M. MOTION
M.1 VECTORS AND SCALARS
M.1.1 Scalars
A
scalar is a quantity which has only a size.eg.temperature,
time, mass
M.1.2 Vectors
A
vector is a quantity which has both a size and a direction.
eg.
displacement, velocity, acceleration, force
34
m north, 3 ms-1 west, 6ms-2 south, 12 N north 10°
west
To
distinguish the scalar F from the vector F we write vectors in one of two
ways.F and 
are
the vector F.
We
represent a vector by an arrow, as such
where the size or magnitude of the vector is represented by the length
of the arrow and the direction is the direction in which the arrow points.
M.1.2.1 Addition of Vectors
If
we have two vectors a and b, to add them we put the vectors head to tail
and the result of the addition is the vector drawn from the starting point
to the finishing point.
M.1.2.2 Subtraction of Vectors
Vector
subtraction is easiest to do if you think of it in terms of addition.
Considera
- b which can be expressed as a + (-b).-b
means the same magnitude, but opposite direction.
Example.Finda
-b
M.1.2.3 Vector Resolutes
Just
as two vectors can be added to give another.A
vector can be split into two or more other vectors.These
are called components or resolutes.
Example
Problem
Set #1:TextPage
303Questions 1 – 8, 10
M.2 KINEMATICS
Kinematics
is the study of the motion of objects.This
involves a study of position, displacement, velocity, acceleration and
time.
M.2.1 Position
The
position of an object tells us where the object is situated.The
symbol is x and the unit is metres (m).
M.2.2 Displacement
The
displacement of an object is the change in its position and the direction
of that change.
Example
Note:The
displacement is independent of how you got there, the distance travelled
or the path taken.
Example
The
mathematical way of writing "change of" is to use the symbol D
(delta).
Thus
change of position = Dx
Since
displacement has a magnitude and a direction it is a vector and is written
as 
or 
.The
unit is metres.
M.2.3 Velocity
Velocity
is a quantity which tells us how fast an object is travelling and also
the direction of travel.Thus velocity
is a vector and is denoted by 
.
The
magnitude part of velocity is called speed and is denoted by s or v.
The
velocity of an object is calculated by using the displacement and the time
taken.
Thus 
This
equation tells us the average velocity over the time Dt.
The
units of velocity are metres per second, written as ms-1
M.2.4 Acceleration
Acceleration
is a quantity which tells us about the change in velocity of an object
and is a vector.Acceleration is
defined as the change in velocity over time.
Thus 
The
units of acceleration are ms-2.
Problem
Set #2:TextPage
304Questions 11 – 35
M.2.5 Graphs of x, v, a and t
These
were dealt with in detail in unit 2.The
following is just a summary.
The
gradient of an x-t graph is the velocity.
The
gradient of an v-t graph is the acceleration.
The
area under an a-t graph is equal to (Dv)
the change in velocity (speed).To
be able to find the actual velocity at any time, we need to know the velocity
at one point in time.Usually the
initial or final.
The
area under an v-t graph is equal to (Dx)
the change in position.To be able
to find the actual position at any time we need to know the position at
one point in time.Usually the initial
or final.
M.2.6 Constant Acceleration Formulae
We
know the shape of a v-t graph when the acceleration is constant.It
is a straight line as shown.
Definitions
a
= acceleration
s
= displacement
v
= final velocity
u
= initial velocity
Dt
= Change in time = t2 -t1
Again
these were dealt with in detail in unit 2.
v
= u + a DtFormula
1
Formula
2
Formula
3
v2
= u2 + 2asFormula 4
These
four formulae are known as the constant acceleration formulae.
Example.
A
bobsled is sliding down a hill with an acceleration of 3 ms-2.It
starts from rest, if we ignore air resistance, find:
a) the
speed after 5 seconds
b)how
far it has travelled in 5 seconds
c)the
time taken to reach 30 ms-1
d)the
distance travelled when it reaches 45 ms-1
Problem
Set #3:TextPage
303Questions 9, 36 – 66
M.3 DYNAMICS
Dynamics
is the study of the causes of motion, in particular the forces acting on
a body.
M.3.1 States of Motion
There
are three states of motion, they are:
1)object
at rest
2)object
travelling at constant velocity
3)object
undergoing acceleration
M.3.2 Galileo's Law of Inertia (Newton's 1st Law)
An
object will remain at rest or in uniform motion in a straight line unless
acted upon by a force.
Note1)this
applies in any particular direction.
2)if
F = 0, v = constant (may be non zero)
M.3.3 Newton's Second Law
When
a constant force acts on a body we notice that the body undergoes constant
acceleration.If we change the size
of the constant force the size of the acceleration changes, such that F a
a.
Applying
a given force to different objects results in differing accelerations.The
property of the object that causes this variation is the inertial mass
(mi) and we have
F
= mi a
which
strictly speaking should be written as
SF
= m a
Force
is a vector and behaves as any vector would.
UnitsForce
has the unit of Newton (N)
mass
has the unit of Kilogram (Kg)
acceleration
has the unit of ms-2
1
Newton »
weight of an apple
Note1)Direction
of a is direction of SF
2)If
a = 0 then SF
= 0
3)If
a = 0 in any direction,
then SF
in that direction = 0
Example
1A force of 6 N accelerates a mass
of 3 Kg.
What
acceleration results?
Example
2A body is acted upon by two forces.3
N north and 4 N east.If its mass
is
6 Kg.What is the acceleration?
M.3.4 Mass and Weight
Mass
is a measure of how much material is contained in an object and is measured
in Kg.
Weight
is a measure of the pull of gravity on an object, it is a force and is
measured in N.
Mass
and weight are not the same thing, however they are related.An
object in a gravitational field of strength g feels a force, called weight
w, that depends on a property of an object called gravitational mass (mg)
and w = mg g.mg
is found to be related to mi by mg = mi.
M.3.5 Types of Forces
1.Weight
The
force by which a mass is attracted to the Earth.It
is proportional to the bodies gravitational mass (mg), the constant
of proportionality being the gravitational field strength (g).(g »9.8
N Kg-1)
W
= m g
2.Reaction
When
a body sits on a surface there is a force acting upwards to prevent it
falling downwards.This force is
called the reaction force.
Example
3.Friction
A
force opposing the motion of a body.eg.
table-floor,car-air
Friction
is actually proportional to the normal reaction force.The
constant of proportionality being the co-efficient of friction (m),
the value of which depends on the shape, surface area, area of contact,
etc.
Friction
=m
N
M.3.6 Newton's Third Law
The
reaction force leads us to Newton's third law, which states:
If
one object exerts a force on another then there is an equal and opposite
force (reaction) on the first object by the second.i.e.
for every action there is an equal and opposite reaction.
Example 1
Example
2
Handout:Worked
Examples
Problem
set #4:TextPage
338Questions 1 – 4, 18 – 23
M.4 Impulse and Momentum
M.4.1 Momentum
If
an object is moving with some speed, then to change its speed or direction
a force must be applied, even then the object will be reluctant to change.This
reluctance of a body to change its velocity is known as momentum.
The
momentum of a body is defined by:
Momentum
= mass of body ´
velocity of body
Momentum
is a vector and should be treated as any vector would.The
units of momentum are Kg ms-1.
Example:
At
what speed must a 60 Kg athlete be running if his or her momentum is to
equal that of a 1000 Kg car travelling at a constant speed of 5 Km/hr.?
Speed
of the car= 5 Km hr-1.
=
=
1.39 ms-1
Momentum
of car
p
= m v
=
1000 ´
1.39
=
1.39 ´
103 Kg ms-1
Athlete
p
= 1.39 ´
103 Kg ms-1m
= 60 Kg
p
= m v
1.39 ´
103 = 60 ´
v
v
= 1.39 ´
103
60
=
23.15 ms-1
Note:In
a closed system (of two or more objects) the momentum of the system remains
the same before and after the collision.
Example
A
railway truck, of mass 80 g, on a model train track is moving with a speed
of 15 cms-1 and collides with a stationary truck of mass 90
g.The two trucks become coupled
together.What is their common speed?
M.4.2 Impulse
To
change a bodies momentum a force must be applied.Usually
this force will be in the form of a collision, of fairly short duration
and its size may vary.This type
of force is called an impulsive force.The
average impulsive force is defined as the rate of change of momentum.
i.e.
or
Impulse
= D
Momentum
I
= DP
orI
= FDtUnit,
Newton second (N S)
Example:
When
served, a tennis ball leaves the racquet with a speed of 45 ms-1.The
impact time between the racquet and the ball is 5 x 10-3 s.The
mass of the ball is 0.06 Kg.Find
the average impulsive force which acts on the ball.The
speed of the ball before being hit is zero.
vf
= 45vi = 0m
= 0.06t = 5 ´
10-3
pf
= 0.06 ´
45 = 2.7
pi
= 0.06 ´
0 = 0
Dp
= pf - pi = 2.7 - 0 = 2.7
Fave
= Dp
Dt
= 
=
540 N
Problem
Set #5:TextPage
338Questions 5, 8 – 15, 24, 26,
27, 29
M.5 Work and Energy
M.5.1 Work
From
the ideas we have about work we can say that work requires some effort.E.g.
pushing a wheel barrow.Thus some
force is being applied over some distance.
From
common sense we get
Work
= Force ´
Displacement
orW
= F ´
d
The
units for work are the Joule (J)
Example:
Find
the work done against a frictional force of 7 N.If
the fridge is kept at a constant velocity of 2 ms-1 for a distance
of 10 m.
M.5.2 Work From Graphs
Suppose
we have a force-distance graph that looks like this
How
do we calculate the work done?
Firstly
we replace the curve by a series of steps of constant force.
Then
what we have is a series of small rectangles, with an area off ´
d.
So
if we add up all the rectangles.
We get, 
orsimplyS
(F ´
d)which is the area under the graph.
So Work
done = area under F-d graph
This
follows for all F-d graphs even those where F is not constant.
Examples:Find
the work done in the following cases
1.
2.
M.6.3 The Scalar Nature of Work
Work
is a scalar but force and displacement are vectors.What
happens if the force applied and the displacement are not in the same direction?We
then need to take the component of the force in the direction of the displacement.
SoWork
= distance ´
component of force in direction of displacement
Work
= d ´
F| |=F
d cosq
Example:
Find
the work done by the force if the distance moved is 6 m.
M.5.4 Energy
When
we do work, such as pushing a wheel barrow, we get tired or use up the
quantity known as energy.Energy
does not disappear, but is either
1)transferred
to another object
or2)transformed
into another kind.
Thus
we formulate the principle of conservation of energy which states that
Energy
is neither created or destroyed
we
say that
Work
done by an object = loss of energy by that object
andWork
done on an object = gain in energy by that object
orW
= DE
The
units for energy are the same as the units for work, the Joule (J).
Example:
If
a man does 200 J of work pushing a wheel barrow, he loses 200 J of energy
and the wheel barrow gains 200 J of energy.
M.5.5 Kinetic Energy
We
define kinetic energy as the energy a body has when it is in motion.We
can derive an expression for kinetic energy.
Consider
an object of mass, m, originally at rest being acted upon by a force of
F N for a distance of d m.No friction.
We
have
Work
done = energy gain = final K.E. (in this case)
\Final
K.E. = F ´
d
=
ma ´
d(eqn 1)
Evaluating
the accn using constant accn formula
v2
= u2 + 2as
vi
= 0u = Vs
= da = a
v2
= 0 + 2ad
V2
= 2ad
Þa
=v2
2d
Now
plug back into eqn 1
Final
K.E.= mv2´d
2d
=
½m v2
soKinetic
Energy (Ek) = ½mv2
In
fact work done = change in kinetic energy = Final K.E. – Initial K.E
DK.E.
= ½ mv2 - ½mu2
Examples:
1.A
body of mass 6 Kg has a speed of 3 ms-1.What
is its K.E.?
2.A
body of mass 4 Kg with a speed of 3 ms-1 accelerates to a speed
of 6 ms-1.What is
a)
the change in K.E.
b)
the work done on the body
M.6 Collisions
There
are two types of collision, they are elastic and inelastic.
For
Elastic collisions, kinetic energy is conserved.This
means that the amount of kinetic energy after the collision is the same
as the amount of kinetic energy before the collision.Momentum
is also conserved.
With
inelastic collisions, the total kinetic energy is not conserved.This
means that the amount of kinetic energy after the collision is less than
the amount of kinetic energy before the collision.Some
of the kinetic energy has been lost , and this has been transferred too
other forms.However momentum is
still conserved.
M.7 Energy Stored in Springs
Springs
can store energy when they are stretched or compressed.We
can store the energy in the spring by applying a force to alter its length,
thus we are doing work on the spring.
we
have
Energy
stored = potential energy of spring = work done on spring
In
about 1675 Robert Hooke noticed that the more you stretch a spring from
it's natural length, the stronger the force needed.
i.e.F µDx
we
write F
= k x Hooke's law
wherek
= spring constant (unit N m-1)
we
get a graph which looks like
Now
P.E. of spring = Work done on it
We
can calculate the work done on a spring in stretching it x metre from the
force-distance graph.
Work
done = area under graph(can't
use w = f ´
d because force)(not
constant)
=
½ F x
But
F = k x
So
work done = ½ k x x
=
½ k x2
\P.E.
of spring = ½ k x2(Joule)
Note:-
For a spring compressed and then released D
P.E. (spring) =D
K.E. (body).Conservation of energy.Etotal
= K.E. + P.E.
Example 1. Find the P.E. of the compressed spring
Example
2.
For
a spring with k = 5 N m-1.Find
a) D
P.E. when compressed from 0à20
cm
b) D
P.E. when compressed from 20à40
cm
c)If
compressed to 20 cm and a body is placed there and let go.What
is the K.E. as it passes zero compression?
Problem
Set #6:TextPage
338Questions 6, 7, 16, 17, 25, 28,
30 – 55
TextPage
383Questions 21 – 34, 61, 62, 78,
79, 85 – 87
M.8 Projectile Motion
A
projectile is an object that is thrown through the air.If
we ignore air resistance, the only force acting on the body is the gravitational
force.We can therefore break the
motion down into two parts, horizontal and vertical.In
the horizontal direction the acceleration equals zero, in the vertical
direction the acceleration is g.
Consider
an object projected horizontally as shown:
In
the horizontal direction we have:
u
= u
a
= 0
s
= X
froms
= ut + ½ at2
X
= ut(1)
In
the vertical direction we have:
u
= 0
a
= -g
s
= y
froms
= ut + ½ at2
y
= -½gt2(2)
combining
equations (1) and (2) we get:
y
= -gx2
2u2
This
is the equation of a parabola.
\the
path of a projectile is in the shape of a parabola.
Now
consider a projectile projected into the air at an angle Ø with
the
ground as shown:
Again
the motion can be split into two components, horizontal and vertical.
In
the horizontal direction we have:
u
= u cosø
v
= u cosø
a
= 0
s
= X
t
= t
froms
= ut + ½ at2
X
= ut cosø(1)
In
the vertical direction we have:
u
= u sinø
v
= changing
a
= -g
s
= y
t
= t
froms
= ut + ½ at2
y
= ut sinø - ½ gt2(2)
combining
equations (1) and (2) we get:
y
= x tanø- gx2.
2u2
cos2ø
Which
is again a parabola.
Example.
An
object is projected horizontally with a velocity of 5.0 ms-1
a)When
will its vertical displacement be 20 m?
b)What
will be its velocity at this time?
a)Vertically,
taking down as negative and the initial height as 0 m.
u
= 0
s
= -20
a
= -10
s
= ut + ½ at2
-20
= 0 + ½ ´
-10 ´
t2
t2
= 4
t
= 2 sec
b)The
vector diagram is as follows:
HorizontallyVh
= 5.0 ms-1
VerticallyVv
= u + at
=
0 - 10 ´
2
=
20 ms-1
from
the diagram
V2
= Vh2 + Vv2
V2
= 5.02 + 202
V2
= 425
V
= 20.6 ms-1 or 21 ms-1
angleØ
= Tan-1 4
=
76°
M.8.1 The Effect of Air Resistance
Air
resistance will have an effect on the motion of objects travelling through
the air.Each time a object hits
a molecule in the air some momentum is transferred to it.The
projectile motion will be changed from being a parabola, very slightly,
but will still be a curve.
Problem
Set #7:
TextPage 380Questions
1 – 11, 35 – 38, 51 –54, 63 – 69,
76, 77
M.9 Circular Motion
M.9.1 Uniform Circular Motion
Prac
#1:EXPT
4.1Circular Motion
Let
us consider an object moving in a circle with a constant speed.
The
speed of the object will be given by:
Speed
=distance
time
=circumference
period
v=2pr
T
The
period (T) is the time for one revolution.
Since
the direction of travel is continually changing the velocity of this object
will also be constantly changing.Because
the velocity is changing there must be an acceleration.
To
find the acceleration we must first find the change in velocity.
DV=
V2 - V1
acceleration=
If
we take a very small time interval then the distance travelled will approximate
a straight line and will be given by vDt.We
will have two triangles as shown below:
and
from the similar triangles we get:
DV=VDt
Vr
DV=V2
Dtr
Hence
the magnitude of the acceleration is
a=v2
r
using
v = 2pr
T
we
get:
a
= 
=
using 
we
geta = 4 p2rf2
The
direction of the acceleration is towards the center of the circle.
The
net force will also be towards the centre of the circle, it is known as
the centripetal force and is given by:
F=ma=m
4p2r
T2
F=mv2
r
M.9.2 Non-Uniform Circular Motion
There
are many cases where the speed is not uniform.The
most common is an object moving in a vertical circle.The
speed will be greater at bottom the of the circle than the top. The
force acting on the object at any time will be given by:
SF=Fc+mg
To
calculate the speed at any particular place on the circle conservation
of energy is used.
Problem
Set #8:TextPageQuestions
12 – 20, 39 – 50, 55 – 60, 70 – 75, 81 – 84