Inertia : Is a property of an object that is its tendency to resist change in its velocity.
The mass of an object is a measure of its inertia.
If an object has a large mass it requires a large force to set it in motion or to change its
velocity.
If an object has a large mass it requires a large force to stop it if it is moving.
FORCE : Force is defined as that which changes or tends to change the state of rest or
uniform motion in a straight line of a body.
OR
Force produces an acceleration which causes a change in the velocity of the object
OR
A body remains in a state of rest or uniform motion in a straight line unless it is acted on
by an external force.
It can be shown that
Force = mass x acceleration
F = m .a
S.I. Units ( Basic )
kg . m.s-2
Which is called Newton
N
The force of gravity on a body is called the weight of the body
Weight = m . g ( g = 9.81 m.s-2 )
Vector quantity which acts vertically down.
The S.I. unit of weight is Newton.
NOTE 1: Force is a vector quantity which is in the same direction as the acceleration
since
F

m. a
=
The two vectors Force and acceleration are in the same direction
2.
Also
Weight

W
=mass x acceleration due to gravity
=
m .

g
The two vectors in the above equation are W and g and act in the same direction.
Therefore W is in the same direction as g i.e. acting vertically downward.
However the acceleration value a is only equal to 9.81 ms-2 if and only if the object is
free falling in the vertically downward direction.
PROBLEM SHEET.
Question 1. A constant force acts on a
5 kg object and reduces its velocity from
7 m.s-1 to 3m.s-1, along the +x direction , in a time of 3 seconds. Calculate the value of
the force.
Question 2. A 6 kg object is to be given an acceleration of 0.7 m.s-2 along the +x
direction calculate the value of the force acting on it.
Question 3. Find the weight of the following masses (a) 10 kg (b) 60 kg
Question 4. Calculate the mass of a body which has a weight of 100 N.
Question 5. Calculate the force on a mass of 10 kg if its velocity is to be increased from
30 km.hr-1 to 60 km.hr-1, along the +x direction , in a time of 10 seconds.
Question 6. Explain the difference between the mass and the weight of a substance
giving the S.I. units of each.
Question 7. A mass of 5 kg starts from rest at the origin. Two Forces act on the mass F1
which is a force of 15 N in the +x direction and F2 a force of 8N at +45o.
(i)
Calculate the total force acting on the object.
(ii)
Calculate the total acceleration.
(iii)
Calculate the total displacement in a time of 10s
(iv)
Calculate the final velocity after 10s.
Question 8. Repeat question 7 if a third force F3 20 N along the +y direction also acts on
the mass.
Momentum. The momentum of a body is defined as the product of its mass multiplied
by its velocity.
Momentum = mass x velocity
Symbol used is p
p= m . v
S.I. Units kg m.s-1
VECTOR in the same direction as the velocity
Conservation of momentum states that when two or more bodies act on each other the
total momentum of the system before the action is equal to the total momentum after the
action provided that no external forces act on the system.
Action is usually a collision between the bodies.
NOTE: Any question which relates the a collision between two or more bodies is solved
using the conservation of momentum.
Example 1. Calculate the momentum of an object of mass of 500 g moving at a velocity
of 20 m.s-1 due East.
Example 2. An object of mass 1 kg moving at velocity 20m.s-1 due east collides with a
stationary object of mass
3 kg. If the two masses couple together calculate the common velocity after the collision.
Example 3. An object moving with a steady velocity of 4 m.s-1 along the + x axis
collides with a second object of double the mass and moving at half the velocity of the
first object. If the two masses couple together calculate the common velocity after the
collision.
Example 4. An object of mass 4 kg moving in the + x direction with a velocity of 10 m.s1
collides with an object of mass 1 kg moving in the – x direction with a velocity of 4 m.s1
. Assuming that the two masses couple together calculate the common velocity after the
collision.
Example 5. An object of mass 4 kg traveling along the + x direction with a velocity of
20m.s-1 collides with an object of mass 2 kg traveling along the – x direction with a
velocity of 15 m.s-1. The two objects bounce off one another and after the collision the
larger mass has a velocity of
15 m.s-1 in the -x direction, calculate the velocity of the smaller mass after the collision.
NEWTONS LAWS OF MOTION
First Law.
Every body remains in a state of rest or uniform motion in a straight line unless acted
upon by a force.
Second Law.
The rate of change of momentum is proportional to the applied force and takes place in
the direction of the force.
Third Law.
When one body exerts a force on a second body, then the second body exerts an equal
and opposite force on the first body.
Action and Reaction are equal and opposite.
Newton’s First Law and circular motion.
Newtons first law which is the definition of force states that a body will remain at rest or
uniform motion if a straight line unless it is acted upon by an external force.
Circular motion or motion in a circle.
When a body moves in a circle then, there must according to Newton’s first law be some
force acting on it which tends to deflect it from its tendency to move in a straight line.
This force is directed towards the centre of the circle around which the object is moving
and is called the CENTRIPETAL FORCE.
If a particle is moving around a circle of radius r (m) with a uniform speed
v (m.s-1), the time take for the object to complete one revolution of the circle is called the
PERIODIC TIME or PERIOD. To distinguish from the general symbol for time t we use
the symbol capital T.


distance
Time 
speed
In one periodic time the distance travelled would be equal to the circumference of the
circle therefore
2r
T
v
where r = radius of the circle
Note although the speed is constant the velocity is changing because the direction is
changing, because the direction of the motion is changing.
Around a circle the direction of the velocity at any point along the circle is the direction
of the tangent to the circle at that point.
Since the velocity is changing, due to the changing direction there is an acceleration.
Acceleration is a vector quantity with both a magnitude and a direction.
The magnitude of the centripetal acceleration can be shown to be

2
v
ac 
r
The direction of the acceleration can be shown be towards the centre of the circle around
which the object is moving.
Given
then
v2
ac 
r
mv
Fc 
r
2
And the direction of acceleration a, and F are the same and so the centripetal force acts
towards the centre of the circle.
Note if an object is moving around a circle the angle at the centre of the circle is also
ch
radians.
radians per second.
And remember that
2 .radians  360
o
v  linear speed ms
-1
  angle measured in radians

t
 angular velocity
=  measured in rad.s
The arc length of the circle
= s
s
r
 =  s  r
s (r )

v 
 r  r
t
t
t
Therfore v  r
where v = linear speed in ms
r  radius of the circle in m
 = angular velocity in rad.s
-1
-1
-1


mv 2 m(r ) 2
Fc 

 mr 2
r
r
This is the equation for the magnitude of the centripetal force and the direction is in the
same direction as ac i.e. towards the centre of the circle.
Example a body of mass 2 kg moves around a circular path of radius 50 cm at a constant
speed of 10 m.s-1 .
Calculate (i) the angular speed and
(ii) the centripetal force.
For circular motion we have shown that
v  r
It can also be shown that
a  r Where a = linear acceleration (ms
and  = angular acceleration (rad.s
-2
)
-2
)
There is also an equivalent set of equations of motion in terms of the angular values
Linear
Angular
v = u + at
 f   i  t
s  ut  1 2 at 2
 =  i  1 2 t 2
v 2  u 2  2as
 f 2   i 2  2
Where
u
i =
r
v
f 
r
f
1m.

a
and  
r
= v except in the case where the radius of the circle is equal to
PROBLEM SHEET.
Question 1. A car which has wheels of radius 30 cm starts from rest and accelerates
uniformly to a speed of 15 m.s-1 in a time of 8seconds. Calculate (a) angular acceleration
(b)time taken to rotate 100 times.
Question 2. A spin drier revolving at 900r.p.m.( revolutions per minute ) slows down
uniformly to 300 r.p.m. while making 50 revolutions.
Calculate (a) angular acceleration and (b) time taken
Question 3. In the hydrogen atom an electron rotates around a circle of radius
0.5 x 10-10 m. The centripetal force being the attraction of the -ve electron to the +ve
nucleus. Calculate the magnitude of the force if the speed of the electron is 2.3 x 106 m.s1
and the mass of the electron is
9 x 10-31 kg.

Newton’s Second Law
Used to derive the equation for Force,
It say that the force is proportional to the rate of change of momentum
i.e.
mv  mu m(v  u)
F

t
t
However
vu
a
t
Then
F  ma  F  kma
Where k is a constant.
It was found by experimental techniques that the constant k is equal to 1
Therefore
F  ma
NEWTON’S THIRD LAW.
Two bodies in contact.
e.g A block of mass m ( kg ) placed on a level surface.
F2
Mass m (kg )
F1
F1 is the weight of the mass m (kg ) acting in the downward vertical direction.
F1 = m . g (Action Force) = W
F1 is the force on the second body i.e. the level surface due to the first body i.e. the mass
By Newton’s Third Law the second body, the plane surface must exert an equal and
opposite force on the mass shown as F2
Therefore
F2 = - F1 = - m . g =
R
Called the Reaction force, symbol R
The total force on the mass in the vertical direction is equal to the sum of the forces
acting in the vertical direction
There are two forces acting in the vertical direction
So
FT = F1 + F2 = m . g + (- m . g) = 0
This is correct since the mass is stationary in the vertical direction the total force in that
direction should equal zero.
NOTES:
1.  F1 
=  F2 
i.e. the magnitude of each of the forces is the same.
2.
F1 and F2 are in opposite directions. In vector notation we use the negative to
indicate opposite direction. So if we use +ve for the vertically down direction then
the vertically upward direction is given a –ve value.
3.
It is also important to remember that the Reaction force between two bodies always
takes place in the direction perpendicular to the direction of the common surface
of the two bodies.

INCLINED PLANE.
Consider now a mass m (kg) placed on a plane inclined at
direction
o
to the horizontal or + x
The force acting on the object is its weight which acts vertically downward
m(kg)
F = W = m .g
i.e.
F  W  mg
To calculate a value for the force tending to pull the mass down the inclined plane we
must find the component of F which acts in that direction i.e. the component of F which
acts along ( or parallel to ) the incline.
We need therefore to resolve F into two components
F1 which is parallel to the incline plane and
F2 which is perpendicular to the incline plane.
Remember once the direction of F1 is chosen the direction of F2 must be perpendicular to
it.
Check resolving vectors into components.
F1
F2
F W

1.
To calculate F1 drop a perpendicular from F to the direction parallel to the incline
plane.
2.
To calculate F2 drop a perpendicular from F to the direction perpendicular to the
incline
direction parallel to the incline, to give the vector F1, is equal to F2. And the
perpendicular dropped to the direction perpendicular to the incline to calculate F2 is equal
to F1
between F and the direction perpendicular to the incline i
Looking at the right angled triangle made by the vectors F , F1 and the vector equal to
F2 we get


opp F1
Sin 

hyp F
Therefore
F1  FSin  mgSin
adj F2
Cos 

hyp F
Therefore
F2  FCos  mgCos
Again be Newtons third law there is also a reaction force which takes place in the
direction perpendicular to the incline i.e.
R   F2   FCos   mgCos
NOTES :
1. F1 and F2 are the components of F in two directions parallel to and perpendicular to
the incline and
F1 + F2
= F
2.
R
mgSin

F
The total force on an object in any direction is equal to the sum of the forces acting on the
object in that direction.

The total force on the mass therefore in the direction parallel to the incline is equal to
FT ( parallelto incline)  F1  mgSin
Therefore the acceleration in that direction may be calculated
acceleration down incline
F1
  gSin
m
Therefore there is an acceleration and so the object moves down the incline.

The total force in the direction perpendicular to the incline is equal to the sum of the
forces in that direction i.e. the sum of the forces F2 and R
FT ( perpendicularto incline)  F2  R
FT ( perpendicularto incline)  mgCos  mgCos
FT ( perpendicularto incline)  0
Correct since the object ( mass ) is NOT moving in the direction perpendicular to the
incline.
For a level surface  = 0
Cos = 1
Sin  = 0
So for a level surface
F1  mgSin  mgSin0  0

F2  mgCos  mgCos0  mg
R  mgCos  mgCos0  mg
As expected F 2  Weight vertically down
R = Reaction Force vertically up
and since F1 = 0 the object does not move along the level surface ALL CORRECT

Question 1. A block of mass 10 kg rests on a level surface. Calculate (i) the weight of
the block (ii) the reaction force on the block due to the level surface.
Question 2. A block of mass 5 kg is placed at rest on an inclined plane, where the angle
between the incline and the +x axis is 50o.
(a)Draw a vector diagram to represent all the forces acting on the block .
(b)
calculate the acceleration of the block down the incline
(c)
the force parallel to the incline
(d)
the reaction force on the block
(e)
how long will it take the block to slide 0.25 m down the incline.
Question 3. A mass of 10kg takes a time of 0.5 s to travel 61.31 cm down an inclined
1 (c) F2 and
(d) R
FRICTION.
Friction is defined as the resistance force which must be overcome when one surface
moves over another surface.
Friction is a force with S.I. Units Newtons and it is vector whose direction is always
opposite to movement.
Friction therefore always opposes motion.
Consider again a block of mass m ( kg) resting on a level surface.
An applied force acts on the block in the horizontal direction
R
FAPP
Ff
W
There are now four forces acting on the block
(i)
its weight acting vertically down
(ii)
the reaction force, R = - W
(iii)
the applied force, in the +x direction and
(iv)
the frictional force, in the – x direction.
It can be shown that
Ff   R
i.e. the value for the magnitude of Ff
reaction force.
NOTE:
1. Ff is the minimum applied force which will move the block and is in the opposite
direction to the applied force
2.
the Reaction force and it is in the direction perpendicular to the plane common to the
two surfaces
3.
ch is determined
by the nature of the two surfaces.
4. Ff is a vector quantity which means that we need to specify both magnitude and
direction the magnitude is given by the equation
Ff   R
and the direction always opposes motion.


On a level surface the total force, in the horizontal direction, is equal to the sum of
the forces in the horizontal direction
FTH  Fapplied  F f
FTH  Fapplied  R
FTH  Fapplied  mg
The total force in the horizontal direction =
Mass of object X acceleration in the horizontal direction
mahorizontal  Fapplied  mg
Total force in the vertical direction = sum of the forces in the vertical direction.
FTvertical  W  R  mg mg  0
Correct since there is no movement in the vertical direction.
FRICTION AND THE INCLINED PLANE
R
F1
F2
Ff W

When a mass m ( kg) is placed on the incline plane with friction there is a
frictional force on the mass opposing motion i.e up the incline
There are two forces parallel to the incline plane
F1 down the incline and Ff up the incline
The total force parallel to the incline is equal to the sum of the two forces along the
incline i.e. F1 + Ff
By Newton’s Third Law
R  mgCos  F2
F f  R  mgCos
Also we know that


F1  mgSin
There the total force parallel to the incline.
FT  F1  F f
FT  mgSin  mgCos
Therefore the acceleration down the incline
aT 
FT mgSin  mgCos

m
m
aT  gSin  gCos

Remember that to calculate the acceleration in any direction divide the total force in
that direction by the mass.
The general equation are
Fdirectionof motion  mgSin  mgCos
R(Reaction Force) = -mgCos 
Ff (Frictional Force) = -mgCos 
For a level surface with no friction
Therefore the equations become


Fdirectionof motion  0
R(Reaction Force) = -mg
Ff (Frictional Force) = -mg

Fdirectionof motion  mgSin
R(Reaction Force) = -mgCos 
Ff (Frictional Force) = 0

coefficien
Fdirectionof motion  mg
R(Reaction Force) = -mg
Ff (Frictional Force) = -mg
PROBLEM SHEET.
Question 1. A wooden block of mass
3 kg rests on a level surface where the coefficient of friction between the block and
the surface is 0.4. Calculate (i) the minimum force required to move the block
(ii) If a force of 60 N is applied to the block in the horizontal direction calculate the
acceleration on the block.
Question 2. A block of mass 6 kg is place on an inclined plane where the angle between the
incline and + x direction is 35o. Draw a vector diagram to show all the forces acting.
Calculate
(i)
the force on the block down the incline
(ii)
the acceleration if the plane is frictionless
(iii)
the acceleration is the coefficient of friction is 0.2
(iv)
If an applied force of 50 N acts on the mass down the incline repeat the above
calculations.
Question 3.(a) How long will it take a wooden block to slide from rest a distance of 2 m
down a slope inclined at 30o to the horizontal if the coefficient of friction between the block
and the surface is 0.25 and the mass of the block is 10 kg.
(b)
Calculate the time if the plane is frictionless.
(c)
Calculate the time if the coefficient of friction is 0.4.
Question 4. A ball rolls down an inclined plane a distance of 4 m in 10 seconds. Calculate
(i)
the acceleration down the incline
(ii) if the incline is at 40o to the + x direction calculate the coefficient of friction between the
incline and the ball.
Question 5. A mass of 80 kg is pulled along a level road by a rope which is inclined at 30o
to the horizontal. The force applied to the rope is 150 N.
Calculate (i) the distance travelled along the road in 1 minute assuming zero friction and
initial velocity zero.
(ii) If the coefficient of friction between the mass and the road is 0.2 calculate the distance
travelled.
Work energy and power.
Definitions:
Work is defined as the product of the force applied by the distance moved in the
direction of the force.
Energy is defined as the ability to do work
Power is defined as the rate of doing work
Work
We get the following equation from the definition
Work  Force x displacement
Work  F.s
Basic S.I. Units kgms
-2
m  Nm
Called the Joule Symbol J

One Joule of work is done when a force on one Newton is moved through a distance of
one metre measured in the direction of the force.
Work done = Energy acquired
Unit of energy and the unit of work are the same therefore the S.I. unit of energy is the
Joule
Conservation of Energy.
Conservation of energy states that energy can neither be created nor destroyed only
converted from one form to another.
The total energy in a closed system is constant.
Potential Energy.
Potential Energy of a body is the energy it possesses due to its position.
Example is the gravitational potential energy stored in a body at a height of h (m) above
the ground or zero of potential energy.
Recall that work done = energy acquired
Therefore the work done in moving the body from the ground ( taking ground as the zero
of potential energy ) to the height h (m) is the energy of the body at that point.
Work  Force.distance
Work done = m.g.h
Potential Energy
= mgh
Kinetic Energy.
The Kinetic energy of a body is the energy it possesses due to its motion.
When a force acting on a body produces movement the force does work and the body
possesses kinetic energy because of its motion.
Again work done = energy acquired
Work done = Force x distance
= ma s
If we assume that for the motion the body started at an initial velocity of zero then u = 0
v 2  2as

Therefore
v2
as 
2a
v 2 
Work done  ma  1 2 mv 2
2a 
Therefore the Kinetic energy of a body moving at a velocity v ( m.s-1) is given by


Kinetic Energy  1 mv 2
2
Note : The zero of P.E is taken as the lowest point of the system and the zero of K.E. is
taken when the body is stationary.
Total Mechanical Energy equals the sum of the potential and the kinetic energy of the
body and for a closed system
Total energy = P.E. + K.E. = Constant
Conservation of Energy example
Consider a body of mass m ( kg) which is at a height h(m) above ground and is allowed
to free-fall from the height to ground.
We will consider the body at three positions
1.
At the maximum height just before it is dropped
2.
At the half way position
3.
At the ground level just before it impacts with the ground
At position 1. The body is stationary
Total Energy  Potential Energy  Kinetic Energy
TE1  PE1  KE1
2
TE1  mgh1  1 mv1
2
Since v1 (velocity before the object is dropped)  0
h1  h
Then the equation becomes
TE1  mgh

At position 2 , the body has dropped a distance of h/2 metres, under acceleration due to
gravity, from the top and is at a height of h/2 metres above ground.
TE 2  PE 2  KE 2
2
TE 2  mgh2  1 2 mv 2
v 2  u 2  2as
General equation
At position 2
h 
h
h2   a = g u = 0 s2 =
2 
2
The equation becomes
v 2  2g
h
 gh
2
h 
h 
TE 2  mg  mg 
2 
2 
TE 2  mgh

At position 3 just before impact with the ground the body has the body has dropped a
distance of h metres, under acceleration due to gravity, from the top and is at a ground
level
2
TE 3  PE 3  KE 3  mgh3  1 2 mv3
At position 3
h3 = 0 u = 0
a = g s3 = h
v 2  u 2  2as  2gh
Equation become TE = mg(0) + 1 2 m2gh 
TE 3  mgh
Therefore the total energy of the system remains constant.

Example 1. A body of mass 5 kg is dropped from the top of a building 100 m high.
Calculate
(i)
The total energy of the body before it is dropped
(ii)
The kinetic energy at 70m from the ground
(iii)
The total energy at 70 m from the ground
(iv)
The velocity at 40m from the ground
(v)
The kinetic energy at 40m from the ground
(vi)
The total energy at 40m from the ground
(vii)
The kinetic energy just before impact with the ground
(viii)
Velocity just before impact with the ground

POWER Defined as the rate of doing work.
Work done
Power 
time taken
Since Work done  Energy aquired
Energy
Power 
time
Joule
S.I. Units
 Js1 called Watt W 
second
Efficiency.
Simple and complex machines that do work have in mechanical parts that move, so some
energy is always lost because of friction, heat, sound etc.
Thus not all of the input energy goes into doing useful work
However remember the total energy is not lost either it is in the heat light sound energy.
Total Input energy = Total output energy
Total Input = Useful output + Heat , light etc
Mechanical Efficiency is essentially a measure of what you get out of a machine that is
useful for the amount of input expressed as a fraction or a percentage.
Efficiency E = Work output x 100
Energy input
S .I. Units NONE it is a % value.

Given that
Work
Power 
time
The equation can also be written as
Power Out(Useful)
Efficiency 
x100
Power In(Rating)
Example 1. An electric pump raises 8,950kg of water from a reservoir whose storage
level is 4 m below the surface to a storage tank 31 m above ground. If the operation takes
one hour to complete find the power rating of the pump if it is 70% efficient.
Question 2. An engine used to pump water out of a shaft 50 m deep discharges it at the
surface with a speed of 20 m.s-1.
If 2 kg of water is removed every second and the engine is 80% efficient what is the
power rating.
Question 3. A 1200 kg car can accelerate from rest to a speed of 25 m.s-1 in 8 s.
Calculate the power of the engine.
Question 4. What power does a person expend on a 100 kg object being dragged down a
slope of angle 20o to the horizontal with a speed of 0.5 m.s-1 if the coefficient of friction
is 0.9
Question 5. A simple pendulum of length
1m is oscillating 30o each side of the normal. What is the velocity of the pendulum bob
at its lowest point.
Question 6. A 400 g mass hangs from the end of a string of length 50 cm. A bullet of
mass 50 g travelling horizontally hits the mass and sticks to it. The system is deflected
through an angle of 40o. Calculate the velocity of the bullet just before impact.
Question 7. A crane of power rating 65 kW can lift a 2000 kg mass to a height of 12 m in
8 s. Calculate the efficiency of the crane.
Question 8. Calculate the power rating of an engine which can accelerate a 50 kg object
from rest to 12 km.hr-1 in 5 s along a frictionless level surface
Question 9. Repeat question 8 if the coefficient of friction between the object and the
plane is 0.4
The Law of Conservation of Energy 2
Energy in a system may take on various forms (e.g. kinetic, potential, heat, light). The
law of conservation of energy states that energy may neither be created nor destroyed.
Therefore the sum of all the energies in the system is a constant.
The most commonly used example is the pendulum:
Position 1 and 3 maximum positions
Position 2 middle position and lowest point of the system
The formula to calculate the potential energy is:
PE = mgh
The mass of the ball = 10kg height, h = 0.2m acceleration due to gravity g = 9.8 m/s2
Substitute the values into the formula and you get:
PE = 19.6J (J = Joules, unit of energy)
At position 1 is the Potential Energy (PE) = 19.6J while the Kinetic Energy (KE) = 0.
As the blue ball is approaching position 2 the PE is decreasing while the KE is
increasing. At exactly halfway between the two positions PE = KE.
At position 2 the Kinetic Energy is at its maximum while the Potential Energy (PE) = 0.
At this point, theoretically, all the PE has transformed into KE.
Therefore now the KE = 19.6J while the PE = 0.
At position 3 Potential Energy (PE) is once again at its maximum and the Kinetic Energy
(KE) = 0.
Total Mechanical Energy = PE + KE
NOTE: This is with the abscence of outside forces such as friction
Using our common sense we know that it's impossible for the pendulum to swing higher
than the height h without giving it a push yourself. If there was no friction, the pendulum
would swing back and forth forever because of the law of conservation of energy.