
=
M2
M1
M1
U2 = 0
U1
M2
M1V1
V2
M1U1 - 0 = M1V1 + M2V2
If two bodies collide with each other, and stick together and move with common
velocity.
Before Collision
After Collision
M1
M2
U1
U2
M1
=
M2
V
m1U1 +m2U2 = (m1+m2)V
V- is the common velocity of the body.
Types of Collision
There are two types of collision namely elastic and inelastic.
For elastic collision, the total momentum is conserved and the kinetic energy is
also conserved
M1
M2
U1
U2
=
M1
M2
V1
V2
41
The total momentum before and after collision
m1U1 +m2U2 = m1V1 +m2V2
The kinetic energy before and after is
1
1
1
1
m1U12 +2m2U22 = 2m1V12 +2m2V22 .
2
Inelastic Collision
The total momentum is conserved but the kinetic energy varies i.e. is not
conserved. This occurs when two colliding bodies stick together after collision:
and move with a common velocity.
Before Collision
M1
M2
U1
U2
M1
+
M2
V
m1U1 +m2U2 = (m1+m2)V
1
1
1
K.e = 2m1U12 +2m2U22 = 2(m1+m2)V2
Application of Newton’s Laws and Conservation of Momentum Laws
Recoil of a Gun
When a gun is fired, the bullet moves forward with a velocity of V, and mass m,
if the gun has mass m, it moves forward with a velocity, V.From conservation of
momentum.
The momentum of the gun and bullet before and after firing is given as
After firing
Before firing the gun
M1V1 = 0
M1U1 = -M2U2
Before firing
m1+V1 = 0
42
After firing
m1U1 +m2U2 = 0
m1U1= -M2V2
M1 = Mass of the gun
M2 = Mass of bullet
U1= Velocity of gun
U2= Velocity of bullet.
Inertia and Inertia Mass
Inertia is the reluctance of a body to change its state of rest or uniform motion
in a straight line. The more massive a body is, the greater the force required to
change its state of rest or of uniform motion and give acceleration to the body,
hence the greater the inertia. Inertia is determined by the quantity of matter
contained in a body.
Since F = ma.
Objects in a Lift
When the lift is ascending with acceleration, a, the tension in the string lifting
the lift is
T = m(g+a)
When the lift is descending, a less than g,
T = m(g-a)
When the lift is moving up with uniform acceleartion,
a = 0,
T = m(g+0)
T = mg.
When the lift is falling under gravity, g = a, T = 0 when the lift is moving down
with acceleration “a” greater than “g”.
T = m(a-g)
= ma + mg.
Evaluation Questions
1. A player hits a ball of mass 24kg moving northwards with a velocity of
15ms-1 causing it to move with a velocity of 20ms-1 southwards. The force
43
acted on the ball for 0.025s. What is the average force exerted on the ball
by the player?
2. A ball of mass 200g travelling with a velocity of 100ms-1 collides with
another ball of mass 800g, moving at 50ms-1 in the same direction. If
they stick together after collision, what is their common velocity?
3. A bullet of mass 0.05kg is fired from a gun of mass 9kg, the bullet
moving with an initial velocity of 200ms-1. Find the backward velocity of
the gun.
44
WEEK 10
TOPIC: MECHANICAL ENERGY AND MACHINES
CONTENT
By the end of the lesson, the students should be able to:





State the law of conservation of mechanical energy
Define machine with examples
Define
(i) Mechanical advantage
(ii) Velocity ratio
(iii) Efficiency, as it relates to a machine
State types of lever
Solve simple problems
Mechanical Energy and Machines
Energy is the ability to do work. There are two types of energy that constitute
mechanical energy, they are: Potential energy and Kinetic energy.
Potential energy, P.E = mgh (Energy of a body by virtue of its
position or rest or stored energy due to height).
m = mass of the body
g = acceleration due to gravity
h = height to which the object is raised
Kinetic energy, K.E =
1
mV2 (Energy of body in motion)
2
m = mass of the body
V = Velocity of the body
The law of conservation of mechanical energy shows that P.E + K.E = constant.
Machine
Machine is a device or a contrivance with which force is applied at one end
(Effort) to overcome load (L) at the other end.
A small force (effort) can be applied to overcome a larger load.
Examples of such machines are Lever, Pulleys, inclined plane, screw, hydraulic
press, etc.
Terms Associated with Machines
Mechanical Advantage of Force Ratio
This is the ratio of load to the effort
45
Load
M.A =
Effort
=
L
E
It has no unit
Velocity Ratio
This is the ratio of distance moved by effort to the distance moved by load
V.R =
distance moved by effort
distance moved by load
=
x
y
When load is greater than effort, M.A>1, the V.R>1.
Efficiency
This is the ratio of work output to the work input of a machine expressed as a
percentage.
Efficiency =
Work output
Work Input
x 100%
But work = Force x distance/displacement
E=
E=
E=
E=
Load x distance moved by load
Effort x distance moved by effort
L
E
x
x
y
L⁄
E
x⁄
y
M.A
V.R
x 100%.
For an ideal machine, the efficiency is 100%, but due to friction, M.A decrease,
and this also affects (reduces) the efficiency of the machine.
Example
A machine with velocity ratio of 5 requires 1200J of work to lift a load of 400N
through a vertical distance of 2.0m. Calculate the efficiency and mechanical
advantage of the machine.
Solution
E=
=
=
M.A
V.R
x 100
400 x 2
1200
800
1200
x 100
x 100
= 66.7%
46
E=
66.7
100
M.A
V.R
=
x 100
M.A
M.A =
5
66.7 x 5
100
= 3.33.
Types of Machines
Lever
A lever is a rigid bar pivotal on a support known as fulcrum or pivot. A force F
is applied at one end of the bar to overcome load L at the other end.
Fulcrum
The distance between the fulcrum and the effort is called effort arm while the
distance between the load and fulcrum is called load arm.
Types of Lever
First Class Lever: For this type of machines, the fulcrum is between the load
and the effort applied e.g. plier, scissors, claw harmers, etc.
Second Class Lever: The load is between effort and the fulcrum.
47
Example is a wheelbarrow, opener, etc.
Third Class Lever: This is when the effort applied is between the load and the
fulcrum e.g. nutcracker, forceps, sugar tong.
Evaluation Questions
1. A machine requires to lift a load 30N through a height of 12m. If the
efficiency of the machine is 50%, how much work is done.
2. Explains the statement: The velocity ratio of a machine is 4.
References
1. Functional Physics by Oluoma, Nelson Publishers, 2011
2. Model Physics for Secondary Schools, J. E. Salihu, Classic Educational
Publishers, 2015
3. New School Physics for Senior Secondary School by M. W. Anyakoha,
AFP Publishers, 2016
2. PULLEYS
A pully is a flat circular metallic or wooden disc, having a grooved rim and capable of evolving
around a fixed point passing through its centre commonly known as axle.
TYPES OF PULLY
1. Single fixed pully: is a pully which has its axis of rotation fixed in a position
3. The Inclined Plane: this is the type of machine used to raised heavy load
such as drum of oil, up a sloping plank to the high floor of lorries. the sloping
plank is an example of an inclined plane.
48