IB Physics HL Study Guide

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Factors that affect how an object absorbs, emits (radiates), and reflects EM radiation incident on them:

1) Nature of the surface: material, shape, texture, etc.

2) Color:

a) Light-colored or silvery objects: absorb little energy, reflect most energy

b) Dark objects: absorb most energy, reflect little energy

When the object is in thermal equilibrium with its surroundings,

energy absorbed = energy radiated 

Pin = Pout

Iin = Iout

An object that acts as a “black-body” will . . . absorb all incoming radiation, not reflect any, then radiate all of it. %ODFNERG\UDGLDWLRQ: radiation emitted by a “perfect” emitter When heated, a low-pressure gas will . . .emit a discrete spectrum

When heated, a solid will . . . emit a continuous spectrum (PLVVLRQ6SHFWUDIRU%ODFN%RGLHV 1. Not all wavelengths of light will be emitted with equal intensity. 2. Emitted wavelength with highest intensity (Ȝmax ) is related to . . . temperature.

3. Area under curve is proportional to . . . total power radiated by body 4. As body heats up, Ȝmax . . . decreases and total power . . . increases 1 

KXPDQFDXVHG@JUHHQKRXVHJDVFRQFHQWUDWLRQV.” (the HQKDQFHGJUHHQKRXVHHIIHFW) (QKDQFHG$QWKURSRJHQLF *UHHQKRXVH(IIHFW± Human activities have released extra carbon dioxide into the atmosphere, thereby enhancing or amplifying the greenhouse effect.

Major cause: the burning/combustion of fossil fuels Possible effect: rise in mean sea-level Outcome: climate change and global warming

11 

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 7KH.HHOLQJ&XUYH: Named after American climate scientist Charles David Keeling, this tracks changes in the concentration of carbon dioxide (CO2) in Earth’s atmosphere at a research station on Mauna Loa in Hawaii. Although these concentrations experience small seasonal fluctuations, the overall trend shows that CO2 is increasing in the atmosphere.

 ,QWHUQDWLRQDO,FH&RUH5HVHDUFK Between 1987 and 1998, several ice cores were drilled at the Russian Antarctic base at Vostok, the deepest being more than 3600 meters below the surface. Ice core data are unique: every year the ice thaws and then freezes again, forming a new layer. Each layer traps a small quantity of the ambient air, and radioactive isotopic analysis of this trapped air can determine mean temperature variations from the current mean value and carbon dioxide concentrations. The depths of the cores obtained at Vostok means that a data record going back more than 420,000 years has been built up through painstaking analysis.

Inspect the graphical representation of the ice core data and draw a conclusion. There is a correlation between Antarctic temperature and atmospheric concentrations of CO2 0HFKDQLVPVWKDWPD\LQFUHDVHWKHUDWHRIJOREDOZDUPLQJ

1. Global warming reduces ice and snow cover, which in turn reduces the albedo. This will result in an increase in the overall rate of heat absorption. 2. Temperature increase reduces the solubility of CO2 in the sea and increases atmospheric concentrations. 3. Continued global warming will increase both evaporation and the atmosphere’s ability to hold water vapor. Water vapor is a greenhouse gas. 4. The vast stretch of permanently frozen subsoil (permafrost) that stretches across the extreme northern latitudes of North America, Europe, and Asia, also known as tundra, are thawing. This releases a significant amount of trapped CO2. 5. Deforestation results in the release of more CO2 into the atmosphere due to “slash-and-burn” clearing techniques, as well as reduces the number of trees available to provide “carbon fixation.”

Smoldering remains of a plot of deforested land in the Amazon rainforest of Brazil. Annually, it is estimated that net global deforestation accounts for about 2 gigatons of carbon emissions to the atmosphere.

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Generally, as the temperature of a liquid rises, it expands. If this is applied to water, then as the average temperature of the oceans increases, they will expand and the mean sea-level will rise. This has already been happening over the last 100 years as the sea level has risen by 20 cm. This has had an effect on island nations and low-lying coastal areas that have become flooded. &RHIILFLHQWRI9ROXPH([SDQVLRQȕ ±fractional change in volume per degree change in temperature

Formula:

'9 E  9R '7 '9

Units:

1 .

E 9R '7

1. The coefficient of volume expansion for water near 20o C is 2 x 10-4 K-1. If a lake is 1 km

deep, how much deeper will it become if it heats up by 20o C? 0.4 m

Precise predictions regarding the rise in sea-levels are hard to make for such reasons as: D  $QRPDORXVH[SDQVLRQRIZDWHU: Unlike many liquids, water does not expand uniformly. From 00C to 40C, it actually contracts and then from 40C upwards it expands. Trying to calculate what happens as different bodies of water expand and contract is very difficult, but most models predict some rise in sea level. E  0HOWLQJRILFH: Floating ice, such as 

the Arctic ice at the North Pole, 

displaces its own mass of water so 

when it melts it makes no difference. 

But melting of the ice caps and 

glaciers that cover land, such as in 

Greenland and mountainous regions 

throughout the world, causes water to 

run off into the sea and this makes the 

sea level rise. 

Glaciers on land melting: raise sea level Sea ice glaciers melting: don’t raise sea level 13 



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3. Greater efficiency of power production. To produce the same amount of power would require less fuel, resulting in reduced CO2 emissions.

4. Replacing the use of coal and oil with natural gas. Gas-fired power stations are more efficient (50%) that oil and coal (30%) and produce less CO2.

5. Use of combined heating and power systems (CHP). Using the excess heat from a power station to heat homes would result in more efficient use of fuel.

6. Increased use of renewable energy sources and nuclear power. Replacing fossil fuel burning power stations with alternative forms such as wave power, solar power, and wind power would reduce CO2 emissions.

1. Use of hybrid vehicles Cars that run on electricity or a combination of electricity and gasoline will reduce CO2 emissions.

2. Carbon dioxide capture and storage (carbon fixation) A different way of reducing greenhouse gases is to remove CO2 from waste gases of power stations and store it underground.

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1. ,QWHUJRYHUQPHQWDO3DQHORQ&OLPDWH&KDQJH,3&&  Established in 1988 by the World Meteorological Organization and the United Nations Environment Programme, its mission is not to carry out scientific research. Hundreds of governmental scientific representatives from more than 100 countries regularly assess the up-to-date evidence from international research into global warming and human induced climate change.

2. .\RWR3URWRFRO This is an amendment to the United Nations Framework Convention on Climate Change. In 1997, the Kyoto Protocol was open for signature. Countries ratifying the treaty committed to reduce their greenhouse gases by given percentages. Although over 177 countries have ratified the protocol by 2007, some significant industrialized nations have not signed, including the United States and Australia. Some other countries such as India and China, which have ratified the protocol, are not currently required to reduce their carbon emissions.

3. $VLD3DFLILF3DUWQHUVKLSRI&OHDQ'HYHORSPHQWDQG&OLPDWH$33&'&  This is a non-treaty agreement between 6 nations that account for 50% of the greenhouse emissions (Australia, China, India, Japan, Republic of Korea, and the United States.) The countries involved agreed to cooperate on the development and transfer of technology with the aim of reducing greenhouse emissions.

14 

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Example:

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0= 1= 2= 3= 4= 5= 6= 7= 8= 9= 10 = 11 = 12 = 13 = 14 = 15 = 16 = 17 = 18 = 19 = 20 =

452 = 4 x 100 + 5 x 10 + 2 x 1

Symbols: 0-9 Place Values: multiples of ten

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Examples: 10011 = 1 x 16 + 0 x 8 + 0 x 4 + 1 x 2 + 1 x 1 = 19

Symbols: 0 and 1 Place Values: multiples of two

Least-significant bit (LSB) Most-significant bit (MSB)

'LJLWDOVLJQDO: potential difference is either High (1) or Low (0) $QDORJXHVLJQDO: potential difference varies continuously with time

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Photocopying

Text or pictures

Optics and electrostatics used to fix powder to paper

Analogue

LPs (vinyl)

Music or speech

Sound variations stored as grooves in vinyl

Analogue

Cassette tapes

Music or speech

Sound variations stored in magnetic fields on tape

Analogue

Floppy disks Hard disks

All forms

Bits stored as variations in magnetic fields on disk

Digital

CD, DVD

All forms

Bits stored as series of optical bumps to be read by laser

Digital 1 

0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1110 1111 10000 10001 10010 10011 10100

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A CD is a fairly simple piece of plastic, about four one-hundredths (4/100) of an inch (1.2 mm) thick. Most of a CD consists of an injection-molded piece of clear polycarbonate plastic. During manufacturing, this plastic is impressed with microscopic bumps arranged as a single, continuous, extremely long spiral track of data. Once the clear piece of polycarbonate is formed, a thin, reflective aluminum layer is sputtered onto the disc, covering the bumps. Then a thin acrylic layer is sprayed over the aluminum to protect it. The label is then printed onto the acrylic. A cross section of a complete CD (not to scale) looks like this:

The elongated bumps that make up the track are each 0.5 microns wide, a minimum of 0.83 microns long and 125 nanometers high. (A nanometer is a billionth of a meter.) Looking through the polycarbonate layer at the bumps, they look something like this: You will often read about "pits" on a CD instead of bumps. They appear as pits on the aluminum side, but on the side the laser reads from, they are bumps. The incredibly small dimensions of the bumps make the spiral track on a CD extremely long. If you could lift the data track off a CD and stretch it out into a straight line, it would be 0.5 microns wide and almost 3.5 miles (5 km) long.

Reflection from bump or pit: Constructive interference Interpreted as 0

Reflection from edge of pit: Destructive interference Interpreted as 1 Condition for destructive interference: height of bump/depth of pit = Ȝ/4 So path difference between two light beams is Ȝ/2

Example: What wavelength of laser light should be used to read the data shown encoded at right? 600 nm

2 

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Output can be virtually indistinguishable from input

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Use of laser ensures that each retrieval is virtually identical since light does not damage surface Very high speed – different section can be accessed randomly

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Miniaturization techniques ensure that large quantities of data can be stored in a small device (eg. flash drives) Easily achieved with little corruption of data (eg. Photoshop)

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$QDORJXH Output can be virtually indistinguishable from input but is more liable to damage or corruption (eg. scratches on LPs) Process of retrieval often affects quality of future retrievals (eg. needle may scratch LP) Slow retrieval speed – data needs to be retrieved in sequential order Storage devices usually take up much more space All manipulation increases possibility of data corruption

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x Information that is potentially problematic can easily be shared (eg. terrorism, crime) x Issues concerning ownership of electronic data (eg. piracy) x Privacy concerns x Use in documenting abuses of human rights x Unequal access to the Internet x Control of information and opinions

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x Quicker access to information needed to make economic decisions (eg. price comparisons) x Rise of new businesses and decline of older ones

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x Reduction in use of paper and other materials used traditionally to stored information x Recycling of electronic junk is problematic due to more dangerous materials used in their manufacture

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&KDUJH&RXSOHG'HYLFH&&'  a silicon chip divided into small area called pixels. Each pixel can be considered to behave as a capacitor (a device that stores charge). The CCD is used to electronically record an image focused onto its surface. ,PDJHFDSWXUH Incident light causes charge to build up within each pixel due to the photoelectric effect. An electrode then measures the potential difference developed across each pixel and converts it into a digital signal. The position of the pixel is also stored. ,PDJHUHWULHYDOSince both the location and potential difference of each pixel are recorded, all the information needed to store and reconstruct the image is saved. Each p.d. can be converted to a digital signal and the digital signals can be converted to an image. The intensity of the light at each pixel, and thus the image, can be reconstructed.

&DSDFLWDQFH: the ratio of charge to potential difference

Formula:

C = Q/V

Units: C/V or farad (F)

Type: Scalar

1. The potential difference measured across a 100 pF pixel is 25 mV. Determine the charge and number of electrons stored in the pixel.

4XDQWXP(IILFLHQF\ ratio of the number of photoelectrons emitted to the number of

photons incident on the pixel

2. If the quantum efficiency of the pixel described above is 90%, how many photons were  incident on it?

4 

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0DJQLILFDWLRQ±ratio of the length of the image on the CCD to the length of the object

3. A digital camera is used to photograph an object that is 3.0 x 10-1 m2 in area. The image that is focused onto the CCD is 4.5 x 10-3 m2. What is the magnification of the camera?

5HVROXWLRQ: the ability to distinguish between two sources of light Applied to CCD: two points on an object may be just resolved on a CCD if the images 

of the points are at least two pixels apart

4. The CCD of a digital camera has a square image collection area that measures 25 mm on each side and a “resolution” of 5.0 megapixels. An object that is photographed by the camera has an area of 4.6 x 10-3 m2. The image formed on the CCD has an area of 1.0 x 10-4 m2. a) Calculate the magnification.

b) Estimate the length of a pixel on the CCD.

c) Two small dots on the object are separated by a distance of 0.20 mm. Deduce whether the images of the dots will be resolved.

5 

IB 12 

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The greater the quantum efficiency, the greater the sensitivity of the CCD.

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A greater magnification means that more pixels are used for a given section of the image. The image will be more detailed.

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The greater the resolution, the greater the clarity of the image and the amount of detail recorded.

Some advantages of using CCDs compared with the use of film: a) Each photo does not require film and is thus cheaper and uses less resources/produces less waste. b) Traditional film has a quantum efficiency of less than 10% while CCDs can have quantum efficiencies of over 90%. This means that very faint images can be photographed with CCDs. c) The image is digital and can be stored and edited more easily. d) Images can be viewed immediately with no processing time delay.  e) Storage, archiving, and retrieving a large number of photos is easy and efficient.  CCDs are used for image capturing in a large range of the electromagnetic spectrum (not just visible light).

'LJLWDO FDPHUDV

Very convenient to take and share photographs, but image quality can be less than that of traditional film unless the camera is of high quality (more expensive).

Digitized images are usually better quality than analogue images stored on magnetic videotape and are easier to store and transport. It is possible to 9LGHR continuously record video without interruption during playback. Searches are FDPHUDV faster and easier to perform. Digital storage is fast and utilizes re-usable media, an advantage for security cameras. Sensitivity of CCDs is better than traditional film and allows for detailed analysis over a range of frequencies. CCDs also allow for remote operation of 7HOHVFRSHV telescopes, both ground-based and in orbit, like the Hubble space telescope. 0HGLFDO ;UD\ LPDJLQJ

Digital X-rays have better contrast and can be processed, allowing for enhancements and detailed study. Information can be quickly shared between hospitals and more easily stored and retrieved.

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In the electric circuit shown below, energy is transferred from the battery to the light bulb by charges that move through a conducting wire because of a potential difference set up in the wire by the battery. The circuit shown contains a typical 9-volt battery. a) What is the emf of the circuit?

b) How much energy does one coulomb of charge carry around the circuit?

Schematic c) How much energy do two coulombs of charge carry around the circuit?

d) How much energy does each coulomb of charge have at point B?

e) How much energy does each coulomb of charge have at point C?

f) What is VB?

What is VC? 

mark potentials at each spot

g) What is ǻVBC?

What is ǻVCD?

What is ǻVDA? 

(OHFWULF&XUUHQW Formula: I = ǻq/ǻt

Units: A (ampere) = C/s

Type: Scalar

Unofficial definition: rate of flow of electric charge 2IILFLDO'HILQLWLRQRI2QH$PSHUH$ RIFXUUHQW±DIXQGDPHQWDOXQLW One ampere is the amount of current flowing in each of two infinitely-long parallel wires of negligible cross-sectional area separated by a distance of one meter in a vacuum that results in a force of exactly 2 x 10-7 N per meter of length of each wire. 6KRUWIRUP– Current is defined in terms of the force per unit length between parallel current-carrying conductors.

Closed circuit: complete pathway for current

Open circuit: incomplete pathway for current – break in circuit – infinite resistance

Short circuit: circuit with little to no resistance – extremely high current – overheating

sketch sketch

1 

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5HVLVWDQFH: ratio of potential difference applied across a piece of material to the current through the material

Formula:

For a wire conductor:

Formula:

A short fat cold wire is the best conductor

R = ȡL/A

Units:

R=V/I

Type:

ohm (ȍ) = V/A

scalar

A long hot skinny wire has the most resistance

Unit: W = J/s

Power: energy per unit time Mechanical Power:

Electrical Power: P = E/t = (ǻq V)/ǻt

Alternate Formulas: Substitute V = IR

P=IV

P = I (IR) = I2 R

P = W/t = F s/t = F v

Type: scalar

P = (V/R)V = V2/ R

Meters in a circuit

Ammeter: measures current

Placement: Must be placed in series to allow current to flow through it Circuit must be broken to insert ammeter

Ideal ammeter: has zero resistance so it will not affect current flowing through it

Schematic diagram

Voltmeter: measures potential difference

Placement: Must be placed in parallel to measure potential difference between two points circuit does not to be broken

Ideal voltmeter: has infinite resistance so it will not allow any current to flow through it and disrupt circuit

2

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IB 12

6HULHV&LUFXLW

3DUDOOHO&LUFXLW

one

More than one

Current

Same everywhere – same for all devices

Current splits – shared among devices

Potential Difference (Voltage)

Voltage shared among devices – voltage splits

Same for all devices

Overall resistance

high

low

Power

low

high

Number of pathways for current

Formulas:

VT = V1 + V2 + …

VT = V1 = V2 = . . .

IT = I1 = I 2 = …

IT = I1 + I 2 + . . .

RT = R1 + R2 + …

1/RT = 1/R1 + 1/R2 + . . .

PT = P1 + P2 + . . .

PT = P1 + P2 + . . .

6HULHV &LUFXLWV

3DUDOOHO &LUFXLWV

Voltage Ratio

Current Ratio

Power Ratio

Power Ratio

3

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IB 12 

Determine the current through and the voltage drop across each resistor in each circuit below.

1.

2.

3.

4.

3RWHQWLDO'LYLGHU Resistors in series act as a “potential (voltage) divider.” They split the potential of the source between them. 5. A 20ȍdevice requires 40 V to operate properly but no 40 V source is available. In each case below, determine the value of added resistor R1 that will reduce the voltage of the source to the necessary 40V for device R2.

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6. A mini light bulb is rated for 0.60 W at 200 mA and is placed in series with a variable resistor. Only a 9.0 volt battery is available to power it. To what value should the variable resistor be set to power the bulb correctly?

Bulb needs only 3 V Bulb has resistance of 15 ȍ at rated power Added resistance should be 30 ohms

4 

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1. /LJKW'HSHQGHQW5HVLVWRU/'5 or /LJKW6HQVRU: A photo-conductive cell made of semiconducting material whose resistance decreases as the intensity of the incident light increases. Describe how the LDR activates the light switch.

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As ambient light decreases, resistance of LDR increases Potential difference across LDR increases Switch needs minimum PD to turn on When light intensity drops to desired level, PD is high enough to turn on switch

2. 1HJDWLYH7HPSHUDWXUH&RHIILFLHQW17& 7KHUPLVWRU or 7HPSHUDWXUH6HQVRU: A sensor made of semiconducting material whose resistance decreases as its temperature increases. Describe how the NTC thermistor activates the fire alarm.

)LUHDODUP

As external temperature increases, resistance of NTC decreases Potential difference across R2 increases Switch needs minimum PD to turn on When temperature increases to desired level, PD is high enough to turn on switch

3. 6WUDLQ*DXJH or)RUFH6HQVRU: A long thin metal wire whose resistance increases as it is stretched since it becomes longer and thinner. Describe how the strain gauge can measure the strain put on a 

section of an airplane body. 

As strain increases, resistance of strain gauge increases 

Potential difference across R2 decreases 

Voltmeter can read change in voltage which can be used

to determine amount of strain on part

5 

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IB 12

1. Determine the current through and the voltage drop across each resistor.

2. The battery has an emf of 12 V and negligible internal resistance and the voltmeter has an internal resistance of 20 kȍ. Determine the reading on the voltmeter.

3. A cell with negligible internal resistance is connected to three resistors as shown. Compare the currents in each part of the circuit.

6

IB 12 

4. Determine the current through and the voltage drop across each resistor.

5. A battery with emf ( and negligible internal resistance is connected in a circuit with three identical light bulbs. a) Determine the reading on the voltmeter when the switch is open and when it is closed. b) State what effect closing the switch has on the current through each bulb and the brightness of each bulb.

7

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IB 12

5HVLVWDQFH: ratio of potential difference applied across a piece of material to the current through the material

Formula: R=V/I

2KP¶V/DZ: for a conductor at constant temperature, the current flowing through it is proportional to the potential difference across it

Relationship:

VĮI

2KPLF'HYLFH a device that obeys Ohm’s law for a wide range of potential differences

Meaning: a device with constant resistance

Example: resistor

1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a resistor. Resistance: 

a) R = V/I at any point 

b) related to slope of graph 

(Reciprocal = resistance) 

2. A resistor is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the resistor.

R = V/I 

R = 7.5 ȍ

b) If the voltage is doubled, what is the new current? V = IR for resistor 

Resistance is constant so double current

8 

1RQ2KPLF'HYLFH a device that does not obey Ohm’s law

Meaning: resistance is not constant

IB 12 

Example: filament lamp

1. On the axes at right, sketch the ,9FKDUDFWHULVWLFV for a filament lamp.

Resistance:

a) R = V/I at any point 

b) as current increases, wire filament heats up and resistance increases

c) Resistance is NOT related to the slope

d) except in initial linear region

2. A flashlight bulb is connected to two 1.5 volt cells and has 0.40 ampere of current flowing through it. a) Calculate the resistance of the bulb.

R = V/I 

R = 7.5 ȍ

b) If the voltage is doubled, what is the new current? V = IR for bulb but resistance is not 7.5 ohms any

more – R increases with T so less than double current

3. Discuss how the resistance varies with increasing potential difference for devices X, Y, and Z.

X: resistance increases - ratio V/I increases Y: resistance is constant – ratio V/I is constant Z: resistance decreases – ratio V/I decreases

9 

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3RWHQWLRPHWHU a type of variable resistor with three contact points Common use: as a potential divider to measure the I-V characteristics of a device The schematic shows how a potentiometer can be used as a potential divider to measure the I-V characteristics of a filament lamp. It is placed in parallel with the lamp and the slider (center contact point) effectively splits the potentiometer into two separate resistors AB and BC. By moving the slider, the ratio of the voltage drops across the resistors AB and BC is varied. Redraw the schematic with an ammeter and a voltmeter correctly placed to measure the I-V characteristics of the filament lamp.

Comment on the circuit characteristics as the slider is moved from A to B to C. Slider at A:

Slider at B:

Slider at C:

10 

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A 6 volt battery is connected to a variable resistor and the current in the circuit and potential difference across the terminals of the battery are measured over a wide range of values of the resistor. The results are shown in the table. 5HVLVWDQFHŸ 

3UHGLFWHG &XUUHQW$ 

$FWXDO &XUUHQW$ 

9ROWDJHDFURVV EDWWHU\9 

2000 200 20 2 0.2 0.02 0.002 0.0002

0.003 0.03 0.3 3 30 300 3000 30000

0.003 0.03 0.29 2.4 8.8 11.5 12.0 12.0

6.00 5.99 5.85 4.80 1.71 0.23 0.02 0.00

Why does the current seem to be limited to a maximum of 12.0 amperes and why does the voltage across the battery not remain constant at 6.0 volts? The battery has some internal resistance. As the 

external resistance decreases, more and more of the 

energy supplied by the battery is used up inside the 

battery. 

(OHFWURPRWLYHIRUFHHPI  total energy per unit charge supplied by the battery

Symbol: İRU( Units: V = J/C

7HUPLQDO9ROWDJH9WHUP  potential difference across the terminals of the battery

Ideal Behavior: Vterm always equals emf since no internal resistance Real Behavior: 1) Think of battery as internal E and tiny internal resistor r

2) Vterm only equals the emf when no current is flowing 3) E is split between R and r 4) When R>>r, Vterm § emf

5) As R decreases, Vr increases and VR decreases 11 

IB 12 5HODWLRQVKLSEHWZHHQHPIDQGWHUPLQDOYROWDJH Treat internal resistance as a series resistor

İ = I RT 

İ = I (R + r) 

İ = IR + Ir 

Note that in the absence of internal resistance, İ = Vterm

1. A resistor is connected to a 12 V source and a switch. With the switch open, a voltmeter reads the potential difference across the battery as 12 V yet with the switch closed, the voltmeter reads only 9.6 V and an ammeter reads 0.40 A for the current through the resistor. Sketch an appropriate circuit diagram and calculate the internal resistance of the source.

2. Discuss the expected I-V characteristics for this battery and how they can be experimentally determined. R can be adjusted from 0 to its max value 

A graph of Vterm vs. I can be drawn 

Specific equation of graph can be compared to math model to derive

internal resistance Emf = Vterm + Ir Vterm = -Ir + emf

so slope = -r and y-intercept = emf

12 

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Direction of magnetic field lines: the direction that the North pole of a small test compass would point if placed in the field (N to S)

Magnetic Field around a Bar Magnet

What is the cause of magnetic fields? Moving electric charges Therefore: current in a wire will produce a magnetic field

The Right Hand Rule for the Magnetic Field around a Wire

7KXPEdirection of conventional current )LQJHUWLSV direction of magnetic field – tangent to circle

a) head-on view

b) side view

Draw concentric circles with increasing spacing and arrows in circular fashion.

Alternate Right Hand Rule for Loops )LQJHUWLSV direction of current

Component fields

Draw concentric circles around wire.

Resultant field

c) side view

Draw dots and crosses

Your turn

7KXPEpoints North Note that a wire loop acts like a:

bar

magnet

Solenoid: coil of wire – many loops

Note that a solenoid bar magnet acts like a:

Draw the magnetic field around this solenoid. Use alternate RHR to find North.

1

0DJQHWLF)RUFHRQD:LUH If a wire with current flowing through it is placed in an external magnetic field, it will experience a force. Why? The Right Hand Rule for the Magnetic Force on a Current-Carrying Conductor in a Magnetic Field

IB 12 

Two magnetic fields – around wire and from external magnet – will either attract or repel

)ODW+DQG thumb and fingers at right angles )LQJHUV external B field – north to south 7KXPE current 3DOP force – “palm pushes” Maximum force occurs when: current is perpendicular to B field No force occurs when: current is parallel to B field

Use the right hand rule for forces to confirm the direction of the force in each case.

0DJQLWXGHRI WKHPDJQHWLF IRUFHRQDZLUH 'HILQLWLRQRIPDJQLWXGH RIPDJQHWLFILHOG  B = F / (IL sin ș)

Find the magnitude and direction of the force on the wire segment confined to the gap between the two magnets as shown when the switch is closed. The strength of the magnetic field in the gap is 1.9 T.

Magnetic field strength Magnetic field intensity Magnetic flux density

) %,/VLQș Where ș is angle between current and B field

Units: Tesla (T)

The ratio of the magnetic force on a wire to the product of the current in the wire, the length of the wire and the sine of the angle between the current and the magnetic field

0.62 N up

2 

0DJQHWLF)RUFHRQD0RYLQJ&KDUJHG3DUWLFOH Why is there a magnetic force on a charged particle as it moves through a magnetic field?

The Hand Rule for the Magnetic Force on a Charge moving in a Magnetic Field

IB 12 

Moving charged particle creates its own magnetic field – two magnetic fields interact

)ODW+DQG thumb and fingers at right angles

)LQJHUV external B field – north to south

5LJKW+DQGpositive charge

7KXPE velocity

/HIW+DQGnegative charge

3DOP force – “palm pushes”

Maximum force occurs when: velocity is perpendicular to B field No force occurs when: velocity is parallel to B field Find the direction of the magnetic force on each particle below as each enters the magnetic field shown.

0DJQLWXGHRIWKHPDJQHWLFIRUFH RQDPRYLQJFKDUJHGSDUWLFOH 'HILQLWLRQRIPDJQLWXGH RIPDJQHWLFILHOG 

) TY%VLQș Where ș is angle between v and B

The ratio of the force on a charged particle moving in a magnetic field to the product of the particle’s charge, velocity and sine of the angle between the direction of the magnetic field and velocity.

B = F / (qv sin ș)

A proton in a particle accelerator has a speed of 5.0 × 106 m/s. The proton encounters a magnetic field whose magnitude is 0.40T and whose direction makes an angle of T = 30.0° with respect to the proton's velocity. Find the magnitude of the magnetic force on the proton and the proton’s acceleration. How would these change if the particle was an electron?

a) 1.6 x 10-13 N b) 9.6 x 1013 m/s2 c) 1.8 x 1017 m/s2, same force but opposite direction 3

0RWLRQRID&KDUJHG3DUWLFOHLQD0DJQHWLF)LHOG

IB 12 

1. A charged particle will follow a circular path in a magnetic field since the magnetic force is always perpendicular to the velocity.

2. The magnetic force does no work on the particle since the magnetic force is always perpendicular to the motion.

3. The particle accelerates centripetally but maintains a constant speed since the magnetic force does no work on it.

5DGLXVRI&LUFXODU3DWK

a) Sketch the paths of a slow and a fast

moving proton at constant speed.

b) Sketch the path of a proton that is slowing

down and one that is speeding up.

Ȉ F = m ac FB = m v2 /r Q v B = m v2 /r r = mv/ qB

c) How would the radius of the path change if the particle were an alpha particle?

&RPSDULQJ(OHFWULFDQG0DJQHWLF)LHOGVDQG)RUFHV

(OHFWULF)LHOG

0DJQHWLF)LHOG

draw paths for stationary, parallel and perp charges

4

(OHFWULF)LHOGVDQG0DJQHWLF)LHOGV 1. A proton is released from rest near the positive plate and leaves through a small hole in the negative plate where it enters a region of constant magnetic field of magnitude 0.10T. The electric potential difference between the plates is 2100 V.

IB 12 

a) Describe the motion of the proton while in the electric field constant acceleration in a straight line

b) Describe the motion of the proton while in the magnetic field constant acceleration and constant speed – circular path

c) Find the speed of the proton as it enters the magnetic field.

Use conserv. Of energy 

a) 6.3 x 105 m/s

d) Find the radius of the circular path of the proton in the magnetic field.

b) 6.6 x 10-2 m

2. A 9HORFLW\6HOHFWRU is a device for measuring the velocity of a charged particle. The device operates by applying electric and magnetic forces to the particle in such a way that these forces balance. a) Determine the magnitude and direction of an electric field that will apply 

and electric force to balance the magnetic force on the proton. 

perp to v and B – from bottom to top of page ȈF = 0 

FB – Fe = 0 

FB = Fe

qvB = Eq

E = vB 

b) What is the resulting speed and trajectory of the proton? v = E/B in a straight line

5 

7KH0DVV6SHFWURPHWHU

IB 12 

A PDVVVSHFWURPHWHU is a device used to measure the masses of isotopes. Isotopes of the same element have the same charge and chemical properties so they cannot be separated by using chemical reactions but have different masses and so can be separated by a magnetic field. A common type of mass spectrometer is known as the %DLQEULGJHPDVVVSHFWURPHWHU and its main parts are shown below.

Ion Source: source of charged isotopes – same charge – different mass

Velocity selector: so all ions have the same speed

Magnetic deflection chamber: radius is proportional to mass

1. A singly charged ion with mass 2.18 x 10-26 kg moves without deflection through a region of crossed magnetic and electric fields then is injected into a region containing only a magnetic field, as shown in the diagram, where it is deflected until it hits a photographic plate. The electric field between the plates of the velocity selector is 950 V/m and the magnetic field in both regions is 0.930 T. Determine the sign of the charge and calculate where the ion lands on the photographic plate.

Sign: could be either in velocity selector Only positive in deflection chamber 

velocity selector v = E/B 

v = 1.0 x 103 m/s

magnetic chamber 

r = mv/qB 

r = 1.5 x 10-4 m d = 3.0 x 10-4 m

2. A hydrogen atom and a deuterium atom (an isotope of hydrogen) move out of the velocity selector and into the region of a constant 0.10 T magnetic field at point S, as shown below. Each has a speed of 1.0 x 106 m/s.

Calculate where they each hit the photographic plate at P. 

Hydrogen = 0.20 m Deuterium = 0.41 m

6 

(OHFWURPDJQHWLF,QGXFWLRQ

IB 12 

In 1819, Hans Christian Oersted discovered that a magnetic compass experiences a force in the vicinity of an electric current – that is, that electric currents produce magnetic fields. Because nature is often symmetric, this led many scientists to believe that magnetic fields could also produce electric currents, a concept known as HOHFWURPDJQHWLFLQGXFWLRQ.

Why does moving a wire through a magnetic field induce a current in the wire?

Free electrons in the wire are charged particles moving through a magnetic field so there is a qvB force on them causing them to move resulting in a current.

'HULYDWLRQRIIRUPXODIRU(0)LQGXFHGLQDPRYLQJZLUH A straight conductor is moved at constant velocity perpendicular to a uniform magnetic field. 1. Electrons in the moving conductor experience a downward magnetic force and migrate to the lower end of the conductor, leaving a net positive charge at the upper end.

FB = qvB

2. As a result of this charge separation, an electric field is built up in the conductor.

E

3. Charge builds up until the downward magnetic force is balanced by the upward electric force due to the electric field. At this point, the charges stop flowing and are in equilibrium.

FB = Fe qvB = Eq E = vB

4. Because of this charge separation, a potential difference is set up across the conductor.

LHR for electrons to show direction of force

ǻV = E d = E Ɛ ǻV= vB Ɛ İ %ƐY

If the conductor is connected to a complete circuit, the induced emf will produce an induced current. … is equivalent to …

$PRXQWRI&XUUHQW

'LUHFWLRQRI&XUUHQW

The amount of induced current in the circuit is given by

The direction of the induced emf and induced current can be found from the right hand rule for forces to find the force on a positive charge in the conductor.

İ %ƐY İ=IR I = BƐv/R

7 

, IB 12

7ZR2SSRVLQJ)RUFHV ,

Y

3DOPSXVKHVFXUUHQWXS )% TY% An applied force (Fapp) in the direction of the velocity induces an emf which causes current to be pushed upwards. 

The magnetic force acts to oppose the applied force, like drag or friction.

)%

3DOPSXVKHVEDUEDFN )% %,O

At a constant speed, Fapp = FB = BIƐ

The induced current now generates a magnetic field around the moving bar that causes a magnetic force (FB) on itself. 

Suppose a rod is moving at a constant speed of 5.0 m/s in a direction perpendicular to a 0.80-T magnetic field as shown. The rod has a length of 1.6m and negligible electrical resistance. The rails also have negligible resistance. The light bulb, however, has a resistance of 96 :. Find: a) the emf produced by the motion of the rod 6.4 V

(b) the magnitude and direction of the induced current in the circuit

e) How much external force is applied to keep the rod moving at this constant speed? 0.086 N

0.067 A CCW

c) the electrical power delivered to the bulb 0.43 W

f) How much work is done by the applied force in 60.0 seconds? 26 J

d) the energy used by the bulb in 60.0 s. 26 J

g) What happens to this work? Converted to electrical energy

8 

0DJQHWLF)OX[

IB 12 

0DJQHWLF)OX[

6\PERO: ĭ

Number of field lines

8QLWVWeber (Wb)

)RUPXOD: B = ĭ/ A

0DJQHWLF)OX['HQVLW\ ILHOGVWUHQJWKLQWHQVLW\ 

or

ĭ=BA

number of field lines per unit area

= T m2

6\PERO: B 8QLWV Wb/m2 = T (tesla)

$QJOH'HSHQGHQFHRI)OX[ What is the amount of magnetic flux if the field lines are not perpendicular to the cross-sectional area? Only the perpendicular component of the magnetic field contributes to the magnetic flux. 1RUPDOOLQH: line perpendicular to plane of cross-sectional area

)RUPXOD: ĭ = (B cos ș) A = B A cos ș $QJOH: ș = angle between normal line and field lines

0DJQHWLF)OX[ĭ  product of the magnetic field strength and a cross-sectional area and the cosine of the angle between the magnetic field and the QRUPDO to the area

)RUPXODĭ %$FRVș

8QLWV7P

0DJQHWLFIOX[OLQNDJHPDJQHWLFIOX[OLQNLQJDFRLO product of magnetic flux through a coil of wire and the number of turns of the wire )RUPXOD1ĭ 1%$FRVș

8QLWV7P

1. A single loop of wire whose cross-sectional area is 0.50 m2 is located in a 0.20 T magnetic field as shown. Calculate the flux through the loop in each case.

a) 0.10 T m2

b) 0.050 T m2

c) 0 2. If the coil of wire in the above example consisted of 50 turns of the wire, calculate the amount of flux linking the coil in each case.

a) 5.0 Tm2

b) 2.5 T m2 c) 0 9 

(0),QGXFHGE\D7LPH&KDQJLQJ)OX[

Moving a magnet towards a coil will increase the magnetic flux linking the coil and will induce an emf and a current in a certain direction.

Holding the magnet stationary will not change the amount of magnetic flux linking the coil and so will not induce an emf or current.

IB 12 

Moving the magnet away from the coil will decrease the magnetic flux linking the coil and will induce an emf and a current in the opposite direction.

Methods of inducing an EMF by a time-changing flux

2. Move magnet or coil

2. Rotate coil

1. Vary magnetic field

)DUDGD\¶V/DZ an induced emf is proportional to the rate of change of the flux linkage

)RUPXOD İ = - N (ǻĭ/ǻt) 1. A coil of area 0.030 m2 with 300 turns of wire rotates as shown in 0.10 second in a magnetic field of constant 0.25 T strength. a) What is the magnitude of the induced emf? 11.3 V

b) What is the magnitude of the induced emf if the coil were stationary at 00 but the field strength changed from 0.25 T to 0.60 T in 0.10 second? 22.5 V

10 

IB 12 

2. A 50 turn coil of wire of area 0.20 m2 is perpendicular to a magnetic field that varies with time as shown by the graph. a) Determine the emf induced in the coil during each time interval. 3 V, 0 V, -1.5 V

b) Sketch a graph of the induced emf vs. time. Emf = - derivative of flux

/HQ]¶V/DZ±)LQGLQJWKH'LUHFWLRQRIWKH,QGXFHGHPI /HQ]¶V/DZ  The direction of an induced emf is such that it produces a magnetic field whose flux opposes the flux change that induced it.

(An emf will be induced so as to keep the net flux constant.)

a) Original flux change – an increasing flux induces an emf and current.

b) Induced flux – opposes increasing flux by pointing in opposite direction thus current is in direction shown.

1. If the magnetic field linking this coil is decreasing with time, in which direction is the induced current?

c) Result - two magnetic fields acting to keep net flux constant.

2. The diagrams show a conducting ring that is placed in a uniform magnetic field. Deduce the direction of the induced current in each case if there is

(a) an increasing B field

(b) a decreasing B field

11

IB 12 3. If the current in the wire is increasing, in which direction will there be an induced current in the rectangular wire loop?

4. If the wire loop moves away from a steady current in the straight wire, in which direction will there be an induced current in the loop?

5. A conducting loop moves at a constant speed into and through a uniform magnetic field as shown in the diagram. Indicate the direction of the induced current. Graph the flux through the loop and the induced emf as a function of time.

6. If a clockwise current through the primary coil is increasing with time, what effect will this have on the secondary coil?

7. Determine the direction of the current in the solenoid in each case.

8. Determine the direction(s) of the induced current as the magnet falls through the loop.

12 

$OWHUQDWLQJ&XUUHQW*HQHUDWRUV

IB 12 

Basic Operation: 1. coil of wire is turned by mechanical means in an external magnetic field

2. emf and current are induced in coil as coil cut flux lines

3. current varies in magnitude and direction as flux linkage changes – current and emf variations are sinusoidal 4. brushes and rings maintain contact with external circuit without getting tangled

5RWDWLRQRID&RLOLQD8QLIRUP0DJQHWLF)LHOGLQGXFHVDQ(0)

As the coil rotates, the flux linking it changes

3RVLWLRQ

0D[LPXP(0)DQG&XUUHQW 1. sides of coil cut field lines perpendicularly

2. plane of coil is parallel to field lines 3. normal to coil is perpendicular to field lines (900)

0LQLPXP(0)DQG&XUUHQW

3RVLWLRQ

Mark when the coil is in positions 1 and 2.

Sketch the graph of the induced current.

1.

sides of coil do not cut field lines perpendicularly (move parallel to them)

2.

plane of coil is perpendicular to field lines

3.

normal to coil is parallel to field lines (00) Sketch a graph of the induced emf for a coil with:

twice the frequency of rotation.

half the frequency of rotation.

13 

$OWHUQDWLQJ&XUUHQW The output of an AC generator is an emf that varies sinusoidally with time.

IB 12 

V0 = peak/ maximum voltage I0 = peak/ maximum current

V = Vo sin Ȧt

(where Ȧ = 2ʌf)

I = V/R so I = (Vo/R) sin Ȧt = Io sin Ȧt

The power output of an AC generator

P=IV P = (Io sin Ȧt)(Vo sin Ȧt) P = Io Vo sin2 Ȧt

Maximum Power Pmax = Io Vo

Average Power Pav = ½ Io Vo = (Io/ rad 2)(Vo / rad2)

Pmax = 2 Pav

Root-Mean-Squared values (RMS):

RMS Values Irms = Io / rad 2 Vrms = Vo / rad 2

= Irms Vrms

The rms value of an alternating current (or voltage) is that value of the direct current (or voltage) that dissipates power in a resistor at the same rate.

1. In the USA, most household voltage is stated as “120 V at 60 Hz.” This is the root-mean-square voltage and the frequency of the AC voltage. Calculate the maximum voltage and mark Vo, Vrms, on the graph. Vo = 170 V

2. In Europe, the “mains electricity” is rated at 230 V. What is the peak household voltage in Europe?

14 

IB 12 Rating: rms values are given as the AC values to be used in calculations, as if they were DC values

Formula:

R = V0/I0 = Vrms/Irms

1. A stereo receiver applies an AC voltage of 34 V to a speaker. The speaker behaves approximately as if it has a resistance of 8.0 :, as the circuit figure indicates. Determine a) the maximum voltage,

b) the rms current,

c) the average power for this circuit.

a) 48 V

b) 4.25 A

c) 145 W

2. A 100 W light bulb is designed to operate from a 120 VAC mains. Determine:

3. A maximum alternating voltage of 170 V is applied across a 50 ȍ resistor. Determine:

a) the maximum power of the light bulb

a) the maximum current through the resistor

b) the maximum current drawn by the bulb

b) the average power dissipated by the resistor

a) 200 W

b) 1.2 A

a) 3.4 A

b) 289 W 15

7KH7UDQVIRUPHU

IB 12

According to Michael Faraday’s original experiment that first produced electromagnetic induction, an emf and current were only induced in the secondary coil when the switch in the primary coil was being opened or closed, that is, when the current in the primary coil was changing (increasing or decreasing). No emf or current was induced in the secondary coil while the switch was stationary in the open or closed position, that is, when the current was steady or off. Therefore, emf can only be induced in the secondary coil when the magnetic field from the current in the primary coil is building up or dying down, that is, while the magnetic flux is changing.

7UDQVIRUPHUa device that increases or decreases AC voltage.

6WUXFWXUHDQGRSHUDWLRQRIDWUDQVIRUPHU

NP, then VS > VP and voltage increases from primary to secondary

İS = VS = -NS (ǻĭ/ǻt) since flux changes are identical

93 96 1316

6WHS'RZQ7UDQVIRUPHU: If NS < NP, then VS < VP and voltage decreases from primary to secondary 16

Voltage and turns in same ratio

IB 12 How can the voltage increase or decrease without violating the conservation of energy principle? The power input at the primary equals the power output at the secondary. (This assumes 100% efficiency

and such a transformer is termed an LGHDOWUDQVIRUPHU.)

,GHDO7UDQVIRUPHU)RUPXOD PP = PS VP IP = VS IS VP/ VS = IS / IP Voltage and current in inverse ratio

1. A 120 VAC wall outlet is used to run a small electronic appliance with a resistance of 2.0 ȍ, as shown in the diagram. a) Is the transformer a step-up or step-down transformer? Cite evidence for your answer.

b) How much voltage does the device need?

c) If the current in the primary coil is 150 mA, how much current does the device use? Assume an ideal transformer.

a) down b) 6 V c) 3 A Real Transformers

Reasons for power losses in real transformers 1. resistance of wires in P and S coils causes heating of coils

Ps < PP eff = Ps / PP

2. not all flux from P coil is linked to S coil

3. core warms up as result of cycles of flux changes (hysteresis)

Modern transformers are up to 99% efficient

4. small currents are induced in core (eddy currents) – reduce by lamination 17

IB 12 

2. The figure shows a step-down transformer used to light a filament lamp with a resistance of 4.0 ȍ under operating conditions. The secondary coil has an effective resistance of 0.2 ȍand the primary current is 150 mA. Calculate: a) the reading on the voltmeter with switch S open 12 V

d) the power taken from the mains supply 36 W

b) the current in the secondary coil with switch S closed 2.86 A

e) the efficiency of the transformer 95%

c) the power dissipated in the lamp and the secondary coil 3.27 W and 1.6 W

+HDOWKDQG6DIHW\&RQFHUQVDVVRFLDWHGZLWK+LJK9ROWDJH3RZHU/LQHV 1. Extra-low-frequency electromagnetic fields, such as those produced by electrical appliances and power lines, induce currents within a human body. Just as AC can induce emfs and currents in secondary coils, so to can they be induced in the human

body since it is a conducting medium

Changing magnetic field induces current in human body

2. Current research suggests that low-frequency fields do not harm genetic material.

f = 60 Hz individual photons of this frequency do not have enough energy to cause ionization in the body childhood leukemia clusters are suspected to have a link to living near overhead power cables 3. The risks attached to the inducing of current in the human body are not well-understood. Risks are likely to be dependent on current density, frequency, and length of exposure

18 

3RZHU7UDQVPLVVLRQ

IB 12

Power loss in transmission lines When current flows through a wire, some energy is lost to the surroundings as the wire heats up due to the collisions between the free electrons in the current and the lattice ions of the wire. This is known as -RXOH KHDWLQJor UHVLVWLYHKHDWLQJ. Since the energy lost per second, or power loss, is proportional to the square of the current (P = I2 R), this energy loss is also know as “I2R loss.”

Methods of reducing I2R loss in power transmission lines 1. Reduce resistance: thicker cables – low resistivity material

Constraints: lengths are fixed, thicker cables are heavier and more expensive

2. Increase voltage: step voltage up to very high levels

Constraints: high voltages are dangerous – must be stepped back down for household use

For economic reasons, there is no ideal value of voltage for electrical transmission. Typical values are shown below. 1. AC power is generated at a power plant at 12,000 V and then stepped up to 240,000 V by step-up transformers. 2. The high-voltage, low-current power is sent via high-voltage transmission lines long distances. 3. In local neighborhoods, the voltage is stepped-down (and current is stepped-up) to 8000 V at substations. 4. This voltage is stepped-down even further at transformers on utility poles on residential streets.

An average of 120 kW of power is delivered to a suburb from a power plant that is 10 km away. The transmission lines have a total resistance of 0.40 ȍ. Calculate the power loss if the transmission voltage is a) 240 V

a) 240,000 V

I = 500 A P = 100 kW

I = 0.50 A P = 0.10 W

19 

(OHFWURVWDWLFV

IB 12

1) electric charge: 2 types of electric charge: positive and negative

2) charging by friction: transfer of electrons from one object to another

3) positive object: lack of electrons

negative object: excess of electrons

4) Types of materials: a) &RQGXFWRUV: materials in which electric charges move freely (e.g. metals, graphite) b) ,QVXODWRUV: materials in which electric charges do not move freely (e.g. plastic, rubber, dry wood, glass, ceramic) c) 6HPLFRQGXFWRUV: materials with electrical properties between those of conductors and insulators (e.g. silicon) d) 6XSHUFRQGXFWRUV: materials in which electrical charges move without resistance (e.g. some ceramics at very low temperatures)

3URSHUWLHVRI$WRPLF 3DUWLFOHV H = elementary unit of charge (magnitude of charge on electron)

Particle

Mass

Electric Charge

Electron

PH [ NJ

T H T [&

Proton

PS [ NJ

T H T [&

Neutron

PQ [ NJ

T  T &

H = 1.60 x 10-19 C

1. A balloon has gained 2500 electrons after being rubbed with wool. What is the charge on the balloon? What is the charge on the wool?

q = -4.0 x 10-16 C q = +4.0 x 10-16 C

2. A rubber rod acquires a charge of -4.5 ȝC. How many excess electrons does this represent?

2.8125 x 1013 e

&RQVHUYDWLRQRI(OHFWULF&KDUJH The total electric charge of an isolated system remains constant.

1

(OHFWULF)RUFH(OHFWURVWDWLF)RUFH&RXORPE)RUFH 

IB 12 

&RXORPE¶V/DZ The electric force between two point charges is directly proportional to the product of the two charges and inversely proportional to square of the distance between them, and directed along the line joining the two charges.

&RXORPE)RUFH

T1T2 )H N 2 U

k = Coulomb constant (electrostatic constant) k = 8.99 x 109 Nǜm2 C -2

NOTE: +-F denotes direction of force not sign of charge Point charge: a charged object that acts as if all its charge is concentrated at a single point Alternate formula for Coulomb force:

)H 

1 T1T2 4SH 0 U2

)H 

T1T2 4SH 0 U 2

k = 1/ 4ʌİ0 İ0 = permittivity of free space = 8.85 x 10-12 C2 N-1 m -2

Use the Coulomb force to estimate the speed of the electron in a hydrogen atom.

2 

7KH3ULQFLSOHRI6XSHUSRVLWLRQ

IB 12 

The net electric force acting on a charged particle is the vector sum of all the electric forces acting on it.

1. Determine the net electrostatic force on charge q1, as shown below.

2. Where can a third charge of +1.0 μC be placed so that the net force acting on it is zero?

3. Three point charges of 

-2.0 μC are arranged as shown. Determine the magnitude and direction of the net force on charge T.

D = 2/3 m

3

(OHFWULF)LHOG

IB 12

Electric field: a region in space surrounding a charged object in which a second charged object experiences an electric force Test charge: a small positive charge used to test an electric field

1. Positively charged sphere

(OHFWULF)LHOG'LDJUDPV 2. Positive point charge

3. Negative point charge

Radial Field: field lines are extensions of radii

5. Two positive charges

8. Oppositely charged parallel plates

6. Two negative charges

7. Two unlike charges

3URSHUWLHVRI(OHFWULF)LHOG/LQHV

1. Never cross 2. Show the direction of force on a small positive test charge 3. Out of positive, into negative 4. Direction of electric field is tangent to the field lines

Uniform Field: field has same intensity at all spots Edge Effect: bowing of field lines at edges

5. Density of field lines is proportional to field strength (density = intensity) 6. Perpendicular to surface 7. Most intense near sharp points

4

(OHFWULF)LHOG6WUHQJWK

IB 12 

(OHFWULF)LHOG6WUHQJWK,QWHQVLW\ electric force exerted per unit charge on a small positive test charge

Electric Field:

)H (  T Units: N/C

Electric Force:

)H (T

Electric Field for a Point Charge:

N



(

4T U 2 N 4  1 4 T U2 4SH 0 U2

Units: N 6SKHULFDO &RQGXFWRU

3RLQW&KDUJH

1. a) Find the magnitude and direction of the electric field at a spot 0.028 meter away from a sphere whose charge is +3.54 microcoulombs and whose radius is 0.60 centimeters. 

2. a) Find the magnitude and direction of the gravitational field at an altitude of 100 km above the surface of the Earth.

b) Find the magnitude and direction of the electric force acting on a -7.02 nC charge placed at this spot.

b)

Find the magnitude and direction of the gravitational force exerted on a 6.0 kg bowling ball placed at this spot.

c) Find the electric field strength at the surface of the sphere.

c) Find the gravitational field strength at the surface of the Earth.

5 

3. a) Find the magnitude and direction of the net electric field halfway between the two charges shown below.

IB 12

b) Determine the electric force on a proton placed at this spot.

4. Two charged objects, $ and %, each contribute as follows to the net electric field at point 3: ($ = 3.00 N/C directed to the right, and (%= downward. What is the net electric field at 3?

E = 3.61 N/C Theta =33.70

5. a) Two positive point charges, T1 = +16 PC and T2 = +4.0 PC, are separated in a vacuum by a distance of 3.0 m. Find the spot on the line between the charges where the net electric field is zero.

6 

6. A proton is released from rest near the positive plate. The distance between the plates is 3.0 mm and the strength of the electric field is 4.0 x 103 N/C. 

IB 12

a) Describe the motion of the proton. 

constant acceleration in a straight line b) Write an expression for the acceleration of the proton.

c) Find the time it takes the proton to reach the negative plate.

d) Find the speed of the proton when it reaches the negative plate.

7. A particle is shot with an initial speed through the two parallel plates as shown. a) Sketch and describe the path it will take if it is a proton, an electron, or a neutron. b) Which particle will experience a greater force?

c) Which particle will experience a greater acceleration?

d) Which particle will experience a greater displacement?

8. In the figure, an electron enters the lower left side of a parallel plate capacitor and exits at the upper right side. The initial speed of the electron is 5.50×106 m/s. The plates are 3.50 cm long and are separated by 0.450 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

7 

(OHFWULF3RWHQWLDO(QHUJ\

IB 12 

*UDYLWDWLRQDO3RWHQWLDO(QHUJ\(3 

High amount of EP

Reason for EP: 1. Test object has mass (test mass = m) 2. Test mass is in a gravitational field (g) caused by larger object (M) 3. Larger object exerts a gravitational force on test mass (Fg = mg) Low amount of EP 4. Test mass has tendency to move to base level due to force 5. Work done moving object between two positions is path independent. Base level where EP = 0

Gravitational potential energy: EP = mgh W = ǻEP = mg ǻh

(OHFWULF3RWHQWLDO(QHUJ\(3  Reason for EP: High amount of EP

1. Test object has charge (test charge = +q) 2. Test charge is in an electric field caused by larger object (Q) 3. Larger object exerts an electric force on test charge (FE = Eq)

Low amount of EP

4. Test charge has tendency to move to base level due to force 5. Work done moving object between two positions is path independent.

Electric potential energy: EP = Eq h W = ǻEP = Eq ǻh

Base level where EP = 0

(OHFWULF3RWHQWLDO(QHUJ\(3  the work done in bringing a small positive test charge in from infinity to that point in the electric field Derivation for Point Charges

EP = 0

(Work done by field)

(3 : )V cosT (3 (T'V  U N4 (3 ³ ( 2 T)GV f V





Electric Potential Energy due to a point charge Formula:

Units:

Type:

J

scalar

U

(3 

N4T V f

(3 

N4T

U

(3

N4T U

8

IB 12 (OHFWULF3RWHQWLDO9  work done per unit charge moving a small positive test charge in from infinity to a point in an electric field. (OHFWULF3RWHQWLDOGXHWRDSRLQWFKDUJH

N4T (3 Formula: 9  U T T N4 4 9  U 4SH 0 U

Units:

(3 T9

% Higher potential

Lower potential

%

1. a) Calculate the potential at a point 2.50 cm away from a +4.8 ȝC charge.



J/C = volts(V)

$ Lower potential

Type: scalar

Zero potential

Zero potential

Higher potential

$

b) How much potential energy will an electron have if it is at this spot?

3. What is the potential where a proton is placed 0.96 m from a -1.2 nC charge?

9

IB 12

3RLQW&KDUJHV

4

4

4

4

(OHFWULF)LHOG

(OHFWULF3RWHQWLDO(QHUJ\

(OHFWULF 3RWHQWLDO

Two objects needed – interaction between the two

One object needed – property of that one object

Two objects needed – quantity possessed by the system

One object needed – property of the field

Magnitude: F = Eq

Magnitude: E = F/q

Magnitude: EP = qV

Magnitude: V = EP/q

(OHFWULF)RUFH

F = kQq/r2

E = kQ/r2

P

= kQq/r

V = kQ/r

Units: N

Units: N/C

Units: J

Units: J/C

Type: vector

Type: vector

Type: scalar (+/-)

Direction: likes repel, unlikes attract

Direction: away from positive, towards negative

Type: scalar (+/-) E Sign: use signs of Q and q

Sign: don’t use when calculating – check frame of reference

Sign: don’t use when calculating – check frame of reference

F = 0 where E = 0

Sign: use sign of Q

EP = 0 where V = 0 10

IB 12 1. a) Calculate the net electric field at each spot (A and B):

b) Calculate the net electric force on a proton placed at each spot.

2. a) Calculate the net electric potential at each spot (A and B):

b) Calculate the electric potential energy of a proton placed at each spot.

11 

(OHFWULF3RWHQWLDODQG&RQGXFWRUV

IB 12

1. all the charge resides on the outside surface 2. the electric field is zero everywhere within

Value at surface = kQ/r2 Electric Potential

For a hollow or solid conductor,

Electric Field Strength

*UDSKVIRUDVSKHULFDOFRQGXFWRU

3. the external electric field acts as if all the charge is concentrated at the center 4. the electric potential is constant ( 0) everywhere within and equal to the potential at the surface

radius

radius Distance

Distance

A spherical conducting surface whose radius is 0.75 m has a net charge of +4.8 ȝC. a) What is the electric field at the center of the sphere?

b) What is the electric field at the surface of the sphere?

c) What is the electric field at a distance of 0.75 m from the surface of the sphere?

d) What is the electric potential at the surface of the sphere?

e) What is the electric potential at the center of the sphere?

f) What is the electric potential at a distance of 0.75 m from the surface of the sphere?

12 

(TXLSRWHQWLDO6XUIDFHV

IB 12 

(TXLSRWHQWLDOVXUIDFH: a surface on which the electric potential is the same everywhere

1. Locate points that are at the same electric potential around each of the point charges shown. 2. Sketch in the electric field lines for each point charge. 3. What is the relationship between the electric field lines and the equipotential surfaces?

Perpendicular 

Field lines point in direction 

of decreasing potential 

(OHFWULF3RWHQWLDO*UDGLHQW

Formula:

The electric field strength is the negative of the electric potential gradient.

'9 (   '[

Units: N/c or V/m For each electric field shown, sketch in equipotential surfaces.

Sketch in equipotential surfaces for the two configurations of point charges below.

13  http://wps.aw.com/aw_young_physics_11/0,8076,898593-,00.html

http://www.surendranath.org/Applets.html

(OHFWULF3RWHQWLDO'LIIHUHQFH

IB 12

(OHFWULF3RWHQWLDO'LIIHUHQFHǻ9 ±work done per unit charge moving a small positive test charge between two points in an electric field

Formula:

'9 '9

Units: J/C = V

:  T '(3 T

'(  3 T'9



+LJKDQG/RZ3RWHQWLDO 1. a) Which plate is at a higher electric potential? positive b) Which plate is at a lower electric potential? negative c) What is the electric potential of each plate? Arbitrary – relative to base level d) What is the potential difference between the plates? Not arbitrary – depends on charge, distance between, strength of electric field, geometry of plates, etc.

Mark plates with example potentials, as well as spots within field Mark “ground” – mark equipotentials

e) Where will: a proton have the most electric potential energy?

an electron?

a neutron?

an alpha particle? Not arbitrary

2. An electron is released from rest near the negative plate and allowed to accelerate until it hits the positive plate. The distance between the plates is 2.00 cm and the potential difference between them is 100. volts. a) Calculate how fast the electron strikes the positive plate.

Eo = Ef

Ee = EK

qV = ½ mv2

v = sqrt (2qV/m) 

v = sqrt(2(1.6 x 10-19)(100 V) 

/(9.11 x 10-31)) 

v = 5.9 x 106 m/s

Formula: qV = ½ mv2 Ve = ½ mv2

b) Calculate the strength of the electric field.

Formula: E = -ǻV/ǻx E = V/d

14 

7KH(OHFWURQYROW

IB 12

(OHFWURQYROW: energy gained by an electron moving through a potential difference of one volt

Derivation:

ǻEe = qǻV ǻEe = (1e)(1 V) = 1 eV

ǻEe = (1.6 x 10-19 C)( 1 V) ǻEe = 1.6 x 10-19 J Therefore: 1 eV = 1.60 x 10-19 J

1. How much energy is gained by a proton moving through a potential difference of 150. V?

150 eV

or 150(1.60 x 10-19) = 2.4 x 10-17 J

2. A charged particle has 5.4 x 10-16 J of energy. How many electronvolts of energy is this?

Factor-label (5.4 x 10-16 J) (1 eV/1.6 x 10-19) = 3375 eV 3. An electron gains 200 eV accelerating from rest in a uniform electric field of 150 N/C. Calculate the final speed of the electron.

4. In Rutherford’s famous scattering experiments (which led to the planetary model of the atom), alpha particles were fired toward a gold nucleus with charge +79H. An alpha particle, initially very far from the gold nucleus, is fired at 2.00 × 107 m/s directly toward the gold nucleus. Assume the gold nucleus remains stationary. How close does the alpha particle get to the gold nucleus before turning around? (the “distance of closest approach”)

2.74 x 10-14 m

15 

(QHUJ\DQG3RZHU

IB 12 

3RZHU*HQHUDWLRQLQDW\SLFDOHOHFWULFDOSRZHUSODQW a) Some fuel is used (coal, natural gas, oil, uranium) to release thermal energy which is used to boil water to make steam. b) *HQHUDWRU'\QDPR - Steam turns turbines attached to coils of wire which turn in a magnetic field inducing an alternating potential difference. c) Potential difference is stepped up by transformers in order to reduce I2R loss of power in transmission lines then stepped down for consumer use.

What are the energy transformations that take place?

Chemical (nuclear) energy in fuel

Thermal energy – hot gases go out chimney/stack

thermal energy in steam

rotational mechanical energy/kinetic energy in turbine

Thermal energy – radiation and convection from boiler

electrical

energy in

generator

Thermal energy – friction in the generator

Degraded energy: energy transferred to the surroundings that is no longer available to do useful work – can’t be converted into other forms

1 

IB 12  Why does the generation of electrical power involve the degradation of energy? 1. Thermal energy can be completely converted to work in a single process.

2. A continuous conversion of thermal energy into work requires a cyclical process.

Example: isothermal expansion Q = ǻU + W ǻU = 0 so Q = W

6HFRQG/DZRI7KHUPRG\QDPLFV 1) The total entropy of the universe is increasing. 2) No cyclical process (engine) is ever 100% efficient. Some energy is transferred out of the system (lost to the surroundings) as unusable energy (degraded energy).

7\SLFDO (IILFLHQF\

)XHO

6DQNH\GLDJUDPVHQHUJ\IORZGLDJUDPV : used to keep track of 

energy transfers and transformations 

Coal

30-35%

1) Thickness of arrow is proportional to amount of energy. Natural Gas

50%

2) Degraded energy points away from main flow of energy. Oil

30-35%

3) Total energy in = total energy out.

Input 100% Chemical energy

Fuel

boiler

Output 30% Electrical energy

dynamo

10% Thermal energy Friction in dynamo

40% Thermal energy 

exhaust gas out 

chimney 

20% Thermal energy 

boiler 

2

)XHOV

IB 12

Fuel: source of energy (in a useful form) How does a fuel work? A fuel releases energy by changing its chemical (or nuclear) structure. Chemical (or nuclear) bonds are broken reducing the fuel’s internal potential energy but increasing the kinetic energy of the substance’s particles which is seen macroscopically as an increase in the temperature of the substance. It is this thermal energy that is used to heat the water that will change to steam to turn the generator’s turbines.

Fossil fuels: coal, oil, natural gas, peat Origins of fossil fuels: organic matter decomposed under conditions of high temperature and pressure over millions of years

Non-renewable fuels: rate of production of fuel is much smaller than rate of usage so fuel will be run out - limited supply

Renewable fuels: resource that cannot be used up or is replaced at same rate as being used

7\SHRIIXHO

5HQHZDEOH"

&2 HPLVVLRQV"

Fossil fuels

No

Yes

Nuclear

No

No

Hydroelectric

Yes

No

Wind

Yes

No

Solar

Yes

No

Wave

Yes

No

This histogram shows the relative proportions of world use of the different types of energy sources, though it will vary from country to country.

NOTE: In most instances, the prime energy source for world energy is . . . the Sun. Exceptions: nuclear, tidal (Moon)

3

)RVVLO)XHO3RZHU3URGXFWLRQ

IB 12

Historical and geographical reasons for the widespread use of fossil fuels: 1. industrialization led to a higher rate of energy usage (Industrial Revolution) 2. industries developed near large deposits of fossil fuels (coal towns) Transportation and storage considerations: 1. Natural gas is usually transported and stored in pipelines. $GYDQWDJHV cost effective 'LVDGYDQWDJHV unsightly, susceptible to leaks, explosions, terrorist activities, political instability (withholding use of pipelines or terminals for political reasons) 2. Many oil refineries are located near the sea close to large cities. Oil is transported via ships, trucks, and pipelines. $GYDQWDJHV workforce and infrastructure in place, easy access to shipping 'LVDGYDQWDJHV oil spills and leakage, hurricanes, terrorist activities

3. Power stations using coal and steel mills are usually located near coal mines. $GYDQWDJHV minimizes shipping costs 'LVDGYDQWDJHV environmental impact (strip mining), mine cave-ins

8VHRIIRVVLOIXHOVIRUJHQHUDWLQJHOHFWULFLW\ $GYDQWDJHV

'LVDGYDQWDJHV

1. high energy density

1. combustion produces pollution, especially SO2 (acid rain)

2. relatively easy to transport

2. combustion produces greenhouse gases (CO2)

3. cheap compared to other sources

3. extraction (mining, drilling) damages environment

4. power stations can be built anywhere

4. nonrenewable

5. can be used in the home

5. coal-fired plants need large amounts of fuel

4

IB 12 

(QHUJ\GHQVLW\RIDIXHOthe ratio of the energy released from the fuel to the mass of the fuel consumed

Formula: De = E/m

)XHO Fusion fuel Uranium-235 Natural gas Gasoline (Petrol) Diesel Biodiesel Crude oil Coal Sugar Wood Cow dung Household waste

Units: J/kg

Use: to compare different types of fuels

How is choice of fuel influenced by energy density? Fuels with higher energy density cost less to transport and store

(QHUJ\'HQVLW\ 0-NJ  300,000,000 90,000,000 53.6 46.9 45.8 42.2 41.9 32.5 17.0 17.0 15.5 10

1. An oil-fired power station produces 1000 MW of power. a) How much energy will the power station produce in one day? ( W

( 3W  3 





( (1000 x 106:)(24 [3600) ( 8.6 [1013 -

b) Estimate how much oil the power station needs each day. HII

useful out (RXW  total in (LQ

'H 

( P - 2.46 [1014 NJ P

8.6 [1013 - .35  (LQ

41.9 [106

(LQ 2.46 [1014 -

P 5.9 [106 NJ



5 

IB 12  2. A 250 MW coal-fired power plant burns coal with an energy density of 35 MJ/kg. Water enters the cooling tower at a temperature of 350 K and leaves at a temperature of 293 K and the water flows through the cooling tower at a rate of 4200 kg/s. a) Calculate the thermal energy removed from the water in the cooling towers each second. 4 PF'7 4 (4200NJ )(4.19 [10 3

)(350 293) NJ.

4 1.0 [109 -

( W 1.0 [109 - 3 

1V

3 1.0 [109: 10000: 3 



b) Assuming the only significant loss of energy is this thermal energy of the water, calculate the energy produced by the combustion of coal each second. (LQ (RXW (LQ 1000 MJ + 250 MJ E LQ 12500-

c) Calculate the mass of coal burned each second. HII

useful out 3RXW  total in 3LQ

2500: HII  12500: HII .20

( 'H  P 1250 [106 - 6 - 35 [10

 NJ P

P 36NJ 

6

1XFOHDU(QHUJ\

IB 12 

Most common source: fissioning of uranium-235 with conversion of some mass into energy Process: a) unstable uranium nucleus is bombarded with a 

neutron and splits into two smaller nuclei and 

some neutrons 

Why use neutrons? Neutral, not repelled by nucleus b) rest mass of products is less than reactants so 

some matter is converted into energy 

Form of energy: KE of products (thermal energy) 235 92

92 1 8 10 Qo141 %D  .U 3 56 36 0Q

c) released neutrons strike other uranium nuclei 

causing further fissions 

1) A particular nuclear reactor uses uranium-235 as its fuel source. When a nucleus of uranium-235 absorbs a neutron, the following reaction can take place: 235 92

90 1 8 10 Qo144 54 ;H 38 6U [0 Q



a) How many neutrons are produced in the reaction? 2 b) Use the information to show that the energy released in the reaction is approximately 180 MeV.

rest mas of

8 2.1895 x 105 MeV c 2

235 92

rest mas of 10 Q 939.56 MeV c 2 2 5 rest mas of 144 54 ;H 1.3408 x 10 MeV c 2 5 rest mas of 90 38 6U 8.3749 x 10 MeV c

7 

IB 12 

2. The energy released by one atom of carbon-12 during combustion is approximately 4 eV. The energy released by one atom of uranium-235 during fission is approximately 180 MeV. a) Based on this information, determine the ratio of the energy density of uranium-235 to that

of carbon-12. (Then, check your answer with the given table of energy densities.)

PDVV=

1 x molar mass 1 $

PDVV=

1 x .235 kg 6.02 [ 1023 mass=3.90x1025 NJ

1 x .012 kg 6.02 [1023 mass=1.99x1026 NJ

mass =

mass =

§ 180 [106 H9 · § 1.60  [1019 -· 11 ¨ ¸¨ ¸ 2.88 [10  - 1 1H9 © ¹© ¹ 

11

' H

2.88 [ 10 7.38 [1013 -/ NJ 25 3.90 [10 NJ 

7.38 [107 0- / NJ





'H U-235 'H C-12

1 x molar mass

1 $

19 §4H9 · §1.60 [10 - · 19 ¸ 6.40 [10 ¨ ¸¨ © 1 ¹ © 1H9 ¹

6.40 [1019 - 'H  3.22 [107 -/ NJ 26 1.99 [10 NJ  32.20-/ NJ



7.38 [107 2.3[106 32.2

b) Based on your answer above, suggest one advantage of uranium-235 compared with fossil fuels. Higher energy density implies that uranium will produce more energy per kilogram – less fuel needed to produce the same amount of energy

8

1XFOHDU)XHODQG5HDFWRUV

IB 12

Naturally Occurring Isotopes of Uranium: 1) 8UDQLXP most abundant (99.3%) but not used for fuel since it has a very small probability of fissioning when it captures a neutron. 2) 8UDQLXP rare (0.3%) but used for fuel since it has a much greater probability of fissioning when captures a neutron but must be a low-energy neutron (thermal neutron). 7KHUPDO1HXWURQ low-energy neutron (§1eV) that favors fission reactions – energy comparable to gas particles at normal temperatures

)XHO(QULFKPHQW process of increasing proportion of uranium-235 in a sample of uranium

1) formation of gaseous uranium (uranium hexafluoride) from uranium ores 2) separated in gas centrifuges by spinning – heavier U-238 moves to outside 3) increases proportion of U-235 to about 3-5% of total (low enrichment) 4) This low enriched hex is compressed and turned into solid uranium-oxide fuel pellets which are packed into tubes called IXHOURGV which will be used in the core of a nuclear reactor. $GYDQWDJH More uranium is available for fission and a chain reaction can be sustained in a reactor to produce nuclear energy. 'LVDGYDQWDJH If the fuel is enriched to a high level (90% = weapons grade) it can be used in the core of a nuclear weapon. Possession of nuclear weapons is seen by many to be a threat to world peace.

9

IB 12 

&KDLQ5HDFWLRQ± neutrons released from one fission reaction go on to initiate further reactions

Uncontrolled Chain Reaction

Uncontrolled nuclear fission: nuclear weapons Controlled nuclear fission: nuclear power production

1) Some material (control rod) absorbs excess neutrons before they strike another nucleus.

Controlled Chain Reaction

2) This leaves only one neutron from each reaction to produce another reaction. 3) If the total mass of uranium used is too small, too many neutrons will escape without causing further fissions so the reaction cannot be sustained. &ULWLFDO0DVV minimum mass of radioactive fuel (uranium) needed for a chain reaction to occur

7KH1XFOHDU5HDFWRU&RUH

)XHO5RGV enriched solid uranium

When neutrons are emitted from a fission reaction in the fuel rods, they have a very high kinetic energy and will pass right out of the fuel rod without colliding with another uranium nucleus to cause more fission. High energy neutrons cannot sustain a chain reaction. Therefore, a material is needed to slow them down. Typically, a material like water or graphite (called a “PRGHUDWRU”) is used to slow down these high-energy neutrons down to “thermal levels” (thermal neutrons § 1 eV) for use in further fission reactions to sustain the chain reaction. The high-energy neutrons slow down when they collide with the atoms in the moderator. To control the rate at which the thermal energy is produced, and therefore to control the temperature of the reactor core, FRQWURO URGV are used to speed up or slow down the chain reaction. These are rods made of a neutron-absorbing substance, like cadmium or boron. They are inserted in between the fuel rods and raised or lowered as needed. If the reaction is proceeding too fast (too hot) the rods are lowered and enough thermal neutrons are absorbed to slow down the reaction to the desired level. Conversely, if the reaction is too slow, the control rods are raised allowing more thermal neutrons to collide with uranium nuclei. 10 

IB 12 

How is the thermal energy released in the fission reactions used to generate electricity? The FRRODQW (which is often the same as the moderator) is fluid circulating around the fuel rods in the reactor core and is heated up by the thermal energy released in the fission chain reaction. This coolant in a closed loop (primary loop) flows through pipes in a tank of water known as the “KHDW H[FKDQJHU” Here the thermal energy of the hot coolant is transferred to cooler water in a secondary loop which turns it to steam. This steam expands against fan blades of turbines and turns a magnet is a coil of wire to generate electricity. 1. State the energy transformations in using nuclear fuels to generate electrical energy:  Nuclear energy in fuel….thermal energy in coolant . . thermal energy in steam in heat exchanger…rotational  mechanical energy/kinetic energy…electrical energy in turbines

2. Sketch a Sankey diagram for a typical nuclear power plant.

11 

IB 12  3. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have to undergo fission to supply the household with electrical energy for a year. State some assumptions made in your calculation.

Assume plant is 100% efficient 

Assume 200 MeV per fission 

12 

IB 12 

4. A fission reaction taking place in a nuclear power station might be:

235 92

92 1

810 Qo141 56 %D 36 .U 3 0 Q

Estimate the initial amount of uranium-235 needed to operate a 600 MW reactor for one year assuming 40% efficiency and 200 MeV released for each fission reaction.

13 

3OXWRQLXPDQG1XFOHDU5HDFWRUV

IB 12

Plutonium-239 is another nuclide used as nuclear fuel because of the energy it releases when it undergoes fission. However, it is not as naturally abundant as uranium and so it typically must be artificially produced as a by-product of uranium fission. In a uranium-fueled reactor, as the U-235 depletes over time, the amount of Pu-239 increases. This plutonium is then extracted (by reprocessing of the uranium fuel rods) for use in a plutonium reactor or in a nuclear warhead. How is plutonium-239 produced in a uranium reactor? It actually is produced from the nonfissionable isotope uranium-238 that occurs in large amounts in fuel rods. Uranium-238 doesn’t undergo nuclear fission but is considered “fertile” since it produces plutonium-239 by the following process. 238 92

239 239 810 Qo92 8 o93 1S 01 H Q 239 1So94 3X01 H Q

239 94

3X Qo %D 6U4 Q 140 56

96 38

239 94

3X 239.052157X

rest mass of 10 Q 1.008665X

239 93

1 0

rest mass of

rest mass of 140 56 %D 139.910581X 1 0

rest mass of 96 38 6U 95.921750X

1. Complete the nuclear reactions listed above. 2. Construct a nuclear energy level diagram for the series of nuclear reactions listed above.

3. Determine the amount of energy released in the fissioning of plutonium-239.

P PǻP 0.19383 u = 180 MeV

Some uranium reactors are even specially designed to produce (or “breed”) large amounts of plutonium and are known as EUHHGHUUHDFWRUV. They are designed so that the fuel rods are surrounded by a blanket of U-238 so that neutrons escaping from the U-235 fissions will induce the conversion of this U-238 to Pu-239. 14 

6DIHW\,VVXHVDQG5LVNVLQWKH3URGXFWLRQRI1XFOHDU3RZHU

IB 12 

8UDQLXP0LQLQJ x RSHQFDVWPLQLQJ environmental damage, radioactive waste rock (tailings) x XQGHUJURXQGPLQLQJ release of radon gas (mines need ventilation), radioactive rock is dangerous for workers, radioactive waste rock (tailings) x OHDFKLQJ Solvents are pumped underground to dissolve the uranium and then pumped back out. This leads to contamination of groundwater. 7KHUPDO0HOWGRZQ Overheating and melting of fuel rods may be caused by a malfunction in the cooling system or the pressure vessel. This overheating may cause the pressure vessel to burst sending radioactive material and steam into atmosphere (as in Chernobyl, Ukraine 1986). Hot material may melt through floor (as in Three Mile Island, Pennsylvania 1979), a scenario dubbed the “China syndrome.” The damage from these possible accidents is often limited by a containment vessel and a containment building. 1XFOHDU:DVWH x

/RZOHYHOZDVWH Radioactive material from mining, enrichment and operation of a plant must be disposed of. It’s often left encased in concrete.

x

+LJKOHYHOZDVWH a major problem is the disposal of spent fuel rods. Some isotopes have ½ lives of thousands of years. Plutonium’s is 240,000 years.

1) Some are stored under water at the reactor site for several years to cool off then sealed in steel cylinders and buried underground. 2) Some are reprocessed to remove any plutonium and useful uranium. The remaining isotopes have shorter ½ lives and the long-term storage need is reduced. 1XFOHDU:HDSRQV0DQXIDFWXUH x Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel grade (3%) x Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent fuel rods &RPSDULQJ1XFOHDU)XHOWR)RVVLO)XHO $GYDQWDJHV 'LVDGYDQWDJHV

1. No global warming effect – no CO2 emissions

1. Storage of radioactive wastes

2. Waste quantity is small compared with fossils fuels

2. Increased cost over fossil fuel plants

3. Higher energy density

3. Greater risks in an accident (due to radioactive contamination)

4. Larger reserves of uranium than oil

15

1XFOHDU)XVLRQ

IB 12

1XFOHDU)XVLRQTwo light nuclei combine to form a more massive nucleus with the release of energy.

Naturally occurring fusion: main source of Sun’s energy – fusion of hydrogen to helium A probable mechanism for the Sun’s fusion is called the SURWRQSURWRQFKDLQ. 1 1

+11 + o12 +10 H Q

1 1

+12 + o3

2 +H  J 

3 2

+H32 +Ho42 +H11 +11 +

This chain is sometimes simplified to 

4+ o+HHQHUJ\

1. If the total mass of four hydrogen nuclei is 6.693 x 10-27 kg and the mass of a helium nucleus

is 6.645 x 10-27 kg, determine the energy released in this simplified fusion reaction.

4.3 x 10-12 J

2. The Sun has a radius 5 of 7.0 x 108 m and emits energy at a rate of 3.9 x 1026 W. The nuclear reactions take place in the spherical core of the Sun of radius  0.255. Determine the number of nuclear reactions occurring per cubic meter 

per second in the core of the Sun. 

4.1 x 1012 m-3 s-1

16 

IB 12 

$UWLILFLDOO\LQGXFHGIXVLRQ Attempts have been underway since the 1950s to build fusion reactors. Experimental reactors have come very close to producing more energy than the amount of energy put in, but a commercial fusion reactor has yet to be built. 3ODVPD The fuel for a fusion reactor is known as a SODVPD. This is a high energy ionized gas in which the electrons and nuclei are separate. If the energy is high enough (that is, the plasma is hot enough), nuclei can collide fast enough to overcome Coulomb repulsion and fuse together. Heating the plasma to the required temperatures (10 million K) is challenging. The nuclei, since they are charged, are accelerated by means of magnetic fields and forces to high kinetic energies (high temperatures). 0DJQHWLFFRQILQHPHQW These charged particles are contained via magnetic fields and travel in a circle in a doughnut shaped ring called a “tokamak” which an acronym of the Russian phrase for “toroidal chamber with magnetic coils” (WRURLGDO QD\DNDPHUDVPDJQLWQ\PL NDWXVKNDPL). 3UREOHPVZLWKFXUUHQWIXVLRQWHFKQRORJ\ x

Maintaining and confining these very high-density and high-temperature plasmas for any length of time is very difficult to do.

x

Experimental reactors that currently can achieve fusion use more energy input than output which makes them not commercially efficient.

Comparison of Nuclear Fission and Nuclear Fusion

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17 

6RODU3RZHU 6RODUKHDWLQJSDQHODFWLYHVRODUKHDWHU  converts light energy from Sun into thermal energy in water run through it

Use: heating and hot water

IB 12 3KRWRYROWDLFFHOOVRODUFHOO  converts light energy from Sun into electrical energy

Use: electricity

Advantages of solar heating panel over solar cell: requires less (storage) area, less cost, more efficient

The amount (intensity) of sunlight varies with: a) time of day b) season (angle of incidence of sunlight – altitude of Sun in sky – Earth’s distance from Sun) c) length of day d) latitude (thickness of atmosphere) Which way should a solar panel or cell be facing in the Northern hemisphere? Why? South to receive Maximum radiation from the sun to provide maximum energy for whole day

$GYDQWDJHV

'LVDGYDQWDJHV 1. Large area needed to collect energy

1. Renewable source of energy 2. Only provides energy during daylight 2. Source of energy is free 3. No global warming effect – no CO2 emissions

3. Amount of energy varies with season, location and time of day

4. No harmful waste products

4. High initial costs to construct/install

18 

IB 12 

1. An active solar heater whose efficiency is 32% is used to heat 1400 kg of water from 200 C to 500 C. The average power received from the Sun in that location is 0.90 kW per m2. a) How much energy will the solar heater need to provide to heat the water?

b) How much energy will be needed from the Sun to heat the water?

c) Calculate the area of the solar heater necessary to heat the water in 2.0 hours.

2. A photovoltaic cell with an area of 0.40 m2 is placed in a position where the intensity of the Sun is 1.0 kW/m2. a) If the cell is 15% efficient, how much power does it produce?

b) If the potential difference across the cell is 5.0 mV, how much current does it produce?

c) Compare placing 10 of these solar cells in series and in parallel.

19

:LQG3RZHU

IB 12

Basic features of a horizontal axis wind turbine:  a) Tower to support rotating blades.  b) Blades that can be rotated to face into the wind.  c) Generator.  d) Storage system or connection to a distribution grid.  Energy transformations:  Solar energy heating Earth . . . Kinetic energy of air . . . kinetic energy of turbine . . electrical energy

1. Determine the power delivered by a wind generator:

2. Reasons why power formula is an estimate: a) Not all KE of wind is transformed into mechanical energy b) Wind speed varies over course of year c) Density of air varies with temperature d) Wind not always directed at 900 to blades

.( 1/ 2PY 2 P= W W 1§P· 2 3 ¨ ¸Y 2© W ¹  1 § U9 · 2  3 ¨ ¸ Y 2© W ¹  1 § U $G · 2 ¨ ¸Y 2 © W ¹ 1 3  UY Y2  2 1 $U Y3 3 2

 3. Why is it impossible to extract this maximum amount of power from the air? a) Speed of air cannot drop to zero after  impact with blades  b) Frictional losses in generator and 

turbulence around blades 

3

4. Why are turbines not placed near one another?



a) Less KE available for next turbine b) Turbulence reduces efficiency of next turbine

20 

IB 12 

1. A wind turbine has a rotor diameter of 40 m and the speed of the wind is 25 m/s on a day when the air density is 1.3 kg/m3. Calculate the power that could be produced if the turbine is 30% efficient.

2. A wind generator is being used to power a solar heater pump. If the power of the solar heater pump is 0.50 kW, the average local wind speed is 8.0 m/s and the average density pf air is 1.1 kg/m3, deduce whether it would be possible to power the pump using the wind generator.

$GYDQWDJHV

1. Renewable source of energy

'LVDGYDQWDJHV

1. Large land area needed to collect energy since many turbines are needed

2. Source of energy is free 2. Unreliable since output depends on wind speed 3. No global warming effect – no CO2 emissions 3. Site is noisy and may be considered unsightly 4. No harmful waste products 4. Expensive to construct 21

:DYH3RZHU

IB 12

Energy can be extracted from water waves in many ways. One such scheme is shown here. 2VFLOODWLQJ:DWHU&ROXPQ2:& ocean wave energy converter:

1. Wave capture chamber is set into rock face on land 

where waves hit the shore. 

2. Tidal power forces water into a partially filled 

chamber that has air at the top. 

3. This air is alternately compressed and decompressed 

by the “oscillating water column.” 

4. These rushes of air drive a turbine which generates 

electrical energy. 

Energy transformations: Kinetic energy of water . . . Kinetic energy of air . . . kinetic energy of turbine . . electrical energy

Determining the energy in each wavelength of the wave and the power per unit length of a wavefront

Energy in each wavelength of the wave PE = mgh PE=mgA

SRZHU

PE = (U V)gA

P

1 PE = (U ( O $/))gA 2 1 PE = $2 O gU / 2

1 2 $ O gU / 3  2 7 1 3  $2 YgU / 2 power per unit length

3( W

3 1 2 $ YgU / 2

How would this power estimate change if the waves were modeled as sine waves instead of square waves? 22 

IB 12 

1. Waves of amplitude 1.5 meter roll onto a beach with a speed of 10 m/s. Calculate: a) how much power they carry per meter of shoreline

b) the power along a 2 km stretch of beach.

2. Waves that are 6.0 meters high with a 100 meter wavelength roll onto a beach at a rate of one wave every 5.0 seconds. Estimate the power of each meter of the wavefront.

$GYDQWDJHV

'LVDGYDQWDJHV

1. Renewable source of energy

1. Can only be utilized in particular areas

2. Source of energy is free

2. High maintenance due to pounding of waves

3. No global warming effect – no CO2 emissions

3. High initial construction costs

4. No harmful waste products

23

+\GURHOHFWULF3RZHU

IB 12 

There are many schemes for using water to generate electrical energy. But all hydroelectric power schemes have a few things in common. Hydroelectric energy is produced by the force of falling water. The gravitational potential energy of the water is transformed into mechanical energy when the water rushes down the sluice and strikes the rotary blades of turbine. The turbine's rotation spins electromagnets which generate current in stationary coils of wire. Finally, the current is put through a transformer where the voltage is increased for long distance transmission over power lines. By far, the most common scheme for harnessing the original gravitational potential energy is by means of VWRULQJZDWHULQODNHV, either natural or artificial, behind a dam, as illustrated in the top picture at right. A second scheme, called WLGDOZDWHUVWRUDJH, takes advantage of big differences between high and low tide levels in bodies of eater such as rivers. A barrage can be built across a river and gates, called sluices, are open to let the high-tide water in and then closed. The water is released at low tide and, as always, the gravitational potential energy is used to drive turbines to produce electrical energy. A third scheme is called SXPSHGVWRUDJHWater is pumped to a high reservoir during the night when the demand, and price, for electricity is low. During hours of peak demand, when the price of electricity is high, the stored water is released to produce electric power. A pumped storage hydroelectric power plant is a net consumer of energy but decreases the price of electricity. Energy transformations: Gravitational PE of water . . . Kinetic energy of water . . . kinetic energy of turbine . . electrical energy

$GYDQWDJHV

'LVDGYDQWDJHV

1. Renewable source of energy

1. Can only be utilized in particular areas

2. Source of energy is free

2. Construction of dam may involve land being buried under water

3. No global warming effect – no CO2 emissions 3. Expensive to construct 4. No harmful waste products 24 

IB 12 

1. A barrage is placed across the mouth of a river at a tidal power station. If the barrage height is 15 meters and water flows through 5 turbines at a rate of 100 kg/s in each turbine, calculate the power that could be produced if the power plant is 70% efficient. 2.6 x 104W Use average height for EP = mgh

2. A reservoir that is 1.0 km wide and 2.0 km long is held behind a dam. The top of this artificial lake is 100 meters above the river where the water is let out at the base of the dam. The top of the intake is 25 meters below the lake’s surface. Assume the density of water is 1000 kg/m3. a) Calculate the energy stored in the reservoir. 4.3 x 1013 J

b) Calculate the power generated by the water if it flows at a rate of 1.0 m3 per second through the turbine. 875 kW

25 

8QLYHUVDO*UDYLWDWLRQ

IB 12 

.HSOHU¶V7KUHH/DZVRI3ODQHWDU\0RWLRQ

/DZ All planets orbit the Sun in elliptical paths with the Sun at one focus.

/DZ An imaginary line joining any planet to the Sun sweeps out equal areas in equal time intervals.

)RUPXOD

7 2DU 3 7 2 NU 3

/DZ The square of the orbital period of any planet is proportional to the cube of its average orbital radius.

7 DU

3 2

1HZWRQ¶V/DZRI8QLYHUVDO*UDYLWDWLRQ Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and that is inversely proportional to the square of the distance between them.

Two approximations used in deriving the law: 1. Masses are considered to be point masses. Point mass: infinitely small object (radius = 0) whose mass is P

2. The force between two spherical masses whose separation is large compared to their radii is the same as if the two spheres were point masses with their masses concentrated at the centers of the spheres. 6XQ

Mean Earth-Sun distance = 1.50 x 1011 m

(DUWK

Mean radius = 6.37 x 106 m Mean radius = 6.96 x 108 m

1 

([WHQGHGVSKHULFDOERG\

3RLQWPDVV

)RUPXOD

P1 ˜ P2 U2 * ˜ P1 ˜ P2 )J = U2 )J D

*UDYLWDWLRQDO&RQVWDQW G = 6.77 x 10-11 N m2/kg2

Re 2Re 3Re 4Re

1HZWRQ¶V'HULYDWLRQRI.HSOHU¶V7KLUG/DZ What provides the centripetal force for orbital motion? gravitation

'HULYDWLRQ

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6)F PD F



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ª1\HDU  365.25GD\V  24KRXUV  3600V ¼º ¬ 3 1.50 [1011 P 0 6

2

4S 2 6.67 [1011 0 V

2.0 [1030 NJ 



accepted value=1.99 [10 NJ 30



*0 § 2S U· ¨ ¸ U © 7 ¹ *0 4S 2 U 2  2 U 7 2 4S 2 7  U 3 *0  2

What is the resultant gravitational force on the Earth from the Sun and Moon, as shown below? Average Earth-Sun distance = 1.50 x 1011 m

Average Earth-Moon distance = 3.84 x 108 m 

6XQ

(DUWK

Mass = 1.99 x 1030 kg

Mass = 5.98 x 1024 kg

0RRQ Mass = 7.36 x 1022 kg

2 

*UDYLWDWLRQDO)LHOG6WUHQJWK *UDYLWDWLRQDOILHOGVWUHQJWK at a point in a gravitational field:

IB 12

the gravitational force exerted per unit mass on a small/point mass

Symbol: g

Deriving formula for gravitational field strength at any point above the surface of a planet

3RLQWPDVV

Formula: g = Fg / m

g = Fg /m g = (GMm/r2)/m g = GM/r2

Units: N/kg (m/s2)

Type: vector

Deriving formula for gravitational field strength at the surface of a planet

1. What is the gravitational field strength of the Earth at its surface?

g = GM/r2 go = G Mp / Rp2

“g” at the surface of the Earth 2. What is the gravitational field strength at an altitude equal to the radius of Earth? 2

go = G ME/Re

([WHQGHGVSKHULFDOERG\

go

“g” ratio g/go = RE2/r2

Re 2Re 3Re 4Re

3 

Average Earth-Moon distance = 3.84 x 108 m

(DUWK

IB 12

0RRQ

Mass = 5.98 x 1024 kg

Mass = 7.36 x 1022 kg

3. a) What is the resultant gravitational field strength at a point midway between the Earth and Moon?

b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?

Average Earth-Moon distance = 3.84 x 108 m

(DUWK Mass = 5.98 x 1024 kg

0RRQ Mass = 7.36 x 1022 kg

4. a) Is there a point where the resultant gravitational field strength of the Earth and Moon is zero? If so, where?

b) What is the resultant gravitational force acting on a 1500. kg space probe at this location?

4

*UDYLWDWLRQDO3RWHQWLDO(QHUJ\

IB 12 

Difference in gravitational potential energy between any two points in a gravitational field:

ǻEP = mgǻh

This difference is SDWKLQGHSHQGHQW. 1. Same ǻEP between any two points no matter what path is taken between them. 2. Work done in moving a mass between two points in a gravitational field is independent of the path taken. 3. ǻEP is zero between any two points at the same level no matter what path is taken. 4. ǻEP is zero for any closed path (a path that begins and ends at same point). “Old” formula for gravitational potential energy: Ep = mgh

discuss 2 problems with definition 1) “g” varies above surface 2) arbitrary base level

Base level: infinity

Gravitational PE at infinity: zero

EP = -400 J

Gravitational Potential Energy of a mass at a point in a gravitational field:

EP = -100 J

EP = 0

the work done in bringing a small point mass in from infinity to that point in the gravitational field Derivation of gravitational potential energy formula

5 

)RUPXOD

3RWHQWLDOHQHUJ\YVGLVWDQFH

IB 12

Ep = - GMm/r EP = -Gm1m2/r Formula not valid inside planet

Ep at surface Ep = -GMpm/Rp

6\PERO9 8QLWV- 7\SHVFDODU What is the gravitational potential energy of a 5000 kg satellite:  a) on the surface of the Earth?

b) orbiting the Earth at an altitude of 200 km?

c) How much does the potential energy of the satellite increase when it is put into this orbit?

*UDYLWDWLRQDO3RWHQWLDO *UDYLWDWLRQDOSRWHQWLDO at a point in a gravitational field:

work done per unit mass to bring a small point mass in from infinity to that point in the gravitational field

)RUPXODV

*UDYLWDWLRQDO3RWHQWLDO YVGLVWDQFH

Difference in gravitational potential: ǻV = W/m

ǻV = ǻEp/m

Gravitational potential at a point: 6\PERO9 V = Ep/m V = -GM/r

8QLWV-NJ 7\SHVFDODU

V at surface Vo = -GMp/Rp 6

IB 12 

1. What is the gravitational potential due to the Earth’s gravitational field: a) at the surface of the Earth?

b) At a location three Earth radii from the center of the Earth?

c) What is the change in potential in moving from the surface to this new location?

d) What is the minimum amount of energy needed to lift a 5000 kg satellite to this location?

2. What is the net gravitational potential at a spot midway between the Earth and the Sun?

6XQ

(DUWK

3. Derive an expression for the gravitational potential at the surface of a planet in terms of the gravitational field strength.

7 

(VFDSH6SHHG

IB 12 

(VFDSH6SHHG: minimum initial speed an object must have at the surface of a planet in order to escape the gravitational attraction of the planet

Travel to infinity Just make it – means velocity is zero at infinity – means EK is zero at infinity as well as EP

Eo = Ep + Ek

Ef = Ep

3ODQHW

Assumptions: planet is isolated – ignore air resistance 'HULYDWLRQ Note: 1. Direction of travel is irrelevant – ǻEp is path independent

2. independent of mass of rocket

3. More speed (EK) is needed in real life since air friction is not negligible at lower altitudes

1. What is the escape speed for Earth?

2. If the Earth became a black hole, how large would it be?

8 

6DWHOOLWH0RWLRQ

1DWXUDO6DWHOOLWHV Period of a satellite:

7 DU 2

72

$UWLILFLDO6DWHOOLWHV

Acceleration of a satellite:

6)F PD F

v = 2ʌ r/T



v = 2ʌ r/T

ac = 4ʌ2 r/T2

3 2

7 DU  “Weightlessness”

Orbital speed of a satellite:

ac = v2/r

3

NU 3

IB 12

ac = g = GM/r2

Free fall

Orbital motion

*0P PY2  U2 U *0 Y2 U *0 Y  U Deep space

1. Compare the motion of satellites A and B. A – faster, less time (smaller period) B – slower, more time (longer period) T2/R3 = constant true for all satellites T = kR3/2

2. What happens to the required orbital speed if: a) the mass of the satellite increases? Nothing – speed is independent of mass

b) the satellite is boosted into a higher orbit? Satellite would orbit at a slower (tangential) speed

3. What would happen to a satellite if it encountered appreciable air resistance? Slow down, drop to lower orbit, and speed up, encounter even more air molecules (denser), cycle continues – spiral to Earth

9

(QHUJ\RI2UELWLQJ6DWHOOLWHV

IB 12

Compare the energies of the two orbiting satellites. Gravitational Potential Energy

Kinetic Energy

(QHUJ\'HULYDWLRQV

Total Energy

*UDSKVRIWKHHQHUJLHVRIDQRUELWLQJVDWHOOLWH

Gravitational Potential Energy

Kinetic Energy

Total Energy

RE

Comparisons:

A 1500 kg satellite is to be put into orbit around the Earth at an altitude of 200 km.

a) How much potential energy will the satellite have at this altitude?

b) How much kinetic energy will the satellite need to orbit at this altitude?

d) What is the orbital speed of the satellite?

e) What is the minimum amount of energy needed to lift the satellite from the surface of the Earth to this altitude?

c) What is the total amount of energy the satellite has at this altitude?

10 

&RPSDULVRQV

IB 12 

(TXLSRWHQWLDOVXUIDFH: a surface on which the potential is the same everywhere

1. The gravitational force does no work as a mass moves on along equipotential surface. 2. The work done in moving a mass between equipotential surfaces is path independent. 3. The work done in moving a mass along a closed path is zero. RQHSRLQWPDVV

WZRSRLQWPDVVHV 11

IB 12 

On the diagram at right: a) Sketch the gravitational field around the point mass.

b) Sketch equipotential surfaces around the point mass. What is the relationship between the gravitational field and the equipotential surfaces? Perpendicular 

Field lines point in direction 

of decreasing potential 

*UDYLWDWLRQDO3RWHQWLDO*UDGLHQW gradient: rate of change with respect to something “slope” or “derivative”

gravitational potential gradient: the gravitational field is the negative gradient of the gravitational potential with respect to distance

)RUPXOD

-70 J/kg

$ %

-80 J/kg

derive g = -ǻV/ǻr

What is the average gravitational field strength between equipotential surfaces A and B if they are 5.0 m apart?

12

3UDFWLFH4XHVWLRQV 1. a) Calculate the gravitational force the Sun exerts on the Earth

2. a) Calculate the strength of the gravitational field of the Sun at a location one million kilometers from the Sun.

3. a) Calculate the strength of the Sun’s gravitational field at the surface of the Earth.

IB 12 

b) Compare this to the gravitational force that the Earth exerts on the Sun.

b) What is the Sun’s gravitational force at this point?

b) Explain why the net gravitational field strength at the surface of the Earth can be approximated as due solely to the Earth’s gravitational field.

13 

IB 12 4. a) Calculate the resultant gravitational field at a spot midway between the Earth and Sun.

b) Compare the contributions from the Sun and the Earth to this resultant field.

c) What is the gravitational force acting on a 5000 kg space probe at this location?

5. A 5000kg satellite orbits Mars at a distance of 1000 km.

Mass of Mars: 6.42 x 1023 kg Mean planetary radius: 3.37 x 106 m

a) What is the gravitational potential at the surface of Mars?

0DUV

b) How much gravitational potential energy does the satellite have on the surface of Mars?

c) What is the gravitational potential at orbiting altitude?

d) How much gravitational potential energy does the satellite have at this altitude?

e) What is the minimum energy needed to lift the satellite to this altitude?

14 

IB 12 

6. A 5000. kg satellite is placed in a low altitude orbit.  a) If the altitude is sufficiently low, what is the approximate radius of the satellite’s orbit?

b) Calculate the satellite’s orbital speed.

c) Calculate the orbital period of the satellite.

d) Calculate the gravitational potential energy of the satellite.

e) Calculate the kinetic energy of the satellite.

f) Calculate the total energy of the satellite.

g) What is the minimum amount of energy needed to lift the satellite into this orbit?

15

3K\VLFVDQG3K\VLFDO0HDVXUHPHQW

IB 12

(YHU\PHDVXUHGYDOXHKDVXQFHUWDLQW\

A child swings back and forth on a swing 10 times in 36.27s ± 0.01 s. How long did one swing take?

notice that multiple trials reduces uncertainty for a single repetition

(36.27 ± 0.01) / 10 = 3.627 s ± 0.001 s

Measurements of time are taken as: 14.23 s, 13.91 s, 14.76 s, 15.31 s. 13.84 s, 14.18 s. What value should be reported?

0HDQ: 14.37 s

*UHDWHVW5HVLGXDO0.94

5HVLGXDOV: 14.37 – 13.84 = 0.53 15.31 – 14.37 = 0.94

5HSRUWHG9DOXH: 14.37 s ± 0.94 s = mean ± greatest residual

1. 0HDVXUHPHQWRecord as many significant figures as the calibration of the measuring instrument allows SOXV one estimated digit.

Voltage ± uncertainty 

V ± ǻV 

2. 8QFHUWDLQW\ Record a reasonable uncertainty estimate that a) has one significant digit, and  b) matches the measurement in place value (decimal place). 

11.6 V ± 0.2 V 

$EVROXWH8QFHUWDLQW\

)UDFWLRQDO8QFHUWDLQW\

3HUFHQWDJH8QFHUWDLQW\

ǻV

ǻV/V

ǻV/V · 100%

0.2 V

0.2 V / 11.6 V

0.2 V / 11.6 V · 100% = 1.7 %

&DOFXODWLRQVZLWK8QFHUWDLQWLHV  $GGLWLRQ6XEWUDFWLRQ5XOH When two or more quantities are added or subtracted, the overall uncertainty is equal to the VXPRIWKHDEVROXWHXQFHUWDLQWLHV. Ex. 1: The sides of a rectangle are measured to be (4.4 ± 0.2) cm and (8.5 ± 0.3) cm. Find the perimeter of the rectangle.

4.4 + 8.5 + 4.4 + 8.5= 25.8 cm 0.2 + 0.3 + 0.2 + 0.3 = 1.0 cm 25.8 cm ± 1.0 cm

1 

IB 12 

 0XOWLSOLFDWLRQ'LYLVLRQ5XOH

When two or more quantities are multiplied or divided, the overall uncertainty is equal to the VXPRIWKHSHUFHQWDJHXQFHUWDLQWLHV. Ex. 2: The sides of a rectangle are measured to be (4.4 ± 0.2) cm and (8.5 ± 0.3) cm. F ind the area of the rectangle.

4.4 x 8.5 = 37.4 cm2

0.2/4.4 = 4.55%

37.4 cm2 ± 8.08%

0.3/8.5 = 3.53%

37.4 cm2 ± 3.02192 cm2

Total = 8.08%

37 cm2 ± 3 cm2

 3RZHU5XOH When the calculation involves raising to a power, PXOWLSO\WKHSHUFHQWDJHXQFHUWDLQW\E\WKHSRZHU. 1

(Don’t forget that [ [2 ) Ex. 3: The radius of a circle is measured to be 3.5 cm ± 0.2 cm. What is the area of the circle with its uncertainty?

DUHD SU 2 DUHD S(3.5)2 38.48FP 2

§0.2 · 2 ¨ u100% ¸ 2  5.71%  11.4% ©3.5 ¹

DUHD 38.48FP2 r11.4% DUHD 38.48FP2 r4.39FP2 DUHD 38FP 2 r4FP 2

2

([HUFLVHV

IB 12

1. Five people measure the mass of an object. The results are 0.56 g, 0.58 g, 0.58 g, 0.55 g, 0.59 g. How would you report the measured value for the object’s mass?

0.57 g ± 0.02 g

(mean ± greatest residual)

2. Juan Deroff measured 8 floor tiles to be 2.67 m ±0.03 m long. What is the length of one floor tile?

0.334 m ± 0.004 m

(0.33375 m ± 0.00375 m)

3. The first part of a trip took 25 ± 3 s, and the second part of the trip took 17 ± 2s. a. How long did the whole trip take?

42 s ± 5 s

b. How much longer was the first part of the trip than the second part? 8s±5s

4. A car traveled 600. m ± 12 m in 32 ± 3 s. What was the speed of the car?

18.75 m/s

12/600 = 2.000% and 3/32 = 9.375 % so total = 11.38% 18.75 m/s ± 11.38% 18.75 m/s ± 2.13375 m/s

Speed = 19 m/s ± 2 m/s 5. The time W it takes an object to fall freely from rest a distance G is given by the formula: where J is the acceleration due to gravity. A ball fell 12.5 m ± 0.3 m. How long did this take?

 ˜    V  W  

1 § 0.3 · 1 u100% ¸   2.4% 1.2% ¨ 2 ©12.5 ¹ 2

W 

G J

W 1.60Vr1.2% W 1.60Vr0.0192V W 1.60Vr0.02V

3 

$QDO\]LQJ'DWD*UDSKLFDOO\

IB 12 

The masses of different volumes of alcohol were measured and then plotted (using *UDSKLFDO$QDO\VLV . Note there are three lines drawn on the graph – the best-fit line, the line of maximum slope, and the line of minimum slope. The slope and y-intercept of the best-fit line can be used to write the specific equation and the slopes and y-intercepts of the max/min lines can be used to find the uncertainties in the specific equation. The specific equation is then compared to a mathematical model in order to make conclusions.

*HQHUDO(TXDWLRQ: y= mx + b 6SHFLILF(TXDWLRQ M = (0.66 g/cm3)V + 0.65 g 8QFHUWDLQWLHV: slope: 0.66 g/cm3 ± 0.11 g/cm3 y-intercept: 0.65 g ± 3.05 g 0DWKHPDWLFDO0RGHO: D = M/V so M = DV

&RQFOXVLRQ3DUDJUDSK: 1. The purpose of the investigation was to determine the relationship between volume and mass for a sample of alcohol.

2. Our hypothesis was that the relationship is linear. The graph of our data supports our hypothesis since a best-fit line falls within the error bars of each data point. 3. The specific equation of the relationship is M = (0.66 g/cm3)V + 0.65 g.

4.

We believe that enough data points were taken over a wide enough range of values to establish this relationship. This relationship should hold true for very small volumes, although if it becomes too small for us to measure with our present equipment we won’t be able to tell, and for very large volumes, unless the mass becomes so large that the liquid will be compressed and change the density.

5. Zero falls within uncertainty range for y-intercept (0.65 g ± 3.05 g) so our results agree with math model and no systematic error is apparent

6. By comparison to the mathematical model we conclude that the slope of the graph represents the density. Therefore the density of the sample is 0.66 g/cm3 ± 0.11 g/cm3. 7. The literature value for the density of this type of alcohol is 0.72 g/cm3. Our results agree with the literature value since the literature value falls within the experimental uncertainty range of 0.66 g/cm3 ± 0.11 g/cm3 .

4

/LQHDUL]LQJ*UDSKV8VLQJ/RJDULWKPV

IB 12

This is a special linearizing (straightening) technique that works with general equations that are SRZHUIXQFWLRQV

3RZHU)XQFWLRQy = c·xn 0HWKRGRIVWUDLJKWHQLQJ graph log y vs. log x

'HULYDWLRQ

“log-log plot” 

y = cxn

Compare to y = mx + b

Take log of both sides

slope = n y-intercept = log c or

n

c = 10b

Log y = log (cx ) Log y = log c + log xn Log y = log c + n log x

([DPSOHV



m=2

y = cx2



y = cx-1

m = -1 or -2

y = kx-2





m = 0.5 y = kx

0.5

m=1 y = kx

1

5

/RJDULWKPLF6WUDLJKWHQLQJ

IB 12 

Why use logarithms? use when you’re not sure what the type of relationship is –

use to check the exponent

5HVHDUFK4XHVWLRQ: 

What is the relationship between kinetic energy and 

speed for a uniformly accelerating object?

)LQGLQJ(UURU%DUVIRUWKH6WUDLJKWHQHG*UDSK 1. Error bars needed on only one axis – choose whichever axis has the most significant uncertainties. 2. Use the greatest residual for the data point with the highest percent uncertainty as the error bars on all data points. KE: log (5-2) = log 3 = .477 log 5 = .699 log (5+2) = log 7 = .845

residuals: .699 - .477 = .222 .845 - .699 = .146

greatest residual = .222 = .22 

use for error bars on log KE axis 

$QDO\VLV general equation: y = cxn slope = 1.980 = n 

ORJ.(YVORJVSHHG

y-intercept= 0.7007 = log c  so c = 100.7007 = 5.0 specific equation: KE = 5.0 v1.98

slope residuals: 1.98 – 1.40 = 0.58 2.60 – 1.98 = 0.62

slope: 1.98 ± 0.62

greatest residual: 0.62

3DUWLDO&RQFOXVLRQ

The purpose of the investigation was to determine the relationship between the kinetic energy and the speed of a uniformly accelerating object. Our hypothesis was that the relationship is quadratic and the graph of our original data supports our hypothesis since a best-fit parabola can be drawn within the error bars of all data points. The data was then linearized using logarithms. Using this graph, the specific equation for the relationship was found to be KE = 5.0 v1.98 . Since a value of 2 falls within the uncertainty range for the exponent of 1.98 ± 0.62, the data is consistent with a quadratic relationship between speed and kinetic energy. However, since the uncertainty range for the exponent is so large (30.%), the relationship might not be quadratic but some other power function.

6

([HUFLVHV±/LQHDUL]LQJ'DWDZLWK/RJDULWKPV

IB 12

In each example below, straighten each graph by logarithms. Then, write the specific equation for each relationship. 

What is the most probable type of relationship in each case?

 7LPHV  “V 

'LVSODFHPHQWP 

“P 

















$FFHOHUDWLRQ PV  “PV

























 'LVWDQFHP  “P 

Specific equation: d = (3.2)t1.9





 0DVVNJ  “NJ

General Equation: y = c xn

)RUFH1  “1 





























Slope: 1.94 Y-intercept: 0.50 = log c c = 3.2

General Equation: y = c xn

Type of relationship: quadratic

Specific equation: a = (12)m-.99

Slope: -0.99 Y-intercept: 1.083 = log c c = 12.1

General Equation: y = c xn

Type of relationship: inverse

Specific equation: F = (15)d-1.9

Slope: -1.9 Type of relationship: inverse quadratic Y-intercept: 1.166 = log c c = 14.7

7

6FDODUVDQG9HFWRUV

IB 12

6FDODUV: quantities that have magnitude only

e.g. - Mass, time, volume, energy, distance, speed

9HFWRUVquantities that have magnitude and direction

e.g. - Velocity, displacement, acceleration, force, momentum, impulse, magnetic field strength, gravitational field strength, electric field strength

Notation: Bold italic ) or arrow hat

G )

$GGLQJ9HFWRUV Find the sum $ + %

6XEWUDFWLQJ9HFWRUV Find the difference $ - %

5HVROYLQJD9HFWRU LQWRLWV&RPSRQHQWV

Sin ș = y/r

Cos ș = x/r

y = r sin ș

x = r cos ș

Practice naming components

A sin ș

A sin 20 3. B cos 35 4.

A cos ș

A cos 20

2. B sin 35 62 sin 35 = 36 m

62 cos 35 = 51 m

8

IB 12 

9 

0HFKDQLFV

IB 12

.LQHPDWLFV±7KH6WXG\RI0RWLRQ

Symbols:

s = distance or displacement

v = final speed or velocity

u = initial speed or velocity

a = average acceleration

Equations:

Y 

V XY  also Y  W 2

'Y Y  X 'W W Y X  DW D

VR V

YW

or

V 



1 V XW   W 2 2 2 2 Y X  2 DV

X Y W 2

Condition for applying equations for uniformly accelerated motion:

must be constant, smooth acceleration equations use average acceleration = instantaneous if acceleration is constant

average vs. instantaneous: over a period of time vs. at one instant

([DPSOH: Two friends bicycle 3.0 kilometers north and then turn to bike 4.0 kilometers east in 25 minutes.

a) What is their average speed?

distance/time 7.0 km / 25 min = 0.28 km/min x 60 min/hr = 16.8 = 17 km/hr

b) What is their average velocity?

displacement/time 5.0 km/ 25 min = 0.20 km/min x 60 min/hr = 12 km/hr Angle: 53o west of north

&RQVWDQW9HORFLW\

Time (s) Distance (m) Velocity (m/s) Acceleration (m/s2)

0 0 25 0

1 25 25 0

2 50 25 0

3 75 25 0

4 100 25 0

IB 12

&RQVWDQW$FFHOHUDWLRQ

Time (s) Distance (m) Velocity (m/s) Acceleration (m/s2)

a) What does the slope of a position-time graph represent? b) What does the slope of a velocity-time graph represent?

0

1

2

3

0 0 5

3 5 5

10 10 5

23 15 5

Instantaneous velocity - derivative Instantaneous acceleration - derivative

c) What does the area under a velocity time graph represent?

displacement - integral

'URSSLQJ 1. A stone is dropped from rest from the top of a tall building. After 3.00 s of free-fall, what is the displacement of the stone? What is its velocity?

s = ut + ½ at2

v = u + at

s = ½ at2 = -45 m

v = -30 m/s

discuss change of frame of reference – downward is positive

How would these graphs change in the presence of air resistance?

Terminal velocity: no acceleration – constant velocity – Fg = Fair

2

IB 12

7KURZLQJ8S A ball is thrown straight up in the air (shown here stretched out for clarity.) Sketch velocity and acceleration vectors at each instant.

A football game customarily begins with a coin toss to determine who kicks off. The referee tosses the coin up with an initial speed of 6.00 m/s. In the absence of air resistance, how high does the coin go above its point of release? How long is it in the air? v2 = u2 + 2 as 02 = (6)2 + 2(-10)s s = 1.8 m

v = u + at 0 = 6.00 + (-10)t t = 0.6 s x 2 = 1.2 s

discuss change of frame of reference – downward is positive

3URMHFWLOH0RWLRQresultant of two independent components of motion

1. Vertical: constant acceleration (in absence of air resistance)

2. Horizontal: constant velocity – no horizontal acceleration

+RUL]RQWDO3URMHFWLOH

A ball is shot horizontally off a cliff that is 100. m high at a speed of 25 m/s. How long does it take to hit the ground? How far away from the base of the cliff does it land?

y-direction:

x-direction:

s = ut + ½ at2 -100 = 0 + ½(-10)t2 t = 4.5 s

s = ut + ½ at2 s = 25 (4.5) + 0 s = 113 m

3

$QJOHG3URMHFWLOH

IB 12

1. Break initial velocity into horiz and vert components

2. maximum height occurs after ½ time 3. maximum range occurs after fulltime and at 450

4. air resistance: not as high nor as far – show on diagram A football was kicked with a speed of 25 m/s at an angle of 30.0o to the horizontal. Determine how high it went and where it landed. Components

time:

Height:

x: cos 300 = vi/25 m/s

v = u + at

s = ut + ½ at2

vi = 21.7 m/s

0 = 12.5 m/s + (-10)(t)

s = (12.5 m/s)(1.25 s) + ½ (-10 m/s2)(1.25 s)2

t = 1.25 s

s = 7.8 m

y: sin 300 = vi/25 m/s total time = 2(1.25) = 2.5 vi = 12.5 m/s

Range: s = ut + ½ at2 s = (21.7 m/s)(2.5 s) + 0 s = 54 m

6WDWLFVDQG'\QDPLFV±7KH6WXG\RI)RUFHV 1HZWRQ¶V/DZVRI0RWLRQ 1. An object at rest remains at rest and an object in motion remains in motion at a constant speed in a straight line (constant velocity) unless acted on by unbalanced forces. (An object continues in uniform motion in a straight line or at rest unless a resultant (net) external force acts on it.) 2. When unbalanced forces act on an object, the object will accelerate in the direction of the resultant (net) force. The acceleration will be directly proportional to the net force and inversely proportional to the object’s mass. (The resultant force on an object is equal to the rate of change of momentum of the object.) 3. For every action on one object, there is an equal and opposite reaction on another object. (When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.) 4

1HZWRQ¶V6HFRQG/DZ

F = ǻp / ǻt

For constant mass

F = ǻ(mv) / ǻt

F = m (ǻv / ǻt)

IB 12

F=ma

QG/DZRUUG /DZ" Net force on ball: not zero so it accelerates – not in equilibrium Fnet = Fg = mg

Action-Reaction pairs: Earth pulls ball down Ball pulls Earth up FEB = -FBE mA = -Ma

Net force on block: zero at rest – in equilibrium Fnet = FN - Fg = 0

Action-Reaction pairs: Earth pulls block down Block pulls Earth up block pushes down on table table pushes up on block

)J

Translational equilibrium: net force acting on object is zero – no acceleration

1. Find the resistive force F caused by the drag of the water on the boat moving at a constant velocity in the diagram shown.

5 

IB 12

2. Find the tension in each cable supporting the 600 N cat burglar pictured.

600 997

796

3. A 20.0-kg floodlight in a park is supported at the end of a horizontal beam of negligible mass that is hinged to a pole, as shown. A cable at an angle of 30.0° with the beam helps to support the light. Find (a) the tension in the cable and (b) the horizontal and vertical forces exerted on the beam by the pole.

7 ƒ

+ 9

4. How would your answers change if the mass of the beam shown above was not negligible? T increase show how reaction force is oriented

a)

400N

b) Rx = 346 Ry = 0 1

G

5. Indicate the direction of the reaction force from the floor and the reaction force from the wall for the situation shown below.

FW

FN Fg1 Fg2 6 Ff

:HLJKW0DVVDQGWKH1RUPDO)RUFH 0DVV 1) the amount of matter in an object 2) the property of an object that determines its resistance to a change in its motion (a measure of the amount of inertia of an object)

IB 12

:HLJKW the force of gravity acting on an object Property: Varies from place to place

Symbol : Fg or W Units : N

Property: Remains constant 

Symbol : m

Units : kg

(OHYDWRUV In each case, the scale will read . . . the normal or reaction force, not the weight

Calculate the acceleration of the man in each case.

ȈF=0

Ȉ F = ma

ȈF=ma

ȈF=ma

FN – F g = 0

FN – F g = m a

FN – Fg = m a

– Fg = m a

FN – 700 = 0

1000 – 700 = 70 a

400 – 700 = 70 a

– 700 = 70 a

FN = 700 N

a = +4.3 m/s2

a = - 4.3 m/s2

a = -10 m/s2

,QFOLQHG3ODQH±Assume the box shown is in equilibrium and draw the . . . Free-body diagram

Head-to-tail vector diagram

Concurrent vector diagram Concurrent vector diagram with perpendicular components

FN Ff

Fg ș ș

7 

IB 12 Calculate the force of friction acting on this box if it accelerates down the incline at a rate of 0.67 m/s2. 4.5 kg

Fnet = ma = 4.5 (0.67) = 3.0 N 

Fg|| = mg sin ș = 15 N 

Ff = 12 N 

200

8QLIRUP&LUFXODU0RWLRQ

8QLIRUP&LUFXODU0RWLRQ constant speed and constant radius

3HULRG time take for one complete cycle symbol: T units: s

1. The direction of the object’s instantaneous velocity is always tangent to the circle in

the direction of motion.

2. Since the direction of the object’s motion is always changing, its velocity is always 

changing therefore the object is always accelerating and is never in equilibrium.

3. Direction of net force – towards the center - centripetal

)RUPXODV

V bar = distance / time V bar = circumference / period V bar = 2ʌr / T

V = 2ʌr / T

V bar = 2ʌr / T ac = v2 / r

ac = (2ʌr / T)2 / r ac = (4ʌ2r2 / T2) / r

F = m a 

Fc = m ac

Fc = mv2 / r

ac = 4 ʌ2 r / T2

The phrase “centripetal force” does not denote a new and separate force created by nature. The phrase merely labels the net force pointing toward the center of the circular path, and this net force is the vector sum of all the force components that point along the radial direction. 1. The model airplane shown has a mass of 0.90 kg and moves at a constant

speed on a circle that is parallel to the ground. Find the tension 7 in the 

guideline (length = 17 m) for a speed of 19 m/s. 

8 

IB 12 2. At amusement parks, there is a popular ride where the floor of a rotating cylindrical room falls away, leaving the backs of the riders “plastered” against the wall. For a particular ride with a radius of 8.0 m and a top speed of 21 m/s, calculate the reaction force and the friction force from the wall acting on a 60. kg rider. Which of these is the centripetal force?

3. A 2100-kg demolition ball swings at the end of a 15-m cable on the arc of a vertical circle. At the lowest point of the swing, the ball is moving at a speed of 7.6 m/s. Determine the tension in the cable.

:RUN3RZHUDQG(IILFLHQF\ :RUN product of force and displacement in the direction of the force )RUPXOD W = (F cos ș) s

8QLWV N m or Joules (J)

W = F s cos ș ș is angle between F and s

3RZHU 1) the rate at which work is done

)RUPXOD P = W/ t = E/t = Q/t

$OWHUQDWH)RUPXOD P=W/t = (F cos șd) / t = F v cos ș

2) the rate at which energy is transferred (IILFLHQF\: 1) ratio of useful work done by a system to the total work done by the system 2) ratio of useful energy output of a system to the total energy input to the system 3) ratio of useful power output of a system to total power input to the system

7\SH Scalar but can be positive or negative 8QLWV J/s = Watts (W) 7\SH Scalar )RUPXOD e = useful out/ total in 9

1. A 45.0-N force is applied to pull a luggage carrier an angle T = 50° for a distance of 75.0 m at a constant speed.IB 12 Find the work done by the applied force. W = F s cos ș WA = (45.0 N)(75 m) cos 50o = 2170 J

2. a) How much work is done dragging the 5.00 kg box to the top of the hill shown if the hill exerts an average friction force of 5.0 N?

b) Compare your answer to the amount of work done lifting the box straight up to the top of the hill.

c)

Calculate the power expended if the box is dragged to the top in 15 seconds.

d) Calculate the efficiency of dragging the box rather than lifting the box.

'HWHUPLQLQJ:RUN'RQH*UDSKLFDOO\

Work = area under curve W=½bh W=½Fs W = (avg force) (displacement)

Work = area under curve W=bh W=fs

1. Work done by a constant force

2. Work done by a constantly varying force ex. stretching a spring

10

(QHUJ\ 1.

2.

3.

IB 12

4.

7\SHVRI(QHUJ\ 1. Kinetic energy

5.

)RUPXODV 1. EK = ½ mv2

(energy of motion)

2. Gravitational Potential energy (energy of position)

2. EP = mgh

3. Elastic potential energy

3. Eelas = ½ kx2

4. Internal energy (thermal energy)

4. Q = mcǻt

5. Chemical Potential energy (stored in chemical bonds) Electrical energy Light energy

5. Ee = Pt = VIt = I2Rt = V2/R t

Q = mL

&RQVHUYDWLRQRI(QHUJ\3ULQFLSOH ,QDQLVRODWHGV\VWHPWKHWRWDODPRXQWRIHQHUJ\UHPDLQVFRQVWDQW 1. A motorcyclist is trying to leap across a canyon by driving horizontally off the cliff at a speed of 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

2. What is the speed of the box at the bottom of the incline if an average frictional force of 15 N acts on it as it slides?

160 N

20 meters 300

11 

/LQHDU0RPHQWXPDQG,PSXOVH /LQHDU0RPHQWXP: the product of an object’s mass and velocity

A 1500. kg car is traveling east at a speed of 25.0 m/s.

inertia (inertial mass) m = 1500 kg scalar

Compare its inertia, momentum, and kinetic energy.

IB 12

)RUPXOD

8QLWV

p = mv

kg m/s

momentum

kinetic energy

p = (1500) (25) = 3.75 x 104 kg m/s, east vector

Ek = ½ (1500)(25)2 = 468750 J = 4.69 x 105 J scalar

Alternate formula for kinetic energy:

How does the momentum of an object change? A net external force acts for a finite amount of time ,PSXOVH: (the change in momentum of a system) the product of the average force and the time interval over which the force acts 

8QLWV

7\SH

Ns or

vector

Usually . . . Force is not instantaneous and is not constant

kg m/s

'HULYDWLRQRI,PSXOVH)RUPXOD

'S 'W

'S )'W  

'(PY)  ) 'W  P'Y  )'W (if mass is constant)  'Y - 'S ) 'W P  ) 

Graphically, the impulse is J = Favg ǻt = area

If force is linear: J = ½ Fmax ǻt

12

%RXQFLQJDQG,PSXOVH

IB 12

A 0.50 kg basketball hits the floor at a speed of 4.0 m/s and rebounds at 3.0 m/s. Calculate the impulse applied to it by the floor. Calculation:

In general:

K K 'S P'Y  K K K 'S P  Y I  YL  K 'S P  Y I  YL 

'S

'S

P'Y   NJ( 4.0  (3.0))  0.50 'S 7.0NJP / V

Velocity vs. time graph for bounce

Force

Velocity

Force vs. time graph for bounce

Time

Time

7KH3ULQFLSOHRI&RQVHUYDWLRQ RI/LQHDU0RPHQWXP

7KHWRWDOPRPHQWXPRIDQLVRODWHGV\VWHPUHPDLQVFRQVWDQW SEHIRUH SDIWHU 7\SHVRI,QWHUDFWLRQV

 %RXQF\

 6WLFN\

([SORVLRQ

13 

&ROOLVLRQV (ODVWLFFROOLVLRQ: a collision in which the total kinetic energy is conserved

IB 12

,QHODVWLFFROOLVLRQ: a collision in which the total kinetic energy is not conserved

Where does some of the mechanical energy go in an inelastic collision? energy of deformation, internal energy, sound energy

1. A freight train is being assembled in a switching yard, and the figure below shows two boxcars. Car 1 has a mass of P1 = 65×103 kg and moves at a velocity of Y01 = +0.80 m/s. Car 2, with a mass of P2 = 92×103 kg and a velocity of Y02 = +1.3 m/s, overtakes car 1 and couples to it. Neglecting friction, find the common velocity Yf of the cars after they become coupled.

2. Is this collision elastic or inelastic? Justify your answer.

3. A EDOOLVWLFSHQGXOXP is sometimes used in laboratories to measure the speed of a projectile, such as a bullet. A ballistic pendulum consists of a block of wood (mass P2 = 2.50 kg) suspended by a wire of negligible mass. A bullet (mass P1 = 0.0100 kg) is fired with a speed Y01. Just after the bullet collides with it, the block (with the bullet in it) has a speed Yf and then swings to a maximum height of 0.650 m above the initial position (see part Eof the drawing). Find the speed Y01 of the bullet, assuming that air resistance is negligible.

14 

1XFOHDU3K\VLFV

IB 12 

1XFOLGH: a particular type of nucleus 1XFOHRQ: a proton or a neutron $WRPLFQXPEHU= SURWRQQXPEHU number of protons in nucleus 0DVVQXPEHU$ QXFOHRQQXPEHU  number of protons + neutrons 1HXWURQQXPEHU1  number of neutrons in nucleus (N = A – Z) ,VRWRSHV: nuclei with same number of protons but different numbers of neutrons x 1 u = 1.661 x 10-27 kg x 1 u = 1 g/mol x 1 u = 931.5 MeV/c2

8QLILHGDWRPLFPDVVXQLWX  1/12th the mass of a carbon-12 nucleus $WRPLFPDVV§ A * u

$ =

;

56 26

)H

27 13

$O

12 6

&

14 6

&

$WRPLF1XPEHU

26

13

6

6

0DVV1XPEHU

56

27

12

14

1HXWURQ1XPEHU

30

14

6

8

$WRPLF0DVV

56 u

27 u

12 u

14 u

0RODU0DVV

56 g

27 g

12 g

14 g

How big are atomic nuclei? 10-15m – 10-14 m How do we know this? Alpha-particle scattering experiments (Geiger-Marsden)

How do we know that neutrons exist? By the existence of isotopes How do we know that isotopes exist? By measurements in the mass spectrometer

Formulas: ET = ET EK = Ee ½ mv2 = qV ½ mv2 = q(kQ/r)

Formulas: v = E/B Note that most nuclei have approximately the same . . .density = 2 x 1017 kg/m3 = 1014 times denser than water

r = mv/qB 1

1XFOHDU6WDELOLW\

IB 12 

What interactions exist in the nucleus? 1. *UDYLWDWLRQDO(long range) attractive but very weak/negligible 2. &RXORPERU(OHFWURPDJQHWLF(long range) repulsive and very strong between protons 3. 6WURQJQXFOHDUIRUFH(short range) attractive and strongest – between any two nucleons 4. :HDNQXFOHDUIRUFH (short range) involved in radioactive decay

Why are some nuclei stable while others are not?

The Coulomb force is a long-range force which means that every 

proton in the nucleus repels every other proton. The strong 

nuclear force is an attractive force between any two nucleons 

(protons and/or neutrons). This force is very strong but is short 

range (10-15 m) which means it only acts between a nucleon and 

its nearest neighbors. At this range, it is stronger than the 

Coulomb repulsion and is what holds the nucleus together. 

Neutrons in the nucleus play a dual role in keeping it stable. They 

provide for the strong force of attraction, through the exchange of 

gluons with their nearest neighbors, and they act to separate 

protons to reduce the Coulomb repulsion. 

Each dot in the plot at right represents a stable nuclide and the 

shape is known as the “band (or valley) of stability.” With few 

exceptions, the naturally occurring stable nuclei have a number 1 of 

neutrons that equals or exceeds the number = of protons. For small 

nuclei (Z < 20), number of neutrons tends to equal number of 

protons (N = Z). 

As more protons are added, the Coulomb repulsion rises faster than the strong force of attraction since the Coulomb force acts throughout the entire nucleus but the strong force only acts among nearby nucleons. Therefore, more neutrons are needed for each extra proton to keep the nucleus together. Thus, for large nuclei (Z > 20), there are more neutrons than protons (N > Z). After Z = 83 (Bismuth), adding extra neutrons is no longer able to counteract the Coulomb repulsion and the nuclei become unstable and decay in various ways. Nuclei above (to the left of) the band of stability have too many neutrons and tend to decay by alpha or beta-minus (electron) emission, both of which reduce the number of neutrons in the nucleus. Nuclei below (to the right of) the band of stability have too few neutrons and tend to decay by betaplus (positron) emission which increases the number of neutrons in the nucleus.

2 

%LQGLQJ(QHUJ\

IB 12

The total mass of a nucleus is always less than the sum of the masses its nucleons. Because mass is another

manifestation of energy, another way of saying this is the total energy of the nucleus is less than the combined

energy of the separated nucleons.

0DVVGHIHFWPDVVGHILFLW ǻP  Difference between the mass of the nucleus and the sum of the masses of its individual nucleons 1XFOHDUELQGLQJHQHUJ\ǻ( 

)RUPXODV

1. energy released when a nuclide is assembled from its individual components 2. energy required when nucleus is separated into its individual components

PQXFOHXV ǻP PQXFOHRQV '( 'P F

Different nuclei have different total binding energies. As a general trend, as the atomic number increases . . . the total binding energy for the nucleus increases. 3DUWLFOH 3URWRQ 1HXWURQ (OHFWURQ

(OHFWULF&KDUJH (OHFWULF&KDUJH H  &  +1 0 -1

+1.60 x 10-19 0 -1.60 x 10-19

5HVW0DVV NJ 

5HVW0DVV X 

5HVW0DVV 0H9F 

1.673 x 10-27 1.675 x 10-27 9.110 x 10-31

1.007276 1.008665 0.000549

938 940 0.511

1. The most abundant isotope of helium has a 24He nucleus whose mass is 6.6447 × 10-27 kg. For this nucleus, find the mass defect and the binding energy.

helium: 0.050404 x 10-27 kg 

4.53636 x 10-12 J 

2. Calculate the binding energy and mass defect for 816O whose measured mass is 15.994915 u.

Oxygen: 0.132613 u 123.5 MeV

3

%LQGLQJ(QHUJ\SHU1XFOHRQ

IB 12 

To see how the nuclear binding energy varies from nucleus to nucleus, it is useful to compare the binding

energy for each nucleus on a per-nucleon basis, as shown in the graph below.