Higher Physics Unit 1 – Our Dynamic Universe Printed Notes

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.

VectorsVectors ScalarsScalars

DisplacementDisplacement DistanceDistance

VelocityVelocity SpeedSpeed

AccelerationAcceleration TimeTime

ForceForce MassMass

MomentumMomentum EnergyEnergy

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).

ExampleA 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?

Solutiona) 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:

Resultant Vector:

Always drawn with two arrows to represent a resultant.

tail nose

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

Her resultant displacement, s, is 50 m at an angle x of 53º East of North.

3 cm(30 m)

4 cm (40 m)

sx

N

E

S

W

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

The plane’s resultant velocity, v, is 130 ms-1 at an angle y of 23º East of North.

5 cm (50 ms-1)

12 cm(120 ms-1)

v

y

N

E

S

W

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

θ

x

xh

xvis equivalent to

sin θ = xv / x

xv = x sin θ

cos θ = xh / x

xh = x cos θ

θ

xxv

xh

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 velocitytime 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

Stopclock

Light gates

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 = accelerationv = velocitys = 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

0 0 0

a v s

t tt

Graphs Showing Constant Velocity

0 0 0

a v s

t tt

Graphs Showing Constant Deceleration

0

0

0

v

a

s

t

tt

The Equations of Motion

1) v = u + at

2) s = ut + ½at²

3) v² = u² + 2as

u – initial velocity at time t = 0v – final velocity at time ta – acceleration of objectt – time to accelerate from u to vs – displacement of object in time t

Example 1

QA 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 + ½at²s = 40 + 10s = 50 m

Example 2

QA 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 - 4s4s = 45² - 15² s = 450 m

Example 3

QA 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 + atv = 20 – 30v = -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 + ½at²s = displacement = area under a v/t graph.

s = = ut + ½(v – u)t= ut + ½(at)t

i.e. s = ut + ½at²

v

u

0 ttime

Velocity

u

v - u

+

3) v² = u² + 2asv = 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 4Ex 4 A ball is launched vertically into the air A ball is launched vertically into the air with an initial velocity of 35 mswith an initial velocity of 35 ms-1-1. What is the . What is the maximum height it will reach?maximum height it will reach?

Ex 5Ex 5 A rock is thrown upwards with an initial A rock is thrown upwards with an initial velocity of 30 msvelocity of 30 ms-1-1. After how long will it be 10 . After how long will it be 10 metres off the ground?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:

4. Choose the best equation to suit the problem.

Up +

Right +

Down -

Left -

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 kga = 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 = F / m= 40 / 20= 2 ms-2

20 kg40 N

a

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?

Fun = 20000 - 9800 - 2000 = 8200 N

a = F / m= 8200 /

1000= 8.2 ms-

2

NASA

W = mg = 1000 x 9.8 = 9800 N

Thrust = 20000 N

Drag= 2000 N

a

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?

a) 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= 40 N

80 kg 60 kgT 70 N

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).

There could also be an unbalanced force, Fun.

Liftof

mass ‘m’

Tension

Weight = mg

or

Fun = ma

There are 6 possible lift situations:

1) Lift at restTension = Weight (no acceleration, no

Fun)2) Lift travelling at constant speed

Tension = Weight (no acceleration, no Fun)

3) Lift accelerating upTension = Weight + Fun

4) Lift decelerating upTension = Weight – Fun

5) Lift accelerating downTension = Weight – Fun

6) Lift decelerating downTension = Weight + Fun

ExampleA lift of mass 4 kg is accelerating upwards at 3 ms-2.What is the tension in the cable?

SolutionT = tension in the cable

Fun = ma

T – mg = ma T = mg + ma T = (4 x 9.8) + (4 x 3) T = 51.2 N

4 kg

T

W = mg

a = 3 ms-2

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.

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.

Friction

Weight

Free-Falling Parachutist

OA – constant acceleration due to gravityAB – decreasing acceleration as frictional force actsBC – terminal velocityCD – non-uniform deceleration (parachute opens)DE – constant speedEF – 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

θ

x

xh

xvis equivalent to

sin θ = xv / x

xv = x sin θ

cos θ = xh / x

xh = x cos θ

θ

xxv

xh

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.

is equivalent to

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.

θ

F

Fh = F cos θ

Fv = F sin θ

ExampleA 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

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

30º

200 N

Friction= 100 N

Fh

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

mgθ

mg sin θ

Reaction Force

θmg cos θ

Component of weight down slope = mg sin θ

Component of weight = mg cos θ perpendicular to slope

ExampleA 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

AimTo compare the total momentum before and after a collisionMethodThe apparatus was set up as shown below:

Linear air track

Microcomputer

Vehicle 1with card

Vehicle 2with card

LightGates

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

Units of m = kgUnits of v = ms-1

Units of p = mv = kgms-1

m1 v1 m2 v2 pbefore m1+m2 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 = Total momentum of all objects before 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:

m1v1 + m2v2 = m1v1’ + m2v2’

m1 m2 m1 m2

v1v2 v1’ v2’

Before After

2. Two masses collide and stick together:

m1v1 + m2v2 = (m1 + m2)v3

m1 m2 m1 + m2

v1v2 v3

Before After

3. An explosion. In this case:

(m1+ m2) v = m1v1 + m2v2

If initially at rest (e.g. gun before firing a bullet), then:

0 = m1v1 + m2v2

m1 m2m

v1v2v

Before After

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

Find the velocity of the trolleys when they stick together after colliding.

Solution

+ 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

3 kg 2 kg

5 ms-1 5 ms-1

ExampleDuring 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

m1v1 + m2v2 = (m1 + m2) v3

(4000 x 4) + (12000 x 2) = (4000 + 12000) v 16000 + 24000 = 16000v

v = 40000 / 16000 v = 2.5 ms-1 in original

direction

4000 kg 12000 kg 4000 kg + 12000 kg

4 ms-1 2 ms-1 v

Before After

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.

Ek = 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 momentumtime

= rate of change of momentum

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 ms-1 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 = mv – muF x 0.05 = (0.2 x 2) – 0

F = 0.4 / 0.05 F = 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.

During contact, the unbalanced force between the club and the ball varies with time as shown in the graph opposite.

F

t0

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 2A 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?

SolutionImpulse = 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

F / N

t / ms0 20

400

Experiment - Impulse

Aim

To calculate the average force exerted by a hockey stick on a ball.

Method

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.

Timer 1 (ms)Timer 2 (ms)

Lightgate infront ofball

Ballwith metalfoil cover

Stick withmetal foilat head

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 = 42.75 g = 0.04275 kgDiameter of ball in metres = 7.5 cm 0.075 m

Time of contact, t (s)

Time to pass through gate

(s)

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.

Before Collision After Collision

By the principle of conservation of momentum:

(2 x 6) + (1 x 0) = 3 x v

v = 4 ms-1

2 kg 1 kg

6 ms-1

At rest

2 kg 1 kg

v

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:

Find:a) the time of flightb) the rangec) the vertical velocity just before

impactd) the horizontal velocity just before

impacte) the actual (resultant) velocity just

before impact

5 ms-1

45 m

range

Solutiona) Vertically,

s= ut + ½at² -45 = 0 – 5t² t² = 45 / 5

= 9 t = 3 s

b) Horizontally, (remember a = 0!!)Range, s = ut

= 5 x 3 s = 15m

hh vv

ss ?? -45-45

uu 5 5 00

vv 55 ??

aa 00 -10-10

tt ?? ??

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, so θ = 80.5º

Actual velocity = 30.4 ms-1 at 80.5º below the horizontal

5

30

v

θ

Oblique Projectiles

Example – Oblique Projection

A projectile is launched with the initial velocity shown.

Find:a) the maximum heightb) the total flight timec) the (horizontal) range

53.1º

30 ms-1

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 + at0 = (30 sin 53.1) – 10t0 = (30 x 0.8) – 10t10t = 24t = 2.4 sTotal flight time = 4.8 s

hh vv

ss ?? ??

uu 30cos53.130cos53.1 30sin53.130sin53.1

vv 30cos53.130cos53.1 00

aa 00 -10-10

tt ?? ??

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:

221

r

mGmF

• 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)

• This question should be broken down into two parts.1. 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.

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)

Time dilation

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 . √1 – v2

c2

t’ = 5 . √1 – 0.62

12

t’ = 6.25 days

Example 2A 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 1An 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

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

Example 2

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 sourcev = 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 = fs 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 observedλr = 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 1Light 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 = vc

0.0737 = v . 3 x 108

v = 2.21 x 107 ms-1

(c) Z is positive therefore galaxy is moving away

Example 2A distant star is travelling directly away from the Earth at

a speed of 2·4 × 107 ms1.(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) Z = v / c = 2.4 x 107 / 3 x 108 = 0.08

(b)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

ExampleWhat 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 = H0d

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 UniverseThe 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.

GammaRays X-rays

Ultra-Violet

VisibleLight

Infra-Red Microwaves

Radio and TVWaves

Short WavelengthHigh Frequency

Long WavelengthLow Frequency

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:

This is called ablackbody spectrum.

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.

Wavelength

Irradiance

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, yellowyellow and finally whitewhite (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.

This matter created stars, galaxies and planets.

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

orPercentage uncertainty e.g. 100 cm +/- 4%

ii. In calculations, the overall uncertainty will be the highest percentage uncertainty.

Prefixes