Marr College - Physics

Marr College - Physics Higher Physics Unit Our Dynamic Universe Key Areas Motion equations and graphs Forces, energy and power Collisions, explosions and impulse Gravitation Special relativity The expanding Universe Big bang theory 1

Key Area Motion equations and graphs 1. Distinguish between distance and displacement (N5 recap). 2. Distinguish between speed and velocity. (N5 recap). 3. Define and classify vector and scalar quantities. (N5 recap). 4. Use scale diagrams, or otherwise, to find the magnitude and direction of the resultant of a number of displacements or velocities. (N5 recap). 5. State what is meant by the resultant of a number of forces. (N5 recap). 6. Use scale diagrams, or otherwise, to find the magnitude and direction of the resultant of a number of forces. (N5 recap). 7. Carry out calculations to find the rectangular components of a vector. 8. State that acceleration is the change in velocity per unit time. 9. Describe the principles of a method for measuring acceleration. 10. State that the gradient of a displacement-time graph gives velocity. 11. State that for a velocity-time graph, the gradient gives the acceleration and the area under the graph gives the displacement. 12. Draw an acceleration time graph using information obtained from velocity time graph for motion with a constant acceleration. 13. Use the terms constant velocity and constant acceleration to describe motion represented in graphical or tabular form. 14. Show how the following relationships can be derived from basic definitions in kinematics: v = u + at, s = ut + ½at 2, v 2 = u 2 + 2as. 15. Carry out calculations using the above kinematic relationships. Red Amber Green 2

Vector and Scalar Quantities (recap on National 5 Physics) A scalar quantity is fully described by its magnitude (size) and unit, e.g. quantity time = 220 s unit magnitude A vector quantity is fully described by its magnitude, unit, and direction, e.g. unit quantity Force = 800 N upwards direction magnitude Distance and Displacement Distance is a scalar quantity. It measures the total distance travelled, no matter in which direction. Displacement is a vector quantity. It is the length measured from the starting point to the finishing point in a straight line. Its direction must be stated. Example A girl walks 3 km due north then turns and walks 4 km due east as shown in the diagram. 4 km FINISH 270 000 090 3 km 180 START 3

Calculate (a) The total distance travelled (b) The girl s final displacement relative to her starting position. Solution (a) Total distance = 3 km + 4 km = 7 km (b) (to calculate displacement, we need to draw a vector diagram) (this solution involves using Phythagoras and trig functions (SOH-CAH=TOA), but you can also solve these types of problems using a scale diagram). 4 km FINISH Magnitude of displacement = 3 km θ START Resultant displacement Θ= 53 = 5 km Resultant displacement = 5 km at 053 Speed and Velocity Speed is a scalar quantity. As discussed above, speed is the distance travelled per unit time. Velocity is a vector quantity (the vector equivalent of speed). Velocity is defined as the displacement per unit time. Since velocity is a vector, you must state its direction. The direction of velocity will be the same as the displacement. 4

Example The girl s walk in the previous example took 2.5 hours. Calculate (a) The average speed (b) The average velocity for the walk, both in km -1 Solution (a) Distance = 7 km Time = 2.5 hours (b) Displacement = 5 km (053 ) Time = 2.5 hours at 053 (When performing calculations on speed and velocity like the above, it is best to write the equations in words as shown, to avoid confusion with symbols. This communicates your understanding in the clearest way). 5

Rectangular Components of a Vector Any vector v can be split up (resolved) into a horizontal component v h and vertical component v v. is equivalent to Example A shell is fired from a cannon as shown. Calculate a) the horizontal component of velocity b) the vertical component of velocity. a) v h = v cos u = 50 cos 60 = 25 m s -1 b) v v = v sin u = 50 sin 60 = 43.3 m s -1 So is equivalent to 6

Acceleration Acceleration is defined as the change in velocity per unit time. The unit is metre per second squared, m s -2. a = v - u t where v = final velocity u = initial velocity t = time taken Measuring acceleration Acceleration is measured by determining the initial velocity, final velocity and time taken. A double mask which interrupts a light gate can provide the data to a microcomputer and give a direct reading of acceleration. Graphs of motion Displacement-time Velocity-time Acceleration-time Interpreting displacement time graphs The gradient of a displacement time graph gives the velocity of the object. Interpreting velocity time graphs The gradient of a velocity time graph gives the acceleration of the object: the steeper the line the greater the acceleration of the object. The area under the graph is the displacement. Interpreting acceleration-time graphs The area under the graph gives us the velocity of the object The following graphs show the relationship between the three graphs 7

Stationary object Constant velocity Acceleration-Time graph 8

Constant acceleration Acceleration-Time graph Example The graph below represents an object moving with a positive velocity of 5 m s 1, which is accelerating at a constant rate. After 300 s the object is moving with velocity of 35 m s 1. A constant acceleration means the velocity is increasing at a constant rate. 9

The acceleration of the object can be found from the gradient of the line. The displacement of the object can be determined by calculating the area under the graph. 2 1 displacement = area under graph = area 1 + area 2 = (300 x 5) + ( ½ x 30 x 300) = 1500 + 4500 = 6000 m 10

Example Sketch of velocity-time graph for a ball thrown upwards 11

Equations of motion where: u - initial velocity of object at time t = 0 v = u + at s = ut + 1 2 at2 v 2 = u 2 + 2as v - final velocity of object at time t a - acceleration of object t - time to accelerate from u to v s - displacement in time t. These equations of motion apply providing: the motion is in a straight line the acceleration is constant. When using the equations of motion, note: the quantities u, v, s and a are all vector quantities a positive direction must be chosen and quantities in the reverse direction must be given a negative sign a deceleration will be negative, for movement in the positive direction. Derivation of equations of motion 1. v=u + at The velocity - time graph for an object accelerating uniformly from u to v in time t is shown below. a = v - u t Changing the subject of the formula gives: v = u + at --------- [1] 12

2. s=ut + ½at 2 Consider the velocity-time graph of an object accelerating from initial velocity u to final velocity v in a time t. The displacement, s, in time t is equal to the area under the velocity time graph. (But from equation 1, v = u+at) velocity / ms -1 v u 0 0 A B t time / s ----------- [2] 3. v 2 = u 2 + 2as (square both sides) (multiply out brackets) (take common factor 2a from 2 nd and 3 rd term) (and from equation[2], s=ut + ½at 2 ) ------------ [3] 13

Objects Launched Upwards At the instant an object is launched upwards, it is travelling at maximum velocity. As soon as the object starts to travel upwards, gravity will accelerate it towards the ground at -9.8 ms -2. As a result, the upward velocity of the object will eventually become 0 ms -1. This happens at its maximum height. Example A spring-powered toy frog is launched vertically upwards from the ground at 4.9 ms -1. (a) What will be the velocity of the toy frog at its maximum height? (b) Calculate: (i) the time taken for the toy frog to reach its maximum height; (ii) the maximum height. Solution s =? u = 4.9 ms -1 v =? a = -9.8 ms -2 t =? a)at maximum height v = 0 ms -1. b)(i) v = u + at 0 = 4.9 + (-9.8t) 0 = 4.9-9.8t 9.8t = 4.9 t = 4.9/9.8 t = 0.5 s (ii) s = ut + ½ at 2 s = (4.9 x 0.5) + (0.5 x -9.8 x 0.5 2 ) s = 2.45 + (-1.225) s = 1.2 m i.e. 1.2 m upwards, so height = 1.2 m 14

Key Area Forces, Energy and Power 1. Define the newton 2. Analyse motion using Newton s First and Second Laws. 3. Carry out calculations using the relationship F = ma in situations where resolution of forces is not required. 4. Use free body diagrams to analyse the forces on an object. Red Amber Green 5. Analyse the velocity-time graph of a falling object when air resistance is taken into account. 6. Analyse the motion of a rocket. 7. Carry out calculations using the relationship F = ma in situations where objects are connected. 8. State what is meant by the resultant of a number of forces. 9. Carry out calculations to find the rectangular components of a force vector. 10. Use scale diagrams, or otherwise, to find the magnitude and direction of the resultant of a number of forces. 11. Analyse the forces acting on an object on a slope. 12. Carry out calculations involving work done, potential energy, kinetic energy and power. 15

Newton s 1st Law of Motion Newton s 1st law of Motion states that an object will remain at rest or travel with a constant speed in a straight line (constant velocity) unless acted on by an unbalanced force. Newton s 2nd Law Newton s 2nd law of motion states that the acceleration of an object: varies directly as the unbalanced force applied if the mass is constant varies inversely as the mass if the unbalanced force is constant. These can be combined to give: (resultant or unbalanced) force (N) F un = m a acceleration (ms -2 ) mass (kg) The unit of force, the newton is defined as the resultant force which will cause a mass of 1kg to have an acceleration of 1 m s -2. Free Body Diagrams Some examples will have more than one force acting on an object. It is advisable to draw a diagram of the situation showing the direction of all forces present acting through one point. These are known as free body diagrams. 16

Examples 1. Forces on a rocket. On take off, the thrust on a rocket of mass 8000 kg is 200 000 N. Find the acceleration of the rocket. Thrust = 200 000 N Weight = mg = 8000 9.8 = 78 400 N Resultant force = 200000 78 400 = 121 600 N F 121600-2 a = = =15.2 m s m 8000 17

2. People in lifts A woman is standing on a set of bathroom scales in a stationary lift (a normal everyday occurrence!). The reading on the scales is 500 N. When she presses the ground floor button, the lift accelerates downwards and the reading on the scales at this moment is 450 N. Find the acceleration of the lift. Weight = 500 N W = 500 N Force upwards = 500 N F = 450 N (reading on scales) Lift is stationary, forces balance W = F Lift accelerates downwards, unbalanced force acts. Resultant Force = Weight - Force from floor = W - F = 500 N = 500 450 = 50 N Resultant Force a = m = 50 50 = 1 m s -2 18

3. Tension A ski tow pulls 2 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. Find a) the acceleration of the system b) the tension in the rope. 60 kg 80kg a) Total mass, m = 140 kg a = F m = 70 140 = 0.5 m s-2 b) Consider the 60 kg skier alone. Tension, T = ma = 60 0.5 = 30 N 19

Resolution of a Force In the previous section, a vector was split into horizontal and vertical components. F v = F sin is equivalent to F h = F cos Example A man pulls a garden roller of mass 100 kg with a force of 200 N acting at 30 0 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? F h = F cos = 200 cos 30 0 =173.2 N Resultant F h = 173.2 - Friction = 173.2-100 = 73.2 N a = F m = 73.2 100 = 0.732 m s-2 20

Force Acting Down a Plane If an object is placed on a slope then its weight acts vertically downwards. A certain component of this force will act down the slope. The weight can be split into two components at right angles to each other. Component of weight down slope = mgsin Component perpendicular to slope = mgcos Example A wooden block of mass 2 kg is placed on a slope at 30 to the horizontal as shown. A frictional force of 4 N acts up the slope. The block slides down the slope for a distance of 3 m. Determine the speed of the block at the bottom of the slope. Component of weight acting down slope = mg sin30 = 2 9.8 0.5 = 9.8 N Resultant force down slope = 9.8 - friction = 9.8-4 = 5.8 N a = F/m v 2 = u 2 + 2as = 5.8 / 2 = 0 + 2 2.9 3 = 2.9 m s -2 = 17.4 v = 4.2 m s -1 21

Conservation of Energy The total energy of a closed system must be conserved, although the energy may change its form. The equations for calculating kinetic energy E k, gravitational potential energy E p and work done are given below. E p = mgh E k = ½ mv 2 E w = F d Energy and work are measured in joules J. Example A trolley is released down a slope from a height of 0.3 m. If its speed at the bottom is found to be 2 m s -1, find a) the energy difference between the E p at top and E k at the bottom. b) the work done by friction c) the force of friction on the trolley a) E p at top = mgh = 1 9.8 0.3 = 2.94 J E k at bottom = ½ mv 2 = ½ 1 4 = 2 J Energy difference = 0.94 J b) Work done by friction = energy difference (due to heat, sound) = 0.94 J c) 0.47 N 22

Power Power is the rate of transformation of energy from one form to another. Power is measured in watts W. 1 Watt = 1 Joule per second i.e. 1 W = 1 Js -1 23

Key Area Collisions, Explosions and Impulse 1. State that momentum is the product of mass and velocity. 2. State that the law of conservation of linear momentum can be applied to the interaction of two objects moving in one dimension, in the absence of net external forces. 3. State that an elastic collision is one in which both momentum and kinetic energy are conserved. 4. State that an inelastic collision is one in which only momentum is conserved. Red Amber Green 5. Carry out calculations concerned with collisions in which the objects move in only one dimension. 6. Carry out calculations concerned with explosions in one dimension. 7. State that impulse = force time. 8. State that impulse = change in momentum. 9. Carry out calculations using the relationship, impulse = change of momentum, i.e. Ft = mv - mu 10. Analyse force-time graphs during contact of colliding objects. 11. Calculate impulse from the area under a force-time graph. 24

Momentum The momentum of an object is given by: Momentum = mass velocity of the object. Momentum = mv kg m s -1 kg m s -1 Note: momentum is a vector quantity. The direction of the momentum is the same as that of the velocity. Conservation of Momentum When two objects collide it can be shown that momentum is conserved provided there are no external forces applied to the system. For any collision: Total momentum of all objects before = total momentum of all objects after. 25

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. Example (a) A car of mass 1200 kg travelling at 10 m s -1 collides with a stationary car of mass 1000 kg. If the cars lock together find their combined speed. (b) By comparing the kinetic energy before and after the collision, find out if the collision is elastic or inelastic. Draw a simple sketch of the cars before and after the collision. BEFORE AFTER a) Momentum = mv Momentum = mv = 1200 10 = (1200 + 1000) v = 12000 kg m s -1 = 2200v kg m s -1 Total momentum before = Total momentum after 12000 = 2200v 12000 2200 = v v = 5.5 m s -1 b) before Ek = 1 2 mv2 after Ek = 1 2 mv2 = 1 2 1200 102 = 1 2 2200 5.52 = 60,000 J = 33,275 J Kinetic energy is not the same, so the collision is inelastic 26

Vector nature of momentum Momentum is a vector quantity, so direction is important. Since the collisions and explosions dealt with will act along the same line (one dimension), then the directions can be simplified by giving: Example momentum to the right a positive sign and momentum to the left a negative sign. Find the unknown velocity below. v = 4m s -1 v =? 8 kg 6 kg (8 + 6 )kg BEFORE Momentum = mv = (8 4) - (6 2) = 20 kg m s -1 AFTER Momentum = mv = (8 + 6)v = 14v Total momentum before = Total momentum after v = 20 14 20 = 14v = 1.43 m s-1 Trolleys will move to the right at 1.43 m s -1 since v is positive. 27

Explosions A single stationary object may explode into two parts. The total initial momentum will be zero. Hence the total final momentum will be zero. Notice that the kinetic energy increases in such a process. Example Two trolleys shown below are exploded apart. Find the unknown velocity. BEFORE Total momentum = mv = 0 AFTER Total momentum = mv = - (2x3) + 1v = - 6 + v Total momentum before = Total momentum after 0 = -6 + v v = 6 m s -1 to the right (since v is positive). 28

Impulse An object is accelerated by a force F for a time, t. The unbalanced force is given by: F = ma = m(v - u) t = mv - mu t Unbalanced force = change in momentum time = rate of change of momentum Ft = mv - mu The term Ft is called the impulse and is equal to the change in momentum. Impulse = change in momentum Note: the unit of impulse, Ns is equivalent to kg m s -1. The concept of impulse is useful in situations where the force is not constant and acts for a very short period of time. One 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 below. 0 0 Since F is not constant the impulse (Ft) is equal to the area under the graph. In any calculation involving impulse the unbalanced force calculated is always the average force and the maximum force experienced would be greater than the calculated average value. Examples Impulse = area under F-t graph 1. In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the rest to a velocity of 2 m s -1 by a force from the cue which lasts 50 ms. What size of force is exerted by the cue? u = 0 v = 2 m s -1 t = 50 ms = 0.05 s m = 0.2 kg F =? Ft = mv - mu F 0.05 = 0.2 2 F = 8 N 29

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 over this period as shown in the graph. Determine the speed of the ball as it leaves the racket. Impulse = Area under graph 0 0 = 1 2 20 10-3 400 = 4 N s u = 0 m = 100 g = 0.1 kg v =? Ft = mv - mu = 0.1v 4 = 0.1 v v = 40 m s -1 3. A tennis ball of mass 0.1 kg travelling horizontally at 10 m s -1 is struck in the opposite direction by a tennis racket. The tennis ball rebounds horizontally at 15 m s -1 and is in contact with the racket for 50 ms. Calculate the force exerted on the ball by the racket. m = 0.1 kg u = 10 m s -1 v = -15 m s -1 (opposite direction to u) t = 50 ms = 0.05 s Ft = mv - mu 0.05 F = 0.1 (-15) - 0.1 10 0.05 F = -1.5-1 0.05 F = -2.5 F = - 2.5 0.05 = -50 N (Negative indicates force in opposite direction to initial velocity) 30

Key Area Gravitation Red Amber Green 1. Explain the curved path of a projectile in terms of the force of gravity. 2. Explain how projectile motion can be treated as two independent motions. 3. Solve numerical problems for an object projected horizontally. 4. Solve numerical problems for an object projected at an angle. 5. Describe how the force between two masses changes with mass and distance. 6. State the Universal Law of Gravitation. 7. Carry out calculations using the relationship. 8. Compare the force of gravity between planets, everyday objects and on a sub-atomic level. 31

Projectile motion A projectile has a combination of vertical and horizontal motions. Various experiments show that these horizontal and vertical motions are totally independent of each other. Closer study gives the following information about each component. Horizontal: constant speed Vertical: constant acceleration downward (due to gravity). Example An aircraft flying horizontally at 200 ms -1, drops a food parcel which lands on the ground 12 seconds later. Calculate: a) the horizontal distance travelled by the food parcel after being dropped b) the height that the food parcel was dropped from. Solution (a useful layout is to present the horizontal and vertical data in a table) (then select the data you need for the question, and clearly indicate whether you are using horizontal or vertical data). initial vertical velocity always 0 for a projectile launched horizontally time of flight the same for both horizontal and vertical Horizontal Vertical d h =? u v = 0 v h = 200 ms -1 v v =? t = 12 s a = 9.8 ms -2 t = 12 s s =? acceleration due to gravity 32

(a) Horizontal: d h =? d h = v h t v h = 200 ms -1 = 200 x 12 t = 12 s = 2400 m (b) Vertical: u v = 0 s = ut + ½ at 2 v v =? = (0x12) + ( ½ x (-9.8) x 12 2 ) a = -9.8 ms -2 t = 12 s s=? = -705.6 m i.e. food parcel is 705.6 m below aircraft after 12 seconds height parcel was dropped from = 705.6 m 33

Projectiles launched at an angle maximum height range When dealing with this type of projectile problem we first need to resolve the resultant velocity into its horizontal and vertical components. The horizontal distance travelled by the projectile is known as its range. When the projectile reaches its maximum height, the vertical velocity is zero. Example A tennis ball is launched from a serving machine with a velocity of 35 ms -1 at an angle of 30 to the horizontal. Determine: (a) The horizontal and vertical components of the initial velocity (b) The maximum height the ball reaches (c) the time taken to reach that height (d) the total time the ball is in the air (e) the horizontal distance the shell travels Solution 30 35 ms -1 (a) v h = v cos = 35 cos 30 o = 30.3 ms -1 v v = v sin = 35 sin 30 o = 17.5 ms -1 a) v 2 = u 2 + 2as vertical motion 0 2 = 17.5 2 + (2 x -9.8 x s) 0 = 306.25-19.6 s 19.6 s = 306.25 s = 306.25/19.6 s = 15.6 m 34

b) v = u + at vertical motion 0 = 17.5 + (-9.8 x t) 0 = 17.5-9.8 t 9.8 t = 17.5 t = 17.5/9.8 t = 1.786 s c) Total time shell is in air = 2 x 1.786 s = 3.57 s d) d = v t horizontal motion d = 30.3 x 3.57 d = 108 m 35

Inverse Square Law of Gravitation Two objects with mass will exert a gravitational force on each other The gravitational force is of attraction, i.e. the force makes one mass move towards the other The size of the gravitational force is o Directly proportional to the mass of each object o Inversely proportional to the square of the distance between the objects This relationship is described in Newton s famous equation Where: F = Gravitational force (N) G = Gravitational constant (6.67x10-11 Nm 2 kg -2 ) Example m 1 and m 2 are the masses of the objects involved (kg) r = the distance between the objects (m) Communications satellites orbit the Earth at a height of 36 000 km. a) How far is this from the centre of the Earth? (radius of Earth, r E = 6.4 x 10 6 m) b) If the satellite has a mass of 250 kg, what is the force of attraction on it from the Earth? (mass of Earth, m E = 6.0 x 10 24 kg) Solution a) (3.6 x 10 7 m + 6.4 x 10 6 m) = 4.24 x 10 7 m from the centre of the Earth. b) 36

Key Area Special Relativity 1. State that the speed of light is constant for all observers 2. State that this led Einstein to propose that measurements of length and time for a moving observer are therefore different to those for a stationary observer Red Amber Green 3. Carry out calculations involving length contraction using 4. Carry out calculations involving time dilation using l' l t' v 1 c t v 1 c 2 2 2 2 37

Special relativity Special relativity was the first theory of relativity Einstein proposed (1905). It was termed as special as it only considers the special case of reference frames moving at constant speed. Later he developed the theory of general relativity which considers accelerating frames of reference. Key Principle of Special relativity (and General relativity) The speed of light in vacuum (c = 3.0 x 10 8 ms -1 ) is constant for all observers no matter what speed you are moving at. This means that no matter how fast you go, you can never catch up with a beam of light, since it always travels at 3 0 10 8 ms 1 relative to you. Relative Motion (known as Frames of Reference) You are sitting on a train relaxing and reading. As far as you are concerned, in your frame of reference, while looking at the book and at your friend sitting next to you, you are stationary. There is no relative motion between you, the book or your friend. You would notice that you were moving if you looked outside the window though and saw a stationary object like a tree. Passengers standing still at a station would see you moving towards them. In their frame of reference, they are stationary and it is you who is moving Consequences of constant speed of light for all observers Lengths measured by a stationary observer will be shorter than lengths measured by the person in the moving object (known as length contraction) Times measured by a stationary observer will be longer than times measured by person in the moving object (known as time dilation) The mass of a moving object will increase as it approaches the speed of light (known as relativistic mass) For these effects to be noticeable, the moving observer must be travelling at a speed close to the speed of light, c. 38

Length contraction The observed decrease in length measured by a stationary observer compared to a moving observer travelling at a speed v is given by: l' l v 1 c 2 2 where: l = decreased length as measured by a stationary observer l = original length and length still measured by moving observer in his/her frame of reference v = speed of moving object c = speed of light in vacuum (3.0 x 10 8 ms -1 ) Example A rocket has a length of 10 m when at rest on the Earth. An observer on Earth watches the rocket passing at a constant speed of 1.5 x 10 8 ms -1. Calculate the length of the rocket as measured by the observer. Solution l = 10 m v = 1.5 x 10 8 ms -1 l' l v 1 c 2 2 c = 3.0 x10 8 ms -1 l = 8.66 m 39

Time dilation The observed increase in time measured by a stationary observer compared to a moving observer travelling at a speed v is given by: where: t = increased time as measured by a stationary observer t = original time and time still measured by moving observer in his/her frame of reference v = speed of moving object c = speed of light in vacuum (3.0 x 10 8 ms -1 ) Example A rocket is travelling at a constant 2.7 10 8 m s 1 compared to an observer on Earth. The pilot measures the journey as taking 240 minutes. How long did the journey take when measured from Earth? Solution: t = 240 minutes v = 2.7 10 8 m s 1 c = 3 10 8 m s 1 t' =? t' t' t 2 v 1 2 c 240 8 (2710 ) 1 8 2 (310 ) 2 t' = 550 minutes Evidence of time dilation The speed of satellites in orbit is sufficient for small increases in time to become significant and seriously affect the synchronisation of global positioning systems (GPS) and TV satellites with users on the Earth. They have to be specially programmed to adapt for the effects of time dilation. Very precise measurements of small increases in time have been performed on fast-flying aircraft and agree with predicted results within experimental uncertainty 40

Key Area - The Expanding Universe Red Amber Green 1. State that the Doppler effect is observed in sound and light. 2. State that the Doppler effect is the change in frequency observed when a source of sound waves is moving relative to an observer. 3. Use the expressions for the apparent frequency detected when a source of sound waves moves relative to a stationary observer, and 4. State that Doppler red-shift is evidence for galaxies moving away from each other. 5. Carry out calculations involving change of wavelength, 6. State that for non-relativistic speeds redshift is the ratio of the velocity of the galaxy to the velocity of light, 7. State that Hubble s law shows the relationship between the recession velocity of a galaxy and its distance from us. 8. Carry out calculations using Hubble s law, v = H o d. 9. State that Hubble s law leads to an estimate of the age of the Universe. 10. State that gravity is the force which slows down expansion. 11. State that the eventual fate of the universe depends on its mass. 12. State that the orbital speed of stars provides a method to determine the mass of galaxies. 13. Describe the evidence for dark matter and dark energy in terms of the evolution of the Universe. 41

Doppler Effect The Doppler Effect is the change in frequency observed when a source of sound waves is moving relative to an observer. Examples of the Doppler effect are: Hearing an increase in a car horn s pitch as the car approaches you Hearing an increase then decrease the pitch of a jet s engine noise as it passes you Sound source moving relative to a stationary observer Sound source moving towards stationary observer frequency is increased: Sounds source moving away from stationary observer frequency is decreased: Example A police car siren emits sound of frequency 40 Hz. The police car travels towards an observer at 20 ms -1. Calculate the frequency heard by the observer. (Speed of sound in air = 340 ms -1 ) Solution f obs =? v = 340 ms -1 v s = 20 ms -1 42

Redshift and Hubble s Law Redshift is an example of the Doppler Effect. It is the term given to the change in frequency of the light emitted by stars, as observed from Earth, due to the stars moving away from us. Redshift has always been present in the light reaching us from stars and galaxies but it was first noticed by astronomer Edwin Hubble, in the 1920 s, when he observed that the light from distant galaxies was shifted to the red end of the spectrum. The extent of redshift is given by: where: z = extent of redshift observed = observed wavelength of light (for example the wavelength seen by an observer on Earth viewing a distant star) rest = wavelength of light if the object is at rest For galaxies moving at non-relativistic speeds, the extent of redshift can also be calculated using: where: z = extent of redshift v = speed of galaxy c = speed of light in vacuum 3x10 8 ms -1 43

Example Light from a distant galaxy is found to contain the spectral lines of hydrogen. One line has a wavelength measured as 466 nm. When the same line is observed from a hydrogen source on Earth it has a wavelength of 434 nm. a) Calculate the extent of redshift, z, for this galaxy. b) Calculate the speed at which the galaxy is moving relative to the Earth. Solution a) b) 44

Hubble s Law Hubble s Law describes the relationship between the velocity of a galaxy, as it moves away from us, and its distance. where: v = velocity of galaxy (ms -1 ) H 0 = Hubble constant (quoted as 2 3 10 18 s 1 ) d = distance between the galaxy and us (m) Example v = H o d A galaxy is moving away from the Earth at a speed of 0 074 c. a) Convert 0 074 c into a speed in ms -1. b) Calculate the approximate distance, in metres, of the galaxy from the Earth. Solution (a) (b) 45

Hubble s Law and Age of the Universe By working back in time it is possible to calculate a time when all the galaxies were at the same point in space. This allows the age of the universe to be calculated. v = speed of galaxy receding from us d = distance of galaxy from us H 0 = Hubble s constant t = time taken for galaxy to reach that distance, i.e. the age of the universe t = d / v but from Hubble s Law, v = H o d So t = d / H o d t = 1 / H o t = 4.347 x 10 17 seconds Convert this time to years (divide by 365x24x60x60) Age of Universe = 1.37 x 10 10 years 13.7 billion years (1 billion = 1 x10 9 ) Gravity and expansion Gravity is an attractive force between two objects with mass You would expect, therefore for gravity to slow down the rate of expansion and pull the galaxies closer together This is not what is observed however. Redshift tells us that the galaxies are moving further apart Other observations lead to the conclusion that the rate of expansion is in fact increasing There must be other factors at work which overcome the force of gravity Fate of the Universe No one is sure about the eventual fate of the Universe It could o Continue expanding forever o The expansion rate could decrease and be left in a steady-state o The expansion could reverse and the galaxies could start moving closer together What is known is that the eventual fate of the Universe will depend on the overall mass 46

Orbital speed of stars and mass of galaxies A star s orbital speed is determined almost entirely by the gravitational pull of matter inside its orbit This means that the mass of a galaxy can be determined by the orbital speeds of stars within it Dark Matter Measurements of the mass of our galaxy and others lead to the conclusion that there is significant mass that cannot be detected This missing mass is called dark matter The majority of mass in the Universe is dark matter Dark Energy As stated above, the expansion rate of the Universe is increasing something is overcoming gravitational attraction The name given to whatever is responsible for the increasing expansion rate is dark energy 47

Key Area Big Bang Theory Red Amber Green 1. State that stellar objects emit radiation over a wide range of wavelengths. 2. State that each stellar object has a peak wavelength. 3. State the relationship between peak wavelength and temperature. 4. State that the temperature of a stellar object can be determined by its peak wavelength. 5. State that there is a relationship between temperature and irradiance (light power per unit area) the hotter the star, the higher the light power per unit area. 6. State that cosmic microwave background radiation (CMB) provides evidence for the Big Bang. 7. State that the Universe cools down as it expands. 8. Describe how the present temperature of the Universe can be determined. The Big Bang The Big Bang Theory postulates that the universe began with a single burst of energy. The early universe was small and incredibly hot. As the universe expanded it cooled and the energy condensed into matter which gradually formed atoms and then more and more complex structures. It took only moments for the first hydrogen and helium nuclei to form from protons and neutrons, but thousands of years for electrons to bind to them to form neutral atoms. The theory therefore predicts that the universe should now have a very cool temperature. If we can measure this temperature we can see if it accords with Big Bang theory. If we can understand stellar temperatures, it can help us know how to find the average temperature of the universe. 48

Temperature of Stellar Objects When you look into the night sky you will see the familiar sight of white pinpoints of light. If you look a little closer, however, you will see that many of them have a colour. The colour of a star tells us the surface temperature of that star, in the same way we can tell the temperature of a flame by its colour: red is a relatively cool flame and blue is very hot. Star Location Colour Temperature / K Betelgeuse Orion Red 3 500 Pollux Gemini Yellow (like Sun) 5 000 Rigel Brightest star in Orion Blue 12 130 This overall colour can be split into the spectrum of the star. Astronomers observe stars through filters to record their brightness in the different wavelengths of their spectrum. Stars emit radiation over a wide range of wavelengths. The graph below is called a thermal emission peak. This shows how the intensity of radiation produced (y-axis) from stars of different temperatures is related to the wavelength of light emitted from the star. Thermal emission peaks allow the temperature of stellar objects to be determined. Three key findings emerge from studying these peaks. 1. Stellar objects emit radiation over the complete range of electromagnetic spectrum wavelengths. 2. Each stellar object has a peak wavelength that depends on its temperature. 3. As the temperature of the star increases: a. the irradiance (light power per unit area) increases b. the peak wavelength decreases 49

Evidence for the Big Bang Cosmic Microwave Background (CMB) Radiation In 1948 it was suggested that if the Big Bang did happen then it would be the biggest single emission of energy in the universe and there should be a measurable peak wavelength associated with it. The universe has cooled considerably since the Big Bang. It was predicted to be at a current temperature of 2 7 K, with an associated peak wavelength in the microwave region. This radiation would be observable in every direction and spread uniformly throughout the universe the cosmic microwave background radiation (CMB). In 1965 the CMB was observed; o Two astronomers, Arno Penzias and Robert Wilson, actually discovered the radiation left over from the Big Bang completely by accident. o There was a source of excess noise in their radio telescope that they could not identify and researchers from another laboratory, who were actively looking for the CMB, recognised it. Those astronomers shared the 1978 Nobel Prize for their accidental discovery of the Big Bang s signature. In 1989 a satellite was launched to study the background radiation- the Cosmic Background Explorer [COBE]. COBE recorded the background radiation of the universe, in all directions, over three years. In 1992 it was announced that COBE had managed to measure fluctuations in the background radiation. This was further evidence to support the Big Bang theory. The uniformity of the CMB is also a very strong indication that this event was the point of origin in the universe. If it took place after this time, somewhere inside an already expanding universe, the radiation would not be spread as evenly in all directions. COBE Map of Cosmic Microwave Background Radiation showing uniformity across all directions 50