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Average Speed
2 variables are measured :
The total journey distance , d, and
The total time for the journey ,t.
Average speed
Metres per
second , m s-1
d
v
t
Metres, m
Seconds , s
Instantaneous Speed
This is the speed at an instant. This is
measured over a very short time interval, as
close to 0 as possible. Hence electronic
timing is used as our reaction time is too
big if stopclocks were used.
d
v
t
Instantaneous Speed
Apparatus
mask
0.025 s
Light gate
Mask cuts light beam and time is recorded
timer
Use s = v x t to calculate instantaneous speed
Where s = width of mask
Vector and Scalar Quantities
Scalar quantities have magnitude only
e.g. temperature
Vector quantities have magnitude and
direction
e.g. force of 10 N to the right
Distance and Displacement
Distance is a scalar quantity but
displacement is a vector and has a direction
associated with it. E.g.
A person walks 3 km north then 4 km east.
Calculate the distance traveled and the
displacement.
Draw a vector diagram
Distance and Displacement
4 km east
north
3 km
north

Draw a line from the start to the finish. This is the
displacement. Use Pythagoras theorem to calculate
the magnitude of the displacement
c2 = a2 + b2
c2 = 32 + 42 = 25
c = 5.
Use Trig to calculate angle Ɵ
cos Ɵ= a/h cos Ɵ = 3/5 =0.6
Ɵ = cos-1 0.6 = 530
Displacement = 5 km , 0530
Distance traveled = 3 + 4 = 7 km
Speed and velocity
Speed is a scalar
Velocity is a vector
dis tan ce
speed 
time
displacement
velocity 
time
Speed and velocity
Calculate the average velocity and speed of
the walker who travels as shown in a time
of 2 hours. 4 km east
North
3 km
north

dis tan ce 7
speed 
  3.5 km h 1
time
2
Acceleration
Unbalanced forces cause vehicles to
accelerate hence it is a VECTOR quantity.
change in velocity
acceleration 
time for change
v u
a
Units m s-1
t
Units m s-2
s
Acceleration
Rearrange the acceleration formula
v  u  at
u  v  at
vu
t
a
The quantity a.t is how much the speed increases by.
Velocity time graphs
Slope = acceleration
Area under graph = displacement
speed time graph
12
speed ( m/ s )
10
8
6
4
2
1
2
3
0
0
5
10
15
20
25
30
time (s )
Displacement = area 1 + area 2 + area 3
35
Area = ½ b x h + b x h + ½ b x h
= ½ x 5 x 10
+ 25 x 10 + ½ x 2 x 10
= 25 + 250 + 10
= 285 m
( we have no indication of direction here )
Change of Direction
Constant Acceleration (Direction Change)
10
Velocity (m/s)
8
6
4
2
0
-2 0
1
2
3
-4
-6
Time (s)
4
5
6
7
• If the line of a velocity/time graph crosses
the “x” axis then the object has changed
direction of travel.
• In the graph above the velocity of the object
changes from negative to positive at 2
seconds.
• This means it “stopped” and changed
direction in an instant of time.
• This graph could represent a ball travelling
up a slope, stopping then travelling
downwards .
• The movement down the slope has been
designated as positive.
• The total displacement after six seconds is
the area under the line.
• S = (½ x 4 x 8) – (½ x 2 x 4) = 12m
Forces
A Force is a push or pull and can change
the speed / shape / and /or direction of an
object.
Newton balance is used to measure pulling
forces
Units are Newtons, N
Mass and Weight
Mass : measure of a body’s inertia, its
resistance to change. Units, kg ,Scalar
Weight : gravitational force acting on an
object, Units , N, vector
Weight = Mass x gravitational field strength
W=mxg
Mass and Weight
Gravitational field strength,g,
= force acting per unit mass ( N kg-1 )
On earth g = 9.8 Nkg-1 i.e. every 1 kg is
pulled towards the centre of the earth by a
force of 9.8 N.
The bigger the mass of a planet the bigger
g is. The smaller the radius the bigger g is.
Friction
This arises when two surfaces rub against
each other.
Can often slow objects down.
Decrease friction via lubrication , use of
bearings , making cars more
streamlined,wearing tight clothing…
Increase friction by increasing area of
tyres / brake pads / parachutes
Balanced Forces
• Same magnitude of force but act in opposite
direction
• Equivalent to NO force acting
500 N friction force
500 N engine force
Balanced Forces
• Newton’s First Law of Motion : an object
remains at rest or continues to move at a
constant velocity unless the forces acting on
it are unbalanced.
Newton’s Second Law
• If the forces acting on an object are
unbalanced it will accelerate.
200 N
friction
force
Unbalanced force, Fun
= 500 -200 = 300 N
500 N
engine
force
Newton’s Second Law
Fun  m x a
Unbalanced force
Newtons
mass x acceleration
kilograms
metres per second
per second
Example.
Calculate the acceleration of a car , 750 kg, when an engine force
of 500 N acts and a frictional force of 250 N acts against the
motion.
500 N
250 N
Fun = 500 – 250
= 250 N
Newton 2
Newton
Defined as ; 1 N is the force required to
accelerate a 1kg mass at 1 ms-2.
Free Body Diagrams
Arrows to show which way the forces act
e.g. a person who weighs 500 N in a lift is
acted on by a 650 N force upwards.
Upward
force =
650 N
Fun = 650 – 500 = 150 N
Mass, m = weight/gravitational field
strength
Mass = 500 / 9.8 = 51 kg.
Weight =
500 N
Gravitational field strength and
acceleration due to gravity
1 kg mass
Fg = m.g
= 1x9.8
Consider a 1kg mass falling vertically
with NO frictional forces acting against
it.
The force acting downward =
gravitational force
This causes the mass to accelerate :
= 9.8 N
Acceleration due to gravity and gravitational
field strength are the same value.
Free Fall
Objects accelerating downwards at the same rate at
which gravity force accelerates them downwards are in
free fall. They appear weightless.
Newton’s Third Law
The rocket exerts a force
on the gases, pushing
them down
The gases exert a force
on the rocket, pushing it
upwards
Newton’s Third Law
• To every force there is an equal and
opposite reaction force. i.e. the rocket
pushes the gases downwards and the gases
push back up on the rocket with the same
size of force. The rocket has a bigger mass
than the gases so the rocket moves up
slower than the gases move downwards.
Work
• Work done is a measure of the energy
transferred. E.g. when lifting a pencil I do
work against the earth’s gravity force,
energy has been transferred: chemical
energy in my body has been turned into
kinetic energy which is turned into potential
energy as the pencil gains height.
Work
Work done = Force x distance
W=Fxd
Joules, J
Newtons, N metres, m
Example
Calculate the work done by the brakes of a
car if a 4.5kN average force are applied
over a distance of 20 m.
Work done
W=Fxd
F = 45 kN = 4.5x104 N, d = 20 m
W = 4.5 x 104 x 20
W = 9.0 x105 J
The car originally had 9.0 x105 J of kinetic energy. To
bring the car to rest the brakes must do the exact same
amount of work, 9.0 x 105 J.
Power
•
•
•
•
•
This is the rate of doing work ( transferring energy )
i.e. work done per second
Units are Watts, W
1 W = 1 J s-1
i.e. a 2000 W heater transfers 2000 J of electrical
energy into heat energy every second
Power
Calculate the work done by a 75kW motor
running for 5 mins.
P = 75 kW = 75 000 W
t = 5 mins = 5 x 60 = 300 s
E
P
t
 E  P x t  75000 x 5 x 60  2.25 x 107 J
Potential Energy
Energy stored in an object as it is lifted of
the ground.
Change in height
Ep  m x g x h
Potential Energy
( joules, J )
mass
( metres, m )
gravitational field strength
(kilograms ,kg ) ( Newtons per kilogram, N kg-1 )
Potential energy
This is really a special case of ‘doing work’
Work done = average force x distance moved
If we lift something vertically at
uniform speed then the forces
acting on it are balanced
Fup = Fgravity = m x g
For a vertical distance we normally write h instead of d
Therefore W = f x d
W=mxgxh
Potential energy
Calculate the potential energy you gain as you climb the
stairs in the school.
m = 78 kg, h = 3.5 m, g = 9.8 N kg-1.
Ep = m x g x h
Ep = 78 x 9.8 x 3.5
Ep = 2675.4 J
Round to 2 sig figs
2.7 x 102 J
Kinetic Energy
Moving objects have kinetic energy, Ek,
E k  0 .5 x m x v
Kinetic energy
( joules, J )
mass
2
speed squared
( kilograms, kg ) ( metres per second
1 2
squared
ms


Kinetic energy
Calculate the speed a 1000 kg van is moving
at if it has 50 000 J of Ek.
Ek = 50 000 J, m = 1000 kg v = ?
E k  0 .5 x m x v 2
2 x Ek
2 x Ek
2 x 50000
v 
v
v
m
m
1000
2
v  100  v  10 m s 1
Cosmology
Signals from Space
We live on a planet called Earth, this is a big
lump of rock that orbits the sun. There are 7 other
planets orbiting the sun. They make up the solar
system along with the moons , asteroids and
dwarf planets.
Signals from Space
The sun is a star, it is our source of heat and
light energy.
Our sun is one of millions of stars that make up
the Milky Way galaxy ( a group of stars ).
The milky way galaxy is one of
millions of galaxies that make
up the universe
Electromagnetic Spectrum
This is a group of radiations ( waves ) that
travel at 3 x 108 ms-1 through air / vacuum.
They are grouped according to their
frequency / wavelength , have different
properties and are detected by different
detectors.
The different signals convey different types
of information e.g. search for ET life uses
radio waves but black holes can be
detected by searching for gamma rays.
Electromagnetic Spectrum
Remember that v  f x 
Radiation
TV and Radio
Microwaves
Infrared
Visible Light
Ultra Violet
X Rays
Gamma rays
Detector
Aerial
Aerial
Photodiode/ (hand)
Eye
Fluorescence of
chemicals
Photographic film
Geiger Muller tube
Increasing
frequency
Electromagnetic Spectrum
Calculate the frequency of 300 m radio
waves. Remember that all members of the
electromagnetic radiation travel at 3 x
108ms-1.
V = 3x108 ms-1 λ = 300 m f = ?
v  f x
f
v

8
3 x 10
6

 1 x 10 Hz
300
Light Year
This is the distance that light would travel
in one year:
Distance = speed x time
d  v x t  3.0 x10 x 365 x 24 x 60 x 60  9.46 x10 m
8
( this is equivalent to going around the
earth 250 million times )
15
moon
Light Year
Time for
Distance
light to
(m)
travel from
earth
1.2 seconds
sun
8 minutes
Object
Next nearest 4.7 years
star
Edge of
100 000
Milky Way years
Galaxy
Telescopes
Used to gather signals from distant
objects ( signals can be any member of
electromagnetic spectrum ):
Spectroscopy
White light can be split into its spectrum by a
prism. The shorter the wavelength of light the
more refraction and bending of the light.
Blue λ = 450 nm
Green λ = 550 nm
Red λ = 650 nm
Red
Green
Blue
Continuous spectrum
• All colours merge into each other , like a rainbow.
• Hot objects emit a continuous spectrum
• Temperature of star can be calculated by looking
at spectrum
• Cool objects emit red light but as the temp
increases , red, green and blue light are emitted :it
glows white
Line Spectrum
•
•
•
•
Emitted by low pressure gases
Chemical composition of stars can be evaluated
Each element has its unique spectrum
These are called emission spectrum
The Big Bang
Big Bang theory states approximately 14 billion years
ago the universe came into existence . It started as a
single point and a rapid expansion occured. Initially the
temperature was very hot and only ‘energy existed’.
As it expanded , it cooled and ‘matter’ was formed .
Initially particles called quarks and electrons were
formed then eventually protons and neutrons. The
simplest elements then followed : Hydrogen then
helium.
Big Bang : The Evidence
Other galaxies are moving away from us , this suggests
that at one time all the ‘matter’ in the universe must
have been at a single point. This time was approximately
13.7 billion years ago.
Cosmic Microwave background radiation is detected
coming from all directions : This is the remnants of the
‘Big Bang’.
Advantages of Space Exploration
Apart from allowing us to better understand ‘where
we come from’ Space Exploration has had a huge
impact on society :
Use of satellites to predict weather/ storms/GPS
Use of sensors to monitor volcanoes/ investigate the
body
Use of new materials in insulation/ replacement body
parts/ scratch resistant lenses
Improvements in computing…………
Space flight
Projectile Motion
Newton’s thought expt
ReentryA heat shield protected the two-man Gemini spacecraft against the enormous heat of reentry into the atmosphere
beginning at a velocity of more than 27,500 kilometers (17,000 miles) per hour. Like those of other early human spacecraft, Gemini's heat
shield derived from ballistic-missile warhead technology. The dish-shaped shield created a shock wave in the atmosphere that held off
most of the heat. The rest dissipated by ablation: charring and evaporation of the shield's surface. Ablative heat shields are not reusable.
The ablative substance of the Gemini heat shield is a paste-like silicone elastomer material which hardens after being poured into a
honeycomb form.
Spacecraft that enter planetary atmospheres are fitted with heat shields to protect them against the high heating loads experienced during
entry. Heat shields fitted to deal with very high speed entry are designed to ablate, that is evaporate in response to the heating loads. The
ablation products carry away heat in the form of latent heat of vaporisation. They also blanket the surface with an insulating layer of gases
and fine particulates which helps protect the surface from further convective and radiative heating. The interaction between the shock
layer flow, the high temperature gases and particles present and the radiation they emit and absorb is highly complex and poorly
understood. A barrier to understanding is the difficulty of simulating the high speed flow, the radiation field and the ablative products
cloud behaviours in the one flow in the laboratory. Experiments carried out with colleagues at the University of Queensland (UQ) have
demonstrated that this can be achieved in the UQ expansion tunnels with models whose exposed surfaces are coated with epoxy resin.
The figure shows the spatial distribution of the visible radiative power density, in arbitrary units, deduced from high speed video images
of the flow round a model of the Japanese Hyabusa spacecraft coated with epoxy resin. UV and IR spectra, integrated along the line of
sight, were also taken along the stagnation streamline in this flow.
Shielding that must be fitted to a spacecraft, such as a manned capsule or the Space Shuttle, if it is to survive the intense heat
generated during reentry. The high heating experienced by a spacecraft when entering the atmosphere is caused by a highpressure bow shock in front of the vehicle (not, as is sometimes supposed, friction with the air). This strong shock wave is caused
by the craft flying at hypersonic speeds, or high supersonic speeds. Hypersonic means greater than Mach 5. The shock wave is
where the atmosphere is rapidly compressed by a factor of 50 to 100, depending on the speed of the vehicle. Because of this rapid
compression the gas is heated to temperatures of 6,000 K or more. This hot gas then impinges on the front of the spacecraft,
transferring heat to the surface.
Ablative heat shields
One way to dissipate this large amount of thermal energy is with a heat shield that works by ablation, that is by parts of it melting
or vaporizing and breaking off in order to carry the heat harmlessly away. This technique was used by reentering Mercury, Gemini,
and Apollo spacecraft. Early manned capsules, which were spherical in shape and not orientated in any special way for reentry,
simply had an all-over ablative covering.
Projectile Motion
This has two components : a constant horizontal
velocity and a
vertical velocity that accelerates uniformly at 9.8 m s-2.
Horizontal velocity remains constant if we ignore
frictional forces and spin.
Vertical velocity changes uniformly as gravitational
force acts on object.
This results in a curved
trajectory :
Projectile motion
Example
20 ms-1
Height of cliff,
h,
Now that’s what
I call a speed
bump.
Range,s,
Calculate the horizontal distance,s, traveled
( range )and the height,h,of the cliff if the car takes 5 s
to hit the ground.
Area under graph = ½ b x h = ½ x 5 x 49 = 122.5 m
Calculate the resultant velocity of the car as it hits the
sea :
Draw a vector diagram:
20 ms-1
Ɵ
49 ms-1
Use Pythagoras to work out hypotenuse
52.9 ms-1
Use trig to work out angle , Ɵ
67.80
Resultant velocity is 52.9 ms-1 , 67.80 below the
horizontal.
Newton’s Thought Expt
The ball is fired
horizontally but
gravitational force
accelerates it towards the
earth. It crashes at point A
If the horizontal velocity is
increased it can ‘reach ‘ a
little further around the
earth to B.
If the horizontal velocity is
increased further it can
travel right round the
Newton’s Thought Expt 3
If the horizontal velocity of the ball is increased
further it flies off into space.
The ball orbits the earth because gravitational force
is pulling it towards the centre. The ball wants to
travel in a straight line but gravitational force pulls it
inwards.
This is why satellites , natural and man made orbit
planets.
Re entry
When the space shuttle re enters the earth’s atmosphere
there are huge frictional forces acting against it.
Some of the shuttle’s kinetic energy is turned into heat
energy.
To stop the craft becoming too hot inside, the underside is
painted black, this is a good emitter of infra red radiation.
The underside is also covered with tiles that have a low
specific heat capacity and a low thermal conductivity. This
ensures that the temperature of the tiles rises quickly( but
lots of heat is radiated to the surroundings ) and a small
amount of heat energy is transferred into the cabin.
Example energy change
The space shuttle, 100 tonnes, slows down from 7 500 ms-1 to
750 m s-1 when it hits the earth’s atmosphere. Assume that all
this change in kinetic energy is turned into heat energy Calculate
the maximum temperature rise if the specific heat capacity of the
thermal tiles, 2000kg, is 50 J kg-10C-1.
Example
Use Ek = 0.5mv2 to calculate kinetic energy
change.
Ek before = 2.8125 x 1012
Ek after = 2.8125 x 1010
Ek change = 2.78 x1012 J
Use Eh = Ek = cm  T
Eh
T 
c xm

2.78 x 1012
50 x 2000
 2.78 x 107 0C
Obviously a lot of the heat energy is re radiated otherwise the
shuttle would melt.
Ablative Heat Shields
When a material turns from a solid to a liquid
or from a liquid to a gas energy is required.
When water evaporates off your skin you cool
down. This idea is used to cool down some
space craft on re entry.
Part of the heat shield is designed to burn away,
the gases produced carry some heat energy
away