Higher Physics
Unit 1 – Our Dynamic
Universe
Printed Notes
Section 1 – Motion
Vectors and Scalars
A scalar quantity is
completely defined
by stating its
magnitude.
A vector quantity is
completely defined
by stating its
magnitude and
direction.
Vectors
Scalars
Displacement Distance
Velocity
Speed
Acceleration
Time
Force
Mass
Momentum
Energy
Distance and Displacement
Distance is the total path length. It is
described by magnitude (size) alone.
Displacement is the direct length from
a starting point to a finishing point.
To describe displacement both
magnitude and direction must be given.
Speed and Velocity
Average Speed = distance
time
Speed has magnitude but no direction.
Average Velocity = displacement
time
Velocity has magnitude AND direction.
Direction of velocity is same as
displacement.
Units for both is ms-1 (metres per
second).
Example
A runner sprints 100 m along a straight track
in 12 s and then takes a further 13 s to jog
20 m back towards her starting point.
a) What distance does she run during the 25 s?
b) What is her displacement from her starting
point after the 25 s?
c) What is her average speed during the 25 s?
d) What is her average velocity during the 25 s?
Solution
a) Distance = 100 m + 20 m = 120 m
b) Displacement = 100 m – 20 m = 80 m in original
direction
c) Av. Speed = 120 / 25 = 4.8 ms-1
d) Av. Velocity = 80 / 25 = 3.2 ms-1 in original
direction
Answers to (b) and (d) have direction as well as
magnitude.
Addition of Vectors
Two or more vectors can be added
(combined) to get a resultant vector.
Magnitude and direction must be taken
into account.
We use a scale diagram for this, add the
vectors ‘nose to tail’, then join the start
and finish point.
Vector:
tail
nose
Resultant Vector:
Always drawn with two arrows to
represent a resultant.
Example – Scale Diagram
A girl walks 30 m North, then 40 m East.
What is her resultant displacement?
To solve:
1) Choose a suitable scale, e.g. 1 cm : 10 m.
2) Using a ruler and protractor, arrange
arrows ‘nose to tail’.
3) Draw in resultant vector, measuring its
length and direction.
Solution
N
4 cm (40 m)
W
3 cm
(30 m)
s
x
E
S
Her resultant displacement, s, is 50 m
at an angle x of 53º East of North.
Example 2
A plane flies North at 120 ms-1 in a wind
blowing East at 50 ms-1. What is the
plane’s resultant velocity?
Hint: Use a scale of 1cm : 10 ms-1
Solution
N
5 cm (50 ms-1)
W
12 cm
(120 ms-1)
E
S
v
y
The plane’s resultant velocity, v, is 130 ms1 at an angle y of 23º East of North.
Resultant of a Number of
FORCES
This is the single force which has the
same effect, in both magnitude and
direction, as the sum of the individual
forces.
We can also use scale diagrams to find
the magnitude and direction of the
resultant of a number of forces.
Resolution (Rectangular
Components) of a Vector
Any vector, x, can be resolved into two
components at right angles to each other.
The horizontal component xh
The vertical component xv
x
xv
is equivalent to
θ
xh
sin θ = xv / x
xv = x sin θ
x
xv
θ
xh
cos θ = xh / x
xh = x cos θ
Velocity
The vertical and horizontal components
of a velocity vector, v, are, respectively:
vv = v sin
θ
vh = v cos θ
Force
The vertical and horizontal components
of a Force vector, F, are, respectively:
Fv = F sin
θ
Fh = F cos θ
Acceleration
Acceleration = change in velocity
time taken
= final velocity – initial velocity
time taken
a = v–u
t
a: acceleration
v: final velocity (ms-1)
u: initial velocity (ms-1
t : time taken (s)
What does this mean?
ACCELERATION IS THE CHANGE IN
VELOCITY PER UNIT TIME.
An acceleration of 2 ms-2 means the
velocity of the body changes by 2 ms-1
every second.
Units are metres per second per second
or ms-2.
Acceleration is a VECTOR.
Experiment to Measure Acceleration
Light gates
Stopclock
Points to Note In Method
• t1 = time for card to pass light gate 1
• t2 = time for card to pass light gate 2
• t3 = time for card to go between light gates
• Card length = s
• u = s / t1
v = s / t2
a = (v - u) / t3
Graphs of Motion
Note that in the following graphs,
a = acceleration
v = velocity
s = displacement
Just as the area under a speed-time
graph gives the distance travelled,
The area under a velocity-time graph
gives the displacement.
Graphs Showing Constant
Acceleration
a
0
v
t
0
s
t
0
t
Graphs Showing Constant Velocity
a
0
v
t
0
s
t
0
t
Graphs Showing Constant
Deceleration
a
v
0
s
t
0
t
0
t
The Equations of Motion
1)v = u + at
2)
s = ut + ½at²
3)
v² = u² + 2as
u – initial velocity at time t = 0
v – final velocity at time t
a – acceleration of object
t – time to accelerate from u to
v
s – displacement of object in
time t
Example 1
Q A space rocket travelling at 20 ms-1
accelerates at 5 ms-2 for 2 s. How far does
the rocket travel during the 2 s?
A
u = 20 ms-1; a = 5 ms-2; t = 2 s; s = ?
Use
s = ut + ½at²
s = 40 + 10
s = 50 m
Example 2
Q A train travelling at 45 ms-1 decelerates
to
15 ms-1 at 2 ms-2. How far does the
train travel while it is decelerating?
A u = 45 ms-1; v = 15 ms-1; a = -2 ms-2; s = ?
(note –ve sign)
Use
v² = u² + 2as
15² = 45² - 4s
4s = 45² - 15²
s = 450 m
Example 3
Q A ball is thrown vertically into the air
with an initial velocity of 20 ms-1. What will
its velocity be after 3 s?
A u = 20 ms-1; t = 3 s; a = -10ms-2; v = ?
Use
v = u + at
v = 20 – 30
v = -10 ms-1
(i.e. the ball is on the way down)
Derivations
The three equations of motion can be
derived as described below.
1) v = u + at
By definition, acceleration is given by:
a = v–u
t
v – u = at
v = u + at
2) s = ut + ½at²
s = displacement = area under a v/t graph.
Velocity
v
v-u
u
u
0
time
+
s=
= ut + ½(v – u)t
= ut + ½(at)t
t
i.e.
s = ut + ½at²
3) v² = u² + 2as
v = u + at
(eqn 1)
Squaring both sides, we get:
v² = (u + at)²
= u² + 2uat + a²t²
= u ² + 2a(ut + ½at²)
i.e.
v² = u² + 2as
These equations only apply to uniform
acceleration in a straight line.
The vector quantities displacement,
velocity and acceleration have
direction associated with them, and so
they will have a positive or negative
sign depending on their direction.
Ex 4 A ball is launched vertically into the air with
an initial velocity of 35 ms-1. What is the
maximum height it will reach?
Ex 5 A rock is thrown upwards with an initial
velocity of 30 ms-1. After how long will it be 10
metres off the ground?
Method for Tackling Problems
1. Write down all the symbols like this:
u=
v=
s=
a= t=
2. Fill in all numbers & values given in question.
3. If there are two directions, use
this diagram for + and – values:
Left -
Up +
Right +
4. Choose the best equation to suit the
problem.
Down -
Section 2 – Forces, Energy
and Power
Newton’s First Law
a) If there are NO forces acting on a
moving object, it will continue to move
at a constant velocity.
b) If BALANCED FORCES act on a
moving object, it will continue to move
at a constant velocity.
Newton’s Second Law
Motion can exist without forces, but for
change of motion to occur, forces must
be involved.
i.e. Unbalanced forces cause
acceleration.
Fun = ma
Fun = Unbalanced Force in N
m = mass in kg
a = acceleration in ms-2
Definition of the Newton:
One Newton is the (unbalanced) force that
gives a 1 kg mass an acceleration of 1 ms-2.
Force is a vector.
An accelerating object has an
unbalanced (resultant) force, F, acting
on it in the same direction as the
acceleration.
If an object is not accelerating then
the unbalanced force F = 0.
Example – Single Force,
Single Mass
A man pushes a
trolley of mass 20
kg along a flat
surface of 40 N.
If the effects of
friction can be
ignored, what is the
acceleration of the
shopping trolley?
a
40 N
20 kg
a =F/m
= 40 / 20
= 2 ms-2
Example 1 – Multiple Force,
Single Mass
A rocket of mass
1000 kg is fired
vertically into the
air.
The rocket motors
provide a thrust of
20 000 N, and
there is a drag
force of 2000 N.
What is the
acceleration of the
missile?
Thrust = 20000 N
Fun = 20000
- 9800
- 2000
= 8200 N
Drag
= 2000 N
a =F/m
= 8200 /
1000
= 8.2 ms-2
N
A
S
A
a
W = mg
= 1000 x 9.8
= 9800 N
Example 2 – Single (External)
Force, Multiple Mass
A ski-tow pulls two
skiers, who are
connected by a thin
nylon rope, along a
frictionless surface.
The tow uses a force
of 70 N and the skiers
have masses of 60 kg
and 80 kg.
a) What is the
acceleration of
the
system?
b) What is the
tension
in the rope?
80 kg
a)
T
60 kg
70 N
a =F/m
= 70 / (80 + 60)
= 70 / 140
= 0.5 ms-2
b) Tension T is a Force.
It is caused by 80 kg
person.
T = ma
= 80 x 0.5
Solving Force Problems
1) Draw a free body diagram showing the
forces acting on the object and the
direction of its acceleration.
2) Deal with one object at a time.
However, when objects are tied
together, they behave like a single
larger object.
3) Use F = ma, but remember F is the
unbalanced force!!
Lift Problems
In all Lift problems, there will always be a
tension in the cable supporting it, and a
weight (W = mg).
Tension
or
Lift
of
mass ‘m’
Fun = ma
Weight = mg
There could also be an unbalanced force, Fun.
There are 6 possible lift situations:
1) Lift at rest
Tension = Weight (no acceleration, no Fun)
2) Lift travelling at constant speed
Tension = Weight (no acceleration, no Fun)
3) Lift accelerating up
Tension = Weight + Fun
4) Lift decelerating up
Tension = Weight – Fun
5) Lift accelerating down
Tension = Weight – Fun
6) Lift decelerating down
Tension = Weight + Fun
Example
A lift of mass 4 kg is accelerating upwards at 3 ms2.
What is the tension in the cable?
a = 3 ms-2
T
4 kg
W = mg
Solution
T = tension in the cable
Fun = ma
T – mg = ma
T = mg + ma
T = (4 x 9.8) + (4 x 3)
T = 51.2 N
Terminal Velocity
Frictional Forces in a Fluid
A fluid is a liquid or a gas.
The motion of a body falling through
any fluid can be divided into three
parts:
1 Initially a body falls with constant
acceleration of 10ms-2 (acceleration
due to gravity) due to its weight.
Friction
2 After a short time the frictional
forces which are building up causes the
acceleration to decrease.
3 Finally the frictional force balances
the weight and the body reaches its
greatest speed. This is the terminal
velocity.
Weight
Free-Falling Parachutist
OA – constant
acceleration due to
gravity
AB – decreasing
acceleration as frictional
force acts
BC – terminal velocity
CD – non-uniform
deceleration (parachute
opens)
DE – constant speed
EF – body hits ground
Resolution (Rectangular
Components) of a Vector
Any vector, x, can be resolved into two
components at right angles to each
other.
The horizontal component xh
The vertical component xv
x
xv
is equivalent to
θ
xh
sin θ = xv / x
xv = x sin θ
x
xv
θ
xh
cos θ = xh / x
xh = x cos θ
Velocity
The vertical and horizontal components
of a velocity vector, v, are, respectively:
vv = v sin
θ
vh = v cos θ
Components of Forces
In the previous section, a vector was split
into horizontal and vertical components.
This can obviously apply to a force.
F
Fv = F sin θ
is equivalent to
θ
Fh = F cos θ
Remember that the resultant of a number of
forces is the single force which has the same
effect, in both magnitude and direction, as the
vector sum of the individual forces.
Example
A man pulls a garden roller of mass 100 kg with a force of
200 N acting at 30º to the horizontal.
If there is a frictional force of 100 N between the roller
and the ground, what is the acceleration of the roller along
the ground?
Solution
200 N
30º
Friction
= 100 N
Fh
Fh = F cos θ = 200 cos 30º = 173.2 N
Fun = 173.2 N – Friction
= 173.2 – 100
= 73.2 N
a = Fun / m = 73.2 / 100 = 0.732 ms-2
Force Acting Down a Plane
If an object is placed on a sloping surface
(plane), its weight acts vertically
DOWNWARDS.
A certain component of this will act down
the slope.
The weight can be split into two components
at right angles to each other.
We can say that the component of weight
acting down the slope is: F = mg sin θ
Reaction Force
mg sin θ
θ
θ
mg cos θ
mg
Component of weight down slope
= mg sin θ
Component of weight
perpendicular to slope
= mg cos θ
Example
A block of wood of mass 2 kg is placed on a slope
which makes an angle of 30º with the horizontal.
A frictional force of 4 N acts on the block.
a) What is the unbalanced force down the slope?
b) What is the acceleration of the block of wood?
Conservation of Energy
The total energy of any closed system is
conserved, although energy may change its
form.
E.g. energy ‘lost’ as it changes to heat, light,
sound, etc.
Ep = mgh
Ek = ½mv²
Ew = Fd
Work Done = Force x Displacement
Energy and Work are measured in Joules (J).
Power
Power is the rate of transformation of
energy from one form to another.
P = energy / time
Power is measured in Watts (W) or Joules
per second (Js-1).
Momentum
Momentum is a vector quantity and is the
product of mass and velocity.
Momentum = mass x velocity
kg
ms-1
The units of momentum are kgms-1.
Momentum is sometimes given the symbol p.
p = mv
The direction of the momentum is
the same as the velocity.
Experiment – Law of Conservation
of Momentum
Aim
To compare the total momentum before and after
a collision
Method
The apparatus was set up as shown below:
Microcomputer
Light
Gates
Vehicle 2
with card
Vehicle 1
with card
Linear air track
The mass in kg of the two vehicles will be
measured with a digital balance.
The computer will be set to measure the
vehicles velocity (note direction!!!) before
and after a collision.
The second vehicle will initially be at rest.
This will allow the momentum before and
after the collision to be calculated.
Results
m1
v1
m2
v2
pbefore m1+m2
Units of m = kg
Units of v = ms-1
Units of p = mv = kgms-1
v3
pafter
Conclusion
Momentum is always conserved in collisions.
However, this will only be the case if the
direction of momentum is taken into
account.
The Law of Conservation of
Momentum
In any collision or explosion free of external
forces, the total momentum remains the same.
This can be applied to the interaction of two
objects moving in one dimension, in the absence of
net external forces.
For any collision:
Total momentum of
all objects before
=
Total momentum of
all objects after
Conservation of Momentum and
Kinetic Energy
In general, there are three types of
problem:
1. Two masses collide and move apart with
different velocities after the collision:
Before
v1
m1
After
v2
v1’
m2
m1
m1v1 + m2v2 = m1v1’ + m2v2’
v2’
m2
2. Two masses collide and stick together:
Before
v1
m1
After
v2
v3
m2
m1 + m2
m1v1 + m2v2 = (m1 + m2)v3
3. An explosion. In this case:
Before
v
After
v1
m
m1
v2
m2
(m1+ m2) v = m1v1 + m2v2
If initially at rest (e.g. gun before firing a
bullet), then:
0 = m1v1 + m2v2
Solving Problems
1. Always make a sketch of the system
before and after the collision or
explosion.
2. Mark all masses and velocities (with
direction!!) on the sketch.
3. You will need to allocate a positive
direction for vector quantities – mark
this also on the sketch.
4. Use the rule: total momentum before =
total momentum after
Example
Solution
5 ms-1
3 kg
5 ms-1
2 kg
Find the velocity
of the trolleys
when they stick
together after
colliding.
+ ve
Tot mom Before = Tot mom After
m1v1 + m2v2 = m3v3
(3 x 5) + (2 x -5) = (3 + 2) v3
15 + (-10) = 5v3
v = 5/5
v = 1 ms-1 to right
Example
During a space mission, it is necessary to ‘dock’ a space probe of mass
4000 kg onto a space ship of mass 12000 kg.
The probe travels at 4 ms-1, and the ship travels at 2 ms-1 ahead of the
probe, but in the same direction.
What is the velocity of the ship after the probe has ‘docked’?
Solution
Before
4 ms-1
2 ms-1
4000 kg
12000 kg
m1v1 + m2v2 =
(4000 x 4) + (12000 x 2)
16000 + 24000 =
v =
v =
After
v
4000 kg + 12000 kg
(m1 + m2) v3
= (4000 + 12000) v
16000v
40000 / 16000
2.5 ms-1 in original direction
Elastic and Inelastic Collisions
An Elastic Collision is one in which both
kinetic energy and momentum are
conserved.
An Inelastic Collision is one in which
only momentum is conserved.
Elastic or Inelastic?
If the collision is elastic, then:
Ek before = Ek after
If collision is inelastic, then there will be
some energy ‘lost’ (due to heat, light, sound,
etc.).
To calculate the energy ‘lost’, find the
difference between Ek before and Ek after.
E = ½mv²
Energy is measured in Joules (J)
Impulse
An object is accelerated by a force, F, for a
time, t.
The unbalanced force is given by:
Funb = ma
= m(v- u)
t
= mv – mu
t
Unbalanced force
= change in momentum
time
= rate of change of
Impulse = change in momentum
Impulse = force x time
Impulse = Ft = mv – mu
Units of Impulse are kgms-1 or Ns.
Impulse is a vector quantity, so DIRECTION
is important.
Example 1
A snooker ball
of mass 0.2 kg
is accelerated
from rest to a
velocity of 2 ms1 by a force
from the cue
which lasts for
50 ms.
What size of
force is exerted
by the cue?
Solution
u = 0, v = 2 ms-1, m = 0.2 kg,
t = 50 ms = 0.05 s, F = ?
Ft
F x 0.05
F
F
=
=
=
=
mv – mu
(0.2 x 2) – 0
0.4 / 0.05
8N
The concept of impulse is useful in situations
where the force is not constant and acts for
a very short period of time.
An example of this is when a golf ball is hit
by a club.
F
During contact, the
unbalanced force
between the club
and the ball varies
with time as shown in
the graph opposite.
0
t
In practical situations the force is not
constant, but comes to a peak and then
decreases.
Impulse = Area under a Force-time graph
In any collision involving impulse, the
unbalanced force calculated is always the
average force and the maximum force
experienced would be greater than the
calculated average value.
Example 2
A tennis ball of
mass 100 g,
initially at rest, is
hit by a racket.
The racket is in
contact with the
ball for 20 ms and
the force of
contact varies as
shown in the
graph.
What was the
speed of the ball
as it left the
racket?
F/N
400
0
Solution
20
t / ms
Impulse = area under graph
= ½ x (20 x 10-3) x
400
= 4 Ns
Ft = mv – mu
4 = 0.1v – 0
v = 4 / 0.1 = 40 ms-1
Experiment - Impulse
Aim
To calculate the average force exerted
by a hockey stick on a ball.
Method
Timer 1 (ms)
Stick with
metal foil
at head
Light
gate in
front of
ball
Ball
with metal
foil cover
Timer 2 (ms)
The mass and diameter of the ball are carefully
measured.
The ball and hockey stick are covered with metal
foil which allows their time of contact to be
measured with timer 1.
The ball is struck from rest through the light
gate, and the time to pass through is displayed on
timer 2.
The millisecond time recorded on timer 2 can
be used with the diameter of the ball (in
metres) to calculate the instantaneous
speed of the ball immediately after being
struck by the stick.
The equation Ft = mv – mu is used to
calculate the force acting on the ball.
Results
Mass of ball in kilograms
0.04275 kg
Diameter of ball in metres
Time of
contact, t (s)
Time to pass
through gate
(s)
= 42.75 g =
= 7.5 cm 0.075 m
Speed
leaving stick,
v (ms-1)
Force acting
on ball, F (N)
Newton’s Third Law and
Momentum
Newton’s Third Law
When two objects interact, the forces
they exert on each other are equal in
size but opposite in direction.
or
For every ACTION there exists an equal
and opposite REACTION.
Consider a 2 kg trolley travelling at 6 ms-1
which collides with a stationary 1 kg trolley.
The two trolleys move off joined together.
6 ms-1
v
At rest
2 kg
1 kg
Before Collision
2 kg
1 kg
After Collision
By the principle of conservation of
momentum:
(2 x 6) + (1 x 0) = 3 x v
v = 4 ms-1
Now consider the gain/loss of momentum of
each trolley during the collision.
The 2 kg trolley lost (12 – 8) = 4 kgms-1 of
momentum.
The 1 kg trolley gained 4 kgms-1 of
momentum.
i.e. The change in momentum of the 2 kg
trolley is equal in size but opposite in
direction to the change in momentum of the
1 kg trolley.
This may be written as:
(Δ mv)2kg = -(Δ mv)1kg
But from Newton’s 2nd Law, the change in momentum of
an object equals the impulse applied to the object:
(Ft)2kg = -(Ft)1kg
But during the collision the contact time (t) is the same
for both trolleys:
(F)2kg = -(F)1kg
Newton’s 3rd Law
This means that during the collision the
force exerted by the 2 kg object on the 1 kg
object is equal in size but opposite in sense
to the force exerted by the 1 kg object on
the 2 kg object.
This is Newton’s 3rd Law.
This Law is applicable to ALL situations
where forces exist.
Projectile Motion
The motion of a projectile consists of two
independent parts or components:
1) constant horizontal velocity (acceleration = 0)
2) constant vertical acceleration (caused by
Earth’s gravitational pull).
These motions can be treated separately or in
combination depending on the circumstances.
Example – Horizontal
Projection
A projectile is fired horizontally
off a cliff as shown:
5 ms-1
Find:
a) the time of flight
range
b) the range
c) the vertical velocity just before impact
d) the horizontal velocity just before
impact
e) the actual (resultant) velocity just
before impact
45 m
Solution
a)
s
u
v
a
t
h
v
?
-45
5
0
5
?
0
-10
?
?
Vertically,
s = ut + ½at²
-45= 0 – 5t²
t²
= 45 / 5
= 9
t= 3 s
b) Horizontally, (remember a = 0!!)
Range, s = ut
= 5x3
s = 15m
c) Vertical velocity,
v = u + at
= 0 + (-10 x 3)
v = -30 ms-1
d) Horizontal velocity, v = 5 ms-1 (constant!)
e)
v²= 5² + 30²
v = 30.4 ms-1
tan θ = 30 / 5,
5
θ
so θ = 80.5º
v
30
Actual velocity = 30.4 ms-1 at
80.5º below the horizontal
Oblique Projectiles
Example – Oblique
Projection
A projectile is launched with
the initial velocity shown.
Find:
a) the maximum height
b) the total flight time
c) the (horizontal) range
30 ms-1
53.1º
s
u
v
a
t
h
v
?
?
30cos53.1 30sin53.1
30cos53.1
0
0
-10
?
?
a) Vertically,
v² = u² + 2as
0 = (30 sin 53.1)² - 20s
0 = (30 x 0.8)² - 20s
20s = 576
s = 28.8 m
b) v = u + at
0 = (30 sin 53.1) – 10t
0 = (30 x 0.8) – 10t
10t = 24
t = 2.4 s
Total flight time = 4.8 s
c) Horizontally,
s = ut
= (30 cos 53.1) x 4.8
= (30 x 0.6) x 4.8
= 86.4 m
Gravitation
Fields
•
A field in Physics is defined to be a region
of space in which certain objects experience
a force.
•
(e.g. magnetic, electrostatic,
gravitational).
Newton’s universal law of
gravitation
•Objects that have mass produce a gravitational
field.
•If an object of mass, m1, is placed in the
gravitational field of another mass, m2, then
there is a force of attraction between the two
masses. This was formulated by Newton in 1665
and the force is found from:
Gm1m2
F
r2
• Where:
– G is universal gravitational constant 6.67 x 10-11
Nm2kg-2
– m1 and m2 are the masses involved (kg)
– r is the distance between the masses (m)
•
This equation is an example of an inverse square law:
the force, F, is inversely proportional to the square of
the distance, r.
Example 1
•What is the force of
attraction between
two pupils of average
mass (60 kg) sitting
1.5 metres apart?
Example 1
•What is the force of
attraction between
two pupils of average
mass (60 kg) sitting
1.5 metres apart?
Value of r
It is important to realise that the value
for r, the distance between two masses,
is the distance between the centre of
the two masses (e.g in planetary
calculations).
Example 2
•Taking the radius of the Earth to be 6.4 x 106 m, find
the force of attraction on a 250kg satellite that is
orbiting at a height of 36 000km above the Earth.
•(mass of Earth = 6.0  1024 kg)
•
1.
This question should be broken down into two parts.
First of all, find the distance, r, between the two
objects.
2. Use Newton’s Universal Law
Example 2
Einstein’s Theory of
Special Relativity (1905)
In Einstein’s Theory of Special Relativity,
the laws of physics are the same for all
observers, whether they be at rest or
moving at a constant velocity with respect
to each other (zero acceleration).
An observer who is at rest or moves at a
constant velocity has their own frame of
reference.
In all frames of reference, the speed of
light (symbol c) remains the same
regardless of whether the source or
observer is in motion.
Einstein’s principles that the laws of
physics and the speed of light are the
same for all observers leads to the
conclusion that moving clocks run slower
(time dilation) and moving objects are
shortened (length contraction).
The two postulates of special
relativity
1. The laws of physics are the same in all
reference frames.
2. The speed of light is a constant value
in every frame of reference.
Time Dilation
When the speed of an object is 0.1c or
more (≥ 10% of the speed of light),
Newtonian mechanics can no longer be
used to describe its motion.
Relativistic calculations must be used.
Time dilation
A clock will “tick” at a slower rate when it moves at
speeds close to the speed of light.
The time for a particular event to occur when
moving near the speed of light is found from:
Where:
– t’ is the observed time of the stationary observer (s)
– t is the observed time of the moving observer (s)
– v is the velocity of the object observed (ms-1)
– c is the speed of light (3x108 ms-1)
Example 1
A spacecraft leaves Earth and travels at a
constant speed of 0.6c to its destination. An
astronaut on board records a flight time of 5
days.
Calculate the time taken for the journey as
measured by an observer on Earth.
t’ =
t’ =
t
.
√1 – v2
c2
5 .
√1 – 0.62
12
t’ = 6.25 days
Example 2
A rocket leaves a planet and travels at a constant
speed of 0.8c to a destination. An observer on the
planet records a time of 20h.
Calculate the time taken for the journey as
measured by the astronaut on board.
t’
=
t
.
√1 – v2
c2
20 =
t .
√1 – 0.82
12
20 x √1 – 0.82 = t
12
t = 20 x (0.6)
t
= 12 h
Length Contraction
The relativistic length in the direction of motion of
an object moving near the speed of light is not
constant, but decreases with speed.
This is called length contraction and is found using:
Where:
l’ is the observed length of the stationary observer (m)
l is the observed length of the moving observer (m)
v is the velocity of the object observed (ms-1)
c is the speed of light (3x108 ms-1)
Example 1
An observer on Earth sees a spaceship travelling
at 0.7c. If the rocket is measured to be 36m in
length when at rest on Earth, how long is the
moving rocket ship as measured by the observer
on Earth?
l’
= l √1 – v2
c2
l’
= 36 √1 – 0.72
12
l’
= 36 x (0.714)
l’ = 25.7m
Example 2
An observer on Earth sees a rocket zoom by at 0.95c.
If the rocket is measured to be 5.5m in length, how
long is the rocket ship as measured by the astronaut
inside the rocket?
l’ = l √1 – v2
c2
5.5
= l √1 – 0.952
12
5.5
= l x (0.312)
l
= 5.5 / 0.312
l = 17.6m
Note:
Relativistic effects only become
noticeable when an object is moving with a
speed of around 0.1c or greater.
If the speed is less than this, the
factor (v2/c2) is extremely small, and
the value of the square root term is
very close to 1.
The Doppler Effect
The Doppler effect is the change in
frequency you notice when a source of
sound waves is moving relative to you.
When the source moves towards you,
more waves reach you per second and
the frequency is increased.
If the source moves away from you,
fewer waves reach you per second and
the frequency is decreased.
Doppler Effect Formula
This is how the formula appears in the
SQA relationships sheet:
Calculating the frequency
Moving towards the source
The observed frequency, fo, is higher:
fo = fs
v .
(v - vs)
fs = frequency of source
v = speed of sound (approx. 340ms-1)
vs = speed of source
Towards = Take away
Calculating the frequency
Moving away from the source
The observed frequency, fo, is lower:
fo = f s
v .
(v + vs)
Away = Add
Example 1
What is the frequency heard by a person as a train
driving at 15 ms-1 towards them sounds its horn
(f = 800 Hz) if the speed of sound in air is 340 ms-1?
fo = fs
v .
(v - vs)
= 800 x
340
.
(340 - 15)
= 800 x 1.04
fo = 837 Hz
Example 2
What frequency would they hear after the train
passes them if it continues at the same speed?
fo = fs
v
(v + vs)
= 800 x
340
(340+15)
= 800 x 0.931
fo = 745 Hz
Redshift
ROYGBIV
White light (light from galaxies and stars) is
broken up into all the colours of the rainbow
Red Orange Yellow Green Blue Indigo Violet
longer λ
shorter λ
All the colours have different frequencies
and wavelengths.
Redshift (also known as Doppler shift) is how much
the frequency of light from a far away object has
moved toward the red end of the spectrum.
It is a measure of how much the ‘apparent’
wavelength of light has been increased.
It has the symbol Z and can be calculated using the
following equation:
Z = λo – λr it can also expressed as: Z = λo - 1
λr
λr
λo = the wavelength observed
λr = the wavelength at rest
Redshift and velocity
We can also work out the redshift if we
know the velocity, v, that the body is
moving at (for slow moving galaxies):
Z=v
c
What is a blueshift?
• When we use the equation for redshift, we
can sometimes end up with a –ve value.
• This means the object is moving closer to
you and is said to be blueshifted.
• It is a measure of how much the ‘apparent’
wavelength of light has been decreased.
Wavelengths
With a redshift, moving away, the
wavelength increases.
With a blueshift, moving towards, the
wavelength decreases.
Example 1
Light from a distant galaxy is found to contain the spectral
lines of hydrogen.
The light causing one of these lines has (an observed)
measured wavelength of 466 nm.
When the same line is observed (at rest) from a hydrogen
source on Earth it has a wavelength of 434 nm.
(a) Calculate the Doppler shift, Z, for this galaxy.
(b) Calculate the speed at which the galaxy is moving
relative to the Earth.
(c) In which direction, towards or away from the Earth, is
the galaxy moving?
(a)
Z
= λo – λr
λr
= 466 – 434
434
Z
= 0.0737
Example 1
(b)
Z=
v
c
0.0737 = v
.
3 x 108
v = 2.21 x 107 ms-1
(c)
Z is positive therefore galaxy is moving
away
Example 2
A distant star is travelling directly away from the Earth at a
speed of 2·4 × 107 ms1.
(a) Calculate the value of Z for this star.
(b) A hydrogen line in the spectrum of light from this star is
measured to be 443 nm. Calculate the wavelength of this
line when it observed from a hydrogen source on the
Earth.
(a)
(b)
Z = v / c = 2.4 x 107 / 3 x 108 = 0.08
Z = λo - 1
λr
0.08 = (443x10-9) – 1
λr
0.08 + 1 = (443x10-9)
λr
λr = (443x10-9)
0.08 + 1
λr
= 410 x 10-9 m / 410 nm
Hubble’s law
Edwin Hubble (1920s) observe that the
light from some distant galaxies was
shifted towards the red end of the
spectrum.
The size of the shift was the same for all
elements coming from the galaxies.
This shift was due to the galaxies moving
away from Earth at speed.
Edwin Hubble (1920s) found that the
further away a galaxy was the faster it
was travelling.
The relationship between the distance, d,
and speed, v, (sometimes called
recessional velocity) of a galaxy is known
as Hubble’s Law:
v = Ho d
Ho = Hubble’s constant = 2.3 x 10-18 s-1
Hubble’s Constant
The value of Ho is given in
data sheet (and is the
value you would use in an
exam) but can vary as
more accurate
measurements are made.
The gradient of the line
in a graph of speed v
distance of galaxies
provides a value for
Hubble’s constant.
Straight line through origin:
directly proportional
Example
What is the speed of a galaxy relative to
Earth that is at an approximate distance
of 4.10 × 1023 m from Earth?
v = Ho d
v = 2.3 x 10-18 x 4.10 x 1023
v = 9.43 x 105 ms-1
Hubble’s Law tells us that the Universe is
expanding.
The Hubble constant can be used to estimate
the minimum age of the Universe:
v = H 0d
v / d = H0
d / v = 1 / H0
t = 1 / H0
(as t = d / v)
So 1 / H0 yields the age of the Universe (assuming
the rate of expansion has been constant).
The Light Year
Sometimes distances can be given in light years.
One light year is the distance travelled by light in
one year.
It can be calculated as follows using d = vt:
3 x 108 (speed of light) x 365 (days) x 24 (hours)
x 60 (mins) x 60 (s)
One light year = 9.46 x 1015 m
Evidence for the expanding
Universe
Expansion rate (3 possibilities)
1. The expansion rate will eventually
reverse due to gravitational attraction
(the Universe will collapse in on itself).
2. The expansion will continue and may
even accelerate (the Universe is torn
apart).
3. The Universe will expand at a constant
rate forever.
Mass of the Universe
The mass of the Universe is estimated by
measuring the speed of the radial motions
of stars and galaxies.
These measurements give us an indication
of how galaxies are moving and interacting;
this gives an indication of their masses.
Greater central mass results in greater
rotational speeds.
Dark Matter
Measurements of these galaxies have
shown that the rates of rotation are too
large for the matter which is observed.
Scientists have concluded there must be
a new type of matter to account for this –
dark matter.
Dark Energy
Modern telescopes have allowed for the
observation of very distant objects.
Scientists have observed that the most
distant are actually accelerating away
from us.
This means that the additional kinetic energy
gained by these objects must be derived
from a source we cannot fully explain.
This is called dark energy.
Matter which makes up
planets, stars and life
accounts for only 4% of
total matter and energy
of the Universe.
Temperatures of Stellar
Objects
We observe stars using the light and
other electromagnetic radiations they
emit.
Short Wavelength
High Frequency
Gamma
Rays
X-rays
Long Wavelength
Low Frequency
UltraViolet
Visible
Light
InfraRed
Microwaves
Radio and TV
Waves
Energy and frequency
The energy, E, of electromagnetic
radiation is directly proportional to its
frequency:
E = hf
h = Planck’s constant (6.63 x 10-34 Js)
f = frequency of the radiation
By measuring the relative amount of each
type of radiation a star emits, it is
possible to make deductions about the
thermal energy (temperature of the
star).
A graph of radiation intensity (irradiance) against
wavelength:
Irradiance
This is called a
blackbody spectrum.
Wavelength
High temperatures correspond to high energies
and higher frequency (lower wavelength) values.
Hotter stars will emit more energy than cooler
stars, so the area under the curve will be greater.
The graph below shows the radiation curves for
various stellar temperatures.
As the temperature of the object increases:
• intensity (irradiance) increases;
• peak intensity wavelength decreases;
• peak intensity frequency increases;
• the energy increases (as E = hf); and
• the colour changes.
Colour change
As an object becomes hotter it starts to glow a
dull red, followed by bright red, then orange,
yellow and finally white (white hot).
At extremely high temperatures it becomes a
bright blue-white colour.
Big Bang Theory
What is the Big Bang Theory?
The Big Bang took place around 13.8 billion years ago.
The Universe was originally very hot and very dense
concentrated in a tiny point known as a singularity
(smaller than an atom).
The Universe expanded suddenly from the singularity
bringing time and space into existence.
Following the Big Bang, temperatures rapidly cooled
and tiny particles of matter began to form.
The first atoms to form were hydrogen and helium.
Evidence for Big Bang
Hubble’s Law (1920s) was first indication
that a starting point for the Universe could
be identified.
There is now other evidence to support the
Big Bang Theory:
1. Large-scale homogeneity
There is a uniform distribution of matter
regardless of where you are in the
Universe.
You would expect this after an initial
explosion – all matter would be
distributed equally in all directions.
2. The abundance of light
elements
The Universe has an abundance of light
elements such as Hydrogen and Helium.
This is consistent with the Big Bang
theory which states that between 3-20
minutes after the Big Bang, nuclear fusion
created these light elements, but as the
Universe cooled no further fusion took
place.
3. Cosmic Microwave
Background Radiation (CMBR)
Scientists discovered that there are
microwaves coming from every direction in
space (CMBR).
This relatively uniform background radiation
is the remains of energy created just after
the Big Bang.
As the universe expands the wavelength of
the radiation emitted increases (less
frequency) to the microwave region.
In 2001, the WMAP (Wilkinson Microwave
Anisotropy Probe) was launched to look at
fluctuations in the CMBR.
The image above reveals 13.77 billion year old
temperature fluctuations (shown as colour
differences) that correspond to the seeds
that grew to become the galaxies. This fine
detail structure is predicted by the Big Bang
Theory.
4. Redshift
The light from the majority of galaxies is
redshifted, rather than blueshifted,
meaning the other galaxies are moving
away from us.
The most likely explanation is that the
whole universe is expanding.
5. The darkness of the sky
(Olbers’ paradox)
The light from the stars in the galaxy
should be and is enough to light our sky at
night, but our sky is dark at night.
The only explanation is that the stars are
moving away from us and their light hasn’t
reached us yet.
Uncertainties
The measurement of ANY physical
quantity is liable to an error or
uncertainty.
1) Systematic Uncertainties
These are caused by some constant factor.
Measurements all affected in the same
way.
E.g.
i) Scale not properly set to zero.
ii) Measuring tape ‘stretched’.
iii) Same mistake made for each
reading.
2) Scale Reading Uncertainties
Indicates how well an instrument scale
can be read.
For analogue scales, uncertainty is
+/- ½ the least division
For digital scales, uncertainty is
+/- 1 in the least significant
digit
3) Random Uncertainties
These show up when you make a series of
measurements of the same quantity.
The mean gives the best estimate of the
true value.
Random Uncertainty = max. reading – min. reading
no. of readings taken
Notes
i.
Uncertainties can be expressed in two
ways:
Absolute uncertainty e.g. (100 +/- 4) cm
or
Percentage uncertainty e.g. 100 cm +/- 4%
ii. In calculations, the overall uncertainty will
be the highest percentage uncertainty.
Prefixes